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string | source_data
string | source_index
int64 |
|---|---|---|
using Microsoft.Extensions.Configuration;
using Castle.MicroKernel.Registration;
using Abp.Events.Bus;
using Abp.Modules;
using Abp.Reflection.Extensions;
using ABP.WebApi.Configuration;
using ABP.WebApi.EntityFrameworkCore;
using ABP.WebApi.Migrator.DependencyInjection;
namespace ABP.WebApi.Migrator
{
[DependsOn(typeof(WebApiEntityFrameworkModule))]
public class WebApiMigratorModule : AbpModule
{
private readonly IConfigurationRoot _appConfiguration;
public WebApiMigratorModule(WebApiEntityFrameworkModule abpProjectNameEntityFrameworkModule)
{
abpProjectNameEntityFrameworkModule.SkipDbSeed = true;
_appConfiguration = AppConfigurations.Get(
typeof(WebApiMigratorModule).GetAssembly().GetDirectoryPathOrNull()
);
}
public override void PreInitialize()
{
Configuration.DefaultNameOrConnectionString = _appConfiguration.GetConnectionString(
WebApiConsts.ConnectionStringName
);
Configuration.BackgroundJobs.IsJobExecutionEnabled = false;
Configuration.ReplaceService(
typeof(IEventBus),
() => IocManager.IocContainer.Register(
Component.For<IEventBus>().Instance(NullEventBus.Instance)
)
);
}
public override void Initialize()
{
IocManager.RegisterAssemblyByConvention(typeof(WebApiMigratorModule).GetAssembly());
ServiceCollectionRegistrar.Register(IocManager);
}
}
}
|
the_stack
| 29,267
|
In many instances, portrayals of exile fall into a familiar narrative. From the Frankfurt School intellectuals who fled Nazi Germany to land in San Diego or Mexico, to Fernando Marcos who was ousted by the historic takeover of Corazon Aquino in the Philippines, exile is usually understood as a political condition borne out of externally imposed departure, usually involving force. The hermeneutics of exile, which is simultaneously a “hermeneutics of hope”, is an interpretive choice borne not only out of force and coercion but a deliberate decision to side against certain orthodoxies, normativities, and calcification of power relations. The many forms of exile require assimilating received information about the known world in order to make it hospitable, which is not the same as centric, idealized, or inward iterations of home but rather ex-centric or outward facing towards the experiential world, leading to what Said was fond of calling “worldliness.”.
|
dclm_baseline
| 1,947,637
|
Good-bye, Ray
I watched Ray Lewis play his last game in Baltimore ("Ray's day," Jan. 7). He never started out as a great man, but he is finishing as one. So somewhere along the way he changed, chose a different destiny, decided on a different life. He was always a master of his trade and now is a master of his beliefs.
So it was with a tear in my eye that that I watched him dance his way out of the tunnel, and it was with one on my cheek and my heart in my mouth that I watched him disappear into the dressing room. Thank you, Ray, for all the wonderful memories, all the incredible days out that you gave me and my son. Those days and you shall never be forgotten,
Joe Devoy, Edgewater
|
dclm_baseline
| 1,506,865
|
Gas Prices Finally Falling To New Lows, But Will It Stay That Way?
By Kristi Eckert | Published
gas tax holiday gas prices
Gas prices, at long last, are finally going down. After months of overly bloated prices, US residents are finally seeing some relief. According to AAA, the national average is currently hovering at $4.16 per gallon. And in 11 states, prices have fallen below that $4 threshold. Texas, Oklahoma, Missouri, Alabama, Georgia, Tennessee, South Carolina, Kentucky, Arkansas, Mississippi, and Louisiana are enjoying prices as low as $3.66 per gallon. Even in California, where prices are still over $5, they are nowhere near the $8 mark that many residents were paying in June. The recent reprieve in gas prices though begs a couple of questions. Why the sudden drop? And can prices be expected to continue on this downward trend?
A big factor driving gas prices down is the fact that global oil demand has significantly receded. This in turn has allowed barrel prices for oil to fall which then translates to lower prices at the pump for the end consumer. Demand for oil is suddenly decreasing due to inflation’s unceasing global hold. And that hold is starting to have an effect on economies worldwide.
In the United States, this is becoming increasingly noticeable, as residents across the country cut back on driving as inflation pounds heavily into their budgets. And according to the Wall Street Journal, it looks like the trend of gas prices heading down is something that consumers can expect to continue barring a few caveats. There are two major factors that could directly affect gas prices going forward, despite the fact that demand has greatly slowed. The first factor is weather-related and the second hinges on what happens overseas between Russia and Ukraine.
In terms of the weather, hurricane season is upon us. Should the Gulf of Mexico, in particular, be ravaged by severe storms, that could adversely affect oil production and supplies in the United States. However, apart from that, Patrick De Haan, head of petroleum analysis at GasBuddy, thinks that the United States is nearly out of the woods. “If nothing goes wrong, we could see prices in October, November, [and] December falling noticeably under $4 a gallon for the national average,” said Haan.
Tom Kloza, who is the global head of energy analysis for OPIS, also pointed to the direction the US economy is headed and how that direction is already impacting gas prices domestically. Kloza noted that “You’ve seen tremendous drops at wholesale prices really in every nook and cranny of the country.” These tremendous drops are indicative of an emerging recession. Kloza said that if/when a recession does fully rear its ugly head gas prices (and many other things) could look very similar to how they did during the Great Recession
Ultimately, with the exception of any unforeseen or mitigating circumstances, it looks like US residents can breathe a sigh of relief. Gas prices are going down, and they are largely expected to continue to head in that much more attractive direction. With inflation’s effects hitting home for a vast number of people combined with the fact that the nation is staring a recession right in the face, there is some small bit of solace to be taken from the fact that at least something is getting cheaper.
|
dclm_baseline
| 1,534,383
|
Time - math word problems - page 30
1. Overload
aircraft-02_10 What overload in g (g-force) has passed the pilot if he accelerated from 0 to 600 km/h in 3 seconds?
2. Grandmother
3. Clock's gears
orloj In the clock machine, three gears fit together. The largest has 168 teeth, the middle 90 teeth, and the smallest 48 teeth. The middle wheel turns around its axis in 90 seconds. How many times during the day do all the gears meet in the starting position?
4. Motorcyclist
motorbike_3 From Trutnov, the motorcyclist started at an average speed of 60km/hour. At 12.30hrs the passenger car was started at a speed of 80km/hour. How many hours and at what distance from Trutnov will car catch a motorcycle?
5. The swimmer
6. Two trains
vlak_8 From station A, the freight train traveled at a speed of 40 km/h in 9h. When he drove 15km, the fast train started from station A in the same direction at a speed of 70km/h. When will it the freight train catch up?
7. Railroad
8. Summer tires
workers_44 Three tire servants have to change the summer tires on 6 cars in 2 hours. Mark's replacement would take 4.5 hours, Jirka would do it in 3 hours and 10 minutes, and Honza in 4 hours. Will they be able to replace all tires at the desired time?
9. Hectares
tractor_8 The tractor plows the first day of 4.5ha, the second day 6.3ha and the third day 5.4ha. It worked whole hours a day, and its hourly performance did not change and was the highest of the possible. How many hectares did it plow in one hour (what is it perfor
10. Velocipedes
11. Father and son
Rodina-01_1 When I was 11, my father was 35 years old. Today, my father has three times more years than me. How old is she?
12. 2 pipes
time_12 2 pipes can fill a tank in 35 minutes. The larger pipe alone can fill the tank in 24 minutes less time than the smaller pipe. How long does each pipie take to fill the tank alone?
13. Driver
cargo_truck_4 The driver of the car at a speed of 100 km/h faced the obstacle and began to brake with a slowing of 5 m/s². What is the path to stopping the car when the driver has registered the obstacle with a delay of 0.7 s?
14. Assembly parts
machine Nine machines produce 1,800 parts on nine machines. How many hours will it produce 2 100 parts on seven such machines?
15. Water flow 2
16. Special watch
clocks2_20 Fero bought a special watch on the market. They have only one (minute) hand and a display that shows which angle between the hour and minute hand. How many hours it was when his watch showed - the minute hand points to number 2; the display shows 125°?
17. Three people
18. Grandfather and grandmother
19. The tourist
bus27_16 The tourist traveled 190km in 5 hours. Part of the journey passed at 5 km/h. The rest he went by bus at a speed of 60 km/h. How long has a bus gone?
20. Scientific notation
|
dclm_baseline
| 160,876
|
Portuguese english translate jobs
My recent searches
Filter by:
Job State
6,898 portuguese english translate jobs found, pricing in USD
I have a catalog of a company to translate. It is made of small text and full of images. Mostly small sentences. 377 pages. I would like price for the translation only. The editing of the file is made by us. Can't upload file because pdf is to big.
$361 (Avg Bid)
$361 Avg Bid
6 bids
Translation 5 days left
Translate some words from English to portuguese
$15 (Avg Bid)
$15 Avg Bid
68 bids
We need a Mozambican national to urgently translate 3 422 words from English to Mozambican Portuguese.
$309 (Avg Bid)
$309 Avg Bid
16 bids
Portuguese translation 4 days left
I need an portuguese who can help me to do proofreading work. I have some articles and need to proofread these files.
$124 (Avg Bid)
$124 Avg Bid
43 bids
I am looking for a professional English writer and Portuguese translator/writer for unique articles and translations, each article must have to be written in professional style, error free and contains between 700-1000 words. I will provide the details about the topic/subject of articles in PM. Time frame: You must have to write and deliver 2 articles
$11 / hr (Avg Bid)
$11 / hr Avg Bid
26 bids
Portuguese language teaching 3 days left
I am looking for experienced Portuguese language teachers (native speakers) to teach Portuguese You will have to provide Portuguese language lessons
$11 / hr (Avg Bid)
$11 / hr Avg Bid
18 bids
I'm looking for a Portuguese speaking virtual assistant to do some research for the Brazilian market.
$117 (Avg Bid)
$117 Avg Bid
11 bids
$143 (Avg Bid)
$143 Avg Bid
20 bids
I need a translation for a paragraph to some of the members of a group I am in. I need a translation in Japanese and Portuguese. Willing to pay $10
$17 (Avg Bid)
$17 Avg Bid
41 bids
English Translation 2 days left
I need to translate 6000 words from English to Portuguese.
$111 (Avg Bid)
$111 Avg Bid
104 bids
I'm looking for someone who can help me. Able to speak fluent Portuguese and English. It would be great if you are living in Europe and North America. This is urgent task and I need reliable man.
$115 (Avg Bid)
$115 Avg Bid
45 bids
We provide PDF files of our print flyers in English language and require DOC files with corresponding Portuguese translations so our graphic designer can replace the texts. I provide one sample file for download. Please native Portuguese speakers only . Please do not bid when you don't have time to finish this within 3 days.
$100 (Avg Bid)
$100 Avg Bid
15 bids
Hi, I need someone who is a native spanish speaker to translater this flyer (in upload area) to portuguese. Gracias!
$18 / hr (Avg Bid)
$18 / hr Avg Bid
52 bids
$18 / hr (Avg Bid)
$18 / hr Avg Bid
6 bids
$21 (Avg Bid)
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36 bids
Portuguese 1 day left
$113 (Avg Bid)
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51 bids
$3 / hr (Avg Bid)
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6 bids
Portuguese content writer 1 day left
$349 (Avg Bid)
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38 bids
$381 (Avg Bid)
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33 bids
community management on social medias
$580 (Avg Bid)
$580 Avg Bid
28 bids
|
dclm_baseline
| 979,025
|
import os
import re
import sublime
import sublime_plugin
import subprocess
def writeSnippet(snippetPath, filePath, componentName):
snippetFile = open(snippetPath, 'r')
isContent = False
lines = []
for line in snippetFile:
if '<content><![CDATA[' in line:
isContent = True
elif ']]></content>' in line:
isContent = False
elif isContent:
lines.append(line.replace('${TM_FILENAME/\\.js$//}', componentName).replace('\n', ''))
snippetFile.close()
file = open(filePath, 'w')
file.write('\n'.join(lines))
file.close()
class EslintFixCommand(sublime_plugin.TextCommand):
def run(self, edit):
if self.view.match_selector(0, 'source.js'):
filename = self.view.file_name()
subprocess.run('export NVM_DIR=$HOME/.nvm && . "/usr/local/opt/nvm/nvm.sh" && eslint --fix ' + filename, cwd=os.path.dirname(filename), shell=True)
class NewReactNativeComponentCommand(sublime_plugin.WindowCommand):
_currentFilename = ''
def onNameSubmit(self, name):
currentFolder = os.path.dirname(self._currentFilename)
componentFolder = os.path.join(currentFolder, name)
componentFile = os.path.join(componentFolder, name + '.js')
stylesFile = os.path.join(componentFolder, name + '.styles.js')
os.makedirs(componentFolder, 0o777, True)
# /Users/rfontes/Library/Application Support/Sublime Text 3/Packages
userPackages = os.path.join(sublime.packages_path(), 'User')
componentSnippet = os.path.join(userPackages, 'ReactFunctionalComponent.sublime-snippet')
stylesSnippet = os.path.join(userPackages, 'ReactStyles.sublime-snippet')
writeSnippet(componentSnippet, componentFile, name)
writeSnippet(stylesSnippet, stylesFile, name)
def run(self):
self._currentFilename = self.window.active_view().file_name()
self.window.show_input_panel('Component name', '', self.onNameSubmit, None, None)
class ReactRelativePathCommand(sublime_plugin.TextCommand):
def run(self, edit):
filename = self.view.file_name()
pathToSrc = re.sub(r'.*(src|App)\/((\w+\/)+)[\w.]+\.js', r'\2', filename)
relative = re.sub(r'\w+', '..', pathToSrc)
cursors = self.view.sel()
firstCursorLocation = cursors[0].begin()
self.view.insert(edit, firstCursorLocation, relative)
|
common_corpus_source
| 98,816
|
cbuffer ConstantBuffer : register(b0)
{
float4x4 viewProjectionMatrix;
};
struct VSOut
{
linear float4 position : SV_POSITION;
linear float4 color : COLOR;
linear float2 texcoord : TEXCOORD;
noperspective float size : SIZE;
noperspective float edgeDistance : EDGE_DISTANCE;
};
struct VSIn
{
float4 positionSize : POSITIONSIZE;
float4 color : COLOR;
};
VSOut main(VSIn input)
{
VSOut output;
float4 inPos = float4(input.positionSize.xyz, 1.f);
float inSize = max(input.positionSize.w, 2.f);
output.position = mul(inPos, viewProjectionMatrix);
output.color = input.color.abgr;
output.size = inSize;
return output;
}
|
the_stack
| 93,711
|
I need help on how to save dictionary elements into a csv file
I intend to save a contact list with name and phone number in a .csv file from user input through a dictionary.
The problem is that only the name is saved on the .csv file and the number is omitted.
contacts={}
def phone_book():
running=True
while running:
command=input('A(dd D)elete L)ook up Q)uit: ')
if command=='A' or command=='a':
name=input('Enter new name: ')
print('Enter new number for', name, end=':' )
number=input()
contacts[name]=number
elif command=='D' or command=='d':
name= input('Enter the name to delete: ')
del contacts[name]
elif command=='L' or command=='l':
name= input('Enter name to search: ')
if name in contacts:
print(name, contacts[name])
else:
print("The name is not in the phone book, use A or a to save")
elif command=='Q' or command=='q':
running= False
elif command =='list':
for k,v in contacts.items():
print(k,v)
else:
print(command, 'is not a valid command')
def contact_saver():
import csv
global name
csv_columns=['Name', 'Phone number']
r=[contacts]
with open(r'C:\Users\Rigelsolutions\Documents\numbersaver.csv', 'w') as f:
dict_writer=csv.writer(f)
dict_writer.writerow(csv_columns)
for data in r:
dict_writer.writerow(data)
phone_book()
contact_saver()
Hello, could you specify your question : what is the problem with your code, and what are you looking for ?
as I am reading your code contacts will look like
{
'name1': '1',
'name2': '2'
}
keys are the names and the value is the number.
but when you did r = [contacts] and iterating over r for data in r that will mess up I guess your code since you are passing dictionary value to writerow instead of a list [name, number]
You can do two things here. parse properly the contacts by:
for k, v in contacts.items():
dict_writer.writerow([k, v])
Or properly construct the contacts into a list with dictionaries inside
[{
'name': 'name1',
'number': 1
}]
so you can create DictWriter
fieldnames = ['name', 'number']
writer = csv.DictWriter(f, fieldnames=fieldnames)
...
# then you can insert by
for contact in contacts:
writer.writerow(contact) # which should look like writer.writerow({'name': 'name1', 'number': 1})
thanks a lot this worked perfectly
for k,v in contacts.items():
dict_writer.writerow([k,v])
|
common_corpus_web
| 126,751
|
Is Trading Right for Me or Should I Stick with Poker?
Discussion in 'Professional Trading' started by The Tripster, Dec 6, 2008.
1. I'm curious to hear how the responses will be different asking this in a trading forum vs. the ones that are posted on poker forums. I've been playing poker for about 4-5 years and have made good money from it while going to a good college; around $275K-$325K. I have an interest in the markets and I should naturallly be able to progress into becoming a good trader because of the similarities in the pyschological side of both of them. I graduate with a Finance major this semester and I'm at the turning point in deciding if I should just stick with poker or become a trader. I've gotten offers from top tier prop trading firms. I have also opened up an account at Tradestation and trade the e-mini S&P and NASDAQ with some minor success and I enjoy doing it. In terms of compensation in both, I would expect to make between 200K-400K a year in poker playing ~30 hours a week, and obv no idea in trading (in terms of profit splits). Other factors that I'm debating are how I would rather be in a social environment vs. at home just playing poker or trading. The negative to that is how I have to have set hours plus a boss and can't take off whenever I want to do random things. The freedom of playing poker is a huge factor for me because I'm not used to being in any other type of environment. I had an internship last year and enjoyed it, and I didn't mind the set schedule, but considering that was for only 3 months vs. a full time job might change my mind. Based on these set of facts, do you think I should stick with poker or give trading at a good firm a shot?
2. DonKee
Obviously a level.
No one who is smart enough or lucky enough to make 6 figures a year playing poker would be dumb enough to ask this question on an open forum at site where 95% of the people who post are "wanna be" traders.
Do you really think that any response would be anything more than verbal diarrhea from people who can't answer your question for you?
"You went full retard."
"Never go full retard."
3. Let the entertainment begin!! LOL :D
4. goto vegas. trade from 6am to 1pm. grab lunch and then play poker rest of the day and night.
5. Joab
Trading is a CAREER for life and should be approached with that mind set.
If you can see yourself playing poker for the rest of your life day in and day out then just do that if not then try trading and see if you like what it offers.
6. Its quite funny to me that numerous people in the past have compared the many parallels of poker and trading. Of course in each "game" you need discipline, a certain understanding of odds, risk managment.... and the list goes on.
However, whats not often mentioned is that you can walk into any poker room and find a group of players who are consistently successful. Everyone knows them by name because of reputation or the amount of hours they spend in a casino..
Conversely, most people can't find even a slim number of traders who are successful over time. Its not that poker players have better skills than traders, but to be successful in trading takes FAR more than just to have to certain inherent skills.
The only people who truly win will always be the house (the casino and the prop firm owners)
7. thedewar
play poker fulltime study markets, then slowly make a transition pending on how successful you become...
200k+ a year is more than alot of traders will make...
in reality you'll just need to gauge how much you think you could make trading based on position size and trading strategy then compare that to what you can make playing poker....
however your earnings can also be compounded trading.... and possibly more readily than they would playing poker...
personally i'd try and balance both if you're only working 30h a week playing poker
8. I second that.
9. it's a life i once dreamed of, I'm happily doomed with fatherhood responsibilities......maybe after my kids goto college????
10. Just to clarify, poker is not the same as odds games like slots, roulette and blackjack where the player plays against the house. The player in poker must pay a rake to the house, to be sure, but in the odds games mentioned above, those who are profitable over a given time period are profitable because of random chance, whereas poker players are profitable because of their skill level relative to the skill levels of their opponents.
The rake is more like a commission. The edge in roulette over the long run belongs to the house over all players. The edge in poker over the long run belongs to the skilled players over the unskilled ones, just like it is in trading.
I trading and in poker, it is not true that the only people that will win in the long run are the houses. The houses and the skilled players will win, the vast majority will lose.
#10 Dec 6, 2008
|
dclm_baseline
| 127,346
|
Understanding the Key Components of Lean Methodology
The Lean Six Sigma methodology consists of principles that can be applied to any business model in order to achieve success. It consists of a unique set of tools, procedures, and practices that have been developed to assist enterprises in better management and an increased bottom line. Understanding the core components of Lean can provide a number of options for different companies looking to improve upon outcomes.
Professionals often seek the Six Sigma process when a company wishes to improve upon the quality of its output. The aim is to reduce the incidence of waste and to improve upon the tools and the knowledge that individuals possess in industry. There are a number of problems that businesses can encounter including poor customer satisfaction, low production, and decreased revenue. This process is implemented in hopes of counteracting these possible problems.
A Six sigma project is often called upon when particular processes are contributing to limitations in the ability to assist the business in achieving its full potential and objectives. Modifications involve attention to detail, identifying strengths and weaknesses, and developing methods to eliminate waste. This will support businesses in working towards the greatest levels of profitability and continued success.
The first principle that is used includes the law of the market. In this instance, the focus is on the satisfaction of the customer and requires that clients are always put first. The organization will need to ensure that all employees and departments understand that such processes always need to be met as the customers are responsible for the ongoing business that is generated.
Flexibility is the second principle that requires companies to incorporate measures that can be easily maneuvered to fit with changes and the growing needs of businesses. The result is the ability to draw on procedures that are not rigid, but instead can be tweaked and altered in an efficient manner where developments emerge. It is important for companies to be able to adapt and respond in the shortest period of time.
The third component includes focus where organizations are required to pay attention to the problems within the business and not on the entire organization. Key personnel are required to recognize the weaker areas of the enterprise and to work on strengthening such aspects to prevent against waste and failure. The downfalls include encountering distractions and not attending to problems efficiently.
Velocity considers the number of processes that a company will be required to implement. When these measures are all running simultaneously, it can hinder performance and the ability to remain ahead of changes or problems. The work input should always be proportional to its output.
The law of complexity is the last component and emphasizes the fact that business success can be achieved if procedures and input are kept simple. The rise in the size and complexity of processes will contribute to confusion among employees, the over use of unnecessary resources, and distractions. The components of lean state that add a number of elements and forget about the value of the customer in industry can prove detrimental for a company.
|
dclm_baseline
| 4,018,307
|
"""
test_teds_access_services
----------------------------------
Tests for `teds` module.
"""
# !/usr/bin/env python
# -*- coding: utf-8 -*-
import unittest
import mock
from ncaplite import ieee1451types as ieee1451
from ncaplite import teds_support
from ncaplite import teds_access_services
from ncaplite import transducer_services_base
class TestTEDSAccessServices(unittest.TestCase):
"""TestCase for TEDS Access Services."""
def test_read_transducer_channel_teds(self):
"""Test service for reading TransducerChannel TEDS."""
def open_mock(tim_id, channel_id):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
trans_comm_id = 1
return {'error_code': error_code, 'trans_comm_id': trans_comm_id}
def close_mock(trans_comm_id):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
return {'error_code': error_code}
def update_teds_cache_mock(trans_comm_id, timeout, teds_type):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
xmlpath = 'tests/SmartTransducerTEDSMock.xml'
xmlns = {'teds': 'http://localhost/1451HTTPAPI'}
teds_types ={teds_support.TEDSType.CHAN_TEDS: "teds:TransducerChannelTEDS",
teds_support.TEDSType.XDCR_NAME: "teds:UserTransducerNameTEDS"}
tc_dict = dict()
for k, v in iter(teds_types.items()):
my_teds = teds_support.teds_element_from_file(v, xmlns, xmlpath)
tc_dict[k] = my_teds[0]
self.tedcash[trans_comm_id] = tc_dict
return {'error_code': error_code}
self.tedcash = dict()
def read_teds_mock(trans_comm_id, timeout, teds_type):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
ted = self.tedcash[trans_comm_id][teds_type]
tc = ieee1451.TypeCode.STRING_TC
value = ted
arg = ieee1451.Argument(tc, value)
arg_array = ieee1451.ArgumentArray()
arg_array.put_by_index(0, arg)
return {'error_code': error_code, 'teds': arg_array}
tdaccs = mock.Mock(spec=transducer_services_base.TransducerAccessBase)
tdaccs.open.side_effect = open_mock
tdaccs.close.side_effect = close_mock
tedsmgr = mock.Mock(spec=transducer_services_base.TedsManagerBase)
tedsmgr.update_teds_cache.side_effect = update_teds_cache_mock
tedsmgr.read_teds.side_effect = read_teds_mock
tedsvc = teds_access_services.TEDSAccessServices()
tedsvc.register_transducer_access_service(tdaccs)
tedsvc.register_teds_manager(tedsmgr)
xmlpath = 'tests/SmartTransducerTEDSMock.xml'
xmlns = {'teds': 'http://localhost/1451HTTPAPI'}
tc_teds_list = teds_support.teds_element_from_file('teds:TransducerChannelTEDS', xmlns, xmlpath)
expected = teds_support.teds_dict_from_xml(tc_teds_list[0])
args = {"ncap_id": 1234,
"tim_id": 1,
"channel_id": 1,
"timeout": ieee1451.TimeDuration(secs=1, nsecs=0)
}
result = tedsvc.read_transducer_channel_teds(**args)
tcted_arg_array = result['transducer_channel_teds']
tcted = tcted_arg_array.get_by_index(0).value
ted_dict = teds_support.teds_dict_from_xml(tcted)
self.assertEqual(ted_dict, expected)
def test_read_user_transducer_name_teds(self):
""" Test reading UserTransducerNameTEDS
"""
def open_mock(tim_id, channel_id):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
trans_comm_id = 1
return {'error_code': error_code, 'trans_comm_id': trans_comm_id}
def close_mock(trans_comm_id):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
return {'error_code': error_code}
def update_teds_cache_mock(trans_comm_id, timeout, teds_type):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
xmlpath = 'tests/SmartTransducerTEDSMock.xml'
xmlns = {'teds': 'http://localhost/1451HTTPAPI'}
teds_types ={teds_support.TEDSType.CHAN_TEDS: "teds:TransducerChannelTEDS",
teds_support.TEDSType.XDCR_NAME: "teds:UserTransducerNameTEDS"}
tc_dict = dict()
for k, v in iter(teds_types.items()):
my_teds = teds_support.teds_element_from_file(v, xmlns, xmlpath)
tc_dict[k] = my_teds[0]
self.tedcash[trans_comm_id] = tc_dict
return {'error_code': error_code}
self.tedcash = dict()
def read_teds_mock(trans_comm_id, timeout, teds_type):
error_code = ieee1451.Error(
ieee1451.ErrorSource.ERROR_SOURCE_LOCAL_0,
ieee1451.ErrorCode.NO_ERROR)
ted = self.tedcash[trans_comm_id][teds_type]
tc = ieee1451.TypeCode.STRING_TC
value = ted
arg = ieee1451.Argument(tc, value)
arg_array = ieee1451.ArgumentArray()
arg_array.put_by_index(0, arg)
return {'error_code': error_code, 'teds': arg_array}
tdaccs = mock.Mock(spec=transducer_services_base.TransducerAccessBase)
tdaccs.open.side_effect = open_mock
tdaccs.close.side_effect = close_mock
tedsmgr = mock.Mock(spec=transducer_services_base.TedsManagerBase)
tedsmgr.update_teds_cache.side_effect = update_teds_cache_mock
tedsmgr.read_teds.side_effect = read_teds_mock
tedsvc = teds_access_services.TEDSAccessServices()
tedsvc.register_transducer_access_service(tdaccs)
tedsvc.register_teds_manager(tedsmgr)
xmlpath = 'tests/SmartTransducerTEDSMock.xml'
xmlns = {'teds': 'http://localhost/1451HTTPAPI'}
tc_teds_list = teds_support.teds_element_from_file('teds:UserTransducerNameTEDS', xmlns, xmlpath)
expected = teds_support.teds_dict_from_xml(tc_teds_list[0])
args = {"ncap_id": 1234,
"tim_id": 1,
"channel_id": 1,
"timeout": ieee1451.TimeDuration(secs=1, nsecs=0)
}
result = tedsvc.read_user_transducer_name_teds(**args)
tcted_arg_array = result['transducer_name_teds']
tcted = tcted_arg_array.get_by_index(0).value
ted_dict = teds_support.teds_dict_from_xml(tcted)
self.assertEqual(ted_dict, expected)
if __name__ == '__main__':
unittest.main()
|
the_stack
| 329,924
|
§ 20. Influence of imperial reform 60-70's on rozvytokNaddnipryanskoyi Ukraine (textbook)
§ 20. Impact imperial reform 60-70's the development of Dnieper Ukraine
Remember: 1. What was the impact of reforms in the Habsburg development of Western lands? 2. When and how was abolished serfdom at Western? 3. Why the existence of serfdom prevent further economic and social development Naddniprianshchyna?
1. Features of Economic Development Naddniprianshchyna reform before 1861
Late 50-ies Dnepr Ukraine was the most developed economic region in the plane Russian Empire. But further economic development is increasingly hindered existence of serfdom. The development industry resisted no market free "labor and narrow demand for industrial products. Existence serfdom also hindered the process of stratification of the peasantry and the formation social strata of rural entrepreneurs and rural laborers. Besides urban entrepreneurs and merchants could not buy land because it was the subject of free buying and selling.
In mid-century gradually okreslyuvatysya started new paths of economic specialization of regions Naddniprianshchyna. In the forefront of the South took the place of sheep wheat for export. On the left bank of the remoteness of markets grain production caused in the southern part of the main transformation source of livestock income in the northern - flax and tobacco industry. Significant gains landowners Left Bank gave distilling. Right Bank in 40-ies become a main area of sugar production. Reduction share of farms in the Right Bank and grain development tsukroburyakovoho production reflects the changes that took place in the economy Naddniprianshchyna.
During the first half Nineteenth century. When accustomed lands of the South, there prevailed sheep, holders Right Bank estates specializing in growing corn, which sold through the Black Sea ports. Grain economy was based on the Right Bank serfs working in the South - dominated the use of free- workforce. With sales of grain to most of the owners estates Right Bank accumulated considerable resources and, at the same time, taking away land from peasants in grain, caused the appearance of a large number of landless serfs. Therefore landlords, based on existing demand, invested in building sugar factories and laid beet plantations, employing landless peasants. The development also contributed to the production tsukroburyakovoho good natural conditions, large forests, which gave fuel for factories and protectional Russian politics government. All this provided saharozavodchika high profits.
At first the sugar industry was mainly landlord. Gradually, her allowing a merchant capital. But merchants were unable to use free labor and serfs were attract laborers. In the mid-nineteenth century. in Dnieper Ukraine merchants, entrepreneurs rented 24 sugar factories and the land around them.
So tsukroburyakova Industry gained market character and their contents become the direct opposite of dominant feudal system of management. The development of sugar industry caused changes in other sectors. Requirements for further development economy called for the elimination of serfdom.
2. Background of the peasant reform
Peasant and other reforms 60-70's in the Russian Empire had the character of delayed modernization carried out under the pressure of historical circumstances. Its implementation empire pushed a few historical conditions:
· further existence of serfdom empire threatened to turn into a secondary country, as eloquently testified her defeat in the Crimean War. Peasant movement Ukrainian provinces in the late war, constant demands release of villagers tsarism by all the opposition organizations, impacts of elimination of serfdom Western Ukraine in 1848 - all of which also contributed transformation problem abolition of serfdom in urgent need of further development Empire;
· serfdom inhibited the rate of economic development. Presence in the Empire regions such as South, convincingly proves the advantages of free- labor, however, caused numerous inconsistencies between the relations which are here existed, and order in other parts of the empire. However, the landlords' farms gave about 50% of commodity grain, and their rapid and simultaneous elimination could produce catastrophic consequences;
· serfdom in form and content too seemed to slavery. Immorality possession "baptized property "zasudzhuvalasya overwhelming majority of representatives of different social groups. Besides the example of European monarchies, which eliminated the ring law, testified that the act does not shake thrones of monarchs, but rather strengthened them and opened the possibility of accelerating social and economic development.
3. Social Movement for Reform
Opinions about the necessity of eliminating serfdom increasingly spread among all strata of the Ukrainian society. Understanding of serfdom as a big social disaster caused the appearance passionate poetry of Taras Shevchenko and prose works of the young Ukrainian writer Mark spinner. They, according to contemporaries, was "one huge The charges against serfdom slavery.
At the same time demands are not heard confined to the release of the peasants. In his letter to the editor of the Russian emigre journal Bell in London Nikolai Kostomarov in general treating positive intentions of the imperial government to abolish serfdom, rose also to ensure that the rights of peasants urivnyaty of nobles and create conditions for unhindered development of Ukrainian language as the basis of Ukrainian cultural solution. He also expressed the need for a Ukrainian state as part of its Single Slavic Union.
Advocated the elimination of serfdom merchants and entrepreneurs, as understood that it is a major obstacle for development of industry and trade. Among the landowners were also a lot of those who held liberal positions. Known for its progressive views Ukrainian leader, owner of large estates in Poltava and Chernihiv Gregory Galagan (1819-1888) was a member commissions to develop projects for reform. Consistently advocating interests of the Ukrainian peasantry, he wrote that "delay this transformation (Abolition of serfdom .- Auto.) May be most harmful effects with all the horrors popular uprisings.
G. Galagan
Rationale for abolition of serfdom squire A. Koshelev (1858)
"Following as landlords farms and feudal life, listening to the words of nobles, peasants and yard people and watching the actions of those and other ... I came to full conviction that it was the deadline for taking decisive action on the abolition of serfdom status ...
Firstly, the discontent of peasants against landowners growing every day ...
Second, the desire of the peasants and yard people appears to be free from the increasing strength and honesty ...
Third, impoverished landlords farmers increases considerably. This factor is unmistakable ... handling reached sizes incredible ... Stock peasant patience is exhausted.
... Seventh, the improvement in Agriculture always require the use of free labor to it. While we have a boon as we can to hire people as you want and moreover, nothing, until these people appear to us with their even any, but costless instruments until the landlord can change every minute peasant plots of land - not until Farming prosunetsya forward.
Eighth, the success of plant and factory industry require increasing the number of available workers. Now many people are trapped in unproductive jobs landowners (in hallway, the barn, etc.). Moreover, various boon, neurochna work absorbs twice, three times more working time than is necessary. All this time, all These workers need to turn to industry, and now feel the labor shortages greatly reduced.
Yet many other circumstances compel much to the early abolition of serfdom ... "
Judge: 1. Why the author considers the abolition of serfdom becomes necessary for further social and economic development Empire? 2. Describe the basic reasons for the abolition of serfdom.
However, most landlords were much more conservative in their views and if agreed to release personal peasants, only without the allocation of land.
Antykriposnytski views and spread among the students. During the 1856-1860 biennium in Kharkov Kiev University and operated a secret political society that is united More than 40 people. Members of the community tried to raise the people to fight for abolition of serfdom. However, his political views they do not go further general political upheaval in the Russian Empire in order to establish constitutional monarchy or republic. The necessity of struggle for national liberation of the Ukrainian people in their views Society members display not found.
4. Features reform 1861 in Dnieper Ukraine
Preparation of peasant reform lasted five years. February 19, 1861 Alexander II signed the "Charter", containing all legislation and reform manifesto to abolish serfdom law. "Regulations" in a wing of 17 acts, most important of which were "total position of farmers, freed from serfdom and four local provisions for certain regions of the empire. Dnieper subject to an three local regulations. In the southern provinces and the southern part of the Kharkov province acted "Great Russian local regulations, on the Right Bank - Individual local situation, "the Left -" Little Russian local regulations.
Emperor Alexander II
Laws peasant Reform decide such basic issues:
· cancel serfdom and defining the new legal status of peasants;
· organization peasant self-government;
· creation Institute conciliator;
· empowerment of peasants land;
· determination duties tymchasovozobov'yazanyh peasants;
· redemption of the land peasants.
Immediately after the publication of the manifesto farmers receive personal freedom. Been announced that henceforth the former serfs rural resident are free and have civil rights - to marry, enter their own property agreements, to act on its behalf in court, open commercial and industrial enterprises to move to other states to acquire real estate and more. However, the estate fully inequality peasants likviduvalasya. They remained lower as assessor, had to bear capitation and other money and natural duties, exposed to corporal punishment, were released from more privileged position.
According to "Regulations" all farmers were to join in the community (community) - a form of peasant governments, which were based on common economic interests. Allotment land was given not to individual farms and communities. On Right Bank and in provinces where there were "littleRussia's position, for payment of fees and taxes in line with each farming separately in other regions of solidarity meet the whole congregation. Fails to pay its tax debt peasant distributed between other community members.
To address economic issues community members gathered on rural stairs. Connecting rural communities cooperating in the parish. In the village elder and parish officers, they elected relied local administrative and household functions: monitor order, organize the requirements of higher authorities and government laws.
Agricultural Activities self and relationships with landlords controlled the peasants Justice of the agents. As conceived by the Russian government were to interfere abuse of landowners. However, most conciliator was itself landlords and protect their own interests first, sometimes even against law.
Central to the reform occupied on the ground. Acts of reform based on the recognition by landowners ownership of all land in the estates, including the farmers' holdings. The farmers declared by users of the land, bound by her vidroblyaty establishing "Regulations" duties - servage or serfdom. To take ownership of their inheritance, a farmer had to buy it landowner. Dimensions duties and plots were established for each area separately, depending on the quality of land. In the South, was installed only "ukazna" rule of inheritance - from 3 to 6.5 acres. On the left bank is higher portion size was 2,7-4,5 tithes, lower - half high. For farmers Right Bank were tied to put in those sizes, marked by "inventory rules "1847-1848 biennium On average, farmers receive tithes 1,9-2,3 land. Landlords, by law, had the right to cut off the peasants "extra" land. Only on the Right Bank peasant holdings were increased. It cause need for the Russian government to win over villagers after suppression Polish uprising of 1863-1864 biennium Overall size of the peasant land use in bank Ukraine was reduced after the reform by 27%.
Given that, according economists to ensure the minimum needs of a farm (the play and pay numerous taxes) needed at least 5 acres land, the vast majority of Ukrainian peasants did not receive this required standard. In addition, many serfs engaged in household landlords dismissed without allocation of land.
As long as peasants vykupaly land for their intended position tymchasovozobov'yazanyh. Dimensions peasant inheritance and duties determined by separate agreements. Their conclusion major concern was a conciliator. Increase duties without larger portion landowners prohibited. But the law did not provide reduction obligations in connection with a reduction in holdings. Therefore, after the interval "Excess" land obligations, the peasants actually increased.
The final stage of peasant reform had become a redemption transaction. Villagers were obliged to pay for their inheritance fifth of the sum of its value. The rest of the state debt provided in peasant who was to return it, together with interest within 49 years. By following almost half a century peasants had to pay the state three times for size of the time value of the land.
Implementation of the redemption transaction completed the separation of farm landlord. At the same time landowners receive money for which to implement the restructuring of their economies a new way.
Cancel serfdom in the Ukrainian lands
5. Changes in the position of state peasants
"Regulations" peasant reform extended to the former serfs. However, significant changes took place in status of state peasants, who accounted for 44% of all peasants Dnieper.
This category of population belonged to the former Cossacks, military settlers, the colonists of the South.
According to the decree in 1866 all ground state peasants, while remaining the property of the state, tied to the rural communities in perpetual use (Podvirna or public). Farmers had to pay for it annually to the treasury "Obrochne submit. State farmers had the right to become owners of plots by the payment of their ransom for 8 years, but the size of their holdings installed no more than 8-15 acres. As a result of this reform exercise which lasted 20 years, the situation of public peasants became slightly better than the former serf. Grant, whom they received were on average twice as large than in the former serfs. In 1886 all state peasants were obliged to redeem their holdings, but the redemption money for them were significantly lower than serfs. Hindered the development of economy state rural communities with the introduction of shared responsibility for payment taxes.
In much better conditions were Southern colonists. Having big to put reform, they have kept them and following it, and got all sorts of benefits. For example, they paid two to three times less "Obrochne submit" than other state farmers. The vast majority of farms colonists was by its nature, large farms, which widely used agricultural equipment and camp follower power, rented, but their plots, large plots of land.
6. The peasant reform
Reform of 1861 caused significant changes in the position of peasants, who at that time were the majority Naddniprianshchyna population. However, because the land here was much better than in other regions of the empire, it is here that most farmers have lost land due to "Segments". The introduction of public forms of land ownership and land use tlumylo individual initiative villagers prevented the formation of market relations in agriculture. The need to pay excessive prices for land prevented farmers develop their own economy. Reform is not created in the rural feeling that they own the land. Under these conditions, redemption payments were seen as new tax, rather than as partial payment received for the land.
Still, despite the large number of defects in general peasant reform of 1861 have a positive value. It eliminated the existing obstacles to socio-economic modernization of the empire.
7. Reform of charge
Abolition of serfdom led to reforms in local government proceedings finance, education, censorship, military affairs. The aim of these measures was to further modernization of steps in other areas of life empire.
To improve the system Local Government 1864 Zemskov reform was carried out. Behind her in counties and provinces zaprovadzhuvalosya Zemskov (local) government. It consisted of a filament (county and province of county fees) and executive (county and provincial management of county) agencies. By county of open meetings elected (MPs) units of different classes - landowners, local owners and farmers. At the county meeting in turn elected vowels province of county fees. Prevailed at the meeting of county landowners.
In the Dnieper Ukraine action of school reform covered only the Left Bank and the South. On the Right Bank was much opposition to the Russian government, the Polish gentry, who participated in the Polish uprising of 1863 Therefore, fearing that activity zemstva contribute to a new revival of Polish national liberation movement in the province, their not created until 1911
Zemstvo Naddniprianshchyna gradually transformed into competent authorities, who conducted a great job in various areas of local governance. Most cases they are funded themselves. Zemstvo created a network of county and provincial agriculturists, organized activities to increase farmers rilnytskoyi culture - delivering seeds, farm machinery, organized courses, exhibitions, reading lectures. Because of their support came peasant cooperative movement, credit institutions for farmers. Zemstvo also took care of construction and maintenance ways, the organization's email service, statistical surveys. They provide food assistance to populations in lean years.
Zemstvo maintained primary and professional school training were free. They repeatedly approached with a request to Government for implementation in schools in the study of Ukrainian language. Unfortunately, all these requests aside, and teachers who tried use the Ukrainian language only as a subsidiary since it was Villagers realized, liberated.
Zemstvo reorganized network health care on principles no charge, the district character create the necessary conditions for inpatient population. One of the biggest achievements of school of medicine was the organization of an effective system to combat epidemics.
Activities zemstva led to in Ukrainian society that they are perceived not only as local governments, but also as an organization, opposition to the policy of Russian Tsars. It is from county agencies left many well-known in future Ukrainian public figures. Zemstvo Naddniprianshchyna repeatedly treated with petitions to the Russian government about the necessity of constitutional reforms.
Activities of county agencies had large positive value for Naddniprianshchyna. On the one hand, they became the school which taught people to the government, on the other - contributed raising national awareness.
8. City and Financial Reform
For urban reform in 1870 governments in the cities varied from birth to bezstanove. According to Act 1875 in all cities Naddniprianshchyna created municipal councils. Suffrage was given to men of 25 years, but only those who were property owners, commercial and industrial enterprises. Workers officials, intellectuals, which together accounted for most of the urban population, but property and had not paid taxes, had no right to vote.
City Duma
Municipal councils, in turn, elected City Council - a permanent executive bodies. He headed the Duma and the administration of Mayor. Activities Council and its chairman controlled the governor, who could cancel any of its decisions. City Duma took care of landscaping, contributed to the development of local trade and industry, health and public education.
Despite the limited reforms municipal government, she was positive, since replaced old city authorities had new leadership, based on bourgeois principle of qualification.
Economic development needs prompted the imperial government to conduct financial reforms. In particular, 1860 , the State Bank, which received preferential right credit trading and industrial enterprises. In all provinces were established independent of the local administration of the Chamber. These monthly reviewed the expenditures of all local institutions. With this move, the government tried to somehow resist abuse and bribery, the extent of which in Russia were high. When the emperor once instructed to check which of its executives is not bribes, he reported that only three of fifty refuse "togifts ", ie not assign local taxes population.
It was canceled a long otkupnyh system for collecting taxes, which the majority of the collected money to the treasury not nahodyla, and pocket leaseholder. Instead installed excise taxes collected by state agencies. However, farmers and townspeople were forced, as continue to pay the poll-tax, from which exempted only the privileged states.
9. Judicial and military reform
Conducted in 1864 Judiciary reform was the most consistent bourgeois character. Because she was made an attempt vyvyschyty imperial justice to the level of contemporary legal Science and judicial practice of European countries. Earlier in the Empire existed is, closed, depending on local court administration with the lack of security charges and red tape. According to new court bezstanovym court statutes became independent from administrative authorities. The meeting of the judiciary were open to the public.
The central part of the new Justice was the District Court in each province. The prosecution maintained prosecutor, defended the interests of the defendant's lawyer. Jurors in the number of 12 people were selected randomly from representatives of all states. They heard both sides, installed, guilty or not guilty of the defendant and the sentence defined by law, the judge and members of the court.
Consideration of minor civil and criminal cases led magistrate. He chose zemstvo or municipal councils. The local administration had the opportunity to remove from office the magistrate or Court of the District Court.
And yet, despite many positive Rice, reform of the judiciary has not been completed. Continued to not linked the general judicial system bower parish court for the peasants, some courts to military and clergy. In court township farmers judged by local law used humiliating punishment beaten. As a result, farmers who were majority of the Dnieper hardly felt the benefits of new imperial judiciary.
The defeat of the Empire in the Crimean War unaochnyla need to reform the army. Due to reorganize military control the country were redistributed 15 county of the three of them - Kiev, Odessa and Kharkiv - Ukrainian were province. The heads of districts were commanders, who led located on their territory and the troops immediately obey military ministry. Government Empire controlled the national composition of the units, because of that units, located in the Dnieper Ukraine, Ukrainian were not 40% of the total composition. However, Ukrainian rozporoshuvaly in parts located in other parts of the empire.
Decisive role in implementation of military reform has played in 1874 instead of recruiting sets a new system of army recruitment through the implementation of general conscription for men who have reached 20 years of age, regardless of birth origin. The term of military service in the Army Troops of 6, the Navy - 7 years. In the late nineteenth century. was found 5-year service life for all generations of troops. Individuals who have received education, serving less time. Military reforms strengthened the imperial army, but in Ukrainian Peasants special joy they have not called, as they recruit changed the total duty military service.
10. Reforms in education and censorship
Economic development needs of the empire caused the need for educational reform. Admitted 1863 new university charter, which among other effects spread to Kharkiv and Kyiv University, gave them a wide autonomy in matters domestic life. University as a whole and each professor had the opportunity freely receive from abroad, any books, send for training young scientists abroad. Training for the majority of students remained paid.
Fourth male House gymnasium. Kiev
Educational reform launched in Empire women's higher education. Although women were not allowed to study in universities introduced them to private institutions of higher women's courses. Thus was opened in Kiev higher rates for women with natural (1870), Physics, Mathematics and History and Philology (1878) Science.
Institute of noble maidens. Kiev
N.Polonska-Vasilenko - himnazystka. 1897
Transformation in the field of secondary education consisted in the reorganization of schools. The right to study in them licked representatives of all states, but high amount of tuition paid unemozhlyvlyuvav obtaining a secondary education of children of common people. Gymnasium divided into real and classic, with a separate study of boys and girls. End of Classical High School gave the opportunity to enter a university, the real - to higher technical school.
Third man home gymnasium. Kiev
Funduklei gymnasium (high school female Kiev)
According to the reform was introduced only primary education system. Schools could establish private persons zemstvo other public institutions, but with government permission. An elementary school was to educate children in the spirit allegiant Empire for the content of education in all types of schools uchylyschnoyu controlled provincial council. Teaching in primary schools (except private) was free.
The imperial government saw a large danger in spreading ideas of freedom-loving printed word. It issued a series of laws that strengthen means of administrative influence on organs of print. Institution of imperial and ecclesiastical censorship in Dnieper Ukraine received wide powers to fight against any attempt to promote press organs awakening of national consciousness. Editors of newspapers and magazines under threat forbidden to touch the closure of acute political issues. Violation of these requirements caused the temporary closure of the issue or even ban it.
11. The consequences of reform 60-70's
Reforms 60-70's had eliminate the vestiges of the feudal imperial reality, but to save same basis of Russian autocracy. In pursuing reform, the empire used the administrative-command methods of management, support bower system in all spheres of social and political life. Above the newly local governments, courts, educational institutions, the press kept control of the imperial administration. But the objective of reforms in some areas has led to unpredictable consequences of empire. Most notable examples This was the reform of charge, which dragged the emergence of effective and authoritative local governance, even to put forward national demands of the imperial government. Reform of charge testified that members of oppressed Empire Ukrainian people have used any possible legal remedies for create their own self-government and improve living conditions.
Reforms 60-70's were Proper time concessions that they made the Russian Empire, and had intended to a full-fledged environment for development of both Ukrainian and any with other nations, oppressed empire.
Questions and Tasks
1. Give facts that show that feudal relations Naddniprianshchyna hindered economic development. (Preparing the response, use the document 1.)
2. What were the conditions of peasant reform?
3. Describe the social movement for the elimination of serfdom in Dnieper Ukraine.
4. Make a table "Features reform in 1861 Dnieper in Ukraine "by the scheme:
The main questions reforms
How are they resolved?
5. How has the situation of former serfs? What rights do they have?
6. What changes have occurred in circumstances state peasants?
7. Describe the effects of peasant reform.
8. Make a table "Impact of Reforms 60-70's the development of Dnieper Ukraine "by the scheme:
Name of Reform
Impact on development Naddniprianshchyna
9. Discover the consequences of reforms 60-70's
|
dclm_baseline
| 1,955,946
|
Santa María Temaxcalapa là một đô thị thuộc bang Oaxaca, México. Năm 2005, dân số của đô thị này là 924 người.
Tham khảo
Đô thị bang Oaxaca
|
common_corpus_web
| 2,020,749
|
Quick Answer: What Is Considered Excessive Idling?
Is excessive idling bad?
As you spend more time idling, the average temperature of the spark plug drops.
This makes the plug get dirty more quickly, which increases fuel con- sumption by 4 to 5 %.
Excessive idling also lets water con- dense in the vehicle’s exhaust, leading to cor- rosion and a reduction of the life of your exhaust system..
Why do cops let their cars idle?
Each time the vehicle is cut off, an officer has to turn off all that equipment or run the risk of draining the vehicle’s battery power in minutes. … During traffic stops, officers are supposed to keep their vehicles running so they can quickly give chase if the person being pulled over decides to flee.
How long can a car idle before overheating?
All cars should be made to be able to idle indefinitely without overheating. Basically after so many minutes the car with adequate cooling should have reached a stable thermal equilibrium and if that’s so then it should idle for an hour, two hours, 8 hours, 12 hours, a day with no further issues.
Why you should not idle your car?
It’s unhealthy. Idling fumes can lead to a number of major health concerns, as they have been linked to asthma, an overall decrease in lung function, cardiac disease, and even cancer. Simply put, they are horrible to breathe in.
Is it OK for a car to idle for a long time?
How long can a car run idle?
How much fuel does a 5.3 burn at idle?
If you drive a Chevrolet Silverado with the 5.3 litre engine, then it uses (5.3 litres engine displacement x 0.6) = 3.2 litres per hour of fuel while idling.
What color gets pulled over the most?
How long can a semi truck idle?
Running the engine allows the air conditioner to run, so the driver can rest comfortably. Some diesel engines require a cooldown period before shutting off. This means the engine may be left to idle for up to 30 minutes after the vehicle is stopped.
Can I leave my car running while I sleep in it?
To prevent such tragedies, check for the smell of exhaust fumes while parking and starting the engine, and take heed of any symptoms of dizziness or exhaustion. … But the most important thing is to avoid sleeping in a parked car with the engine and air-conditioner running.
Is it better to idle in park or neutral?
Shift to Neutral When Stopped If you’re not moving but your engine is running, you’re getting zero miles per gallon. … This shift is even more important when the air conditioner is running, so the engine doesn’t have to strain so hard while idling. A manual transmission should be shifted to neutral at every stop.
How long is excessive idling?
Ten seconds of idling can burn more fuel than turning off and restarting the engine. Plus, excessive idling can damage your engine’s components, including spark plugs, cylinders and exhaust systems.
What is excessive idling?
How do police cars stay running?
1)Keeping the AC/heater on during the hot/cold months. 2)The vehicles act as a repeater with their radios. While they have batter life, keeping the vehicle running keeps them from drawing on the battery reserves.
What causes engine idle problems?
Is idling with AC on bad?
Is it bad to leave your car running for 30 minutes?
Letting your car idle while in park might not seem like a big deal. … However, letting your car idle is actually detrimental to the modern automotive engine, wastes gasoline, and causes environmental damage. Modern engines do not, in fact, need more than a few seconds or idling time before they can be driven safely.
How many miles is an hour of idling equal to?
30 milesMileage Is Inadequate to Measure Engine Wear Idling is the main reason to track engine hours instead of mileage. When a vehicle idles for hours each day using a PTO, mileage isn’t being recorded by the odometer, but one hour of idling is equal to 25 to 30 miles of driving.
|
dclm_baseline
| 4,172,789
|
Slashdot is powered by your submissions, so send in your scoop
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Comment: Re:If they can do it to Google, they can do it to (Score 1) 341
by Kashif Shaikh (#33417420) Attached to: The Case For Oracle
I heard that Sun used to charge royaltees for mobile phones that licensed the J2ME platform - complaint or certified implementations. And this kinda proves this: From Q: Does deploying Java SE for embedded devices or purposes require a royalty? A: Yes. Sun's license for Java SE enables it to be freely used for general purpose desktops or servers. If Java SE is bundled as part of a dedicated solution that involves or controls hardware of some kind, then it's likely an embedded application and is subject to modest royalty payments. I believe Google java implementation is custom - not part of SE or ME - so they avoid having pay any royaltees to Oracle. This is why I think Oracle is pissed.
|
dclm_baseline
| 3,593,777
|
Hi, me again doing another MWPP fic. This one's not going to be a oneshot though. I thought one with a bit more plot and structure would be better.
Disclaimer: I'm from Missouri. I you don't know, that's in America so…don't own. Wish I did though…lots of money.
"My god Moony, wake up."
Remus Lupin slowly raised his head from his arm. They were in the Great Hall eating breakfast. Peter was eating; Sirius was turned around mouthing words to his Ravenclaw girlfriend. James sat right across from him, staring over his glass of pumpkin juice.
"What's it to you?"
"Well we just got down here…WAKE UP!"
Remus jolted back awake and glared at James. Before the argument could even start, Sirius turned around. "Ease up Prongs. You know how he is…this time of month."
"Shut your mouth Sirius! Besides, it's not that time of month. Check the calendar."
"Moony calm down…Moony? Remus? For the love of god! WAKE UP!"
Remus glared at James. "If you don't stop yelling I swear by Merlin's pants I'll crucio you into next week."
Peter looked up at the two. Sirius smoothly got out of his seat and went to sit with the Ravenclaws. Remus was glaring at James and vise-versa. In an attempt to ease the tension, Peter said: "Why exactly do we swear by Merlin and his clothes? I mean…"
James and Remus both shouted at him: "Shut-up Wormtail!"
Remus, having had enough, got up, grabbed his bag, and left for Divination. Sirius looked after him for a bit before giving his girlfriend a peck on the cheek and going over to where he had been sitting. James grunted at him.
"Nicely handled Prongs."
"Shut-up. Not my fault he's such a prat."
"Don't start." James sighed, "Sorry. I'm just worried about him. He doesn't just fall asleep, not even the day after."
Sirius nodded and grabbed his bag. James and Peter followed suit and the three went to the North Tower to join Remus.
Reviews are loved so give a bitch some love huh? Click it…why are you still reading this? CLICK THE BUTTON. LOL take your time.
|
dclm_baseline
| 2,789,276
|
Two Zero Electric Motorcycles Catch Fire While Charging In Hong Kong
Zero S Police Motorcycle
Zero S Police Motorcycle
Two years ago, Hong Kong’s police force ordered 52 Zero S ZF9 electric motorcycles.
To the best of our knowledge, no major issues have been reported in regards to the use of these Zero motorcycles, that is until just recently when two charging Zeros burst into flames.
As the South China Morning Post reports:
“Fifty electric motorcycles from the police force have been pulled from daily operation after another two were damaged in a blaze that broke out at Ngau Tau Kok police station…”
“The two motorbikes were being charged at the police station’s car park on Siu Yip Street when one of them burst into flames…”
The one that caught fire the ignited the second electric bike, as well as a nearby police van.
South China Morning Post adds:
“Firemen fought the blaze with a hose and doused the flames… No one was injured, and firefighters found no suspicious cause.”
An investigation is ongoing, but the cause of the blaze is believed to be a short circuit, likely within the motorcycle.
Until the investigation is complete, all 52 Zero motorcycles have been removed from service.
Source: South China Morning Post
Categories: Bikes
Tags: ,
Leave a Reply
13 Comments on "Two Zero Electric Motorcycles Catch Fire While Charging In Hong Kong"
newest oldest most voted
Way to take a vacation.
Or way to play 52 pick-up !!
They should just be more “AWARE” maybe pick a new charging location and leave the bikes in service. This could only be an isolated incident. Why throw the baby away with the bath water???
Because lawyers exist. And have created the ultimate “Better be safe than sorry” society.
The rest of the world and China especialy is not like the US where you can spil your coffe on yourself and sue the resturant…
That case isn’t nearly as cut and dry as the mainstream media led everyone to believe.
“That case isn’t nearly as cut and dry…”
YES, it IS! Basically, the upshot of this and a plethora of other similarly stupid cases is, that if you are a moron and do something moronic that causes you injury, you can sue whoever it was that gave you the means to demonstrate the fact that you are a moron.
“During discovery, McDonalds produced documents showing more than 700 claims by people burned by its coffee between 1982 and 1992. … This history documented McDonalds’ knowledge about the extent and nature of this hazard.” It also documents that those 700 people (at least) are morons.
Coffe is HOT (who knew!) Why on *earth* would you try to hold a just-brewed plastic cup of the stuff *between you thighs!!!!!*, to put milk etc in it? Seriously, the idea is completely insane and it should have been thrown out of court immediately.
Stop bad mouthing lawyers! Can you imagine a world without lawyers?
I have a 2015 Zero S and I had an issue with my onboard charger overheating and melting. Luckily my breaker tripped before a fire started. I dropped it off at the dealer June 29th and I’m still waiting for it to be returned. Last I heard the new charger was still overheating and they were awaiting a new BMS board to fix the problem.
I spoke to the nearest ZERO dealer near me (east of Rochester), and asked about the optional battery pack (this was a year ago). He said they haven’t ordered any since they caused an ‘interference problem’.
Was the internal charger, the typical 10 amp, 120 volt unit that overheated on you?
Then don’t plug a 100 into a 220.
You do know that the onboard chargers are rated for both ranges, right? 110-240 VAC, 50/60 Hz
Who cares if it will start minor unscheduled oxidation reaction consuming a shack or few of them some night :/ Brave people don’t care about rules!
|
dclm_baseline
| 3,218,069
|
/* -------------------------------------------------------------------------- */
/* Copyright 2002-2020, OpenNebula Project, OpenNebula Systems */
/* */
/* Licensed under the Apache License, Version 2.0 (the "License"); you may */
/* not use this file except in compliance with the License. You may obtain */
/* a copy of the License at */
/* */
/* http://www.apache.org/licenses/LICENSE-2.0 */
/* */
/* Unless required by applicable law or agreed to in writing, software */
/* distributed under the License is distributed on an "AS IS" BASIS, */
/* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. */
/* See the License for the specific language governing permissions and */
/* limitations under the License. */
/*--------------------------------------------------------------------------- */
package goca
import (
"encoding/xml"
"errors"
"github.com/OpenNebula/one/src/oca/go/src/goca/parameters"
"github.com/OpenNebula/one/src/oca/go/src/goca/schemas/cluster"
)
// ClustersController is a controller for Clusters
type ClustersController entitiesController
// ClusterController is a controller for Cluster entity
type ClusterController entityController
// Clusters returns a Clusters controller.
func (c *Controller) Clusters() *ClustersController {
return &ClustersController{c}
}
// Cluster returns a Cluster controller.
func (c *Controller) Cluster(id int) *ClusterController {
return &ClusterController{c, id}
}
// ByName returns a Cluster ID from name
func (c *ClustersController) ByName(name string) (int, error) {
var id int
clusterPool, err := c.Info(false)
if err != nil {
return -1, err
}
match := false
for i := 0; i < len(clusterPool.Clusters); i++ {
if clusterPool.Clusters[i].Name != name {
continue
}
if match {
return -1, errors.New("multiple resources with that name")
}
id = clusterPool.Clusters[i].ID
match = true
}
if !match {
return -1, errors.New("resource not found")
}
return id, nil
}
// Info returns a cluster pool. A connection to OpenNebula is
// performed.
func (cc *ClustersController) Info(decrypt bool) (*cluster.Pool, error) {
response, err := cc.c.Client.Call("one.clusterpool.info", decrypt)
if err != nil {
return nil, err
}
clusterPool := &cluster.Pool{}
err = xml.Unmarshal([]byte(response.Body()), clusterPool)
if err != nil {
return nil, err
}
return clusterPool, nil
}
// Info retrieves information for the cluster.
func (cc *ClusterController) Info() (*cluster.Cluster, error) {
response, err := cc.c.Client.Call("one.cluster.info", cc.ID)
if err != nil {
return nil, err
}
cluster := &cluster.Cluster{}
err = xml.Unmarshal([]byte(response.Body()), cluster)
if err != nil {
return nil, err
}
return cluster, nil
}
// Create allocates a new cluster. It returns the new cluster ID.
func (cc *ClustersController) Create(name string) (int, error) {
response, err := cc.c.Client.Call("one.cluster.allocate", name)
if err != nil {
return -1, err
}
return response.BodyInt(), nil
}
// Delete deletes the given cluster from the pool.
func (cc *ClusterController) Delete() error {
_, err := cc.c.Client.Call("one.cluster.delete", cc.ID)
return err
}
// Update adds cluster content.
// * tpl: The new cluster contents. Syntax can be the usual attribute=value or XML.
// * uType: Update type: Replace: Replace the whole template.
// Merge: Merge new template with the existing one.
func (cc *ClusterController) Update(tpl string, uType parameters.UpdateType) error {
_, err := cc.c.Client.Call("one.cluster.update", cc.ID, tpl, uType)
return err
}
// AddHost adds a host to the given cluster.
// * hostID: The host ID.
func (cc *ClusterController) AddHost(hostID int) error {
_, err := cc.c.Client.Call("one.cluster.addhost", cc.ID, int(hostID))
return err
}
// DelHost removes a host from the given cluster.
// * hostID: The host ID.
func (cc *ClusterController) DelHost(hostID int) error {
_, err := cc.c.Client.Call("one.cluster.delhost", cc.ID, int(hostID))
return err
}
// AddDatastore adds a datastore to the given cluster.
// * dsID: The datastore ID.
func (cc *ClusterController) AddDatastore(dsID int) error {
_, err := cc.c.Client.Call("one.cluster.adddatastore", cc.ID, int(dsID))
return err
}
// DelDatastore removes a datastore from the given cluster.
// * dsID: The datastore ID.
func (cc *ClusterController) DelDatastore(dsID int) error {
_, err := cc.c.Client.Call("one.cluster.deldatastore", cc.ID, int(dsID))
return err
}
// AddVnet adds a vnet to the given cluster.
// * vnetID: The vnet ID.
func (cc *ClusterController) AddVnet(vnetID int) error {
_, err := cc.c.Client.Call("one.cluster.addvnet", cc.ID, int(vnetID))
return err
}
// DelVnet removes a vnet from the given cluster.
// * vnetID: The vnet ID.
func (cc *ClusterController) DelVnet(vnetID int) error {
_, err := cc.c.Client.Call("one.cluster.delvnet", cc.ID, int(vnetID))
return err
}
// Rename renames a cluster.
// * newName: The new name.
func (cc *ClusterController) Rename(newName string) error {
_, err := cc.c.Client.Call("one.cluster.rename", cc.ID, newName)
return err
}
|
common_corpus_source
| 1,669,469
|
import numpy as np
import pandas as pd
from loguru import logger
from scipy import stats
from sklearn import preprocessing
from sklearn.decomposition import PCA
from statsmodels.sandbox.stats.multicomp import multipletests
from .constants import GROUP_COL, IDS, PKS
class WebOmicsInference(object):
def __init__(self, data_df, design_df, data_type, remove_cols=None, min_value=0, replace_mean=True):
data_df = data_df.copy()
# remove all the default columns from dataframe if nothing provided
if remove_cols is None:
remove_cols = ['padj_', 'FC_', 'significant_', 'obs', PKS[data_type], IDS[data_type]]
remove_cols = tuple([x.lower() for x in remove_cols])
to_drop = list(filter(lambda x: x.lower().startswith(remove_cols), data_df.columns))
df = data_df.drop(to_drop, axis=1)
# remove rows that are all NAs and all 0s
df = df.dropna(how='all')
df = df.loc[~(df == 0).all(axis=1)]
self.data_df = df
self.design_df = design_df
self.data_type = data_type
# data imputation:
# - if all zero in group then replace with min_value
# - replace any other zeros with mean of group
self._impute_data(min_value, replace_mean=replace_mean)
# def heatmap(self, N=None, standardize=True, log=False):
# if standardize:
# data_df = self.standardize_df(self.data_df, log=log)
# else:
# data_df = self.data_df
# if N is not None:
# plt.matshow(data_df[0:N])
# else:
# plt.matshow(data_df)
# plt.colorbar()
def standardize_df(self, data_df, log=False, axis=1):
if data_df.empty:
return data_df
data_df = data_df.copy()
# log only if all the values are positive
# assume if there's a negative value, the data has been pre-processed, so don't do it again
data_arr = np.array(data_df)
if np.all(data_arr >= 0) and log:
data_arr = np.log(data_arr)
# center data to have 0 mean and unit variance for heatmap and pca
scaled_data = preprocessing.scale(data_arr, axis=axis)
# set the values back to the dataframe
sample_names = data_df.columns
data_df[sample_names] = scaled_data
return data_df
def run_deseq(self, keep_threshold, case, control):
logger.info('DeSEQ2 support is not available, please install rpy2 extra package')
def run_ttest(self, case, control):
logger.info('t-test case is %s, control is %s' % (case, control))
count_data = self.data_df
col_data = self.design_df
sample_group = col_data[col_data[GROUP_COL] == case]
case_data = count_data[sample_group.index]
sample_group = col_data[col_data[GROUP_COL] == control]
control_data = count_data[sample_group.index]
nrow, _ = count_data.shape
pvalues = []
lfcs = []
indices = []
# if there's a negative number, assume the data has been logged, so don't do it again
log = np.all(count_data.values >= 0)
# T-test for the means of two independent samples
for i in range(nrow):
case = case_data.iloc[i, :].values
control = control_data.iloc[i, :].values
idx = count_data.index[i]
# remove 0 values, which were originally NA when exported from PiMP
case = case[case != 0]
control = control[control != 0]
# log the data if it isn't already logged
if log:
case_log = np.log2(case)
control_log = np.log2(control)
else:
case_log = case
control_log = control
statistics, pvalue = stats.ttest_ind(case_log, control_log)
if not np.isnan(pvalue):
lfc = np.mean(case_log) - np.mean(control_log)
pvalues.append(pvalue)
lfcs.append(lfc)
indices.append(idx)
# correct p-values
reject, pvals_corrected, _, _ = multipletests(pvalues, method='fdr_bh')
result_df = pd.DataFrame({
'padj': pvals_corrected,
'log2FoldChange': lfcs
}, index=indices)
return result_df
def run_limma(self, case, control):
logger.info('limma support is not available, please install rpy2 extra package')
def get_pca(self, rld_df, n_components, plot=False):
df = rld_df.transpose()
pca = PCA(n_components=n_components)
X = pca.fit_transform(df)
# if plot:
# fig, ax = plt.subplots()
# ax.scatter(X[:, 0], X[:, 1])
# for i, txt in enumerate(df.index):
# ax.annotate(txt, (X[i, 0], X[i, 1]))
# plt.tight_layout()
# fn = '{uuid}.png'.format(uuid=uuid.uuid4())
# plt.save(fn)
cumsum = np.cumsum(pca.explained_variance_ratio_)
return X, cumsum
def _impute_data(self, min_value, replace_mean=True):
if self.design_df is not None:
grouping = self.design_df.groupby('group')
for group, samples in grouping.groups.items():
# If all zero in group then replace with minimum
temp = self.data_df.loc[:, samples]
temp = (temp == 0).all(axis=1)
self.data_df.loc[temp, samples] = min_value
if replace_mean:
# replace any other zeros with mean of group
subset_df = self.data_df.loc[:, samples]
self.data_df.loc[:, samples] = subset_df.mask(subset_df == 0, subset_df.mean(axis=1), axis=0)
else:
# replace all 0s with min_value
self.data_df = self.data_df.replace(0, min_value)
|
the_stack
| 340,167
|
World's Largest Sheet Music Selection
Messe de Minuit pour Noel
Mitternachtsmesse zu Weihnachten H 9
By Marc-Antione Charpentier
Be the first! Write a Review
SST(A)TB vocal soli, ST(A)TB choir
Mitternachtsmesse zu Weihnachten H 9. Composed by Marc-Antione Charpentier. Edited by Hans Ryschawy. This edition: urtext. French Sacred Music. Sacred vocal music, Latin Masses; Christmas. Choral score. 24 pages. Duration 25 minutes. Published by Carus Verlag (CA.2102905).
Item Number: CA.2102905
ISBN M-007-17196-4. Size: A4 inches. Language: Latin.
Marc-Antoine Charpentier's Christmas mass, the "Messe de minuit", is throughly in the French tradition, celebrating the joyful news of the Birth of Christ with cheerful happiness. In this midnight mass, composed in 1694, ten traditional French Christmas hymns have been rearranged, lending the work dancelike energy, as well as charming tonal color. In independent movements the rather small instrumental ensemble consisting of two flutes, strings and organ plays a significant role in the musical events of this mass, alternating with various combinations of the voices. Solo voices are only used together in small ensembles and, in accordance with the practices of the period, the boundaries between solo ensemble and choir are fluid; thus these solo passages can also be sung by members of the choir.
Close X
I am a music teacher.
|
dclm_baseline
| 1,114,297
|
Marital Fidelity in a MySpace World
From the first relationship in the Garden itself, it's clear spouses don't always have their partner's best interest in mind.
You know the story – the serpent tempted Eve who in turn coaxed Adam who was with her to bite into the forbidden fruit. Adam, of course, was Eve's husband, her helpmate, her friend, her confidant. And yet, there's nothing in the Genesis account that would indicate Eve struggled with her decision to involve Adam. Did she not care about him? Did she not love him? Certainly, but isn't it interesting that "love" just wasn't enough of a protection in this particular situation?
In a nutshell, Eve bit hook, line and sinker when the devil convinced her that the real reason God didn't want her to eat the fruit was because he was trying to withhold from them something exceedingly good. In essence, the serpent was calling God a liar. All of a sudden Eve went from someone completely vested in her husband to one who contributed to his downfall (which is not to say Adam didn't have the responsibility for his own actions).
It doesn't take a degree in Marriage and Family to realize that husbands and wives have for thousands of years been making similarly selfish decisions, many with lifelong negative consequences. One thing that is different, however, is that in today's world, couples face technologically-driven temptations along this line as well. Old enticements often come repackaged with a new twist using today's computers and the anonymity that comes within cyberworld.
A MySpace World
Obviously, the Internet has introduced pornography into many, many homes. Although a huge problem, I want to focus on less obvious temptations affecting some marriages. For instance, the other day while doing research about social networking sites, I stumbled upon a MySpace profile that I'm sure represents untold thousands. Here was a man who described himself as married with several children. But he had built his entire online profile describing himself as some type of, well, I'll use the term "sex-god." He had posted photos of women in various sensual poses, often with very little clothing. In an online poll he had pasted into his site, he had asked sexually-oriented questions in inappropriate fashion.
I couldn't help but wonder why a married man would do such a thing? Nor could I help but wonder what his wife or children would think if they discovered his secret online "life." (Perhaps his wife does know, which presents a whole other set of problems). Because the Internet world is considered a private domain where the user can be anonymous, this man apparently felt safe to try and make himself out to be some sort of Casanova.
Whatever the reason, he's certainly not alone. Just ask Sue Hoogestraat. According to an article in The Wall Street Journal, she was understandably upset when she discovered her husband had a virtual marriage in Second Life, a fictional cyberworld where online visitors participate in commerce, sex and relationships, and build their own imaginary world. Although Ric Hoogestraat had never met, or even spoken with, Janet Speilman, the woman who controlled her online character, the two constantly spent time "together" in cyberspace – eventually "marrying." "It's really devastating," expressed Sue. "You talk to someone or bring them a drink, and they'll be having sex with a cartoon." Ric can't understand why his wife would have a problem with his online escapades. In his mind, it's all a big game. But is it?
The Dangers of Anonymity
Brad Paisley's hit country single, "Online," humorously describes a pudgy loner who boasts of chatting with several women online (or at least he thinks they're women). As he chats, he claims to be taller and thinner than he actually is. Although the song intentionally pokes fun at how people can and do represent themselves via the anonymity of the Internet, it's no laughing matter when a married person goes online intentionally looking for someone to fill a need that only his or her spouse should satisfy.
Some may argue that having an edgy social networking profile, carrying on online relationships and chatting within cyberspace is relatively innocent stuff because there's no intention of really getting involved physically. I'd argue to the contrary. It's not just getting involved physically that crosses the line, it's the very desire to play around online in an area involving a certain amount of intimacy.
For some married individuals who feel under-appreciated and under-respected, the anonymity of the Web allows for a chance to flirt and pretend. What could be wrong with that? Dr. David Jeremiah, commenting about a list of "hedges" that individuals need to put up to guard their marriages, had this to say about flirting: "Never flirt, even in jest. Never flirt with someone other than your [spouse]." Of course, flirting used to be something that primarily occurred face-to-face. Not any more. Flirting can now take place through chat, text message, IM, online gaming and profiles. Still, Dr. Jeremiah's advice is valid.
The Gift of Intimacy
The author of Proverbs offers incredible advice along these lines when he says: "Drink water from your own cistern, running water from your own well. ...May you rejoice in the wife of your youth...may her breasts satisfy you always. ...Why be captivated, my son, by an adulteress?" (Proverbs 5:15-20, NIV) Clearly, the writer is not referring to actual water but to the intimacy that occurs in marriage. Furthermore, this water is more than just the sexual aspect. It includes the closeness, the oneness and the entire sense of unity that a husband and wife can and do experience when they do things God's way.
One of the greatest gifts the Lord ever gave us was the gift of intimacy. As described in the Bible, "For this reason a man will leave his father and mother and be united to his wife, and the two will become one flesh" (Matthew 19:5, NIV). Jesus added, "So they are no longer two, but one" (Matthew 19:6, NIV). It's that oneness, that intimacy, that water that married individuals are called upon by marriage's Creator to guard closely, pray about regularly and fight for diligently.
1. {{ footnote.fullText }}
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|
dclm_baseline
| 4,555,900
|
Icon Key
Bookmark and Share
Job Search
Use sales letters after you have sent letters to personal friends, business acquaintances, and recruiters.
The letter to friends is like the grand opening for a store. It's the first step in your marketing campaign. It announces your candidacy to the world and launches your job search.
Write to recruiters second, because they are in the employment business. Although they don't work for you, and they fill a relatively small number of positions, they match candidates with employment opportunities all day long. There just might be a fit.
If you skip the friendship and recruiter letters and start with direct mail to companies, you're skipping two important building blocks in a campaign, and you may lengthen your job search unnecessarily.
There are some exceptions to this, of course. If you were an Executive Vice President for MicroSoft, and if you know Bill Gates personally, for example, most other software companies would probably want to talk to you immediately.
Sales letters are letters of self-introduction. They make "cold calls" for you and open doors that appear closed. Use them to scan the market to see who might be interested in talking further. Save yourself from being beaten up on the telephone (although you must follow most letters with a call). Calling to follow up on a well-written letter is much easier than calling cold.
Sales Letters That Open Doors
William S. Frank, M.A.,
25 Reasons I love consulting.
by William S. Frank
5. Variety. Every day is different.
|
dclm_baseline
| 1,790,850
|
//
// StopOrderModel.h
// AWBTCChinaAPIDemo
//
// Created by mafengwo on 15/3/19.
// Copyright (c) 2015年 Aldaron. All rights reserved.
//
#import <Foundation/Foundation.h>
#import "AWBaseModel.h"
/**
* 止损止盈订单Model
*/
@interface StopOrderModel : AWBaseModel
/** 止损止盈订单号. **/
@property (nonatomic, readonly) NSInteger stopOrderID;
/** bid | ask **/
@property (nonatomic, readonly) NSString *type;
/** 触发止损止盈订单的BTC/LTC单价. 如果设置了追单波动价格数量/率，触发单价将由系统自动调节. **/
@property (nonatomic, readonly) NSNumber *stop_price;
/** 追单波动数量，设置后可动态决定止损止盈单价. **/
@property (nonatomic, readonly) NSNumber *trailing_amount;
/** 追单波动率，设置后可动态决定止损止盈单价. **/
@property (nonatomic, readonly) NSNumber *trailing_percentage;
/** 由止损止盈订单触发的订单中，所下订单的BTC/LTC单价. **/
@property (nonatomic, readonly) NSNumber *price;
/** [BTCCNY|LTCCNY|LTCBTC] **/
@property (nonatomic, readonly) NSString *market;
/** 由止损止盈订单触发的订单中，所下订单的BTC/LTC数量. **/
@property (nonatomic, readonly) NSNumber *amount;
/** 止损止盈订单下单时间. **/
@property (nonatomic, readonly) NSInteger date;
/** [ open | closed | cancelled | error ] **/
@property (nonatomic, readonly) NSString *status;
/** 由止损止盈订单触发的订单的订单号，如果止损止盈订单尚未触发或者已被取消则值为null. **/
@property (nonatomic, readonly) NSInteger order_id;
@end
|
common_corpus_source
| 331,625
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\begin{document}
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\title{On rigid plane curves} \author{Victor Kulikov and Eugenii Shustin} \date{}
\maketitle
\begin{abstract} In the article, we exhibit a series of new examples of rigid plane curves, that is, curves, whose collection of singularities determines them almost uniquely up to a projective transformation of the plane. \end{abstract}
{\small {\bf Keywords:} plane curves, rigid curves, equisingular families of curves
{\bf2010 Mathematics Subject Classification:} 14H10, 14H20, 14H50 }
\section*{Introduction} \subsection{Background and motivation} We work over the complex field, though most of results can be stated over any algebraically closed field of characteristic zero.
The space ${\mathcal C}_d$ of plane curves of degree $d$ can be identified with $\mathbb P^{d(d+3)/2}$. It has a natural {\bf equisingular stratification} with the strata determined by the collection of degrees and multiplicities of irreducible components and by the collection of topological singularity types of the considered curves (see \cite{Z1,Z2}; below, the strata will be called the {\it families of equisingular curves}). Properties of this stratification have been studied by algebraic geometers since 19th century, attracting attention of leading experts like Zeuthen, Severi, Segre, Zariski and others (see, for example, \cite{GLS} for a modern survey in this area).
In this paper we focus on the minimal equisingular families, that is, those which are formed by reduced curves and contain only finitely many orbits of the action of the group of projective transformations of the plane ${\operatorname{Aut}}(\mathbb P^2)$. The curves belonging to these families are called {\bf rigid} curves; the corresponding families we also call {\bf rigid} (see Definition \ref{rigid} below). The study of rigid curves is motivated by their appearance in several important problems. First of all, finite coverings of the projective plane branched along rigid curves are used to obtain examples of rigid, so called, Miyaoka -- Yau surfaces (see, for example, \cite{Hir}, \cite{K-H}). Next (cf. \cite{Pa}), celebrated Belyi's Theorem \cite{Be} says that each projective curve defined over $\overline{\mathbb Q}$ can be represented as a finite covering of $\mathbb P^1$ branched at three points. Note that any three points in $\mathbb P^1$ are rigid in the sense of Definition given below. Therefore we can hope that for any field $F$ of transcendence degree two over $\overline{\mathbb Q}$ there are a projective model $X$ defined over $\overline{\mathbb Q}$ with the field of rational functions $\overline{\mathbb Q}(X)\simeq F$ and a finite morphism $f:X\to \mathbb P^2$ branched along a rigid plane curve. There are interesting relations to the geometry of line arrangements (see \cite{Hir}) and to rational cuspidal curves (see \cite{ZO}).
The goal of our note is to exhibit examples of rigid curves of any degree and any genus, and rigid families covered by arbitrarily many orbits of the ${\operatorname{Aut}}(\mathbb P^2)$-action.
\subsection{Definitions and main results} Throughout the paper we consider isolated plane curve singular points up to topological equivalence, briefly calling any class of topologically equivalent singular points a {\it singularity type}. Given a singularity type $S$, the number of irreducible components of singular curve germs of type $S$ is an invariant, which we denote $m_S$.
Cardinality of a finite set $F$ will be denoted by $|F|$.
\subsubsection{Irreducible rigid curves} Let $S_1,...,S_r$ be a sequence of distinct singularity types, $n_1,...,n_r$ a sequence of positive integers, $r\ge1$. Introduce the formal sum ${\mathbf S}=\sum_{i=1}^rn_iS_i$. Given a positive integer $d$, denote by $V(d;g;{\mathbf S})$ the (equisingular) family of reduced, irreducible plane curves of degree $d$ having precisely $\sum_{i=1}^rn_i$ singular points and such that $n_i$ singular points are of type $S_i$, $i=1,...,r$. Here $$g=\frac{(d-1)(d-2)}{2}-\sum_{i=1}^rn_i\delta(S_i)$$ is the geometric genus of the considered curves.
Such a family is a locally closed union of quasiprojective subvarieties of ${\mathcal C}_d$ (cf. \cite{GLS}). It is, of course, invariant with respect to the action of ${\operatorname{Aut}}({\mathbb P}^2)$, and hence consists of entire orbits of the ${\operatorname{Aut}}({\mathbb P}^2)$-action.
\begin{df} \label{rigid-irr} We say that a non-empty equisingular family $V(d,g,{\mathbf S})$ is {\bf $k$-rigid} if it is the union of $k$ distinct orbits of the ${\operatorname{Aut}}({\mathbb P}^2)$-action in ${\mathcal C}_d$ for some $k\in {\mathbb N}$. If $k=1$ then we say that $V$ is {\bf strictly rigid.} The curves belonging to a $k$-rigid family of plane curves are called {\bf rigid.} \end{df}
\subsubsection{Reducible rigid curves} Considering reduced, irreducible curves, we, first, introduce families of reducible curves with numbered components, then identify families obtained from each other by permutation of components.
Let ${\bf d}=(d_1,\dots, d_N)$ and ${\bf g}=(g_1,\dots, g_N)$ be two collections of integers, $d_i\geq 1$ and
$g_i\geq 0$ for $i=1,\dots, N$. To encode the distribution of singularity types among components and the distribution of local branches centered at singular points that are intersection points of components, we do the following. For a fixed singularity type $S$ and fixed $N$ denote by ${\bf J}_S=\{ J_{S,k}\}$ the set of all nonempty subsets $J_{S,k}$ of $\{1,2,...,N\}$, $1\leq k\leq\sum_{j=1}^{m_S} \binom{N}{j}$, such that $|J_{S,k}|\leq m_{S}$. Let $V({\bf d};{\bf g}; \sum_{\{ S_j\}}\sum_{\{ J_{S_j,k}\}}n_{J_{S_j,k}}S_j)$ be the family of plane reduced curves $\overline C=C_1\cup\dots\cup C_N\subset {\mathbb P}^2$ such that $C_i$ are irreducible curves of degree $\deg C_i=d_i$ and genus $g_i$, and for each type $S_j$ of plane singularities the intersection $\cap_{i\in J_{S_j,k}}C_i$ contains exactly $n_{J_{S_j,k}}$ singular points of $\overline C$ of the type $S_j$ which do not lie in $C_l$ for $l\not\in J_{S_j,k}$. This is a locally closed union of quasiprojective subvarieties of ${\mathcal C}_d$ (cf. \cite{GLS}).
The sum ${\bf S}=\sum_{\{ S_j\}}\sum_{\{ J_{S_j,k}\}}n_{J_{S_j,k}}S_j$ is called the {\bf singularity type} of the curves $\overline C\in V({\bf d};{\bf g};{\bf S})$. We identify the families $V({\bf d};{\bf g};{\bf S})$ obtained by permutations of the curves $C_1,\dots,C_N$ and compatible permutations of ${\bf S}$.
A singularity type ${\bf S}$ splits into two parts, ${\bf S}={\bf S}^{ess}+{\bf S}^{non-ess}$, as follows. For fixed ${\bf d}$ and ${\bf g}$, we say that ${\bf S}^{ess}$ is an {\bf essential part} of the singularity type ${\bf S}$ (and resp. ${\bf S}^{non-ess}$ is a {\bf non-essential part} of the singularity type ${\bf S}$) if the family $V({\bf d;\bf g; \bf S})$ is determined uniquely by ${\bf d}$, ${\bf g}$, and the property that the curves $\overline C$ have the singularities ${\bf S}^{ess}$ among the all singularities of $\overline C$. If ${\bf S}^{ess}$ is an essential part of a singularity type ${\bf S}$, then we will use notation $V({\bf d;\bf g;\bf S}^{ess}+\dots)$ to denote the family $V({\bf d;\bf g;\bf S})$. \begin{df} \label{rigid} We say that a family $V=V({\bf d;\bf g; \bf S})$ is {\bf $k$-rigid} if it is the union of $k$ distinct orbits of the ${\operatorname{Aut}}({\mathbb P}^2)$-action in ${\mathcal C}_d$ for some $k\in {\mathbb N}$. If $k=1$ then we say that $V$ is {\bf strictly rigid.} The curves belonging to a $k$-rigid family of plane curves are called {\bf rigid.} \end{df}
Note that the number of irreducible components of a $k$-rigid family $V({\bf d;\bf g;\bf S})$ is less or equal $k$ (since it can (and does) happen that some orbits can lie in the closure of another) and, in particular, $V({\bf d;\bf g;\bf S})$ is irreducible if it is strictly rigid. Note also that if a family $V({\bf d};{\bf g};{\bf S})$ is rigid, then \begin{equation} \label{dim} \dim V({\bf d};{\bf g};{\bf S})\leq \dim PGL(\mathbb C,3)=8.\end{equation}
\subsubsection{Main results} Our results are as follows: \begin{itemize} \item in Theorem \ref{d4}, the complete list of rigid curves of degree $\leq 4$ is given; \item in Theorem \ref{cl2}, we give an infinite series of examples of strictly rigid families of irreducible rational curves $V(d;0;{\bf S})$; \item in Theorem \ref{t-oe1}, for each $g\geq 1$, we prove the existence of strictly rigid irreducible plane curves of genus $g$; \item examples of irreducible $2$-rigid families of irreducible curves are given in Theorems \ref{add2} and \ref{add3}, and Theorem \ref{rigit} provides examples of $k$-rigid families $V({\bf d};{\bf g};{\bf S})$ consisting of $k$ irreducible components for each $k\in {\mathbb N}$. \end{itemize}
We do not know answers to the following questions, which seem to be interesting:
\begin{que} Do there exist irreducible $k$-rigid families $V({\bf d};{\bf g};{\bf S})$ with $k> 2${\rm ?}
Do there exist irreducible $2$-rigid families $V(d;g;{\bf S})$ with $g\geq 1${\rm ?}\end{que}
Through the paper, we use the following notations for singularity types of plane curves: \begin{itemize} \item $T_{m,n}$, $2\leq m\leq n$, is the type of singularity given by equation $x^m+y^{n}=0$; but if $m=2$ then a singularity of type $T_{2,n}$, as usual, will be denoted by $A_{n-1}$ and the singularities of types $T_{m,m}$ will be called {\it simple.} \item $T^m_{m,n}$, $2\leq m< n$, is the type of singularity given by equation $y(x^m+y^{n})=0$. \item $T^n_{m,n}$, $2\leq m< n$, is the type of singularity given by equation $x(x^m+y^{n})=0$. \item $T^{m,n}_{m,n}$, $1\leq m< n$, is the type of singularity given by equation $xy(x^m+y^{n})=0$. \end{itemize}
{\small {\bf Acknowledgements}. The first author has been supported by grants of RFBR 14-01-00160 and 15-01-02158, and by the Government of the Russian Federation within the framework of the implementation of the 5-100 Programme Roadmap of the National Research University Higher School of Economics, AG Laboratory. The second author has been supported by the Hermann-Minkowski-Minerva Center for Geometry at the Tel Aviv University and by the German-Israeli Foundation grant no. 1174-197.6/2011. Main ideas behind this work have appeared during the visit of the second author to the Higher School of Economics (Moscow). We are grateful to these institutions for support and excellent working conditions.} \section{Rigid curves of small degree} In the following Theorem, we provide the complete list of rigid reduced curves of degree $\leq 4$.
\begin{thm} \label{d4} Let $\overline C$ be a rigid reduced curve of degree $\leq 4$. Then $\overline C$ belongs to one of the following families: \begin{itemize} \item[$(I)$] strongly rigid families: \begin{itemize} \item[$(I_1)$] $V(1;0;\emptyset)$; \item[$(I_2)$] $V((1,1);(0,0);A_1)$; \item[$(I_3)$] $V(2;0;\emptyset)$; \item[$(I_4)$] $V((1,1,1);(0,0,0);3A_1)$; \item[$(I_5)$] $V((1,1,1);(0,0,0);T_{3,3})$; \item[$(I_6)$] $V((2,1);(0,0);2A_1)$; \item[$(I_7)$] $V((2,1);(0,0);A_3)$; \item[$(I_8)$] $V(3;0;A_1)$; \item[$(I_9)$] $V(3;0;A_2)$; \item[$(I_{10})$] $V((1,1,1,1);(0,0,0,0);6A_1)$; \item[$(I_{11})$] $V((1,1,1,1);(0,0,0,0);T_{3,3}+3A_1)$; \item[$(I_{12})$] $V((2,1,1);(0,0,0);2A_3+A_1)$; \item[$(I_{13})$] $V((2,1,1);(0,0,0);A_3+3A_1)$; \item[$(I_{14})$] $V((2,1,1);(0,0,0);T_{2,4}^2+A_1)$; \item[$(I_{15})$] $V((2,2);(0,0);A_5+A_1)$; \item[$(I_{16})$] $V((2,2);(0,0);A_7)$; \item[$(I_{17})$] $V((3,1);(0,0);A_5+A_1)$; \item[$(I_{18})$] $V((3,1);(0,0);T_{2,4}^2)$; \item[$(I_{19})$] $V((3,1);(0,0);T_{2,3}^3)$; \item[$(I_{20})$] $V((3,1);(0,0);A_{2}+A_5)$; \item[$(I_{21})$] $V((3,1);(0,0);A_3+A_2+A_1)$; \item[$(I_{22})$] $V(4;0;3A_2)$; \item[$(I_{23})$] $V(4;0;A_4+A_2)$; \item[$(I_{24})$] $V(4;0;A_{6})$;\end{itemize} \item[$(II)$] irreducible $2$-rigid families: \begin{itemize}\item[$(II_{1})$] $V((3,1);(0,0);T_{2,3}^2+A_1)$; \item[$(II_{2})$] $V(4;0;T_{3,4})$. \end{itemize} \end{itemize} \end{thm} {\bf Proof.} It is well known that if $\deg \overline C\leq 3$ then only smooth cubics are not rigid. All other families $V({\bf d};{\bf g}; {\bf S})$ of curves of degree $\leq 3$ are listed in $(I_1)$ -- $(I_9)$. Therefore we assume below that $\deg \overline C=4$.
Again, it is well known that if $\overline C$ consists of four lines or two lines and a quadric, then $\overline C$ is not rigid if and only if $\overline C$ consists of four lines having a common point or $\overline C$ consists of a quadric $Q$ and two lines in general position with respect to $Q$. The rigid families in the case when $\overline C$ consists of four lines or two lines and a quadric is listed in $(I_{10})$ -- $(I_{14})$.
Consider the case when $\overline C$ consists of two irreducible components: either $\overline C=Q_0\cup Q_1$, where $Q_0$ and $Q_1$ are smooth quadrics, or $\overline C=C\cup L$, where $C$ is a cubic and $L$ is a line.
Consider the case when $\overline C_1=Q_0\cup Q_1\in V((2,2);(0,0);{\bf S})$. We have ${\bf S}=m_1A_1+m_3A_3+m_5A_5+m_7A_7$, where $m_1+2m_3+3m_5+4m_7=4$. The singularity type ${\bf S}$ consists of $k=m_1+m_3+m_5+m_7$, $1\leq k\leq 4$, singular points and if $m_1=4$, that is, ${\bf S}=4A_1$, then $V((2,2);(0,0);4A_1)$ is not rigid by inequality \eqref{dim}, since $\dim V((2,2);(0,0);4A_1)=10$. Similarly, if $m_1=2$, that is, ${\bf S}=A_3+2A_1$, then $V((2,2);(0,0);A_3+2A_1)$ is not rigid, since $\dim V((2,2);(0,0);A_3+2A_1)=9$. So, we can assume that $m_1\leq 1$.
Consider the pencil of quadrics $Q_{\lambda}$ defined by quadrics $Q_0$ and $Q_1$. It is easy to see that in the cases $m_1=m_5=1$ or $m_7=1$ it contains the unique degenerate element $Q_{\infty}$ consisting of two (coinciding if $m_7=1$) lines $L_1\cup L_2$ and in the case $m_3=2$ it contains two degenerate elements one of which, $Q_{\infty}$, consists of two coinciding lines $L_1=L_2$ and the other one consists of two different lines. Let $f(z_1,z_2,z_3)=0$ be an equation of the quadric $Q_0$ and $l_i(z_1,z_2,z_3)=0$, $i=1,2$, be an equation of the line $L_i$. Then, without less of generality, we can assume that $$F_{\lambda}(z_1,z_2,z_3):= f(z_1,z_2,z_3)+\lambda l_1(z_1,z_2,z_3)l_2(z_1,z_2,z_3)=0$$ is the equation of $Q_{\lambda}$. Moreover, applying a projective transformation, we can assume that $f(z_1,z_2,z_3)=z_1^2-z_2z_3$ and $l_1(z_1,z_2,z_3)=z_3$, $l_1(z_1,z_2,z_3)= z_1$ in the case $m_1=m_5=1$; $l_1(z_1,z_2,z_3)=l_2(z_1,z_2,z_3)=z_1$ in the case $m_3=2$; and $l_1(z_1,z_2,z_3)=l_2(z_1, z_2,z_3)=z_3$ in the case $m_7=1$. Therefore we can assume that \begin{equation}\label{pen1} F_{\lambda}(z_1,z_2,z_3)= (z_1^2-z_2z_3)+\lambda z_1z_3 \end{equation} in the case $m_1=m_5=1$, \begin{equation}\label{pen3} F_{\lambda}(z_1,z_2,z_3)= (z_1^2-z_2z_3)+\lambda z_3^2 \end{equation} in the case $m_7=1$, and \begin{equation}\label{pen2} F_{\lambda}(z_1,z_2,z_3)= (z_1^2-z_2z_3)+\lambda z_1^2 \end{equation} in the case $m_3=2$.
Note that if $m_3=2$, then the degenerate element $Q_{-1}$ of the pencil $Q_{\lambda}$ consists of two lines $z_2=0$ and $z_3=0$ (see equation \eqref{pen2}).
In the case $m_1=m_5=1$ (case $(I_{15}))$ the strong rigidity of the curve $\overline C_1$ follows from equality $\overline C_{\lambda}=Q_0\cup Q_{\lambda}=h_{\lambda}(\overline C_1)$ for each $\lambda\in \mathbb C^*$, where the automorphism $h_{\lambda}$ acts as follows: $h_{\lambda}(z_1:z_2:z_3)=(\lambda z_1: \lambda^2 z_2:z_3)$; and in the case $m_7=1$ (case $(I_{16}))$ the strong rigidity of the curve $\overline C_1$ follows from equality $\overline C_{\lambda^2}=Q_0 \cup Q_{\lambda^2}=h_{\lambda}(\overline C_1)$ for each $\lambda\in \mathbb C^*$.
Let us show that $\overline C_1$ is not rigid if $m_3=2$. Indeed, assume that $\overline C_1$ is rigid. Then for any two elements $\lambda_{1},\lambda_{2}\in \mathbb C^*\setminus \{ -1\})$, $\lambda_{1}\neq \lambda_{2}$, there is a projective transformation $h_{{\lambda}_1,{\lambda}_2}\in{\operatorname{Aut}}(\mathbb P^2)$ such that $h_{{\lambda}_1, {\lambda}_2}(\overline C_{{\lambda}_1})=\overline C_{{\lambda}_2}$, where $\overline C_{\lambda }=Q_{0}\cup Q_{\lambda}$. The automorphism $h_{{\lambda}_1,{\lambda}_2}$ leaves invariant the pencil $Q_{\lambda}$. In particular, it leaves invariant the singular elements $Q_{\infty}$ and $Q_{-1}$ of the pencil $Q_{\lambda}$, $$h_{{\lambda}_1,{\lambda}_2}(Q_{\infty})=Q_{\infty}\, \, \, \text{and}\, \, \, h_{{\lambda}_1,{\lambda}_2}(Q_{-1})=Q_{-1}.$$ Let us show that $h_{\lambda_1,\lambda_2}(Q_0)\neq Q_0$ for $\lambda_1\neq \lambda_2$. Indeed, if $h_{\lambda_1,\lambda_2}(Q_0)= Q_0$, then in the case $m_3=2$ we have either $h^*_{\lambda_1,\lambda_2}(z_1) = az_1$, $h^*_{\lambda_1,\lambda_2}(z_2)= bz_2$, and $h^*_{\lambda_1,\lambda_2}(z_3)= cz_3$ or $h^*_{\lambda_1, \lambda_2}(z_1)= az_1$, $h^*_{\lambda_1,\lambda_2}(z_2)= bz_3$, and $h^*_{\lambda_1,\lambda_2}(z_3)= cz_2$ for some $a,b,c$ such that $a^2=bc$, since $h_{\lambda_1,\lambda_2}(Q_{\infty})= Q_{\infty}$ and $h_{\lambda_1,\lambda_2} (Q_{-1})= Q_{-1}$. Therefore $h_{\lambda_1,\lambda_2}(Q_{\lambda})= Q_{\lambda}$ for all $\lambda \in \mathbb C$ that is possible only if $\lambda_1=\lambda_2$. Therefore we must have $h_{\lambda_1,\lambda_2}(Q_{0})= Q_{\lambda_2}$ and $h_{\lambda_1,\lambda_2}(Q_{\lambda_1})= Q_{0}$. But, for three pairwise different $\lambda_1,\lambda_2,\lambda_ 3\in \mathbb C^*\setminus \{ -1\}$ we obtain that $h_{\lambda_1,\lambda_3}(Q_0)=h_{\lambda_2,\lambda_3}\circ h_{\lambda_1,\lambda_2}(Q_0)=Q_0$ and hence $\lambda_1=\lambda_3$. Contradiction.
Consider the case when $\overline C=C\cup L$, where $C$ is a cubic and $L$ is a line. It is easy to see that $\overline C$ is rigid only if $C$ is a rational curve. Therefore we have two cases: $C$ is a nodal cubic or $C$ is a cuspidal cubic.
Let $C$ be a nodal cubic. Then (the case $(I_8)$) $V(3;0;A_1)$ is strongly rigid and $\dim V(3;0;A_1)=8$. Therefore $L$ must be a "very special line" with respect to $C$, that is, $L$ is either the tangent line of one of
two branches of the node of $C$ or $L$ is the tangent line of $C$ at a flex point of $C$. In both two cases it is easy to see that $\overline C$ is strictly rigid and we have two cases: $(I_{17})$ -- $(I_{18})$.
If $C$ is a cuspidal cubic, then $L$ is a tangent line to $C$ or it passes through the cusp of $C$, since $V(3;0;A_2)$ is strongly rigid and $\dim V(3;0;A_2)=7$. If $L$ is a tangent line to $C$ at its cusp, then we have case ($I_{19}$) and if $L$ is the tangent line at the flex point of $C$, then we have case $(I_{20})$. It is easy to see that in both cases $\overline C$ is strongly rigid.
If $L$ is a tangent line to $C$ and it does not pass through the cusp of $C$, let us choose homogeneous coordinates $(z_1:z_2:z_3)$ such that $z_3=0$ is an equation of $L$, $z_1=0$ is an equation of the line tangent to $C$ at its cusp, and the line given by equation $z_2=0$ passes through the cusp of $C$ and the tangent point of $L$ and $C$. Then there is a parametrization of $C$ in $\mathbb P^2$ of the following form \begin{equation} \label{parx} z_1=t^3,\quad z_2=t^2,\quad z_3=1+at,\end{equation} where $a\neq 0$. Denote by $C_a$ a curve in $\mathbb P^2$ given by parametrization \eqref{parx}. Now, the strong rigidity of $V((3,1);(0,0);A_3+A_2+A_1)$ (case $(I_{21})$) follows from equality $h(C_a\cup L)=C_1\cup L$, where the linear transformation $h\in {\operatorname{Aut}}(\mathbb P^2)$ acts as follows, $h(z_1:z_2:z_3)=(z_1:a^{-1}z_2:a^{-3}z_3)$.
Let $L$ be not the tangent line at the cusp of $C$ and it pass through the cusp of $C$. Then if $z_2=0$ is an equation of $L$, $z_1=0$ is the line tangent to $C$ at its cusp, and the line given by equation $z_3=0$ is tangent to $C$ at its nonsingular point $p$ of the intersection $C\cap L$, then again we can assume that $C$ is given by parametrization \eqref{parx} and there are two possibilities: either $a=0$ or $a\neq 0$. Denote by $C_a$ a curve given by parametrization \eqref{parx}. It is easy to see that $p\in C_0\cap L$ is the flex point of $C_0$ if $a=0$ and $p\in C_a\cap L$ is not the flex point of $C_a$ if $a\neq 0$. Therefore there is not a linear transformation $h\in {\operatorname{Aut}}(\mathbb P^2)$ such that $h(C_0\cup L)=C_a\cup L$. On the other hand, the linear transformation $h:(z_1:z_2:z_3)\rightsquigarrow (z_1:a^{-1}z_2:a^{-3}z_3)$ sends $C_a\cup L$ to $C_1\cup L$. Therefore $V((3,1);(0,0), T_{2,3}^2)$ (case $(II_1)$) is an irreducible $2$-rigid family.
Now, consider the case when $\overline C$ is an irreducible curve of degree four. All possible type of singularities ${\bf S}$ of irreducible curves of degree four is given in \cite{Wal}: $$ {\bf S}=m_1A_1+m_2A_2+\dots m_6A_6+m_7T_{3,3}+m_8T_{2,3}^2+m_9T_{3,4},$$ where $m_i$, $1\leq i\leq 9$, are non-negative integers satisfying the following inequality: \begin{equation} \label{type} m:=m_1+m_2+2m_3+2m_4+3m_5+3m_6+3m_7+3m_8+3m_9\leq 3.\end{equation} By \cite[Theorem 2]{Sh}, we have \begin{equation} \label{dimV} \begin{array}{l} \dim V(4;3-m;m_1A_1+m_2A_2+\dots+ m_6A_6+m_7T_{3,3}+m_8T_{2,3}^2+m_9T_{3,4})= \\ 14 - (m_1+2m_2+3m_3+4m_4+5m_5+6m_6+4m_7+5m_8+6m_9). \end{array}\end{equation} Inequality \eqref{dim} applied to \eqref{dimV} gives rise to the inequality \begin{equation} \label{di} 14 - (m_1+2m_2+3m_3+4m_4+5m_5+6m_6+4m_7+5m_8+6m_9)\leq 8. \end{equation} It is easy to see that inequalities \eqref{type} and \eqref{di} have only the following non-negative integer solutions: \begin{itemize} \item[$1)$] $m_2=3$ and $m_i=0$ for $i\neq 2$, \item[$2)$] $m_2=m_4=1$ and $m_i=0$ for $i\neq 2,4$, \item[$3)$] $m_6=1$ and $m_i=0$ for $i\neq 6$, \item[$4)$] $m_9=1$ and $m_i=0$ for $i\leq 8$. \end{itemize} Therefore only the following irreducible curves $\overline C$ of degree four can be rigid: either $\overline C\in V(4;0;3A_2)$, or $\overline C\in V(4;0;A_2+A_4)$, or $\overline C\in V(4;0;A_6)$, or $\overline C\in V(4;0;T_{3,4})$.
The strict rigidity of a curve $\overline C\in V(4;0;3A_2)$ (case $(I_{22})$) is well known.
To show that a curve $\overline C\in V(4;0;A_2+A_4)$ (case $(I_{23})$) is strictly rigid, let us choose a non-homogeneous coordinate $t$ in ${\mathbb P}^1$ and homogeneous coordinates $(z_1:z_2:z_3)$ in ${\mathbb P}^2$ such that $f(t=0)=(0:0:1)$ is the singular point of $\overline C$ of type $A_4$ and the line $z_1=0$ is tangent to $\overline C$ at this point, the image $f(t=\infty)=(1:0:0)$ is the singular point of $\overline C$, and the line $z_3=0$ is tangent to $C$ at this point. Then (after a suitable linear change of the coordinate $t$) the morphism $f$ is given by \begin{equation} \label{nu1} z_1=t^4, \, \, z_2=t^2, \, \, z_3=t+1, \end{equation} that is, $\overline C$ is strictly rigid.
To show that a curve $\overline C\in V(4;0;A_6)$ (case $(I_{24})$) is strictly rigid, note that, by Pl\"ucker's formulas, $\overline C$ must have a flex point. As above, let $f:{\mathbb P}^1\to{\mathbb P}^2$ be a morphism such that $\overline C=f({\mathbb P}^1)$. We can choose a non-homogeneous coordinate $t$ in ${\mathbb P}^1$ and homogeneous coordinates $(z_1:z_2:z_3)$ in ${\mathbb P}^2$ such that $f(t=0)=(0:0:1)$ is the singular point of $C$, the line $z_1=0$ is the tangent line to $C$ at its singular point, the image $f(t=\infty)=(1:0:0)$ is a flex point of $C$, and, moreover, the line $z_3=0$ is the tangent line to $C$ at this point. Then (after a suitable linear change of the coordinate $t$) the morphism $f$ is given by \begin{equation} \label{nu2} z_1=t^4, \, \, z_2=t^2(t-1), \, \, z_3=F(t):=a_1t+a_2, \end{equation} where $a_2\neq 0$. It is easy to see that the polynomial $F(t)-a_2(t-1)^2$ must be divisible by $t^2$, since $C$ has a singularity of type $A_6$ at $(0:0:1)$. Therefore we have $a_1=-2a_2$ and we can put $a_2=1$, that is, $V(4;0;A_6)$ is strictly rigid.
Let us show that the family $V(4;0;T_{3,4})$ (case $(II_{2})$) is irreducible and $2$-rigid. Indeed, let $\overline C\in V(4;0;T_{3,4})$, then, by Pl\"ucker's formulas, $\overline C$ must have a flex point. Let $f:{\mathbb P}^1\to{\mathbb P}^2$ be a morphism such that $\overline C=f({\mathbb P}^1)$. We can choose non-homogeneous coordinate $t$ in ${\mathbb P}^1$ and homogeneous coordinates $(z_1:z_2:z_3)$ in ${\mathbb P}^2$ such that $f(t=0)=(0:0:1)$ is the singular point of $\overline C$, the line $z_1=0$ is the tangent line to $\overline C$ at its singular point, the image $f(t=\infty)=(1:0:0)$ is a flex point of $\overline C$, and, moreover, the line $z_3=0$ is the tangent line to $\overline C$ at this point. Then (maybe, after change of the coordinate $t$) the morphism $f$ is given by \begin{equation} \label{nu3} z_1=t^4, \, \, z_2=t^3, \, \, z_3=at+1. \end{equation} Therefore $V(4;0;T_{3,4})$ is an irreducible family.
As in case $(II_1)$, there are two possibilities: either $a=0$ or $a\neq 0$.
Denote by $\overline C_a\in V(4;0;T_{3,4})$ a curve given by equation \eqref{nu3}. It is easy to see that $\overline C_0$ has the unique flex point, namely, $p_1=(1:0:0)$, and one can check that the curve $\overline C_1$ has two flex points, $p_1$ and $p_2=f(t=-2)=(16:-8:-1)$. Hence there is no projective transformation $h\in {\operatorname{Aut}}({\mathbb P}^2)$ such that $\overline C_0=h(\overline C_1)$.
If we perform the change $t_1=at$, where $a\neq 0$, then we get $h(\overline C_1)=\overline C_a$, where $$h(z_1:z_2:z_3)=(a^{4}z_1:a^{3}z_2:z_3)\ ,$$ that is, $V(4;0;T_{3,4})$ is $2$-rigid.
$\Box$
\begin{rem} {\rm According to Theorem \ref{d4}, the families $V((1,1,1,1);(0,0,0,0);T_{4,4})$ and $V((2,2);(0,0); 2A_3)$ are not rigid and it is easy to see that $$\dim V((1,1,1,1);(0,0,0,0);T_{4,4})=6\, \, \, \, \text{and} \, \, \, \, \dim V((2,2);(0,0); 2A_3)=8,$$ that is, the inequality \eqref{dim} is not sufficient for $V({\bf d};{\bf g};{\bf S})$ to be a rigid family.} \end{rem}
\section{Strictly rigid rational curves of degree $\geq 5$} \begin{thm} \label{cl2} For each $n\geq 2$ the family $V_n=V(2n; 0;3T_{n,n+1}+\dots)$ is strictly rigid. The non-essential part of singularities of $C\in V_n$ consists of simple singularities. \end{thm}
To prove Theorem \ref{cl2} we need in the following \begin{thm} \label{cl3} Let ${\bf d}_n=(n,1,1,1)$, ${\bf g}=(0,0,0,0)$, and $n_{(1,2)}=n_{(1,3)}=n_{(1,4)}=1$. Then the family $$\hat{V}_n=V({\bf d}_n;{\bf g};(n_{(1,2)}+n_{(1,3)}+ n_{(1,4)})A_{2n-1}+\dots)$$ is strictly rigid for each $n\geq 2$. The non-essential part of singularities of $\overline C\in \hat{V}_n$ consists of simple singularities. \end{thm}
We prove Theorems \ref{cl2} and \ref{cl3} simultaneously. \newline {\bf Proof.} Theorem \ref{cl2} in the case $n=2$ and Theorem \ref{cl3} in the case $n\leq 3$ are well known, and we prove the general case by induction on $n$.
Assume that for $n\leq n_0$ Theorem \ref{cl2} is true. Then the number of virtual cusps of $C\in V_n$ ($n \leq n_0$) is equal to $3(n-1)$ and the number of its virtual nodes is equal to $2(n-1)(n-2)$ (see the definition of the numbers of virtual cusps and virtual nodes, for example, in \cite{Ku}). Let $\hat{C}$ be the dual curve to $C$. By Pl\"ucker's formulas, we have $\deg \hat{C}=n+1$ and the number of virtual cusps of the curve $\hat{C}$ is zero. Therefore the irreducible branches of the singular points of $\hat{C}$ are smooth and hence $\hat C$ has only simple singularities, since $C$ has the only simple singularities and three singularities of type $T_{n,n+1}$. Note also that $\hat{C}$ has three flex points, say $p_1$, $p_2$, and $p_3$. Let $L_i$, $i=1,2,3$, be the tangent line to $\hat{C}$ at the point $p_i$. It is easy to see that $$\overline C=\hat{C}\cup L_1\cup L_2\cup L_3\in \hat{V}_{n+1}=V({\bf d}_{n+1}; {\bf g};(n_{(1,2)}+n_{(1,3)}+n_{(1,4)})A_{2n+1}+\dots).$$ Moreover, if $n\leq n_0$ and the curve $\overline C_1=C_1\cup L'_1\cup L'_2\cup L'_3$ belongs to $\hat{V}_{n+1}$, then the curve $C_1$ must have only simple singularities, since the dual curve $\hat{C}_1$ belongs to $V_{n}$. Therefore the assumption that Theorem \ref{cl2} is true for $n\leq n_0$ implies the statement of Theorem \ref{cl3} for $n\leq n_0+1$.
$\Box$
\begin{rem} \label{rem} {\rm For $n\leq n_0+1$, it follows from strong rigidity that if $\overline C_i=C_i\cup L_1\cup L_2\cup L_3\in \hat{V}_{n}$, $i=1,2$, are two curves such that $C_1$ and $C_2$ have two common flex points, say $p_1$ and $p_2$, then their third flex points $p_3=C_1\cap L_3$ and $q_3=C_2\cap L_3$ should coincide, $p_3=q_3$, and consequently $C_1=C_2$.} \end{rem}
To complete the proof of Theorems \ref{cl2} and \ref{cl3}, note that if $\overline C=C \cup L_1\cup L_2\cup L_3$ belongs to $\hat{V}_{n}$ for $n\leq n_0+1$, then $\sigma(C)\in V_{n+1}$, where $\sigma$ is a quadratic transformation of the plane with centers at the vertices of the triangle $L_1\cup L_2\cup L_3$. Therefore Theorem \ref{cl2} in the case $n=n_0+1$ follows from Theorem \ref{cl3} and Remark \ref{rem}.
$\Box$
The following theorem provides an infinite series of strictly rigid families parameterizing the unions of two rational curves.
\begin{thm} \label{add1} The families $V((2n+1,1);(0,0);n_{(1,2)}(T_{2n-1,2n}^{2n}+A_1)+n_{(1)}A_{4n-2})$, where $n_{(1,2)}=n_{(1)}=1$, are strictly rigid for $n\geq 2$. \end{thm} {\bf Proof.} Let $$C\cup L\in V((2n+1,1);(0,0);(0,0);n_{(1,2)}(T_{2n-1,2n}^{2n}+A_1)+n_{(1)}A_{4n-2}).$$ Then $C\in V(2n+1;0;T_{2n-1,2n}+A_{4n-2})$ and $L$ is the line tangent to $C$ at its singular point $p_1$ of type $T_{2n-1,2n}$. Let $f:{\mathbb P}^1\to{\mathbb P}^2$ be a morphism such that $C=f({\mathbb P}^1)$. We can choose non-homogeneous coordinate $t$ in ${\mathbb P}^1$ and homogeneous coordinates $(z_1:z_2:z_3)$ in ${\mathbb P}^2$ such that $f(t=\infty)=p_1=(1:0:0)$ is the singular point of $C$ of type $T_{2n-1,2n}$ and the line $L$ is given by equation $z_3=0$, $f(t=0)=p_2=(0:0:1)$ is the singular point of $C$ of type $A_{4n-2}$ and the line $z_1=0$ is the tangent line to $C$ at $p_2$. Then (maybe, after change of the coordinates in ${\mathbb P}^2$ and ${\mathbb P}^1$) the morphism $f$ is given by \begin{equation} \label{par2} z_1=t^4P_{2n-3}(t), \, \, z_2=t^{2}, \, \,z_3=t-1 ,\end{equation} where $P_{2n-3}(t)=t^{2n-3}+\sum_{i=0}^{2n-4}a_it^{i}$.
To prove Theorem \ref{add1}, it suffices to show that the polynomial $P_{2n-3}(t)$ is defined uniquely by the property that the point $p_2$ is a singular point of $C$ of type $A_{2(2n-1)}$. For this put $x=z_2/z_3$, $y=z_1/z_3$ and $x_2=x$, $y_2=y/x^2$. Then the germ of singularity $(C,0)\subset({\mathbb C}^2,o)$, given by parametrization (at $t=0$): $$x_2=t^{2}/(t-1), \, \, y_2=(t-1)P_{2n-3}(t),$$ has singularity type $A_{2(2n-3)}$ at the point $o=(0,-a_0)$.
Consider a sequence of polynomials $A_k(t)=t^k+\sum_{i=0}^{k-1}a_{i,k}t^i$, $k\in {\mathbb N}$, where $A_1(t) = t+1$ and \begin{equation} \label{seq} \displaystyle A_{k+1} = \frac{t^2A_k(t)-A_k(1)}{t-1} \end{equation} for each $k\geq 1$.
Now, Theorem \ref{add1} follows from \begin{lem} \label{l} Let a germ of singularity $(C,0)\subset({\mathbb C}^2,o)$ of type $A_{2k}$, $k\geq 1$, be given by parametrization of the form (at $t=0$): $$ x=t^{2}/(t-1), \, \, y=(t-1)P_{k}(t),$$ where $P_{k}(t)=t^k+\sum_{i=0}^{k-1}a_it^i$ and $o=(0,-a_0)$. Then the polynomial $P_{k}(t)$ is defined uniquely and, moreover, $P_k(t)=A_k(t)$. \end{lem} {\bf Proof.} If $k=1$ then it is easy to see that $P_1(t)=t+1$. Assume that for $k<k_0$ Lemma is true and prove it in the case $k=k_0$.
If $(C,0)\subset({\mathbb C}^2,o)$ is a germ of singularity of type $A_{2k_0}$, then the polynomial $(t-1)P_{k_0}(t)+a_0$ is divisible by $t^2$. Therefore $(t-1)P_{k_0}(t)+a_0 =t^2Q_{k_0}(t)$, where $Q_{k_0}(t)=t^{k_0-1}+\sum_{i=0}^{k_0-2}b_it^i$ is a polynomial of degree $k_0-1\geq 1$ and the singularity given by $$ x_1=t^{2}/(t-1), \, \, y_1=(t-1)Q_{k_0-1}(t)$$ is of type $A_{2(k_0-1)}$. Then, by assumption, $Q_{k_0-1}(t)=A_{k_0-1}(t)$ and hence $$(t-1)P_{k_0}(t)+a_0 =t^2A_{k_0-1}(t).$$ In particular, $a_0=A_{k_0-1}(1)$, that is, $$\displaystyle P_{k_0}(t) = \frac{t^2A_{k_0-1}(t)-A_{k_0-1}(1)}{t-1}.$$
$\Box$
\section{Strictly rigid curves of positive genera} The following theorem states that the strongly rigid family $V(4;0;3A_2)$ (see case $(I_{22})$ in Theorem \ref{d4}) is the first member in an infinite sequence of strictly rigid families.
\begin{thm}\label{t-oe1} For any $n\ge2$, the family $V(2n,\left[\frac{n}{2}\right]-1,A_n+2T_{n,2n-1})$ is non-empty and strictly rigid. \end{thm} {\bf Proof.} Let $C$ be a plane curve of degree $2n$ with singularities $A_n+2T_{n,2n-1}$. Choose projective coordinates $(z_1:z_2:z_3)$ so that the singular points of type $T_{n,2n-1}$ are located at $(0:1:0)$ and $(0:0:1)$ with tangent lines $z_3=0$ and $z_2=0$, respectively, and the remaining intersection points of $C$ with these lines are $(1:-1:0)$ and $(1:0:-1)$, respectively. Then, in the affine coordinates $x=z_2/z_1$, $y=z_3/z_1$, the curve $C$ is given by an equation \begin{equation}1+x+y+\sum_{i=1}^na_ix^iy^i=0,\quad a_n\ne0\ , \label{e-oe1}\end{equation} the following lemma completes the proof of Theorem:
\begin{lem}\label{l-oe1} There exists a unique curve in ${\mathbb C}^2$, given by equation \eqref{e-oe1} and having a singularity of type $A_n$ in $({\mathbb C}^*)^2$. It is irreducible and has genus $\left[\frac{n}{2}\right]-1$. \end{lem} {\bf Proof.} Note that a singularity of type $T_{n,2n-1}$ is analytically irreducible, $\delta(T_{n,2n-1})=(n-1)^2$, and if a plane curve $C'$ has a singular point of type $T_{n,2n-1}$ then $\deg C'\geq 2n-1$. Therefore the curve $C$ in question is irreducible, since if a curve $C'$ of degree $2n-1$ has two singular points of type $T_{n,2n-1}$, then $$g(C')\leq \frac{(2n-2)(2n-3)}{2}-2\delta(T_{n,2n-1})=-(n-1)<0,$$ a contradiction.
Under assumption that $C$ has no other singularities in ${\mathbb C}^2$, the genus value follows from the fact that $\delta(T_{n,2n-1})=(n-1)^2$ and $\delta(A_n)=\left[\frac{n+1}{2}\right]$.
To find a curve given by \eqref{e-oe1} and having a singularity of type $A_n$, substitute $(x,y)$ for $(x/y,y)$ in \eqref{e-oe1} and multiply by $y$: $$y^2+y\left(1+\sum_{i=1}^na_ix^i\right)+x=0\ .$$ This is a quadratic equation in $y$ with the discriminant $$ \Delta(x)=\left( 1+\sum_{i=1}^na_ix^i\right)^2-4x\ .$$ The existence of a singularity of type $A_n$ on the considered curve in $({\mathbb C}^*)^2$ is equivalent to the condition that $\Delta(x)$ has a root of multiplicity $n+1$. We shall show that this condition has a unique solution $(a_1,...,a_n)$ such that $a_n\ne0$, and that for this solution the other roots of $\Delta$ are simple. So, into the relation \begin{equation}\Delta(x)=\left(1+\sum_{i=1}^na_ix^i\right)^2-4x\quad \text{is divisible by}\quad (1+tx)^{n+1},\ t\ne0\ , \label{e-oe2}\end{equation} we substitute $x=z^2$, $t=-\tau^2$, and then get an equivalent condition \begin{equation}\left(1+\sum_{i=1}^na_iz^{2i}-2z\right)\left(1+\sum_{i=1}^na_iz^{2i}+2z\right) \ \text{is divisible by}\ (1-\tau^2 z^2)^{n+1},\ \tau\ne0\ .\label{e-oe5}\end{equation} Since the difference $4z$ of the factors in the former product in \eqref{e-oe5} is not divisible neither by $1+\tau z$, nor by $1-\tau z$ for any $\tau\ne0$, we reduce \eqref{e-oe5} (possibly replacing $\tau$ by $-\tau$) to the conditions: for $k=1,2$, $$F_k(z)=1+\sum_{i=1}^na_iz^{2i}+(-1)^k2z\quad \text{is divisible by}\quad(1+(-1)^{k+1}\tau z)^{n+1},\quad \tau\ne0\ ,$$ which are equivalent to the following combinations: \begin{equation}\begin{cases}&F_k(z),F_k'(z)\ \text{are divisible by}\ 1+(-1)^{k+1}\tau z,\\ &F_k''(z)\ \text{is divisible by}\ (1+(-1)^{k+1}\tau z)^{n-1},\end{cases}\quad\tau\ne0\ .\label{e-oe3}\end{equation} The latter relations in \eqref{e-oe3} yield that $$F_1''(z)=F''_2(z)=\sum_{i=1}^n2i(2i-1)a_iz^{2i-2}\quad \text{is divisible by}\quad (1-\tau^2z^2)^{n-1}\ ,$$ which results in \begin{equation}a_i=\frac{(-1)^{i-1}}{i(2i-1)}\binom{i-1}{n-1}a_1\tau^{2i-2},\quad i=2,...,n\ , \label{e-oe4}\end{equation} whereas the former relations in \eqref{e-oe3} lead to the system \begin{equation} -2+\frac{a_1}{\tau}\sum_{i=1}^n\frac{2(-1)^i}{2i-1}\binom{i-1}{n-1}=1+\frac{2}{\tau} +\frac{a_1}{\tau^2}\sum_{i=1}^n\frac{(-1)^{i-1}}{i(2i-1)}\binom{i-1}{n-1}=0\ .\label{e-oe6}\end{equation} Observe that $$\sum_{i=1}^n\frac{2(-1)^i}{2i-1}\binom{i-1}{n-1}=-2I_1,\quad I_1=\int_0^1(1-\xi^2)^{n-1}d\xi\ ,$$ where $I_1>0$, and $$\sum_{i=1}^n\frac{(-1)^{i-1}}{i(2i-1)}\binom{i-1}{n-1}=2I_2, \quad I_2=\int_0^1\left(\int_0^\xi(1-\eta^2)^{n-1}d\eta\right)d\xi\ ,$$ where $0<I_2<I_1$. Hence system \eqref{e-oe6} has a unique solution, which in view of \eqref{e-oe4} implies the uniqueness of the sought curve $C$.
The last step in the proof of Theorem \ref{t-oe1} is to show that $C$ has no other singularity in ${\mathbb C}^2$, or, equivalently, that the polynomial $\Delta(x)$ has no multiple root other than $1/\tau^2$ if $(a_1,\dots, a_n)$ is the solution of equations \eqref{e-oe4} and \eqref{e-oe6}. Indeed, let $x=\theta^2$, $\theta^2\neq 1/\tau^2$, be a root of $\Delta(x)$ of multiplicity $m\geq 2$. Then $z=(-1)^k\theta$ is a root of polynomial $F_k(z)$ of multiplicity $m$. Consider the curve $C'$ given by the equation $w-1-\sum_{i=1}^na_iz^{2i}=0$. The curve $C'$ is rational and it has singularity of type $T_{2n-1,2n}$ at infinity. In the pencil of lines $\lambda w+\mu z=0$, there are at least three lines which intersect $C'$ in the number of points less than $2n=\deg C'$: by assumption, each of the lines given by $w=2z$ and $w=-2z$ is tangent to $C'$ at least at two points with multiplicities $n+1$ and $m$, respectively, and the line given by $z=0$ intersects $C'$ at its singular point with multiplicity $2n-1$. Let $\overline C'$ be the normalization of $C'$. By Hurwits formula, applied to morphism $f:\overline C'\to{\mathbb P}^1$ of degree $2n$ defined by the pencil of lines $\lambda w+\mu z=0$, we get the following inequality: $$-2\geq -4n+ 2(n+m-1)+(2n-2)$$ which breaks down for $m\geq 2$.
$\Box$
Consider the Fermat curve $C_n\subset {\mathbb P}^2$ of degree $n\geq 3$ given by equation $z_1^n+z_2^n+z_3^n=0$. By Pl\"ucker formulas, the dual curve $\hat C$ to $C$ has degree $n(n-1)$ and the following type of singularities: $${\bf S}(\hat C_n)= 3nT_{n-1,n} +{\bf S}_{F_n},$$ where ${\bf S}_{F_n}$ is a sum of singularity types of the simple singularities of the curve dual to the Fermat curve $C_n$.
The following theorem will be used in the proof of Theorem \ref{rigit} (see subsection \ref{sec6}). \begin{thm} \label{rigi} Let ${\bf d}_n=(n(n-1),1,1,1)$, ${\bf g}_n=(\frac{(n-1)(n-2)}{2},0,0,0)$, and $n_{(1,2)}=n_{(1,3)}=n_{(1,4)}=n$, $n_{(1)}=n_{(2,3)}=n_{(2,4)}=n_{(3,4)}=1$. Then the family $$V_n=V({\bf d}_n;{\bf g}_n;(n_{(1)}{\bf S}_{F_n}+(n_{(1,2)}+n_{(1,3)}+ n_{(1,4)})T^{n-1}_{n-1,n}+(n_{(2,3)}+n_{(2,4)}+n_{(3,4)})A_1)$$ is strictly rigid for each $n\geq 3$. \end{thm} {\bf Proof.} Consider a curve $\overline C=C_1\cup C_2\cup C_3\cup C_4\in V_n$ and denote the components $C_i$ of $\overline C$ as follows: $C_1=C$ and $C_i=L_{i-1}$ for $i\geq 2$. The curves $L_1$, $L_2$, $L_3$ are lines and $C$ is a curve of degree $n(n-1)$. The curve $C$ has $3n$ singular points of the singularity type $T_{n-1,n}$, since $n_{(1,2)}=n_{(1,3)}=n_{(1,4)}=n$, and the set of all other singularities of $C$ is ${\bf S}_{F_n}$, that is, the singularity type of $C$ is the same as the singularity type of the dual curve to the Fermat curve of degree $n$. Therefore, the dual curve $\hat C$ to $C$ is non-singular and $\deg \hat C=n$, and to prove Theorem, it suffices to show that there are homogeneous coordinates in ${\mathbb P}^2$ such that in these coordinate system the curve $\hat C$ is the Fermat curve.
Let $(z_1:z_2:z_3)$ be homogeneous coordinates in $\hat{{\mathbb P}}^2$ such that $p_1=(1:0:0)$, $p_2=(0:1:0)$, and $p_3=(0:0:1)$ are the points in $\hat{{\mathbb P}}^2$ dual, respectively, to the lines $L_1$, $L_2$, and $L_3$, and let $$F(z_1,z_2,z_3)=\sum_{i_1+i_2+i_3=n}a_{\overline i}z^{\overline i}$$ be an equation of the curve $\hat C$, where $\overline i=(i_1,i_2,i_3)$ and $z^{\overline i}=z_1^{i_1}z_2^{i_2}z_3^{i_3}$. Without loss of generality, we can assume that $a_{n,0,0}=a_{0,n,0}=a_{0,0,n}=1$ since the points $p_1$, $p_2$, and $p_3$ does not belong to the curve $\hat C$.
The curve $\hat C$ has $3n$ flex points given in local coordinates by equation $y=x^n$ and for $i=1,2,3$ there are $n$ lines $L_{i,1},\dots, L_{i,n}$ from the pencil of lines passing through $p_i$ such that they are tangent to $\hat C$ at its flex points.
Let $q_j=L_{1,j}\cap \hat C=(q_{1,j}:q_{2,j}:q_{3,j})$, $1\leq j\leq n$, and let us rewrite the equation of the curve $\hat C$ in the form \begin{equation} \label {eq-pr} F(z_1,z_2,z_3)=z_1^n+ \sum_{i=1}^{n}\binom{n}{i} H_{i}(z_2,z_3)z_1^{n-i}, \end{equation} where $H_i(z_2,z_3)$ are homogeneous polynomials in $z_2$, $z_3$ of degree $i$. Then for $i=1,\dots,n-1$ the homogeneous polynomial $H_1^i(z_2,z_3)-H_i(z_2,z_3)$ of degree $i$ has $n>i$ different roots, namely, $(q_{2,1}:q_{3,1}),\dots, (q_{2,n}:q_{3,n})$. Therefore $H_i(z_2,z_3)=H_1^i(z_2,z_3)$ and hence the polynomial $F(z_1,z_2,z_3)$ has the form \begin{equation} \label{eq-pr1} F(z_1,z_2,z_3)=(z_1+H_{1}(z_2,z_3))^{n}+\widetilde H_n(z_2,z_3), \end{equation} where $\widetilde H_n(z_2,z_3)$ is a homogeneous polynomial of degree $n$, $$\widetilde H_n(z_2,z_3)=\alpha\prod_{j=1}^n(q_{3,j}z_2-q_{2,j}z_3)$$ with some $\alpha\in\mathbb C$.
By the same arguments, we have \begin{equation} \label{eq-pr2} F(z_1,z_2,z_3)=(z_2+G_{1}(z_1,z_3))^{n}+\widetilde G_n(z_1,z_3) \end{equation} and \begin{equation} \label{eq-pr10} F(z_1,z_2,z_3)=(z_3+P_{1}(z_1,z_2))^{n}+\widetilde P_n(z_1,z_2), \end{equation} where $G_1(z_1,z_3)$, $P_1(z_1,z_2)$ and $\widetilde G_n(z_1,z_3)$, $\widetilde P_n(z_1,z_2)$ are homogeneous polynomials of degree one and $n$ respectively.
It follows from \eqref{eq-pr2} that $\widetilde H_n(z_2,z_3)=(z_2+G_{1}(0,z_3))^{n}+\widetilde G_n(0,z_3)-H_1^n(z_2,z_3)$. Therefore \begin{equation} \label{eq-pr3} F(z_1,z_2,z_3)=(z_1+az_2+bz_3)^n-(az_2+bz_3)^n+(z_2+cz_3)^n+(1-c^n)z_3^n \end{equation} for some $a,b,c\in{\mathbb C}$ (remind that, by assumption, $a_{n,0,0}=a_{0,n,0}=a_{0,0,n}=1$).
It follows from \eqref{eq-pr2} and \eqref{eq-pr3} that for $1\leq k\leq n-1$ \begin{equation} \label{eq-pr4} G^k_1(z_1,z_3)=a^{n-k}((z_1+bz_3)^k-b^kz_3^k)+c^kz_3^k .\end{equation} In particular, $G_1(z_1,z_3)=a^{n-1}z_1+cz_3$ and hence \begin{equation} \label{eq-pr5} (a^{n-1}z_1+cz_3)^k=a^{n-k}((z_1+bz_3)^k-b^kz_3^k)+c^kz_3^k .\end{equation} It follows from \eqref{eq-pr5} that \begin{equation} \label{eq-pr6} a^{j(n-1)}c^{k-j}=a^{n-k}b^{k-j} \end{equation} for $1\leq k\leq n-1$ and $1\leq j\leq k$. In particular, if we put $j=k$ in \eqref{eq-pr6}, then we obtain that $a^n=1$ if $a\neq 0$.
Similarly, it follows from \eqref{eq-pr10} and \eqref{eq-pr3} that for $1\leq k\leq n-1$ \begin{equation} \label{eq-pr7} P^k_1(z_1,z_2)=b^{n-k}((z_1+az_2)^k-a^kz_2^k)+c^{n-k}z_2^k .\end{equation} In particular, $P_1(z_1,z_3)=b^{n-1}z_1+cz_2$ and hence \begin{equation} \label{eq-pr8} (b^{n-1}z_1+cz_2)^k=b^{n-k}((z_1+az_2)^k-a^kz_2^k)+c^{n-k}z_2^k .\end{equation} It follows from \eqref{eq-pr8} that \begin{equation} \label{last} \begin{array}{rll} b^{j(n-1)}c^{k-j} & = & b^{n-k}a^{k-j}, \\ c^k & =& c^{n-k} \end{array} \end{equation} for $1\leq k\leq n-1$ and $1\leq j\leq k$. In particular, if $c\neq 0$ then we obtain that $c^2=c^n=1$, that is, $c=\pm 1$ and $n$ is an even number if $c=-1$. If we put $j=k$ in \eqref{last}, then we obtain that $b^n=1$ if $b\neq 0$.
Let us show that the case when $abc\neq 0$ is impossible. Indeed, if $abc\neq 0$, then $a^n=b^n=1$, $c=\pm 1$, and $c=-1$ only if $n$ is even. If we apply again \eqref{eq-pr6} and \eqref{last} we obtain that $c^{k-j}=(\frac{a}{b})^{k-j}=(\frac{b}{a})^{k-j}$ for $1\leq k\leq n-1$ and $1\leq j\leq k$. Therefore, $c=\frac{a}{b}=\frac{b}{a}$ and hence $$F(z_1,z_2,z_3)=(z_1+az_2+acz_3)^n-(az_2+acz_3)^n+(z_2+cz_3)^n=(z_1+az_2+acz_3)^n,$$ since $a^n=c^n=1$. This contradicts the assumption that $C$ is an irreducible reduced curve.
It easily follows from \eqref{eq-pr6} and \eqref{last} that the case, when the only one number either $a$ or $b$, or $c$ is equal to zero, is also impossible.
In the case when $a=b=0$ and $c\neq 0$ we have $F(z_1,z_2,z_3)=z_1^n+(z_2\pm z_3)^n$. But, it is impossible since $C$ is an irreducible curve.
The cases $a=c=0$, $b\neq 0$ and $b=c=0$, $a\neq 0$ are also impossible since in these cases we have, respectively, that $F(z_1,z_2,z_3)=(z_1+bz_3)^n+ z_2^n$ since $b^n=1$ or $F(z_1,z_2,z_3)=(z_1+az_2)^n+ z_3^n$ since $a^n=1$. As a result, we obtain that $a=b=c=0$, that is $F(z_1,z_2,z_3)=z_1^n+ z_2^n+z_3^n$.
$\Box$
\section{$k$-Rigid curves with $k\geq 2$} \subsection{$2$-Rigid irreducible families of equisingular plane curves of degree $\geq 5$} \label {sec5} An infinite series of irreducible $2$-rigid families is given in the following two theorems. \begin{thm} \label{add2} The families $V(2n+1;0;T_{n+1,2n+1}+T_{n,2n+1})$ are $2$-rigid and irreducible for $n\geq 2$. \end{thm} {\bf Proof.} Let $C\in V(2n+1;0;T_{n+1,2n+1}+T_{n,2n+1})$ and let $f:{\mathbb P}^1\to{\mathbb P}^2$ be a morphism such that $C=f({\mathbb P}^1)$. We can choose non-homogeneous coordinate $t$ in ${\mathbb P}^1$ and homogeneous coordinates $(z_1:z_2:z_3)$ in ${\mathbb P}^2$ such that $f(t=0)=p_1=(0:0:1)$ is the singular point of $C$ of type $T_{n+1,2n+1}$, the line $z_1=0$ is the tangent line to $C$ at $p_1$, $f(t=\infty)=p_2=(1:0:0)$ is the singular point of $C$ of type $T_{n,2n+1}$, the line $L_3$ given by $z_3=0$ is the tangent line to $C$ at $p_2$. Then (maybe, after change of the coordinates in ${\mathbb P}^2$) the morphism $f$ is given by \begin{equation} \label{par} z_1=t^{2n+1}, \, \, z_2=t^{n+1}, \, \, z_3=at+1\end{equation} for some $a\in {\mathbb C}$. Let $C_a$ has parametrization \eqref{par}. The intersection number of $C_a$ and $L_3$ at the point $p_2$ is $$(C_0,L_3)_{p_2}=\left\{ \begin{array}{ll} 2n+1\,\, & \text{if}\, a=0\\ 2n\,\, & \text{if}\, a\neq 0. \end{array}\right. $$ Hence there is no projective transformation $h\in {\operatorname{Aut}}({\mathbb P}^2)$ such that $h(C_0)=C_1$. On the other hand, if $a\neq 0$, we make change $t_1=at$, and then the projective transformation $h((z_1:z_2:z_3))=(a^{2n+1}z_1:a^{n+1}z_2:z_3)$ sends $C_1$ to $C_a$.
$\Box$
\begin{thm} \label{add3} The families $V(4n;0;T_{2n-1,4n}+T_{2n+1,4n})$ are $2$-rigid and irreducible for $n\geq 3$. \end{thm} {\bf Proof.} Let $C\in V(4n;0;T_{2n-1,4n}+T_{2n+1,4n})$ and let $f:{\mathbb P}^1\to{\mathbb P}^2$ be a morphism such that $C=f({\mathbb P}^1)$. As in the proof of Theorem \ref{add2}, it is easy to show that we can choose non-homogeneous coordinate $t$ in ${\mathbb P}^1$ and homogeneous coordinates $(z_1:z_2:z_3)$ in ${\mathbb P}^2$ such that $f(t=0)=p_1=(0:0:1)$ is the singular point of $C$ of type $T_{2n-1,4n}$, the line $z_1=0$ is the tangent line to $C$ at $p_1$, $f(t=\infty)=p_2=(1:0:0)$ is the singular point of $C$ of type $T_{2n+1,4n}$, and the line $z_3=0$ is the tangent line to $C$ at $p_2$. Then (maybe, after change of the coordinates in ${\mathbb P}^2$ and ${\mathbb P}^1$) the morphism $f$ is given by \begin{equation}\label{par1} z_1=(t^2-a^2)t^{4n-2}, \, \, z_2=t^{2n-1}, \, \,z_3=1 \end{equation} for some $a\in {\mathbb C}$. Let $C_a$ has parametrization \eqref{par1}. The intersection number of $C_a$ and $L_1$ at the point $p_1$ is $$(C_0,L_1)_{p_1}=\left\{ \begin{array}{ll} 4n\,\, & \text{if}\, a=0\\ 4n-2\,\, & \text{if}\, a\neq 0. \end{array}\right. $$ Hence there is no projective transformation $h\in {\operatorname{Aut}}({\mathbb P}^2)$ such that $h(C_0)=C_1$. On the other hand, if $a\neq 0$, we make change $t_1=a^{-1}t$, and then the projective transformation $h((z_1:z_2:z_3))=(a^{-4n}z_1:a^{1-2n}z_2:z_3)$ sends $C_a$ to $C_1$.
$\Box$
\subsection{$k$-Rigid families parameterizing curves with $k$ connected components} \label {sec6} \begin{thm} \label{rigit} For each $k\in {\mathbb N}$ there is a $k$-rigid family of equisingular plane curves consisting of $k$ irreducible components. \end{thm} {\bf Proof.} Let $n=2k+1$ and let $n_{(1,2)}=n-2$, $n_{(1,3)}=n_{(1,4)}=n$, and $n_{(1)}=n_{(2,4)}=n_{(3,4)}=n_{(1,2,5)}=n_{(1,2,6)}=n_{(3,4,5,6)}=1$. Consider the family $\overline V_n=V(\overline {\bf d}_n;\overline {\bf g}_n;\overline {\bf S}_n)$, where
$$\overline {\bf d}_n=(n(n-1),1,1,1,1,1),\qquad \overline {\bf g}_n=(\frac{(n-1)(n-2)}{2},0,0,0,0,0),$$ and $$\begin{array}{ll} \overline {\bf S}_n = & (n_{(1,2)}+n_{(1,3)}+ n_{(1,4)})T^{n-1}_{n-1,n}+(n_{(1,2,5)}+n_{(1,2,6)})T^n_{n-1,n} \\ & +n_{(1)}{\bf S}_{F_n}+n_{(3,4,5,6)}T_{4,4}+(n_{(2,4)}+n_{(3,4)})A_1),\end{array}$$ where ${\bf S}_{F_n}$ is a sum of singularity types of the simple singularities of the curve dual to the Fermat curve given by equation $z_1^n+z_2^n+z_3^n=0$.
Consider a curve $\widetilde C=\bigcup_{i=1}^6C_i\in \overline V_n$. Denote the curve $C_1$ by $C$ and the curve $C_i$ by $L_{i-1}$ for $i\geq 2$. The curves $L_1, \dots, L_5$ are lines and $C$ is a curve of degree $n(n-1)$ and it is easy to see that $\overline C=C\cup L_1\cup L_2\cup L_3\in V_n$, where $V_n$ is the family of plane curves from Theorem \ref{rigi}. By Theorem \ref{rigi}, we can assume that $C$ is the curve dual to the Fermat curve of degree $n$ and $L_i$ is given by equation $z_i=0$ for $i=1,2,3$. Then it follows from the singularity type of the curve $\widetilde C$ that $L_4$ and $L_5$ have, respectively, equations $z_2+\varepsilon^{m_1}z_3=0$ and $z_2+\varepsilon^{m_2}z_3=0$, where $m_1\not\equiv m_2(mod\, n)$ and $\varepsilon$ is a primitive root of the equation $x^n+1=0$.
Consider two curves $\widetilde C_i=\overline C\cup L_{4,i}\cup L_{5,i}\in \overline V_n$, $i=1,2$, where $L_{4,i}$ is given by equation $z_2+\varepsilon^{m_{1,i}}z_3=0$ and $L_{5,i}$ is given by equation $z_2+\varepsilon^{m_{2,i}}z_3=0$. It is easy to see that a projective transformation $h\in {\operatorname{Aut}}({\mathbb P}^2)$ such that $\widetilde C_2=h(\widetilde C_1)$ exists if and only if $m_{1,1}-m_{1,2}\equiv \pm(m_{2,1}-m_{22}) \mod n$. Hence $\overline V_n$ is a $(\frac{n-1}{2})$-rigid family of plane curves consisting of $\frac{n-1}{2}=k$ irreducible components.
$\Box$
{\ncsc Steklov Mathematical Institute \\[-21pt]
Gubkina str., 8 \\[-21pt]
Moscow, Russia. \\[-21pt]
{\it E-mail address}: {\ntt [email protected]}
\vskip10pt
{\ncsc School of Mathematical Sciences \\[-21pt]
Raymond and Beverly Sackler Faculty of Exact Sciences\\[-21pt]
Tel Aviv University \\[-21pt]
Ramat Aviv, 69978 Tel Aviv, Israel} \\[-21pt]
{\it E-mail address}: {\ntt [email protected]}
\end{document}
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math_pile
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Psalm 72
Aramaic Bible in Plain English
1God, give your judgment to the King and your righteousness to the King's son
2That he shall judge your people in righteousness and your poor ones in judgment.
3The mountains will bring peace to your people, and the hill of your righteousness.
4For he will judge the poor of the people and he will save the afflicted children and abase the oppressors.
5They shall revere you with the sun and before the moon to a generation of generations.
6He will descend like rain upon the grass and like gentle showers that descend upon the Earth.
7The righteous shall spring up in his days and the abundance of peace until the moon passes away.
8He shall rule from sea to sea and from the rivers unto the ends of the Earth.
9The islands shall bow before him and his enemies will lick up the dust.
10Kings of Tarshish and of the islands will bring him offerings; the Kings of Sheba and of Seba shall bring him offerings.
11All the Kings of the Earth will worship him and all the nations shall serve him.
12Because he delivers the afflicted from him who is stronger than he, and the poor who has no helper.
13He shows pity upon the poor and upon the afflicted and he saves the souls of the poor.
14He redeems their souls from fraud and from evil; their blood is precious in his eyes.
15He shall live and gold of Sheba will be given to him and we shall turn to him always, and they shall bless him all day.
16He shall be like the abundance of grain in the Earth and its fruit will spring up in the top of the mountains like Lebanon and they will sprout from the city like the grass of the Earth.
17And his Name shall be blessed to eternity and his Name is before the sun; all the nations will be blessed in him and they all will glorify him.
18Blessed is Lord Jehovah, God of Israel, who alone does great wonders!
19And blessed is his honorable name to eternity! Let his honor fill all the Earth! Amen and amen!
Psalm 71
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Mixed Dataset Summary
Generated on 2025-11-12 21:12:06 UTC.
- Total samples: 3,264,000
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| Source | Dataset ID | Samples |
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