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guacamole LDAP creation

Nov 18, 2022AustinLeath
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Append to a file

Jun 1, 2023CodeCatch

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filename = "data.txt"
data = "Hello, World!"
with open(filename, "a") as file:
file.write(data)

Distinct Primes Finder > 1000

Nov 18, 2022AustinLeath

0 likes • 3 views

primes=[]
products=[]
def prime(num):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
return False
else:
primes.append(num)
return True
for n in range(30,1000):
if len(primes) >= 20:
break;
else:
prime(n)
for previous, current in zip(primes[::2], primes[1::2]):
products.append(previous * current)
print (products)

Caesar Encryption

Mar 10, 2021Skrome

0 likes • 1 view

import string
def caesar(text, shift, alphabets):
def shift_alphabet(alphabet):
return alphabet[shift:] + alphabet[:shift]
shifted_alphabets = tuple(map(shift_alphabet, alphabets))
final_alphabet = "".join(alphabets)
final_shifted_alphabet = "".join(shifted_alphabets)
table = str.maketrans(final_alphabet, final_shifted_alphabet)
return text.translate(table)
plain_text = "Hey Skrome, welcome to CodeCatch"
print(caesar(plain_text, 8, [string.ascii_lowercase, string.ascii_uppercase, string.punctuation]))

when predicate lambda

Nov 19, 2022CodeCatch

0 likes • 6 views

def when(predicate, when_true):
return lambda x: when_true(x) if predicate(x) else x
double_even_numbers = when(lambda x: x % 2 == 0, lambda x : x * 2)
print(double_even_numbers(2)) # 4
print(double_even_numbers(1)) # 1

Multiply Two Matrices

May 31, 2023CodeCatch

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# Function to multiply two matrices
def multiply_matrices(matrix1, matrix2):
# Check if the matrices can be multiplied
if len(matrix1[0]) != len(matrix2):
print("Error: The number of columns in the first matrix must be equal to the number of rows in the second matrix.")
return None
# Create the result matrix filled with zeros
result = [[0 for _ in range(len(matrix2[0]))] for _ in range(len(matrix1))]
# Perform matrix multiplication
for i in range(len(matrix1)):
for j in range(len(matrix2[0])):
for k in range(len(matrix2)):
result[i][j] += matrix1[i][k] * matrix2[k][j]
return result
# Example matrices
matrix1 = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
matrix2 = [[10, 11],
[12, 13],
[14, 15]]
# Multiply the matrices
result_matrix = multiply_matrices(matrix1, matrix2)
# Display the result
if result_matrix is not None:
print("Result:")
for row in result_matrix:
print(row)

Using logic with sets

Nov 18, 2022AustinLeath

0 likes • 1 view

#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}
'''