def generate_floyds_triangle(num_rows):
    triangle = []
    number = 1
    for row in range(num_rows):
        current_row = []
        for _ in range(row + 1):
            current_row.append(number)
            number += 1
        triangle.append(current_row)
    return triangle


def display_floyds_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 Floyd's Triangle: "))

# Generate Floyd's Triangle
floyds_triangle = generate_floyds_triangle(num_rows)

# Display Floyd's Triangle
display_floyds_triangle(floyds_triangle)

# 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))

# Python Program to calculate the square root

num = float(input('Enter a number: '))

num_sqrt = num ** 0.5
print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))

# 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)

import pandas as pd



x = pd.read_excel(FILE_NAME)

print(x)

# Python binary search function
def binary_search(arr, target):
    left = 0
    right = len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

# Usage
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 7
result = binary_search(arr, target)
if result != -1:
    print(f"Element is present at index {result}")
else:
    print("Element is not present in array")