Arka fon:

Aslında dün gece bu soruyu gönderdim ve belirsizliğine tepki gösterdim. O zamandan beri sadece problemin ifadesini değil aynı zamanda karmaşıklığını (O (1) değil) ilgili birçok personele de başvurdum. Bu programlama problemi, Amazon röportajı sorusundaki kötü bir dönüş.

Soru:

Rastgele birleştirilmiş tamsayılardan oluşan bir Dize [0, 250), 0 - 250 münhasır, göz önüne alındığında, sırada BİR numara eksik. İşiniz bu eksik sayıyı hesaplayacak bir program yazmak. Dizinin içinde başka hiçbir eksik numara yok ve bu sorunu bu kadar zor ve muhtemelen hesaplama açısından zorlaştıran da budur.

Bu problemi elle yapmak, aşağıdaki örnek 1 ve 2 gibi daha küçük stringler üzerinde çok kolaydır. Tersine, üç ya da dört basamaklı sayıları içeren inanılmaz büyük veri kümelerinde eksik bir sayı hesaplamak inanılmaz derecede zor olacaktır. Bu sorunun arkasındaki fikir, bu süreci sizin için yapacak bir program inşa etmektir.

Önemli bilgi:

Dün gece bu problemi gönderdiğimde oldukça kafa karıştırıcı görünen şey şuydu: tam olarak ne eksik bir sayı olarak tanımlanırdı. Eksik bir sayı, yukarıda belirtilen aralığın INSIDE sayısıdır; Mutlaka basamak değil. Örnek 3'te, sırayla görünmesine rağmen eksik sayının 9 olduğunu göreceksiniz. DIGIT 9'un [0, 30): “9”, “19” ve “29” dizisinde görüneceği 3 yer var. Amacınız, bunlar arasında ayrım yapmak ve 9'un eksik NUMARA olduğunu keşfetmek (örnek 3'ün içinde). Başka bir deyişle, zor kısım hangi basamak dizilerinin tamamlandığını ve hangilerinin diğer sayılara ait olduğunu bulmakta yatar.

Giriş:

Girdi, 0 ile 249 arası dahil tam sayıları veya 0 ile 250 arasında özel (bir başka deyişle [0, 250)) içeren bir String S'dir. Bu tamsayılar, yukarıda belirtildiği gibi, rastgele bir sıra oluşturmak için karıştırılır. NO sınırlayıcıları (“42, 31, 23, 44”) veya 0 'ları (003076244029002); sorunlar tam olarak örneklerde açıklandığı gibidir. Gerçek sorunlarda yalnızca 1 çözüm bulunması garanti edilir. Bunlar için birden fazla çözüme izin verilmiyor.

Kazanma Kriterleri:

Kim en hızlı ve en düşük hafızaya sahip olan kazanır. Bir zamanın bağlandığı mucizevi olayda, zaman kırıcı için daha düşük bellek kullanılacaktır. Lütfen Yapabiliyorsanız Büyük O'yu listeleyiniz!

Örnekler:

Örnek 1 ve 2'nin aralığı [0, 10)

Örnek 3 ve 4'ün bir aralığı [0, 30)

(Örnek 1-4 yalnızca tanıtım amaçlıdır. Programınızın bunları yürütmesi gerekmez.)

Örnek 5, [0, 250) aralığına sahiptir.

1. 420137659    
- Missing number => 8

2. 843216075    
- Missing number => 9  

3. 2112282526022911192312416102017731561427221884513 
- Missing number => 9

4. 229272120623131992528240518810426223161211471711
- Missing number => 15

5. 11395591741893085201244471432361149120556162127165124233106210135320813701207315110246262072142253419410247129611737243218190203156364518617019864222241772384813041175126193134141008211877147192451101968789181153241861671712710899168232150138131195104411520078178584419739178522066640145139388863199146248518022492149187962968112157173132551631441367921221229161208324623423922615218321511111211121975723721911614865611197515810239015418422813742128176166949324015823124214033541416719143625021276351260183210916421672722015510117218224913320919223553222021036912321791591225112512304920418584216981883128105227213107223142169741601798025
- Missing number => 71

Test Data: 

Problem 1: 6966410819610521530291368349682309217598570592011872022482018312220241246911298913317419721920718217313718080857232177134232481551020010112519172652031631113791105122116319458153244261582135510090235116139611641267691141679612215222660112127421321901862041827745106522437208362062271684640438174315738135641171699510421015199128239881442242382361212317163149232839233823418915447142162771412092492141987521710917122354156131466216515061812273140130240170972181176179166531781851152178225242192445147229991613515911122223419187862169312013124150672371432051192510724356172282471951381601241518410318414211212870941111833193145123245188102

Problem 2: 14883423514241100511108716621733193121019716422221117630156992324819917158961372915140456921857371883175910701891021877194529067191198226669314940125152431532281961078111412624224113912011621641182322612016512820395482371382385363922471472312072131791925510478122073722091352412491272395020016194195116236186596116374117841971602259812110612913254255615723013185162206245183244806417777130181492211412431591541398312414414582421741482461036761192272120204114346205712198918190242184229286518011471231585109384415021021415522313136146178233133168222201785172212108182276835832151134861116216716910511560240392170208215112173234136317520219

Problem 3: 1342319526198176611201701741948297621621214122224383105148103846820718319098731271611601912137231471099223812820157162671720663139410066179891663131117186249133125172622813593129302325881203242806043154161082051916986441859042111711241041590221248711516546521992257224020174102234138991752117924457143653945184113781031116471120421331506424717816813220023315511422019520918114070163152106248236222396919620277541101222101232171732231122301511263822375920856142187182152451585137352921848164219492411071228936130762461191564196185114910118922611881888513917712153146227193235347537229322521516718014542248813617191531972142714505519240144

Problem 4: 2492402092341949619347401841041875198202182031161577311941257285491521667219229672211881621592451432318618560812361201172382071222352271769922013259915817462189101108056130187233141312197127179205981692121101632221732337196969131822110021512524417548627103506114978204123128181211814236346515430399015513513311152157420112189119277138882021676618323919018013646200114160165350631262167910238144334214230146151171192261653158161213431911401452461159313720613195248191505228186244583455139542924222112226148941682087115610915344641782142472102436810828123731134321131241772242411722251997612923295223701069721187182171471055710784170217851

Clingo , ≈ 0.03 saniye

Bu, doğru bir şekilde ölçülmek için çok hızlı - 250'de yapay olarak durmak yerine daha büyük giriş vakalarına izin vermeniz gerekecek.

% cat(I) means digits I and I+1 are part of the same number.
{cat(I)} :- digit(I, D), digit(I+1, E).

% prefix(I, X) means some digits ending at I are part of the same
% number prefix X.
prefix(I, D) :- digit(I, D), not cat(I-1), D < n.
prefix(I, 10*X+D) :- prefix(I-1, X), digit(I, D), cat(I-1), X > 0, 10*X+D < n.

% Every digit is part of some prefix.
:- digit(I, D), {prefix(I, X)} = 0.

% If also not cat(I), then this counts as an appearance of the number
% X.
appears(I, X) :- prefix(I, X), not cat(I).

% No number appears more than once.
:- X=0..n-1, {appears(I, X)} > 1.

% missing(X) means X does not appear.
missing(X) :- X=0..n-1, {appears(I, X)} = 0.

% Exactly one number is missing.
:- {missing(X)} != 1.

#show missing/1.

Örnek giriş

Giriş (listesidir k , k inci basamak) çiftleri. İşte sorun 1:

#const n = 250.
digit(0,6;1,9;2,6;3,6;4,4;5,1;6,0;7,8;8,1;9,9;10,6;11,1;12,0;13,5;14,2;15,1;16,5;17,3;18,0;19,2;20,9;21,1;22,3;23,6;24,8;25,3;26,4;27,9;28,6;29,8;30,2;31,3;32,0;33,9;34,2;35,1;36,7;37,5;38,9;39,8;40,5;41,7;42,0;43,5;44,9;45,2;46,0;47,1;48,1;49,8;50,7;51,2;52,0;53,2;54,2;55,4;56,8;57,2;58,0;59,1;60,8;61,3;62,1;63,2;64,2;65,2;66,0;67,2;68,4;69,1;70,2;71,4;72,6;73,9;74,1;75,1;76,2;77,9;78,8;79,9;80,1;81,3;82,3;83,1;84,7;85,4;86,1;87,9;88,7;89,2;90,1;91,9;92,2;93,0;94,7;95,1;96,8;97,2;98,1;99,7;100,3;101,1;102,3;103,7;104,1;105,8;106,0;107,8;108,0;109,8;110,5;111,7;112,2;113,3;114,2;115,1;116,7;117,7;118,1;119,3;120,4;121,2;122,3;123,2;124,4;125,8;126,1;127,5;128,5;129,1;130,0;131,2;132,0;133,0;134,1;135,0;136,1;137,1;138,2;139,5;140,1;141,9;142,1;143,7;144,2;145,6;146,5;147,2;148,0;149,3;150,1;151,6;152,3;153,1;154,1;155,1;156,3;157,7;158,9;159,1;160,1;161,0;162,5;163,1;164,2;165,2;166,1;167,1;168,6;169,3;170,1;171,9;172,4;173,5;174,8;175,1;176,5;177,3;178,2;179,4;180,4;181,2;182,6;183,1;184,5;185,8;186,2;187,1;188,3;189,5;190,5;191,1;192,0;193,0;194,9;195,0;196,2;197,3;198,5;199,1;200,1;201,6;202,1;203,3;204,9;205,6;206,1;207,1;208,6;209,4;210,1;211,2;212,6;213,7;214,6;215,9;216,1;217,1;218,4;219,1;220,6;221,7;222,9;223,6;224,1;225,2;226,2;227,1;228,5;229,2;230,2;231,2;232,6;233,6;234,0;235,1;236,1;237,2;238,1;239,2;240,7;241,4;242,2;243,1;244,3;245,2;246,1;247,9;248,0;249,1;250,8;251,6;252,2;253,0;254,4;255,1;256,8;257,2;258,7;259,7;260,4;261,5;262,1;263,0;264,6;265,5;266,2;267,2;268,4;269,3;270,7;271,2;272,0;273,8;274,3;275,6;276,2;277,0;278,6;279,2;280,2;281,7;282,1;283,6;284,8;285,4;286,6;287,4;288,0;289,4;290,3;291,8;292,1;293,7;294,4;295,3;296,1;297,5;298,7;299,3;300,8;301,1;302,3;303,5;304,6;305,4;306,1;307,1;308,7;309,1;310,6;311,9;312,9;313,5;314,1;315,0;316,4;317,2;318,1;319,0;320,1;321,5;322,1;323,9;324,9;325,1;326,2;327,8;328,2;329,3;330,9;331,8;332,8;333,1;334,4;335,4;336,2;337,2;338,4;339,2;340,3;341,8;342,2;343,3;344,6;345,1;346,2;347,1;348,2;349,3;350,1;351,7;352,1;353,6;354,3;355,1;356,4;357,9;358,2;359,3;360,2;361,8;362,3;363,9;364,2;365,3;366,3;367,8;368,2;369,3;370,4;371,1;372,8;373,9;374,1;375,5;376,4;377,4;378,7;379,1;380,4;381,2;382,1;383,6;384,2;385,7;386,7;387,1;388,4;389,1;390,2;391,0;392,9;393,2;394,4;395,9;396,2;397,1;398,4;399,1;400,9;401,8;402,7;403,5;404,2;405,1;406,7;407,1;408,0;409,9;410,1;411,7;412,1;413,2;414,2;415,3;416,5;417,4;418,1;419,5;420,6;421,1;422,3;423,1;424,4;425,6;426,6;427,2;428,1;429,6;430,5;431,1;432,5;433,0;434,6;435,1;436,8;437,1;438,2;439,2;440,7;441,3;442,1;443,4;444,0;445,1;446,3;447,0;448,2;449,4;450,0;451,1;452,7;453,0;454,9;455,7;456,2;457,1;458,8;459,1;460,1;461,7;462,6;463,1;464,7;465,9;466,1;467,6;468,6;469,5;470,3;471,1;472,7;473,8;474,1;475,8;476,5;477,1;478,1;479,5;480,2;481,1;482,7;483,8;484,2;485,2;486,5;487,2;488,4;489,2;490,1;491,9;492,2;493,4;494,4;495,5;496,1;497,4;498,7;499,2;500,2;501,9;502,9;503,9;504,1;505,6;506,1;507,3;508,5;509,1;510,5;511,9;512,1;513,1;514,1;515,2;516,2;517,2;518,2;519,3;520,4;521,1;522,9;523,1;524,8;525,7;526,8;527,6;528,2;529,1;530,6;531,9;532,3;533,1;534,2;535,0;536,1;537,3;538,1;539,2;540,4;541,1;542,5;543,0;544,6;545,7;546,2;547,3;548,7;549,1;550,4;551,3;552,2;553,0;554,5;555,1;556,1;557,9;558,2;559,5;560,1;561,0;562,7;563,2;564,4;565,3;566,5;567,6;568,1;569,7;570,2;571,2;572,8;573,2;574,4;575,7;576,1;577,9;578,5;579,1;580,3;581,8;582,1;583,6;584,0;585,1;586,2;587,4;588,1;589,5;590,1;591,8;592,4;593,1;594,0;595,3;596,1;597,8;598,4;599,1;600,4;601,2;602,1;603,1;604,2;605,1;606,2;607,8;608,7;609,0;610,9;611,4;612,1;613,1;614,1;615,1;616,8;617,3;618,3;619,1;620,9;621,3;622,1;623,4;624,5;625,1;626,2;627,3;628,2;629,4;630,5;631,1;632,8;633,8;634,1;635,0;636,2).

Örnek çıktı

$ clingo missing.lp problem1.lp 
clingo version 5.2.2
Reading from missing.lp ...
Solving...
Answer: 1
missing(148)
SATISFIABLE

Models       : 1+
Calls        : 1
Time         : 0.032s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s)
CPU Time     : 0.032s

C ++, 6.1 saniyede 5000 rastgele test vakası

Bu pratik olarak hızlı, ancak onu yavaşlatan bazı testisler olabilir. Karmaşıklık bilinmiyor.

Birden fazla çözüm varsa, hepsini yazdıracak. Örnek .

Açıklama:

1. Rakamların oluşumunu sayın.

2. Tüm olası cevapları listele.

3. Bir adayın geçerli bir cevap olup olmadığını kontrol edin:

   3-1. Dize (ler) yalnızca bir kez meydana gelen sayılara bölün ve aday hariç, tanımlandığı şekilde işaretlensin.
   Örneğin 2112282526022911192312416102017731561427221884513, sadece bir tane var 14, 211228252602291119231241610201773156ve ve içine bölünebilir 27221884513.

   3-2. Herhangi bir bölünmüş dizenin uzunluğu 1 ise, tanımlandığı şekilde işaretleyin.
   Herhangi bir çelişki olursa (bir kereden fazla tanımlandıysa), aday geçerli değildir.
   Adayı ipte bulamazsak, aday geçerlidir.

   3-3. Herhangi bir bölünme yapılırsa, adım 3-1'i tekrarlayın. Aksi takdirde, adayın geçerli olup olmadığını kontrol etmek için kaba bir kuvvet araştırması yapın.

#include <cmath>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

const int VAL_MAX = 250;
const int LOG_MAX = log10(VAL_MAX - 1) + 1;
using bools = std::bitset<VAL_MAX>;

std::pair<size_t, size_t> count(const std::string& str, const std::string& target)
{
    size_t ans = 0, offset = 0, pos = std::string::npos;
    for (; (offset = str.find(target, offset)) != std::string::npos; ans++, pos = offset++);
    return std::make_pair(ans, pos);
}

bool dfs(size_t a, size_t b, const std::vector<std::string>& str, bools& cnt, int t)
{ // input: string id, string position, strings, identified, candidate
    if (b == str[a].size()) a++, b = 0;
    if (a == str.size()) return true;   // if no contradiction on all strings, the candidate is valid

    int p = 0;
    for (int i = 0; i < LOG_MAX; i++) { // assume str[a][b...b+i] is a number
        if (str[a].size() == b) break;
        p = p * 10 + (str[a][b++] ^ '0');
        if (p < VAL_MAX && !cnt[p] && p != t) { //if no contradiction
            cnt[p] = true;
            if (dfs(a, b, str, cnt, t)) return true; // recursively check
            cnt[p] = false;
        }
    }
    return false;
}

struct ocr {
    int l, r, G;
    bool operator<(const ocr& i) const { return l > i.l; }
};

int cal(std::vector<std::string> str, bools cnt, int t)
{ // input: a list of strings, whether a number have identified, candidate
  // try to find numbers that only occur once in those strings
    int N = str.size();
    std::vector<ocr> pos;

    for (int i = 0; i < VAL_MAX; i++) {
        if (cnt[i]) continue;             // try every number which haven't identified
        int flag = 0;
        std::string target = std::to_string(i);
        ocr now;
        for (int j = 0; j < N; j++) {     // count occurences
            auto c = count(str[j], target);
            if ((flag += c.first) > 1) break;
            if (c.first) now = {c.second, c.second + target.size(), j};
        }
        if (!flag && t == i) return true; // if cannot find the candidate, then it is valid
        if (i != t && flag == 1) pos.push_back(now), cnt[i] = true;
        // if only occur once, then its position is fixed, mark as identified
    }
    if (!pos.size()) { // if no number is identified, do a brute force search
        std::sort(str.begin(), str.end(), [](const std::string& a, const std::string& b){return a.size() < b.size();});
        return dfs(0, 0, str, cnt, t);
    }

    std::sort(pos.begin(), pos.end());
    std::vector<std::string> lst;
    for (auto& i : pos) {      // split strings by identified numbers
        if ((size_t)i.r > str[i.G].size()) return false;
        std::string tmp = str[i.G].substr(i.r);
        if (tmp.size() == 1) { // if split string has length 1, it is identified
            if (cnt[tmp[0] ^ '0']) return false; // contradiction if it is identified before
            cnt[tmp[0] ^ '0'] = true;
        }
        else if (tmp.size()) lst.push_back(std::move(tmp));
        str[i.G].resize(i.l);
    }
    for (auto& i : str) { // push the remaining strings; same as above
        if (i.size() == 1) {
            if (cnt[i[0] ^ '0']) return false;
            cnt[i[0] ^ '0'] = true;
        }
        else if (i.size()) lst.push_back(std::move(i));
    }
    return cal(lst, cnt, t); // continue the split step with new set of strings
}

int main()
{
    std::string str;
    std::vector<ocr> pos;
    std::vector<int> prob;
    std::cin >> str;

    int p[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    for (int i = 0; i < VAL_MAX; i++)
        for (char j : std::to_string(i)) p[j ^ '0']++;
    for (char i : str) p[i ^ '0']--; // count digit occurrences
    {
        std::string tmp;
        for (int i = 0; i < 10; i++)
            while (p[i]--) tmp.push_back(i ^ '0');
        do {           // list all possible candidates (at most 4)
            int c = std::stoi(tmp);
            if (c < VAL_MAX && tmp[0] != '0') prob.push_back(c);
        } while (std::next_permutation(tmp.begin(), tmp.end()));
    }
    if (prob.size() == 1) { std::cout << prob[0] << '\n'; return 0; }
                       // if only one candidate, output it
    for (int i : prob) // ... or check if each candidate is valid
        if (cal({str}, bools(), i)) std::cout << i << '\n';
}

Çevrimiçi deneyin!

Temiz , ~ 0.3 s

Algoritmada büyük bir hata düzeltildi, şimdi yeniden optimize etmeniz gerekiyor.

module main
import StdEnv
import StdLib
import System.CommandLine

maxNum = 250
sample = "11395591741893085201244471432361149120556162127165124233106210135320813701207315110246262072142253419410247129611737243218190203156364518617019864222241772384813041175126193134141008211877147192451101968789181153241861671712710899168232150138131195104411520078178584419739178522066640145139388863199146248518022492149187962968112157173132551631441367921221229161208324623423922615218321511111211121975723721911614865611197515810239015418422813742128176166949324015823124214033541416719143625021276351260183210916421672722015510117218224913320919223553222021036912321791591225112512304920418584216981883128105227213107223142169741601798025"
case1 = "6966410819610521530291368349682309217598570592011872022482018312220241246911298913317419721920718217313718080857232177134232481551020010112519172652031631113791105122116319458153244261582135510090235116139611641267691141679612215222660112127421321901862041827745106522437208362062271684640438174315738135641171699510421015199128239881442242382361212317163149232839233823418915447142162771412092492141987521710917122354156131466216515061812273140130240170972181176179166531781851152178225242192445147229991613515911122223419187862169312013124150672371432051192510724356172282471951381601241518410318414211212870941111833193145123245188102"
case2 = "14883423514241100511108716621733193121019716422221117630156992324819917158961372915140456921857371883175910701891021877194529067191198226669314940125152431532281961078111412624224113912011621641182322612016512820395482371382385363922471472312072131791925510478122073722091352412491272395020016194195116236186596116374117841971602259812110612913254255615723013185162206245183244806417777130181492211412431591541398312414414582421741482461036761192272120204114346205712198918190242184229286518011471231585109384415021021415522313136146178233133168222201785172212108182276835832151134861116216716910511560240392170208215112173234136317520219"
case3 = "1342319526198176611201701741948297621621214122224383105148103846820718319098731271611601912137231471099223812820157162671720663139410066179891663131117186249133125172622813593129302325881203242806043154161082051916986441859042111711241041590221248711516546521992257224020174102234138991752117924457143653945184113781031116471120421331506424717816813220023315511422019520918114070163152106248236222396919620277541101222101232171732231122301511263822375920856142187182152451585137352921848164219492411071228936130762461191564196185114910118922611881888513917712153146227193235347537229322521516718014542248813617191531972142714505519240144"
case4 = "2492402092341949619347401841041875198202182031161577311941257285491521667219229672211881621592451432318618560812361201172382071222352271769922013259915817462189101108056130187233141312197127179205981692121101632221732337196969131822110021512524417548627103506114978204123128181211814236346515430399015513513311152157420112189119277138882021676618323919018013646200114160165350631262167910238144334214230146151171192261653158161213431911401452461159313720613195248191505228186244583455139542924222112226148941682087115610915344641782142472102436810828123731134321131241772242411722251997612923295223701069721187182171471055710784170217851"

failing = "0102030405060708090100101102103104105106107108109110120130140150160170180190200201202203204205206207208209210220230240249248247246245244243242241239238237236235234233232229228227226225224223222221219218217216215214213212211199198197196195194193192191189188187186185184183182181179178177176175174173172171169168167166165164163162161159158157156155154153152151149148147146145144143142141139138137136135134133132131129128127126125124123122121119118117116115114113112111999897969594939291898887868584838281797877767574737271696867666564636261595857565554535251494847464544434241393837363534333231292827262524232221191817161514131211987654321"

dupes = "19050151158951391658227781234527110196235731198137214733126868520474181772192213718517314542182652441211742304719519143231236593134207203121171237201705111617211824810013324511511436253946122155201534113626129692410611318356178791080921122151321949681166200188841675156120546124912883216212189712281541382202411041372421642917614416870223753814121124318415710310515010682172099012716167102179894920613516297239186222232225635312262134019719915382229399107111802082341491811011604815220291125247641482401691871755205639495788414314011714616366130175601931092467744819271230159131158714761192105218019822421812423322919341426216523821428232"

:: Position :== [Int]
:: Positions :== [Position]
:: Digit :== (Char, Int)
:: Digits :== [Digit]
:: Number :== ([Char], Positions)
:: Numbers :== [Number]
:: Complete :== (Numbers, [Digits])

numbers :: [[Char]]
numbers = [fromString (toString n) \\ n <- [0..(maxNum-1)]]

candidates :: [Char] -> [[Char]]
candidates chars
    = moreCandidates chars []
where
    moreCandidates :: [Char] [[Char]] -> [[Char]]
    moreCandidates [] nums
        = removeDup (filter (\num = isMember num numbers) nums)
    moreCandidates chars []
        = flatten [moreCandidates (removeAt i chars) [[c]] \\ c <- chars & i <- [0..]]
    moreCandidates chars nums
        = flatten [flatten [moreCandidates (removeAt i chars) [ [c : num] \\ num <- nums ]] \\  c <- chars & i <- [0..]]

singletonSieve :: Complete -> Complete
singletonSieve (list, sequence)
    | (list_, sequence_) == (list, sequence)
        = reverseSieve (list, sequence)
    = (list_, sequence_)
where
    singles :: Numbers
    singles 
        = filter (\(_, i) = length i == 1) list
    list_ :: Numbers
    list_
        = [(a, filter (\n = not (isAnyMember n (flatten [flatten b_ \\ (a_, b_) <- singles | a_ <> a]))) b) \\ (a, b) <- list]
    sequence_ :: [Digits]
    sequence_
        = foldr splitSequence sequence (flatten (snd (unzip singles)))

reverseSieve :: Complete -> Complete
reverseSieve (list, sequence)
    # sequence
        = foldr splitSequence sequence (flatten (snd (unzip singles)))
    # list
        = [(a, filter (\n = not (isAnyMember n (flatten [flatten b_ \\ (a_, b_) <- singles | a_ <> a]))) b) \\ (a, b) <- list]
    # list
        = [(a, filter (\n = or [any (isPrefixOf n) (tails subSeq) \\ subSeq <- map (snd o unzip) sequence]) b) \\ (a, b) <- list]
    = (list, sequence)
where
    singles :: Numbers
    singles
        = [(a, i) \\ (a, b) <- list, i <- [[subSeq \\ subSeq <- map (snd o unzip) sequence | isMember subSeq b]] | length i == 1]


splitSequence :: Position [Digits] -> [Digits]
splitSequence split sequence
    = flatten [if(isEmpty b) [a] [a, drop (length split) b] \\ (a, b) <- [span (\(_, i) = not (isMember i split)) subSeq \\ subSeq <- sequence] | [] < max a b]

indexSubSeq :: [Char] Digits -> Positions
indexSubSeq _ []
    = []
indexSubSeq a b
    # remainder
        = indexSubSeq a (tl b)
    | startsWith a b
        = [[i \\ (_, i) <- take (length a) b] : remainder]
    = remainder
where
    startsWith :: [Char] Digits -> Bool
    startsWith _ []
        = False
    startsWith [] _
        = False
    startsWith [a] [(b,_):_]
        = a == b
    startsWith [a:a_] [(b,_):b_]
        | a == b
            = startsWith a_ b_
        = False

missingNumber :: String -> [[Char]]
missingNumber string
    # string
        = [(c, i) \\ c <-: string & i <- [0..]]
    # locations
        = [(number, indexSubSeq number string) \\ number <- numbers]
    # digits
        = [length (indexSubSeq [number] [(c, i) \\ c <- (flatten numbers) & i <- [0..]]) \\ number <-: "0123456789"]
    # missing
        = (flatten o flatten) [repeatn (y - length b) a \\ y <- digits & (a, b) <- locations]
    # (answers, _)
        = hd [e \\ e <- iterate singletonSieve (locations, [string]) | length (filter (\(a, b) = (length b == 0) && (isMember a (candidates missing))) (fst e)) > 0]
    # answers
        = filter (\(_, i) = length i == 0) answers
    = filter ((flip isMember)(candidates missing)) ((fst o unzip) answers)


Start world
    # (args, world)
        = getCommandLine world
    | length args < 2
        = abort "too few arguments\n"
    = flatlines [foldr (\num -> \str = if(isEmpty str) num (num ++ [',' : str]) ) [] (missingNumber arg) \\ arg <- tl args]

İle derleyin clm -h 1024M -s 16M -nci -dynamics -fusion -t -b -IL Dynamics -IL Platform main

Bu, dizginin içermesi gereken her sayıyı alarak ve dizgede istenen sayı dizisinin bulunduğu yerlerin sayısını sayarak çalışır. Daha sonra tekrar tekrar bu adımları yapar:

• Eğer sayının muhtemel konumları yoksa, cevap budur.
• Her numarayı bir olası pozisyon ile kaldırın (bunları arayın singles)
• Önceden kaldırılmış numaralardan (a singles) herhangi bir konumla çakışan kalan tüm numaralardan her pozisyonu kaldır