def bsgs(g, h, p):
N = 1+(floor(sqrt(p)))
#Store big steps in bTree so that we can search in logn time
L1 = BinaryTree({pow(g, i, p) : i for i in range(N)})
# Precompute via Fermat's Little Theorem
c = pow(g, N * (p - 2), p)
#Store Little Steps
L2 = BinaryTree({(h * pow(c, j, p)) % p: j for j in range(N)})
for key in L2.keys(): #O(n)
if L1.__contains__(key): #computed in log(n) time
x= (L2.get(key) * N + L1.get(key))
print(x, "is a solution for x")
return "Done"
from bintrees import BinaryTree
data = {3:'White', 2:'Red', 1:'Green', 5:'Orange', 4:'Yellow', 7:'Purple', 0:'Magenta'}
tree = BinaryTree(data)
tree.update({6: 'Teal'})
def displayKeyValue(key, value):
print('Key: ', key, ' Value: ', value)
tree.foreach(displayKeyValue)
print('Item 3 contains: ', tree.get(3))
print('The maximum item is: ', tree.max_item())
