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Nov 18, 2022AustinLeath
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Differentiate Between type() and instance()

May 31, 2023CodeCatch

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class Rectangle:
pass
class Square(Rectangle):
pass
rectangle = Rectangle()
square = Square()
print(isinstance(rectangle, Rectangle)) # True
print(isinstance(square, Rectangle)) # True
print(isinstance(square, Square)) # True
print(isinstance(rectangle, Square)) # False

Convert Decimal to Binary and Hexadecimal

May 31, 2023CodeCatch

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# Prompt user for a decimal number
decimal = int(input("Enter a decimal number: "))
# Convert decimal to binary
binary = bin(decimal)
# Convert decimal to hexadecimal
hexadecimal = hex(decimal)
# Display the results
print("Binary:", binary)
print("Hexadecimal:", hexadecimal)

two-digit integer

Feb 26, 2023wabdelh

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#You are given a two-digit integer n. Return the sum of its digits.
#Example
#For n = 29 the output should be solution (n) = 11
def solution(n):
return (n//10 + n%10)

bruteforce password cracker

Nov 18, 2022AustinLeath

0 likes • 4 views

import itertools
import string
import time
def guess_password(real):
chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuation
attempts = 0
for password_length in range(1, 9):
for guess in itertools.product(chars, repeat=password_length):
startTime = time.time()
attempts += 1
guess = ''.join(guess)
if guess == real:
return 'password is {}. found in {} guesses.'.format(guess, attempts)
loopTime = (time.time() - startTime);
print(guess, attempts, loopTime)
print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")
val = input("Enter a password you want to crack that is 9 characters or below: ")
print(guess_password(val.lower()))

Return Letter Combinations

Nov 18, 2022AustinLeath

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# @return a list of strings, [s1, s2]
def letterCombinations(self, digits):
if '' == digits: return []
kvmaps = {
'2': 'abc',
'3': 'def',
'4': 'ghi',
'5': 'jkl',
'6': 'mno',
'7': 'pqrs',
'8': 'tuv',
'9': 'wxyz'
}
return reduce(lambda acc, digit: [x + y for x in acc for y in kvmaps[digit]], digits, [''])

CSCE 2100 Question 3

Nov 18, 2022AustinLeath

0 likes • 11 views

# question3.py
from itertools import product
V='∀'
E='∃'
def tt(f,n) :
xss=product((0,1),repeat=n)
print('function:',f.__name__)
for xs in xss : print(*xs,':',int(f(*xs)))
print('')
# this is the logic for part A (p\/q\/r) /\ (p\/q\/~r) /\ (p\/~q\/r) /\ (p\/~q\/~r) /\ (~p\/q\/r) /\ (~p\/q\/~r) /\ (~p\/~q\/r) /\ (~p\/~q\/~r)
def parta(p,q,r) :
a=(p or q or r) and (p or q or not r) and (p or not q or r)and (p or not q or not r)
b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
c= a and b
return c
def partb(p,q,r) :
a=(p or q and r) and (p or not q or not r) and (p or not q or not r)and (p or q or not r)
b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
c= a and b
return c
print("part A:")
tt(parta,3)
print("part B:")
tt(partb,3)