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Bubble sort

0 likes • Nov 19, 2022
Python
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hex to rgb

CodeCatch
0 likes • Nov 19, 2022
Python
def hex_to_rgb(hex):
return tuple(int(hex[i:i+2], 16) for i in (0, 2, 4))
hex_to_rgb('FFA501') # (255, 165, 1)

Propositional logic with itertools

AustinLeath
0 likes • Nov 18, 2022
Python
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)

Using logic with sets

AustinLeath
0 likes • Nov 18, 2022
Python
#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}
'''

Factorial of N

CodeCatch
0 likes • Nov 19, 2022
Python
import math
def factorial(n):
print(math.factorial(n))
return (math.factorial(n))
factorial(5)
factorial(10)
factorial(15)

Find URL in string

CodeCatch
0 likes • Nov 19, 2022
Python
# 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))

collect dictionary

CodeCatch
0 likes • Nov 19, 2022
Python
from collections import defaultdict
def collect_dictionary(obj):
inv_obj = defaultdict(list)
for key, value in obj.items():
inv_obj[value].append(key)
return dict(inv_obj)
ages = {
'Peter': 10,
'Isabel': 10,
'Anna': 9,
}
collect_dictionary(ages) # { 10: ['Peter', 'Isabel'], 9: ['Anna'] }