fzz-buzz-2-200-points.md

Fzz Buzz 2 - 200 points

Oh no! Two of my keys are broken! Please help me make the same Fzz Buzz program, sans that one letter and queston marks.

As a side note, use of eval() and exec() is also frowned upon and will be marked invalid.

Solution

This writeup is a python solution

The goal of this challenge was to write a fizz buzz program without using the letter i or the question mark character ?. You must read an integer n from stdin and print that many lines of fizz buzz. For example, if n = 17, output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17

How to get input / print

In order to get user input, you must call input() and likewise, to print you must call print(). Therefore, we need some way to alias these functions without using the character i.

Solution:

Both input and print belong to the __builtins__ module that is loaded automatically on startup. Therefore, we can find a reference to these functions by using getattr on the __builtins__ module.

f = getattr(globals()['__builtins__'],'input')
p = getattr(globals()['__builtins__'],'print')

Then by simply escaping the i character we get this:

f = getattr(globals()['__bu\x69lt\x69ns__'],'\x69nput')
p = getattr(globals()['__bu\x69lt\x69ns__'],'pr\x69nt')

Now f will call input and p will call print.

How to loop

Since the user provides the number of lines to print, we must find some way to loop from 1 to n.

Solution:

Let's use recursion. We will read n from the input and then define a function go(k) that prints the text for line k and calls itself with k+1 if k < n. Finally, we will start it off by calling go(1) That looks like this:

n = f() # Using our alias

def go(k):
	# Do some print magic here
	if (k < n):
		go(k+1)

go(1) # Start the chain

How to use conditionals?

We ran into a problem in our last task: we can't use if since it has an i. Therefore, we must find some other way to get conditional behavior.

Solution:

How about short-circuiting? In python (and many other languages) a conditional will stop evaluation if one side can determine the entire expression. For example, in the following expression, the conditional inside print checks if a == 1. Since that is False, there is no way for the expression to be true (because False AND x == False regardless of x) Therefore, the expression b == 1 isn't even evaluated.

a = 5
b = 3

print(a == 1 and b == 1)

We can use this to our advantage. How about if we put the check as the first conditional and the print as our second. Therefore, if the check succeeds, we print the line. Here is what that looks like:

a = ((k % 15 == 0) and p('F\x69zzBuzz'))

In the previous line, we first check if k (the line number) is divisible by 3 and 5 (or just 15). If it is not, the conditional short-circuits and the right side is not evaluated. If it is, we call p('F\x69zzBuzz') which prints FizzBuzz. (The a = is just so that it is a valid expression).

Putting it all together

Therefore, using all of these ideas, we can come up with the following program:

# Create aliases
f = getattr(globals()['__bu\x69lt\x69ns__'],'\x69nput')
p = getattr(globals()['__bu\x69lt\x69ns__'],'pr\x69nt')

# Get user input
n = f()

# Prints text for line k and calls itself with the next line
def go(k):
	a = ((k % 15 == 0) and p('F\x69zzBuzz'))
	a = ((k % 3 != 0 and k % 5 == 0) and p('Buzz'))
	a = ((k % 3 == 0 and k % 5 != 0) and p('F\x69zz'))
	a = ((k % 3 != 0 and k % 5 != 0) and p(k))
	a = ((k < n) and go(k + 1))

go(1)

Writeup by Harrison Green

External Writeups

• (none)