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Binary search algorithm

Nov 19, 2022CodeCatch
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Using logic with sets

Nov 18, 2022AustinLeath

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#Sets
U = {0,1,2,3,4,5,6,7,8,9}
P = {1,2,3,4}
Q = {4,5,6}
R = {3,4,6,8,9}
def set2bits(xs,us) :
bs=[]
for x in us :
if x in xs :
bs.append(1)
else:
bs.append(0)
assert len(us) == len(bs)
return bs
def union(set1,set2) :
finalSet = set()
bitList1 = set2bits(set1, U)
bitList2 = set2bits(set2, U)
for i in range(len(U)) :
if(bitList1[i] or bitList2[i]) :
finalSet.add(i)
return finalSet
def intersection(set1,set2) :
finalSet = set()
bitList1 = set2bits(set1, U)
bitList2 = set2bits(set2, U)
for i in range(len(U)) :
if(bitList1[i] and bitList2[i]) :
finalSet.add(i)
return finalSet
def compliment(set1) :
finalSet = set()
bitList = set2bits(set1, U)
for i in range(len(U)) :
if(not bitList[i]) :
finalSet.add(i)
return finalSet
def implication(a,b):
return union(compliment(a), b)
###########################################################################################
###################### Problems 1-6 #######################################
###########################################################################################
#p \/ (q /\ r) = (p \/ q) /\ (p \/ r)
def prob1():
return union(P, intersection(Q,R)) == intersection(union(P,Q), union(P,R))
#p /\ (q \/ r) = (p /\ q) \/ (p /\ r)
def prob2():
return intersection(P, union(Q,R)) == union(intersection(P,Q), intersection(P,R))
#~(p /\ q) = ~p \/ ~q
def prob3():
return compliment(intersection(P,R)) == union(compliment(P), compliment(R))
#~(p \/ q) = ~p /\ ~q
def prob4():
return compliment(union(P,Q)) == intersection(compliment(P), compliment(Q))
#(p=>q) = (~q => ~p)
def prob5():
return implication(P,Q) == implication(compliment(Q), compliment(P))
#(p => q) /\ (q => r) => (p => r)
def prob6():
return implication(intersection(implication(P,Q), implication(Q,R)), implication(P,R))
print("Problem 1: ", prob1())
print("Problem 2: ", prob2())
print("Problem 3: ", prob3())
print("Problem 4: ", prob4())
print("Problem 5: ", prob5())
print("Problem 6: ", prob6())
'''
Problem 1: True
Problem 2: True
Problem 3: True
Problem 4: True
Problem 5: True
Problem 6: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
'''

Untitled

Sep 14, 2024rgannedo-6205

0 likes • 2 views

https://codecatch.net/post/06c9f5b7-1e00-40dc-b436-b8cccc4b69be

LeetCode Flood Fill

Oct 15, 2022CodeCatch

0 likes • 0 views

class Solution(object):
def floodFill(self, image, sr, sc, newColor):
R, C = len(image), len(image[0])
color = image[sr][sc]
if color == newColor: return image
def dfs(r, c):
if image[r][c] == color:
image[r][c] = newColor
if r >= 1: dfs(r-1, c)
if r+1 < R: dfs(r+1, c)
if c >= 1: dfs(r, c-1)
if c+1 < C: dfs(r, c+1)
dfs(sr, sc)
return image

Calculate Square Root

Nov 18, 2022AustinLeath

0 likes • 4 views

# Python Program to calculate the square root
num = float(input('Enter a number: '))
num_sqrt = num ** 0.5
print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))

clamp number

Nov 19, 2022CodeCatch

0 likes • 3 views

def clamp_number(num, a, b):
return max(min(num, max(a, b)), min(a, b))
clamp_number(2, 3, 5) # 3
clamp_number(1, -1, -5) # -1

curry function

Nov 19, 2022CodeCatch

0 likes • 1 view

from functools import partial
def curry(fn, *args):
return partial(fn, *args)
add = lambda x, y: x + y
add10 = curry(add, 10)
add10(20) # 30