寻找 DNA Java 的超序列

我正在努力研究“查找超序列”算法。

输入用于一组字符串

String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";

结果将是正确对齐的字符串集（下一步应该是合并）

String E = "ca ag cca  cc ta    cat  c a";
String F = "c gag ccat ccgtaaa g  tt  g";
String G = " aga acc tgc  taaatgc t a ga";

感谢您的任何建议（我在这项任务上呆了一天多）

合并后超字符串将是

cagagaccatgccgtaaatgcattacga

“这种情况”中超序列的定义类似于

当且仅当字符串 R 中的所有字符按照它们在输入序列 R 中出现的顺序出现在超级序列 S 中时，字符串 R 才包含在超级序列 S 中。

我尝试过的“解决方案”（这又是错误的做法）是：

public class Solution4
{
    static  boolean[][] map = null;
    static int size = 0;

    public static void main(String[] args)
    {
        String A = "caagccacctacatca";
        String B = "cgagccatccgtaaagttg";
        String C = "agaacctgctaaatgctaga";

        Stack data = new Stack();
        data.push(A);
        data.push(B);
        data.push(C);


        Stack clone1 = data.clone();
        Stack clone2 = data.clone();

        int length  =  26;
        size        =  max_size(data);

        System.out.println(size+" "+length);
        map = new boolean[26][size];

        char[] result = new char[size];

        HashSet<String> chunks = new HashSet<String>();
        while(!clone1.isEmpty())
        {
            String a = clone1.pop();

            char[] residue = make_residue(a);

            System.out.println("---");
            System.out.println("OLD     : "+a);
            System.out.println("RESIDUE : "+String.valueOf(residue));


            String[] r = String.valueOf(residue).split(" ");

            for(int i=0; i<r.length; i++)
            {
                if(r[i].equals(" ")) continue;
                //chunks.add(spaces.substring(0,i)+r[i]);
                chunks.add(r[i]);
            }
        }

        for(String chunk : chunks)
        {
            System.out.println("CHUNK   : "+chunk);
        }
    }

    static char[] make_residue(String candidate)
    {
        char[] result = new char[size];
        for(int i=0; i<candidate.length(); i++)
        {
            int pos = find_position_for(candidate.charAt(i),i);
            for(int j=i; j<pos; j++) result[j]=' ';
            if(pos==-1) result[candidate.length()-1] = candidate.charAt(i);
            else        result[pos] = candidate.charAt(i);
        }
        return result;
    }

    static int find_position_for(char character, int offset)
    {
        character-=((int)'a');

        for(int i=offset; i<size; i++)
        {
        //  System.out.println("checking "+String.valueOf((char)(character+((int)'a')))+" at "+i);
            if(!map[character][i])
            {
                map[character][i]=true;
                return i;
            }
        }
        return -1;
    }

    static String move_right(String a, int from)
    {
        return a.substring(0, from)+" "+a.substring(from);  
    }


    static boolean taken(int character, int position)
    { return map[character][position]; }

    static void take(char character, int position)
    {
        //System.out.println("taking "+String.valueOf(character)+" at "+position+" (char_index-"+(character-((int)'a'))+")");
        map[character-((int)'a')][position]=true;
    }

    static int max_size(Stack stack)
    {
        int max=0;
        while(!stack.isEmpty())
        {
            String s = stack.pop();
            if(s.length()>max) max=s.length();
        }

        return max;
    }

}

找到任何公共超序列并不是一件困难的任务：

在您的示例中，可能的解决方案类似于：

公共类 SuperSequenceTest {

public static void main(String[] args) {
    String A = "caagccacctacatca";
    String B = "cgagccatccgtaaagttg";
    String C = "agaacctgctaaatgctaga";

    int iA = 0;
    int iB = 0;
    int iC = 0;

    char[] a = A.toCharArray();
    char[] b = B.toCharArray();
    char[] c = C.toCharArray();


    StringBuilder sb = new StringBuilder();

    while (iA < a.length || iB < b.length || iC < c.length) {
        if (iA < a.length && iB < b.length && iC < c.length && (a[iA] == b[iB]) && (a[iA] == c[iC])) {
            sb.append(a[iA]);
            iA++;
            iB++;
            iC++;
        }
        else if (iA < a.length && iB < b.length && a[iA] == b[iB]) {
            sb.append(a[iA]);
            iA++;
            iB++;
        }
        else if (iA < a.length && iC < c.length && a[iA] == c[iC]) {
            sb.append(a[iA]);
            iA++;
            iC++;
        }
        else if (iB < b.length && iC < c.length && b[iB] == c[iC]) {
            sb.append(b[iB]);
            iB++;
            iC++;
        } else {
            if (iC < c.length) {
                sb.append(c[iC]);
                iC++;
            }
            else if (iB < b.length) {
                sb.append(b[iB]);
                iB++;
            } else if (iA < a.length) {
                sb.append(a[iA]);
                iA++;
            }
        }
    }
    System.out.println("SUPERSEQUENCE " + sb.toString());
}

}

然而，真正要解决的问题是找到已知问题的解决方案最短公共超序列 http://en.wikipedia.org/wiki/Shortest_common_supersequence http://en.wikipedia.org/wiki/Shortest_common_supersequence, 这并不容易。

有很多与该主题相关的研究。

例如参见：

http://www.csd.uwo.ca/~lila/pdfs/Towards%20a%20DNA%20solution%20to%20the%20Shortest%20Common%20Superstring%20Problem.pdf http://www.csd.uwo.ca/~lila/pdfs/Towards%20a%20DNA%20solution%20to%20the%20Shortest%20Common%20Superstring%20Problem.pdf

http://www.ncbi.nlm.nih.gov/pubmed/14534185 http://www.ncbi.nlm.nih.gov/pubmed/14534185