• Nov 19, 2022 •CodeCatch
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# Python program for Plotting Fibonacci # spiral fractal using Turtle import turtle import math def fiboPlot(n): a = 0 b = 1 square_a = a square_b = b # Setting the colour of the plotting pen to blue x.pencolor("blue") # Drawing the first square x.forward(b * factor) x.left(90) x.forward(b * factor) x.left(90) x.forward(b * factor) x.left(90) x.forward(b * factor) # Proceeding in the Fibonacci Series temp = square_b square_b = square_b + square_a square_a = temp # Drawing the rest of the squares for i in range(1, n): x.backward(square_a * factor) x.right(90) x.forward(square_b * factor) x.left(90) x.forward(square_b * factor) x.left(90) x.forward(square_b * factor) # Proceeding in the Fibonacci Series temp = square_b square_b = square_b + square_a square_a = temp # Bringing the pen to starting point of the spiral plot x.penup() x.setposition(factor, 0) x.seth(0) x.pendown() # Setting the colour of the plotting pen to red x.pencolor("red") # Fibonacci Spiral Plot x.left(90) for i in range(n): print(b) fdwd = math.pi * b * factor / 2 fdwd /= 90 for j in range(90): x.forward(fdwd) x.left(1) temp = a a = b b = temp + b # Here 'factor' signifies the multiplicative # factor which expands or shrinks the scale # of the plot by a certain factor. factor = 1 # Taking Input for the number of # Iterations our Algorithm will run n = int(input('Enter the number of iterations (must be > 1): ')) # Plotting the Fibonacci Spiral Fractal # and printing the corresponding Fibonacci Number if n > 0: print("Fibonacci series for", n, "elements :") x = turtle.Turtle() x.speed(100) fiboPlot(n) turtle.done() else: print("Number of iterations must be > 0")
• Feb 26, 2023 •wabdelh
0 likes • 1 view
#You are given a two-digit integer n. Return the sum of its digits. #Example #For n = 29 the output should be solution (n) = 11 def solution(n): return (n//10 + n%10)
• Jun 1, 2023 •CodeCatch
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bytes_data = b'Hello, World!' string_data = bytes_data.decode('utf-8') print("String:", string_data)
def print_pyramid_pattern(n): # outer loop to handle number of rows # n in this case for i in range(0, n): # inner loop to handle number of columns # values changing acc. to outer loop for j in range(0, i+1): # printing stars print("* ",end="") # ending line after each row print("\r") print_pyramid_pattern(10)
• Sep 14, 2024 •rgannedo-6205
https://codecatch.net/post/06c9f5b7-1e00-40dc-b436-b8cccc4b69be
#84 48 13 20 61 20 33 97 34 45 6 63 71 66 24 57 92 74 6 25 51 86 48 15 64 55 77 30 56 53 37 99 9 59 57 61 30 97 50 63 59 62 39 32 34 4 96 51 8 86 10 62 16 55 81 88 71 25 27 78 79 88 92 50 16 8 67 82 67 37 84 3 33 4 78 98 39 64 98 94 24 82 45 3 53 74 96 9 10 94 13 79 15 27 56 66 32 81 77 # xor a list of integers to find the lonely integer res = a[0] for i in range(1,len(a)): res = res ^ a[i]