Python Recursive addition

Question

Recursive Multiplication Design a recursive function that accepts two arguments into the parameters x and y. The function should return the value of x times y. Remember, multiplication can be performed as repeated addition as follows: 7 × 4 = 4 + 4 + 4 + 4 + 4 + 4 + 4

Yes, this is my home work from like three weeks ago, but i just cant figure this thing out.

def mult(x, y):

     x = int(input("enter x"))
     y = int(input("enter y"))
     i = 0
     for i in range(y):
             x = x + (x - 1)
             i = i + 1
             print(x)
 mult(x, y)

This is what i have, but if i put 10 for x and 10 for y i get 1024 instead of 100. what is going on here?

Not sure how you get 1024; your posted code would produce 9217 for x = 10 and y = 10.. It'll print out another 9 numbers before that point, but none of those is ever equal to 1024 either. — Martijn Pieters
– Martijn Pieters, Commented Apr 23, 2016 at 18:02

Martijn Pieters · Accepted Answer · 2016-04-23 17:59:39Z

2

You are incrementing x each time you add x - 1 to it. So for 10 times 10 you start with 10, then add 10 - 1 = 9, making 19, and store that back in x. The next iteration you add 19 - 1 = 18 to x = 19, so now you have x = 37. Then you add 37 - 1 = 36 to 37, so now you have x = 73. Next you'd add 72, and you are already way past the real answer of 100, in just 4 steps; there are another 6 to go!

Keep your sum in a separate variable, and start that at zero. Add y to that each time:

total = 0
for i in range(x):
    total = total + y

print(total)

There is no need to subtract 1 from x, the for loop takes care of counting for you.

None of this is actually using recursion. This is a recursive version:

def mult(x, y):
    if x == 0:
        return 0
    return y + mult(x - 1, y)

So if x is 0, there is no multiplication and you return 0, otherwise you sum y to the multiplication of y and x - 1. Yes, here you do subtract one.

edited Apr 23, 2016 at 17:59

answered Apr 23, 2016 at 17:42

Martijn Pieters

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9 Comments

Richard Bermudez Kenja Trading Over a year ago

x = 0 y = 0 def mult(x, y): x = int(input("enter x")) y = int(input("enter y")) total = mult(x, 1) if x == 0: return 0 return y + mult(x - 1, y) print(total) mult(x, y)

Martijn Pieters Over a year ago

@RichardBermudezKenjaTrading move the input() calls out of the function. You don't want to ask for new values each recursive call. So instead of x = 0, use x = int(input('Enter x: '), same for y, then call the mult(x, y) function.

Richard Bermudez Kenja Trading Over a year ago

x = int(input("enter x")) y = int(input("enter y")) def mult(x, y): total = mult(x, 1) if x == 0: return 0 return y + mult(x - 1, y) print(mult(x,y)) mult(x, y)

Richard Bermudez Kenja Trading Over a year ago

it goes in to an infinite loop and then says its reached its maximum recursion

Martijn Pieters Over a year ago

@RichardBermudezKenjaTrading your version does not match mine, does it. You call mult() on the first line of the function, passing in x and 1. That'll only lead to the function forever calling itself. Why did you add that call?

|

trans1st0r · Accepted Answer · 2016-04-23 17:42:14Z

1

The recursion is:

  mult(x, y) = x + mult(x, y-1)
  mult(x, 1) = x  // This is the base condition

Now, you should be able to implement it

answered Apr 23, 2016 at 17:42

trans1st0r

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Comments

Leonel · Accepted Answer · 2016-04-23 22:52:22Z

0

Here is the working code, I normally use C++ print "welcome to recursive function" x = input("Enter the x: " ) y = input("Enter the y: " )

CONSTANT = x def recursive(x, y): if y == 0: x = 0 return x if y == 1: return x if y > 1: x = x + recursive(x, y-1) return x x = recursive(x, y) print "The resulting value is ", x

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answered Apr 23, 2016 at 22:52

Leonel

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