23
\$\begingroup\$

Given an image, output the [width in pixels of a full vertical section]1 (if one exists). If no vertical section exists, output 0.

Input may be provided as a local file or a nested array. If you choose to take input as a nested array, white pixels should be represented by a truthy value while non-white pixels should be represented by a falsey value.

1. the number of contiguous, all-white columns


You can assume that

  • no image will be larger than 1000 square pixels

  • there will be no more than one full vertical section per image


Examples

Inputs:

Outputs:

50
57
0
0

Here are the first two examples, highlighted (in yellow) to show their sections:

\$\endgroup\$
  • \$\begingroup\$ Can there be islands of black in the middle so that there's multiple vertical sections? \$\endgroup\$ – xnor Feb 29 '16 at 22:41
  • \$\begingroup\$ @xnor: There will only ever be one full vertical section per image. I'll add that to the spec. \$\endgroup\$ – Zach Gates Feb 29 '16 at 22:43
  • \$\begingroup\$ My code is outputting 50 for the first test case, but correct numbers for the last 3, with the vertical slice from columns 233 to 282 (=50 pixels across). Can you confirm 48 is the correct number? \$\endgroup\$ – David Feb 29 '16 at 23:03
  • \$\begingroup\$ @David: I see the correct slice from columns 232 to 282 (exclusive). I believe you're right. \$\endgroup\$ – Zach Gates Feb 29 '16 at 23:10
  • 2
    \$\begingroup\$ I don't think anyone's having problems, but it might be worth explicitly mentioning that you're looking for the number of contiguous, all-white columns. It's clear from the example, but it's generally preferred to not rely on examples or test cases. \$\endgroup\$ – MichaelS Mar 1 '16 at 2:04

12 Answers 12

36
\$\begingroup\$

Jelly, 2 bytes

PS

Try it here!

If I encode an image like so:

0000111111111100000
0000000111111111000
0000000001111100000
0000000011111000000
0001111111111111100
0000001111110000000
0000000111111110000
0000111111111100000

Into a nested array like this:

[[0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0],...]

Then P takes the element-wise product of all of the row vectors, and S sums all of the ones in the result, yielding the length of the vertical slice. (This works only because there’s guaranteed to be only one contiguous slice.) In our case, the answer is 3.

\$\endgroup\$
  • 21
    \$\begingroup\$ ಠ_ಠ This level of golf astounds me. \$\endgroup\$ – Addison Crump Feb 29 '16 at 23:09
  • \$\begingroup\$ What is the output for when there is no contiguous slices? (valid input) \$\endgroup\$ – Addison Crump Feb 29 '16 at 23:12
  • 3
    \$\begingroup\$ ps works in MATL too! \$\endgroup\$ – David Feb 29 '16 at 23:13
  • \$\begingroup\$ Then there will be no column of all 1s, meaning the result of P will be [0,0,0...0], of which the Sum is 0, as expected. \$\endgroup\$ – Lynn Feb 29 '16 at 23:13
  • \$\begingroup\$ @David Post it then? You might need Xps, though, if the image can be a single row (or ask the OP if there's a minimum size) \$\endgroup\$ – Luis Mendo Mar 1 '16 at 0:16
7
\$\begingroup\$

APL, 4 bytes

+/×⌿

Try it here.

This is my first APL answer!

Thanks to @jimmy23013 and @NBZ for saving bytes!

\$\endgroup\$
  • \$\begingroup\$ This is not a function. (+/×/⍉) doesn't work. \$\endgroup\$ – jimmy23013 Mar 1 '16 at 10:40
  • 1
    \$\begingroup\$ But you can use (+/×⌿) and that's 1 byte shorter. \$\endgroup\$ – jimmy23013 Mar 1 '16 at 10:41
  • \$\begingroup\$ Save another 2 bytes by removing the parentheses. Lots of other APL answers have just an anonymous function train that must named or parenthesized to be used: +/×⌿ f←+/×⌿ f picture \$\endgroup\$ – Adám Mar 2 '16 at 1:23
6
\$\begingroup\$

Bash + common utilities, 17

rs -Tc|grep -vc 0

If you're not using grep for , then you're doing it wrong ;-).

This uses the rs utility to do transposition. rs is bundled in OSX, but will need installing in most linux with something like sudo apt-get install rs.

Input columns are TAB separated, and rows are newline separated:

0   0   0   0   1   1   1   1   1   1   1   1   1   1   0   0   0   0   0   
0   0   0   0   0   0   0   1   1   1   1   1   1   1   1   1   0   0   0   
0   0   0   0   0   0   0   0   0   1   1   1   1   1   0   0   0   0   0   
0   0   0   0   0   0   0   0   1   1   1   1   1   0   0   0   0   0   0   
0   0   0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   0   
0   0   0   0   0   0   1   1   1   1   1   1   0   0   0   0   0   0   0   
0   0   0   0   0   0   0   1   1   1   1   1   1   1   1   0   0   0   0   
0   0   0   0   1   1   1   1   1   1   1   1   1   1   0   0   0   0   0

If you like, you can preprocess the example input images into this format with imagemagick and (GNU) sed. E.g:

$ for img in "AmXiR.jpg" "vb2Yt.jpg" "1V7QD.jpg" "MqcDJ.jpg" ; do
>     convert -depth 1 "$img" xpm:- | \
>     sed -nr '/pixels/{:l;n;/}/q;s/^"(.*)",?$/\1/;y/ ./01/;s/./&\t/g;p;bl}' | \
>     rs -Tc|grep -vc 0
> done
50
57
0
0
$
\$\endgroup\$
6
\$\begingroup\$

Perl, 21 22 bytes

Fixed version

Includes +2 for -lp (-l can be omitted and would still be a valid solution, but it's ugly without the final newline)

Give sequences of 1's and 0's on 0 or more lines on STDIN. You can add spaces or commas or whatever between the digits if you want as long as the usage is.consistent on all lines.

$a|=~$_}{$_=$a=~y;\xce;

This works as shown, but replace \xce by the literal byte value to get the claimed score

If there are multiple vertical sections this returns the sum of all section widths. If you want the width of a vertical section use

$a|=~$_}{$a=~/\xce+/;$_="@+"-"@-"

Old version

I originally misunderstood the challenge and implemented a program that gives true or false based on if a vertical line exists at all. Code and explanation here are for this old version

$a|=~$_}{$_|=~$a=~1

If only I could add 1=~ at the left for almost perfect symmetry... I suppose the closest would be

1=>$a|=~$_}{$_|=~$a=~1

Explanation

$a|=~$_     The bitwise operators in perl (&, |, ^, ~) also work on strings by 
            working on the sequence of byte values. The digits "0" and "1" happen
            to have the same ASCII value differing only in the last bit which is
            0 for "0" and 1 for "1". So I would really like to do an "&" here.
            Unfortunately "&" of two different length strings shortens the result
            to the shortest of the strings and my accumulator starts as an empty 
            string. The "|" of two strings however extends to the longest string.
            So instead I will apply De Morgan's law and use "|" on the
            complemented byte string 
}{          Standard perl golf trick. "-p code" transforms to (simplified)
            "while (<>) { code; print }". So if code is "code1 } { code2" this
            becomes "while (<>) { code1 } {code2; print }". So you can use code1
            for the loop operation, use code2 for the final calculation and get a
            free print by assigning to $_
$_|=~$a=~1  I would like to match the accumulator with the bit complement of "1",
            but $a=~~1 doesn't work because the 1 is not a string but a number.
            $a=~~"1" would work but is too long. Next up is complementing $a back
            and matching with 1, so $_=~$a=~1. That also doesn't work since the
            first =~ will be interpreted as a string match insteads of equals
            followed by complement. Easily solved by writing it as $_= ~a=~1. But
            if I am going to give up a byte I can at least have some fun with it.
            Using $_|= also makes the parse work and has the advantage that the
            failure case will give 0 instead of an empty string, which looks
            nicer. It also makes the code look very symmetric. I can also bring
            out the symmetry more by putting 1=> in front (which evaluates 1
            before the assignment and then immediately discards it)
\$\endgroup\$
4
\$\begingroup\$

Python 2, 30 bytes

There is a surprisingly elegant solution using many of my favourite built-in functions chained together.

lambda c:sum(map(all,zip(*c)))

Using the test image from @Lynn:

>>> image = [[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0]]
>>> func = lambda c:sum(map(all,zip(*c)))
>>> func(image)
3
\$\endgroup\$
4
\$\begingroup\$

Pyth, 5

s*VFQ

Try it here

This uses Lynn's algorithm, but I decided to post it just to show how to golf vector operations in Pyth. The trick here is to chain the "sugar" syntax helpers V and F so that the fold is applied as a vector operation. The operator that is folded is of course multiplication, and then the result is summed to get the final answer.

\$\endgroup\$
4
\$\begingroup\$

JavaScript (ES6), 54 45 43 bytes

a=>a[s=0].map((_,i)=>s+=a.every(b=>b[i]))|s
a=>a[s=0].map((_,i)=>s+=!a.some(b=>b[i]))|s

Based on @Lynn's Jelly answer, though since golfed by using every or some instead of reduce. The first version encodes black = 0 while the second encodes black = 1.

Edit: Saved 2 further bytes thanks to @edc65.

\$\endgroup\$
  • 3
    \$\begingroup\$ Try using map \$\endgroup\$ – CalculatorFeline Mar 1 '16 at 1:51
  • \$\begingroup\$ It's 45 in my count. And you did not tried hard enough because it could be 43. \$\endgroup\$ – edc65 Mar 1 '16 at 22:19
  • \$\begingroup\$ a=>a[s=0].map((_,i)=>s+=!a.some(b=>b[i]))|s \$\endgroup\$ – edc65 Mar 1 '16 at 22:35
  • 1
    \$\begingroup\$ @edc65 Well, you know the two hard problems of computing are cache invalidation, naming and off-by-one errors... \$\endgroup\$ – Neil Mar 2 '16 at 0:26
4
\$\begingroup\$

J, 5 6 bytes

Takes Boolean matrix as argument.

[:+/*/

This is my first J answer! (was wrong for 1½ years…)

*/ columnwise product

+/ sum

[: cap (serves as placeholder since +/ should not take a left argument)

Try it online!

\$\endgroup\$
3
\$\begingroup\$

CJam, 7 bytes

q~:.*:+

Try it online!

q~      e# read input and evaluate: push nested array
:.*     e# fold vectorized product over nested array: element-wise product of rows
:+      e# fold addition over array: compute its sum
\$\endgroup\$
2
\$\begingroup\$

Mathematica 24

Length@Cases[Total@#,0]&

Takes an array in the following form:

{{1, 0, 0, 0, 1, 0},
{1, 0, 0, 1, 1, 1},
{1, 1, 0, 0, 0, 0},
{1, 1, 0, 0, 1, 1},
{1, 0, 0, 1, 1, 1}}

And in this case outputs:

1
\$\endgroup\$
  • \$\begingroup\$ Or Length[Total@#~Cases~0]& but same byte count \$\endgroup\$ – CalculatorFeline Mar 1 '16 at 1:51
  • \$\begingroup\$ 1 and 0 are not truthy or falsy in Mathematica (and if they were the assignment would probably be the other way round). \$\endgroup\$ – Martin Ender Mar 1 '16 at 12:46
1
\$\begingroup\$

𝔼𝕊𝕄𝕚𝕟, 7 chars / 9 bytes

⨭МƟïⓜ⨴$

Try it here (Firefox only).

This is @Lynn's great algorithm, but I found it independently. (I thought there was a builtin for this somewhere, still looking :P)

Explanation

МƟï transposes the input array, ⓜ⨴$ turns each inner vector into its product, and sums the resulting array.

\$\endgroup\$
1
\$\begingroup\$

Japt, 6 4 bytes

Takes input as an array of rows, with 1 being white and 0 being black.

y xe
  • 2 bytes saved thanks to ETH.

Test it


Explanation

y xe
          :Implicit input of array U.
y         :Transpose.
   e      :Map over each sub-array, checking if every element is truthy.
  x       :Reduce by summing, converting booleans to 1 or 0.
          :Implicit output of resulting integer.
\$\endgroup\$
  • \$\begingroup\$ I think you can do y x_× for 5. Actually, e works as well as ×, so y xe for 4 :-) \$\endgroup\$ – ETHproductions Aug 13 '17 at 2:41
  • \$\begingroup\$ Missed that comment over the weekend, @ETHproductions - thanks :) \$\endgroup\$ – Shaggy Aug 15 '17 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.