Справжнім скарбом були цифри, які ми зробили по дорозі


23

Ваше завдання - написати програму, функцію або фрагмент (так, фрагменти дозволені), які просто виводять ціле число. Однак ви повинні бути в змозі розділити ваше подання на префікси, які також створюють окремі цілі числа. Ви не можете використовувати жодні байти, що з’явились у попередніх префіксах. Наприклад, у нас можуть бути префікси:

1         # 1    (Now we can't use 1)
1-6       # -5   (Now we can't use - or 6)
1-6/3     # -1   (Now we can't use / or 3)
1-6/3+0xA # 9    Final submission

Правила

  • Ваша мета - створити, щоб спробувати створити найбільш унікальні цілі числа, зберігаючи їх близько до нуля.
    • Система підрахунків балів ((number of unique integers)**3)/(sum of absolute values), де чим вище ваш бал, тим краще. Наведений вище приклад оцінки (43)/(1+|5|+|1|+9)=64/16=4 .
  • Повинно бути не менше двох унікальних цілих чисел (без ділення на нуль!)
  • Будь ласка, відформатуйте свою відповідь аналогічно:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Згадайте, якщо ваше повідомлення складається з фрагментів, які оцінюють за значенням, функціями або повноцінними програмами.
  • Перелічіть кожен із префіксів та (необов'язково) пояснення, як вони працюють.

2
Здається, що підсумковий рядок подання замінив 3 на 2 (я не редагував це, щоб виправити це, якщо я зрозумів неправильно)
trichoplax

4
Мені здається, ви можете отримати майже оптимальну оцінку в Python, написавши print 123-len(open(__file__).read())після #цього один байт, який не використовувався досі (крім деяких контрольних символів), з префіксами, що рубають ці байти. Налаштуйте 123 під будь-який центр, діапазон виходів становить близько 0. Для кращого результату частину програми без коментарів можна переписати, використовуючи лише 7 різних символів, які є універсальними для Python. Чи все це дійсно?
xnor

1
@JoKing Я включаю мінуси, коли різниця негативна.
xnor

1
@JoKing Цей метод також можна виконати мовами, де рядок можна записати без завершальної цитати.
xnor

2
Я не можу повірити, що мало хто з інших людей помітив це і підтримав попередній коментар. : PI щойно помітив, що @trichoplax теж помітив друкарські помилки.
Пітер Кордес

Відповіді:


19

TI-BASIC (TI-84 + С), 2383/14161=952.0

118-length("+×/ABCDEFGHIJK...

256 байт TI-BASIC розбивається таким чином:

  • 240 байт - однобайтові лексеми, які можна цитувати
  • 11 - початок двобайтних жетонів
  • 2 не використовуються
  • Решта 3 є ", символи і символ нового рядка , які порушують строки.

Рядок містить усі 237 допустимих байтів у першій категорії, які ще не є 118-length( .

Програма обчислює 238 різних цілих чисел, починаючи від 118 вниз до -119. Таким чином, оцінка становить

2383n=119118|n|=952.


1
Чи не повинно рядок закінчуватися "який байт вже використаний? Або чому б ні? Я трохи збентежений цим і в інших відповідях, здається, це трапляється, напевно, я щось пропускаю"
findusl

2
@findusl TI-BASIC має тенденцію до автоматичного заповнення дужок та рядків для вас. щоб ви могли написати mean({1,2,3, що б завершити mean({1,2,3}). рядки також завершені: Disp "Hi!завершує до Disp "Hi!". Тому остаточна кінцева пропозиція не потрібна
Conor O'Brien

13

vim, 14³ / 53 = 51,77

i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7

Спробуйте в Інтернеті! (усі префікси)

Зламатися:

i1                  1
0                   10
<esc>X              0
<c-x>               -1  <c-x> decrements
r2                  -2
hx                  2   hx is X, but we've already used that
<c-a>               3   <c-a> increments
P                   -3  paste the - from two steps ago
a<del>4             -4
<c-c>d^             4   <c-c> is an <esc> alternative
s5                  5
<c-o>I-<c-o>$<c-o>  -5  <c-o> allows a single normal mode input
C6                  -6  trailing <c-o> from previous line used here
<c-h>7              -7  <c-h> = backspace

ви не використовуєте <байт? Або це заповнювач для іншого байта?
findusl

1
@findusl ні - <esc>являє собою ключ втечі та <c-char>представляє ctrl + char
Doorknob

11

Lenguage , 256³ / 16384 = 1024

Вихідний код рішення можна описати так:

183662808731984092                                         0x00 bytes
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3461439604291557962702078999341076235513663562296408699075 0xFF bytes

Кожен запуск однакових байтів у поєднанні з попередніми запусками становить префікс, а повна програма - всі рядки, описані вище, об'єднані.

Це оптимальний бал. (Програми друкують всі номери від -127 до 128 включно.)

Щоб створити цю відповідь, спершу я скопіював тут відповідь Java для створення програм, що ведуть мозку, для друку рядка. Я керував цим mainфункцією:

public static void main(String[] args) {
    for (int i = -127; i <= 128; ++i) generate(""+i);
}

Потім я запустив цей сценарій Ruby, щоб створити текст, прочитаний тут:

#!/usr/bin/ruby
nums = File.open('f').map do |line|
    line.chomp.sub(/^\+/, '>+').chars.map{|ch|
        '+-><.,[]'.index(ch).to_s(3).rjust(3,?0)
    }.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"\n"

Найбільша програма - це 3594571896764310192036774345469579167648804468538578264427 байт завдовжки (і відбитки -90).


1
Реквізит для швидкості та практичного вирішення проблем!
lirtosiast

8

(1113)/2530 рік=540.5656126482213

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)[!+[]+!+[]+[+[]]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]])().$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}

(Завдяки цій цікавій утиліті для конверсії.) Вищезазначене ефективно оцінює наступний код:

(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}

... який створює проксі, який перетворює довжину імені властивості в число. Перший префікс закінчується одним символом після .та використовує лише 6 унікальних символів, виключаючи .$. Наступні 63 фрагменти додають 63 інші однобайтові символи, які дійсні в іменах властивостей JS об'єкта. Наступні 44 фрагменти додають двобайтові дійсні символи, що складаються з невикористаних байтів. Кінцеві скибочки додати &'', |""**""і ,~{}. Перший зріз повертається -2, другий 2, потім -3, потім 3, -4, 4, -5і т.д. аж до останніх трьох скибочок , які 0, 1і -1.


4

43/4=16

Для цього ми використовуємо фрагменти мозкового спалаху, тому ми переймаємось значенням, а не тим, що вони роблять.

Повна програма:

<>(())[[][][]]{}

Префікси та вихідні дані

 0 : <>
 1 : <>(())
-2 : <>(())[[][][]]
-1 : <>(())[[][][]]{}

Спробуйте в Інтернеті!

Пояснення

Це, мабуть, найкраща відповідь на роздуми, оскільки ми використали всі символи, про які піклується Мозок, і ми не можемо об'єднати числа ближче до нуля.

Перший фрагмент є <>, <>змінює стек , але в іншому випадку має значення нуль, тому ми починаємо з нуля. Звідси ми додаємо, що (())це робить одне. Ми могли би щойно зробити (), тому що це теж одне, але якщо ми зараз натиснемо його, ми зможемо згадати його згодом за допомогою {}та []. Це майже єдиний спосіб {}і []може створити цінність, тому ми це зробимо. Оскільки на стеку є одна річ, []вважається 1, тому ми віднімаємо 3 із загальної [[][][]]величини, приводячи нас до -2. Нарешті, ми використовуємо {}для того, щоб вискочити той, який ми натиснули раніше, довівши нас до -1.


4

(113)/32=41.59375

Після двох годин проб і помилок , я , нарешті , отримав всі цілі числа від -4до +57).

0^cmp('','_')&3+1/~9*2-[-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]][[]==[[]]]|6<<7<<7>>8>>8and 5

Це фрагмент із такими фрагментами префікса:

# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/4) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
#  0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True

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1
Ви можете змінити 2*2на 22&7(-3 і 6 замість -3 і 4), щоб виправити повторення *.
Кевін Кройсейсен

@KevinCruijssen дякую за пропозицію.
ов

Ви можете звільнити 'і _використовуючи cmp(cmp,())замість cmp('','_')Якщо це корисно.
Пшеничний майстер

Ви також можете зробити розділ списку трохи коротшим і набагато простіше зрозуміти, виконуючи це --[-1-1-1-1-1], але це не дуже покращує ваш результат.
Пшеничний майстер

3

Партія, 176³ / 7744 = 704

@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖרÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

Перший префікс включає двокрапку і виводить число -88. Подальші префікси виводять висхідні числа та кінцеві результати програми 87.


3

Pyth, (2493)/15500=996,0

-C\|l"\x01\x02\x03\x04\x05\x06\x07\x08\t\n\x0b\x0c\r\x0e\x0f\x10\x11\x12\x13\x14\x15\x16\x17\x18\x19\x1a\x1b\x1c\x1d\x1e\x1f !#$%&\'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstuvwxyz{}~\x80\x81\x82\x83\x84\x85\x86\x87\x88\x89\x8a\x8b\x8c\x8d\x8e\x8f\x90\x91\x92\x93\x94\x95\x96\x97\x98\x99\x9a\x9b\x9c\x9d\x9e\x9f\xa0¡¢£¤¥¦§¨©ª«¬\xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖרÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

Усі виділені символи є одиничними байтами - я представив програму в цій формі для зручності копіювання та вставлення. Це посилання на повну програму без уникнутих символів; як видно, є багато заповнювачів для використовуваних недрукованих символів.

Використовуваний підхід схожий з відповіддю TI-BASIC ліртосіаста , хоча я написав свою відповідь перед переглядом інших записів. Шнур працює наступним чином:

-C\|l"...   
  \|        "|"
 C          Character code of the above, i.e. 124
-           Subtract from the above...
    l"...   ... the length of the trailing string - the closing quote is inferred

Рядок складається з кожного однобайтового символу, відмінного від 0 (NULL), 127 (backspace) та символів на екрані котла (-C\|l" ). Таким чином, кожен символ рядка може бути видалений по одному, щоб створити рядки довжиною 248 до 0. Ці програми дають результати в діапазоні [-124 - 124] з сумою абсолютних значень 15500.

Код був створений за допомогою цієї програми :

+J"-C\\|l\""sCM-U256+CMJ,Z127   
 J"-C\\|l\""                    Assign the boilerplate code to J
                     CMJ        Get the character code of each character in J
                    +   ,Z127   Append 0 and 127
               -U256            Get values in range [0-256) which are not in the above
             CM                 Get each character with code points in the above
            s                   Concatenate into string
+J                              Prepend boilerplate

І ця кількість, згенерована з кожної з усічених програм, була знайдена за допомогою цієї програми :

sm.a.vd>._+J"-C\\|l\""sCM-U256+CMJ,Z127 5   
          +J"-C\\|l\""sCM-U256+CMJ,Z127     The full program string, as above
        ._                                  Get the prefixes of the above
       >                                5   Discard the first 5 (those prefixes which don't include all the boilerplate code)
 m                                          Map each of the above to:
    .vd                                       Evaluate as Pyth code
  .a                                          Take the absolute value of the above
s                                           Take the sum

3

Желе , 984

LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑרŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘

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Ви можете побачити префікси (розділені подвійним новим рядком) тут .

Надруковані номери в порядку:

0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123

2

Пітон 2, 113/53=25.1132075472

015-8*2+6/3&9%7if[]else 4^ord('\b')

Префікси:

0                                   # 0

01                                  # 1

015                                 # 13 (octal)

015-8                               # 5

015-8*2                             # -3

015-8*2+6                           # 3

015-8*2+6/3                         # -1

015-8*2+6/3&9                       # 9 (-1 bitwise-AND 9 = 9)

015-8*2+6/3&9%7                     # 2

015-8*2+6/3&9%7if[]else 4           # 4

015-8*2+6/3&9%7if[]else 4^ord('\b') # 12 (4 bitwise-XOR 8 -- OR would also work)

Не оптимально, але найкраще, що я міг особисто знайти саме в цій структурі.


2

Пітон 2 63/9=24

False-False**0&1+[[]<[[]]][[]<[]]^~3>>2

Ось префікси та що вони виводять

 0 : False-False
-1 : False-False**0
 1 : False-False**0&1
 2 : False-False**0&1+[[]<[[]]][[]<[]]
-2 : False-False**0&1+[[]<[[]]][[]<[]]^~3
-3 : False-False**0&1+[[]<[[]]][[]<[]]^~3>>2

2

Чисто , 403/ 400 = 160

import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \\ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`

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Визначає, nщо виробляє -20, -19, ..., 18, 19залежно від довжини типу G.

Найменший префікс Gвиглядає :: G = G, передуючи різному імпорту та визначенням. Виглядає
найбільший префікс , подібний до цього.G:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`


Я думаю, що я щось пропускаю в цьому виклику, але ... ця відповідь (і жменька інших) повторно використовує "байти, що з'явилися в попередніх префіксах", чи не так?
Blackhole

4
@Blackhole Як я (і я припускаю, що інші) це зрозуміли, ви можете використовувати байти кілька разів у тому ж префіксі, але не в наступних.
Οurous

2

Japt ,2453/15006=980.0163268026123

_Ê-122}$($"	
 !#%&')*+,./03456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^`abcdefghijklmnopqrstuvwxyz|~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉËÌÍÎÏÐÑÒÓÔÕÖרÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

Спробуйте в Інтернеті!

Як і інші записи, використовує довжину рядка. Символи в початковому фрагменті, а також \r(textareas не грає добре з цим) та {(символ інтерполяції рядка) виключаються з рядка.

Перша програма полягає в тому _Ê-122}$($", що кожна програма на один символ довше, за винятком випадків додавання \\.


Не зовсім впевнений, що це дійсно, не перевіряли, що в рядку немає символів фрагмента
лише ASCII

Я збирався запропонувати заміну 122з , #zале це не поліпшить ваш рахунок.
Кудлатий

1

Пітон 3, (63)/12=18

0x1-2+3|True*~4

Це фрагмент.

Індивідуальні фрагменти префікса:

0          # gives 0

0x1        # gives 1 (hex syntax)

0x1-2      # gives -1 (1 minus 2)

0x1-2+3    # gives 2 (-1 plus 3)

0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)

0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)

Бітові оператори Python розглядають негативні цілі числа як доповнення двох з необмеженим 1s праворуч, тому -5трактується як ...11111011, що при ORed з ...000010дає ...11111011незмінний.


1

Excel, (63)/12=18

=0              0
=01             1
=01*2           2
=01*2-5         -3
=01*2-5+8       5
=01*2-5+8/4     -1

Prefix Проблема не добре піддається структурі формули Excel


1

Рунічні Enchantments , 117 3 /3422 = 468.034

>"ul'B-$;	 !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijkmnopqrstvwxyz{|}~€

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І префікс .

Відбитки [-58,58] включно.

Використовує аналогічну тактику, як і інші записи, використовуючи поступово довший рядок, отримуючи його довжину, віднімаючи B(десяткове значення 66) і друкуючи результат. Використовує символ, що не друкується (відображається в блоці коду, тому що, мабуть, це те, що &#128;оцінюється в HTML), щоб використовувати значення байтів0x80 (і споживаючи інакше непридатні 0x00) та набрати ще одне значення. Тільки байтові значення до 127 , які не можуть бути представлені будь-яким способом , є 10і 13, тому що вони змушують програму два рядки в висоту і не бути правильно прочитаний покажчик інструкції.


1

Деревне вугілля , 95³ / 2256 = 380.04211 96³ / 2304 = 384

I⁻⁴⁸L´´ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~

Перший префікс - перші 7 байт і виводить число 47. Спробуйте в Інтернеті! Подальші префікси виводять спадні числа та кінцеві результати програми -48. Спробуйте в Інтернеті! Редагування: Збільшена оцінка на 3,995789 завдяки @ ASCII.


Я думаю, ви повинні мати можливість уникнути символу, який не є ASCII, на початку або в кінці рядка? (використовуючи символ втечі, той, що схожий на перевернутий могильний акцент)
лише ASCII
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