介绍

每个人都知道游戏井字游戏，但是在这个挑战中，我们将介绍一些小技巧。我们将只使用十字架。第一个连续放置三个十字架的人输了。一个有趣的事实是，有人输掉前的最大十字架数等于6：

X X -
X - X
- X X

这意味着对于3 x 3的面板，最大数量为6。因此对于N = 3，我们需要输出6。

另一个示例，对于N = 4或4 x 4板：

X X - X
X X - X
- - - -
X X - X

这是一个最佳解决方案，您可以看到最大的十字架数量等于9。12 x 12板的最佳解决方案是：

X - X - X - X X - X X -
X X - X X - - - X X - X
- X - X - X X - - - X X
X - - - X X - X X - X -
- X X - - - X - - - - X
X X - X X - X - X X - -
- - X X - X - X X - X X
X - - - - X - - - X X -
- X - X X - X X - - - X
X X - - - X X - X - X -
X - X X - - - X X - X X
- X X - X X - X - X - X

结果为74。

任务

任务很简单，给定一个大于0的整数，输出可以沿着行，列或对角线在一行中没有相邻三个X的情况下放置的最大十字架数量。

测试用例

N     Output
1     1
2     4
3     6
4     9
5     16
6     20
7     26
8     36
9     42

可以在https://oeis.org/A181018上找到更多信息。

规则

• 这是代码高尔夫球，因此以最少的字节提交为准！
• 您可以提供功能或程序。

Pyth，57 51 49字节

L.T.e+*]Ykbbsef!s.AMs.:R3ssmyBdsm_BdCBcTQsD^U2^Q2

就像@PeterTaylor的CJam解决方案一样，这是蛮力的，所以它在O（n ² 2 ^{n 2}）。n = 4时，在线解释器不会在一分钟内完成。

在这里尝试N <4。

尝试对角线功能。

L.T.e+*]Ykbb         y(b): diagonals of b (with some trailing [])
s e                  sum of the last (with most ones) array such that
f                    filter lambda T:
 ! s .AM                none of the 3 element sublists are all ones               
   s .:R3               all 3 element sublists
   s s                  flatten
   myBd                 add the diagonals
   sm_B d               add the vertically flipped array and transpose
   CBcTQ                array shaped into Q by Q square, and its transpose
 sD ^U2 ^Q2             all binary arrays of length Q^2 sorted by sum

CJam（58 56字节）

2q~:Xm*{7Yb#W=}:F,Xm*{ee{~0a@*\+}%zS*F},_Wf%:z&Mf*1fb:e>

这非常慢，而且占用大量内存，但这对您来说是代码高尔夫。

解剖

2q~:Xm*        e# Read input into X and find Cartesian product {0,1}^X
{7Yb#W=}:F,    e# Filter with a predicate F which rejects arrays with a 111
Xm*            e# Take the Cartesian product possible_rows^X to get possible grids
{              e# Filter out grids with an anti-diagonal 111 by...
  ee{~0a@*\+}% e#   prepending [0]*i to the ith row
  zS*F         e#   transposing, joining on a non-1, and applying F
},
_Wf%:z         e# Copy the filtered arrays and map a 90 degree rotation
&              e# Intersect. The rotation maps horizontal to vertical and
               e# anti-diagonal to diagonal, so this gets down to valid grids
Mf*            e# Flatten each grid
1fb            e# Count its 1s
:e>            e# Select the maximum

$\Theta(a^X)$$a$$1.83928675\ldots$$\Theta(a^{X^2})$$\Theta(a^{X^4})$。

$O(X a^{3X})$

public class A181018 {
    public static void main(String[] args) {
        for (int i = 1; i < 14; i++) {
            System.out.format("%d:\t%d\n", i, calc(i));
        }
    }

    private static int calc(int n) {
        if (n < 0) throw new IllegalArgumentException("n");
        if (n < 3) return n * n;

        // Dynamic programming approach: given two rows, we can enumerate the possible third row.
        // sc[i + rows.length * j] is the greatest score achievable with a board ending in rows[i], rows[j].
        int[] rows = buildRows(n);
        byte[] sc = new byte[rows.length * rows.length];
        for (int j = 0, k = 0; j < rows.length; j++) {
            int qsc = Integer.bitCount(rows[j]);
            for (int i = 0; i < rows.length; i++) sc[k++] = (byte)(qsc + Integer.bitCount(rows[i]));
        }

        int max = 0;
        for (int h = 2; h < n; h++) {
            byte[] nsc = new byte[rows.length * rows.length];
            for (int i = 0; i < rows.length; i++) {
                int p = rows[i];
                for (int j = 0; j < rows.length; j++) {
                    int q = rows[j];
                    // The rows which follow p,q cannot intersect with a certain mask.
                    int mask = (p & q) | ((p << 2) & (q << 1)) | ((p >> 2) & (q >> 1));
                    for (int k = 0; k < rows.length; k++) {
                        int r = rows[k];
                        if ((r & mask) != 0) continue;

                        int pqrsc = (sc[i + rows.length * j] & 0xff) + Integer.bitCount(r);
                        int off = j + rows.length * k;
                        if (pqrsc > nsc[off]) nsc[off] = (byte)pqrsc;
                        if (pqrsc > max) max = pqrsc;
                    }
                }
            }

            sc = nsc;
        }

        return max;
    }

    private static int[] buildRows(int n) {
        // Array length is a tribonacci number.
        int c = 1;
        for (int a = 0, b = 1, i = 0; i < n; i++) c = a + (a = b) + (b = c);

        int[] rows = new int[c];
        int i = 0, j = 1, val;
        while ((val = rows[i]) < (1 << (n - 1))) {
            if (val > 0) rows[j++] = val * 2;
            if ((val & 3) != 3) rows[j++] = val * 2 + 1;
            i++;
        }

        return rows;
    }
}

C， 460 456 410 407 362 351 318字节

这是一个非常糟糕的答案。这是一种极其缓慢的蛮力方法。我正在尝试通过组合for循环来进一步打高尔夫球。

#define r return
#define d(x,y)b[x]*b[x+y]*b[x+2*(y)]
n,*b;s(i){for(;i<n*(n-2);++i)if(d(i%(n-2)+i/(n-2)*n,1)+d(i,n)+(i%n<n-2&&d(i,n+1)+d(i+2,n-1)))r 1;r 0;}t(x,c,l,f){if(s(0))r 0;b[x]++;if(x==n*n-1)r c+!s(0);l=t(x+1,c+1);b[x]--;f=t(x+1,c);r l>f?l:f;}main(c,v)char**v;{n=atol(v[1]);b=calloc(n*n,4);printf("%d",t(0,0));}

测试用例

$ ./a.out 1
1$ ./a.out 2
4$ ./a.out 3
6$ ./a.out 4
9$ ./a.out 5
16$

不打高尔夫球

n,*b; /* board size, board */

s(i) /* Is the board solved? */
{
    for(;i<n*(n-2);++i) /* Iterate through the board */
            if(b[i%(n-2)+i/(n-2)*n]&&b[i%(n-2)+i/(n-2)*n+1]&&b[i%(n-2)+i/(n-2)*n+2] /* Check for horizontal tic-tac-toe */
                    || b[i] && b[i+n] && b[i+2*n] /* Check for vertical tic-tac-toe */
                    || (i%n<n-2
                            && (b[i] &&b [i+n+1] && b[i+2*n+2] /* Check for diagonal tic-tac-toe */
                                    || b[i+2*n] && b[i+n+1] && b[i+2]))) /* Check for reverse diagonal tic-tac-toe */
                    return 1;
    return 0;
}

t(x,c,l,f) /* Try a move at the given index */
{
    if(s(0)) /* If board is solved, this is not a viable path */
            return 0;
    b[x]++;
    if(x==n*n-1) /* If we've reached the last square, return the count */
            return c+!s(0);

    /* Try it with the cross */
    l=t(x+1,c+1);

    /* And try it without */
    b[x]--;
    f=t(x+1,c);

    /* Return the better result of the two */
    return l>f?l:f;
}

main(c,v)
char**v;
{
    n=atol(v[1]); /* Get the board size */
    b=calloc(n*n,4); /* Allocate a board */
    printf("%d",t(0,0)); /* Print the result */
}

编辑：将int变量声明为未使用的参数；删除y坐标，只使用索引；将变量移到参数列表而不是全局，修复传递给的不必要的参数s()；合并循环，删除不必要的括号；替换&&用*，||用+; 宏化3合一检查

Ç263 264 283 309

编辑 @Peter Taylor保存了几个字节-比我希望的要少。然后2个字节用于分配更多的内存，现在我可以尝试更大的大小，但是这变得非常耗时。

注意在添加说明时，我发现我在浪费字节，将网格保留在R数组中-这样您就可以看到找到的解决方案...不需要挑战！
我在高尔夫版本中将其删除

一个高尔夫C程序，可以在合理的时间内实际找到n = 1..10的答案。

s,k,n,V[9999],B[9999],i,b;K(l,w,u,t,i){for(t=u&t|u*2&t*4|u/2&t/4,--l; i--;)V[i]&t||(b=B[i]+w,l?b+(n+2)/3*2*l>s&&K(l,b,V[i],u,k):b>s?s=b:0);}main(v){for(scanf("%d",&n);(v=V[i]*2)<1<<n;v%8<6?B[V[k]=v+1,k++]=b+1:0)V[k]=v,b=B[k++]=B[i++];K(n,0,0,0,k);printf("%d",s);}

我的测试：

7-> 26于10秒
8-> 36于18秒
9-> 42于1162秒

少打高尔夫球，试图解释

#include <stdio.h>

int n, // the grid size
    s, // the result
    k, // the number of valid rows 
    V[9999], // the list of valid rows (0..to k-1) as bitmasks
    B[9999], // the list of 'weight' for each valid rows (number of set bits)
    R[99],  // the grid as an array of indices pointing to bitmask in V
    b,i; // int globals set to 0, to avoid int declaration inside functions

// recursive function to fill the grid
int K(
  int l, // number of rows filled so far == index of row to add
  int w, // number of crosses so far
  int u, // bit mask of the preceding line (V[r[l-1]])
  int t, // bit mask of the preceding preceding line (V[r[l-2]])
  int i) // the loop variables, init to k at each call, will go down to 0
{
  // build a bit mask to check the next line 
  // with the limit of 3 crosses we need to check the 2 preceding rows
  t = u&t | u*2 & t*4 | u/2 & t/4; 
  for (; i--; )// loop on the k possibile values in V
  {
    R[l] = i; // store current row in R
    b = B[i] + w; // new number of crosses if this row is accepted
    if ((V[i] & t) == 0) // check if there are not 3 adjacent crosses
      // then check if the score that we can reach from this point
      // adding the missing rows can eventually be greater
      // than the current max score stored in s
      if (b + (n + 2) / 3 * 2 * (n - l - 1) > s)
        if (l > n-2) // if at last row
          s = b > s ? b : s; // update the max score
        else  // not the last row
          K(l + 1, b, V[i], u, k); // recursive call, try to add another row
  }
}

int main(int j)
{
  scanf("%d", &n);

  // find all valid rows - not having more than 2 adjacent crosses
  // put valid rows in array V
  // for each valid row found, store the cross number in array B
  // the number of valid rows will be in k
  for (; i<1 << n; V[k] = i++, k += !b) // i is global and start at 0
    for (b = B[k] = 0, j = i; j; j /= 2) 
      b = ~(j | -8) ? b : 1, B[k] += j & 1;
  K(0,0,0,0,k); // call recursive function to find the max score
  printf("%d\n", s);
}

Ruby，263个字节

这也是一个蛮力解决方案，面临的问题与科尔·卡梅隆（Cole Cameron）提出的C答案相同，但由于它是红宝石而不是C，因此速度甚至更慢。但是，它更短。

c=->(b){b.transpose.all?{|a|/111/!~a*''}}
m=->(b,j=0){b[j/N][j%N]=1
x,*o=b.map.with_index,0
c[b]&&c[b.transpose]&&c[x.map{|a,i|o*(N-i)+a+o*i}]&&c[x.map{|a,i|o*i+a+o*(N-i)}]?(((j+1)...N*N).map{|i|m[b.map(&:dup),i]}.max||0)+1:0}
N=$*[0].to_i
p m[N.times.map{[0]*N}]

测试用例

$ ruby A181018.rb 1
1
$ ruby A181018.rb 2
4
$ ruby A181018.rb 3
6
$ ruby A181018.rb 4
9
$ ruby A181018.rb 5
16

不打高尔夫球

def check_columns(board)
  board.transpose.all? do |column|
    !column.join('').match(/111/)
  end
end

def check_if_unsolved(board)
  check_columns(board) && # check columns
    check_columns(board.transpose) && # check rows
    check_columns(board.map.with_index.map { |row, i| [0] * (N - i) + row + [0] * i }) && # check decending diagonals
    check_columns(board.map.with_index.map { |row, i| [0] * i + row + [0] * (N - i) }) # check ascending diagonals
end

def maximum_crosses_to_place(board, index=0)
  board[index / N][index % N] = 1 # set cross at index
  if check_if_unsolved(board)
    crosses = ((index + 1)...(N*N)).map do |i|
      maximum_crosses_to_place(board.map(&:dup), i)
    end
    maximum_crosses = crosses.max || 0
    maximum_crosses + 1
  else
    0
  end
end

N = ARGV[0].to_i
matrix_of_zeros = N.times.map{ [0]*N }

puts maximum_crosses_to_place(matrix_of_zeros)

Haskell，143个字节

在某些方面，这还没有完成，但是我很开心，所以这里是：

• 因为如果在不同行上应用检查水平“获胜”模式将返回无效，则N <3的输入将返回0
• “数组”是分解成位的整数，因此很容易枚举
• [[i！
• 由于未检查边界，因此它会针对每个边界检查81 * 4 = 324个模式可能的最大值，从而导致N = 3占用了我的笔记本电脑9秒钟，N = 5占用了我太长时间而无法完成
• 1/0上的布尔逻辑用于T / F以节省空间，例如（*）是&&，（1-x）是（not x），等等。
• 因为它检查整数而不是数组，所以需要（div p1 L）==（div p2 L）以确保不在不同的行上检查模式，其中L是行长，p1，p2是位置
• 可能的最大值是汉明重量

这是代码：

r=[0..81]
(%)=div
s=sum
i!x|i<0=(*0)|0<1=(*mod(x%(2^i))2)
f l=maximum[s[i!x$1-s[s[1#2,l#(l+l),(l+1)#(l+l+2),(1-l)#(2-l-l)]|p<-r,let  a#b=p!x$(p+a)!x$(p+b)!x$s[1|p%l==(p+mod b l)%l]]|i<-r]|x<-[0..2^l^2]]