16
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Let's start by re-defining a reflection of a character in a 2-d array of characters:

Given a square 2-d array of characters with distinct lowercase alphabetical letters, define a reflection of a letter in the matrix as swapping it with the character directly across from it through the center of the square.

Thus, a reflection of the letter c in

abcde
fghij
klmno
pqrst
uvwxy

would result in the configuration

abwde
fghij
klmno
pqrst
uvcxy

because the c and the w have been switched.

Some more examples (with the same original configuration as above):

Reflecting the character e would form

 abcdu
 fghij
 klmno
 pqrst
 evwxy

Reflecting the character m would make

 abcde
 fghij
 klmno
 pqrst
 uvwxy

Reflecting the character b would form

 axcde
 fghij
 klmno
 pqrst
 uvwby

The Challenge

Given a 2-d array of characters with distinct lowercase letters, go through each character in a given string and "reflect" it in the matrix.

Clarifications: The letters in the string are from a-z, the letters are unique, and the array is at least 1x1 and at most 5x5 (obviously, as there are only 26 characters in the English alphabet.) The characters in the string are guaranteed to be in the 2-d array. The string is at most 100 characters long.

Input

An string s, an integer N, and then a NxN array of characters.

Example

Input:

ac
2
ab
cd

Output:

dc
ba

*Reason: First, reflect the a with the d. Then, reflect the c with the b because c is the second letter in the input string.


Scoring

  • Output can be given by any convenient method.
  • Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
  • Standard loopholes are forbidden.
  • This is so all usual golfing rules apply, and the shortest code (in bytes) wins.

Current Winner

<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 163084; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 12012; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>

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  • 9
    \$\begingroup\$ Half an hour is not really enough time to get any meaningful use out of the sandbox. \$\endgroup\$ – Sriotchilism O'Zaic Apr 23 '18 at 4:22
  • 3
    \$\begingroup\$ No problem, it looks good. \$\endgroup\$ – user202729 Apr 23 '18 at 5:12
  • 1
    \$\begingroup\$ (also we have a Stack snippet leaderboard) \$\endgroup\$ – user202729 Apr 23 '18 at 5:12
  • 6
    \$\begingroup\$ 50 hours is a bit too short of a time span to accept a winner; typically you'd want to wait a week or so. However, on PPCG, it's common practice to not accept any answers because that discourages future answers, and we want to keep challenges open forever. \$\endgroup\$ – HyperNeutrino Apr 23 '18 at 5:25
  • 2
    \$\begingroup\$ All you examples have the characters in alphabetical order. I assume that's not an assumption we can make? Also, do we have to take N as an input if we don't need it? \$\endgroup\$ – Stewie Griffin Apr 23 '18 at 7:43
3
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Octave, 85 68 66 bytes

Using eval, with a loop inside it saved a lot of bytes! I got the inspiration from this answer by Luis Mendo!

@(c,N,A)eval"for C=c,A(flip(k))=A(k=[x=find(A==C),N^2+1-x]);end,A"

Try it online!

Explanation:

f=@(c,N,A)          % Anonymous function that takes the three input variables
eval"...          % Evaluate the string and run it:
 for C=c,          % Loop over the character list 'c'
  .. x=find(A==C)   % Find the index where the character C is in A, and store it as 'x'
  .. k=[x,N^2+1-x]  % Create a vector with the index of C, and the index of its reflection
   A(flip(k))=A(k)  % Swap the elements in these positions
  end               % End loop
  A"                % Display the new 'A'
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2
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Python 2, 76 bytes

lambda s,a:[a[[i,~i][(s.count(c)+s.count(a[~i]))%2]]for i,c in enumerate(a)]

Try it online!

Takes input:

  • s: string
  • N: ignored
  • a: joined string of chars

Returns a flat list of chars


If I have to take the array as a list:

Python 2, 111 108 107 104 bytes

lambda s,n,a:[[x[sum(map(s.count,x))%2]for i in range(n)for x in[[a[j][i],a[~j][~i]]]]for j in range(n)]

Try it online!

Takes input:

  • s: string
  • n: int
  • a: 2D lists of chars

Returns a 2D list of chars

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1
\$\begingroup\$

Java 10, 126 123 116 bytes

(s,n,m)->{for(var c:s)for(int N=n*n,i,j;N-->0;)if(m[i=N/n][j=N%n]==c){m[i][j]=m[n+~i][n+~j];m[n+~i][n+~j]=c;break;}}

Modifies the input character-matrix instead of returning a new one to save bytes.

Try it online.

Explanation:

(s,n,m)->{        // Method with the three parameters and no return-type
  for(var c:s)    //  Loop over the characters given
    for(int N=n*n,i,j;N-->0;)
                  //   Inner loop over the matrix
      if(m[i=N/n][j=N%n]==c){
                  //    If the current character and matrix-value are equals:
        m[i][j]=m[n+~i][n+~j];m[n+~i][n+~j]=c;
                  //     Swap the values in the matrix at indices [i, j] and [n-i-1, n-j-1]
        break;}}  //     Stop the inner loop (go to next iteration of the outer loop)
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  • \$\begingroup\$ Dang, coding in java!??!?! +1 Def \$\endgroup\$ – NL628 Apr 23 '18 at 23:06
1
\$\begingroup\$

Python 3, 122 111 bytes

lambda l,n,A:[[[A[a][b],A[~b][~a]][sum(map(l.count,[A[a][b],A[~b][~a]]))%2]for b in range(n)]for a in range(n)]

Try it online!

Returns a 2D array of chars.

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  • \$\begingroup\$ You should be able to golf this further.. No way modifying the input matrix is shorter in Java than in Python.. ;p To get you started (and making it 1 byte shorter instead of 1 byte longer than my Java answer): Removing the a and b and using n+~x and n+~y directly saves 2 bytes: a,b=n+~x,n+~y;A[x][y],A[a][b]=A[a][b],A[x][y] to A[x][y],A[n+~x][n+~y]=A[n+~x][n+~y],A[x][y] \$\endgroup\$ – Kevin Cruijssen Apr 23 '18 at 9:43
1
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R, 74 61 bytes

-13 bytes thanks to Giuseppe.

function(s,n,m){for(x in s)m[v]=rev(m[v<-(i=x==m)|rev(i)]);m}

Try it online!

Inputs a vector of characters to search as s, size of the matrix as n and matrix itself as m. If it is absolutely necessary to take the first argument as string, that would pretty much spoil the fun.

\$\endgroup\$
  • \$\begingroup\$ Using logical rather than numeric indices, I golfed down to 61 bytes \$\endgroup\$ – Giuseppe Apr 23 '18 at 13:07
  • \$\begingroup\$ Wow, that's a hefty improvement, thanks a lot. \$\endgroup\$ – Kirill L. Apr 23 '18 at 13:22
0
\$\begingroup\$

Jelly, 15 14 bytes

FW;Ṛi,C$¥¦/ṁḷY

Try it online!

Full program.

Explanation:

FW;Ṛi,C$¥¦/ṁḷY       Main link. Input: ['ab','cd'] (left), 'ac' (right).
FW                   Flatten and Wrap it in a list. Current value = ['abcd'].
  ;                  Concatenate it with right argument. ['abcd','a','c']
          /          Reduce from left:
   Ṛ     ¦             Apply everse at...
    i   ¥              the index (of right argument in left argument)...
     ,C$               and its complement index.

The last operation needs more explanation. Denote f = Ṛi,C$¥¦, then for value ['abcd','a','c'] it computes ('abcd' f 'a') f 'c', which expands to:

Ṛi,C$¥¦    Function f. Assume left argument = 'abcd' and right argument = 'a'
Ṛ          First, compute the reverse. Get 'dcba'.
 i   ¥     To compute the indices to apply to, first the index of 'a' in 'abcd'
             is ('abcd' i 'a') = 1. (first index)
  ,C$      Then pair with (1 C) = 0. (last index)
      ¦    Apply 'dcba' to 'abcd' at indices 0 and 1:

              abcd
              dcba
              ^1 ^0
              ====
              dbca
\$\endgroup\$
0
\$\begingroup\$

Retina 0.8.2, 96 bytes

+`^(.)(.*¶(.|¶)*)((.)((.|¶)*))?\1(?(4)|(((.|¶)*)(.))?)((?<-3>.|¶)*$(?(3).))
$2$11$9$1$6$5$12
1A`

Try it online! Takes the string s and the array of characters as a newline-delimited string without the integer N. Explanation: Each character c of s is processed in turn. The regex matches two positions equidistant from the ends of the array, of which one is c and the other is its mirror m. These characters are swapped and c is removed from s.

+`

Processes each character of s in turn.

^(.)

$1 captures c.

(.*¶(.|¶)*)

$3 captures a stack of characters in the array prefixing one of c or m. $2 captures the rest of s plus all of these characters.

((.)((.|¶)*))?

If m precedes c, $4 has a value, $5 captures m and $6 captures the characters between m and c. $7 captures $6 as a stack but its value is not used.

\1

c is now matched in the array itself.

(?(4)|(((.|¶)*)(.))?)

If m wasn't already matched, then $8 optionally captures a value, in which case $9 captures the characters from c to m, $10 captures $9 as a stack which is unused and $11 captures m. The value is optional in case c and m are the same character.

((?<-3>.|¶)*$(?(3).))

$12 captures the characters suffixing the other of c and m. A balancing group is used to ensure $12 is as long as $3 was deep, i.e. the prefix and suffix are the same length.

$2$11$9$1$6$5$12

The pieces are then put back together - first the rest of s and the prefix of the array, then if c preceeded m then m then the middle, then c, then if m preceeded c then the middle then m, then the suffix.

1A`

Now that s is empty it is deleted.

\$\endgroup\$
0
\$\begingroup\$

JavaScript, 85 bytes

Takes string S and an array A as joined string.

([...S],[...A])=>S.map(c=>[A[j],A[i]]=[A[i=A.indexOf(c)],A[j=A.length+~i]])&&A.join``

f=
([...S],[...A])=>S.map(c=>[A[j],A[i]]=[A[i=A.indexOf(c)],A[j=A.length+~i]])&&A.join``

console.log(
  f('abcd','ac')
)

\$\endgroup\$
0
\$\begingroup\$

Perl 5 -lpF, 97 bytes

$i=0|<>;$_=join'',<>;s/\W//g;for$k(@F){/$k/g;$t=substr$_,-pos,1;eval"y/$t$k/$k$t/"}s%.{$i}%$&$/%g

Try it online!

\$\endgroup\$

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