• 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")
• Nov 18, 2022 •AustinLeath
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#question1.py def rose(n) : if n==0 : yield [] else : for k in range(0,n) : for l in rose(k) : for r in rose(n-1-k) : yield [l]+[r]+[r] def start(n) : for x in rose(n) : print(x) #basically I am printing x for each rose(n) file print("starting program: \n") start(2) # here is where I call the start function
def max_n(lst, n = 1): return sorted(lst, reverse = True)[:n] max_n([1, 2, 3]) # [3] max_n([1, 2, 3], 2) # [3, 2]
• Dec 24, 2025 •CodeCatch
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string
• Apr 21, 2023 •sebastianagauyao2002-61a8
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print("hellur")
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def sum_of_powers(end, power = 2, start = 1): return sum([(i) ** power for i in range(start, end + 1)]) sum_of_powers(10) # 385 sum_of_powers(10, 3) # 3025 sum_of_powers(10, 3, 5) # 2925