1–255 sayıları için Brainfuck oluştur


34

1'den 255'e kadar olan sayılar için verilen sayıyı dizinin bir byte'ında üretecek olan BF kodunu ve yeni bir satır yazdıracak bir program yazın.

Örneğin, çıktının ilk dört satırı olabilir (ve büyük olasılıkla olacaktır):

+
++
+++
++++

Kazanan en küçüğü olacak: source code + output(bayt cinsinden).

Açıklamalar ve Revizyonlar :

  • BF programları sarma hücreleri kullanır.

  • Çıkış BF programı, sayıyı içeren hücre olan yalnızca sıfır olmayan hücre ile sona ermelidir.

  • Programlar artan sırada verilmelidir.

  • 0 için bir program çıktısı isteğe bağlıdır.

  • Negatif veri işaretçilerine izin verilmiyor. <ilk işaretçi hiçbir şey yapmaz. (fırlatıp atması daha uygunsa yorum yapın)


1
@JoKing Tüm çıktı sayılır.
Mason

2
Oh, görüyorsunuz, kodun çıkış hücresinde bitmesi gerekmediğini söylüyorsunuz
Jo King

1
“BF” nin tam olarak sizin bağlamınızda neyi kastettiğine dair referanslar elde etmek faydalı olurdu, örneğin esolangs.org/wiki/Brainfuck_constants veya başkası, vs.
HolyAvengerOne

2
@Mason +>++++++++++.giriş için geçerli bir program olur 1mu?
Jonathan Frech

6
BF çıkışını golf oynamakla dengelemek zorunda olduğunuz iki kısıtlı bir mücadele için +1, BF çıkış kodunu golfe çevirmek. İlginç bir twist ☺
Chronocidal

Yanıtlar:


15

Perl 6 , 224 + 3964 = 5834 4188 bayt

map {say (.[0]~'['~.[3]~'>'~.[1]~'<]')x?.[1],'>'x?.all,.[2]}o*.min({$_>>.abs.sum+6*?.[1]})>>.&{<- +>[.sign>0]x.abs},classify({0+|(grep(*%%1,(((256 X*^4)X+.[0]%256)X/-.[3]))[0]*.[1]+.[2])%256},[X] |(^27-13 xx 3),-7..-1){^256}

Çevrimiçi deneyin! (mayıs zaman aşımı. değiştirin ^27-13için^25-12 bazı ekstra çıktı pahasına hafifçe hızlandırmak için)

*>[*>*<]>*Her birinin *belirli bir +s veya -s olduğu formdaki en kısa kodu verir . Gerekmediği takdirde halkayı çıkarmak gibi bazı ekstra adımlar da vardır> .

Söyleyebileceğim kadarıyla, çıktı bu özel format için en çok kullanılan olanı.

Açıklama:

([X] |(^27-13 xx 3),-7..-1)        # Define the search space as the cross product of:
                                        # -13 to 13 for:
                                            # Initialisation     +++>
                                            # Change in target   [*>+++<]
                                            # Last change        >+++
                                        # And -7 to -1 for the change in start [-->*<]
  .classify({                  })  # Group them by calculating
                  (256 X*^4)                         # Each of the multiples of 256
                 (          X+.[0]%256)              # Plus the initialisation
                (                      X/-.[3])      # Divided by the change in start
      grep(*%%1,                               )     # Filter out the whole numbers
                                                [0]  # And take the first value
          # This is the amount of times the inner loop will execute
          # Being Nil, converted to 0 if it is an infinite loop
      *.[1]              # Multiply by the change to the target cell
           +.[2]         # And add the final section
     (          )%256    # And modulo the whole lot by 256
                     +|0 # And floor it just to keep the .0 out
classify(                   ){^256}     # Take the corresponding groups in order
   .map(                             )  # And map each to
        *.min({                    })   # Find the minimum by:
               $_>>.abs.sum             # The sum of the absolute values    
                           +6*?.[1]     # Plus 6 if it loops
      >>.*{                   }    # Then map each value to
           <- +>[.sign>0]          # + or - depending on the sign
                         x.abs     # Repeated by the absolute value 
   {                    }o              # And pass this to the next code block
    say                       # Print
        (.[0]~'['~.[3]~'>'~.[1]~'<]')             # The loop section
                                     x?.all       # If it is needed
                                           ,.[2]  # And the final part

12

Malbolge , 28 743 bayt + 7 166 çıktı

Çok yaratıcı değil, değil mi? Bu kötü çocuğa golf oynamaya çalışacağım.

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BAe(D=<`:9>7[;43270/St2+*N.-m%*#G'&}|#zy?>=<;\rwvon4Ukping-kdLbg`&%$#"!~}|\[ZYXQ9OTMq4JOHlFEJIHGF?cCB;_">7<;:3Wx65.3,+O).'&%I#"'&%|BA@?>|^]sxq7unVrqj0QPle+chg`_%]E[`_XWVzZSXWVUNrqpo2NGLEJIHGF?c=BA:?8\};43WVUTSRQP0p.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Plkjchg`&G]\aZ~^@?UyY;WPUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I)(h&}$#z@~}v{zs9876tVUqpong-kjihgfe^F\[!_AWVzyxwvutTS54JINGkEDCBfFE'C<;:^!=<54381Uvu32+*N(n&J*#i!&}$#"y?`_{zsxwp6543210/mfed*Kaf_^$Ea`_^WVUyYXQ9OTMqQ3IHlLKDIBA@dcba`@"!=6Z4z2165.R210/(LKJIH"!~}|Bz!x}|{zs9wYutsrk1oQgled*Kg`ed]#[`Y^WVUyYXQ9OTMqQ3IHGLKJCgfedcba`_?>=6Z49816/St210/.'K+*#i!&}$#"y?wv{zs9wvXnm3qSohgfe+*)('Hd]\aZ~^W?[TSXWVONrRKJOHMLEihgfe(>bBA@?>=6Z4z870/.R210/on&Jk)"!~}C#zb~}v{zyxwpo5slqpih.Okdihg`&%$#"!~}|\[ZYXWVUTSRQPO10LEDh+AF?cC<;:^>=6|4X87654-s+*Non,%*#GFEDCB"y~}|{zyxwpo5Vlkjing-kdLbg`&%$#"!~}@?UyY;WVONrLKJn10LEDh+AF?cC%$@?>=6Z:9876v4321*Non,%*#GFEDC#c!xw=uzsr8potslqpong-eMibaf_%]E[`_XWVzZSXWVUNrqponm0/KDIHAF?c=BA:?8\};43WVUTSRQP0p.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/POejcb(Ie^$bD`YX]VzT<XQVUNrqpon10LEDh+AF?cC<;:^>=6|4Xyx05.RsP*/(',+$HGFEDe{z!~w=^]sxq7unmlk10/.-,+cha`_%cb[Z_X|\[=Sw:PUTMq4JOHl/KDhBAFED=a$@9>7<54XWVUTSR210/on&J$)"!~%$#z@awv{zyxwp6Wmlk1ong-Njibgf_%cba`_AWVzyxwvVUTS5Ko2NGkKDCBfedcba`@?>=6|4X8765.Rs10/.'K+k#('~D$#"!x}|{zs9wYotsrqji/mfNjiha'e^Fba`Y^W{zyxwvuUNSRKJIm0FEiIBGF?>=<`@?>=6|4X87654-s+*Non,%*#GFEDC{"!~}|{zyxwpo5Vlkjing-kdLbg`&%$#"!_^@?UyY;WVONrLKJn10LEDh+AFED=<`@?>=6|4Xyx05.RQPONMLK+k#('~D$#"!x}|{zs9wYutsrk1Rhmlejib(fe^$Ea`Y^WV[Tx;QPONMRKonmlk.JIBAF?c=BA:?8\};43WVUTSR21q/.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Pfkdihg`&^F\a`_X]VzT<XQVUNrqponmlk.JCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I)(h&}$#z@~}v{zs98765VUqpong-kjihgfe^F\[!_AWVzyxwvu8TMLKPINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:9wvoWm3qSing-Ndiha`&^F\a`_X]VzT<XQVUNrR4o2NMFEDCgfedcC%$@?>=6Z4z2165.R210/(LKJIHG'&}e#"y?wv{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponmlk.JCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#"'&%|BAya}|{ts9wvXtsrkj0/.Oejchg`&^F\[!_AWVzyxwvutsr54JINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:98voWm3qSing-Ndiha`&^F\a`_X]VzZYR:u8TMRQPImlkjiIBAF?c=BA:?8\};43WVUTSRQ1q/.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Pfkdihg`&^F\a`_X]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#('~%|B"b~}|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqponml/KDCHAFED=aA@">=65Yzy70/.R210/on&J$#"!E}e#"y?`_{zsxwp6543210ng-kjcb(I_%cbD`YX]VzZYR:u8TMRQPImlkjiIBAF?c=BA:?8\};43WVUTSRQ1q/.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Pfkdihg`&^F\a`_X]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#"'&%|B"b~}|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqponm0/KDCBf@ED=<`:9>7[;4381Uvu32+*N(n&J*#i!&}$#"y?`_{zsxwp6543210/mfed*Kaf_^$Ea`_^WVUyYXQ9OTMq4JOHlFEDIHG@d'&<A@?>7[ZYXWx65.3,+O/on&Jk)"!~}CBA@?>|{z\xwpun4rkj0QPle+ihgfe^F\[!_AWVzZ<XWPtTS54JINGk.DIBA@dcba`_^]=6Z49816/St210/.'K+*#i!&}$#"y?`|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqpon10FKDIBf@ED=<`:9>7[;4381U/S-2+0/.-,%I)ih~}${z@?>=<;:987oWm3qSing-Ndiha`&^F\a`_X]Vz=YRQVOsSR4o2NMFEDCgfedcC%$@?>=6Z4z2165.R210/(LKJIH"!~}|Bz!x}|{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponmlkjiIH*)?D=<`@">=65Yzy70/.R2+0/.'&%I)(h&}$#z@~}v{zs987654rTpong-kjihgfe^F\[!_AWVzyxwvu8TMLKPINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:xwvoWm3qSing-Ndiha`&^F\a`_X]Vz=YRQVOsSR4o2NMFEDCgfedcb%$@?>=6Z4z2165.R210/(LKJIHG'&}e#"y?wv{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I)(h&}$#z@~}v{zs987654rTpong-kjihgfe^F\[!_AWVzyxwvu8TMLKPINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:xwvoWm3qSing-Ndiha`&^F\a`_X]VzT<XQVUNrR4o2NMFEDCgfedcb%$@?>=6Z4z2165.R210/(LKJIHG'&}e#"y?wv{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#"'&%|BAya}|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqponmlkjiIHG@(D=a;_9>7654X81w/.-,+O/(L,+k#"!E}e#"y?`_{zsxwp65432pong-kjcb(I_%cbD`YX]Vzg

Çevrimiçi deneyin!


Ama neden yyyyyyyy?
Joshua,

@Joshua bowling, oğlum
Krzysztof Szewczyk

Bu gerçekten golf değil, ama yine de bir oy var. Kaynak kod aracı yardım edildi mi / oluşturuldu mu?
Ethan

/ * Aslında cevapları karıştırdım * / Evet.
Krzysztof Szewczyk

Bunun arkasındaki fikir de bununla
Krzysztof Szewczyk

12

Brainfuck, 77 75 73 + 32894 = 32967 32969 32971 bayt

++++++[->+++++++<]>+>++++++++++>+[>+[-<<<.>>>]<<.>[->+>+<<]>>[-<<+>>]<<+]

Çevrimiçi deneyin!

çıkış mümkün olan en basit

+
++
+++
++++
...

açıklama:

++++++[->+++++++<]>+ set cell 2 to 43 (ascii of plus)
>++++++++++ set cell 3 to 10 (ascii of new line)
>+ set cell 4 to 1
[
    >+ increment cell 5
    [
        -<<<.>>> decrement cell 5 and print a plus (content of cell 2)
    ] until cell 5 == 0
    <<.> print a new line (content of cell 3)
    [
        ->+>+<< move value of cell 4 to cell 5 & 6, setting cell 4 to 0
    ]
    >> goto cell 6
    [
        -<<+>> move it's value to cell 4, setting cell 6 to 0
    ]
<<+ increment cell 4
] exit when cell 4 goes beyond 255 because cell contains C uchar meaning 255 + 1 == 0

10
Görmek istediğim de buydu - başka bir BF kodu üreten bir BF kodu!
Dhruv Saxena

10

Stax , skor 4751 4783 (812 bayt + 3971)

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Koş ve hata ayıkla

En iyi yayınlanan programlar ile başladım .
Bazı regex-fu'ları kullandım, en fazla 2 hücreyi kullanan en kısa programlarla sınırlandırdım. Sonra izleri <veya >karakterleri kırptım . Bunun, program sonlandırmasında yabancı olmayan sıfır hücre olmadığından emin olmanın muhtemelen muhafazakar bir yol olduğunu düşünüyorum. Sonra sabit kolmogorov tipi çıktı için stax programları oluşturmak için yazdığım deneysel bir stax programında araştırdım.
Bu program tekrar tekrar dize değiştirmeleri uygulayarak çalışır. Her adımda, en sık oluşan> 1 uzunluk alt dizini arar ve kullanılmayan bir karakterle değiştirir.


@JoKing: Sanırım yabancı olmayan sıfır hücrelerini ele aldım. Brainfuck program boyutunda 200 bayttan fazlaya mal oldu, ancak neredeyse hepsini sıkıştırılabilirlikle geri kazandım.
özyinelemeyle

7

Mangal kömürü , 707 698 410 + 3627 = 4334 4325 4037 bayt

UT≔”}⊞J5±)↷γ²⁼⎇⦃<✂f^⊗L…¬⁻←«θ↥v⊙^≔¶υSψVτ16⁷·9I⌕↘;⦃@Pmt↙ |TL ‹.bE^↷Am⟧←⪫✂«GIχ¤⟲V⁻PÀ$χ¹'$↙‖%S³6◧N=$kHIpQ×ïu|%÷I↖➙⁸≔Wλ¹ê8⌕dNK‽3H∨↥γh➙↘⊙⊕“~Oj↨-⬤…⊟⁺§◨CB℅P⌕KNEAR№K⬤X"¬S⎇⧴V⁻±6⁼✂kι×CÀ⊞‴≡w↓γ=`→P5η1C⊖OSoNυs⊘$M↙êαη↖φ¡¿:θ-γ“rJW%E(7<w¤Uφ´ρHπ←SX↔τ↧%<Tº⎇0gθμ↓⌕;σw⌈pL;Y↘YΠ⊙>ξLzλ↓⁸ι⎚|⌕ΠP″M³⧴⬤¦➙⟧⌕/δ;↥⁻ºJK⌊≡<⊖λ✳Jκ⟲➙ξ⭆|^Σ*βMπ⍘⊟;ÀU÷‹⭆◧�ωκ?σηkYOδO/Bº?lAnaK{*Kaκ◨+↧aSφ0q‖B/φx⊘⌕«³ψü✂‹º≡/yc⁴&J↙S²~⎇z§‖$SP≧”θG↘←¹⁴+⮌⪪θ⸿↑Fθ§⁺+-ι⌕-+ιG→↙¹⁴-

Çevrimiçi deneyin! Bağlantı, kodun ayrıntılı bir versiyonudur. Açıklama:

UT

Boşluk dolgusunu kapatın.

≔”...”θ

İçin JonathanAllen en cevapları @ oluşan geniş bir sıkıştırılmış dize atama -128.. -15fakat +ve -aktarılmamıştır işaretler.

G↘←¹⁴+

+Sıkıştırılmış dizideki geri dönüş, bir sonraki çıktıyı bir sonraki satıra taşıyacak olmasına rağmen, imleç alt köşede bırakılmış olmasına rağmen, 14. tarafın bir üçgenini çizin .

⮌⪪θ⸿

Dönen karakterlerde büyük dizgiyi bölün ve her alt dizgiyi ters sırada yazdırın; böylece sonuçlar 15'den 128'e çıkarılır.

128 sonucunun üzerine yazılan 128 sonucunun üzerine yazılacak şekilde bir satır yukarı kaydırın.

Fθ§⁺+-ι⌕-+ι

Dize aktarma yoluyla döngü + sırasında -tekrar dönün ve tekrar geri dönün, böylece 128 ila 241 için doğru sonuçları üretirler.

G→↙¹⁴-

-242 - 255 için doğru sonuçları üreten, 14. tarafın bir üçgenini çizin .


5

Jöle , 1224 + 3716 = 4940 bayt

⁾+-ẋ€Ɱ14ZY€U0¦j“6VⱮ×ė7¬(Ị¢ẋṀⱮM⁵Ѭkbvœ⁸½ẋƓ0⁽ṖçḟŻßɓẉḷ0Ƙ¥@ⱮZĊⱮ{ṫṇØ"ỵðẓ⁵!ḳqḄƬiỴƥṇØm@ɗẆḅƥƲ⁴ŀ-5¦€ÑɓZĖ/gPṄḌ!ẹ$ḞıƒĿỵ⁷£Q.%¦ẊiUı-M⁹ƈxṁ,CsḲtÆƇỴṄĿiæEṛⱮẒʠþƘ%ƘƙṾ ('ȥ€½⁵ḥ+,þ@ẇ&ạV|ĊuAYḃfṖƘLƥQtPƬivxHj)Ṇɓ5JṘØẓæĿøɗjḥrñþa®OṅḍṪ¥=ɼġċṫßṬỌƈrUẉçŻ½\=]€ʂ_ⱮṖ¥Ƥȥ6SṡÆcạdn;ṅⱮDɦ⁹ṢAy)~Ḷ`ẒẓMTİṂḋ|ẉ]Wɠ¿⁾Ṣ|ḷ6hẸƒⱮQ1ẏƝC@Ŀ!ʠ⁽ṃ@ƓŒQ3@ƝḊñçcZ\¥3Z¤~çD>ċọuⱮȦAẈⱮ%L3Æ¢ḞtĖė!ƇtñṪɓẓ¥Fṅ⁵shB'wṪẸ¦ṄÞṭ³ʂḶƊ³iȧṂRœŒƤ\r1Çwi6ŀỵɼḃa⁵Ṣ_Q⁸Ẹ'{|\+Æ®|ḤcʂÑ/Ɓz¶ɦÄ!ʂ"Ẋ ẓĠĠ⁷⁵QƝ¶%ṙƇḋ[^j&W×*°ḳçʂSżḊⱮ⁻IȦṄXȥlẋḅ7;⁺ḃİÞÆðLX¢1K£€Ä&X½VȮ(;Q£ḞḢ¹zG+ṅ¹LḥW³ḅd@^ẊḶJ¹T8ṛ($ȧṢzq,Ṫ⁻ȥ{Ṛ"Ḍ®Ä8QḋþɼȮhỵB"Ḍ⁶ȧZ⁵ẒNɓḃȧ¶Ƒð$Ẇ/"Eṭ*I:ØḃL}<KȦ+ṣ¥x&Ṇ£Œṫḋġ0lİḍ¿H£(ỌƝ×^Ḃ°⁽⁼UƭĠḥkQð7ṫƤżȷƘxjƑRḣqƒ$HƬ7ḳ-JµnṇṣðXŻİẉbSu×]bṾ0ƊHßçQh⁸°ƒɦSCñ_⁾ʂC⁼Ġø⁵SAʋƊİ¡⁴ÄḋẸḶwȧZẈĠ7rṀẏẉṖa¤ɱELƝȧẈṣṄk]d⁹øṇÞṡ.ạtƥṢḅ⁺ṂLpÑƘṄṡḍ⁵,Ǥ$Ọ8ṛuṚvAṖÑ1!vƤD3߶ʂа]EÞUĠ€ḋḲ⁸¬r`YḊ0ṙ5ċmṅȯ*ɲU÷pƭẉṭȦB¹ɦSNɱ)]ĠṾʋ³Øḟ23ṭð#ẆuẎṬṫVɠ(ỊỊQɼF}ịƒ$Ẏ_Ṡ'ḳOLc?ṾŀẊẎṆ⁵p"VẏAȮ⁴ⱮȦ®e®Ɱi"ÇJẊ4ñḍḲY]ḌḌẓ⁺ƙ"iṄḅoLṙfOS&}HGɼĖİĠḷuḃ³ṡıḳỊẹzq⁶ƈ£ċHZɱ.#⁶ḟUṗŀȮṘḶḲ]@¶+ḊĖ8ĖṆɗçŻŀ®ṭẇƓḄḷıM@⁷²36ɲṗ¡ḂḊ'Ṿ⁵ėƙṘ-⁺µʠṡṂ[_¤ḥṢ]ṘÐḤ½ḟ4ȷ}E¹Ṙb⁹ḅḢ¹hƬZ§Ẏẹ÷Æ$ḅoĖẉ⁹ịJ.ȥḊẋʋṄȯ1<ẎṄḲṛœ"æ)ḥ8)ḤlñA⁾%⁶LỴ⁶M4Ṙ\`ỵƊȥŀƒ⁷ḌƬƙḳƑ⁴vʂ⁻ðQpñḷḳṄœ>ṪỴƭƙɓ3[&Ḅzḅ<⁾µİṪȧ⁹C>ẹ{ẈÐlC&j?LṆṛ⁽æȤið<⁽$Ḋ7⁻FṡḅɓɱḂJoPŻẆṃṛḂ¹ẓð[1eƘ2T⁶ḟɼ7P~©ṚṙE8RƒṬẹLœẇẸịøḷ*⁾²ÄƓy€VƈɱNSẏẓѶpƲḞḅX⁹ọaœ<aỴTĠ^ðƑṙẊḅOḥŀG4ị¤ÑėÐịʠɗ=YṚċẋĠżẉịṪṁtḳṪ{ṬṃıızD/ĊvȤpḣðСþfÞ⁶ỵỊµṅḷÄ÷Vẇ\Ạ$-§OẠn^ȯfẎlḊd⁶ni¥ẓɱn¶’ṃ“¶><-][+”¤⁷

Tam bir program.

Çevrimiçi deneyin!

Nasıl?

Neredeyse bütünüyle mevcut en iyilerin sadece sıfır olmayan tek bir iz bırakan izolatörlerde sıkıştırılması ve takip bandı hareketleri kaldırılmış durumda. BF programlarının bir alt kümesini, bu naif programı yenecek en kısa çözümleri sonlandıracakları ve verebilecekleri şekilde değerlendirmenin bir yolu vardır. Bunu daha akıllı patern tabanlı veya faktoring oluşturma programı ile yenmenin bir yolu da olabilir.

⁾+-ẋ€Ɱ14ZY€U0¦j“ ... ’ṃ“¶><-][+”¤⁷ - Link: no arguments
⁾+-                                - list of characters ['+','-']
    €                              - for each:
   ẋ                               -   repeat
     Ɱ14                           -   mapped across [1..14]
        Z                          - transpose
         Y€                        - join each with newline characters
           U0¦                     - reverse the rightmost
                                   -   (now we have ["+\n++\n+++\n ...","... \n---\n--\n-"]
                                ¤  - nilad followed by link(s) as a nilad:
               “ ... ’             -   a really big number compressed as base 250
                       “¶><-][+”   -   list of characters ['\n','>','<','-',']','[','+']
                      ṃ            -   decompress - use as base 7 digits [1,2,3,4,5,6,0]
              j                    - join (the list ["+\n++...","...--\n-"]) with that
                                   - implicit print
                                 ⁷ - a newline character
                                   - implicit print

Suçluların düzeltilmesi (ve arka bant hareketlerinin kaldırılması)
Jonathan Allan,

5

SuperMarioLang , 231 + 32894 bayt

)
))++>(>+)*>[!((&(>[!*>-)-[!([!
===+"="==="=#===="=#="====#==#
+++<(       )    !+< !  ( <
+===+ (   - .    #=" #===="
>[!+( (   !(<
"=#++ (   #="
- (++ !.))    )))            <
) +++ #======================"
+ +++
+ ++!
!+<=#
#="

Çevrimiçi deneyin!

Bu kesinlikle daha fazla golf oynayabilir, çünkü çıktı beyin fırtınası için en temel olanıdır, ancak bu cevabı yazmam bütün günümü aldı (üç çocuğum boş zamanımı bırakmam için biraz zaman bıraktı) ve en azından başarmayı başardığım için gurur duyuyorum bu.


4

Python 2,70 + 8428 = 8498

-2 Bayt A__ sayesinde!
-20 Bayt Jonathan Allan'a Teşekkürler!
-229 Bayt, sayıyı
16'dan 9'a değiştirerek ikinci hücreye -1000ish bayt koyarak

p='+'
i=1
exec"print[p*i,i/9*p+'[>'+p*9+'<-]>'+i%9*p][i>20];i+=1;"*255

Çevrimiçi Deneyin!

Çıktı


4

Ruby 271 + 5363 = 5634

1.upto(255){|n|r=n>(o=n>128?256-n:n)??-:?+;puts o>20?(s=o.to_s(i=(3..9).find{|i|!(s=o.to_s i)[1..-2][s[0]]}).bytes;s[-1]+=s[0]%8;(s[1,9].reverse.map{|c|(c-=s[0])<0??-*-c:c>0??+*c:?-}*?>+'[>'+?+*(s[0]%8)).tr(n>o ?'+-':'','-+')+'[-<'+?+*i+'>]<<]'+(s[-1]>s[0]?'':?>+r)):r*o}

Çevrimiçi deneyin!

Her değeri, başka bir yerde ilk basamağını sıfır içermeyen en küçük tabana dönüştürür ve ardından o tabandan dönüştürür. 127'den büyük değerler ters olarak hesaplanır.


Ambalajsız, 221 + 5888 = 6109

1.upto(255){|n|puts n>20?(s=n.to_s(i=(3..9).find{|i|!(s=n.to_s i)[1..-2][s[0]]}).bytes;s[-1]+=s[0]%8;s[1,9].reverse.map{|c|(c-=s[0])<0??-*-c:c>0??+*c:?+}*?>+'[>'+?+*(s[0]%8)+'[-<'+?+*i+'>]<<]'+(s[-1]>s[0]?'':?>+?-)):?+*n}

Sarılmayan hücrelerde, aynı yaklaşımı kullanarak.

Çevrimiçi deneyin!


4

JavaScript (Node.js) , 691 + 3627 = 4318

@ Neil'in Kömür Cevabı ile aynı yaklaşımı kullanarak ve bu nedenle @ JonathanAllan'ın Jelly cevabına da dayanarak kullanın .

_=>(a=require('zlib').inflateRawSync(Buffer('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','base64'))+'')+`
--[>-<--]>-
`+[...a.split`
`].reverse().map(s=>s.replace(/[-+]/g,c=>c>','?'+':'-')).join`
`

Çevrimiçi deneyin!


2

Resmi olmayan fıçı 16 + 32895 = 32911 bayt

Bir golf dili için temel bir çözüm. Bu düşünebildiğim en basit şey.

ÿï((:|\+$;)_\
')

Çevrimiçi deneyin!


Durduğu gibi, bunun tersine dönmesi gerekiyor (aşağıdan çıkış yapıp
Jonathan Allan

1
OP sorgunuza cevap verene kadar bekleyeceğim.
Bir

2

Ruby 23 + 32895 = 32918 bayt

256.times{|n|puts ?+*n}

Temel olarak. Bu, aklıma gelen en basit çözüm.


0Muhtemelen olmalı 1(Sıfır için de çıktı isteyip istemediğim sorulsa da)
Jonathan Allan

-3: 256.times{|n|puts ?+*n}çok fazla önemi yok ...
primo


1

Scala , 95 + 16639 = 16734 bayt

object M extends App{(1 to 127).map(x=>println("+"*x));(0 to 127).map(x=>println("-"*(128-x)))}

Çevrimiçi deneyin!

Belli ki kazanmayacak olan basit bir cevap. Yalnızca -operatörün (baytları azaltarak) 255'e geri döndüğü gerçeğini kullanır .


Scala en iyi golf dili değil, ancak burada tipik golf kuralları altında izin verilen bütün bir uygulama yerine sadece bir yöntem yazarak birkaç bayttan tasarruf edebilirsiniz. Ek olarak, sonucu yazdırmak yerine, sadece listeyi geri döndürmek daha verimlidir. Çevrimiçi Denemede, alakasız baytları saymadan yönteminizi çalıştırmak için tüm nesne genişletme uygulama türü öğelerini üstbilgi ve altbilgiye koyabilirsiniz.
Ethan

Bu durumda, 36 bayt ile tasarruf edebilirsiniz: tio.run/…
Ethan

1

05AB1E , skor: 4848 (1219 bayt) kaynak kodu + 3629 bayt çıkış)

'+14L×»Â'+'-:•тômG‚ΣP;e3₃ìèÕwƵÜè-½;¨Z±µΛé±V™NkKJžšë₅ušΘ(M₄+ܧ‘мoÕθÚzÇYï#J×¢θýει™₃tQØËв¿U®GƵ´GZ’¯ε¨jjØÛλÄ₅X∍µxθÆvËjS¹∊f˜«VÐZ<ÇĆ’Š2&ØÍäßÍĆlΓV₆ëßê©Œ‡ÛiyĆ=*÷Í´¢‹j,3½íµ'ž4‘û29ôãζм§x…1P|ÛéΣ=~çš5Œ±€Ô“q òǝ?ó¬Æí5¢G‘°êóÿв4LFÍK&zζb2Ó∍æïι8₃4XƵÜÙôt₁‘,Ö…6₅ÞαÇø†c÷Ûλ9…F;ĆA¬iмéλ8ä¶×ƶYΔè¡aû
v=M„ûñ]C₅Õ¶Þ*Ú`Úˆ/₃UιΩW¾eTεvˆ£nYõ¶S¼ÿ{õN9Ω¨£1w‚Ï”Xd;¹OýŒéDнĀvÌ–d=±ΛΣÃÊîD—GR>~ºD‹K¥‘l×yz.éFE1Í©ØM/ƒœOαU‘KΓO‰∍Aм‚œ2нƶþøÌ×¼āHgΩC'Λê¡-߅̾Ā–м–¿<₂δ¡áтgö¬Í~θFíнā‹°ü8[À(xï¸.›*W©¹º₅ÇмδçΛλÉFÕL4†EćÛ´ǝ{тÀ¯†ª™ŽćÉuè¬ƵÀSìFÙη¶1ȸ֛GÜlRv˜jy5mfè∞_åEηŠyo‡xÐ/™¥òÜ#Áx#м6r&₁cÿX۬ƄÌƵ₅∊бγ²Θj∞;6o·¼ýŠΩÚò›c[>ö₅¥=—ªÃ±¿ecSBÐ6Ê!ú¢E¡âìþ߃¿;Ò;„Xoƶ*∍Σǝñ"Tµ†8s®βµ4ìA|«÷γt³+<B¤špTp¸ï7Ëo[>–îiTôó檂?É8zн²ìC1ãl6+ƶå4sЌÚb(°·8ˆ´ˆŸ²ÚÌY3ŸËîÿ‘àUāçh9im„ÝĆm3ŠC×η“åX¨₄|ëPô3O<6Mþ'Ì-s{e`ζQΔ¹œP@l%¥‹èδcsÎcΘÂþ®i₅∞ð¡@`¸¿…BÎN2н>g;ΛSníÐ^Rαθ₆ΣÕ3¹ÐÔCfrQ¦7¨gfŒ|v||þÚÜvz≠pệT˜ǝ=ß·„®¡xи™#?†-Aʒ2åβ₃A¬Ão6ºтõ}Ë.&QηÕ~Δ4€@-5î^a̬.»Èõ4áL¾ò¥n
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₆ŽsLвQ”ÖÜvGõƶiò{÷ÀPy/‹θÑè}¿Á5º˜¯sëØSËƶK_ÍyX∊3Øå4IOθ I+∊ÌñÙçakÞŸŒʒ椱,mεjæ‰O%<ÅtƒVöV=³ÇƶƒC¬‰xðȬM4Ïóä)∍Êfa§õØÂ,“X¾₆₄Ö¦ÈJµÿmȾÎ∍=¡YнŸV!¨J£ü|&¢cUg4e±6w™¼“fÊÙ ,Ž|šP·ùèd}ãŠÅ#GγhYÇN´¼ÁÌMGʒ§Æ1¸‚Δ:j7ΩƵAqá¢<äò´Θ•“-+>][<
“ÅвJs»

Çıktı limanıdır @Neil'in Kömür cevabının , bu yüzden onu da güçlendirdiğinizden emin olun!

Çevrimiçi deneyin.

Açıklama:

'+                      '# Push a "+"
  14L                    # Push a list in the range [1,14]
     ×                   # Repeat the "+" that many times as string: ["+","++","+++",...]
      »                  # Join these strings by newlines
       Â                 # Bifurcate it (short for Duplicate & Reverse copy)
        '+'-:            # Replace all "+" for "-"
•тôm...ò´Θ•              # Push compressed integer 18302226724133383998250107335646038608225046109581810887431446835557987256955354954509163336111304735021044106950262344427892947550841899099611054599885158084492762836812161427050275372983896356189873217422270707048679161884382784973706990123491668808316983431947218815813441209357230471947480445527653281307616982417034289994948061000591427114479102114229222423495882782326672492922269629953210111953959859902281658658439835047182218017657439552630082372181376525413759195763958434475193943488791777228373958162363214252781530693967200164833437881609482421594458966138936433283311419810119896020066082377462298326514652481546557215787238749539873039910952003326954299252586309028025200870623285261199142261807190771369911425142504345271105103035478661301795311828767848235694787283635190364512722037791815037475799545052058894119573664059402985074146226606245848663046901585891882552845134633210352731812274795773552227786140415336764040421001184646630833787917147474644077938952053956874031774587527717793206158934471919975714697099518810712871948398923739276321843455690477328633199064849928974478179435369018512187592263559949835435473650276637191671401061097340919482725489354844550472281209666291367830643727358624135371626379451084552903536762775083445643853806852513122856150361979701049267928548063465967555886420646898485890108420374549485423234679327438138302730692296669063696581268627535131608200283731275951916433249161017999011290215932205767570177905442947203826039265793694687731078121685736352831955773450680945121984143563963149079990880719573067270197057276219243821370885160589340870891346257233778661271435191351926058080186177296974642815621539128350975752011448032262905976766027084285390087966682234081285502231618383962136055937741758125210487103109250885525370548106186539295203084216890820183575639032509902729248016346072449636148699098049659529168757116706057794418245039559549604674043961198447420311513558044229534569679723496972989178091506175996419296780639212192856671882116470677803387276814324094247508763467887301684211112080372036284371596072213153957411329532202432808677726223798116216330275138697515009114689489577370759238857602332613821627667530873656034962827810927061440822808985980383150080767015247752949877604372029666921293343149038246728649404223795601960991061986482063744094221616603849190547637439116347239768975065217383194655478092271791087679802480625740835053103772632489195507735140119501008503485917456615266596210333924964188989678201446160111091052524780358620148464886929989973412559470628329156848340802659185674541202787279386158230228148429451357621709967247567009904339076971643378255946241011579618610095231079053137553024558887196808709177094386352264708730475553352082713138948975317023830903305435434148828341201637230241697870602236452176330225025183518037443992277303117971849493548326433875
           "-+>][<\n"    # Push string "-+>][<\n"
                     Åв  # Convert the integer with this string as custom base
                       J # Join all characters together to a single string
s                        # Swap so the triangle of "-" we created it as the top of the stack
 »                       # Join the strings on the stack by newlines
                         # (and output implicitly as result)

Madenin (bölümün bu 05AB1E ucunu bakın ne kadar büyük tamsayılar sıkıştırmak için ) anlamak için •тôm...ò´Θ•olduğunu 183...875.

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