Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
342 views
in Technique[技术] by (71.8m points)

c# - Creating a power set of a Sequence

I am trying to create a program that is a base for creating possible combinations of a sequence, string or a number. This is some sort of encryption / decryption program. I am using Visual Studio 2013 and C#. What I am trying to do is to generate a power set out of a sequence, but I am little bit confused and can't proceed any further. Here is the code.

public static void randomSeq()
{
    int temp = 0;
    string seq = "1234";

    var sb = new StringBuilder();
    char[] bits = seq.Select((char c) => c).ToArray();

    Console.Write("Given Sequence: ");
    Console.Write(seq);
    Console.WriteLine();
    Console.WriteLine("Generated possiblities");

    foreach (char item in bits)
        Console.WriteLine(item);
    do
    {
        if (temp <= 2)
        {
            for (int i = temp + 1; i < bits.Length; i++)
            {
                 sb.Append(bits[temp]);
                 sb.Append(bits[i]);
                 Console.WriteLine(sb);
                
                 sb.Clear();
            }
        }
        else if (temp > 2)
        {
            for (int k = 0; k < temp; k++)
                sb.Append(bits[k]);

            for (int l = temp + 1; l < bits.Length; l++)
            {
                sb.Append(bits[temp]);
                sb.Append(bits[l]);
                Console.WriteLine(sb);

                sb.Clear();
            }
        }
        temp++;
    }
    while (temp != bits.Length);
}

I want this code to be generic i.e. I pass any sequence and it generates a power set for me. Then I want to reuse it further in my program. I can do the rest simply I am stuck in generating the power set. Can someone help me?.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Power set is easy to generate if one is familiar with bits. For the set of N elements, there will be 2^N subsets which will go to power set (including empty set and initial set). So each element will be either IN or OUT (1 or 0 in other words).

Taking this into consideration, it is easy to represent subsets of the set as bit masks. Then enumerating through all possible bit masks, it is possible construct the whole power sets. In order to do this we need to examine each bit in bit mask and take element of input set if there is 1 in that place. Below is example for string (collection of chars) as input. It can be easily rewritten to work for collection of any type values.

private static List<string> PowerSet(string input)
{
    int n = input.Length;
    // Power set contains 2^N subsets.
    int powerSetCount = 1 << n;
    var ans = new List<string>();

    for (int setMask = 0; setMask < powerSetCount; setMask++)
    {
        var s = new StringBuilder();
        for (int i = 0; i < n; i++)
        {
            // Checking whether i'th element of input collection should go to the current subset.
            if ((setMask & (1 << i)) > 0)
            {
                s.Append(input[i]);
            }
        }
        ans.Add(s.ToString());
    }

    return ans;
}

Example

Suppose you have string "xyz" as input, it contains 3 elements, than there will be 2^3 == 8 elements in power set. If you will be iterating from 0 to 7 you will get the following table. Columns: (10-base integer; bits representation (2-base); subset of initial set).

0   000    ...
1   001    ..z
2   010    .y.
3   011    .yz
4   100    x..
5   101    x.z
6   110    xy.
7   111    xyz

You can notice that third column contains all subsets of initial string "xyz"


Another approach (twice faster) and generic implementation

Inspired by Eric's idea, I have implemented another variant of this algorithm (without bits now). Also I made it generic. I believe this code is near to fastest of what can be written for Power Set calculation. Its complexity is the same as for bits approach O(n * 2^n), but for this approach constant is halved.

public static T[][] FastPowerSet<T>(T[] seq)
{
    var powerSet = new T[1 << seq.Length][];
    powerSet[0] = new T[0]; // starting only with empty set

    for (int i = 0; i < seq.Length; i++)
    {
        var cur = seq[i];
        int count = 1 << i; // doubling list each time
        for (int j = 0; j < count; j++)
        {
            var source = powerSet[j];
            var destination = powerSet[count + j] = new T[source.Length + 1];
            for (int q = 0; q < source.Length; q++)
                destination[q] = source[q];
            destination[source.Length] = cur;
        }
    }
    return powerSet;
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...