


"""simple docstring"""


from math import cos, sin






def   a__			(					__lowercase )						->						Optional[int]:
				_A       	=      float(_UpperCAmelCase )
				return [[scaling_factor * int(x == y ) for x in range(2 )] for y in range(2 )]






def   a__			(					__lowercase )						->						Dict:
				_A       	=      cos(_UpperCAmelCase ), sin(_UpperCAmelCase )
				return [[c, -s], [s, c]]






def   a__			(					__lowercase )						->						Dict:
				_A       	=      cos(_UpperCAmelCase ), sin(_UpperCAmelCase )
				_A       	=      c * s
				return [[c * c, cs], [cs, s * s]]






def   a__			(					__lowercase )						->						Optional[int]:
				_A       	=      cos(_UpperCAmelCase ), sin(_UpperCAmelCase )
				_A       	=      c * s
				return [[2 * c - 1, 2 * cs], [2 * cs, 2 * s - 1]]


print(f'''    {scaling(5) = }''')
print(f'''  {rotation(45) = }''')
print(f'''{projection(45) = }''')




print(f'''{reflection(45) = }''')





"""simple docstring"""

# This Python program implements an optimal binary search tree (abbreviated BST)
# building dynamic programming algorithm that delivers O(n^2) performance.
#
# The goal of the optimal BST problem is to build a low-cost BST for a
# given set of nodes, each with its own key and frequency. The frequency
# of the node is defined as how many time the node is being searched.
# The search cost of binary search tree is given by this formula:
#
# cost(1, n) = sum{i = 1 to n}((depth(node_i) + 1) * node_i_freq)
#
# where n is number of nodes in the BST. The characteristic of low-cost
# BSTs is having a faster overall search time than other implementations.
# The reason for their fast search time is that the nodes with high
# frequencies will be placed near the root of the tree while the nodes
# with low frequencies will be placed near the leaves of the tree thus
# reducing search time in the most frequent instances.
import sys
from random import randint

class 							lowercase   :







				def __init__(				self :       Optional[int]   ,  _lowerCamelCase :       str   ,  _lowerCamelCase :       Optional[Any]							):



					"""simple docstring"""

					A_	:					Union[str, Any]						 =		key
					A_	:					str						 =		freq







				def __str__(				self :       Optional[Any]							):



					"""simple docstring"""

					return F"""Node(key={self.key}, freq={self.freq})"""

def 		lowercase_   (		_UpperCAmelCase	,   _UpperCAmelCase	,   _UpperCAmelCase	,   _UpperCAmelCase	,   _UpperCAmelCase	,   _UpperCAmelCase						):




	"""simple docstring"""






	if i > j or i < 0 or j > len(_UpperCAmelCase						) - 1:
		return

	A_	:					Union[str, Any]						 =		root[i][j]
	if parent == -1:  # root does not have a parent
		print(f"""{key[node]} is the root of the binary search tree."""						)
	elif is_left:
		print(f"""{key[node]} is the left child of key {parent}."""						)
	else:
		print(f"""{key[node]} is the right child of key {parent}."""						)

	print_binary_search_tree(_UpperCAmelCase	,   _UpperCAmelCase	,   _UpperCAmelCase	,   node - 1	,   key[node]	,   _UpperCAmelCase						)
	print_binary_search_tree(_UpperCAmelCase	,   _UpperCAmelCase	,   node + 1	,   _UpperCAmelCase	,   key[node]	,   _UpperCAmelCase						)

def 		lowercase_   (		_UpperCAmelCase						):




	"""simple docstring"""






	nodes.sort(key=lambda _UpperCAmelCase						: node.key						)

	A_	:					Any						 =		len(_UpperCAmelCase						)

	A_	:					int						 =		[nodes[i].key for i in range(_UpperCAmelCase						)]
	A_	:					List[str]						 =		[nodes[i].freq for i in range(_UpperCAmelCase						)]

	# This 2D array stores the overall tree cost (which's as minimized as possible);
	# for a single key, cost is equal to frequency of the key.
	A_	:					List[str]						 =		[[freqs[i] if i == j else 0 for j in range(_UpperCAmelCase						)] for i in range(_UpperCAmelCase						)]
	# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
	# array
	A_	:					Optional[int]						 =		[[freqs[i] if i == j else 0 for j in range(_UpperCAmelCase						)] for i in range(_UpperCAmelCase						)]
	# stores tree roots that will be used later for constructing binary search tree
	A_	:					List[Any]						 =		[[i if i == j else 0 for j in range(_UpperCAmelCase						)] for i in range(_UpperCAmelCase						)]

	for interval_length in range(2	,   n + 1						):
		for i in range(n - interval_length + 1						):
			A_	:					List[Any]						 =		i + interval_length - 1

			A_	:					List[Any]						 =		sys.maxsize  # set the value to "infinity"
			A_	:					Union[str, Any]						 =		total[i][j - 1] + freqs[j]

			# Apply Knuth's optimization
			# Loop without optimization: for r in range(i, j + 1):
			for r in range(root[i][j - 1]	,   root[i + 1][j] + 1						):  # r is a temporal root
				A_	:					Tuple						 =		dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
				A_	:					List[str]						 =		dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
				A_	:					Optional[int]						 =		left + total[i][j] + right

				if dp[i][j] > cost:
					A_	:					Dict						 =		cost
					A_	:					List[str]						 =		r

	print('''Binary search tree nodes:'''						)
	for node in nodes:
		print(_UpperCAmelCase						)

	print(f"""\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}."""						)
	print_binary_search_tree(_UpperCAmelCase	,   _UpperCAmelCase	,   0	,   n - 1	,   -1	,   _UpperCAmelCase						)

def 		lowercase_   (		):




	"""simple docstring"""






	A_	:					List[Any]						 =		[Node(_UpperCAmelCase	,   randint(1	,   50						)						) for i in range(10	,   0	,   -1						)]
	find_optimal_binary_search_tree(_UpperCAmelCase						)


if __name__ == "__main__":
		main()





