43
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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.

Challenge

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.

Examples

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


Rules

  • 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.
Laikoni
26.4k7 gold badges54 silver badges116 bronze badges
asked Jan 29, 2017 at 10:12
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9
  • 1
    \$\begingroup\$ Will the haiku-w always contain 3 lines? \$\endgroup\$ Commented Jan 29, 2017 at 10:31
  • 1
    \$\begingroup\$ Yes. If the input contains more than or fewer than 3 lines, the program should return false. \$\endgroup\$ Commented Jan 29, 2017 at 10:33
  • 6
    \$\begingroup\$ Will there ever be leading or trailing spaces on any line? Or multiple spaces separating words? \$\endgroup\$ Commented Jan 29, 2017 at 10:34
  • 9
    \$\begingroup\$ By the way, clarifications like this are a primary reason to post proposed questions in the Sandbox first. :) \$\endgroup\$ Commented Jan 29, 2017 at 10:43
  • 12
    \$\begingroup\$ Bonus points for submissions where the code itself is a haiku-w. \$\endgroup\$ Commented Jan 30, 2017 at 7:22

44 Answers 44

1
2
25
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JavaScript (ES6), (削除) 73 (削除ここまで) (削除) 72 (削除ここまで) (削除) 64 (削除ここまで) (削除) 63 (削除ここまで) (削除) 54 (削除ここまで) (削除) 42 (削除ここまで) 39 bytes

Thanks to Neil for saving 13 bytes

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

Explanation

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.

answered Jan 29, 2017 at 10:17
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6
  • 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\$ Commented Jan 29, 2017 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\$ Commented Jan 29, 2017 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\$ Commented Jan 29, 2017 at 10:55
  • \$\begingroup\$ Hehe, you're right. I should've thought of that... \$\endgroup\$ Commented Jan 29, 2017 at 10:56
  • \$\begingroup\$ You still don't need the +'' - == stringifies if the other argument is a string. \$\endgroup\$ Commented Jan 29, 2017 at 11:13
14
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AWK (GNU Awk), (削除) 24 (削除ここまで), (削除) 30 (削除ここまで), (削除) 28 (削除ここまで), 20 bytes

Golfed

517253==0ドル=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ドル} 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)

0ドル=q

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

517253==0ドル

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ドル) 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)

Test

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

Try It Online !

answered Jan 29, 2017 at 11:59
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4
  • 2
    \$\begingroup\$ This fails if the input consists of one line containing 575 words, or two lines containing 57 and 5 words, etc. \$\endgroup\$ Commented Jan 29, 2017 at 12:51
  • \$\begingroup\$ @Lynn, true, putting on hold, until this is fixed. \$\endgroup\$ Commented Jan 29, 2017 at 13:09
  • \$\begingroup\$ @Lynn, should be fixed now \$\endgroup\$ Commented Jan 29, 2017 at 15:47
  • 1
    \$\begingroup\$ Very clever fix! :) \$\endgroup\$ Commented Jan 29, 2017 at 15:55
10
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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.

answered Jan 29, 2017 at 11:33
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1
  • \$\begingroup\$ If you're okay with supporting only Python 2, you can save two bytes: map(str.count,l,' '). \$\endgroup\$ Commented Jan 31, 2017 at 17:07
8
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Jelly, (削除) 10 (削除ここまで) 9 bytes

ċ6ドル="\©\‘

Try it online!

Explanation

ċ6ドル="\©\‘ 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
answered Jan 29, 2017 at 10:43
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3
  • \$\begingroup\$ Also ċ6ドル=4,6,4 and ċ6ドル=464D¤... I can’t find anything shorter, though. (Oh, you can flip it, too: 464D=ċ6ドル$) \$\endgroup\$ Commented Jan 29, 2017 at 13:01
  • \$\begingroup\$ ċ6ドルḌ=464 works fine for 8. \$\endgroup\$ Commented Jan 29, 2017 at 13:42
  • \$\begingroup\$ Actually, no it does not, sorry. \$\endgroup\$ Commented Jan 29, 2017 at 13:58
7
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Batch, 102 bytes

@echo off
call:c&&call:c 2||exit/b
:c
set/an=%1+5
set/ps=
for %%W in (%s%)do set/an-=1
exit/b%n%

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

answered Jan 29, 2017 at 10:25
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1
  • 1
    \$\begingroup\$ ......well crap \$\endgroup\$ Commented Jan 29, 2017 at 11:40
7
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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

Length/@StringSplit/@#=={5,7,5}&

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

answered Jan 29, 2017 at 10:36
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6
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Retina, 12 bytes

M%` 
^4¶6¶4$

(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.

answered Jan 29, 2017 at 12:07
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0
5
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Haskell, (削除) 34 (削除ここまで) 33 bytes

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

Try it online!.

Edit: thanks to @xnor for a byte!

answered Jan 29, 2017 at 11:49
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1
  • \$\begingroup\$ Pointful is shorter: f l=[sum[1|' '<-c]|c<-l]==[4,6,4]. \$\endgroup\$ Commented Jan 29, 2017 at 22:40
5
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PowerShell, 43 bytes

"$args"-replace'\S'-match'^( )
1円 
1円$'

Try it online!

Explanation

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円 refers to that. Newlines are embedded in the string. Note the second line of the regex contains two spaces after the backreference.

answered Jan 30, 2017 at 0:58
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1
  • 1
    \$\begingroup\$ That is an interesting aproach! \$\endgroup\$ Commented Jan 31, 2017 at 9:28
4
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Python, (削除) 58 (削除ここまで) 44 bytes

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

-14 by tbodt

answered Jan 29, 2017 at 10:27
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3
  • \$\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\$ Commented Jan 29, 2017 at 10:58
  • \$\begingroup\$ 44 bytes: lambda h:[len(l.split())for l in h]==[5,7,5] \$\endgroup\$ Commented Jan 29, 2017 at 11:40
  • \$\begingroup\$ Someone make this shorter to reach crossed out 44 \$\endgroup\$ Commented Feb 2, 2017 at 20:56
4
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Perl, 26 bytes

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

$m.=@F}{$_=$m==575&$.==3

Try it online!

answered Jan 29, 2017 at 13:23
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4
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Pyth, 9 bytes

qj464T/R;

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
answered Jan 29, 2017 at 18:13
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4
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Japt, 11 bytes

Saved lots of bytes thanks to @ETHproductions

This takes an array of three strings as input.

® ̧lÃ\"5,7,5

Run it online!

answered Jan 29, 2017 at 21:36
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4
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Pyke, (削除) 11 (削除ここまで) 9 bytes

dL/uq

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

answered Jan 29, 2017 at 12:03
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0
3
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J, 12 bytes

4 6 4=#@;:@>

The input is a boxed list of strings.

Explanation

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 (#).

answered Jan 29, 2017 at 17:47
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3
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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))
answered Jan 30, 2017 at 0:10
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1
  • \$\begingroup\$ First time I've ever seen el before, thanks for that. \$\endgroup\$ Commented Jan 30, 2017 at 5:48
3
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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};
 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);
}

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

A positive testcase:

enter image description here

Pokechu22
2133 silver badges10 bronze badges
answered Jan 29, 2017 at 15:45
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3
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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)
 cout<<'1';
 else cout << '0';
 return 0;
}
\$\endgroup\$
1
  • 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\$ Commented Jan 29, 2017 at 23:16
3
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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)
 lengths.eql?([5,7,5])
end

or...

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.

answered Feb 1, 2017 at 9:58
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2
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Python 2, (削除) 57 (削除ここまで) 64 bytes

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

i,j=input,''
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.

answered Jan 29, 2017 at 12:37
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3
  • 4
    \$\begingroup\$ This will fail in cases like this one. \$\endgroup\$ Commented Jan 29, 2017 at 14:46
  • \$\begingroup\$ I'll give you that @Dada. That is one serious test case :) \$\endgroup\$ Commented Jan 29, 2017 at 14:49
  • \$\begingroup\$ Corrected and tested with your test case. \$\endgroup\$ Commented Jan 29, 2017 at 14:53
2
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MATL, 16 bytes

"@Y:Ybn&h][ACA]=

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

Try it Online

Explanation

 % 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
answered Jan 29, 2017 at 15:33
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2
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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

answered Jan 29, 2017 at 15:56
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2
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Perl 6, 25 bytes

{.lines».words~~(5,7,5)}
answered Jan 29, 2017 at 16:07
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2
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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. :)

answered Jan 29, 2017 at 23:24
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2
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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..

Ungolfed:

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);
 }
}
answered Jan 30, 2017 at 10:26
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2
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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.

Ungolfed:

{@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.

Ungolfed:

{@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}
answered Jan 30, 2017 at 12:44
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2
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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.')

Explanation

[' '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)
answered Jan 30, 2017 at 13:58
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2
\$\begingroup\$

05AB1E, 9 bytes

€#€g575SQ

Try it online!

Explanation

€# # split each line on spaces
 €g # get length of each line
 575 # push the number 575
 S # split into list of digits
 Q # compare for equality
answered Jan 31, 2017 at 10:10
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5
  • \$\begingroup\$ I couldn't get away from vy and 10 bytes, nice solution. Using ð¢ and 464 instead was my approach, unfortunately 464 is still 2 bytes compressed, obviously. \$\endgroup\$ Commented Jan 31, 2017 at 19:41
  • \$\begingroup\$ @carusocomputing: That's an interesting idea. I think you could post that as ð¢464SïQ in 2sable for 8 bytes. \$\endgroup\$ Commented Jan 31, 2017 at 20:41
  • \$\begingroup\$ You go ahead, that's more your answer than mine in all honesty (I think it's a winning answer too!) \$\endgroup\$ Commented Jan 31, 2017 at 21:03
  • \$\begingroup\$ @carusocomputing: ð¢ is the main part of that answer and it was your idea :) \$\endgroup\$ Commented Jan 31, 2017 at 21:20
  • \$\begingroup\$ ahhh, just give me a shoutout. If it ends up having a flaw I won't know how to fix it haha. \$\endgroup\$ Commented Jan 31, 2017 at 21:24
2
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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.

answered Jan 29, 2017 at 19:16
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5
  • \$\begingroup\$ What if you have only two lines as input, or four? \$\endgroup\$ Commented Jan 30, 2017 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\$ Commented Jan 30, 2017 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\$ Commented Jan 30, 2017 at 18:11
  • \$\begingroup\$ @Pokechu22 yeah I tried that, still ended up being longer. \$\endgroup\$ Commented Jan 30, 2017 at 21:18
  • \$\begingroup\$ You can use & instead of && to save two bytes \$\endgroup\$ Commented Jan 31, 2017 at 9:53
2
\$\begingroup\$

SmileBASIC, (削除) 96 (削除ここまで) 94 bytes

INPUT A,ドルB,ドルC$?C(A,4ドル)*C(B,6ドル)*C(C,4ドル)DEF C(S,E)WHILE""<S
INC N,POP(S)<"!
WEND
RETURN N==E
END
answered Jan 30, 2017 at 4:04
\$\endgroup\$
1
2

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