• Nov 18, 2022 •AustinLeath
0 likes • 14 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)
• May 5, 2026 •CodeCatch
0 likes • 9 views
import re def find_pattern(text, pattern): match = re.search(pattern, text) if match: return match.group() else: return 'No match found' text = 'Hello, world! This is a test.' pattern = r'world!.*test.' result = find_pattern(text, pattern) print(result)
• Nov 19, 2022 •CodeCatch
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list_1 = [1,2,3,4,5,6,7,8,9] cubed = map(lambda x: pow(x,3), list_1) print(list(cubed)) #Results #[1, 8, 27, 64, 125, 216, 343, 512, 729]
0 likes • 15 views
• May 31, 2023 •CodeCatch
0 likes • 6 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)
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# Python code to find the URL from an input string # Using the regular expression import re def Find(string): # findall() has been used # with valid conditions for urls in string regex = r"(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'\".,<>?«»“”‘’]))" url = re.findall(regex,string) return [x[0] for x in url] # Driver Code string = 'My Profile: https://codecatch.net' print("Urls: ", Find(string))