Question

我在网上发现了一篇关于越南三年级数学的文章。我想看看我们如何解决这个问题会很有趣。以数学或程序方式。

我为此编写了一个测试函数，可以随意使用任何您熟悉的编程语言：

var answer = [1,1,1,1,1,1,1,1,1,1];
var isCorrect = function (answer) {
    return ((((((((((((a[0]+13)*a[1])/a[2])+a[3])+12)*a[4])-a[5])-11)+a[6])*a[7])/a[8])-10) === 66;
};

规则是数组中不应有可重复的数字。接受的号码是1-9。

Answer 1

from constraint import *

def formula(a, b, c, d, e, f, g, h, i):
    return abs(a + (13 * (b / c)) + d + (12 * e) - f -
               11 + ((g * h) / i) - 10 - 66) < 0.001

if __name__ == '__main__':
    all_vals = [float(i) for i in xrange(1, 10)]
    p = Problem()
    [p.addVariable(l, all_vals) for l in 'abcdefghi']

    p.addConstraint(AllDifferentConstraint())
    p.addConstraint(FunctionConstraint(formula), 'abcdefghi')
    result = p.getSolution()
    print [s for s in sorted(result.items(), key=lambda (k,v): k)]

答案是：

[('a', 9.0), ('b', 8.0), ('c', 6.0), 
 ('d', 2.0), ('e', 4.0), ('f', 1.0), 
 ('g', 5.0), ('h', 7.0), ('i', 3.0)]

Answer 2

Maraca，您的MATLAB代码运行良好，但应更改S行以获得正确的算术结果。新代码是：

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例如，一个测试解决方案是（a b c d e f g h i）=（3 2 1 5 4 7 9 8 6）

在代码中使用此更改，有128种独特的解决方案。

请参阅link。

Answer 3

Matlab / Octave（蛮力）：

p = perms(1:9) ;
S = (((((((((((p(:,1)+13).*p(:,2))./p(:,3)+p(:,4))+12).*p(:,5))-p(:,6))-11)+p(:,7)).*p(:,8))./p(:,9))-10) == 66 ;
sum(S)
p(S,:)

我在八度音乐在线执行它，它表明有 144个解决方案只需要一小部分时间来计算。（http://octave-online.net初始弹出窗口看起来很可怕，但它只是试图帮助，点击它并将整个程序代码复制到底部的控制台行中并点击[返回]以查看所有144个解决方案，有时你得到错误＆＃34;有效载荷太大＆＃34;但它只显示，数学在后台工作正常。）

<强>解释

对于数学问题，显然一种旨在解决这些问题的编程语言是一个很好的选择，并且可以从你的肩膀上完成很多工作，所以我使用Octave虽然我是Java程序员，这里是该计划的作用：

1:9会提供一个数字为1到9（1 x 9）的向量
perms()计算此向量的所有排列并将其作为矩阵返回（9！x 9）
;会抑制输出
p =置换矩阵存储在p
p(:,n)访问n的{{1}}列，因此所有解决方案都会立即计算
p和.*确保单元格乘法而不是标准矩阵乘法
结果向量与./的比较会返回一个逻辑向量，该向量存储在66中，当方程为真且{{1}时，它就像一个S的向量否则（9！x 1）
1输出解决方案的数量
0返回之前的等式为真的所有行，解决方案（144 x 9）

Answer 4

我想可能有一种更有效的方法解决这个问题但是，scince只有9个变量，其中有9个可能值，而且我们有计算机，蛮力。

我们的想法是为每个排列尝试check（）函数，即你可以将数字1到9放在数组a []中。这意味着9！迭代= 362880我的笔记本电脑可以在几秒钟内完成。

我构建了recursove permutate（）：对于每个级别它，inmutate函数循环数字1到9，将它们中的每一个放在数组的级别位置（但仅当数字不在先前位置时）并且自己调用下一级别。在级别9中，数组已满，因此可以调用检查功能。

这是java中的代码：

public class Test {
    static float[] a = { 0, 0, 0, 0, 0, 0, 0, 0, 0};
private static int countSolutions = 0;
public static void main(String[] args){
    permutate(0);
    System.out.println(countSolutions);
}

public static boolean permutate( int level){
    boolean isCorrect = false;
    if(level == 9){
        isCorrect = check();
        if(!isCorrect){
        }
    }
    else 
        for(int i = 1; i <= 9; i++){
            int j = 0;
            for(; j < level && a[j]!=i; j++);
            if( j == level){
                a[level] = i;
                isCorrect = permutate( level + 1);
            }
        }
    return isCorrect;
}

private static boolean check() {
    float result = ((((((((((((a[0]+13)*a[1])/a[2])+a[3])+12)*a[4])-a[5])-11)+a[6])*a[7])/a[8])-10);
    if ( result == 66){
        countSolutions ++;
        for (int i = 0; i < 9; i++) {
            System.out.print(a[i]+",");
        }
        System.out.println();
        return true;
    }else{
        return false;
    }
}

}

结果（144）：

1.0 2.0 4.0 7.0 5.0 8.0 3.0 6.0 9.0

1.0 2.0 7.0 5.0 3.0 4.0 9.0 8.0 6.0

...

流行测验：解决数学难题？

4 个答案: