Magic Square是一个二维数组，必须以某种方式填充1 to n^2中的数字，以便所有行，列和对角线的总和必须为(n*(n^2 + 1) / 2)。所以在3x3阵列中它是3*(3*3 + 1) / 2) = 3*10 / 2 = 15，在4x4阵列中它是4*17 / 2 = 34等。
如果我没记错的话，3x3有8个可能的解决方案，4x4有7000个以上，5x5有数十亿个。
我的任务是在合理的时间内找到所有可能的解决方案。 3x3是即时8解决方案。然而，在4x4中，我只能在17到20秒内从7000+解决方案中获得3456（如果我将打印移除到屏幕，则为6+）。
#include <iostream>
#include <windows.h>
#include <ctime>
using namespace std;
//n is the size of the square
//k is n^2
int n = 0, k = 0;
//nr is the number of solutions
unsigned int nr = 0;
//The magic Square
int y[20][20];
bool member[20];
bool dbug = false;
//SUM of the ROW is (n*(n*n + 1) / 2)
bool rowAccept(int p) {
int sum = 0;
for (int i = 1; i <= n; i++)
sum += y[p][i];
return (sum == (n*(k + 1) / 2));
}
//SUM of the COLUMN is (n*(n*n + 1) / 2)
bool colAccept(int i, int j) {
int sum = 0;
for (int p = 1; p <= n - 1; p++)
sum += y[p][i];
sum += j;
return (sum == (n*(k + 1) / 2));
}
/*
x x y y
x x y y
z z t t
z z t t
Each of these sub squares will have the SUM of (n*(n*n + 1) / 2) if it's a valid square
Also the middle sub square:
x y
z t
*/
inline bool subAccept(int p, int i, int j) {
return (y[p - 1][i - 1] + y[p - 1][i] + y[p][i - 1] + j == 34);
}
/*
These corners have to have the SUM of (n*(n*n + 1) / 2) if it's a valid square
x _ x _ _ x _ x
_ _ _ _ _ _ _ _
x _ x _ _ x _ x
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
x _ x _ _ x _ x
_ _ _ _ _ _ _ _
x _ x _ _ x _ x
*/
inline bool miniCorners(int p, int i, int j) {
return (y[p - 2][i - 2] + y[p - 2][i] + y[p][i - 2] + j == 34);
}
//Check if the SUM of the diagonal is (n*(n*n + 1) / 2)
bool diag(int j) {
int sum = 0;
for (int k = 1; k < n; k++)
sum += y[k][k];
sum += j;
return (sum == 34);
}
//Check if the SUM of the other diagonal(sorry I don't know what it is called in english) is (n*(n*n + 1) / 2)
bool revDiag(int j) {
int sum = 0;
for (int k = 1; k < n; k++)
sum += y[k][n - k + 1];
sum += j;
return (sum == 34);
}
//Print to screen
void show() {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
if (y[i][j] > 9)
cout << " " << y[i][j];
else
cout << " " << y[i][j];
cout << endl;
}
cout << endl;
if (dbug) {
cin.get();
}
}
//Call all the above functions to see if it's a Magic Square
inline bool magic(int p, int i, int j) {
//3x3 Magic Sqaure
//These are only patterns, only the colAccept(i, j) is called
//This works fine
if (n == 3) {
if (p == 2 && i == 1 && j != 1 && j != 9 && y[1][2] != 1 && y[1][2] != 9)
return false;
if (p == 2 && i == 3 && j != 1 && j != 9 && y[1][2] != 1 && y[1][2] != 9 && (y[2][1] != 1 || y[2][1] != 9))
return false;
if (p == 3 && i == 2 && j != 1 && j != 9 && y[1][2] && y[1][2] && y[2][1] != 1 && y[2][1] != 9 && y[2][3] != 1 && y[2][3] != 9)
return false;
if (p == 2 && i == 2 && j != 5)
return false;
if (p == 3 && !colAccept(i, j))
return false;
return true;
}
//4x4 Magic Sqaure
//All the functions are needed and some more patterns checking
//3456 solutions in 17-20 seconds instead of 7000+ (around 7008 or 7080, can't remember and can't find it anywhere)
else if (n == 4)
{
// SubAccept#1
//------------------------------------------
if (p == 2 && i == 2 && !subAccept(p, i, j))
return false;
if (p == 2 && i == 4 && !subAccept(p, i, j))
return false;
if (p == 3 && i == 3 && !subAccept(p, i, j))
return false;
//------------------------------------------
/*
_ _ _ _
x _ _ x
x _ _ x
_ _ _ _
SUM(x) = (n*(n*n + 1) / 2)
*/
if (p == 3 && i == 4)
return ((y[2][1] + y[2][4] + y[3][1] + j) == 34);
// MiniCorners
//--------------------------------------------
if (p == 3 && i == 3 && !miniCorners(p, i, j))
return false;
if (p == 3 && i == 4 && !miniCorners(p, i, j))
return false;
if (p == 4 && i == 3 && !miniCorners(p, i, j))
return false;
if (p == 4 && i == 4 && !miniCorners(p, i, j))
return false;
//--------------------------------------------
/*
x _ _ _
_ _ _ y
_ _ _ y
x _ _ _
x + x = y + y
*/
if (p == 4 && i == 1)
return ((j + y[1][1]) == (y[2][4] + y[3][4]));
/*
_ _ _ x
_ y _ _
_ _ y _
x _ _ _
x + x = y + y
*/
if (p == 4 && i == 1)
return ((y[1][4] + j) == (y[2][2] + y[3][3]));
/*
x _ _ x
_ _ _ _
_ _ _ _
_ y y _
x + x = y + y
*/
if (p == 4 && i == 3)
return ((j + y[4][2]) == (y[1][4] + y[1][1]));
/*
_ x x _
_ _ _ _
_ _ _ _
_ x x _
SUM(x) = (n*(n*n + 1) / 2)
*/
if (p == 4 && i == 3)
return ((y[1][2] + y[1][3] + y[4][2] + j) == 34);
//other diagonal
if (p == 4 && i == 1 && !revDiag(j))
return false;
//If in last ROW check for COLUMN SUM = (n*(n*n + 1) / 2)
if (p == 4 && !colAccept(i, j))
return false;
//diagonal
if (p == 4 && i == 4 && !diag(j))
return false;
// SubAccept#2
//------------------------------------------
if (p == 4 && i == 2 && !subAccept(p, i, j))
return false;
if (p == 4 && i == 4 && !subAccept(p, i, j))
return false;
//------------------------------------------
return true;
}
else
{
cout << "You only have time for 3x3 and 4x4.\n";
exit(0);
}
}
//Basically permutations with backtracking
void perm(int i, int p) {
for (int j = 1; j <= k; j++) {
//If the number is not already in the square AND if the criteria for magic square is met
if (!member[j] && magic(p, i, j)) {
y[p][i] = j; //Add to the magic square
member[j] = true; //It is now a member
if (i < n) {
perm(i + 1, p); //If FIRST ROW isn't full go again
}
else if (rowAccept(p)) {//If the SUM of the row is (n*(n*n + 1) / 2) {3x3: 15; 4x4: 34} continue
if (p == n) { //If we are in the last ROW
show(); //Show the solution
p = 1; //back to 1st ROW
nr++; //Count the solutions
}
else {
perm(1, p + 1); //If it isn't the last ROW go to the next one
}
}
member[j] = false; //Number is no longer a member
} //I'm not sure this is in the right place but I no longer know where to move it
}
}
int main() {
//Make sure there are no members
for (int j = 1; j <= k; j++)
member[j] = false;
//3x3 or 4x4; else exit(0)
cout << "Magic Square size 3x3 or 4x4 ?\n";
cin >> n;
k = n*n;
//Stop at every solution ?
cout << "Debug ?(0/1)\n";
cin >> dbug;
//Start timing and GO
int start_s = clock();
perm(1, 1);
int stop_s = clock();
cout << "Result: " << nr << endl;
cout << "Time: " << (stop_s - start_s) / double(CLOCKS_PER_SEC) << " seconds\n";
}
我知道这很长，我的意大利面不容易理解，但我试着尽可能地评论。另外我很抱歉发布了整个内容，但我不知道哪个部分搞砸了解决方案的数量。
答案 0 :(得分：0)
解决了！
事实证明，有两种类型的Magic Squares。
一方面它们被限制为数字n ^ 2（我的情况），另一方面它们可以包含任何数字，只要rowSUM，colSUM和diagSUM相等。
我的magic（...）函数中关于subAccept（），miniCorners（）和其他中断条件的条件不适用于我的情况。我拿出来但现在代码速度较慢但结果是正确的。