SSP算法长度为k的最小子集

时间:2018-05-29 23:50:01

标签: subset-sum pari

假设S是一个具有模数为n的t个元素的集合。确实有2 ^ t个任意长度的子集。说明PARI / GP程序,其找到不同元素的最小子集U(就长度而言),使得U中所有元素的总和为0模n。编写一个通过强力搜索的程序很容易,但是当t和n变大时,暴力是不可行的,所以希望能帮助编写一个不使用强力来解决subset sum problem这个实例的程序。

2 个答案:

答案 0 :(得分:1)

动态方法:

def isSubsetSum(st, n, sm) :

    # The value of subset[i][j] will be
    # true if there is a subset of 
    # set[0..j-1] with sum equal to i
    subset=[[True] * (sm+1)] * (n+1)

    # If sum is 0, then answer is true
    for i in range(0, n+1) :
        subset[i][0] = True

    # If sum is not 0 and set is empty,
    # then answer is false
    for i in range(1, sm + 1) :
        subset[0][i] = False

    # Fill the subset table in botton 
    # up manner
    for i in range(1, n+1) :
        for j in range(1, sm+1) :
            if(j < st[i-1]) :
                subset[i][j] = subset[i-1][j]
            if (j >= st[i-1]) :
                subset[i][j] = subset[i-1][j] or subset[i - 1][j-st[i-1]]

    """uncomment this code to print table
    for i in range(0,n+1) :
        for j in range(0,sm+1) :
            print(subset[i][j],end="")
    print(" ")"""

    return subset[n][sm];

答案 1 :(得分:0)

我从here获得了此代码。我不知道天气似乎有效。

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function getSummingItems(a,t){
  return a.reduce((h,n) => Object.keys(h)
                                 .reduceRight((m,k) => +k+n <= t ? (m[+k+n] = m[+k+n] ? m[+k+n].concat(m[k].map(sa => sa.concat(n)))
                                                                                      : m[k].map(sa => sa.concat(n)),m)
                                                                 :  m, h), {0:[[]]})[t];
}
var arr = Array(20).fill().map((_,i) => i+1), // [1,2,..,20]
    tgt = 42,
    res = [];

console.time("test");
res = getSummingItems(arr,tgt);
console.timeEnd("test");
console.log("found",res.length,"subsequences summing to",tgt);
console.log(JSON.stringify(res));
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