Python beginner recursion

Question

I am trying to get a simple function which takes n and prints

If n > 0:
    print((n*'*')+(n*'!'), end=' ')

and trying to get the same solution but recursively. I am a beginner to recursion, and often I get the "higher level of thinking", but I am having trouble understanding the code that must follow.

My base case is that when n is 0 it prints nothing. When n is more than 1 it will print n copies of * + n copies of!

def repeat(n):
    if n <= 0:
        pass
    else:
        repeat(n-1)
        print((n*'*')+(n*'!'), end=' ')

right now it prints n, and then n-1 successively until 0. I have tried breaking it up into two print statements and using more than one recursion .. but it becomes a messy pattern.

I am also not allowed to use loops. This one is driving me insane; I have come up with several solutions to it besides the easy one line statement, but none that use recursion.

Hint: you have an if n <= 0 clause, but you don't use it. BTW, lose the "n*'*'". — Brian Cain
– Brian Cain, Commented Oct 26, 2012 at 21:09
Another hint: docs.python.org/reference/simple_stmts.html#index-23 — Brian Cain
– Brian Cain, Commented Oct 26, 2012 at 21:11

Peter Mortensen · Accepted Answer · 2016-09-17 16:02:10Z

3

It's simpler if you build and return a string and print it outside of the function, like this:

def printPattern(n):
    if n <= 0:
        return ''
    return '*' + printPattern(n-1) + '!'

Or as a one-liner:

def printPattern(n):
    return '*' + printPattern(n-1) + '!' if n > 0 else ''

Either way, this works:

print printPattern(5)
> *****!!!!!

edited Sep 17, 2016 at 16:02

Peter Mortensen

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answered Oct 26, 2012 at 21:24

Óscar López

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6 Comments

Rohit Jain Over a year ago

I think, OP doesn't want this. He said, when n is more than 1, then print n copies of * + n copies of !.

Rohit Jain Over a year ago

So, for n = 5 :- 5 * and 5 !, then n = 4 : - 4 * and 4!, and so on, till n = 0.

Rohit Jain Over a year ago

Or did I misunderstood the question? :(

Óscar López Over a year ago

@RohitJain I believe my answer is correct, "n copies of *" plus "n copies of !" means just that, say for n==3: ***!!!

jwl4 Over a year ago

Thank you Oscar! That was it, I was using that technique but not in an iteration before! This was it. Thank you!

|

Reinstate Monica · Accepted Answer · 2012-10-26 21:23:48Z

1

Assume you have a solution for n - 1. Prepend * and append !.

def repeat(n):
    if n > 0:
        print("*", end=" ")
        repeat(n - 1)
        print("!", end=" ")

answered Oct 26, 2012 at 21:23

Reinstate Monica

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Comments

Peter Mortensen · Accepted Answer · 2016-09-17 16:04:42Z

1

The following does what you appear to want.

def repeat(n):
  def stars(n):
    return '*'+stars(n-1)+'!' if n > 0 else ''
  print stars(n)

For example, repeat(5) prints *****!!!!! and repeat(8) prints ********!!!!!!!!.

edited Sep 17, 2016 at 16:04

Peter Mortensen

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answered Oct 26, 2012 at 21:27

James Waldby - jwpat7

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Comments

Peter Mortensen · Accepted Answer · 2016-09-17 16:02:43Z

0

I don't actually know what you're asking... If there's a more efficient or better way of doing this? This would be quite obvious:

def repeat(n):
    if n >= 0:
        print((n*'*')+(n*'!'), end=' ')
        return repeat(n-1)

edited Sep 17, 2016 at 16:02

Peter Mortensen

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answered Oct 26, 2012 at 21:18

Nathan

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Comments

Peter Mortensen · Accepted Answer · 2016-09-17 16:04:10Z

0

I would use two strings here, and return a concatenated string of those two strings when n<=0 and use return instead of printing inside function:

def repeat(n, strs1="", strs2=""):     # The Default value of strings is ""

    if n <= 0:
        return strs1 + strs2   # if n<=0 then concatenate the two strings and return them
    else:
        strs1 += "*"          # Add * to strs1
        strs2 += "!"          # Add ! to strs2
        return repeat(n-1, strs1, strs2)  # Pass n-1, as well as the two strings to a recursive call

print(repeat(5))
print(repeat(3))

Output:

*****!!!!!
***!!!

edited Sep 17, 2016 at 16:04

Peter Mortensen

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answered Oct 26, 2012 at 21:18

Ashwini Chaudhary

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