FizzBuzz in Bash with no Modulus or for/while loops

08/07/2017 • 5 minute read

A few days ago on #bash, people were sharing their least favorite interview questions. During that discussion, FizzBuzz inevitably came up. If you don’t know what that is, here are the rules:

Write a short program that prints each number from 1 to 100 on a new line.

For each multiple of 3, print “Fizz” instead of the number.

For each multiple of 5, print “Buzz” instead of the number.

For numbers which are multiples of both 3 and 5, print “FizzBuzz” instead of the number.

The most common solutions to this problem hinge upon knowing the Modulus Operator to test for the remainder of a number after division. In this case, testing 3, 5, and 15 for remainders and then looping through to print.

That is boring though, so let’s do it with no modulus, no for/while loops, and trying to save as many characters as possible.

You can jump to my final solution if you don’t want to see the steps to get there.

Step 1: Printing Numbers 1-100

Reaching for echo {1..100} or a printf variant would be an obvious choice here, but remember that I am trying to save characters, so let’s use seq 100.

Step 2: Finding a way to test for divisibility:

I can’t use a modulus operator, so that throws out the conventional methods for these kind of tests.

With that in mind, let me introduce factor:

factor [NUMBER]...

Print the prime factors of each specified integer NUMBER.

Some people question why this command is even in coreutils, but it is perfect for our purpose.

Since 3 and 5 are both prime, then we can skate by using factor on each number, like in this shorter example:

$ seq 10 | factor

2
3
2 2
5
2 3
7
2 2 2
3 3
2 5
11
2 2 3
13
2 7
3 5

I’ll handle the non-prime (15) in the next step.

Step 3: Formatting our lines

First off, I want to say thank you to /u/neilmoore for fixing some issues with my first script.

Alright, we have our list and our factors. Let’s start FizzBuzzing.

I chose sed over my preferred tool for this kind of job (awk) because of the modulus operator issue again.

All I am doing with sed is a simple replace. Let’s start with 15 since I will need to pipe that other sed substitutions later:

$ seq 15 |factor | sed 's/.*3 * 5.*/FizzBuzz/g'

2
3
2 2
5
2 3
7
2 2 2
3 3
2 5
11
2 2 3
13
2 7
FizzBuzz

Note: I am kind of cheating here. Since 15 isn’t a prime number, it can’t be included in factor, but instead I can look to see if 3 and 5 appear on the same line. If these were any other number pairs I might have to worry, but since anything ending with a 3 is divisible by 3 and the same for 5, then I am good to just ignore the ??: output from factor.

Alright, let’s add in the tests for 3 and 5:

$ seq 100 | factor | sed 's/.*3 * 5.*/FizzBuzz/g; s/.* 5\( .*\|$\)/Buzz/g; s/.* 3\( .*\|$\)/Fizz/g'
1:
2: 2
Fizz
4: 2 2
Buzz
Fizz
7: 7
8: 2 2 2
Fizz
Buzz
11: 11
Fizz
13: 13
14: 2 7
FizzBuzz
#### Continued to 100 ####

Note: If you didn’t know, sed commands can be chained together using ; or -e with separate commands, but I am trying to save space so I went for the less readable ;. Learn more in the sed manual.

We’re getting there now. The last remaining issue is to get rid of the factor output of primes after the : sign with one more chained sed command:

$ seq 100 | factor | sed 's/.*3 * 5.*/FizzBuzz/g; s/.* 5\( .*\|$\)/Buzz/g; s/.* 3\( .*\|$\)/Fizz/g; s/:.*//g'
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
#### Continued to 100 ####

Perfect. This is a nice little one-liner, but I bet we can make it smaller.

(Somewhat) Cheating Solution

Although I love factor and will always defend it, there is a simpler way that doesn’t use it at all.

While reading through the sed manual for the earlier script, I stumbled upon the section on Selecting lines by numbers. The syntax of sed 'first~step' lets me select lines by multiples of a certain number. You can probably guess where this is going.

Without having to use factor at all, I am left with this:

$ seq 100|sed '0~5s/.*/Buzz/;0~3s/.*/Fizz/;0~15s/.*/FizzBuzz/'
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
#### Continued to 100 ####

Instead of a find and replace based on prime factors, I am running through line numbers with multiples of 3, 5, or 15 and replacing the whole line.

The sed section differs only slightly from my other solution in that the search for 15 happens last so it doesn’t get overwritten by the 3 or 5 replacements.

This is only 64 60 characters, but I called this somewhat cheating because I don’t really feel that it is in the spirit of the question to just hard-code values, but it works.

Conclusion

This was a fun exercise in a challenge everyone loves to hate. Hopefully I get a chance to use it in the future.

Please let me know if you find a smaller solution, I would love to share it!