Standart altı sapma

Question

Bir popülasyonun ortalaması \$(x_1,\dots,x_n)\$\ olarak tanımlanır$\bar x=\frac1n\sum_{i=1}^n x_i\$. (Düzeltilmemiş) standart sapma nüfusun olarak tanımlanır \$\sqrt{\frac1n\sum (x_i-\bar x)^2}\$. Popülasyonun ne kadar dağınık olduğunu ölçer: büyük bir standart sapma, değerlerin birbirinden çok uzak olduğunu gösterir; düşük bir standart sapma, yakın olduklarını gösterir. Tüm değerler aynıysa, standart sapma 0'dır.

Negatif olmayan tam sayıların (boş olmayan) listesini girdi olarak alan ve standart sapmasını çıkaran bir program veya işlev yazın. Ama puanlama kuralını kontrol edin, çünkü bu kod golfü değildir !

Giriş çıkış

Giriş / Çıkış esnektir. Cevabınız en az 2 ondalık basamağa kadar doğru olmalıdır (yuvarlama veya kesme). Girişin yalnızca 0 ile 255 arasındaki tam sayıları içermesi ve boş olmaması garanti edilir.

Puanlama

Puanınızı hesaplamak için kodunuzu tam sayı kod noktalarına dönüştürün (ASCII veya diliniz için standart olan herhangi bir kod sayfasını kullanarak) ve standart sapmayı hesaplayın. Puanınız, kodunuzdaki bayt sayısının standart sapma ile çarpımıdır. Daha düşük puan daha iyidir. Bu nedenle, aynı zamanda (a) kısa ve (b) yakın kod noktalarına sahip karakterler kullanan kodu hedeflemelisiniz.

İşte puanınızı hesaplamak için çevrimiçi bir hesap makinesi (ASCII kullandığınızı varsayarak).

Test durumları

Input              | Output
77 67 77 67        | 5
82                 | 0
73 73 73           | 0
83 116  97 116 115 | 13.336

Yerleşikler hakkında bir uyarı: dilinizde yerleşik bir yer varsa, sorun değil (ve yalnızca bir karakter kullanıyorsa sizin için iyi!) Ama kullandığından emin olun \$n\$ve değil \$n-1\$ formüldeki payda olarak yoksa cevabınız geçerli olmayacaktır.

Answer 1

MATL , puan 65.30697

tYmhZs

Çevrimiçi deneyin! Veya tüm test durumlarını doğrulayın .

Nasıl çalışır

ZsVarsayılan değeriyle (1 giriş, 1 çıkış) yerleşik işlev , düzeltilmiş standart sapmayı hesaplar :

\$\sqrt{\frac 1 {n-1}\sum (x_i-\bar x)^2}\$

Düzeltilmemiş standart sapma 2-girişi versiyonu ile elde edilebilir Zs: 1&Zs, burada 1ikinci giriş aracı düzeltilmemiştir olarak. lveya puanı düşürmek Tyerine kullanılabilir 1, ancak &diğer karakterlerden çok uzaktır. 2$veya H$yerine kullanılabilir &, ancak $daha da uzundur.

Bu nedenle, Zsgirdideki (düzeltilmiş standart sapma) varsayılan sürümünü , ortalaması eklenerek kullanmak daha iyidir . Bu, girdi uzunluğunu artırır ve payda 1katkıda bulunur 0, bu da düzeltilmiş standart sapmanın düzeltilmemiş olmasına neden olur.

t     % Implicit input: numeric vector. Duplicate
Ym    % Mean
h     % Concatenate the input vector with its mean
Zs    % Corrected standard deviation

Answer 2

J , 19 bayt, Puan 119.8249

- ~ 1 Bubbler sayesinde

0x23 ile 0x2F arasında çoğu karaktere sahip olmaya çalışır #$%&'()*+,-./, :biraz daha uzakta olur.

(+/%$)&.:*:&(-+/%#)

Çevrimiçi deneyin!

Nasıl çalışır

(+/%$)&.:*:&(-+/%#) (-+/%#) x - sum divided by length *:& and squared (+/%$)&.:           mean of that
      &.:*:         reverse square -> square root

Answer 3

Google E-Tablolar, Puan 142.6885

=STDEVP(F:F

Google E-Tablolar, parantezleri otomatik olarak kapatır ve Fgiriş sütunu olarak kullanılması standart sapmayı en aza indirir. Bu, Excel'in düzeltilmemiş standart sapmasına göre bir bayt kazandırır, çünkü Excel STDEV.Pbunun yerineSTDEVP

Answer 4

R , 34 bayt 24 bayt, puan 789.5923 723.4687 722.6112

sd(c(scan()->J,mean(J)))

Çevrimiçi deneyin!

Düzenleme: popülasyon sd'yi ( burada bulduğum ) hesaplamak için daha kısa bir formüle geçildi , bu artık yalnızca önceki sürüm için aşağıda ana hatlarıyla verilen golflerden en iyi değişken adını seçmekten fayda sağlıyor.

Düzenleme2: Robin Ryder sayesinde puan 0,8575 azaldı

(Önceki) ungolfed kodu olduğunu idi: x=scan();sqrt(mean((x-mean(x))^2))(1104.484 bir puan almış olan)

Bundan, sıralı puan artırıcı golfler:

• x=scan();`?`=mean;sqrt(?(x-?x)^2)= mean()tek karakterli tek operatör olarak yeniden tanımla (puan 983.8933)
• x=scan();`?`=mean;(?(x-?x)^2)^.5= Değişimi sqrt()için ()^.5(skor 918,6686)
• H=scan();`?`=mean;(?(H-?H)^2)^.5= Değişimi xiçin Holan ve böylece standart sapmanın azaltılmasına yardımcı programın ortalamaya en yakın kod noktası değeri (801.4687 puan)
• I=scan();`?`=mean;I=I-?I;(?I^2)^.5= x-mean(x)parantez sayısını azaltmak için (bunlar ASCII aralığının en uç noktasındadır ve dolayısıyla standart sapmayı arttırmak için) önce ayrı ayrı hesaplayın ve değişken adını olarak yeniden ayarlayın I. Bu kod uzunluğunu 2 karakter artırsa da puanı 789,5923'e düşürüyor.

---

R + multicon , 15 bayt, skor 273.5032

multicon::popsd

Kitaplıktan yerleşik popsdişlev kullanan önemsiz çözüm multicon. TIO'da
kurulu değil, ancak bu kodu kopyalayıp yapıştırarak rdrr.io'da deneyebilirsiniz :

x=c(67,77,67,77)  # data
multicon::popsd(x)

Answer 5

Wolfram Dili (Mathematica) , Puan 537.0884

A@((#-A@#)^2)^.5&;A=Mean

Çevrimiçi deneyin!

@att 17.6142 puan tasarruf etti

Answer 6

Python 3, Puan 680.5175

Golfçü çözümün en iyi olmadığı yerde. Yerleşik olmayanların daha iyi olabileceğinden şüpheliyim ama yanılıyor olabilirim.

import statistics;statistics.pstdev

Çevrimiçi deneyin!

Python 3, Puan 733.6818

from statistics import*;pstdev

Python 3, Puan 798.5587

__import__('statistics').pstdev

Answer 7

05AB1E , Puan: 531.168 431.516 360.278 ( 10 15 14 bayt )

Osg/nsn-Osg/(t

05AB1E kodlama sayfasını kullanır . Kullanılan karakterlerin kod noktaları vardır [79,73,103,47,110,73,110,45,68,79,73,103,47,40,116].

Çevrimiçi deneyin veya tüm test durumlarını doğrulayın .

Açıklama:

                # Get the arithmetic mean of the (implicit) input-list by:
O               #  Summing the (implicit) input-list
 I              #  Push input-list again
  g             #  Pop and push its length
   /            #  Divide the sum by this length
                #  (which gives a better score than the builtin `ÅA`)
    n           # Square it
     I          # Push the input again (better score than `s` or `¹`)
      n         # Square each value in the input as well
       -        # Subtract each from the squared mean
                # Take the arithmetic mean of that list again by:
        O       #  Summing it
         Ig     #  Push the input-list again, and pop and push its length
           /    #  Divide the sum by this length
            (   # Negate it
             t  # And take its square-root
                # (after which the result is output implicitly)

Answer 8

JavaScript (ES7), Puan  1359 1228  1156.077

@ Edc65 sayesinde 72 puan kazandılar

D=>D[F='map'](C=>B-=(C+E/A)**2/A,D[F](C=>E+=--A?C:9,A=B=E=0))&&B**.5

Çevrimiçi deneyin!

Karakter dökümü

 char. | code | count
-------+------+-------
   0   |  48  |   1
   2   |  50  |   1
   5   |  53  |   1
   9   |  57  |   1
   &   |  38  |   2
   '   |  39  |   2
   (   |  40  |   3
   )   |  41  |   3
   *   |  42  |   4
   +   |  43  |   2
   ,   |  44  |   2
   -   |  45  |   3
   .   |  46  |   1
   /   |  47  |   2
   :   |  58  |   1   <-- mean ≈ 59.43
   =   |  61  |   9
   >   |  62  |   3
   ?   |  63  |   1
   A   |  65  |   4
   B   |  66  |   3
   C   |  67  |   4
   D   |  68  |   3
   E   |  69  |   3
   F   |  70  |   2
   [   |  91  |   2
   ]   |  93  |   2
   a   |  97  |   1
   m   | 109  |   1
   p   | 112  |   1

Answer 9

Arn , puan = 925.3172 655.6836 602.7985 123.2274

sdev:s

Yerleşik standart sapma işlevini kullanır. Daha ilginç bir cevap için eski cevaba gidin

Eski Cevap

Standart Sapma çok daha yüksek olacağı için sıkıştırmıyorum. Bu cevabı çok daha kısa bir yöntem bulduğum için güncelledim (14 baytta oturarak). Buraya bağlantı verin (bu, puanın başvurduğu programdır). Orijinal programı gelecek nesillerin hatırı için bırakacağım

:/(+v{:*v-(:s.mean}\)/((:s)#

Dene!

Açıkladı

$$\large\sqrt {\frac1n \sum(x_i-\bar x)^2}$$Formülü kullandım. :/sqrt öneki, :*kare önek, +v{:*v-(:s.mean}\Blokla eşlendikten +sonra (toplama) ile kıvrımlar v{:*v-(:s.mean}. vgeçerli giriştir, :sboşlukta böler (hiçbir değişken sağlanmaz, bu nedenle değişkeni _, yani STDIN'i varsayar ). Sonra uzunluğa ( #sonek) göre böler .

Answer 10

Io , puan = 1454.7164672196433912

-19.58295474318379 @ManishKundu'ya teşekkürler

method(:,:map(Z,(Z- :average)squared)average sqrt)

Çevrimiçi deneyin!

Answer 11

Jelly , (14 bayt ), puan 218.314

^{(218.31399405443526)}

+/÷LN+*2+/÷L*.

Çevrimiçi deneyin! Veya bir öz değerlendirmeye bakın .

Bayt kodu: 2b 2f 1c 4c 4e 2b 2a 32 2b 2f 1c 4c 2a 2e

Nasıl?

Saf bir program _Æm²Æm½348.47 olacaktır (her birinin ortalamasını çıkarın, her birinin karesini alın, ortalamasını alın ve sonra karekökünü alın).

ÆmKod noktaları birbirinden oldukça farklı olan ( 0x0dve 0x6d) iki baytlık monaddan kurtulmak için aşağıdakilerden birini yapmamız gerektiğini biliyoruz:

• ÷( 0x1c) kullanarak bölmek veya
• çarp, ×( 0x11) ve ters çevir, İ( 0xc6)

Ancak son baytlar da oldukça farklıdır, bu nedenle bu yanıt ÷( 0x1c) 'a yakın baytları kullanmaya çalışır .

+/÷LN+*2+/÷L*. - Link: list of numbers, A
 /             - reduce (A) by:
+              -   addition             -> sum(A)
   L           - length (A)
  ÷            - divide                 -> mean(A)
    N          - negate
     +         - add (to A, vectorised) -> [mean(A)-v for v in A]
       2       - two
      *        - exponentiate           -> [(mean(A)-v)² for v in A]
         /     - reduce by:
        +      -   addition             -> sum((mean(A)-v)² for v in A)
           L   - length (A)
          ÷    - divide                 -> sum((mean(A)-v)² for v in A)/n
             . - a half
            *  - exponentiate           -> √(sum((mean(A)-v)² for v in A)/n)

Answer 12

Wolfram Dili (Mathematica) , 31 bayt, puan 478.3451

a[a_]=RootMeanSquare[a-Mean[a]]

Çevrimiçi deneyin!

Answer 13

Kömür , 15 bayt, stddev 46.741654, puan 701.12481

Ｉ₂∕ΣＸ⁻θ∕ΣθＬθ²Ｌθ

Çevrimiçi deneyin! Bağlantı, kodun ayrıntılı sürümüne yöneliktir. Bağlantı testi durumu, kodun Kömür kod sayfasındaki bayt değerleridir. Açıklama:

         θ      Input `x`
        Σ       Summed
       ∕  Ｌθ    Divided by `n`
     ⁻θ         Vectorised subtracted from `x`
    Ｘ       ²   Squared
   Σ            Summed
  ∕          Ｌθ Divided by `n`
 ₂              Square rooted
Ｉ               Cast to string
                Implicitly printed

Standart sapma için alternatif formül, \$ \sqrt{\bar{x^2}-\bar x^2} \$biraz daha küçük bir standart sapmaya sahipken, 17 bayt alır ve bu nedenle 755.6 gibi daha yüksek bir puanla sonuçlanır.

Answer 14

Setanta , puan: 2728.508

gniomh(g){f:=0h:=0e:=fad@g le i idir(0,e){d:=g[i]f+=d h+=d*d}toradh freamh@mata((h-f*f/e)/e)}

Burada deneyin!

Answer 15

C (GCC) , ^{107 104} 99 bayt, stddev ^{25.25 \$\cdots\$25.32} 25.00, ^2702.01 puan ^{\ $\cdots\$ 2634.27} 2475.426270

Ceilingcat sayesinde 3 bayt ve 46.95288 puan kazandınız !!! Att
sayesinde 5 bayt ve 158.848632 puan tasarruf etti !!!

E;float D,G,H;float F(F,C)int*C;{E=F;for(H=G=0;E>-F;0>E?G+=D*D:(H+=*C++))D=H/F-C[--E];G=sqrt(G/F);}

Çevrimiçi deneyin!

Answer 16

Yüksek oranlarda konuşma , bir sürü bayt, 0 Skor

Sorumluluk reddi: Hala hatalar olabilir (şu anda vardır; sqrt rutininde b / = 2 bozuk)

Giriş: ASCII Karakteri olarak n, ardından n Numaralar da ASCII'ye dönüştürülür

Çıktı: Bit 15 ve 16 arasındaki ondalık nokta ile 32 bit sabit nokta değerini gösteren 8 ASCII karakteri (cevabı almak için sadece 8 ASCII karakterini 32 bit tam sayı olarak yorumlayın ve 1 << 16 ile bölün)

Çalışma Zamanı: Dürüst olmak gerekirse test değerlerini test etmedim. Çok uzun sürüyor. İlk parçayı karekök ve tüm test durumları amaçlandığı gibi çalışana kadar kontrol ettim, ancak çalışma süresi zaten bir saatten fazla sürüyor. Sqrt bağımsız olarak çok küçük sayılarda test edildi ve hala 9'un karekökü için kabaca 10 dakika sürüyor. Yani toplam çalışma süresi muhtemelen birkaç saattir ...

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herhangi bir karakter. Bu \$ 10^{7168} \$Baytlar, \$ 10^{7078} \$Gözlemlenebilir Evrenler veya \$ 10^{7091} \$sadece verileri tutmak için güneş kütlesinde kara delikler "Kod" beyin fırtınasından dönüştürüldü

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

hangi yaklaşık ~ 8000 Talimat ile devreye girer. Aslında Brainf * ck'i ilk kez kullanıyordum ve onun temelde bir turing makinesi olduğunu görmem biraz zaman aldı ama sonra bilgisayar bilimi derecem sonunda işe yaradı: D Başlangıçta, özellikle Ortalama. Ama sonra tıkladı ve tüm hesaplamayı yapmak için durum makineleri yapmaya başladım. İşte kullandığım durum makinelerinden bazıları:

Kareköke ulaşana kadar her şey eğlenceli ve oyundu; Daha küçük bir bölüm olacağını düşünmüştüm, ancak kodun çoğunu aldı. Christophe Meesen tarafından bir algoritmayı benimsedim ve bazı kısıtlamaları beyin f * ck'de iyi oynaması için değiştirdim.

Sözdizimi vurgulama için VSCode'da ve Hata Ayıklama için Visual Brainf * ck'de geliştirme yapıldı

İşte bazı yorumlar içeren kod:

,                       // Read how many input terms we will get       {n_}
[                       // Copy value twice                            {0; 0; 0; 0; 0; 0; 0; 0; 0; 0; n_; n}          
    ->>>>>>>>>>+>+
    <<<<<<<<<<<
]>>>>>>>>>>

[                       // Check if n≫0 and read n values & calculate their sum
    -<<<<<<<<<<,        // read xi; decrement running counter           {xi_; 0; 0; 0; 0; 0; 0; 0; 0; 0; n—i; n}
    [->+>+<<]           // copy xi twice                                {0_; xi; xi; 0; 0; 0; 0; 0; 0; 0; n—i; n}
    >[-<+>]             // move first xi back                           {xi; 0_; xi; 0; 0; 0; 0; 0; 0; 0; n—i; n}
    >[                  // add byte to 32bit integer sum                {•; •; xi; sum{4}; sum{3}; sum{2}; sum{1}; loopbreak; carry; 0; •; •}
        ->>>>+>+>+              // add number
        <<[>-]>[-<]>>[>]<       // set(clear) carry flag for 1st byte
        [<<<+>>+>               // apply carry flag
        <<<[>>-<]>>[-<]>>[>]<   // set(clear) carry flag for 2nd byte
        [<<<<+>>>>+>            // apply carry flag
        <<<<<[>>>-<]>>[-<]>>[>]<// set(clear) carry flag for 3rd byte
        [-<<<<<+>>>>->]         // apply carry and clear all flags
        ]
        ]                       // works by trial and error; no idea how it works exactly but it passes all tests;;;
        <<<<<<
    ]                   //                                              {xi; 0; 0_; sum{4}; sum{3}; sum{2}; sum{1}; 0; 0; n—i; n}
    >>>>>>>>>
    [->+<]              // move n one right
    <[->+<]             // move n—i one right
    <<<
    <[->+<]             // move sum one right
    <[->+<]
    <[->+<]
    <[->+<]
    >>>>>>>>
]                       // now we have {x0; •••; xi; 0; 0; 0; sum{4}; sum{3}; sum{2}; sum{1}; 0; 0; 0; 0_ ; n}
// move data for Divisionmodule
>[->>+<<]
>>>>++++++

//In—Place Divisionmodule 4 src bytes {(x1); x2; x3; x4; 0; carry; temp; rem; q; flag; n; 0; bytes_}
[
    -<<<                  Decrement bytecount
    +<<<<<<<<<            Set copyflag
    [[->>>>>>+<<<<<<]     Copy Value
        >>>>>>>>>-<<<<<<<<<]  Clear copyflag
    >>>>>>>>>
    [-<<<<[->>>>+<<<]>>>[-<<<->>>   Apply carry if no copy
        >-<+<<+>                        Adjust n & remainder
        >>[<-]<[->+<<<-<<<<<<<+>>>>>>>>]if n=0; inc Q on target 
        ]>              
    ]
    <<<
    [[                    Div_
            ->+>>+>-[<-]>[>]<     Remainder
            <[-<+<[->>>+<<<]>>]   Overflow to Q
            <<<
        ]
        >>[-<<<<<<<<+>>>>>>>>] Copy back
        <<<[->>>>+<<<]>>>[-<<<-Check Carry
            >+>>+>-                Adjust n & Remainder
            [<-]<[->+<<<-<<<<<<<+>>>>>>>>] if n=0; inc Q on target
        ]                       Q_
        <<
    ]
    >[-<<+>>>>>+<<<]       Copy Remainder to Carry & n
    >>>>>[->+<]            Move Acc
    <<[->+<]
    <<<<<[->+<]
    >>>>>>>>
]<<
// Div end {x1; x2; x3; x4; d1; d2; 0; 0; 0; 0; 0; carry; 0; rem; 0; 0; n_}
// mean is guaranteed to be max 3 bytes long (math)
// we now have: inputs x1•••xi; n; mean x

// Prepare state machine for derivation sum
// {xi; 0; 0; 0; 0; (m4)0; (m3)0; (m2)0; m1; mD1; mD2; 0; 0; 0; 0; 0; c; 0; r; 0; 0; n_} to
// {xi; 0{28}; n—i; n_; m1; mD1; mD2; sum1; sum2; sum3; sum4; sumD1; sumD2; sumD3; sumD4}
<<<[-]<<[-]>>>>>                                // clear carry & rem
[->>>>>>>>>>>>+>+<<<<<<<<<<<<<]                 // move n & mean
<<<<<<<<<<<[->>>>>>>>>>>>>>>>>>>>>>>>>>>+<<<<<<<<<<<<<<<<<<<<<<<<<<<]
<[->>>>>>>>>>>>>>>>>>>>>>>>>>>+<<<<<<<<<<<<<<<<<<<<<<<<<<<]
<[->>>>>>>>>>>>>>>>>>>>>>>>>>>+<<<<<<<<<<<<<<<<<<<<<<<<<<<]
>>>>>>>>>>>>>>>>>>>>>>>>> n—i

// state machine layout:
// {xi/o1; o2; o3; o4; o5; o6; o7; o8;
        {0; 0; 0;
            {0; 0; 0; add_tmp; carry_flag; 0; add; mul; mul_backup; carry; bytes_add(3)}
        (xi—xm)1; xim2; xim3; 0; bytes_mul(3)}
    0; xim1; xim2; xim3; 0; n—i_; n; m1; mD1; mD2; •••}
[-
    >>[-<<<+<<<+<<<<<<+>>>>>>>>>>>>]         // copy medain m1•mD2 to xim1•xim3
    <<<[->>>+<<<]
    >>>>[-<<<<+<<+<<<<<<+>>>>>>>>>>>>]
    <<<<[->>>>+<<<<]
    >>>>>[-<<<<<+<+<<<<<<+>>>>>>>>>>>>]
    <<<<<[->>>>>+<<<<<]
    <<< xim1(2nd)
        
    // xi — mean {xi; 0; f; sum; •••}; first count down: then up
    [-<<<<<<-<<<<<<<<<<<<<<<<<<<+>>>>>>>>>>>>>>>>>>>>>>>>>]
    <<<<<<<<<<<<<<<<<<<<<<<<<<<<[>>+>[-<-<<->>]<[-<<[->>>+<<<]>]<]
    >>>[->>>>>>>>>>>>>>>>>>>+>>>>>>+<<<<<<<<<<<<<<<<<<<<<<<<<]
    >>>>>>>>>>>>>>>>>>>>>>>>> xim1(2nd)

    // pow multiplicator
    <<+++             // setup mul byte counter (3)
    [-  // multiplication byte loop
        // setup mul count & copy active byte
        <<<<<+++
        >>>>>>>>>[-<<<<<+<<<<<<+>>>>>>>>>>>]
        <<<<<[->>>>>+<<<<<]
        <<<< bytes_add 
        [-  // add byte loop
            <[-<<<<<<+>>>>>>]       // process carry
            <[-<+>]                 // setup counter
            <<<<<<<<<<<<[->>>>>>>+<<<<<<<] // copy target byte to temp
            >>>>>>>>>>> mul
            [-  // add multiplier loop
                >+>>>>>[-<<<<<<<+>>>>>>>]            // copy source byte to temp
                <<<<<<< add
                [- // add adder loop
                    >>>>>>>+        // restore source byte
                    <<<<<<<<<+      // set carry flag
                    <+              // add number
                    [>-]>[>]<       // clear carry flag
                    [->>>>>+<<<<<]  // apply carry if flag
                    >>
                ]>
            ]
            <<<<[-<<<<<<<+>>>>>>>]    // restore target byte from temp
            >>>>>[-<+>]>[-<+>]>[-<+>] // move substatemachine to next byte
            <
        ]
        <[-<<<<<<<<<<<+>>>>>>>>>>>]   // apply remaining carry
        <[-]
        >>>>>>[-<+>]>[-<+>]>[-<+>]>>[-<+>]     // move statemachine to next byte
        <
    ]
    <<[-]<[-]<[-]>>>>>>>>>[-]>[-]>[-] b3       // delete statemachine
    
    //In—Place Divisionmodule 8 src bytes {(x1); x2; x3; x4; d1; d2; d3; d4; 0; carry; temp; rem; q; flag; n; 0; bytes_}
    >>>[-<<+<<<<<<<<<<<<<<<<<+>>>>>>>>>>>>>>>>>>>]  // copy n
    <<[->>+<<]                                      // restore n
    <<<<<<<<<<<<<<<++++++++                         // set bytecount
    // same as before; just for 8 src bytes
    [-<<<+<<<<<<<<<<<<<[[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>-<<<<<<<<<<<<<]>>>>>>>>>>>>>[-<<<<[->>>>+<<<]>>>[-<<<->>>>-<+<<+>>>[<-]<[->+<<<-<<<<<<<<<<<+>>>>>>>>>>>>]]>]<<<[[->+>>+>-[<-]>[>]<<[-<+<[->>>+<<<]>>]<<<]>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<[->>>>+<<<]>>>[-<<<->+>>+>-[<-]<[->+<<<-<<<<<<<<<<<+>>>>>>>>>>>>]]<<]>[-<<+>>>>>+<<<]>>>>>[->+<]<<[->+<]<<<<<[->+<]>>>>>>>>]<<

    [-]<<<[-]<<[-]      // clear n; rem & carry
    <<<<<<<<++++++++      // move data for easy add
    // {s1; s2; s3; s4; d1; d2; d3; d4; 0; 8_; 0{22}; n—i; n; m1; mD1; mD2; sum1; sum2; sum3; sum4; sumD1; sumD2; sumD3; sumD4} to
    // {•••n—i; n; mD1—mD3; sum1—sumD4; 0{17}; s1—d4}
    [-
    <<[->>>>>>
    >>>>>>>>>>
    >>>>>>>>>> n
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>+    new d4
    <<<<<<
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<< n
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<]
    >>[-<+>]<
    ]
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>
    >> n
    
    // 32bit adding state machine
    // {s1•••s8; 0{8}; 0; add≪; carry flag; 0; add≫; carry; 0; bytes_; 0; s1•••s8}
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>++++++++
    [-
        <<<<<<<<<<
        <<<<<< s8≪
        [->>>>>>>>>>+<<<<<<<<<<]
        >>>>>>>>>>>>>> carry [-<+>]< add≫
        [-<<+<+[>-]>[>]<[->>>+<<<]>>]
        >>>>>>>>>>>> s8≫
        [-<<<<<<<<<<<<+>>>>>>>>>>>>]
        <<<<<<<<<<<<
        [-<<+<+[>-]>[>]<[->>>+<<<]>>]
        <<< add≪
        [-<<<<<<<<<<+>>>>>>>>>>]
        >>>>[-<+>]>>[-<+>]<
    ]<<<<<<<<<<<<<<<
    <<<<< n—i

    // move to next input number 
    [-<+>]>[-<+>]>[-<+>]>[-<+>]>[-<+>]
    >[-<+>]>[-<+>]>[-<+>]>[-<+>]>[-<+>]>[-<+>]>[-<+>]>[-<+>]
    <<<<<<<<<<<<
]
>[-]>[-]>[-]>[-]    // clear mean & n
>


// Square root algorithm; adopted from github com/chmike/fpsqrt ≫ sqrtF2F
// changes: b check is bound to 0 instead of 0x40 because easier
//      — q≫=8 changed to q≫=6 because otherwise the decimal point shifts two places
// { r(8)_; b(8); 0; loopflag; temp; 0; q(8); t(8); state machines}
>>>>>>>> b(0)
++++++++++
++++++++++
++++++++++
++++++++++
++++++++++
++++++++++
++++ 0x40 = 64
>>>>>>>>>+
[[-]
<<<<<<<<<    // while b≫0
1 [->>>>>>>>>>+<<<<<<<<<<]
>>>>>>>>>>[<<+>]>[>]<
[-<<<<<<<<<<+>>>>>>>>>>]
2 <<<<<<<<<[->>>>>>>>>+<<<<<<<<<]
>>>>>>>>>[<<+>]>[>]<
[-<<<<<<<<<+>>>>>>>>>]
3 <<<<<<<<[->>>>>>>>+<<<<<<<<]
>>>>>>>>[<<+>]>[>]<
[-<<<<<<<<+>>>>>>>>]
4 <<<<<<<[->>>>>>>+<<<<<<<]
>>>>>>>[<<+>]>[>]<
[-<<<<<<<+>>>>>>>]
5 <<<<<<[->>>>>>+<<<<<<]
>>>>>>[<<+>]>[>]<
[-<<<<<<+>>>>>>]
6 <<<<<[->>>>>+<<<<<]
>>>>>[<<+>]>[>]<
[-<<<<<+>>>>>]
7 <<<<[->>>>+<<<<]
>>>>[<<+>]>[>]<
[-<<<<+>>>>]
8 <<<[->>>+<<<]
>>>[<<+>]>[>]<
[-<<<+>>>]>> q(0)

>>>>>>>>[-]>[-]>[-]++++++++ t(2)   // t = q
[-
    <<<<<<<<<<[->>>>>>>>+>+<<<<<<<<<]
    >>>>>>>>>[-<<<<<<<<<+>>>>>>>>>]
    >[->+<]>
] t(7)⊕3

++++++++    // t ⊕= b
[-
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<<   b(n)
    [->>>>>>>>>>
    >>>>>>>>>> t(n)
    >>>>>>>>>+>
    >>>>>>>>>>
    >>>>>+<<<<<
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<<]
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>
    [-<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<<+
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>]
    >
    [->+<]>
] t(7)⊕4
>>>>>++++++++
// 8 bit adder; exits on t(0)
[-<<<<<<<<<<<<<<<<[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>>[-<+>]<[-<<+<+[>-]>[>]<[->>>+<<<]>>]>>>>>>>>>>>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<<<<<<<<<<[-<<+<+[>-]>[>]<[->>>+<<<]>>]<<<[-<<<<<<<<<<+>>>>>>>>>>]>>>>[-<+>]>>[-<+>]<]<<<<<<<<<<<<<<<
✔

// r ≫= t (t ≪ r)
>>>>>>>>>++++++++ t(7)⊕2
[-
<<<<<<<<<<
<<<<<<<<<<
<<<<<<<<<<
<<<<<<< r(n)
[->>>>>>>>
>>>>>>>>>>
>>>>>>>>>> t(n)
>>>>>>>>+>>
>>>>>>>>
>>>>>>>>+
<<<<<<<<
<<<<<<<<
<<<<<<<<<< t(n)
<<<<<<<<<<
<<<<<<<<<<
<<<<<<<<]
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>[-<<<<<<
<<<<<<<<<<
<<<<<<<<<<
<<<<<<<<<<+
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>]>[->+<]>]
++++++++
<+>
// 8 bit compare ≪
// {s1•••s8;0; temp/≪ = 0; byte≪; byte≫; ≫ = 0; 0; ≪ is less; exit comp temp; exit compare; bytes_; 0(8); r1•••r8}
[-<[-<+>
    <<<<<<<<<<<<<<<<[->>>>>>>>>+>+<<<<<<<<<<] // copy s(n)
    >>>>>>>>>[-<<<<<<<<<+>>>>>>>>>]           // restore s(n)
    >>>>>>>>>>>>>>>>>[-<<<<<<<<<<<<<<<+>>>>>>>>>>>>>>>] // copy r(n)
    <<<<<<<<<<<<<<<
    <+>
    [-
        >+<[>-]>[>]<<       // set »=0 on 0 in byte»
        <<+>-[<-]<[<]>      // set «=0 on 0 in byte«
        [>>>[-<[-]<[-]<[-]>>>>>+>-<<<]<<<[->>>>>+>-<<<[-]<[-]<<]]
        >>>[-<[-]<[-]>>]<<+>
    ]<-[->>>>>[-]<<<<<]>
    >>>>>]
    <[->+<]>>[->+<]<[->+<]<<[->+<]>>>>
]>[-]>[-]>[-]>[-]>[-]>[-]>[-]<<<<<<<<[-]<<
[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<<<<<<<<<<
// t(7) ⊕ 3
// if t ≪ r
[-<<<<<<<<<< t(0)
    // q = t ⊕ b
    <<<<<<<<[-]>[-]>[-]>[-]>[-]>[-]>[-]>[-]>
    >>>>>>>>>>++++++++ t(7)⊕3
    [-<<<<<<<<<<[->>>>>>>>+>+<<<<<<<<<]
    >>>>>>>>>[-<<<<<<<<<+>>>>>>>>>]>[->+<]>]

    ++++++++[-
        <<<<<<<<<< t'(n)
        <<<<<<<< t(n)
        <<<<<<<<<<
        <<<<<<<<<< b(n)
        [-
            >>>>>>>>>>>>+ q(n)
            >>>>>>>> t(n)
            >>>>>>>> t'(n)
            >>>>>>>>>>
            >>>>>>>>>>
            >>>>>+<<<<<
            <<<<<<<<<<
            <<<<<<<<<<
            <<<<<<<<
            <<<<<<<<
            <<<<<<<<<<<<  
        ]
        >>>>>>>>>>>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]
        >>>>>>>> t(n)
        >>>>>>>> t'(n)
        >>>>>>>>>> [->+<]>        
    ]
    >>>>>++++++++
    // 8 bit adder; exits on t'(0) = new q (q')
    [-<<<<<<<<<<<<<<<<[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>>[-<+>]<[-<<+<+[>-]>[>]<[->>>+<<<]>>]>>>>>>>>>>>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<<<<<<<<<<[-<<+<+[>-]>[>]<[->>>+<<<]>>]<<<[-<<<<<<<<<<+>>>>>>>>>>]>>>>[-<+>]>>[-<+>]<]<<<<<<<<<<<<<<<
    // move q' to q
    >>>>>>>>++++++++[-<<<<<<<<[-<<<<<<<<<<<<<<<<+>>>>>>>>>>>>>>>>]>>>>>>>>[-<+>]<]<<<<<<<< t(0)
    // r —= t copy t to second add reg
    >>>>>>>>>>++++++++ t(7)⊕2
    [-<<<<<<<<<<[-
        >>>>>>>> a(n)
        >>>>>>>>>>
        >>>>>>>>>>
        >>>>>> b(n⊕1)
    ]>>>>>>>>>>[->+<]>]>>>>>>>> b(1)
    [-<->]>[-<->]>[-<->]>[-<->]>[-<->]>[-<->]>[-<->]>[-<->]<<<<<<<< // build inverse
    <<<<<<<<<<<<<<<<<<<<<<<<< a(0)
    >++++++++[-
        <<<<<<<<<<
        <<<<<<<<<<
        <<<<<<<<<<
        <<<<<<<[-
            >>>>>>>>>>
            >>>>>>>>>>
            >>>>>>>>>>
            >>>>>>+<<<<<<
            <<<<<<<<<<
            <<<<<<<<<<
            <<<<<<<<<<
        ]>>>>>>>
        >>>>>>>>>>
        >>>>>>>>>>
        >>>>>>>>>>[->+<]>
    ]>>>>>>>>>>>>>>>++++++++
    // 8 bit adder; exits on bitcounter
    [-<<<<<<<<<<<<<<<<[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>>[-<+>]<[-<<+<+[>-]>[>]<[->>>+<<<]>>]>>>>>>>>>>>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<<<<<<<<<<[-<<+<+[>-]>[>]<[->>>+<<<]>>]<<<[-<<<<<<<<<<+>>>>>>>>>>]>>>>[-<+>]>>[-<+>]<]
    >>>>>>>>>+<<<<<<<<<++++++++ add missing 1 from inverse
    // 8 bit adder; exits on a(0) = r'(0) = t(7) ⊕ 1
    [-<<<<<<<<<<<<<<<<[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>>[-<+>]<[-<<+<+[>-]>[>]<[->>>+<<<]>>]>>>>>>>>>>>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<<<<<<<<<<[-<<+<+[>-]>[>]<[->>>+<<<]>>]<<<[-<<<<<<<<<<+>>>>>>>>>>]>>>>[-<+>]>>[-<+>]<]<<<<<<<<<<<<<<<
    >>>>>>>>++++++++ move r' ≫ r
    [-<[-
        <<<<<<<< t(n)
        <<<<<<<< q(n)
        <<<<<<<<<<<< b(n)
        <<<<<<<<+>>>>>>>>
        >>>>>>>>>>>>
        >>>>>>>>
        >>>>>>>>
    ]>[-<+>]<]<<<<<<< t(0)
>>>>>>>>>>] // endif t ≪ r
<<<<<<<<<< t(0)

// r = r ⊕ r
>>>>>>>>>>++++++++
[-
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<[-
        >>>>>>>>>>
        >>>>>>>>>>
        >>>>>>>>>>
        >>>>>>+>>>>>
        >>>>>>>>>>
        >>>>>>>>>>+
        <<<<<<<<<<
        <<<<<<<<<<
        <<<<< <<<<<<
        <<<<<<<<<<
        <<<<<<<<<<
        <<<<<<<<<<
    ]>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>[->+<]>
]
>>>>>>>>>>>>>++++++++
// 8 bit adder; exits on r'(0)
[-<<<<<<<<<<<<<<<<[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>>[-<+>]<[-<<+<+[>-]>[>]<[->>>+<<<]>>]>>>>>>>>>>>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<<<<<<<<<<[-<<+<+[>-]>[>]<[->>>+<<<]>>]<<<[-<<<<<<<<<<+>>>>>>>>>>]>>>>[-<+>]>>[-<+>]<]<<<<<<<<<<<<<<<
>>>>>>>>++++++++[-<[-
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<+>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>
]>[-<+>]<]<<<<<<< t(0)

// b /= 2
>>>>>>>>>++++++++[-
    <<<<<<<<<<
    <<<<<<<<<<
    <<<<<<<<<[-
        >>>>>>>>>>
        >>>>>>>>>>
        >>>>>>>>+<<<<<<<<
        <<<<<<<<<<
        <<<<<<<<<<
    ]>>>>>>>>>
    >>>>>>>>>>
    >>>>>>>>>>[->+<]>
]>>>>++>>++++++++
// in place div
[-<<<+<<<<<<<<<<<<<[[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>-<<<<<<<<<<<<<]>>>>>>>>>>>>>[-<<<<[->>>>+<<<]>>>[-<<<->>>>-<+<<+>>>[<-]<[->+<<<-<<<<<<<<<<<+>>>>>>>>>>>>]]>]<<<[[->+>>+>-[<-]>[>]<<[-<+<[->>>+<<<]>>]<<<]>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<[->>>>+<<<]>>>[-<<<->+>>+>-[<-]<[->+<<<-<<<<<<<<<<<+>>>>>>>>>>>>]]<<]>[-<<+>>>>>+<<<]>>>>>[->+<]<<[->+<]<<<<<[->+<]>>>>>>>>]<<
[-]<<<[-]<<[-]<<<<<<<<<++++++++
[-<[-
    <<<<<<<< t(n)
    <<<<<<<< q(n)
    <<<<<<<<<<<<+ b(n)
    >>>>>>>>>>>>
    >>>>>>>>
    >>>>>>>>
]>[-<+>]<]<<<<<<<< t(0)
<<<<<<<<<<<
] // next iteration

// q ≫ 6 = q / 64
>>>> q(0) >>>>>>>> t(0) [-]>[-]>[-]>[-]>[-]>[-]>[-]>[-]<
++++++++++
++++++++++
++++++++++
++++++++++
++++++++++
++++++++++
++++
>>++++++++
[-<<<+<<<<<<<<<<<<<[[->>>>>>>>>>+<<<<<<<<<<]>>>>>>>>>>>>>-<<<<<<<<<<<<<]>>>>>>>>>>>>>[-<<<<[->>>>+<<<]>>>[-<<<->>>>-<+<<+>>>[<-]<[->+<<<-<<<<<<<<<<<+>>>>>>>>>>>>]]>]<<<[[->+>>+>-[<-]>[>]<<[-<+<[->>>+<<<]>>]<<<]>>[-<<<<<<<<<<<<+>>>>>>>>>>>>]<<<[->>>>+<<<]>>>[-<<<->+>>+>-[<-]<[->+<<<-<<<<<<<<<<<+>>>>>>>>>>>>]]<<]>[-<<+>>>>>+<<<]>>>>>[->+<]<<[->+<]<<<<<[->+<]>>>>>>>>]<<
[-]<<<[-]<<[-]<<<<<<<<<<<<<<<<<.>.>.>.>.>.>.>.

Answer 17

MATLAB / Oktav, 12 bayt, puan 336,32

(Kurallara uymak için Giuseppe'nin rehberliğine göre değiştirildi)

@(A)std(A,1)

İsimli argüman, Apuan için en düşük sapmayı sağlar, standart çıktı değişkenine çıktı verir Ansve aslında komut penceresine yazılır.
Çevrimiçi deneyin!

stdyerleşik bir işlevdir. Varsayılan olarak \$N-1\$şeytan çıkarıcı olarak ancak 1ikinci argüman olarak geçerek \ olarak değiştirildi$N\$.