22
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Consider a square block of text, N characters wide by N tall, for some odd integer N greater than 1.

As an example let N = 5 and the text be:

MLKJI
NWVUH
OXYTG
PQRSF
ABCDE

Notice that this is the alphabet (besides Z) spiraled around counter-clockwise from the lower left corner. It's kind of like a rolled up carpet.

Spiral Alphabet

"Unrolling" the text by one quarter turn clockwise so FGHI are on the same level as ABCDE results in:

     PONM
     QXWL
     RYVK
     STUJ
ABCDEFGHI

This unrolling can be done 7 more times until the text is a single line:

         SRQP
         TYXO
         UVWN
ABCDEFGHIJKLM

             UTS
             VYR
             WXQ
ABCDEFGHIJKLMNOP

                WVU
                XYT
ABCDEFGHIJKLMNOPQRS

                   XW
                   YV
ABCDEFGHIJKLMNOPQRSTU

                     YX
ABCDEFGHIJKLMNOPQRSTUVW

                       Y
ABCDEFGHIJKLMNOPQRSTUVWX

ABCDEFGHIJKLMNOPQRSTUVWXY

Challenge

The challenge is to write a program that is an N×N block of text that outputs the number of times it has "unrolled" by a quarter turn when it is rearranged into the unrolling patterns and run.

There are really two contests here: (hopefully it won't be too messy)

  1. Do this with the smallest N. (down to a limit of N = 3)
  2. Do this with the largest N. (no limit)

There will not be an accepted answer but the winner in each of these categories will receive at least 50 bounty rep from me. In case of ties the oldest answers win.

Example

If your code block is

MyP
rog
ram

running it as is should output 0.

Running

   rM
   oy
ramgP

should output 1.

Running

     or
ramgPyM

should output 2.

Running

       o
ramgPyMr

should output 3.

Finally, running ramgPyMro should output 4.

Details

  • The output should be printed to stdout (or the closest alternative) by itself. There is no input.
  • You may only use printable ASCII (hex codes 20 to 7E, that includes space) in your code.
  • Spaces fill the empty space in the unrolling arrangements. (Unless you're unrolling to the left.)
  • Only the arrangements from completely square to completely flat need to have valid output. No other arrangements will be run.
  • You may not read your own source.
  • You may use comments.
  • N = 1 is excluded since in many languages the program 0 would work.
  • If desired you may unroll to the left rather than the right. So e.g.

    MyP
    rog
    ram
    

    becomes

    Pg
    yo
    Mrram
    

    and so on. No extra spaces are added when rolling this way. The lines just end

(Related: Write a Rectangular Program that Outputs the Number of Times it was Rotated)

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  • \$\begingroup\$ Before I read the "challenge" paragraph, I was expecting a challenge to write a program that outputs itself unrolled \$\endgroup\$ – John Dvorak Oct 18 '14 at 7:04
  • 1
    \$\begingroup\$ why does N have to be odd? \$\endgroup\$ – John Dvorak Oct 18 '14 at 7:05
  • 1
    \$\begingroup\$ @JanDvorak I suppose N didn't have to be odd, but it makes the spirals more standardized. It's staying that way but feel free to post an N = 2 as a comment if you find one. \$\endgroup\$ – Calvin's Hobbies Oct 18 '14 at 7:11
  • 8
    \$\begingroup\$ Just an idea: Unrolling the "carpet" to the right creates many lines starting with whitespace, eliminating languages like Python. If you'd allow unrolling to the left, there'd be no need for additional whitespace and Python is (theoretically) possible. \$\endgroup\$ – Falko Oct 18 '14 at 10:27
  • 5
    \$\begingroup\$ Do you have a magic book with infinite great challenge ideas? How else do you keep coming up with such interesting challenges? \$\endgroup\$ – Justin Oct 18 '14 at 16:55
27
+50
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Golfscript, N <- [5,7..]

.   .
 . . 
 ..  
.  .#
],9\-

Fully unrolled:

],9\-#  .   .  .  . . ...

Explanation:

  • . (multiple times) - duplicate the input
  • ] - collect the stack into a single array
  • , - take its length
  • 9\- - subtract it from 9
  • # - line comment

Whitespace is a NOP, but any other NOP would have worked just as well.

Fully rolled up, it uses nine copies of the input (contents ignored) as the stack; 9 - 9 = 0; it has not been unrolled.

Each unroll hides one more dot (duplicate) behind the comment, shrinking the stack once, incrementing the output.

Fully unrolled, it uses only the input (contents ignored) as the stack; 9 - 1 = 8; it has been unrolled 8 times.

The same approach works for any N > 4: Change 9 to the appropriate value of 2*N+1, then extend the pattern of dots (duplicate) using the same spiral pattern that ensures exactly one dot gets unrolled during each unroll.

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  • \$\begingroup\$ Well, unless someone finds N = 3, this will be the winning answer in both categories. \$\endgroup\$ – Calvin's Hobbies Oct 18 '14 at 17:36
  • 3
    \$\begingroup\$ @Calvin'sHobbies should I be a total dick and post a left-unrolling solution as well? :-) \$\endgroup\$ – John Dvorak Oct 18 '14 at 17:39
  • \$\begingroup\$ Why not. Another answer seems unlikely otherwise :P \$\endgroup\$ – Calvin's Hobbies Oct 18 '14 at 17:45
  • 1
    \$\begingroup\$ Why not go for one which can unroll in both directions? :) \$\endgroup\$ – Beta Decay Oct 18 '14 at 19:59
  • \$\begingroup\$ @BetaDecay hmm... :-) \$\endgroup\$ – John Dvorak Oct 18 '14 at 20:19
13
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GolfScript, N = 4

This one right rolls as original spec.

.. . 
...# 
.#.~
],8-

Here are the unrolls:

    ...
    #..
    ..
],8-~#.

       .#.
       ...
],8-~#. ..

          ..
          .#
],8-~#. ....

            ..
],8-~#. ....#.

              .
],8-~#. ....#..

],8-~#. ....#...

Try it here

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  • \$\begingroup\$ How did you think of this arrangement? \$\endgroup\$ – proud haskeller Oct 18 '14 at 22:31
  • 3
    \$\begingroup\$ @proudhaskeller It's better if you don't know ... \$\endgroup\$ – Optimizer Oct 18 '14 at 22:32
  • 8
    \$\begingroup\$ Did you brute-search a solution? \$\endgroup\$ – proud haskeller Oct 18 '14 at 22:33
  • \$\begingroup\$ Special challenge: can you make one out of .s and #s? \$\endgroup\$ – John Dvorak Oct 19 '14 at 6:29
  • \$\begingroup\$ I like the trailing ~. Maybe I can steal it for N=3? \$\endgroup\$ – John Dvorak Oct 19 '14 at 6:31
9
+100
\$\begingroup\$

APL, N = 3

201
340
5|0

Unrolled:

   32
   40
5|001

     43
5|00102

       4
5|001023

5|0010234

Try it online.

It calculates the remainder of that number divided by 5. Only the result of last line is printed.

APL, N = 2

⍬∞
≡0

Unrolled:

  ⍬
≡0∞

≡0∞⍬

Try it online.

returns the depth (not to be confused with the dimension or length) of an array:

  • 0 is not an array. So the depth is 0.
  • 0∞ is an array with two items 0 and (infinity). It has depth 1.
  • 0∞⍬ has another item , which is an empty array with depth 1. So 0∞⍬ has depth 2.

These two programs also works in the online interpreter. I'm not sure if the later one is syntactically correct.

⍬0
≡∞


⍬¯
≡0

APL, for any N >= 4

For N = 4:

∞  ∞
 ∞∞
∞ ∞
⍴1↓∞

Fully unrolled:

⍴1↓∞  ∞  ∞ ∞ ∞∞∞

For N = 5:

∞   ∞
 ∞ ∞
 ∞∞
∞  ∞
⍴1↓ ∞

Fully unrolled:

⍴1↓ ∞   ∞   ∞  ∞  ∞ ∞ ∞∞∞

1↓ removes an item in the array. It also returns the empty array if the argument is scalar. gets the array length.

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  • \$\begingroup\$ Any explanations? \$\endgroup\$ – proud haskeller Oct 20 '14 at 6:48
  • \$\begingroup\$ @proudhaskeller Edited. \$\endgroup\$ – jimmy23013 Oct 20 '14 at 7:42
  • \$\begingroup\$ You can ideally use the same depth logic for any N. Thanks to APL \$\endgroup\$ – Optimizer Oct 20 '14 at 8:52
  • \$\begingroup\$ @Optimizer It is not that easy. Things before the last line still have to be syntactically correct. So I can't use most of the functions or other punctuation characters like ()[] as they will appear at some unwanted place. \$\endgroup\$ – jimmy23013 Oct 20 '14 at 9:02
  • \$\begingroup\$ I meant like: ` ⍬ \n⍬⍬0\n≡ ∞` (Not exactly that, but you get the idea) \$\endgroup\$ – Optimizer Oct 20 '14 at 9:04

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