Waste less time on Facebook — follow Brilliant.
×

Foo function

Recently I took the test for Computer Science and I got a Level 3... I couldn't figure out the Level 4 Question and I found it quite interesting and could anyone explain me how it works..?

How many '1' s are printed out when you call foo(5)?

   # python code start
   def foo(n):
        count = 1
        if n > 0:
            count += foo(n - 1) + foo(n - 1)
        print '1'
        return count

Note by Muzaffar Ahmed
3 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

the answer is exactly 28

Shivam Banka - 3 years ago

Log in to reply

Have you tried doing it for \(n=1\) and \(n=2\)?

Josh Silverman Staff - 3 years, 8 months ago

Log in to reply

Nope.. I'm not so good at programming so can you please explain me this one from the beginning..?

Muzaffar Ahmed - 3 years, 8 months ago

Log in to reply

If you call foo(1), it is greater than zero and, so, the function passes the if statement. It the adds 2*foo(0) (which needs to be called) to count and prints 1.

When it calls foo(0), 0 is not greater than 0 and, so, the if statement is not passed. The function then runs the code that appears after the if statement, where it returns the current value of count which is 1 and prints 1. However, foo(0) was called twice (count += foo(0) + foo(0)), so it prints 1 twice.

So, for foo(1), we print 1 a total three times.

If we called foo(2), this same kind of branching process would occur. I.e., we'd print 1, then we'd call foo(1) twice, which would print 1 twice, then for each foo(1), we'd call foo(0) twice (a total of 4 foo(0)) calls, which would print 1 four times. Making a total of 7. In general, calling foo(n), we print 1 a total of \(2^{n+1}-1\) times.

Josh Silverman Staff - 3 years, 8 months ago

Log in to reply

@Josh Silverman Ohh.. I got it.. ! Thanks bro :-)

Muzaffar Ahmed - 3 years, 8 months ago

Log in to reply

@Muzaffar Ahmed No problem. It was foon to explain.

Josh Silverman Staff - 3 years, 8 months ago

Log in to reply

@Josh Silverman Lol :D

Muzaffar Ahmed - 3 years, 8 months ago

Log in to reply

:) There's a trick.

Zi Song Yeoh - 3 years, 8 months ago

Log in to reply

So tell me please :) I know we can directly execute it but I want to understand how it works..

Muzaffar Ahmed - 3 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...