A haiku is a poem with three lines, with a 5/7/5 syllable count, respectively.

A haiku-w is poem with three lines, with a 5/7/5 word count, respectively.


Write a program that will return true if the input is a haiku-w, and false if not.

A valid haiku-w input must consist of 3 lines, separated by a newline.

  • Line 1 must consist of 5 words, each word separated by a space.
  • Line 2 must consist of 7 words, each word separated by a space.
  • Line 3 must consist of 5 words, each word separated by a space.


The man in the suit
is the same man from the store.
He is a cool guy.

Result: True

Whitecaps on the bay:
A broken signboard banging
In the April wind.

Result: False


  • This is , so the shortest answer in bytes wins.
  • Standard code-golf loopholes apply. Cheating is prohibited.
  • Other boolean return values, such as 1 and 0, are acceptable.
  • A length-3 list of strings as an input is also acceptable.
  • Valid haiku-w inputs should not have leading or trailing spaces, or multiple spaces separating words.
  • 1
    \$\begingroup\$ Will the haiku-w always contain 3 lines? \$\endgroup\$ – user41805 Jan 29 '17 at 10:31
  • 1
    \$\begingroup\$ Yes. If the input contains more than or fewer than 3 lines, the program should return false. \$\endgroup\$ – DomTheDeveloper Jan 29 '17 at 10:33
  • 5
    \$\begingroup\$ Will there ever be leading or trailing spaces on any line? Or multiple spaces separating words? \$\endgroup\$ – Greg Martin Jan 29 '17 at 10:34
  • 8
    \$\begingroup\$ By the way, clarifications like this are a primary reason to post proposed questions in the Sandbox first. :) \$\endgroup\$ – Greg Martin Jan 29 '17 at 10:43
  • 11
    \$\begingroup\$ Bonus points for submissions where the code itself is a haiku-w. \$\endgroup\$ – Glorfindel Jan 30 '17 at 7:22

42 Answers 42


JavaScript (ES6), 73 72 64 63 54 42 39 bytes

Thanks to Neil for saving 13 bytes

a=>a.map(b=>b.split` `.length)=='5,7,5'


This is a fat-arrow function that takes an array of strings as argument. It replaces each line by its word count. If it is a haiku-w, a now contains an array of a five, a seven and a five again. Since JavaScript doesn't allow us to compare 2 arrays at the same time, the array is converted to a string first, and then compared. This results in a boolean that is returned.

  • 1
    \$\begingroup\$ % and * have the same precedence, so you don't need the ()s, although I think (d*2|5) might also work. Also you can get away with a single &, although I think you can improve on even that by using (b...).length==3>b.some(...length-...). \$\endgroup\$ – Neil Jan 29 '17 at 10:29
  • \$\begingroup\$ Thanks for the tip about the parentheses. Also, I changed my approach, so I don't explicitly check for length anymore. \$\endgroup\$ – Luke Jan 29 '17 at 10:46
  • 1
    \$\begingroup\$ Ah, in that case, don't bother with the calculation, just use a=>a.map(c=>c.split .length)=='5,7,5'. \$\endgroup\$ – Neil Jan 29 '17 at 10:55
  • \$\begingroup\$ Hehe, you're right. I should've thought of that... \$\endgroup\$ – Luke Jan 29 '17 at 10:56
  • \$\begingroup\$ You still don't need the +'' - == stringifies if the other argument is a string. \$\endgroup\$ – Neil Jan 29 '17 at 11:13

AWK (GNU Awk), 24, 30, 28, 20 bytes


517253==$0=q=q NF NR

Will output "517253" for True, and empty string for False.

In awk, any nonzero numeric value or any nonempty string value is true. Any other value (zero or the null string, "") is false

The GNU Awk User's Guide

How It Works

Each awk statement (rule) consists of a pattern (or expression) with an associated action:

pattern {action}

Awk will read the input line by line (record by record) and evaluate pattern expression to see if a corresponding action is to be invoked.

The code above is a standalone Awk expression (pattern) w/o the action block, which is implied to be {print $0} in that case.

It should be read right-to-left:

q=q NF NR

Append a Number of Fields (words) and Number of Records (i.e. the current line number), to the variable q.

This way, when processing a proper Haiku-w, q will be set to:

  • 51 - on line #1 (5 words)
  • 5172 - on line #2 (5 words + 7 words)
  • 517253 - on line #3 (5 words + 7 words + 5 words)


Assign the newly computed value of q to $0 (which holds the whole current line/record by default).


Compare it with a "signature" for a proper Haiku-w (517253), if there is a match, the whole expression evaluates to "true" and a corresponding action (implicit print $0) is run, sending "517253" to stdout (True), otherwise output will be empty (False).

Note that this will properly recognize a Haiku-w, even if it is followed by an arbitrary number of garbage lines, but I believe that is ok, as:

A length-3 list of strings as an input is also acceptable.

(i.e. we can assume the input to be 3 lines long)


>awk '517253==$0=q=q NF NR'<<EOF
The man in the suit
is the same man from the store.
He is a cool guy.


Try It Online !

  • 2
    \$\begingroup\$ This fails if the input consists of one line containing 575 words, or two lines containing 57 and 5 words, etc. \$\endgroup\$ – Lynn Jan 29 '17 at 12:51
  • \$\begingroup\$ @Lynn, true, putting on hold, until this is fixed. \$\endgroup\$ – zeppelin Jan 29 '17 at 13:09
  • \$\begingroup\$ @Lynn, should be fixed now \$\endgroup\$ – zeppelin Jan 29 '17 at 15:47
  • 1
    \$\begingroup\$ Very clever fix! :) \$\endgroup\$ – Lynn Jan 29 '17 at 15:55

Python, 42 bytes

lambda l:[s.count(' ')for s in l]==[4,6,4]

Try it online!

Takes input as a list of lines, with the words separated by single spaces.

As we're guaranteed there'll be no leading or trailing spaces, and only single spaces will seperate each word, we can verify a w-haiku by simply counting the spaces in each line.

We do this in a list comprehension, to create a list of the space-counts. If it is a correct haiku, it should look like [4, 6, 4], so we compare it with this and return the result.

  • \$\begingroup\$ If you're okay with supporting only Python 2, you can save two bytes: map(str.count,l,' '). \$\endgroup\$ – vaultah Jan 31 '17 at 17:07

Jelly, 10 9 bytes


Try it online!


ċ€⁶⁼“¥©¥‘  Input: length-3 list of strings
 €         For each string
ċ ⁶          Count the number of spaces
    “¥©¥‘  Convert string to code page indices, makes [4, 6, 4]
   ⁼       Match
  • \$\begingroup\$ Also ċ€⁶⁼4,6,4 and ċ€⁶⁼464D¤… I can’t find anything shorter, though. (Oh, you can flip it, too: 464D⁼ċ€⁶$) \$\endgroup\$ – Lynn Jan 29 '17 at 13:01
  • \$\begingroup\$ ċ€⁶Ḍ=464 works fine for 8. \$\endgroup\$ – Jonathan Allan Jan 29 '17 at 13:42
  • \$\begingroup\$ Actually, no it does not, sorry. \$\endgroup\$ – Jonathan Allan Jan 29 '17 at 13:58

Batch, 102 bytes

@echo off
call:c&&call:c 2||exit/b
for %%W in (%s%)do set/an-=1

Exits with non-zero errorlevel as soon as it reads a line with the wrong number of words.

  • 1
    \$\begingroup\$ ......well crap \$\endgroup\$ – tbodt Jan 29 '17 at 11:40

Mathematica, 21 bytes

{4,6,4}==Count@" "/@#&

Unnamed function taking a list of lists of characters as input and returning True or False. Simply counts how many spaces are in each list of characters, which under the rules of the challenge correlate perfectly with the number of words in each line.

Previous submission:

Mathematica, 31 bytes


Unnamed function taking a list of strings as input and returning True or False.


Haskell, 34 33 bytes

f l=[sum[1|' '<-c]|c<-l]==[4,6,4]

Try it online!.

Edit: thanks to @xnor for a byte!

  • \$\begingroup\$ Pointful is shorter: f l=[sum[1|' '<-c]|c<-l]==[4,6,4]. \$\endgroup\$ – xnor Jan 29 '17 at 22:40

Retina, 12 bytes


(there's a trailing space after the first line)

Try it online!

  • M%`  - Count the number of spaces in each line.

    • M - Match mode - print the number of matches.
    • % - for each line
    • ` - separate configuration and regex pattern
    • - just a space.
  • ^4¶6¶4$ - There should be 4, 6, and 4 spaces, and exactly three lines.

    • matches newlines. The rest is a simple regular expression.

Prints 1 for valid input, 0 for invalid.


Python, 58 44 bytes

lambda h:[len(l.split())for l in h]==[5,7,5]

-14 by tbodt

  • \$\begingroup\$ You're allowed to take the input as a length 3 list of strings. You can save the bytes spent using split("\n"). \$\endgroup\$ – miles Jan 29 '17 at 10:58
  • \$\begingroup\$ 44 bytes: lambda h:[len(l.split())for l in h]==[5,7,5] \$\endgroup\$ – tbodt Jan 29 '17 at 11:40
  • \$\begingroup\$ Someone make this shorter to reach crossed out 44 \$\endgroup\$ – Charlie Feb 2 '17 at 20:56

Perl, 26 bytes

24 bytes of code + 2 bytes for -ap flags.


Try it online!


Pyth, 9 bytes


A program that takes input of a list of "quoted strings" and prints True or False as appropriate.

Test suite

How it works

qj464T/R;   Program. Input: Q
qj464T/R;Q  Implicit variable fill
     T      Are the base-10
 j          digits
  464       of 464
q           equal
      /     to the number
        ;   of spaces
       R    in each string
         Q  in the input?
            Implicitly print

Japt, 11 bytes

Saved lots of bytes thanks to @ETHproductions

This takes an array of three strings as input.


Run it online!


PowerShell, 43 bytes

"$args"-replace'\S'-match'^(    )

Try it online!


Takes input as a newline separated string.

Removes all non-whitespace, then checks to see that it matches "4 spaces, newline, 6 spaces, newline, 4 spaces newline" exactly, using a regex.

The capture group matches 4 spaces, the backreference \1 refers to that. Newlines are embedded in the string. Note the second line of the regex contains two spaces after the backreference.

  • 1
    \$\begingroup\$ That is an interesting aproach! \$\endgroup\$ – Ismael Miguel Jan 31 '17 at 9:28

Pyke, 11 9 bytes


Try it here!

dL/       -  map(i.count(" "), input)
        q - ^ == V
   u  -  [4, 6, 4]

After the u byte there are the following bytes: 0x03 0x04 0x06 0x04


J, 12 bytes

4 6 4=#@;:@>

The input is a boxed list of strings.


This is a fork with a constant left tine. This checks the result of the right tine, #@;:@>, for equality with 4 6 4. The right time unboxes each (>), then (@) converts each string to words (;:), then (@) takes the length of each (#).


R, 48 bytes

all(stringr::str_count(scan(,"")," ")==c(4,6,4))

Reads a 3-length character vector from stdin and works by counting the number of spaces. To count the number of spaces we use the str_count from the stringr package which can count occurrences based on a regex pattern.

An alternative approach without using packages could be:

all(sapply(scan(,""),function(x)length(el(strsplit(x," "))))==c(5,7,5))
  • \$\begingroup\$ First time I've ever seen el before, thanks for that. \$\endgroup\$ – BLT Jan 30 '17 at 5:48

C 142 bytes

void f(){char *c;l=3;s[3]={0};while(l>0){if(*c==' ')s[l-1]++;if((*c=getchar())=='\n'){l--;}}printf("%d",(s[0]==4 && s[1]==6 && s[2]==4)?1:0);}

Ungolfed version:

void f()
  char *c;
  c = (char *)malloc(sizeof(char)); 
  int l=3;
  int s[3]= {0};

    if(*c==' ')

    if( (*c=getchar())=='\n')
  printf("%d",(s[0]==4 && s[1]==6 && s[2]==4)?1:0);

Returns 1 for 5/7/5 sequence else 0.

A positive testcase:

enter image description here


C++, 357 bytes

Sort of new to code golf, but this is the best I could do quickly

#include <iostream>
using namespace std;
int n(std::string s)
    int b = 0;
    for(char c: s)
        if(c == ' ') b++;
    cout << "C = " << b;
    return b;
int main()
    string a, b, c;
    getline(cin, a);
    getline(cin, b);
    getline(cin, c);
    if(n(a)==4 && n(b)==6 && n(c)==4)
    else cout << '0';
    return 0;
  • 4
    \$\begingroup\$ Welcome to PPCG! The goal of code golf is to make your code as short as possible; a good first step would be to remove all unnecessary whitespace. \$\endgroup\$ – ETHproductions Jan 29 '17 at 23:16

Ruby 1.9.3

Not golfed, but is itself a haiku-w

def haiku_w(input)
  lines = input.split("\n")
  lengths = lines.map(&:split).map(&:length)


lines equals input split (newlines)
lengths equals lines map split map length
lengths equals five seven five?

Unfortunately doesn't work with leading whitespace on any lines, but does cope with trailing.


Python 2, 57 64 bytes

Edit Corrected with the addition of 7 bytes after feedback from @Dada. Thanks!

for x in i():j+=`len(x.split())`+' '
i(j=='5 7 5 ')

Try it online!

Not the shortest Python answer by a long way but just wanted to use the new trick I learned recently of using input() to display the output and save a print statement. Takes a list of lines as input. Requires Ctrl C (or any other key-press for that matter) to terminate the program (with an exception) in a terminal but works fine without on TIO.

  • 4
    \$\begingroup\$ This will fail in cases like this one. \$\endgroup\$ – Dada Jan 29 '17 at 14:46
  • \$\begingroup\$ I'll give you that @Dada. That is one serious test case :) \$\endgroup\$ – ElPedro Jan 29 '17 at 14:49
  • \$\begingroup\$ Corrected and tested with your test case. \$\endgroup\$ – ElPedro Jan 29 '17 at 14:53

MATL, 16 bytes


The input is a cell array of strings and returns a truthy or falsey array.

Try it Online


        % Implicitly grab input
"       % For each element in the cell array
@Y:Yb   % Split it on spaces
n       % Count the number of elements
&h      % Horizontally concatenate everything on the stack
[ACA]   % Create the array [5 7 5]
=       % Perform an element-wise equality
        % Implicitly display the truthy/falsey array

MATLAB / Octave, 38 bytes

@(x)cellfun(@(y)sum(y==32),x)==[4 6 4]

This solution accepts a cell array of strings as input, counts the number of spaces in each line and then compares the result to the array [4 6 4] and yields a truthy (all values are 1) or falsey (any value is zero) array.

Online demo


Perl 6, 25 bytes


Clojure, 44 bytes

#(=(for[l %](count(filter #{\ }l)))'(4 6 4))

Input is list of strings. Function finds only spaces and counts them. This explanation is a Haiku. :)


Java 7, 154 bytes

class M{public static void main(String[]a){System.out.print(a.length==3&&a[0].split(" ").length==5&a[1].split(" ").length==7&a[2].split(" ").length==5);}}

The program requirement and potential of having less or more than three lines, not too mention Java's verbosity itself, causes this 'golfed' code to be pretty big..


Try it here.

class M{
  public static void main(String[] a){
    System.out.print(a.length == 3
        && a[0].split(" ").length == 5
         & a[1].split(" ").length == 7
         & a[2].split(" ").length == 5);

SimpleTemplate, 77 bytes

Sadly, the regular expression aproach is the shortest.

{@if"@^([^\s]+ ?){5}\s([^\s]+ ?){7}\s([^\s]+ ?){5}+$@"is matchesargv}{@echo1}

Requires that the text is given as the first argument, with *NIX-style newlines. This won't work with Windows-style newlines.


{@if "@^([^\s]+ ?){5}\s([^\s]+ ?){7}\s([^\s]+ ?){5}+$@"is matches argv}
    {@echo 1}

Non-regex based, 114 byes

{@setR 1}{@eachargv asL keyK}{@php$DATA[L]=count(explode(' ',$DATA[L]))!=5+2*($DATA[K]&1)}{@set*R R,L}{@/}{@echoR}

This requires that each line is given as an argument to the function.


{@set result 1}
{@each argv as line key k}
    {@php $DATA['line'] = count(explode(' ', $DATA['line'])) != 5+2*( $DATA['k'] & 1 )}
    {@set* result result, line}
{@echo result}

Stacked, 22 bytes

[' 'eq sum]map(4 6 4)=

Takes input from the top of the stack as a list of character strings, as such:

($'The man in the suit' $'is the same man from the store.' $'He is a cool guy.')


[' 'eq sum]map(4 6 4)=
[         ]map          map the following function over each item
 ' 'eq                  vectorized equality with ' '
       sum              summed
              (4 6 4)=  is equal to (4 6 4)

Java (OpenJDK), 82 bytes

-2 bytes thanks to @corvus_192!

s->s[0].split(" ").length==5&s[2].split(" ").length==5&s[1].split(" ").length==7

Try it online!

It looks so golfable but without a builtin map function, I can't find a good way. Iterating through the array is a few bytes longer, as is writing a map function using streams.

Lambda expression takes an array of Strings and returns a Boolean.

  • \$\begingroup\$ What if you have only two lines as input, or four? \$\endgroup\$ – Kevin Cruijssen Jan 30 '17 at 10:30
  • \$\begingroup\$ @KevinCruijssen according the op, "A length-3 list of strings as an input is also acceptable". My program takes a length-3 list of lines. \$\endgroup\$ – Pavel Jan 30 '17 at 17:40
  • \$\begingroup\$ You could get a map function if you called Arrays.stream, but that's long enough that it may not be worth using (especially if you need to import Arrays) \$\endgroup\$ – Pokechu22 Jan 30 '17 at 18:11
  • \$\begingroup\$ @Pokechu22 yeah I tried that, still ended up being longer. \$\endgroup\$ – Pavel Jan 30 '17 at 21:18
  • \$\begingroup\$ You can use & instead of && to save two bytes \$\endgroup\$ – corvus_192 Jan 31 '17 at 9:53

SmileBASIC, 96 94 bytes

INPUT A$,B$,C$?C(A$,4)*C(B$,6)*C(C$,4)DEF C(S,E)WHILE""<S

R, 100 bytes

f<-function(a){all(sapply(a,function(x){length(grep(" ",as.list(strsplit(x,"")[[1]])))})==c(4,6,4))}

Takes as an argument a length-3 list of strings. Probably won't be golfed further since further golfing turns it into @Billywob's answer.


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