For those who wish a lot more challenge then the old Spanish alphabetical order, let's take a look at how the Hungarian alphabet is ordered.

a, á, b, c, cs, d, dz, dzs, e, é, f, g, gy, h, i, í, j, k, l, ly, m, n, ny, o, ó, ö, ő, p, q, r, s, sz, t, ty, u, ú, ü, ű, v, w, x, y, z, zs

actually, q, w, x and y are not used in Hungarian words, but they are included for loanwords and foreign names. Foreign accented characters which are not part of the Hungarian alphabet (like ñ), have the same priority as the non-accented ones, but we disregard them for this challenge.

The rules, summarized:

  • Digraphs (cs, sz, etc.) and the trigraph (dzs) are considered as they were letters on their own.
  • If the same digraph or trigraph occurs twice directly after each other in a word, they are written in a simplified way: ssz instead of szsz, ddzs instead of dzsdzs but for the alphabetical order the non-simplified order is used. For example kasza < kaszinó < kassza, because kassza is used as k+a+sz+sz+a for the sake of ordering. Sometimes you can find the non-contracted version in a word, in case of compound words.
  • capitalization doesn't matter, with the exception when the two words would be exactly the same without capitalization, in which case the lower case letter has priority
  • The short and long versions of accented vowels have the same priority (a - á, e -é, i - í, o - ó, ö - ő, u - ú ü - ű), with a single exception: if the two words would otherwise be exactly the same, the short vowel has priority over the long vowel. Note, that the vowels with umlaut (ö and ü) are completely different characters from o and u.
  • Hyphens or spaces (for example, in compound words, names, etc.) are completely ignored
márvány sírkő

The task

Your program/function receives strings, composed of characters from the Hungarian alphabet (both lower- and upper-case), but a string might contain spaces or hyphens. For simplicity's sake, the minus sign (ASCII 45) can be used as a hyphen. Note that some characters (like the ő) are not part of ASCII. You can use any encoding you wish, if it supports all the required characters.

You have to order the lines correctly and display/return the result.

You can use any randomly ordered subset of the above examples for testing.


Please don't use any built-in or other way which already knows the Hungarian alphabetical ordering by itself. It would make the competition pointless, and take all the challenge from finding the best regular expression or the best code golfing tricks.


To clear a clarification asked by isaacg: "two strings that only differ by capitalization and long vs. short vowels, but differs in both ways" : Although no rule in the official document explicitly addresses this question, an example found within points to the length of the vowel having more importance than the capitalization.

  • \$\begingroup\$ @FryAmTheEggman Where do you see that? \$\endgroup\$ Mar 11, 2016 at 17:57
  • 10
    \$\begingroup\$ Man, I can't even memorize our proper alphabetical order. How am I going to program this? ;) \$\endgroup\$ Mar 11, 2016 at 18:13
  • 1
    \$\begingroup\$ I've been trying to come up with a bound-to-fail counterexample, where an apparent digraph is actually two letters, such as malacsült or nyílászáró. I wonder if there are any (but you'd need a vocabulary to check for that, which is presumably not part of this challenge) \$\endgroup\$ Mar 11, 2016 at 21:19
  • 2
    \$\begingroup\$ There is no example containing dzs \$\endgroup\$ Mar 11, 2016 at 21:58
  • 2
    \$\begingroup\$ Grammar. Level: Hungarian! \$\endgroup\$
    – sergiol
    Jul 28, 2017 at 0:51

3 Answers 3


Java 8, 742 bytes

Could reduce by another 3 bytes naming the function s instead of sort or another 16 bytes if not counting class-definition.

public class H{String d="cs|dzs?|gy|ly|sz|ty|zs";void sort(java.util.List<String>l){l.sort((a,b)->{String o="-a-á-b-cs-dzs-e-é-f-gy-h-i-í-j-k-ly-m-ny-o-ó-ö-ő-p-q-r-sz-ty-u-ú-ü-ű-v-w-x-y-zs-";int i=c(r(a),r(b),r(o));return i!=0?i:(i=c(a,b,o))!=0?i:b.charAt(0)-a.charAt(0);});}String r(String a){for(int i=0;i<8;i++)a=a.toLowerCase().replace("ááéíóőúű".charAt(i),"aaeioöuü".charAt(i));return a;}int c(String a,String b,String o){a=n(a);b=n(b);while(!"".equals(a+b)){int i=p(a,o),j=p(b,o);if(i!=j)return i-j;a=a.substring(i%4);b=b.substring(j%4);}return 0;}int p(String a,String o){a=(a+1).replaceAll("("+d+"|.).*","-$1");return o.indexOf(a)*4+a.length()-1;}String n(String a){return a.toLowerCase().replaceAll("(.)(?=\\1)("+d+")| |-","$2$2");}}

Can be used like this:

new H().sort(list);


public static void main(String[] args) {
    test(Arrays.asList("cudar", "cukor", "cuppant", "csalit", "csata"));
    test(Arrays.asList("kasza", "kaszinó", "kassza", "kaszt", "nagy", "naggyá", "nagygyakorlat", "naggyal",
    test(Arrays.asList("jácint", "Jácint", "Zoltán", "zongora"));
    test(Arrays.asList("Eger", "egér", "író", "iroda", "irónia", "kerek", "kerék", "kérek", "szúr", "szül"));
    test(Arrays.asList("márvány", "márványkő", "márvány sírkő", "Márvány-tenger", "márványtömb"));

private static void test(final List<String> input) {
    final ArrayList<String> random = randomize(input);
    System.out.print(input + " -> " + random);
    new H().sort(random);
    System.out.println(" -> " + random + " -> " + input.equals(random));

private static ArrayList<String> randomize(final List<String> input) {
    final ArrayList<String> temp = new ArrayList<>(input);
    final ArrayList<String> randomOrder = new ArrayList<>(input.size());
    final Random r = new Random();
    for (int i = 0; i < input.size(); i++) {
    return randomOrder;


[cudar, cukor, cuppant, csalit, csata] -> [csata, cudar, cuppant, csalit, cukor] -> [cudar, cukor, cuppant, csalit, csata] -> true
[kasza, kaszinó, kassza, kaszt, nagy, naggyá, nagygyakorlat, naggyal, nagyít] -> [naggyá, kassza, kaszinó, nagygyakorlat, nagyít, nagy, kaszt, kasza, naggyal] -> [kasza, kaszinó, kassza, kaszt, nagy, naggyá, nagygyakorlat, naggyal, nagyít] -> true
[jácint, Jácint, Zoltán, zongora] -> [Zoltán, jácint, zongora, Jácint] -> [jácint, Jácint, Zoltán, zongora] -> true
[Eger, egér, író, iroda, irónia, kerek, kerék, kérek, szúr, szül] -> [egér, Eger, kerék, iroda, író, kerek, kérek, szúr, irónia, szül] -> [Eger, egér, író, iroda, irónia, kerek, kerék, kérek, szúr, szül] -> true
[márvány, márványkő, márvány sírkő, Márvány-tenger, márványtömb] -> [márványtömb, márványkő, Márvány-tenger, márvány sírkő, márvány] -> [márvány, márványkő, márvány sírkő, Márvány-tenger, márványtömb] -> true


public class HungarianOrder {

    String d = "cs|dzs?|gy|ly|sz|ty|zs";

    void sort(java.util.List<String> l) {
        l.sort((a, b) -> {
            String o = "-a-á-b-cs-dzs-e-é-f-gy-h-i-í-j-k-ly-m-ny-o-ó-ö-ő-p-q-r-sz-ty-u-ú-ü-ű-v-w-x-y-zs-";
            int i = c(r(a), r(b), r(o));
            return i != 0 ? i
                    : (i = c(a, b, o)) != 0 ? i
                            : b.charAt(0) - a.charAt(0);

    // toLower + remove long accent
    String r(String a) {
        for (int i = 0; i < 8; i++)
            a = a.toLowerCase().replace("ááéíóőúű".charAt(i), "aaeioöuü".charAt(i));
        return a;

    // iterate over a and b comparing positions of chars in o
    int c(String a, String b, String o) {
        a = n(a);
        b = n(b);
        while (!"".equals(a + b)) {
            int i = p(a, o), j = p(b, o);
            if (i != j)
                return i - j;
            a = a.substring(i % 4);
            b = b.substring(j % 4);
        return 0;

    // find index in o, then looking if following characters match
    // return is index * 4 + length of match; if String is empty or first character is unknown -1 is returned
    int p(String a, String o) {
        a = (a+1).replaceAll("("+d+"|.).*", "-$1");
        return o.indexOf(a) * 4 + a.length() - 1;

    // expand ddz -> dzdz and such
    String n(String a) {
        return a.toLowerCase().replaceAll("(.)(?=\\1)("+ d +")| |-", "$2$2");

I am using Java's List-type and the order()-function of it, but the comparator is all mine.

  • \$\begingroup\$ Impressive! I imagine you should be able to drop the list type specifier <String> and save a few characters at the cost of a few warnings? \$\endgroup\$
    – Josh
    Mar 11, 2016 at 21:22
  • \$\begingroup\$ @Josh nah, it would produce two casts as Java would infer Object as type of a and b then. I could probably get away defining a class-generic-parameter extending String, though. Also I am not expecting to have the shortest code. ;-) \$\endgroup\$ Mar 11, 2016 at 21:25

Perl, 250

Includes +11 for -Mutf8 -CS.

use Unicode::Normalize;$r="(?=cs|zs|dz|sz|[glnt]y)";print map/\PC*
/g,sort map{$d=$_;s/d\Kd(zs)|(.)\K$r\2(.)/\L$+\E$&/gi;s/d\Kzs/~$&/gi;s/$r.\K./~$&/gi;s/(\p{Ll}*)(\w?)\s*-*/\U$1\L$2/g;$c=$_;$b=$_=NFD lc;y/̈̋/~~/d;join$;,$_,$b,$c,$d}<>

Uses the decorate-sort-undecorate idiom (AKA Schwartzian Transform), and multilevel sorting, where the levels are:

  • L1: compare base letters, ignore diacritics, case, and some punctuation.
  • L2: compare base letters and diacritics, ignore case and some punctuation.
  • L3: compare base letters, diacritics and case, ignore some punctuation.
  • Ln: tie-breaking byte-level comparison.

Internally, (ASCII 0x1C Field Separator — whose value is less than any character in the alphabet for this challenge) is used as a level separator.

This implementation has many limitations, amongst them:

  • No support for foreign characters.
  • Cannot disambiguate between contracted geminated (long) digraphs/trigraphs, and consonant+digraph/trigraph, e.g: könnyű should collate as <k><ö><ny><ny><ű>, while tizennyolc should collate as <t><i><z><e><n><ny><o><l><c>; házszám 'address = house (ház) number (szám)' should collate as <h><á><z><sz><á><m> and not as *<h><á><zs><z><á><m>.
  • Collation for contracted long digraphs is not that consistent (but it is stable): we disambiguate at the identical level (ssz <n szsz, ..., zszs <n zzs ); glibc collates the short forms before the full forms (ssz < szsz, ..., zzs < zszs ), ICU collates the long forms before the short forms starting at L3 Case and Variants (szsz <3 ssz, ..., zszs <3 zzs )

Expanded version:

use Unicode::Normalize;

$r="(?=cs|zs|dz|sz|[glnt]y)";   # look-ahead for digraphs

print map/\PC*\n/g,             # undecorate
  sort                          # sort
  map{                          # decorate

          $d=$_;                # Ln: identical level

          # expand contracted digraphs and trigraphs

          # transform digraphs and trigraphs so they 
          #  sort correctly

          # swap case, so lower sorts before upper
          # also, get rid of space, hyphen, and newline

          $c=$_;                # L3: Case

          $b=$_=NFD lc;         # L2: Diacritics

          # transform öő|üű so they sort correctly
          # ignore diacritics (acute) at this level

                                # L1: Base characters

†. Some well-known multi-level collation algorithms are the Unicode Collation Algorithm (UCA, Unicode UTS#10), ISO 14651 (available at the ISO ITTF site) the LC_COLLATE parts at ISO TR 30112 (draft available at the ISO/IEC JTC1/SC35/WG5 home) which obsoletes ISO/IEC TR 14652 (available at the ISO/IEC JTC1/SC22/WG20 home) and LC_COLLATE at POSIX.

‡. Doing this correctly would require a dictionary. ICU treats weirdly capitalized groups as non-contractions/non-digraphs/non-trigraphs, e.g: ccS <3 CcS <3 cCs <3 cCS <3 CCs <3 cS <3 cs <3 Cs <3 CS <3 ccs <3 Ccs <3 CCS

  • \$\begingroup\$ You should be able to save some bytes using my expansion RegExp. \$\endgroup\$ Mar 19, 2016 at 14:48

Python 3, 70

Saved 8 bytes thanks to shooqie.

I love Python. :D

Expects a list of strings.

from locale import*;setlocale(0,'hu')
f=lambda x:sorted(x,key=strxfrm)
  • 3
    \$\begingroup\$ Isn't this a standard loophole? \$\endgroup\$
    – vsz
    Mar 11, 2016 at 18:13
  • 1
    \$\begingroup\$ @vsz Not as far as I know. Using built-ins is part of lots of challenges. \$\endgroup\$ Mar 11, 2016 at 18:14
  • 1
    \$\begingroup\$ @vsz It has too low an up- to down-vote ratio on the standard loopholes post to be counted as a standard, you'd have to explicitly ban it. \$\endgroup\$ Mar 11, 2016 at 18:15
  • 1
    \$\begingroup\$ Ok, done it. I considered explicitly banning it but i though it should be obvious that it would make the whole challenge a moot point. I'm sorry for the inconvenience. \$\endgroup\$
    – vsz
    Mar 11, 2016 at 18:16
  • 2
    \$\begingroup\$ This isn't a standard loophole. It's a built-in. Built-ins are allowed by default. This shouldn't have a single downvote in my opinion. \$\endgroup\$
    – mbomb007
    Mar 23, 2016 at 19:41

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