Çarpık satranç tahtası


20

Bu zorluk, kare boyutunun, tahta boyunca sabit olmak yerine, aşağıda açıklandığı gibi, azalmayan belirli bir sırayı takip ettiği bir satranç tahtası inşa etmekle ilgilidir.

Kurul yinelemeli olarak tanımlanır. Büyüklükte bir kart n×n olacak şekilde genişletilir (n+k)×(n+k) boyutu karelerinden oluşan bir "tabaka" ile aşağı genişleterek ve sağa k , k büyük bölen bir n değil aşan n . Diyagonaldeki kareler her zaman aynı renktedir.

Özellikle, olarak temsil renklerle kurulu düşünün #ve +.

  1. Satranç tahtasını

    #
    
  2. Kart şu ana kadar 1×1 boyutuna sahip . Tek böleni 1 olan 1 ve aşmadığı 1 . Aldığımız Yanik=1ve boyutlarda kareler bir katman eklenerek kurulu uzatmak1ile#diyagonal içinde:

    #+
    +#
    
  3. Şimdiye kadar inşa edilen tahta 2×2 boyutunda . Arasında bölenler 2 olan 1,2 , ve maksimum bölen aşmayan 2 ,1. Yani yinek=1ve tahta

    #+#
    +#+
    #+#
    
  4. Boyut 3×3 . k=1 . Şuraya genişlet:

    #+#+
    +#+#
    #+#+
    +#+#
    
  5. Boyut 4×4 . Şimdi k=2 , çünkü 2 , 4 geçmeyen maksimum bölücü 4 . Diyagonalrenkte,2×2büyüklüğünde karelerden oluşan birkalınlık tabakası2ile genişletin:2×2#

    #+#+##
    +#+###
    #+#+++
    +#+#++
    ##++##
    ##++##
    
  6. Boyut 6×6 . Şimdi k=2 . 8×8 boyutuna uzatın . Şimdi k=2 . 10×10 boyutuna uzatın . Şimdi k=2 . 12×12 boyutuna uzatın . Şimdi k=3 . 15 bedene kadar genişlet :

    #+#+##++##++###
    +#+###++##++###
    #+#+++##++#####
    +#+#++##++##+++
    ##++##++##+++++
    ##++##++##+++++
    ++##++##++#####
    ++##++##++#####
    ##++##++##++###
    ##++##++##+++++
    ++##++##++##+++
    ++##++##++##+++
    ###+++###+++###
    ###+++###+++###
    ###+++###+++###
    

En son eklenen 3×3 boyutundaki karelerin, önceden eklenen 2×2 boyutundaki karelerle kısmen çakışan kenarlara sahip olduğunu unutmayın .

k değerlerinin oluşturduğu dizi azalmaz:

1 1 1 2 2 2 2 3 3 3 3 4 4 4 6 6 6 6 6 6 ...

ve OEIS'te görünmüyor. Bununla birlikte, tahtanın boyut sırası olan kümülatif versiyonu A139542'dir ( fark etmek için @Arnauld sayesinde ).

Meydan okuma

Girdi : tahtadaki katman sayısını temsil eden pozitif bir tamsayı Sİsterseniz, giriş olarak S yerine S1 alabilirsiniz ( 0 - endeksli); aşağıya bakınız.S0

Çıktı : S katmanları olan bir tahtanın ASCII-art gösterimi .

  • Çıktı STDOUT veya bir işlev tarafından döndürülen bir argüman aracılığıyla olabilir. Bu durumda, satırsonu olan bir dize, bir 2B karakter dizisi veya bir dize dizisi olabilir.

  • Tahtayı temsil etmek için iki karakteri tutarlı bir şekilde seçebilirsiniz .

  • Büyüme yönünü tutarlı bir şekilde seçebilirsiniz . Yani, yukarıdaki gösterimler yerine (aşağı ve sağa doğru büyür), yansıyan veya döndürülmüş sürümlerinden herhangi birini üretebilirsiniz.

  • Boşluk tahta için kullanılan iki karakterden biri olmadığı sürece, sondaki veya öndeki boşluğa izin verilir (çıktı STDOUT üzerinden ise).

  • İsteğe bağlı olarak " 0 indeksli " girişi kullanabilirsiniz; yani, S katmanları olan bir kartı belirten S-1 girişi olarak alın .S

Bayt cinsinden en kısa kod kazanır.

Test senaryoları

1:

#

3:

#+#
+#+
#+#

5:

#+#+##
+#+###
#+#+++
+#+#++
##++##
##++##

6:

#+#+##++
+#+###++
#+#+++##
+#+#++##
##++##++
##++##++
++##++##
++##++##

10:

#+#+##++##++###+++
+#+###++##++###+++
#+#+++##++#####+++
+#+#++##++##+++###
##++##++##+++++###
##++##++##+++++###
++##++##++#####+++
++##++##++#####+++
##++##++##++###+++
##++##++##+++++###
++##++##++##+++###
++##++##++##+++###
###+++###+++###+++
###+++###+++###+++
###+++###+++###+++
+++###+++###+++###
+++###+++###+++###
+++###+++###+++###

15:

#+#+##++##++###+++###+++####++++####
+#+###++##++###+++###+++####++++####
#+#+++##++#####+++###+++####++++####
+#+#++##++##+++###+++#######++++####
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++##++##++#####+++###+++++++####++++
++##++##++#####+++###+++++++####++++
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25:

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Çıkış olarak bir tamsayı matrisine izin veriliyor mu (örn. 0s ve 1s) veya dizeler / karakterler mi olmalı?
Nick Kennedy

@Nick Chars olmalı, üzgünüm
Luis Mendo

2
Çok iyi yazılmış bir soru!
Greg Martin

@GregMartin Hey, teşekkürler!
Luis Mendo

Yanıtlar:


3

Jöle , 40 31 bayt

1SÆD>Ðḟ½ƊṀṭƲ³¡Äż$Ḷ:Ḃ^þ`ʋ/€ḷ""/Y

Çevrimiçi deneyin!

S1

İz bırakmadan Y , tamsayıların bir listesini döndürür, ancak bu meydan okuma için spesifikasyon dışındadır.

açıklama

Bu program üç aşamada çalışır.

  1. kk
  2. Bunların her biri için döşeme boyutu ile bir dama tahtası oluşturunk
  3. Her seferinde bir sonraki panonun sol üst bölümünü mevcut panelle değiştirerek dama tahtası listesinde çalışın.

1. Aşama

1                 | Start with 1
           Ʋ³¡    | Loop through the following the number of times indicated by the first argument to the program; this generates a list of values of k
 S                | - Sum
        Ɗ         | - Following three links as a monad 
  ÆD              |   - List of divisors
    >Ðḟ½          |   - Exclude those greater than the square root
         Ṁ        |   - Maximum
          ṭ       | - Concatenate this to the end of the current list of values of k 
              Äż$ | Zip the cumulative sum of the values of k with the values

2. aşama

      ʋ/€ | For each pair of k and cumulative sum, call the following as a dyad with the cumulative sum of k as the left argument and k as the right (e.g. 15, 3)
Ḷ         | - Lowered range [0, 1 ... , 13, 14]
 :        | - Integer division by k [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
  Ḃ       | - Mod 2 [0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0]
   ^þ`    | - Outer product using xor function and same argument to both side

Sahne 3

   /  | Reduce using the following:
ḷ""   | - Replace the top left portion of the next matrix with the current one
    Y | Finally join by newlines

Bence asker aslında #ve +karakterleri istiyor . Ama vay, hala etkileyici, yarısını anladım. Ve böyle bir dilde nasıl programlıyorsunuz? Sadece kopyaladığınız bir karakter tablosu ve anlamları var mı?
Fabian Röling

@ FabianRöling OP herhangi iki karakterin kabul edilebilir olduğunu belirtir. Jelly için github.com/DennisMitchell/jelly adresinde iyi bir tanıtım eğitimi var . Python'u biliyorsanız, kaynak da oldukça okunabilir. Öğretici ve atom ve Quicks listelerinde bile, biraz zaman aldı ve bazı durumlarda kafamı döndürmek için kaynağa atıfta bulundum.
Nick Kennedy


4

Python 2 , 217 215212 bayt

def f(x):
 b=['1'];n=1
 for i in range(x):P=max(j*(n%j<(j<=n**.5))for j in range(1,1+n));n+=P;b=[l+P*`j/P%2^i%2`for j,l in enumerate(b)];s=len(b[0]);b+=[((v*P+`1^int(v)`*P)*s)[:s]for v in b[0][len(b):]]
 return b

Çevrimiçi deneyin!

0 dizinli, karakterler olarak 0ve kullanır1


1
@LuisMendo 2 bayt kurtardı: D
Rod

3

Python 2 , 184 178 176 169 bayt

def h(j,a=['1'],R=range):
 for i in R(j):L=len(a);k=max(x for x in R(1,L+1)if(x*x<=L)>L%x);a=[a[m]+k*`(i+m/k)%2`for m in R(L)]+[((`i%2`*k+`~i%2`*k)*L)[:L+k]]*k
 return a

Çevrimiçi deneyin!

Kullanımları 1, 0için #, -; 0-indexing kullanır .


2

JavaScript (ES7), 164 bayt

Giriş 0 dizinlidir. İle bir matris çıkarır0için #ve1için +.

n=>(b=[1],g=(a,w,d=w**.5|0)=>b[n]?a:w%d?g(a,w,d-1):g(a.concat(Array(d).fill(b.push(d)&&i++)),w+d))([0],i=1).map((_,y,a)=>a.map((_,x)=>(x/b[v=a[x>y?x:y]]^y/b[v])&1))

Çevrimiçi deneyin!


2

Kömür , 37 bayt

FN«≔⊕⌈Φ₂⊕Lυ¬﹪Lυ⊕κηFη«PL⭆⊞Oυω§#+÷⁻κμη↙

Çevrimiçi deneyin! Bağlantı, kodun ayrıntılı versiyonudur. 1 endeksli. Çıktı aşağı ve sol büyür (aşağı ve sağ ekstra bir bayt maliyeti, ancak aynı bayt sayısı için büyüyebilir). Açıklama:

FN«

döngü S zamanlar.

≔⊕⌈Φ₂⊕Lυ¬﹪Lυ⊕κη

Hesaplamak kn+1. Bu sadecen=0 bu durumda bu formül k=1.

Fη«

döngü k kez, her yeni satır ve sütun için bir kez.

PL⭆⊞Oυω§#+÷⁻κμη

Çıktı satır ve sütun arasındaki alternatif emin olmak #ve +bu şekilde karakterler #(diyagonal dışarıya dan çünkü bizler çizim) dizenin sonunda bir sınır olduğunu ancak her zaman ilk karakterdir. ⊞Oυωher satırı her seferinde bir karakter daha uzun yapar, bu dan uzunluk olarak.

Bir sonraki sıra için aşağı ve sola hareket edin.


2

05AB1E , 43 42 bayt

$G©ÐX‚ˆÑʒ®>t‹}àDU+}¯εÝ`θ÷ɨDδ^}RζεðKζðδK€θ

Esinlenerek @NickKennedy Jelly yanıt ve arka kısmı, ζεðKζðδK€θbir liman @Emigna 'burada s 05AB1E cevap .

Bir matris döndürür 0yerine #ve 1yerine +.

Çevrimiçi deneyin veya ilkini çıkararak çevrimiçi deneyin[2,n]sonuçlar ( J,altbilgi ve --no-lazybayrak sonuçtaki matrisi güzel yazdırmak içindir).

Açıklama:

$                # Push 1 and the input
 G               # Loop the input - 1 amount of times:
  ©              #  Store the top of the stack in variable `r` (without popping)
   Ð             #  And triplicate the top as well
    X           #  Pair it with variable `X` (which is 1 by default)
      ˆ          #  And pop and store this pair in the global array
    Ñ            #  Get the divisors of the integer we triplicated
     ʒ         #  Get the highest divisor which is truthy for:
                #   Where the divisor integer is smaller than
      ®>t        #   the square root of `r+1`
            DU   #  Store a copy of this largest filtered divisor as new variable `X`
              +  #  And add it to the triplicated integer
               # After the loop: push the global array
   ε             # Map each pair to:
    Ý θ          #  Convert the first value in the pair to a list in the range [0,n]
     `           #  and push both this list and the second value to the stack
       ÷         # Integer-divide each value in the list by the second value
        É        # Check for each value if it's even (1 if even; 0 if odd)
         ¨       # Remove the last item
          Dδ     # Loop double vectorized over this list:
            ^    #  And XOR the values with each other
   }R            # After the map: reverse the list of digit-matrices
     ζ           # Zip/transpose; swapping rows and columns, with a space as filler
      ε          # map each matrix to:
       ðK        #  Remove all spaces from the current matrix
         ζ       #  Zip/transpose with a space as filler again
          ðδK    #  Deep remove all spaces
             €θ  #  Then only leave the last values of each row
                 # (after which the resulting matrix of 0s and 1s is output implicitly)

1

Haskell'in 149 146 bayt

(iterate g["#"]!!)
g b|let e=(<$[1..d]);l=length b;d=last[i|i<-[1..l],i*i<=l,mod l i<1];m="+#"++m=(e$take(l+d)$e=<<'#':m)++zipWith(++)(e=<<e<$>m)b

Bu 0 indekslenir, bir dize listesi döndürür ve yukarı ve sola doğru büyür.

Çevrimiçi deneyin!

(iterate g["#"]!!)                    -- start with ["#"], repeatedly add a layer
                                      -- (via function 'g'), collect all results in
                                      -- a list and index it with the input number

g b | let                             -- add a single layer to chessboard 'b'

 l=length b                           -- let 'l' be the size of 'b'
 d=last[i|i<-[1..l],i*i<=l,mod l i<1] -- let 'd' be the size of the new layer
 e=(<$[1..d])                         -- let 'e' be a functions that makes 'd'
                                      --   copies of it's argument
 m="#+"++m                            -- let 'm' be an infinite string of "+#+#+..."

 =                                    -- return
              zipWith(++)             --   concatenate pairwise
                         (e=<<e<$>m)  --   a list of squares made by expanding each
                                      --   char in 'm' to size 'd'-by-'d'
                                    b --   and 'b' (zipWith truncates the infinite
                                      --   list of squares to the length of 'b')
                                      --
           ++                         --   and prepend
                                      --
(e$take(l+d)$e=<<'#':m)               --   the top layer, i.e. a list of 'd' strings
                                      --   each with the pattern 'd' times '#'
                                      --   followed by 'd' times '+', etc., each
                                      --   shortened to the correct size of 'l'+'g'

1

Perl 6 , 156 144 155 154 bayt

+ 11 nimi tarafından bildirilen bir hatayı düzeltmek için.

{$!=-1;join "
",(1,{my \k=max grep $_%%*,1.. .sqrt;++$!;flat .kv.map(->\i,\l {l~($!+i/k)%2+|0 x k}),substr(($!%2 x k~1-$!%2 x k)x$_,0,$_+k)xx k}...*)[$_]}

Kabaca Chas Brown'un Python çözümüne dayanıyor . S'yi sıfır indeksli olarak alır. Çıktılar 0ve 1.

Çevrimiçi deneyin!


Sabit. Şimdi köşeler aynı rengi paylaşmalı.
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