Ini adalah urutan A054261

The $n$ th nomor penahanan utama adalah jumlah terendah yang berisi pertama $n$ bilangan prima sebagai substring. Misalnya, angka $235$ adalah angka terendah yang berisi 3 bilangan prima pertama sebagai substring, menjadikannya bilangan penampung prima ke-3.

Sepele untuk mengetahui bahwa empat nomor penahanan prima pertama adalah $2$ , , dan , tetapi kemudian semakin menarik. Karena prime berikutnya adalah 11, nomor kontainmen prime berikutnya bukan , tetapi$23$$235$$2357$$235711$$112357$ karena terkecil dengan properti.

Namun, tantangan sebenarnya datang ketika Anda melampaui 11. Nomor penahanan utama berikutnya adalah . Perhatikan bahwa dalam nomor ini, substring dan tumpang tindih. Nomor$113257$11133 juga tumpang tindih dengan nomor tersebut 13.

Mudah untuk membuktikan bahwa urutan ini meningkat, karena nomor berikutnya harus memenuhi semua kriteria nomor sebelum itu, dan memiliki satu substring lagi. Namun, urutannya tidak meningkat secara ketat, seperti yang ditunjukkan oleh hasil untuk n=10dan n=11.

Tantangan

Tujuan Anda adalah menemukan sebanyak mungkin nomor penahanan utama. Program Anda harus menampilkannya secara berurutan, mulai dari 2 dan naik.

Aturan

1. Anda diizinkan untuk nomor perdana hard-code.
2. Anda tidak diizinkan untuk nomor penahanan utama kode keras (2 adalah satu-satunya pengecualian), atau angka ajaib yang membuat tantangan sepele. Tolong berbaik hatilah.
3. Anda dapat menggunakan bahasa apa pun yang Anda inginkan. Harap sertakan daftar perintah untuk menyiapkan lingkungan untuk mengeksekusi kode.
4. Anda bebas menggunakan CPU dan GPU, dan Anda dapat menggunakan multithreading.

Mencetak gol

Skor resmi akan dari laptop saya (dell XPS 9560). Tujuan Anda adalah menghasilkan sebanyak mungkin angka penahanan utama dalam 5 menit.

Spesifikasi

• 2.8GHz Intel Core i7-7700HQ (boost 3.8GHz) 4 core, 8 thread.
• 16GB 2400MHz DDR4 RAM
• NVIDIA GTX 1050
• Linux Mint 18.3 64-bit

Jumlah yang ditemukan sejauh ini, bersama dengan prime terakhir ditambahkan ke nomor:

 1 =>                                                       2 (  2)
 2 =>                                                      23 (  3)
 3 =>                                                     235 (  5)
 4 =>                                                    2357 (  7)
 5 =>                                                  112357 ( 11)
 6 =>                                                  113257 ( 13)
 7 =>                                                 1131725 ( 17)
 8 =>                                               113171925 ( 19)
 9 =>                                              1131719235 ( 23)
10 =>                                            113171923295 ( 29)
11 =>                                            113171923295 ( 31)
12 =>                                           1131719237295 ( 37)
13 =>                                          11317237294195 ( 41)
14 =>                                        1131723294194375 ( 43)
15 =>                                      113172329419437475 ( 47)
16 =>                                     1131723294194347537 ( 53)
17 =>                                   113172329419434753759 ( 59)
18 =>                                  2311329417434753759619 ( 61)
19 =>                                231132941743475375961967 ( 67)
20 =>                               2311294134347175375961967 ( 71)
21 =>                              23112941343471735375961967 ( 73)
22 =>                             231129413434717353759619679 ( 79)
23 =>                           23112941343471735359619678379 ( 83)
24 =>                         2311294134347173535961967837989 ( 89)
25 =>                        23112941343471735359619678378979 ( 97)
26 =>                      2310112941343471735359619678378979 (101)
27 =>                    231010329411343471735359619678378979 (103)
28 =>                 101031071132329417343475359619678378979 (107)
29 =>              101031071091132329417343475359619678378979 (109)
30 =>              101031071091132329417343475359619678378979 (113)
31 =>           101031071091131272329417343475359619678378979 (127)
32 =>           101031071091131272329417343475359619678378979 (131)
33 =>         10103107109113127137232941734347535961967838979 (137)
34 =>      10103107109113127137139232941734347535961967838979 (139)
35 =>   10103107109113127137139149232941734347535961967838979 (149)
36 => 1010310710911312713713914923294151734347535961967838979 (151)

Terima kasih kepada Ardnauld, Ourous, dan japh karena telah memperpanjang daftar ini.

Perhatikan bahwa n = 10dan n = 11adalah angka yang sama, karena $113171923295$ adalah angka terendah yang berisi semua angka $[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]$ , tetapi juga berisi$31$ .

Untuk referensi, Anda dapat menggunakan fakta bahwa skrip Python asli yang saya tulis untuk menghasilkan daftar di atas menghitung 12 istilah pertama dalam waktu sekitar 6 menit.

Aturan tambahan

Setelah hasil pertama masuk, saya menyadari bahwa ada peluang bagus bahwa hasil teratas dapat berakhir dengan skor yang sama. Dalam hal jika seri, pemenang akan menjadi orang dengan waktu tersingkat untuk menghasilkan hasilnya. Jika dua atau lebih jawaban menghasilkan hasil yang sama cepatnya, itu hanya akan menjadi kemenangan yang diikat.

Catatan akhir

5 menit runtime hanya dilakukan untuk memastikan skor yang adil. Saya akan sangat tertarik melihat apakah kita dapat mendorong urutan OEIS lebih lanjut (sekarang ini berisi 17 angka). Dengan kode Ourous ', saya telah menghasilkan semua angka sampai n = 26, tetapi saya berencana membiarkan kode berjalan untuk jangka waktu yang lebih lama.

Papan angka

1. Python 3 + Google OR-Tools : 169
2. Scala : 137 (tidak resmi)
3. Pemecah TSP Concorde : 84 (tidak resmi)
4. Perakitan C ++ (GCC) + x86 : 62
5. Bersih : 25
6. JavaScript (Node.js) : 24

Python 3 + Google OR-Tools , skor 169 dalam 295 detik (skor resmi)

Bagaimana itu bekerja

Setelah membuang bilangan prima berlebihan yang terkandung dalam bilangan prima lainnya, gambarlah grafik berarah dengan tepi dari setiap prime ke setiap sufiksnya, dengan jarak nol, dan edge ke setiap prime dari masing-masing awalannya, dengan jarak yang ditentukan oleh jumlah digit tambahan . Kami mencari jalur terpendek leksikografis pertama melalui grafik mulai dari awalan kosong, melewati setiap prime (tetapi tidak harus melalui setiap awalan atau akhiran), dan berakhir di akhiran kosong.

Misalnya, berikut adalah tepi jalur optimal ε → 11 → 1 → 13 → 3 → 31 → 1 → 17 → ε → 19 → ε → 23 → ε → 29 → ε → 5 → ε untuk n = 11, sesuai ke string keluaran 113171923295.

Dibandingkan dengan pengurangan langsung ke masalah salesman keliling , perhatikan bahwa dengan menghubungkan bilangan prima secara tidak langsung melalui simpul sufiks / awalan ekstra ini, alih-alih secara langsung satu sama lain, kami telah secara dramatis mengurangi jumlah tepian yang perlu kami pertimbangkan. Tetapi karena node tambahan tidak perlu dilalui tepat sekali, ini bukan lagi contoh TSP.

Kami menggunakan pemecah kendala CP-SAT tambahan dari Google OR-Tools, pertama-tama untuk meminimalkan panjang total jalur, kemudian untuk meminimalkan setiap kelompok digit tambahan secara berurutan. Kami menginisialisasi model dengan hanya kendala lokal: setiap prime mendahului satu sufiks dan berhasil satu awalan, sementara setiap sufiks / awalan mendahului dan berhasil dalam jumlah bilangan prima yang sama. Model yang dihasilkan dapat berisi siklus terputus; jika demikian, kami menambahkan batasan konektivitas tambahan secara dinamis dan jalankan kembali solver.

Kode

import multiprocessing
from ortools.sat.python import cp_model


def superstring(strings):
    def gen_prefixes(s):
        for i in range(len(s)):
            a = s[:i]
            if a in affixes:
                yield a

    def gen_suffixes(s):
        for i in range(1, len(s) + 1):
            a = s[i:]
            if a in affixes:
                yield a

    def solve():
        def find_string(s):
            found_strings.add(s)
            for i in range(1, len(s) + 1):
                a = s[i:]
                if (
                    a in affixes
                    and a not in found_affixes
                    and solver.Value(suffix[s, a])
                ):
                    found_affixes.add(a)
                    q.append(a)
                    break

        def cut(skip):
            model.AddBoolOr(
                skip
                + [
                    suffix[s, a]
                    for s in found_strings
                    for a in gen_suffixes(s)
                    if a not in found_affixes
                ]
                + [
                    prefix[a, s]
                    for s in unused_strings
                    if s not in found_strings
                    for a in gen_prefixes(s)
                    if a in found_affixes
                ]
            )
            model.AddBoolOr(
                skip
                + [
                    suffix[s, a]
                    for s in unused_strings
                    if s not in found_strings
                    for a in gen_suffixes(s)
                    if a in found_affixes
                ]
                + [
                    prefix[a, s]
                    for s in found_strings
                    for a in gen_prefixes(s)
                    if a not in found_affixes
                ]
            )

        def search():
            while q:
                a = q.pop()
                for s in prefixed[a]:
                    if (
                        s in unused_strings
                        and s not in found_strings
                        and solver.Value(prefix[a, s])
                    ):
                        find_string(s)
            return not (unused_strings - found_strings)

        while True:
            if solver.Solve(model) != cp_model.OPTIMAL:
                raise RuntimeError("Solve failed")

            found_strings = set()
            found_affixes = set()
            if part is None:
                found_affixes.add("")
                q = [""]
            else:
                part_ix = solver.Value(part)
                p, next_affix, next_string = parts[part_ix]
                q = []
                find_string(next_string)
            if search():
                break

            if part is not None:
                if part_ix not in partb:
                    partb[part_ix] = model.NewBoolVar("partb%s_%s" % (step, part_ix))
                    model.Add(part == part_ix).OnlyEnforceIf(partb[part_ix])
                    model.Add(part != part_ix).OnlyEnforceIf(partb[part_ix].Not())
                cut([partb[part_ix].Not()])
                if last_string is None:
                    found_affixes.add(next_affix)
                else:
                    find_string(last_string)
                q.append(next_affix)
                if search():
                    continue

            cut([])

    solver = cp_model.CpSolver()
    solver.parameters.num_search_workers = 4
    affixes = {s[:i] for s in strings for i in range(len(s))} & {
        s[i:] for s in strings for i in range(1, len(s) + 1)
    }
    prefixed = {}
    for s in strings:
        for a in gen_prefixes(s):
            prefixed.setdefault(a, []).append(s)
    suffixed = {}
    for s in strings:
        for a in gen_suffixes(s):
            suffixed.setdefault(a, []).append(s)
    unused_strings = set(strings)
    last_string = None
    part = None

    model = cp_model.CpModel()
    prefix = {
        (a, s): model.NewBoolVar("prefix_%s_%s" % (a, s))
        for a in affixes
        for s in prefixed[a]
    }
    suffix = {
        (s, a): model.NewBoolVar("suffix_%s_%s" % (s, a))
        for a in affixes
        for s in suffixed[a]
    }
    for s in strings:
        model.Add(sum(prefix[a, s] for a in gen_prefixes(s)) == 1)
        model.Add(sum(suffix[s, a] for a in gen_suffixes(s)) == 1)
    for a in affixes:
        model.Add(
            sum(suffix[s, a] for s in suffixed[a])
            == sum(prefix[a, s] for s in prefixed[a])
        )

    length = sum(prefix[a, s] * (len(s) - len(a)) for a in affixes for s in prefixed[a])
    model.Minimize(length)
    solve()
    model.Add(length == solver.Value(length))

    out = ""
    for step in range(len(strings)):
        in_parts = set()
        parts = []
        for a in [""] if last_string is None else gen_suffixes(last_string):
            for s in prefixed[a]:
                if s in unused_strings and s not in in_parts:
                    in_parts.add(s)
                    parts.append((s[len(a) :], a, s))
        parts.sort()
        part = model.NewIntVar(0, len(parts) - 1, "part%s" % step)
        partb = {}
        for part_ix, (p, a, s) in enumerate(parts):
            if last_string is not None:
                model.Add(part != part_ix).OnlyEnforceIf(suffix[last_string, a].Not())
            model.Add(part != part_ix).OnlyEnforceIf(prefix[a, s].Not())
        model.Minimize(part)
        solve()
        part_ix = solver.Value(part)
        model.Add(part == part_ix)
        p, a, last_string = parts[part_ix]
        unused_strings.remove(last_string)
        out += p
    return out


def gen_primes():
    yield 2
    n = 3
    d = {}
    for p in gen_primes():
        p2 = p * p
        d[p2] = 2 * p
        while n <= p2:
            if n in d:
                q = d.pop(n)
                m = n + q
                while m in d:
                    m += q
                d[m] = q
            else:
                yield n
            n += 2


def gen_inputs():
    num_primes = 0
    strings = []

    for new_prime in gen_primes():
        num_primes += 1
        new_string = str(new_prime)
        strings = [s for s in strings if s not in new_string] + [new_string]
        yield strings


with multiprocessing.Pool() as pool:
    for i, out in enumerate(pool.imap(superstring, gen_inputs())):
        print(i + 1, out, flush=True)

Hasil

Berikut adalah 1000 nomor penahanan prima pertama , dihitung dalam 1½ hari pada sistem 8-core / 16-thread.

C ++ (GCC) + rakitan x86, skor _{32 36} 62 dalam 259 detik (resmi)

Hasil dihitung sejauh ini. Komputer saya kehabisan memori setelah 65.

1 2
2 23
3 235
4 2357
5 112357
6 113257
7 1131725
8 113171925
9 1131719235
10 113171923295
11 113171923295
12 1131719237295
13 11317237294195
14 1131723294194375
15 113172329419437475
16 1131723294194347537
17 113172329419434753759
18 2311329417434753759619
19 231132941743475375961967
20 2311294134347175375961967
21 23112941343471735375961967
22 231129413434717353759619679
23 23112941343471735359619678379
24 2311294134347173535961967837989
25 23112941343471735359619678378979
26 2310112941343471735359619678378979
27 231010329411343471735359619678378979
28 101031071132329417343475359619678378979
29 101031071091132329417343475359619678378979
30 101031071091132329417343475359619678378979
31 101031071091131272329417343475359619678378979
32 101031071091131272329417343475359619678378979
33 10103107109113127137232941734347535961967838979
34 10103107109113127137139232941734347535961967838979
35 10103107109113127137139149232941734347535961967838979
36 1010310710911312713713914923294151734347535961967838979
37 1010310710911312713713914915157232941734347535961967838979
38 1010310710911312713713914915157163232941734347535961967838979
39 10103107109113127137139149151571631672329417343475359619798389
40 10103107109113127137139149151571631672329417343475359619798389
41 1010310710911312713713914915157163167173232941794347535961978389
42 101031071091131271371391491515716316717323294179434753596181978389
43 101031071091131271371391491515716316723294173434753596181917978389
44 101031071091131271371391491515716316717323294179434753596181919383897
45 10103107109113127137139149151571631671731792329418191934347535961978389
46 10103107109113127137139149151571631671731791819193232941974347535961998389
47 101031071091271313714915157163167173179181919321139232941974347535961998389
48 1010310710912713137149151571631671731791819193211392232941974347535961998389
49 1010310710912713137149151571631671731791819193211392232272941974347535961998389
50 10103107109127131371491515716316717317918191932113922322722941974347535961998389
51 101031071091271313714915157163167173179181919321139223322722941974347535961998389
52 101031071091271313714915157163167173179181919321139223322722923941974347535961998389
53 1010310710912713137149151571631671731791819193211392233227229239241974347535961998389
54 101031071091271313714915157163167173179211392233227229239241819193251974347535961998389
55 101031071091271313714915157163167173179211392233227229239241819193251972574347535961998389
56 101031071091271313714915157163167173179211392233227229239241819193251972572634347535961998389
57 101031071091271313714915157163167173179211392233227229239241819193251972572632694347535961998389
58 101031071091271313714915157163167173179211392233227229239241819193251972572632694347535961998389
59 1010310710912713137149151571631671731792113922332277229239241819193251972572632694347535961998389
60 101031071091271313714915157163167173211392233227722923924179251819193257263269281974347535961998389
61 1010310710912713137149151571631671732113922332277229239241792518191932572632692819728343475359619989
62 10103107109127131371491515716316717321139223322772293239241792518191932572632692819728343475359619989
63 1010307107109127131371491515716316717321139223322772293239241792518191932572632692819728343475359619989
64 10103071071091271311371391491515716316721173223322772293239241792518191932572632692819728343475359619989
65 10103071071091271311371491515716313916721173223322772293239241792518191932572632692819728343475359619989

Semua ini setuju dengan hasil dari pemecah berbasis Concorde , sehingga mereka memiliki peluang bagus untuk menjadi benar.

Changelog:

• Perhitungan yang salah untuk panjang konteks yang diperlukan. Versi sebelumnya adalah 1 terlalu besar, dan juga memiliki bug. Nilai: 32 34

• Menambahkan pengoptimalan dengan konteks yang sama. Nilai: 34 36

• Merombak algoritma untuk menggunakan string bebas konteks dengan benar, ditambah beberapa optimasi lainnya. Nilai: 36 62

• Menambahkan penulisan yang tepat.

• Menambahkan varian bilangan prima.

Bagaimana itu bekerja

Peringatan: ini adalah dump otak. Gulir ke akhir jika Anda hanya menginginkan kodenya.

Singkatan:

• PCN : masalah nomor penahanan utama yaitu tantangan ini
• SCS : superstring umum terpendek
• TSP : masalah salesman keliling

Program ini pada dasarnya menggunakan algoritma pemrograman dinamis buku teks untuk TSP.

1. Ditambah pengurangan dari PCN / SCS, masalah yang sebenarnya kami selesaikan, menjadi TSP.
2. Plus menggunakan konteks item, bukan semua digit di setiap item.
3. Plus membagi masalah berdasarkan bilangan prima yang tidak dapat tumpang tindih dengan ujung bilangan prima lainnya.
4. Ditambah penggabungan perhitungan untuk bilangan prima dengan angka awal / akhir yang sama.
5. Ditambah tabel pencarian precomputed dan tabel hash khusus.
6. Ditambah beberapa prefetching dan bit-packing tingkat rendah.

Itu banyak bug potensial. Setelah bermain-main dengan entri anselmus dan gagal membujuk salah hasil dari itu, saya setidaknya harus membuktikan bahwa pendekatan saya secara keseluruhan benar.

Meskipun solusi berbasis Concorde (jauh, jauh) lebih cepat, itu didasarkan pada pengurangan yang sama, jadi penjelasan ini berlaku untuk keduanya. Selain itu, solusi ini dapat diadaptasi untuk OEIS A054260 , urutan bilangan prima yang mengandung prime; Saya tidak tahu bagaimana menyelesaikannya secara efisien dalam kerangka TSP. Jadi masih agak relevan.

Pengurangan TSP

Mari kita mulai dengan benar-benar membuktikan bahwa mengurangi menjadi TSP benar. Kami memiliki serangkaian string, katakanlah

A = 13, 31, 37, 113, 137, 211

dan kami ingin menemukan superstring terkecil yang berisi barang-barang ini.

Mengetahui panjangnya sudah cukup

Untuk PCN, jika ada beberapa string terpendek, kita harus mengembalikan yang terkecil secara leksikografis. Tetapi kita akan melihat masalah yang berbeda (dan lebih mudah).

• SCS : Diberikan awalan awal dan satu set item, temukan string terpendek yang berisi semua item sebagai substring, dan mulai dengan awalan itu.
• SCS-Length : Cukup temukan panjang SCS.

Jika kita dapat menyelesaikan Panjang SCS, kita dapat merekonstruksi solusi terkecil dan mendapatkan PCN. Jika kami tahu bahwa solusi terkecil dimulai dengan awalan kami, kami mencoba memperluasnya dengan menambahkan setiap item, dalam urutan leksikografis, dan menyelesaikannya untuk panjangnya lagi. Ketika kami menemukan item terkecil yang panjang solusinya sama, kami tahu bahwa ini harus menjadi item berikutnya dalam solusi terkecil (mengapa?), Jadi tambahkan dan kembalikan pada item yang tersisa. Metode untuk mencapai solusi ini disebut pengurangan diri .

Turing grafik tumpang tindih maksimal

Misalkan kita mulai memecahkan SCS untuk contoh di atas dengan tangan. Kami mungkin akan:

• Singkirkan 13dan 37, karena mereka sudah substring dari barang-barang lainnya. Setiap solusi yang mengandung 137, misalnya, juga harus mengandung 13dan 37.
• Mulai mengingat kombinasi 113,137 → 1137, 211,113 → 2113dll

Ini sebenarnya adalah hal yang benar untuk dilakukan, tetapi mari kita buktikan demi kelengkapan. Ambil solusi SCS; misalnya, superstring terpendek Aadalah

2113137

dan itu dapat didekomposisi menjadi gabungan dari semua item di A:

211
 113
   31
    137

(Kami mengabaikan item yang berlebihan 13, 37.) Perhatikan bahwa:

1. Posisi awal dan akhir setiap item meningkat setidaknya 1.
2. Setiap item tumpang tindih dengan item sebelumnya sejauh mungkin.

Kami akan menunjukkan bahwa setiap superstring terpendek dapat diurai seperti ini:

1. Untuk setiap pasangan item yang berdekatan x,y, ymulai dan berakhir di posisi lebih lambat dari x. Jika ini tidak benar, maka itu xadalah substring yatau sebaliknya. Tapi kami sudah menghapus semua item yang merupakan substring, sehingga tidak bisa terjadi.

2. Misalkan item yang berdekatan dalam urutan memiliki tumpang tindih kurang dari maksimal, misalnya 21113bukannya 2113. Tapi itu akan membuat ekstra 1berlebihan. Tidak ada item yang lebih baru membutuhkan inisial 1(seperti dalam 2 1 113), karena itu terjadi lebih awal dari 113, dan semua item yang muncul setelah 113tidak dapat mulai dengan digit sebelumnya 113(lihat poin 1). Argumen serupa mencegah ekstra terakhir 1(seperti dalam 211 1 3) yang digunakan oleh item apa pun sebelumnya 211. Tapi superstring terpendek kami , menurut definisi, tidak akan memiliki angka yang berlebihan, sehingga tumpang tindih non-maksimal tidak akan terjadi.

Dengan properti ini, kami dapat mengonversi masalah SCS menjadi TSP:

1. Hapus semua item yang merupakan substring dari item lain.
2. Buat grafik terarah yang memiliki satu simpul untuk setiap item.
3. Untuk setiap pasangan item x, y, menambahkan tepi dari xke yyang berat badannya jumlah simbol tambahan ditambahkan dengan menambahkan yuntuk xdengan tumpang tindih maksimal. Sebagai contoh, kami akan menambahkan tepi dari 211dengan 113dengan bobot 1, karena 2113menambahkan satu digit lagi 211. Ulangi untuk ujung dari yke x.
4. Tambahkan simpul untuk awalan awal, dan tepi dari itu ke semua item lainnya.

Jalur apa pun pada grafik ini, dari awalan awal, sesuai dengan gabungan maksimal-tumpang tindih semua item di jalur itu, dan bobot total jalur sama dengan panjang string gabungan. Karena itu, setiap tur berbobot terendah, yang mengunjungi semua item setidaknya satu kali, sesuai dengan superstring terpendek.

Dan itulah reduksi dari SCS (dan SCS-Length) ke TSP.

Algoritma pemrograman dinamis

Ini adalah algoritma klasik, tetapi kami akan memodifikasinya sedikit, jadi inilah pengingat cepat.

(Saya telah menulis ini sebagai algoritme untuk SCS-Length alih-alih TSP. Mereka pada dasarnya setara, tetapi kosakata SCS membantu ketika kita sampai pada optimasi spesifik SCS.)

Panggil set item input Adan awalan yang diberikan P. Untuk setiap- kelemen subset Sdi A, dan setiap elemen edari S, kami menghitung panjang string terpendek yang dimulai dengan P, berisi semua S, dan diakhiri dengan e. Ini melibatkan menyimpan tabel dari nilai (S, e)ke SCS-Lengths mereka.

Ketika kita sampai ke setiap subset S, tabel harus sudah berisi hasil S - {e}untuk semua edalam S. Karena tabel bisa menjadi cukup besar, saya menghitung hasilnya untuk semua- ksubset elemen, lalu k+1, dll. Untuk ini, kita hanya perlu menyimpan hasilnya untuk kdan k+1pada satu waktu. Ini mengurangi penggunaan memori dengan faktor kasar sqrt(|A|).

Satu detail lagi: alih-alih menghitung panjang minimum SCS, saya benar-benar menghitung total tumpang tindih maksimum antara item. (Untuk mendapatkan Panjang SCS, cukup kurangi total tumpang tindih dari jumlah panjang item.) Menggunakan tumpang tindih membantu beberapa optimasi berikut.

[2.] Konteks barang

Sebuah konteks adalah akhiran terpanjang item yang dapat tumpang tindih dengan item berikut. Jika barang kami adalah 113,211,311, maka 11konteks untuk 211dan 311. (Ini juga merupakan konteks awalan untuk 113, yang akan kita lihat di bagian [4.])

Dalam algoritme DP di atas, kami melacak solusi SCS yang diakhiri dengan setiap item, tetapi kami tidak benar-benar peduli item mana yang diakhiri SCS. Yang perlu kita ketahui hanyalah konteksnya. Jadi, misalnya, jika dua SCS untuk set yang sama berakhir pada 23dan 43, setiap SCS yang melanjutkan dari satu juga akan bekerja untuk yang lain.

Ini adalah optimasi yang signifikan, karena bilangan prima non-sepele hanya berakhir pada digit 1 3 7 9. Keempat konteks satu digit 1,3,7,9(ditambah konteks kosong) sebenarnya cukup untuk menghitung PCNs untuk bilangan prima hingga131 .

[3.] Item bebas konteks

Yang lain telah menunjukkan bahwa banyak bilangan prima dimulai dengan angka 2,4,5,6,8, seperti 23,29,41,43.... Tak satu pun dari ini dapat tumpang tindih dengan bilangan prima sebelumnya (selain dari 2dan 5, bilangan prima tidak dapat berakhir dengan angka-angka ini; 2dan 5sudah akan dihapus sebagai berlebihan). Dalam kode, ini disebut sebagai string bebas konteks .

Jika input kami memiliki item bebas konteks, setiap solusi SCS dapat dipecah menjadi blok

<prefix>... 23... 29... 41... 43...

dan tumpang tindih di setiap blok tidak tergantung pada blok lainnya. Kami dapat mengocok blok atau menukar item di antara blok yang memiliki konteks yang sama, tanpa mengubah panjang SCS.

Jadi, kita hanya perlu melacak kemungkinan multiset konteks, satu untuk setiap blok.

Contoh lengkap: untuk bilangan prima kurang dari 100, kami memiliki 11 item bebas konteks dan konteksnya:

23 29 41 43 47 53 59 61 67 83 89
 3  9  1  3  7  3  9  1  7  3  9

Konteks multiset awal kami:

1 1 3 3 3 3 7 7 9 9 9

Kode mengacu pada ini sebagai konteks gabungan , atau ccontext . Kemudian, kita hanya perlu mempertimbangkan himpunan bagian dari item yang tersisa:

11 13 17 19 31 37 71 73 79 97

[4.] Penggabungan konteks

Begitu kita mencapai bilangan prima dengan 3 digit atau lebih, ada lebih banyak redundansi:

 101 151 181 191 ...
 107 127 157 167 197 ...
 109 149 1009 ...

Kelompok-kelompok ini memiliki konteks awal dan akhir yang sama (biasanya — itu tergantung pada bilangan prima lainnya yang ada di input), sehingga mereka tidak bisa dibedakan ketika tumpang tindih item lainnya. Kami hanya peduli tumpang tindih, sehingga kami dapat memperlakukan bilangan prima dalam kelompok konteks yang sama ini sebagai tidak bisa dibedakan. Sekarang himpunan bagian DP kami diringkas menjadi multisubset

4 × 1_1
5 × 1_7
3 × 1_9

(Ini juga mengapa pemecah memaksimalkan panjang tumpang tindih alih-alih meminimalkan panjang SCS: optimasi ini mempertahankan panjang tumpang tindih.)

Ringkasan: optimasi tingkat tinggi

Berjalan dengan INFOoutput debug akan mencetak statistik suka

solve: N=43, N_search=26, ccontext_size=18, #contexts=7, #eq_context_groups=16

Baris khusus ini adalah untuk SCS-Length dari 62 bilangan prima pertama, 2untuk 293.

• Setelah menghapus item yang mubazir, kita memiliki 43 bilangan prima yang tidak saling substring.
• Ada 7 konteks unik :1,3,7,11,13,27 ditambah string kosong.
• 17 dari 43 bilangan prima yang bebas konteks : 43,47,53,59,61,89,211,223,227,229,241,251,257,263,269,281,283. Ini dan awalan yang diberikan (dalam hal ini, string kosong) membentuk dasar dari konteks gabungan awal .
• Dalam 26 item yang tersisa ( N_search), ada 16 grup dengan konteks sama nontrivial .

Dengan mengeksploitasi struktur ini, perhitungan SCS-Length hanya perlu memeriksa 8498336 (multiset, ccontext)kombinasi. Pemrograman dinamis langsung akan mengambil 43×2^43 > 3×10^14langkah - langkah, dan dengan kasar memaksa permutasi akan mengambil 6×10^52langkah - langkah. Program masih perlu menjalankan SCS-Length beberapa kali lagi untuk merekonstruksi solusi PCN, tetapi itu tidak memakan waktu lebih lama.

[5., 6.] Optimalisasi tingkat rendah

Alih-alih melakukan operasi string, pemecah Panjang SCS bekerja dengan indeks item dan konteks. Saya juga menghitung jumlah tumpang tindih antara setiap konteks dan pasangan item.

Kode awalnya menggunakan GCC unordered_map, yang tampaknya merupakan tabel hash dengan bucket daftar tertaut dan ukuran hash utama (yaitu divisi mahal). Jadi saya menulis tabel hash saya sendiri dengan probing linier dan kekuatan dua ukuran. Ini menghasilkan 3x speedup dan 3x pengurangan memori.

Setiap status tabel terdiri dari beberapa item, konteks gabungan, dan jumlah yang tumpang tindih. Ini dikemas ke dalam entri 128-bit: 8 untuk jumlah tumpang tindih, 56 untuk multiset (sebagai bitet dengan enkode run-length), dan 64 untuk ccontext (RLE 1-dibatasi). Pengkodean dan penguraian ulang ccontext adalah bagian tersulit dan saya akhirnya menggunakan yang baruPDEP instruksi (ini sangat baru, GCC belum memiliki intrinsik untuk itu).

Akhirnya, mengakses tabel hash benar-benar lambat ketika Nmenjadi besar, karena tabel tersebut tidak muat di cache lagi. Tetapi satu-satunya alasan kami menulis ke tabel hash, adalah untuk memperbarui jumlah tumpang tindih yang paling dikenal untuk setiap negara. Program membagi langkah ini menjadi antrian prefetch, dan loop bagian dalam mengambil setiap tabel pencarian beberapa iterasi sebelum benar-benar memperbarui slot itu. Speedup 2 × lain di komputer saya.

Bonus: peningkatan lebih lanjut

AKA Bagaimana Concorde begitu cepat?

Saya tidak tahu banyak tentang algoritma TSP, jadi di sini adalah tebakan kasar.

Concorde menggunakan metode branch-and-cut untuk menyelesaikan TSP.

• Ini mengkodekan TSP sebagai program linear integer
• Ini menggunakan metode pemrograman linier, serta heuristik awal, untuk mendapatkan batas bawah dan atas pada jarak tur optimal
• Batasan ini kemudian dimasukkan ke dalam cabang dan algoritma rekursif terikat yang mencari solusi optimal. Sebagian besar pohon pencarian dapat dipangkas, jika batas bawah yang dihitung untuk subtree melebihi batas atas yang diketahui
• Ini juga mencari memotong pesawat untuk mempererat relaksasi LP dan mendapatkan batas yang lebih baik. Biasanya, pemotongan ini menyandikan pengetahuan tentang fakta bahwa variabel keputusan harus bilangan bulat

Gagasan yang jelas dapat kami coba:

• Memangkas dalam pemecah Panjang SCS, terutama ketika merekonstruksi solusi PCN (pada saat itu, kita sudah tahu berapa panjang larutan itu)
• Turunkan beberapa batas bawah yang mudah dihitung untuk SCS, yang dapat digunakan untuk membantu pemangkasan
• Menemukan lebih banyak simetri atau redundansi dalam distribusi bilangan prima untuk dieksploitasi

Namun, kombinasi branch-and-cut sangat kuat, jadi kita mungkin tidak bisa mengalahkan pemecah canggih seperti Concorde, untuk nilai yang besar N.

Bonus bonus: bilangan prima penahanan utama

Berbeda dengan solusi berbasis Concorde, program ini dapat dimodifikasi untuk menemukan bilangan prima yang mengandung terkecil ( OEIS A054260 ). Ini melibatkan tiga perubahan:

1. $1/\ln(n)$

2. Ubah kode pemecah Panjang SCS untuk mengategorikan solusi berdasarkan apakah jumlah digitnya dapat dibagi dengan 3. Ini melibatkan penambahan entri lain, digit jumlah mod 3, ke masing-masing negara DP. Ini sangat mengurangi kemungkinan solver utama terjebak dengan permutasi non-prima. Ini adalah perubahan yang saya tidak tahu bagaimana menerjemahkan ke TSP. Itu dapat dikodekan dengan ILP, tetapi kemudian saya harus belajar tentang hal ini yang disebut "ketimpangan subtour" dan bagaimana menghasilkan itu.

3. Bisa jadi semua PCN terpendek dapat dibagi oleh 3. Dalam hal itu, prime prime kontainment terkecil harus setidaknya satu digit lebih panjang dari PCN. Jika pemecah SCS-Length kami mendeteksi ini, kode rekonstruksi solusi memiliki opsi untuk menambahkan satu digit tambahan pada titik mana pun dalam proses. Ia mencoba menambahkan setiap kemungkinan digit 0..9dan setiap item yang tersisa ke awalan solusi saat ini, dalam urutan leksikografis seperti sebelumnya.

Dengan perubahan ini, saya bisa mendapatkan solusinya hingga N=62. Kecuali untuk 47, di mana kode rekonstruksi macet dan menyerah setelah 1 juta langkah (saya belum tahu mengapa, belum). Prima penahanan utama adalah:

1 2
2 23
3 523
4 2357
5 112573
6 511327
7 1135217
8 1113251719
9 11171323519
10 113171952923
11 113171952923
12 11131951723729
13 11317237419529
14 1131723294375419
15 113172329541947437
16 1131723294195343747
17 1113172329419434753759
18 11231329417437475361959
19 231132941743475375967619
20 2311294134347175967619537
21 23112941343471735967619537
22 231129413434717359537679619
23 23112941343471735375961983679
24 11231294134347173535961967983789
25 23112941343471735359679837619789
26 2310112941343471735359619783789679
27 231010329411343471735359619678379897
28 101031071132329417343475359619798376789
29 101031071091132329417343475359619767898379
30 101031071091132329417343475359619767898379
31 1010310710911131272329417343475359619678979837
32 1010310710911131272329417343475359619678979837
33 10103107109113127137232941734347535978961967983
34 10103107109113127137139232941734347535961967838979
35 10103107109113127137139149232941734347535961976798389
36 1010310710911312713713914923294151734347535976198389679
37 1010310710911312713713914915157232941734347535967619798389
38 10103107109111312713713914915157163232941734347535967897961983
39 10103107109113127137139149151571631672329417343475961979838953
40 10103107109113127137139149151571631672329417343475961979838953
41 10103107109111312713713914915157163167173232941794347535976198983
42 1010310710911131271371391491515716316717323294179434761819535989783
43 1010310710911131271371391491515716316723294173434753596181917989783
44 101031071091131271371391491515716316717323294179434753836181919389597
45 10103107109113127137139149151571631671731792329418191934347538961975983
46 101031071091113127137139149151571631671731791819193232941974347535989836199
47 (failed)
48 1010310710912713137149151571631671731791819193211392232941974347895359836199
49 10103107109112713137149151571631671731791819193211392232272941974347619983535989
50 10103107109127131371491515716316717317918191932113922322722941974347595389836199
51 101031071091271313714915157163167173179181919321139223322722941974347595389619983
52 101031071091271313714915157163167173179181919321139223322722923941974347538361995989
53 10103107109112713137149151571631671731791819193211392233227229239241974347619983538959
54 101031071091271313714915157163167173179211392233227229239241819193251974347619953835989
55 1010310710911271313714915157163167173179211392233227229239241819193251974325747596199538983
56 101031071091271313714915157163167173179211392233227229239241819193251972572634347619959895383
57 101031071091271313714915157163167173179211392233227229239241819193251972572632694359538983619947
58 101031071091271313714915157163167173179211392233227229239241819193251972572632694359538983619947
59 1010310710912713137149151571631671731792113922332277229239241819193251972572632694347535983896199
60 1010310710911271313714915157163167173211392233227722923924179251819193257263269281974347535961998389
61 1010310710912713137149151571631671732113922332277229239241792518191932572632692819728343538947619959
62 10103107109127131371491515716316717321139223322772293239241792518191932572632692819728343534759896199

Kode

Kompilasi dengan

g++ -std=c++14 -O3 -march=native pcn.cpp -o pcn

Untuk versi bilangan prima, tautkan juga dengan GMPlib, mis

g++ -std=c++14 -O3 -march=native pcn-prime.cpp -o pcn-prime -lgmp -lgmpxx

Program ini menggunakan instruksi PDEP, yang hanya tersedia pada prosesor x86 terbaru (Haswell +). Baik komputer saya dan maxb mendukungnya. Jika milik Anda tidak, program akan dikompilasi dalam versi perangkat lunak yang lambat. Peringatan kompilasi akan dicetak saat ini terjadi.

#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <unordered_map>
#include <string>
#include <algorithm>
#include <array>

using namespace std;

void debug_dummy(...) {
}

#ifndef INFO
//#  define INFO(...) fprintf(stderr, __VA_ARGS__)
#  define INFO debug_dummy
#endif

#ifndef DEBUG
//#    define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#  define DEBUG debug_dummy
#endif

bool is_prime(size_t n)
{
    for (size_t d = 2; d * d <= n; ++d) {
        if (n % d == 0) {
            return false;
        }
    }
    return true;
}

// bitset, works for up to 64 strings
using bitset_t = uint64_t;
const size_t bitset_bits = 64;

// Find position of n-th set bit of x
uint64_t bit_select(uint64_t x, size_t n) {
#ifdef __BMI2__
    // Bug: GCC doesn't seem to provide the _pdep_u64 intrinsic,
    // despite what its manual claims. Neither does Clang!
    //size_t r = _pdep_u64(ccontext_t(1) << new_context, ccontext1);
    size_t r;
    // NB: actual operand order is %2, %1 despite the intrinsic taking %1, %2
    asm ("pdep %2, %1, %0"
         : "=r" (r)
         : "r" (uint64_t(1) << n), "r" (x)
         );
    return __builtin_ctzll(r);
#else
#  warning "bit_select: no x86 BMI2 instruction set, falling back to slow code"
    size_t k = 0, m = 0;
    for (; m < 64; ++m) {
        if (x & (uint64_t(1) << m)) {
            if (k == n) {
                break;
            }
            ++k;
        }
    }
    return m;
#endif
}

#ifndef likely
#  define likely(x) __builtin_expect(x, 1)
#endif
#ifndef unlikely
#  define unlikely(x) __builtin_expect(x, 0)
#endif

// Return the shortest string that begins with a and ends with b
string join_strings(string a, string b) {
    for (size_t overlap = min(a.size(), b.size()); overlap > 0; --overlap) {
        if (a.substr(a.size() - overlap) == b.substr(0, overlap)) {
            return a + b.substr(overlap);
        }
    }
    return a + b;
}

vector <string> dedup_items(string context0, vector <string> items)
{
    vector <string> items2;
    for (size_t i = 0; i < items.size(); ++i) {
        bool dup = false;
        if (context0.find(items[i]) != string::npos) {
                dup = true;
        } else {
            for (size_t j = 0; j < items.size(); ++j) {
                if (items[i] == items[j]?
                    i > j
                        : items[j].find(items[i]) != string::npos) {
                    dup = true;
                    break;
                }
            }
        }
        if (!dup) {
            items2.push_back(items[i]);
        }
    }
    return items2;
}

// Table entry used in main solver
const size_t solver_max_item_set = bitset_bits - 8;
struct Solver_entry
{
    uint8_t score : 8;
    bitset_t items : solver_max_item_set;
    bitset_t context;

    Solver_entry()
    {
        score = 0xff;
        items = 0;
        context = 0;
    }
    bool is_empty() const {
        return score == 0xff;
    }
};

// Simple hash table to avoid stdlib overhead
struct Solver_table
{
    vector <Solver_entry> t;
    size_t t_bits;
    size_t size_;
    size_t num_probes_;

    Solver_table()
    {
        // 256 slots initially -- this needs to be not too small
        // so that the load factor formula in update_score works
        t_bits = 8;
        size_ = 0;
        num_probes_ = 0;
        resize(t_bits);
    }
    static size_t entry_hash(bitset_t items, bitset_t context)
    {
        uint64_t h = 0x3141592627182818ULL;
        // Add context first, since its bits are generally
        // less well distributed than items
        h += context;
        h ^= h >> 23;
        h *= 0x2127599bf4325c37ULL;
        h ^= h >> 47;
        h += items;
        h ^= h >> 23;
        h *= 0x2127599bf4325c37ULL;
        h ^= h >> 47;
        return h;
    }
    size_t probe_index(size_t hash) const {
        return hash & ((size_t(1) << t_bits) - 1);
    }
    void resize(size_t t2_bits)
    {
        assert (size_ < size_t(1) << t2_bits);
        vector <Solver_entry> t2(size_t(1) << t2_bits);
        for (auto entry: t) {
            if (!entry.is_empty()) {
                size_t h = entry_hash(entry.items, entry.context);
                size_t mask = (size_t(1) << t2_bits) - 1;
                size_t idx = h & mask;
                while (!t2[idx].is_empty()) {
                    idx = (idx + 1) & mask;
                    ++num_probes_;
                }
                t2[idx] = entry;
            }
        }
        t.swap(t2);
        t_bits = t2_bits;
    }
    uint8_t update_score(bitset_t items, bitset_t context, uint8_t score)
    {
        // Ensure we can insert a new item without resizing
        assert (size_ < t.size());

        size_t index = probe_index(entry_hash(items, context));
        size_t mask = (size_t(1) << t_bits) - 1;
        for (size_t p = 0; p < t.size(); ++p, index = (index + 1) & mask) {
            ++num_probes_;
            if (likely(t[index].items == items && t[index].context == context)) {
                t[index].score = max(t[index].score, score);
                return t[index].score;
            }
            if (t[index].is_empty()) {
                // add entry
                t[index].score = score;
                t[index].items = items;
                t[index].context = context;
                ++size_;
                // load factor 4/5 -- ideally 2-3 average probes per lookup
                if (5*size_ > 4*t.size()) {
                    resize(t_bits + 1);
                }
                return score;
            }
        }
        assert (false && "bug: hash table probe loop");
    }
    size_t size() const {
        return size_;
    }
    void swap(Solver_table table)
    {
        t.swap(table.t);
        ::swap(size_, table.size_);
        ::swap(t_bits, table.t_bits);
        ::swap(num_probes_, table.num_probes_);
    }
};

/*
 * Main solver code.
 */
struct Solver
{
    // Inputs
    vector <string> items;
    string context0;
    size_t context0_index;

    // Mapping between strings and indices
    vector <string> context_to_string;
    unordered_map <string, size_t> string_to_context;

    // Items that have context-free prefixes, i.e. prefixes that
    // never overlap with the end of other items nor context0
    vector <bool> contextfree;

    // Precomputed contexts (suffixes) for each item
    vector <size_t> item_context;
    // Precomputed updates: (context, string) to overlap amount
    vector <vector <size_t>> join_overlap;

    Solver(vector <string> items, string context0)
        :items(items), context0(context0)
    {
        items = dedup_items(context0, items);
        init_context_();
    }

    void init_context_()
    {
        /*
         * Generate all relevant item-item contexts.
         *
         * At this point, we know that no item is a substring of
         * another, nor of context0. This means that the only contexts
         * we need to care about, are those generated from maximal join
         * overlaps between any two items.
         *
         * Proof:
         * Suppose that the shortest containing string needs some other
         * kind of context. Maybe it depends on a context spanning
         * three or more items, say X,Y,Z. But if Z ends after Y and
         * interacts with X, then Y must be a substring of Z.
         * This cannot happen, because we removed all substrings.
         *
         * Alternatively, it depends on a non-maximal join overlap
         * between two strings, say X,Y. But if this overlap does not
         * interact with any other string, then we could maximise it
         * and get a shorter solution. If it does, then call this
         * other string Z. We would get the same contradiction as in
         * the previous case with X,Y,Z.
         */
        size_t N = items.size();
        vector <size_t> max_prefix_overlap(N), max_suffix_overlap(N);
        size_t context0_suffix_overlap = 0;
        for (size_t i = 0; i < N; ++i) {
            for (size_t j = 0; j < N; ++j) {
                if (i == j) continue;
                string joined = join_strings(items[j], items[i]);
                size_t overlap = items[j].size() + items[i].size() - joined.size();
                string context = items[i].substr(0, overlap);
                max_prefix_overlap[i] = max(max_prefix_overlap[i], overlap);
                max_suffix_overlap[j] = max(max_suffix_overlap[j], overlap);

                if (string_to_context.find(context) == string_to_context.end()) {
                    string_to_context[context] = context_to_string.size();
                    context_to_string.push_back(context);
                }
            }

            // Context for initial join with context0
            {
                string joined = join_strings(context0, items[i]);
                size_t overlap = context0.size() + items[i].size() - joined.size();
                string context = items[i].substr(0, overlap);
                max_prefix_overlap[i] = max(max_prefix_overlap[i], overlap);
                context0_suffix_overlap = max(context0_suffix_overlap, overlap);

                if (string_to_context.find(context) == string_to_context.end()) {
                    string_to_context[context] = context_to_string.size();
                    context_to_string.push_back(context);
                }
            }
        }
        // Now compute all canonical trailing contexts
        context0_index = string_to_context[
                           context0.substr(context0.size() - context0_suffix_overlap)];
        item_context.resize(N);
        for (size_t i = 0; i < N; ++i) {
            item_context[i] = string_to_context[
                                items[i].substr(items[i].size() - max_suffix_overlap[i])];
        }

        // Now detect context-free items
        contextfree.resize(N);
        for (size_t i = 0; i < N; ++i) {
            contextfree[i] = (max_prefix_overlap[i] == 0);
            if (contextfree[i]) {
                DEBUG("  contextfree: %s\n", items[i].c_str());
            }
        }

        // Now compute all possible overlap amounts
        join_overlap.resize(context_to_string.size(), vector <size_t> (N));
        for (size_t c_index = 0; c_index < context_to_string.size(); ++c_index) {
            const string& context = context_to_string[c_index];
            for (size_t i = 0; i < N; ++i) {
                string joined = join_strings(context, items[i]);
                size_t overlap = context.size() + items[i].size() - joined.size();
                join_overlap[c_index][i] = overlap;
            }
        }
    }

    // Main solver.
    // Returns length of shortest string containing all items starting
    // from context0 (context0's length not included).
    size_t solve() const
    {
        size_t N = items.size();

        // Length, if joined without overlaps. We try to improve this by
        // finding overlaps in the main iteration
        size_t base_length = 0;
        for (auto s: items) {
            base_length += s.size();
        }

        // Now take non-context-free items. We will only need to search
        // over these items.
        vector <size_t> search_items;
        for (size_t i = 0; i < N; ++i) {
            if (!contextfree[i]) {
                search_items.push_back(i);
            }
        }
        size_t N_search = search_items.size();

        /*
         * Some groups of strings have the same context transitions.
         * For example "17", "107", "127", "167" all have an initial
         * context of "1" and a trailing context of "7", no other
         * overlaps are possible with other primes.
         *
         * We group these strings and treat them as indistinguishable
         * during the main algorithm.
         */
        auto eq_context = [&](size_t i, size_t j) {
            if (item_context[i] != item_context[j]) {
                return false;
            }
            for (size_t ci = 0; ci < context_to_string.size(); ++ci) {
                if (join_overlap[ci][i] != join_overlap[ci][j]) {
                    return false;
                }
            }
            return true;
        };
        vector <size_t> eq_context_group(N_search, size_t(-1));
        for (size_t si = 0; si < N_search; ++si) {
            for (size_t sj = si-1; sj+1 > 0; --sj) {
                size_t i = search_items[si], j = search_items[sj];
                if (!contextfree[j] && eq_context(i, j)) {
                    DEBUG("  eq context: %s =c= %s\n", items[i].c_str(), items[j].c_str());
                    eq_context_group[si] = sj;
                    break;
                }
            }
        }

        // Figure out the combined context size. A combined context has
        // one entry for each context-free item plus one for context0.
        size_t ccontext_size = N - N_search + 1;

        // Assert that various parameters all fit into our data types
        using ccontext_t = bitset_t;
        assert (context_to_string.size() + ccontext_size <= bitset_bits);
        assert (N_search <= solver_max_item_set);
        assert (base_length < 0xff);

        // Initial combined context.
        unordered_map <size_t, size_t> cc0_full;
        ++cc0_full[context0_index];
        for (size_t i = 0; i < N; ++i) {
            if (contextfree[i]) {
                ++cc0_full[item_context[i]];
            }
        }
        // Now pack into unary-encoded bitset. The bitset stores the
        // count for each context as <count> number of 0 bits,
        // followed by a 1 bit.
        ccontext_t cc0 = 0;
        for (size_t ci = 0, b = 0; ci < context_to_string.size(); ++ci, ++b) {
            b += cc0_full[ci];
            cc0 |= ccontext_t(1) << b;
        }

        // Map from (item set, context) to maximum achievable overlap
        Solver_table k_solns;
        // Base case: cc0 with empty set
        k_solns.update_score(0, cc0, 0);

        // Now start dynamic programming. k is current subset size
        size_t eq_context_groups = 0;
        for (size_t g: eq_context_group) eq_context_groups += (g != size_t(-1));
        if (context0.empty()) {
            INFO("solve: N=%zu, N_search=%zu, ccontext_size=%zu, #contexts=%zu, #eq_context_groups=%zu\n",
                 N, N_search, ccontext_size, context_to_string.size(), eq_context_groups);
        } else {
            DEBUG("solve: context=%s, N=%zu, N_search=%zu, ccontext_size=%zu, #contexts=%zu, #eq_context_groups=%zu\n",
                  context0.c_str(), N, N_search, ccontext_size, context_to_string.size(), eq_context_groups);
        }
        for (size_t k = 0; k < N_search; ++k) {
            decltype(k_solns) k1_solns;

            // The main bottleneck of this program is updating k1_solns,
            // which (for larger N) becomes a huge table.
            // We use a prefetch queue to reduce memory latency.
            const size_t prefetch = 8;
            array <Solver_entry, prefetch> entry_queue;
            size_t update_i = 0;

            // Iterate every k-subset
            for (Solver_entry entry: k_solns.t) {
                if (entry.is_empty()) continue;

                bitset_t s = entry.items;
                ccontext_t ccontext = entry.context;
                size_t overlap = entry.score;

                // Try adding a new item
                for (size_t si = 0; si < N_search; ++si) {
                    bitset_t s1 = s | bitset_t(1) << si;
                    if (s == s1) {
                        continue;
                    }
                    // Add items in each eq_context_group sequentially
                    if (eq_context_group[si] != size_t(-1) &&
                        !(s & bitset_t(1) << eq_context_group[si])) {
                        continue;
                    }
                    size_t i = search_items[si]; // actual item index

                    size_t new_context = item_context[i];
                    // Increment ccontext's count for new_context.
                    // We need to find its delimiter 1 bit
                    size_t bit_n = bit_select(ccontext, new_context);
                    ccontext_t ccontext_n =
                        ((ccontext & ((ccontext_t(1) << bit_n) - 1))
                         | ((ccontext >> bit_n << (bit_n + 1))));

                    // Select non-empty sub-contexts to substitute for new_context
                    for (size_t ci = 0, bit1 = 0, count;
                         ci < context_to_string.size();
                         ++ci, bit1 += count + 1)
                    {
                        assert (ccontext_n >> bit1);
                        count = __builtin_ctzll(ccontext_n >> bit1);
                        if (!count
                            // We just added new_context; we can only remove an existing
                            // context entry there i.e. there must be at least two now
                            || (ci == new_context && count < 2)) {
                            continue;
                        }

                        // Decrement ci in ccontext_n
                        bitset_t ccontext1 =
                            ((ccontext_n & ((ccontext_t(1) << bit1) - 1))
                             | ((ccontext_n >> (bit1 + 1)) << bit1));

                        size_t overlap1 = overlap + join_overlap[ci][i];

                        // do previous prefetched update
                        if (update_i >= prefetch) {
                            Solver_entry entry = entry_queue[update_i % prefetch];
                            k1_solns.update_score(entry.items, entry.context, entry.score);
                        }

                        // queue the current update and prefetch
                        Solver_entry entry1;
                        size_t probe_index = k1_solns.probe_index(Solver_table::entry_hash(s1, ccontext1));
                        __builtin_prefetch(&k1_solns.t[probe_index]);
                        entry1.items = s1;
                        entry1.context = ccontext1;
                        entry1.score = overlap1;
                        entry_queue[update_i % prefetch] = entry1;

                        ++update_i;
                    }
                }
            }

            // do remaining queued updates
            for (size_t j = 0; j < min(update_i, prefetch); ++j) {
                Solver_entry entry = entry_queue[j];
                k1_solns.update_score(entry.items, entry.context, entry.score);
            }

            if (context0.empty()) {
                INFO("  hash stats: |solns[%zu]| = %zu, %zu lookups, %zu probes\n",
                     k+1, k1_solns.size(), update_i, k1_solns.num_probes_);
            } else {
                DEBUG("  hash stats: |solns[%zu]| = %zu, %zu lookups, %zu probes\n",
                      k+1, k1_solns.size(), update_i, k1_solns.num_probes_);
            }
            k_solns.swap(k1_solns);
        }

        // Overall solution
        size_t max_overlap = 0;
        for (Solver_entry entry: k_solns.t) {
            if (entry.is_empty()) continue;
            max_overlap = max(max_overlap, size_t(entry.score));
        }
        return base_length - max_overlap;
    }
};

// Wrapper for Solver that also finds the smallest solution string
string smallest_containing_string(vector <string> items)
{
    items = dedup_items("", items);

    size_t soln_length;
    {
        Solver solver(items, "");
        soln_length = solver.solve();
    }
    DEBUG("Found solution length: %zu\n", soln_length);

    string soln;
    vector <string> remaining_items = items;
    while (remaining_items.size() > 1) {
        // Add all possible next items, in lexicographic order
        vector <pair <string, size_t>> next_solns;
        for (size_t i = 0; i < remaining_items.size(); ++i) {
            const string& item = remaining_items[i];
            next_solns.push_back(make_pair(join_strings(soln, item), i));
        }
        assert (next_solns.size() == remaining_items.size());
        sort(next_solns.begin(), next_solns.end());

        // Now try every item in order
        bool found_next = false;
        for (auto ns: next_solns) {
            size_t i;
            string next_soln;
            tie(next_soln, i) = ns;
            DEBUG("Trying: %s + %s -> %s\n",
                  soln.c_str(), remaining_items[i].c_str(), next_soln.c_str());
            vector <string> next_remaining;
            for (size_t j = 0; j < remaining_items.size(); ++j) {
                if (next_soln.find(remaining_items[j]) == string::npos) {
                    next_remaining.push_back(remaining_items[j]);
                }
            }

            Solver solver(next_remaining, next_soln);
            size_t next_size = solver.solve();
            DEBUG("  ... next_size: %zu + %zu =?= %zu\n", next_size, next_soln.size(), soln_length);
            if (next_size + next_soln.size() == soln_length) {
                INFO("  found next item: %s\n", remaining_items[i].c_str());
                soln = next_soln;
                remaining_items = next_remaining;
                // found lexicographically smallest solution, break now
                found_next = true;
                break;
            }
        }
        assert (found_next);
    }
    soln = join_strings(soln, remaining_items[0]);

    return soln;
}

int main()
{
    string prev_soln;
    vector <string> items;
    size_t p = 1;
    for (size_t N = 1;; ++N) {
        for (++p; items.size() < N; ++p) {
            if (is_prime(p)) {
                char buf[99];
                snprintf(buf, sizeof buf, "%zu", p);
                items.push_back(buf);
                break;
            }
        }

        // Try to reuse previous solution (this works for N=11,30,32...)
        string soln;
        if (prev_soln.find(items.back()) != string::npos) {
            soln = prev_soln;
        } else {
            soln = smallest_containing_string(items);
        }
        printf("%s\n", soln.c_str());
        prev_soln = soln;
    }
}

Cobalah online!

Dan versi utama hanya pada TIO . Maaf, tapi saya tidak golf program ini dan ada batas panjang posting.

JavaScript (Node.js) , skor 24 dalam 241 detik

Hasil

• $a(1)$$a(21)$
• $a(22)=231129413434717353759619679$
• $a(23)=23112941343471735359619678379$
• $a(1)$$a(24)$

Algoritma

Ini adalah pencarian rekursif yang mencoba semua cara yang memungkinkan untuk menggabungkan angka, dan akhirnya mengurutkan daftar yang dihasilkan dalam urutan leksikografis ketika simpul daun tercapai.

$x$$y$$k$$x$$k$$y$$k$$y$$k$$x$

Di awal setiap iterasi, setiap entri yang dapat ditemukan di entri lain dihapus dari daftar.

Percepatan signifikan dicapai dengan melacak node yang dikunjungi, sehingga kami dapat membatalkan lebih awal ketika operasi yang berbeda mengarah ke daftar yang sama.

Sebuah percepatan kecil dicapai dengan memperbarui dan mengembalikan daftar jika memungkinkan daripada menghasilkan salinan, seperti yang disarankan oleh pengguna anonim Neil.

Contoh

$n=7$$[2,3,5,7,11,13,17]$

[]                        // start with an empty list
[ 2 ]                     // append 2
[ 2, 3 ]                  // append 3
[ 2, 3, 5 ]               // append 5
[ 2, 3, 5, 7 ]            // append 7
[ 2, 3, 5, 7, 11 ]        // append 11
[ 2, 3, 5, 7, 11, 13 ]    // append 13
[ 2, 5, 7, 11, 13 ]       // remove 3, which appears in 13
  [ 2, 5, 7, 113, 13 ]    //   try to merge 11 and 13 into 113
  [ 2, 5, 7, 113 ]        //   remove 13, which now appears in 113
  [ 2, 5, 7, 113, 17 ]    //   append 17
  [ 2, 5, 113, 17 ]       //   remove 7, which appears in 17
  --> leaf node: 1131725  //   new best result
[ 2, 5, 7, 11, 13, 17 ]   // append 17
[ 2, 5, 11, 13, 17 ]      // remove 7, which appears in 17
  [ 2, 5, 113, 13, 17 ]   //   try to merge 11 and 13 into 113
  [ 2, 5, 113, 17 ]       //   remove 13, which now appears in 113
                          //   abort because this node was already visited
                          //   (it was a leaf node anyway, so we don't save much here)
  [ 2, 5, 117, 13, 17 ]   //   try to merge 11 and 17 into 117
  [ 2, 5, 117, 13 ]       //   remove 17, which now appears in 117
  --> leaf node: 1171325  //   not better than the previous one
--> leaf node: 11131725   // not better than the previous one

Kode

Cobalah online!

let f = n => {
  let visited = {},
      a, d, k, best, search;

  // build the list of primes, as strings
  for(a = [ '2' ], n--, k = 3; n; k++) {
    for(d = k; k % (d -= 2);) {}
    d == 1 && n-- && a.push(k + '');
  }

  best = a.join('');

  // recursive search function
  (search = (a, n = 0, r = []) => {
    let x, y, i, j, k, s;

    // remove all entries in r[] that can be found in another entry
    r = r.filter((p, i) => !r.some((q, j) => i != j && ~q.indexOf(p)));

    // abort early if this node was already visited
    if(visited[r]) {
      return;
    }

    // otherwise, mark it as visited
    visited[r] = 1;

    // walk through all distinct pairs (x, y) in r[]
    for(i = 0; i < r.length; i++) {
      for(j = i + 1; j < r.length; j++) {
        x = r[i];
        y = r[j];

        // try to merge x and y if:
        // 1) the first k digits of x equal the last k digits of y
        for(k = 1; x.slice(0, k) == y.slice(-k); k++) {
          r[i] = y + x.slice(k);
          search(a, n, r);
        }

        // or:
        // 2) the first k digits of y equal the last k digits of x
        for(k = 1; y.slice(0, k) == x.slice(-k); k++) {
          r[i] = x + y.slice(k);
          search(a, n, r);
        }
        r[i] = x;
      }
    }

    if(x = a[n]) {
      // there are other primes to process, so go on with the next one
      search(a, n + 1, [...r, x]);
    }
    else {
      // this is a leaf node: see if we've improved our current score
      s = r.join('');

      if(s.length <= best.length) {
        s = r.sort().join('');

        if(s.length < best.length || s < best) {
          best = s;
        }
      }
    }
  })(a);

  return best;
}

Pemecah Concorde TSP , skor 84 dalam 299 detik

Yah ... saya merasa konyol karena baru menyadari ini sekarang.

Semua ini pada dasarnya adalah masalah salesman keliling . Untuk setiap pasangan bilangan prima pdan q, tambahkan tepi yang beratnya adalah jumlah digit yang ditambahkan oleh q(menghapus digit yang tumpang tindih). Juga, tambahkan tepi awal ke setiap prime p, yang beratnya adalah panjangnyap . Jalur salesman keliling terpendek cocok dengan panjang nomor penahanan perdana terkecil.

Kemudian pemecah TSP tingkat industri, seperti Concorde , akan menyelesaikan masalah ini.

Entri ini mungkin harus dianggap tidak bersaing.

Hasil

Solver mencapai N=350sekitar 20 jam CPU. Hasil lengkapnya terlalu panjang untuk satu posting SE, dan OEIS tidak menginginkan begitu banyak istilah. Inilah 200 yang pertama:

1 2
2 23
3 235
4 2357
5 112357
6 113257
7 1131725
8 113171925
9 1131719235
10 113171923295
11 113171923295
12 1131719237295
13 11317237294195
14 1131723294194375
15 113172329419437475
16 1131723294194347537
17 113172329419434753759
18 2311329417434753759619
19 231132941743475375961967
20 2311294134347175375961967
21 23112941343471735375961967
22 231129413434717353759619679
23 23112941343471735359619678379
24 2311294134347173535961967837989
25 23112941343471735359619678378979
26 2310112941343471735359619678378979
27 231010329411343471735359619678378979
28 101031071132329417343475359619678378979
29 101031071091132329417343475359619678378979
30 101031071091132329417343475359619678378979
31 101031071091131272329417343475359619678378979
32 101031071091131272329417343475359619678378979
33 10103107109113127137232941734347535961967838979
34 10103107109113127137139232941734347535961967838979
35 10103107109113127137139149232941734347535961967838979
36 1010310710911312713713914923294151734347535961967838979
37 1010310710911312713713914915157232941734347535961967838979
38 1010310710911312713713914915157163232941734347535961967838979
39 10103107109113127137139149151571631672329417343475359619798389
40 10103107109113127137139149151571631672329417343475359619798389
41 1010310710911312713713914915157163167173232941794347535961978389
42 101031071091131271371391491515716316717323294179434753596181978389
43 101031071091131271371391491515716316723294173434753596181917978389
44 101031071091131271371391491515716316717323294179434753596181919383897
45 10103107109113127137139149151571631671731792329418191934347535961978389
46 10103107109113127137139149151571631671731791819193232941974347535961998389
47 101031071091271313714915157163167173179181919321139232941974347535961998389
48 1010310710912713137149151571631671731791819193211392232941974347535961998389
49 1010310710912713137149151571631671731791819193211392232272941974347535961998389
50 10103107109127131371491515716316717317918191932113922322722941974347535961998389
51 101031071091271313714915157163167173179181919321139223322722941974347535961998389
52 101031071091271313714915157163167173179181919321139223322722923941974347535961998389
53 1010310710912713137149151571631671731791819193211392233227229239241974347535961998389
54 101031071091271313714915157163167173179211392233227229239241819193251974347535961998389
55 101031071091271313714915157163167173179211392233227229239241819193251972574347535961998389
56 101031071091271313714915157163167173179211392233227229239241819193251972572634347535961998389
57 101031071091271313714915157163167173179211392233227229239241819193251972572632694347535961998389
58 101031071091271313714915157163167173179211392233227229239241819193251972572632694347535961998389
59 1010310710912713137149151571631671731792113922332277229239241819193251972572632694347535961998389
60 101031071091271313714915157163167173211392233227722923924179251819193257263269281974347535961998389
61 1010310710912713137149151571631671732113922332277229239241792518191932572632692819728343475359619989
62 10103107109127131371491515716316717321139223322772293239241792518191932572632692819728343475359619989
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165 223070101032270907229241072510925712631126911271281128340140929342113137274314433173344914574613941463467347487503494750952149952337353541515475155756356953587576015760761796419193593796439647971976536596619967683676980982163823977398277829853838577859986389881816778778839887
166 22307010103227090722924107251092571263112691127128112834014092934211313727431443317334491457461394146346734748750349475095214995233735354151547515575635695358757601576076179641919359379643964797197653659661996768367698098216382397739827782983838538577859986389881816778778839887
167 223070101032270907229241072510925712631126911271281128340140929342113137274314433173344914574613941463467347487503494750952149915152337353541547515575635695358757601576076179641919359379643964797197653659661996768367698098216382397739827782983838538577859986389881816778778839887
168 2230701010322709072292410725109257126311269112712811283401409293421131372743144331733449145746139414634673474875034947509521499151523373535415475155756356953587576015760761796419193593796439647971976536596619967683676980982163823977398277829838385385778599786389881816778778839887
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170 22307010090709227101310322924107251092571263112691127272811283401409293421134431373317344914574334613941463467347487503494750952149915152337515415475575635356953587576015760761796419193593796439647971976536596619967683676980982163823977398277829838385385778599786389881816778778839887
171 22307010090709227101310191032292410725109257126311269112727281128340140929342113443137331734491457433461394146346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
172 22307010090709227101310191021032292410725109257126311269112727281128340140929342113443137331734491457433461394146346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
173 223070100907092271013101910210310722924109251257126311269112727281128340140929342113443137331734491457433461394146346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
174 223070100907092271013101910210310331107229241092512571263132691127272811283401409293421137334431734491457433461394146346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
175 223070100907092271013101910210310331103922924107251092571263132691127272811283401409293421137334431734491457433461394146346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
176 223070100907092271013101910210310331103922924104910725109257126313269112727281128340140929342113733443173449414574334613946346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
177 223070100907092271013101910210310331103922924104910510725109257126313269112727281128340140929342113733443173449414574334613946346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
178 223070100907092271013101910210310331103922924104910510610725109257126313269112727281128340140929342113733443173449414574334613946346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
179 223070100907092271013101910210310331103922924104910510610631325107257109263269112727281128340140929342113733443173449414574334613946346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
180 223070100907092271013101910210310331103922924104910510610631325106911072571092632692811272728340140929342113733443173449414574334613946346734748750349475095214991935233751515415475575635356953587576015760761796419643964796535937971976596619967683676980982163823977398277829838385385778599786389881816778778839887
181 223070100907092271013101910210310331103922924104910510610631325106911072571087263269281092834012727409293421137334431734494145743346139463467347487503494750952149919352337515154154755756353569535875760157607617964196439647965359379719765966199676836769809821638239773982778298383853857785997863898811816778778839887
182 2230701009070922710131019102103103311039229241049105106106313251069107257108726326928109112727283401409293421137334431734494145743346139463467347487503494750952149919352337515154154755756353569535875760157607617964196439647965359379719765966199676836769809821638239773982778298383853857785997863898811816778778839887
183 2230701009070922710131019102103103311039229241049105106106313251069107257108726326928109110932834012727409293421137334431734494145743346139463467347487503494750952149919352337515154154755756353569535875760157607617964196439647965359379719765966199676836769809821638239773982778298383853857785997863898811816778778839887
184 2230701009070922710131019102103103311039229241049105106106313251069107257108726326928109110932834010971929340941272742113733443173449457433461394634673474875034947509521499193523375151541547557563535695358757601576076179641976439647965359379765966199676836769809821638239773982778298383853857785997863898811816778778839887
185 2230701009070922710131019102103103311039229241049105106106313251069107257108726326928109110932834010971929340941272742113733443173449457433461394634673474875034947509521499193523375151541547557563535695358757601576076179641976439647965359379765966199676836769809821638239773982778298383853857785997863898811816778778839887
186 2230701009070922710131019102103103311039229241049105106106313251069107257108726326928109110932834010971929340941272742113733443173449457433461394634673474875034947509521499193523375151541547557563535695358757601576076179641976439647965359379765966199676836769809821638239773982778298383853857785997863898811816778778839887
187 223070100907092271013101910210310331103922924104910510610631325106910725710872632692810911093283401097192934094127274211173344317433449457461373463467347487503494750952149919352337515154157547557563535695358757601635937960761796419764396479765365966199676836769809821677397782398277829838385385778599786389881811398839887787
188 223070100907092271013101910210310331103922924104910510610631325106910725710872632692810911093283401097192934094111727421123344317334494574337346137463467347487503494750952127514991935235354151575475575635695358757601635937960761796419764396479765365966199676836769809821677397782398277829838385385778599786389881811398839887787
189 1009070101307092232271019102103103310491051061063110392292410691072510872571091109326326928109719283401117274092934211233443131733449411294574337346137463467347487503494750952127514991935235354151575475575635695358757601635937960761796419764396479765365966199676836769809821677397782398277829838385385778599786389881811398839887787
190 10090701013070922322710191021031033104910510610631103922924106910725108725710911093263269281097192834011172740929342112334431317334494112945743373461374634673474875034947509521139523535412751499193547557563569535875760157607617964197643964796535937976596619967683676980982163823977398277829838385385778599786389881151816778778839887
191 100907010130709101910210310331049105106106311039223227106910722924108725109110932571097192632692811172728340112334092934211294113137334431734494574337461394634673474875034947509521151153523535412751499193547557563569535875760157607617964197643964796535937976596619967683676980982163823977398277829838385385778599786389881816778778839887
192 1009070101307091019102103103310491051061063110392232271069107229241087251091109325710971926326928111727283401123340929342112941131373344317344945743374613946346734748750349475095211511535235354116354751275575635695358757601499193593796076179641976439647976536596619967683676980982157739778239827782983838538578599786389881816778778839887
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195 100907010130709101910210310331049105106106311039223227106910722924108725109110932571097111719263269281123283401129293409411313727421151153443173344945743346139463467347487503494750952116352337353541181187512754755756356953587576014991935937960761796419764396479765365966199676836769809821577397782398277829838385385785997863898816778778839887
196 100907010130709101910210310331049105106106310691072231103922710872292410911093251097111711232571926326928112928340113137274092934211511534431733449411634574334613946346734748750349475095211811875119352337353541275475575635695358757601499196076179641976439647965359379765966199676836769809821577397782398277829838385385785997863898816778778839887
197 100907010130709101910210310331049105106106310691072231103922710872292410911093251097111711232571926326928112928340113137274092934211511534431733449411634574334613946346734748750349475095211811875119352337353541201275475575635695358757601499196076179641976439647965359379765966199676836769809821577397782398277829838385385785997863898816778778839887
198 1009070101307091019102103103310491051061063106910710872231103922710911093229241097111711232511292571926326928113132834011511534092934211634431733449411811872743345746137346346734748750349475095211935233751201213953535412754755756356958757601499196076179641976439647965359379765966199676836769809821577397782398277829838385385785997863898816778778839887
199 10090701013070910191021031033104910510610631069107108710911039223110932271097111711232292411292511313257192632692811511532834011634092934211811872743173344334494119345746137346346734748750349475095212012139523375121754127547557563535695358757601499196076179641976439647965359379765966199676836769809821577397782398277829838385385785997863898816778778839887
200 100907010130709101910210310331049105106106310691071087109109311039110971117112322711292292411313251151153257192632692811632834011811872740929342119344317334494120121373457433461394634673474875034947509521217512233752353541275475575635695358757601499196076179641976439647965359379765966199676836769809821577397782398277829838385385785997863898816778778839887

Kode

Berikut ini adalah skrip Python 3 untuk memanggil pemecah Concorde berulang kali hingga membangun solusi.

Concorde gratis untuk penggunaan akademis. Anda dapat mengunduh biner yang dapat dieksekusi dari Concorde yang dibangun dengan paket pemrograman liniernya sendiri QSopt, atau jika Anda entah bagaimana memiliki lisensi untuk IBM CPLEX, Anda dapat membangun Concorde dari sumber untuk menggunakan CPLEX.

#!/usr/bin/env python3
'''
Find prime containment numbers (OEIS A054261) using the Concorde
TSP solver.

The n-th prime containment number is the smallest natural number
which, when written in decimal, contains the first n primes.
'''

import argparse
import itertools
import os
import sys
import subprocess
import tempfile

def join_strings(a, b):
  '''Shortest string that starts with a and ends with b.'''
  for overlap in range(min(len(a), len(b)), 0, - 1):
    if a[-overlap:] == b[:overlap]:
      return a + b[overlap:]
  return a + b

def is_prime(n):
  if n < 2:
    return False
  d = 2
  while d*d <= n:
    if n % d == 0:
      return False
    d += 1
  return True

def prime_list_reduced(n):
  '''First n primes, with primes that are substrings of other
     primes removed.'''
  primes = []
  p = 2
  while len(primes) < n:
    if is_prime(p):
      primes.append(p)
    p += 1

  reduced = []
  for p in primes:
    if all(p == q or str(p) not in str(q) for q in primes):
      reduced.append(p)
  return reduced

# w_med is an offset for actual weights
# (we use zero as a dummy weight when splitting nodes)
w_med = 10**4
# w_big blocks edges from being taken
w_big = 10**8

def gen_tsplib(prefix, strs, start_candidates):
  '''Generate TSP formulation in TSPLIB format.

     Returns a TSPLIB format string that encodes the length of the
     shortest string starting with 'prefix' and containing all 'strs'.

     start_candidates is the set of strings that solution paths are
     allowed to start with.
     '''
  N = len(strs)

  # Concorde only supports symmetric TSPs. Therefore we encode the
  # asymmetric TSP instances by doubling each node.
  node_in = lambda i: 2*i
  node_out = lambda i: node_in(i) + 1
  # 2*(N+1) nodes because we add an artificial node with index N
  # for the start/end of the tour. This node is also doubled.
  num_nodes = 2*(N+1)

  # Ensure special offsets are big enough
  assert w_med > len(prefix) + sum(map(len, strs))
  assert w_big > w_med * num_nodes

  weight = [[w_big] * num_nodes for _ in range(num_nodes)]
  def edge(src, dest, w):
    weight[node_out(src)][node_in(dest)] = w
    weight[node_in(dest)][node_out(src)] = w

  # link every incoming node with the matching outgoing node
  for i in range(N+1):
    weight[node_in(i)][node_out(i)] = 0
    weight[node_out(i)][node_in(i)] = 0

  for i, p in enumerate(strs):
    if p in start_candidates:
      prefix_w = len(join_strings(prefix, p))
      # Initial length
      edge(N, i, w_med + prefix_w)
    else:
      edge(N, i, w_big)
    # Link every str to the end to allow closed tours
    edge(i, N, w_med)

  for i, p in enumerate(strs):
    for j, q in enumerate(strs):
      if i != j:
        w = len(join_strings(p, q)) - len(p)
        edge(i, j, w_med + w)

  out = '''NAME: prime-containment-number
TYPE: TSP
DIMENSION: %d
EDGE_WEIGHT_TYPE: EXPLICIT
EDGE_WEIGHT_FORMAT: FULL_MATRIX
EDGE_WEIGHT_SECTION
''' % num_nodes

  out += '\n'.join(
    ' '.join(str(w) for w in row)
    for row in weight
  ) + '\n'

  out += 'EOF\n'
  return out

def parse_tour_soln(prefix, strs, text):
  '''This constructs the solution from Concorde's 'tour' output format.
     The format simply consists of a permutation of the graph nodes.'''
  N = len(strs)
  node_in = lambda i: 2*i
  node_out = lambda i: node_in(i) + 1
  nums = list(map(int, text.split()))

  # The file starts with the number of nodes
  assert nums[0] == 2*(N+1)
  nums = nums[1:]

  # Then it should list a permutation of all nodes
  assert len(nums) == 2*(N+1)

  # Find and remove the artificial starting point
  start = nums.index(node_out(N))
  nums = nums[start+1:] + nums[:start]
  # Also find and remove the end point
  if nums[-1] == node_in(N):
    nums = nums[:-1]
  elif nums[0] == node_in(N):
    # Tour printed in reverse order
    nums = reversed(nums[1:])
  else:
    assert False, 'bad TSP tour'
  soln = prefix
  for i in nums:
    # each prime appears in two adjacent nodes, pick one arbitrarily
    if i % 2 == 0:
      soln = join_strings(soln, strs[i // 2])
  return soln

def scs_length(prefix, strs, start_candidates, concorde_path, concorde_verbose):
  '''Find length of shortest containing string using one call to Concorde.'''
  # Concorde's small-input solver CCHeldKarp, tends to fail with the
  # cryptic error message 'edge too long'. Brute force instead
  if len(strs) <= 5:
    best = len(prefix) + sum(map(len, strs))
    for perm in itertools.permutations(range(len(strs))):
      if perm and strs[perm[0]] not in start_candidates:
        continue
      soln = prefix
      for i in perm:
        soln = join_strings(soln, strs[i])
      best = min(best, len(soln))
    return best

  with tempfile.TemporaryDirectory() as tempdir:
    concorde_path = os.path.join(os.getcwd(), concorde_path)
    with open(os.path.join(tempdir, 'prime.tsplib'), 'w') as f:
      f.write(gen_tsplib(prefix, strs, start_candidates))

    if concorde_verbose:
      subprocess.check_call([concorde_path, os.path.join(tempdir, 'prime.tsplib')],
                            cwd=tempdir)
    else:
      try:
        subprocess.check_output([concorde_path, os.path.join(tempdir, 'prime.tsplib')],
                                cwd=tempdir, stderr=subprocess.STDOUT)
      except subprocess.CalledProcessError as e:
        print('Concorde exited with error code %d\nOutput log:\n%s' %
              (e.returncode, e.stdout.decode('utf-8', errors='ignore')),
              file=sys.stderr)
        raise

    with open(os.path.join(tempdir, 'prime.sol'), 'r') as f:
      soln = parse_tour_soln(prefix, strs, f.read())
    return len(soln)

# Cache results from previous N's
pcn_solve_cache = {} # (prefix fragment, strs) -> soln

def pcn(n, concorde_path, concorde_verbose):
  '''Find smallest prime containment number for first n primes.'''
  strs = list(map(str, prime_list_reduced(n)))
  target_length = scs_length('', strs, strs, concorde_path, concorde_verbose)

  def solve(prefix, strs, target_length):
    if not strs:
      return prefix

    # Extract part of prefix that is relevant to cache
    prefix_fragment = ''
    for s in strs:
      next_prefix = join_strings(prefix, s)
      overlap = len(prefix) + len(s) - len(next_prefix)
      fragment = prefix[len(prefix) - overlap:]
      if len(fragment) > len(prefix_fragment):
        prefix_fragment = fragment
    fixed_prefix = prefix[:len(prefix) - len(prefix_fragment)]
    assert fixed_prefix + prefix_fragment == prefix

    cache_key = (prefix_fragment, tuple(strs))
    if cache_key in pcn_solve_cache:
      return fixed_prefix + pcn_solve_cache[cache_key]

    # Not in cache, we need to calculate it.
    soln = None

    # Try strings in ascending order until scs_length reports a
    # solution with equal length. That string will be the
    # lexicographically smallest extension of our solution.
    next_prefixes = sorted((join_strings(prefix, s), s)
                           for s in strs)

    # Try first string -- often works
    next_prefix, _ = next_prefixes[0]
    next_prefixes = next_prefixes[1:]
    next_strs = [s for s in strs if s not in next_prefix]
    next_length = scs_length(next_prefix, next_strs, next_strs,
                             concorde_path, concorde_verbose)
    if next_length == target_length:
      soln = solve(next_prefix, next_strs, next_length)
    else:
      # If not, do a weighted binary search on remaining strings
      while len(next_prefixes) > 1:
        split = (len(next_prefixes) + 2) // 3
        group = next_prefixes[:split]
        group_length = scs_length(prefix, strs, [s for _, s in group],
                                  concorde_path, concorde_verbose)
        if group_length == target_length:
          next_prefixes = group
        else:
          next_prefixes = next_prefixes[split:]
      if next_prefixes:
        next_prefix, _ = next_prefixes[0]
        next_strs = [s for s in strs if s not in next_prefix]
        check = True
        # Uncomment if paranoid
        #next_length = scs_length(next_prefix, next_strs, next_strs,
        #                         concorde_path, concorde_verbose)
        #check = (next_length == target_length)
        if check:
          soln = solve(next_prefix, next_strs, target_length)

    assert soln is not None, (
      'solve failed! prefix=%r, strs=%r, target_length=%d' %
      (prefix, strs, target_length))

    pcn_solve_cache[cache_key] = soln[len(fixed_prefix):]
    return soln

  return solve('', strs, target_length)

parser = argparse.ArgumentParser()
parser.add_argument('--concorde', type=str, default='concorde',
                    help='path to Concorde binary')
parser.add_argument('--verbose', action='store_true',
                    help='dump all Concorde output')
parser.add_argument('--start', type=int, metavar='N', default=1,
                    help='start at this N')
parser.add_argument('--end', type=int, metavar='N', default=1000000,
                    help='stop after this N')
parser.add_argument('--one', type=int, metavar='N',
                    help='solve for a single N and exit')

def main():
  opts = parser.parse_args(sys.argv[1:])

  if opts.one is not None:
    opts.start = opts.one
    opts.end = opts.one

  prev_soln = ''
  for n in range(opts.start, opts.end+1):
    primes = map(str, prime_list_reduced(n))
    if all(p in prev_soln for p in primes):
      soln = prev_soln
    else:
      soln = pcn(n, opts.concorde, opts.verbose)

    print('%d %s' % (n, soln))
    sys.stdout.flush()
    prev_soln = soln

if __name__ == '__main__':
  main()