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Shuffle Deck of Cards

May 31, 2023CodeCatch
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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}
'''

Hello, python

Jan 20, 2021Ntindle

0 likes • 2 views

print(“Hello World”)

Factorial of N

Nov 19, 2022CodeCatch

0 likes • 0 views

import math
def factorial(n):
print(math.factorial(n))
return (math.factorial(n))
factorial(5)
factorial(10)
factorial(15)

radians to degrees

Nov 19, 2022CodeCatch

0 likes • 0 views

from math import pi
def rads_to_degrees(rad):
return (rad * 180.0) / pi
rads_to_degrees(pi / 2) # 90.0

Bitonic sort

Nov 19, 2022CodeCatch

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# Python program for Bitonic Sort. Note that this program
# works only when size of input is a power of 2.
# The parameter dir indicates the sorting direction, ASCENDING
# or DESCENDING; if (a[i] > a[j]) agrees with the direction,
# then a[i] and a[j] are interchanged.*/
def compAndSwap(a, i, j, dire):
if (dire==1 and a[i] > a[j]) or (dire==0 and a[i] > a[j]):
a[i],a[j] = a[j],a[i]
# It recursively sorts a bitonic sequence in ascending order,
# if dir = 1, and in descending order otherwise (means dir=0).
# The sequence to be sorted starts at index position low,
# the parameter cnt is the number of elements to be sorted.
def bitonicMerge(a, low, cnt, dire):
if cnt > 1:
k = cnt/2
for i in range(low , low+k):
compAndSwap(a, i, i+k, dire)
bitonicMerge(a, low, k, dire)
bitonicMerge(a, low+k, k, dire)
# This funcion first produces a bitonic sequence by recursively
# sorting its two halves in opposite sorting orders, and then
# calls bitonicMerge to make them in the same order
def bitonicSort(a, low, cnt,dire):
if cnt > 1:
k = cnt/2
bitonicSort(a, low, k, 1)
bitonicSort(a, low+k, k, 0)
bitonicMerge(a, low, cnt, dire)
# Caller of bitonicSort for sorting the entire array of length N
# in ASCENDING order
def sort(a,N, up):
bitonicSort(a,0, N, up)
# Driver code to test above
a = [3, 7, 4, 8, 6, 2, 1, 5]
n = len(a)
up = 1
sort(a, n, up)
print ("\n\nSorted array is")
for i in range(n):
print("%d" %a[i]),

Compute all the Permutation of a String

May 31, 2023CodeCatch

0 likes • 2 views

import itertools
def compute_permutations(string):
# Generate all permutations of the string
permutations = itertools.permutations(string)
# Convert each permutation tuple to a string
permutations = [''.join(permutation) for permutation in permutations]
return permutations
# Prompt the user for a string
string = input("Enter a string: ")
# Compute permutations
permutations = compute_permutations(string)
# Display the permutations
print("Permutations:")
for permutation in permutations:
print(permutation)