

def  __lowerCAmelCase (  A    ,       A      ):
	# "Simpson Rule"
	# int(f) = delta_x/2 * (b-a)/3*(f1 + 4f2 + 2f_3 + ... + fn)
	UpperCAmelCase_    			=     (boundary[1] - boundary[0]) / steps
	UpperCAmelCase_    			=     boundary[0]
	UpperCAmelCase_    			=     boundary[1]
	UpperCAmelCase_    			=     make_points(A    ,       A    ,       A      )
	UpperCAmelCase_    			=     0.0
	y += (h / 3.0) * f(A      )
	UpperCAmelCase_    			=     2
	for i in x_i:
		y += (h / 3) * (4 - 2 * (cnt % 2)) * f(A      )
		cnt += 1
	y += (h / 3.0) * f(A      )
	return y



def  __lowerCAmelCase (  A    ,       A    ,       A      ):
	UpperCAmelCase_    			=     a + h
	while x < (b - h):
		yield x
		UpperCAmelCase_    			=     x + h



def  __lowerCAmelCase (  A      ):  # enter your function here
	UpperCAmelCase_    			=     (x - 0) * (x - 0)
	return y



def  __lowerCAmelCase (  ):
	UpperCAmelCase_    			=     0.0  # Lower bound of integration
	UpperCAmelCase_    			=     1.0  # Upper bound of integration
	UpperCAmelCase_    			=     10.0  # define number of steps or resolution
	UpperCAmelCase_    			=     [a, b]  # define boundary of integration
	UpperCAmelCase_    			=     method_a(A    ,       A      )
	print(F"y = {y}"      )


if __name__ == "__main__":
				main()


# 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   __UpperCamelCase      :




	def __init__(						self		:  Optional[Any]						,		lowerCAmelCase		:  Any						,		lowerCAmelCase		:  Optional[Any]							):


		'''simple docstring'''







		UpperCAmelCase_    			=     key
		UpperCAmelCase_    			=     freq




	def __str__(						self		:  List[str]							):


		'''simple docstring'''







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



def  __lowerCAmelCase (  A    ,       A    ,       A    ,       A    ,       A    ,       A      ):
	if i > j or i < 0 or j > len(A      ) - 1:
		return

	UpperCAmelCase_    			=     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(A    ,       A    ,       A    ,       node - 1    ,       key[node]    ,       A      )
	print_binary_search_tree(A    ,       A    ,       node + 1    ,       A    ,       key[node]    ,       A      )



def  __lowerCAmelCase (  A      ):
	nodes.sort(key=lambda A      : node.key      )

	UpperCAmelCase_    			=     len(A      )

	UpperCAmelCase_    			=     [nodes[i].key for i in range(A      )]
	UpperCAmelCase_    			=     [nodes[i].freq for i in range(A      )]

	# 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.
	UpperCAmelCase_    			=     [[freqs[i] if i == j else 0 for j in range(A      )] for i in range(A      )]
	# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
	# array
	UpperCAmelCase_    			=     [[freqs[i] if i == j else 0 for j in range(A      )] for i in range(A      )]
	# stores tree roots that will be used later for constructing binary search tree
	UpperCAmelCase_    			=     [[i if i == j else 0 for j in range(A      )] for i in range(A      )]

	for interval_length in range(2    ,       n + 1      ):
		for i in range(n - interval_length + 1      ):
			UpperCAmelCase_    			=     i + interval_length - 1

			UpperCAmelCase_    			=     sys.maxsize  # set the value to "infinity"
			UpperCAmelCase_    			=     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
				UpperCAmelCase_    			=     dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
				UpperCAmelCase_    			=     dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
				UpperCAmelCase_    			=     left + total[i][j] + right

				if dp[i][j] > cost:
					UpperCAmelCase_    			=     cost
					UpperCAmelCase_    			=     r

	print("Binary search tree nodes:"      )
	for node in nodes:
		print(A      )

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



def  __lowerCAmelCase (  ):
	# A sample binary search tree
	UpperCAmelCase_    			=     [Node(A    ,       randint(1    ,       50      )      ) for i in range(10    ,       0    ,       -1      )]
	find_optimal_binary_search_tree(A      )


if __name__ == "__main__":
				main()