Fizzbuzz Problem

The Fizzbuzz problem was created as a test for prospective programmers.
Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”.

An algorithm

Jim Fuller 2012-02-24

XQuery

for $n in (1 to 100)
let $fizz := if ($n mod 3) then () else "fizz"
let $buzz := if ($n mod 5) then () else "buzz"
return
  if ($fizz or $buzz) then concat($fizz,$buzz) else $n           

Result

1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz fizz 22 23 fizz buzz 26 fizz 28 29 fizzbuzz 31 32 fizz 34 buzz fizz 37 38 fizz buzz 41 fizz 43 44 fizzbuzz 46 47 fizz 49 buzz fizz 52 53 fizz buzz 56 fizz 58 59 fizzbuzz 61 62 fizz 64 buzz fizz 67 68 fizz buzz 71 fizz 73 74 fizzbuzz 76 77 fizz 79 buzz fizz 82 83 fizz buzz 86 fizz 88 89 fizzbuzz 91 92 fizz 94 buzz fizz 97 98 fizz buzz

An XPath algorithm

Dimitre Novatchev 2012-02-26

XQuery

for $n in (1 to 100),
    $fizz in not($n mod 3),
    $buzz in not($n mod 5)
  return  
    concat("fizz"[$fizz], "buzz"[$buzz], $n[not($fizz or $buzz)])

Result

1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz fizz 22 23 fizz buzz 26 fizz 28 29 fizzbuzz 31 32 fizz 34 buzz fizz 37 38 fizz buzz 41 fizz 43 44 fizzbuzz 46 47 fizz 49 buzz fizz 52 53 fizz buzz 56 fizz 58 59 fizzbuzz 61 62 fizz 64 buzz fizz 67 68 fizz buzz 71 fizz 73 74 fizzbuzz 76 77 fizz 79 buzz fizz 82 83 fizz buzz 86 fizz 88 89 fizzbuzz 91 92 fizz 94 buzz fizz 97 98 fizz buzz

Optimised by factoring a repeated subsequence

Rob Whitby 2012-02-26

XQuery

let $max := 100
let $seq := (for $i in (1 to 14) return if ($i mod 3 eq 0) then "fizz" else if ($i mod 5 eq 0) then "buzz" else "", "fizzbuzz")
for $s at $i in fn:subsequence(for $i in (1 to fn:ceiling($max div 15)) return $seq, 1, $max) 
return if ($s eq "") then $i else $s            
            

Result

compilation error

Optimised by factoring a repeated subsequence - modified

Rob Whitby 2012-02-26

For eXist at least, the result of fn:ceiling() had to be coerced to xs:integer

XQuery

let $max := 100
let $seq := (for $i in (1 to 14) return if ($i mod 3 eq 0) then "fizz" else if ($i mod 5 eq 0) then "buzz" else "", "fizzbuzz")
for $s at $i in fn:subsequence(for $i in (1 to xs:integer(fn:ceiling($max div 15))) return $seq, 1, $max) 
return if ($s eq "") then $i else $s            
            

Result

1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz fizz 22 23 fizz buzz 26 fizz 28 29 fizzbuzz 31 32 fizz 34 buzz fizz 37 38 fizz buzz 41 fizz 43 44 fizzbuzz 46 47 fizz 49 buzz fizz 52 53 fizz buzz 56 fizz 58 59 fizzbuzz 61 62 fizz 64 buzz fizz 67 68 fizz buzz 71 fizz 73 74 fizzbuzz 76 77 fizz 79 buzz fizz 82 83 fizz buzz 86 fizz 88 89 fizzbuzz 91 92 fizz 94 buzz fizz 97 98 fizz buzz

An algorithm

Mark Lawson 2012-02-26

XQuery

for $x in (1 to 100)
let $y := number($x mod 3 = 0) +2 * number($x mod 5 = 0)
return if($y =0) then $x else ('Fizz','Buzz','Fizz/Buzz')[$y]            

Result

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 Fizz/Buzz 16 17 Fizz 19 Buzz Fizz 22 23 Fizz Buzz 26 Fizz 28 29 Fizz/Buzz 31 32 Fizz 34 Buzz Fizz 37 38 Fizz Buzz 41 Fizz 43 44 Fizz/Buzz 46 47 Fizz 49 Buzz Fizz 52 53 Fizz Buzz 56 Fizz 58 59 Fizz/Buzz 61 62 Fizz 64 Buzz Fizz 67 68 Fizz Buzz 71 Fizz 73 74 Fizz/Buzz 76 77 Fizz 79 Buzz Fizz 82 83 Fizz Buzz 86 Fizz 88 89 Fizz/Buzz 91 92 Fizz 94 Buzz Fizz 97 98 Fizz Buzz

A highly optimised algorithm

Dimitre Novatchev 2012-02-26

XQuery

for $k in 1 to 100 idiv 15 +1,
          $start in 15*($k -1) +1,
          $end in min((100, $start + 14))
          return
            let $results :=
                      ($start, $start+1,
                      'fizz',
                      $start+3,
                      'buzz', 'fizz',
                      $start+6, $start+7, 
                      'fizz',
                      'buzz',
                      $start+10,
                      'fizz',
                      $start+12, $start+13,
                      'fizzbuzz')
               return
                      subsequence($results, 1, $end -$start +1)        

Result

1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz fizz 22 23 fizz buzz 26 fizz 28 29 fizzbuzz 31 32 fizz 34 buzz fizz 37 38 fizz buzz 41 fizz 43 44 fizzbuzz 46 47 fizz 49 buzz fizz 52 53 fizz buzz 56 fizz 58 59 fizzbuzz 61 62 fizz 64 buzz fizz 67 68 fizz buzz 71 fizz 73 74 fizzbuzz 76 77 fizz 79 buzz fizz 82 83 fizz buzz 86 fizz 88 89 fizzbuzz 91 92 fizz 94 buzz fizz 97 98 fizz buzz

Strings and their modulus values parameterised to support ease of modification

Chris Wallace 2007-11-26

David Paterson pointed out in his blog that algorithms with fixed values are limited and looked at generalising the problem using an XML configuration file. This is an XQuery version.

XQuery

let $config :=
<fizzbuzz>
<range min="1" max="100"/>
<test>
   <mod value="3" test="0">Fizz</mod>
   <mod value="5" test="0">Buzz</mod>
</test>
</fizzbuzz>

return 
 
   for $i in ($config/range/@min to $config/range/@max)
   let $s :=  
       for $mod in $config/test/mod
       return
         if ($i mod $mod/@value = $mod/@test)
         then string($mod)
         else ()
   return
     if (exists($s))
     then string-join($s,'')
     else $i
            

Result

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz Fizz 22 23 Fizz Buzz 26 Fizz 28 29 FizzBuzz 31 32 Fizz 34 Buzz Fizz 37 38 Fizz Buzz 41 Fizz 43 44 FizzBuzz 46 47 Fizz 49 Buzz Fizz 52 53 Fizz Buzz 56 Fizz 58 59 FizzBuzz 61 62 Fizz 64 Buzz Fizz 67 68 Fizz Buzz 71 Fizz 73 74 FizzBuzz 76 77 Fizz 79 Buzz Fizz 82 83 Fizz Buzz 86 Fizz 88 89 FizzBuzz 91 92 Fizz 94 Buzz Fizz 97 98 Fizz Buzz

An algorithm

Chris Wallace 2012-03-02

XQuery

for $n in 1 to 100
let $x :=  ( if($n mod 3) then () else "Fizz",
             if($n mod 5) then () else "Buzz"
           )
return
   if (exists($x)) then string-join($x,"") else $n 

Result

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz Fizz 22 23 Fizz Buzz 26 Fizz 28 29 FizzBuzz 31 32 Fizz 34 Buzz Fizz 37 38 Fizz Buzz 41 Fizz 43 44 FizzBuzz 46 47 Fizz 49 Buzz Fizz 52 53 Fizz Buzz 56 Fizz 58 59 FizzBuzz 61 62 Fizz 64 Buzz Fizz 67 68 Fizz Buzz 71 Fizz 73 74 FizzBuzz 76 77 Fizz 79 Buzz Fizz 82 83 Fizz Buzz 86 Fizz 88 89 FizzBuzz 91 92 Fizz 94 Buzz Fizz 97 98 Fizz Buzz

An algorithm

Unknown 2010-09-01

see Pastebin

XQuery

for $n in (1 to 100)
let $fizz := if ($n mod 3) then () else "fizz"
let $buzz := if ($n mod 5) then () else "buzz" 
return
  if ($fizz or $buzz) then concat($fizz,$buzz) else $n

Result

1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz fizz 22 23 fizz buzz 26 fizz 28 29 fizzbuzz 31 32 fizz 34 buzz fizz 37 38 fizz buzz 41 fizz 43 44 fizzbuzz 46 47 fizz 49 buzz fizz 52 53 fizz buzz 56 fizz 58 59 fizzbuzz 61 62 fizz 64 buzz fizz 67 68 fizz buzz 71 fizz 73 74 fizzbuzz 76 77 fizz 79 buzz fizz 82 83 fizz buzz 86 fizz 88 89 fizzbuzz 91 92 fizz 94 buzz fizz 97 98 fizz buzz