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Binary search algorithm

0 likes • Nov 19, 2022
Python
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LeetCode Flood Fill

CodeCatch
0 likes • Oct 15, 2022
Python
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

clamp number

CodeCatch
0 likes • Nov 19, 2022
Python
def clamp_number(num, a, b):
return max(min(num, max(a, b)), min(a, b))
clamp_number(2, 3, 5) # 3
clamp_number(1, -1, -5) # -1

CSCE 2100 Question 3

AustinLeath
0 likes • Nov 18, 2022
Python
# 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)

Nodes and Trees

AustinLeath
0 likes • Nov 18, 2022
Python
import random
class Node:
def __init__(self, c):
self.left = None
self.right = None
self.color = c
def SetColor(self,c) :
self.color = c
def PrintNode(self) :
print(self.color)
def insert(s, root, i, n):
if i < n:
temp = Node(s[i])
root = temp
root.left = insert(s, root.left,2 * i + 1, n)
root.right = insert(s, root.right,2 * i + 2, n)
return root
def MakeTree(s) :
list = insert(s,None,0,len(s))
return list
def MakeSet() :
s = []
count = random.randint(7,12)
for _ in range(count) :
color = random.randint(0,1) == 0 and "Red" or "White"
s.append(color)
return s
def ChangeColor(root) :
if (root != None) :
if (root.color == "White") :
root.SetColor("Red")
ChangeColor(root.left)
ChangeColor(root.right)
def PrintList(root) :
if root.left != None :
PrintList(root.left)
else :
root.PrintNode()
if root.right != None :
PrintList(root.right)
else :
root.PrintNode()
t1 = MakeTree(MakeSet())
print("Original Colors For Tree 1:\n")
PrintList(t1)
ChangeColor(t1)
print("New Colors For Tree 1:\n")
PrintList(t1)
t2 = MakeTree(MakeSet())
print("Original Colors For Tree 2:\n")
PrintList(t2)
ChangeColor(t2)
print("New Colors For Tree 2:\n")
PrintList(t2)
t3 = MakeTree(MakeSet())
print("Original Colors For Tree 3:\n")
PrintList(t3)
ChangeColor(t3)
print("New Colors For Tree 3:\n")
PrintList(t3)

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)

Radix sort

CodeCatch
0 likes • Nov 19, 2022
Python
# Python program for implementation of Radix Sort
# A function to do counting sort of arr[] according to
# the digit represented by exp.
def countingSort(arr, exp1):
n = len(arr)
# The output array elements that will have sorted arr
output = [0] * (n)
# initialize count array as 0
count = [0] * (10)
# Store count of occurrences in count[]
for i in range(0, n):
index = (arr[i]/exp1)
count[int((index)%10)] += 1
# Change count[i] so that count[i] now contains actual
# position of this digit in output array
for i in range(1,10):
count[i] += count[i-1]
# Build the output array
i = n-1
while i>=0:
index = (arr[i]/exp1)
output[ count[ int((index)%10) ] - 1] = arr[i]
count[int((index)%10)] -= 1
i -= 1
# Copying the output array to arr[],
# so that arr now contains sorted numbers
i = 0
for i in range(0,len(arr)):
arr[i] = output[i]
# Method to do Radix Sort
def radixSort(arr):
# Find the maximum number to know number of digits
max1 = max(arr)
# Do counting sort for every digit. Note that instead
# of passing digit number, exp is passed. exp is 10^i
# where i is current digit number
exp = 1
while max1/exp > 0:
countingSort(arr,exp)
exp *= 10
# Driver code to test above
arr = [ 170, 45, 75, 90, 802, 24, 2, 66]
radixSort(arr)
for i in range(len(arr)):
print(arr[i]),