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Dictionary Sort

Nov 18, 2022AustinLeath
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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)

LeetCode Flood Fill

Oct 15, 2022CodeCatch

0 likes • 0 views

class Solution(object):
def floodFill(self, image, sr, sc, newColor):
R, C = len(image), len(image[0])
color = image[sr][sc]
if color == newColor: return image
def dfs(r, c):
if image[r][c] == color:
image[r][c] = newColor
if r >= 1: dfs(r-1, c)
if r+1 < R: dfs(r+1, c)
if c >= 1: dfs(r, c-1)
if c+1 < C: dfs(r, c+1)
dfs(sr, sc)
return image

Bubble sort

Nov 19, 2022CodeCatch

0 likes • 0 views

# Python program for implementation of Bubble Sort
def bubbleSort(arr):
n = len(arr)
# Traverse through all array elements
for i in range(n-1):
# range(n) also work but outer loop will repeat one time more than needed.
# Last i elements are already in place
for j in range(0, n-i-1):
# traverse the array from 0 to n-i-1
# Swap if the element found is greater
# than the next element
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]
# Driver code to test above
arr = [64, 34, 25, 12, 22, 11, 90]
bubbleSort(arr)
print ("Sorted array is:")
for i in range(len(arr)):
print ("%d" %arr[i]),

Hello World

Sep 9, 2023AustinLeath

0 likes • 19 views

print("test")

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)

lambda example

Nov 19, 2022CodeCatch

0 likes • 3 views

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]