FizzBuzz Ruby one-liner

Question

Rosettacode.org has this excellent one-line FizzBuzz solution in Ruby.

1.upto(100){|n|puts'FizzBuzz '[i=n**4%-15,i+13]||n}

The trouble is, I don’t understand it. The part that puzzles me is the ”n to the power of 4 modulo -15”. Does anyone have an explanation or a reference to an explanation? I want to use this way of selecting substrings in other problems. For more information on FizzBuzz, see [https://rosettacode.org/wiki/FizzBuzz]

Answer 1

I don't know how they discovered to raise to the fourth power, but the -15 is because FizzBuzz deals with multiples of 3 or multiples of 5 or multiples of both 3 and 5 (ie, multiples of 15)...then negating it ends up working with negative indices quite well. We can see that it works with Modular Exponentiation. The Memory-efficient method section there says:

c mod m = (a ⋅ b) mod m
c mod m = [(a mod m) ⋅ (b mod m)] mod m

In our case, the c is our n, so we have

c ** 4 % m

using the law of exponents, we know that (c ** e1) * (c ** e2) = c ** (e1 + e2), so c ** 4 = (c ** 2) * (c ** 2), so we now have an a and a b, which are both c ** 2. Thus:

(c ** 4) % m = ((c ** 2) * (c ** 2)) % m
             = (((c ** 2) % m) * ((c ** 2) % m)) % m
             = (((c ** 2) % m) ** 2) % m

and following the same steps, again:

(c ** 2) % m = (c * c) % m
             = ((c % m) * (c % m)) % m
             = ((c % m) ** 2) % m

and finally:

(c ** 4) % m = ((((c % m) ** 2) % m) ** 2) % m

When m = -15, the only values for c % m are (-14..0) and we can build a simple table to look at. Since we only ever operate on the result of a modulo, we only need to be able to prove these 15 numbers work:

c%m    **2     %m    **2     %m
-14 => 196 => -14 => 196 => -14
-13 => 169 => -11 => 121 => -14
-12 => 144 => -06 =>  36 => -09
-11 => 121 => -14 => 196 => -14
-10 => 100 => -05 =>  25 => -05
-09 =>  81 => -09 =>  81 => -09
-08 =>  64 => -11 => 121 => -14
-07 =>  49 => -11 => 121 => -14
-06 =>  36 => -09 =>  81 => -09
-05 =>  25 => -05 =>  25 => -05
-04 =>  16 => -14 => 196 => -14
-03 =>   9 => -06 =>  36 => -09
-02 =>   4 => -11 => 121 => -14
-01 =>   1 => -14 => 196 => -14
 00 =>   0 =>  00 =>   0 =>  00

Now, looking at our table, the values for all multiples of 3 are -09, the values for all multiples of 5 are -05, and things that are multiples of 3 and 5 are set to 00; everything else is -14 (If we had used 15 instead of -15, we'd have 6, 10, 0, and 1, respectively, and would need a look up to turn that into string indices). Plugging those in for the start parameter of String#[] with the string 'FizzBuzz ' gives us:

'FizzBuzz '[-9] # => 'F'
'FizzBuzz '[-5] # => 'B'
'FizzBuzz '[0]  # => 'F'
'FizzBuzz '[-14]# => nil

and adding 13 to those numbers to get the length:

'FizzBuzz '[-9, 4]   # => "Fizz"
'FizzBuzz '[-5, 8]   # => "Buzz "
'FizzBuzz '[0, 13]   # => "FizzBuzz "
'FizzBuzz '[-14, -1] # => nil

Answer 2

Quite tricky.

Modulus is a periodic function. You can get more periodic function with the same schema, changing the exponent (k) and divisor (h):

y = x**k % h

Or just see the x,y pairs for the case:

h = 4 # exponent
k = -15 # divisor

xy = []
1.upto 100 do |n|
  i= n**h % k
  xy << [n, i]
end
p xy

The periodicity is evident choosing a basic example of y = x % 2: k = 1 and h = 2. You get a series of 1, 0, 1, 0, 1, ...

To visualise the function used in this case, you can plot in ruby for example using gnuplot gem.

require 'gnuplot' 

Gnuplot.open do |gp|
  Gnuplot::Plot.new( gp ) do |plot|

    plot.title  "Periodic function for FizzBuzz"

    x = (0..100).collect { |v| v }
    p y = x.collect { |v| v ** 4 % -15 }

    plot.data << Gnuplot::DataSet.new( [x, y] ) do |ds|
      ds.with = "linespoints"
    end
  end
end

Answer 3

I'll try to add a simpler explanation to @Simple Lime's excellent answer. If n is a multiple of 3, let's denote it as 3k, Now:

(3k)^4  == 81(k^4)

81 % 15 == 6 and let's subtract 15 (since it's modulo -15) to get -9.

Likewise when n is a multiple of 5, it's 625(k^4) and 625 % 15 == 10 and after subtracting we get -5.

Otherwise, n might be a multiple of 2, 7, 11, and 13. In all these cases n^4 % 15 will be 1 (see Simple Lime's table) and -15 will get us -14.