10

so as fun i decided to write a simple program that can solve the 8 Out of 10 Cats Does Countdown number puzzle, link is form Countdown, but same rules. so my program simply goes through a all possible combinations of AxBxCxDxExF, where letters are numbers and "x" are +, -, / and *. here is the code for it:

private void combineRecursive( int step, int[] numbers, int[] operations, int combination[]){
    if( step%2==0){//even steps are numbers
        for( int i=0; i<numbers.length; i++){
            combination[ step] = numbers[ i];
            if(step==10){//last step, all 6 numbers and 5 operations are placed
                int index = Solution.isSolutionCorrect( combination, targetSolution);
                if( index>=0){
                    solutionQueue.addLast( new Solution( combination, index));
                }
                return;
            }
            combineRecursive( step+1, removeIndex( i, numbers), operations, combination);
        }
    }else{//odd steps are operations
        for( int i=0; i<operations.length; i++){
            combination[ step] = operations[ i];
            combineRecursive( step+1, numbers, operations, combination);
        }
    }
}

and here is what i use to test if the combination is what i want to not.

public static int isSolutionCorrect( int[] combination, int targetSolution){
    double result = combination[0];
    //just a test
    if( Arrays.equals( combination, new int[]{100,'*',7,'-',8,'*',3,'+',7,'+',50})){
        System.out.println( "found");
    }
    for( int i=1; i<combination.length; i++){
        if(i%2!=0){
            switch( (char)combination[i]){
                case '+': result+= combination[++i];break;
                case '-': result-= combination[++i];break;
                case '*': result*= combination[++i];break;
                case '/': result/= combination[++i];break;
            }
        }
        if( targetSolution==result){
            return i;
        }
    }       
    return targetSolution==result?0:-1;
}

so in last episode i found a problem with my code. this was the solution to one of the puzzles.

(10*7)-(8*(3+7))

i noticed that i do find this combination "10*7-8*3+7" (twice), but because i check for solutions by doing the operation left to right i actually don't find all answers. i only check for solutions like this ((((10*7)-8)*3)+7). so even though i found the combination i don't have the right order.

so now the question is how do i test all possible math orders, like (10*7)-(8*(3+7)), (10*7)-((8*3)+7) or 10*(7-8)*(3+7)? i though i can use a balance tree with operations as balancing nodes. but still i have not idea how to go through all possible combinations without moving around the formula.

(10*7)-(8*(3+7))
          -
     /        \
    *         *
  /   \      /  \
700   7     8    +
                / \
              7    3

(10*7)-((8*3)+7)
          -
     /        \
    *         +
  /   \      /  \
700   7     *    7
           / \
          8  3

10*(7-8)*(3+7)

                 *
           /           \
          *
     /        \         10
    -          +
  /   \      /  \
7     8     3    7

how do i do this in code? not looking for solved code more of how i should change perspective to fix it. i don't know why i am stumped at it.

about me: 4th year computer science, not new or noob at programming (i like to believe at least ;))

4

1 回答 1

4

使用表示表达式的专用类而不是数组更容易解决这个问题。然后你可以简单地枚举所有可能的树。我为类似任务编写的另一个答案和一个显示如何生成所有二叉树的答案的混合给出了这个：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class NumberPuzzleWithCats
{
    public static void main(String[] args)
    {
        List<Integer> numbers = Arrays.asList(10,7,8,3,7);
        solve(numbers);
    }

    private static void solve(List<Integer> numbers)
    {
        List<Node> nodes = new ArrayList<Node>();
        for (int i=0; i<numbers.size(); i++)
        {
            Integer number = numbers.get(i);
            nodes.add(new Node(number));
        }
        System.out.println(nodes);
        List<Node> all = create(nodes);
        System.out.println("Found "+all.size()+" combinations");


        for (Node node : all)
        {
            String s = node.toString();
            System.out.print(s);
            if (s.equals("((10*7)-(8*(3+7)))"))
            {
                System.out.println(" <--- There is is :)");
            }
            else
            {
                System.out.println();
            }
        }
    }

    private static List<Node> create(Node n0, Node n1)
    {
        List<Node> nodes = new ArrayList<Node>();
        nodes.add(new Node(n0, '+', n1));
        nodes.add(new Node(n0, '*', n1));
        nodes.add(new Node(n0, '-', n1));
        nodes.add(new Node(n0, '/', n1));
        nodes.add(new Node(n1, '+', n0));
        nodes.add(new Node(n1, '*', n0));
        nodes.add(new Node(n1, '-', n0));
        nodes.add(new Node(n1, '/', n0));
        return nodes;
    }

    private static List<Node> create(List<Node> nodes)
    {
        if (nodes.size() == 1)
        {
            return nodes;
        }
        if (nodes.size() == 2)
        {
            Node n0 = nodes.get(0);
            Node n1 = nodes.get(1);
            return create(n0, n1);
        }
        List<Node> nextNodes = new ArrayList<Node>();
        for (int i=1; i<nodes.size()-1; i++)
        {
            List<Node> s0 = create(nodes.subList(0, i));
            List<Node> s1 = create(nodes.subList(i, nodes.size()));
            for (Node n0 : s0)
            {
                for (Node n1 : s1)
                {
                    nextNodes.addAll(create(n0, n1));
                }
            }
        }
        return nextNodes;
    }

    static class Node
    {
        int value;
        Node left;
        Character op;
        Node right;

        Node(Node left, Character op, Node right)
        {
            this.left = left;
            this.op = op;
            this.right = right;
        }
        Node(Integer value)
        {
            this.value = value;
        }

        @Override
        public String toString()
        {
            if (op == null)
            {
                return String.valueOf(value);
            }
            return "("+left.toString()+op+right.toString()+")";
        }
    }
}

它将打印创建的所有组合，包括您一直在寻找的组合：

[10, 7, 8, 3, 7]
Found 16384 combinations
(10+(7+(8+(3+7))))
(10*(7+(8+(3+7))))
...
((10*7)+(8*(3+7)))
((10*7)*(8*(3+7)))
((10*7)-(8*(3+7))) <--- There is is :)
((10*7)/(8*(3+7)))
((8*(3+7))+(10*7))
...
((7/3)-((8/7)/10))
((7/3)/((8/7)/10))

当然，通过比较String表示来检查是否找到了正确的解决方案是......“非常务实”，这样说，但我认为这一代的实际方法在这里很重要。

（我希望这真的是您一直在寻找的 - 我无法查看您链接到的网站...）

于 2015-07-23T22:38:39.293 回答