四阶素数幻方问题
用1到16构成一个四阶幻方,要求任意相邻两个方格中的数字之各均为素数?
(原帖见:http://topic.csdn.net/u/20070830/18/1f1957c1-5e66-4c3b-8883-d7eef64c8da1.html)
NowCan 网友的解法:
直接递归搜，4阶很快的。以下这个程序就是这样的思路，结果未经过验证。
/*
将1-N^2这N^2个数添如N*N的方格中，每个方格填一个整数，使所有相邻两个方格内的两个整数之和为质数。
例如N=3时如下，无解。N=4时，2992解？N=5时，917848解？算了半个小时
A0 A1 A2
A3 A4 A5
A6 A7 A8
PRIME(A0+A1)
PRIME(A0+A3)
PRIME(A1+A2)
PRIME(A1+A4)
PRIME(A2+A5)
PRIME(A3+A4)
PRIME(A3+A6)
PRIME(A4+A5)
PRIME(A4+A7)
PRIME(A5+A8)
PRIME(A6+A7)
PRIME(A7+A8)
*/
#include <stdio.h >
#include <math.h >
#include <windows.h >
#define MAX_NUM 30
#define _PRINT_ 0
unsigned long Result[MAX_NUM * MAX_NUM], ResultNum, Used[MAX_NUM * MAX_NUM]={0};
bool PrimeTable[MAX_NUM * MAX_NUM * 2]={ false };
unsigned long N, QN;
/* */
void CreatePrimeTable(void)
{
PrimeTable[0]=false;
PrimeTable[1]=false;
PrimeTable[2]=true;
PrimeTable[3]=true;
for(unsigned long j=5; j <= MAX_NUM * MAX_NUM * 2; j+=2)
{
PrimeTable[j]=true;
for(unsigned long i=3; i <= sqrt((double)j); i+=2)
{
if(j % i == 0)
{
PrimeTable[j]=false;
break;
}
}
}
}
/* */
inline bool IsPrime(unsigned long n)
{
return PrimeTable[n];
}
/* */
bool CheckIt(unsigned long Deep)
{
if(Deep == 0)
{
return true;
}
else if(Deep < N)
{
return IsPrime(Result[Deep] + Result[Deep - 1]);
}
else if(Deep % N == 0)
{
return IsPrime(Result[Deep] + Result[Deep - N]);
}
else
{
return(IsPrime(Result[Deep] + Result[Deep - 1]) && IsPrime(Result[Deep] + Result[Deep - N]));
}
}
/* */
void go(unsigned long Deep)
{
if(Deep == QN)
{
ResultNum++;
#if (_PRINT_)
printf("Find it! No.%lu/n", ResultNum);
for(unsigned long i=0; i < QN; i++)
{
printf("%lu/t", Result[i]);
if(i % N == N - 1)
{
printf("/n");
}
}
#else
printf("/rFind:%lu", ResultNum);
#endif
}
else
{
for(unsigned long i=1; i <= QN; ++i)
{
if(!Used[i])
{
Result[Deep]=i;
if(CheckIt(Deep))
{
Used[i]=1;
go(Deep + 1);
Used[i]=0;
}
}
}
}
}
/* */
int main(void)
{
DWORD tim;
ResultNum=0;
printf("Input N:");
scanf("%lu", &N);
QN=N * N;
tim=GetTickCount();
CreatePrimeTable();
go(0);
printf("/n/nN=%lu/n", N);
printf("Total=%lu/n", ResultNum);
printf("Time=%lu/n", GetTickCount() - tim);
return 0;
}
几乎未加任何处理,转成VB代码如下:
Dim prime() As Byte, result() As Long, used() As Byte, n As Long, resultcount As Long
Function checkit(ByVal level As Long) As Byte
If level = 0 Then checkit = 1: Exit Function
If level < n Then checkit = prime(result(level) + result(level - 1)): Exit Function
If level Mod n = 0 Then checkit = prime(result(level) + result(level - n)): Exit Function
checkit = prime(result(level) + result(level - 1)) * prime(result(level) + result(level - n))
End Function
Sub run(Optional ByVal level As Long = 0)
Dim i As Long, j As Long
If level = n * n Then
resultcount = resultcount + 1
Debug.Print "No." & resultcount & vbCrLf & String(4 * n, "-")
For i = 0 To n - 1
For j = 0 To n - 1
Debug.Print Left(result(i * n + j) & " ", 4);
Next
Debug.Print
Debug.Print
Next
Else
For i = 1 To n * n
If used(i) = 0 Then
result(level) = i
If checkit(level) = 1 Then
used(i) = 1
run level + 1
used(i) = 0
End If
End If
Next
End If
Close #1
End Sub
Sub main()
Dim i As Long, tt As Double
n = 4
ReDim prime(n * n * 2)
ReDim result(n * n)
ReDim used(n * n)
ReDim s(1 To 1000000)
prime(2) = 1
prime(3) = 1
For i = 5 To 2 * n * n Step 2
For j = 3 To Int(Sqr(i))
If i Mod j = 0 Then Exit For
Next
If j = Int(Sqr(i)) + 1 Then prime(i) = 1
Next
tt = Timer
run
MsgBox "共找到" & resultcount & "组解，用时" & Timer - tt & "秒钟!"
End Sub
返回:
No.1
----------------
1 2 11 12
4 9 8 5
7 10 3 14
6 13 16 15
No.2
----------------
1 2 11 12
4 9 8 5
13 10 3 14
6 7 16 15
No.3
----------------
1 2 11 12
4 15 8 5
7 16 3 14
6 13 10 9
No.4
----------------
1 2 11 12
4 15 8 5
13 16 3 14
6 7 10 9
No.5
----------------
1 2 11 12
10 3 8 5
7 16 15 14
6 13 4 9
No.6
----------------
1 2 11 12
10 3 8 5
13 16 15 14
6 7 4 9
No.7
----------------
1 2 11 12
10 9 8 5
7 4 3 14
6 13 16 15
No.8
----------------
1 2 11 12
10 9 8 5
7 4 15 14
6 13 16 3
No.9
----------------
1 2 11 12
10 9 8 5
13 4 3 14
6 7 16 15
No.10
----------------
1 2 11 12
10 9 8 5
13 4 15 14
6 7 16 3
No.11
----------------
1 2 11 12
16 3 8 5
7 10 9 14
6 13 4 15
No.12
----------------
1 2 11 12
16 3 8 5
13 10 9 14
6 7 4 15
No.13
----------------
1 2 11 12
16 15 8 5
7 4 3 14
6 13 10 9
No.14
----------------
1 2 11 12
16 15 8 5
7 4 9 14
6 13 10 3
No.15
----------------
1 2 11 12
16 15 8 5
13 4 3 14
6 7 10 9
No.16
----------------
1 2 11 12
16 15 8 5
13 4 9 14
6 7 10 3
.......
共计2992组解法
