Friday, February 19, 2016

LintCode 30- Dice Sum


http://www.lintcode.com/en/problem/dices-sum/
Throw n dices, the sum of the dices' faces is S. Given n, find the all possible value of S along with its probability.
 Notice
You do not care about the accuracy of the result, we will help you to output results.
Given n = 1, return [ [1, 0.17], [2, 0.17], [3, 0.17], [4, 0.17], [5, 0.17], [6, 0.17]].
二维数组dp[i][j]保存投掷i次得到和为j的概率,当前位置的概率为当前投掷1 ~ 6的情况下的前序概率和
  1.     public List<Map.Entry<Integer, Double>> dicesSum(int n) {  
  2.         List<Map.Entry<Integer, Double>> res = new LinkedList();  
  3.         double[][] dp = new double[n + 1][6 * n + 1];  
  4.         for (int i = 1; i <= 6; i++) {  
  5.             dp[1][i] = 1.0 / 6.0;  
  6.         }  
  7.         for (int i = 2; i <= n; i++) {  
  8.             for (int j = i; j <= 6 * i; j++) {  
  9.                 for (int k = 1; k <= 6; k++) {  
  10.                     if (j > k) {  
  11.                         dp[i][j] += dp[i - 1][j - k];  
  12.                     }  
  13.                 }  
  14.                 dp[i][j] /= 6.0;  
  15.             }  
  16.         }  
  17.         for (int i = n; i <= 6 * n; i++) {  
  18.             res.add(new AbstractMap.SimpleEntry<Integer, Double>(i, dp[n][i]));  
  19.         }  
  20.         return res;  
  21.     } 
https://zhengyang2015.gitbooks.io/lintcode/dices_sum_20.html
用dp解决。dp[i][j]表示i个骰子一共得到j点的概率。要得到dp[i][j]可以考虑若最后一个筛子的点数为k(1~6),则前i-1个筛子一共得到的点数为j-k(因为i-1个筛子至少得到i-1点,所以j-k >= i - 1 => k <= j - i + 1)。所以只要把最后一个筛子为k的各种情况加起来最后再除以6即可(每多一个筛子概率要多除以一个6)。
状态函数:
dp[i][j] = dp[i-1][j-k] (i <= j <= 6 * i, k >= 1 && k <= j - i + 1 && k <= 6)

    public List<Map.Entry<Integer, Double>> dicesSum(int n) {
        List<Map.Entry<Integer, Double>> result = new ArrayList<Map.Entry<Integer, Double>>();
        if(n < 1){
            return result;
        }
        //初始化n=1的情况
        double[][] matrix = new double[n + 1][6 * n + 1];
        for(int i = 1; i <= 6; i++){
            matrix[1][i] = 1.0/6;
        }

        for(int i = 2; i <= n; i++){
        //i个筛子至少得到i点,至多得到6 * i点
            for(int j = i; j <= 6 * i; j++){
            //k表示最后一个筛子能取的点数
                for(int k = 1; k <= 6; k++){
                    if(k <= j - i + 1){
                        matrix[i][j] += matrix[i - 1][j - k];
                    }
                }
                //相对i-1个筛子多了一个筛子,因此加和的每一项都要除以6
                matrix[i][j] /= 6.0;
            }
        }

        for(int i = n; i <= 6 * n; i++){
            result.add(new AbstractMap.SimpleEntry<Integer, Double>(i, matrix[n][i]));
        }

        return result;
    }

Buttercola: Indeed: Sum Dice Possibility
这个地里面经没看到过,写一个函数float sumPossibility(int dice, int target),就是投dice个骰子,求最后和为target的概率。因为总共的可能性是6^dice,所以其实就是combination sum,求dice个骰子有多少种组合,使其和为target。先用brute force的dfs来一个O(6^dice)指数复杂度的,然后要求优化,用dp,最后结束代码写的是两者结合的memorized search吧,面试官走的时候还说了句such a good solution。

    public static float sumPossibility(int n, int target) {
        if (n <= 0 || target <= 0) {
            return 0;
        }
       
        int total = (int) Math.pow(6, n);
       
        int[][] dp = new int[n + 1][target + 1];
        dp[0][0] = 1;
       
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= target; j++) {
                for (int k = 1; k <= 6; k++) {
                    if (j >= k) {
                        dp[i][j] += dp[i - 1][j - k];
                    }
                }
            }
        }
       
        return (float) dp[n][target] / total;
    }
Read full article from Buttercola: Indeed: Sum Dice Possibility

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