Skip to main content

Palindrome checker

Nov 19, 2022CodeCatch
Loading...

More Python Posts

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)

Create a Pascal’s Triangle

May 31, 2023CodeCatch

0 likes • 1 view

def generate_pascals_triangle(num_rows):
triangle = []
for row in range(num_rows):
# Initialize the row with 1
current_row = [1]
# Calculate the values for the current row
if row > 0:
previous_row = triangle[row - 1]
for i in range(len(previous_row) - 1):
current_row.append(previous_row[i] + previous_row[i + 1])
# Append 1 at the end of the row
current_row.append(1)
# Add the current row to the triangle
triangle.append(current_row)
return triangle
def display_pascals_triangle(triangle):
for row in triangle:
for number in row:
print(number, end=" ")
print()
# Prompt the user for the number of rows
num_rows = int(input("Enter the number of rows for Pascal's Triangle: "))
# Generate Pascal's Triangle
pascals_triangle = generate_pascals_triangle(num_rows)
# Display Pascal's Triangle
display_pascals_triangle(pascals_triangle)

Print pyramid pattern

Nov 19, 2022CodeCatch

0 likes • 0 views

def print_pyramid_pattern(n):
# outer loop to handle number of rows
# n in this case
for i in range(0, n):
# inner loop to handle number of columns
# values changing acc. to outer loop
for j in range(0, i+1):
# printing stars
print("* ",end="")
# ending line after each row
print("\r")
print_pyramid_pattern(10)

Propositional logic with itertools

Nov 18, 2022AustinLeath

0 likes • 5 views

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('')
# p \/ (q /\ r) = (p \/ q) /\ (p \/ r)
def prob1(p,q,r) :
x=p or (q and r)
y= (p or q) and (p or r)
return x==y
tt(prob1,3)
# p/\(q\/r)=(p/\q)\/(p/\r)
def prob2(p,q,r) :
x=p and ( q or r )
y=(p and q) or (p and r)
return x==y
tt(prob2,3)
#~(p/\q)=(~p\/~q)
def prob3(p,q) :
x=not (p and q)
y=(not p) or (not q)
return x==y
tt(prob3,2)
#(~(p\/q))=((~p)/\~q)
def prob4(p, q):
x = not(p or q)
y = not p and not q
return x == y
tt(prob4, 2)
#(p/\(p=>q)=>q)
def prob5(p,q):
x= p and ( not p or q)
return not x or q
tt(prob5,2)
# (p=>q)=((p\/q)=q)
def prob6(p,q) :
x = (not p or q)
y=((p or q) == q)
return x==y
tt(prob6,2)
#((p=>q)=(p\/q))=q
def prob7(p,q):
if ((not p or q)==(p or q))==q:
return 1
tt(prob7,2)
#(p=>q)=((p/\q)=p)
def prob8(p,q):
if (not p or q)==((p and q)==p):
return 1
tt(prob8,2)
#((p=>q)=(p/\q))=p
def prob9(p,q):
if ((not p or q)==(p and q))==p:
return '1'
tt(prob9,2)
#(p=>q)/\(q=>r)=>(p=>r)
def prob10(p,q,r) :
x = not ((not p or q) and (not q or r)) or (not p or r)
return x
tt(prob10, 3)
# (p = q) /\ (q => r) => (p => r)
#answer 1
def prob11(p,q,r) :
x = not((p is q) and (not q or r)) or (not p or r)
return x
tt(prob11, 3)
#(p=q)/\(q=>r)=>(p=>r)
#answer 2
def prob11(p,q,r):
x=(p==q) and (not q or r)
y=not p or r
return not x or y
tt(prob11,3)
#((p=>q)/\(q=r))=>(p=>r)
def prob12(p,q,r):
x=(not p or q) and ( q==r )
y=not p or r
return not x or y
tt(prob12,3)
#(p=>q)=>((p/\r)=>(q/\r))
def prob13(p,q,r):
x=not p or q
y=(not(p and r) or ( q and r))
return not x or y
tt(prob13,3)
#Question#2----------------------------------------
#(p=>q)=>r=p=>(q=>r)
def prob14(p,q,r):
x=(not(not p or q) or r)
y=(not p or (not q or r))
return x==y
tt(prob14,3)
def prob15(p, q):
x = not(p and q)
y = not p and not q
return x == y
tt(prob15, 2)
def prob16(p, q):
x = not(p or q)
y = not p or not q
return x == y
tt(prob16, 2)
def prob17(p):
x = p
y = not p
return x == y
tt(prob17, 1)

Shuffle Deck of Cards

May 31, 2023CodeCatch

0 likes • 1 view

import random
# Define the ranks, suits, and create a deck
ranks = ['Ace', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'Jack', 'Queen', 'King']
suits = ['Hearts', 'Diamonds', 'Clubs', 'Spades']
deck = [(rank, suit) for rank in ranks for suit in suits]
# Shuffle the deck
random.shuffle(deck)
# Display the shuffled deck
for card in deck:
print(card[0], "of", card[1])

delay time lambda

Nov 19, 2022CodeCatch

0 likes • 0 views

from time import sleep
def delay(fn, ms, *args):
sleep(ms / 1000)
return fn(*args)
delay(lambda x: print(x), 1000, 'later') # prints 'later' after one second