Соответствующий алгоритм в Javascript

Question

Я ищу реализацию алгоритма Blossom для JavaScript или чего-то подобного.

У меня есть набор пар

A - B

А - С

B - D

и мне нужно выбрать пары, при условии, что каждая буква может только в конечном итоге один раз на выходе. В приведенном выше случае, правильный результат был бы

А - С

В - D

потому, что А, В, С и D все в конечном итоге в результатах. Неверный результат будет

А - В

, который оставил бы C и D вне.

Answer 1

Несомненно, почему бы и нет?

/* 
    Edmonds's maximum matching algorithm 
    Complexity: O(v^3) 
    Written by Felipe Lopes de Freitas 
    Adapted to JavaScript from C++ (http://pastebin.com/NQwxv32y) by גלעד ברקן 
*/ 

var MAX = 100, 
    undef = -2, 
    empty = -1, 
    noEdge = 0, 
    unmatched = 1, 
    matched = 2, 
    forward = 0, 
    reverse = 0; 

           //Labels are the key to this implementation of the algorithm. 
function Label(){    //An even label in a vertex means there's an alternating path 
     this.even = undefined; //of even length starting from the root node that ends on the 
     this.odd = new Array(2); //vertex. To find this path, the backtrace() function is called, 
};        //constructing the path by following the content of the labels. 
           //Odd labels are similar, the only difference is that base nodes 
           //of blossoms inside other blossoms may have two. More on this later. 

function elem(){    //This is the element of the queue of labels to be analyzed by 
     this.vertex = undefined; 
     this.type = undefined; //the augmentMatching() procedure. Each element contains the vertex 
};        //where the label is and its type, odd or even. 

var g = new Array(MAX);   //The graph, as an adjacency matrix. 
for (var i=0; i<MAX; i++){ 
    g[i] = new Array(MAX); 
} 
           //blossom[i] contains the base node of the blossom the vertex i 
var blossom = new Array(MAX); //is in. This, together with labels eliminates the need to 
           //contract the graph. 

           //The path arrays are where the backtrace() routine will 
var path = new Array(2); 
for (var i=0; i<2; i++){ 
    path[i] = new Array(MAX); 
} 
var endPath = new Array(2); //store the paths it finds. Only two paths need to be 
           //stored. endPath[p] denotes the end of path[p]. 
var match = new Array(MAX); //An array of flags. match[i] stores if vertex i is in the matching. 
        //label[i] contains the label assigned to vertex i. It may be undefined, 
var label = new Array(MAX); //empty (meaning the node is a root) and a node might have even and odd 
        //labels at the same time, which is the case for nonbase nodes of blossoms 
for (var i=0; i<MAX; i++){ 
    label[i] = new Label(); 
} 
var queue = new Array(2*MAX);   //The queue is necessary for efficiently scanning all labels. 
var queueFront,queueBack; //A label is enqueued when assigned and dequeued after scanned. 
for (var i=0; i<2*MAX; i++){ 
    queue[i] = new elem(); 
} 

function initGraph(n){ 
    for (var i=0; i<n; i++) 
     for (var j=0; j<n; j++) g[i][j]=noEdge; 
} 

function readGraph(){ 

    var n = graph.n, 
     e = graph.e; 

    //int n,e,a,b; 
    //scanf(" %d %d",&n,&e);  //The graph is read and its edges are unmatched by default. 
    initGraph(n);    //Since C++ arrays are 0..n-1 and input 1..n , subtractions 
    for (var i=0; i<e; i++){ //are made for better memory usage. 
     //scanf(" %d %d",&a,&b); 
     var a = graph[i][0], 
      b = graph[i][1]; 
     if (a!=b) 
      g[a-1][b-1]=g[b-1][a-1]=unmatched; 
    } 
    return n; 
} 

function initAlg(n){    //Initializes all data structures for the augmentMatching() 
    queueFront=queueBack=0;  //function begin. At the start, all labels are undefined, 
    for (var i=0; i<n; i++){ //the queue is empty and a node alone is its own blossom. 
     blossom[i]=i; 
     label[i].even=label[i].odd[0]=label[i].odd[1]=undef; 
    } 
} 

function backtrace (vert, pathNum, stop, parity, direction){ 
    if (vert==stop) return;   //pathNum is the number of the path to store 
    else if (parity==0){    //vert and parity determine the label to be read. 
     if (direction==reverse){ 
      backtrace(label[vert].even,pathNum,stop,(parity+1)%2,reverse); 
      path[pathNum][endPath[pathNum]++]=vert; 
     }        //forward means the vertices called first enter 
     else if (direction==forward){ //the path first, reverse is the opposite. 
      path[pathNum][endPath[pathNum]++]=vert; 
      backtrace(label[vert].even,pathNum,stop,(parity+1)%2,forward); 
     } 
    } 
    /* 
     stop is the stopping condition for the recursion. 
     Recursion is necessary because of the possible dual odd labels. 
     having empty at stop means the recursion will only stop after 
     the whole tree has been climbed. If assigned to a vertex, it'll stop 
     once it's reached. 
    */ 
    else if (parity==1 && label[vert].odd[1]==undef){ 
     if (direction==reverse){ 
      backtrace(label[vert].odd[0],pathNum,stop,(parity+1)%2,reverse); 
      path[pathNum][endPath[pathNum]++]=vert; 
     } 
     else if (direction==forward){ 
      path[pathNum][endPath[pathNum]++]=vert; 
      backtrace(label[vert].odd[0],pathNum,stop,(parity+1)%2,forward); 
     } 
    } 
    /* 
     Dual odd labels are interpreted as follows: 
     There exists an odd length alternating path starting from the root to this 
     vertex. To find this path, backtrace from odd[0] to the top of the tree and 
     from odd[1] to the vertex itself. This, put in the right order, will 
     constitute said path. 
    */ 
    else if (parity==1 && label[vert].odd[1]!=undef){ 
      if (direction==reverse){ 
      backtrace(label[vert].odd[0],pathNum,empty,(parity+1)%2,reverse); 
      backtrace(label[vert].odd[1],pathNum,vert,(parity+1)%2,forward); 
      path[pathNum][endPath[pathNum]++]=vert; 
      } 
      else if (direction==forward){ 
       backtrace(label[vert].odd[1],pathNum,vert,(parity+1)%2,reverse); 
       backtrace(label[vert].odd[0],pathNum,empty,(parity+1)%2,forward); 
       path[pathNum][endPath[pathNum]++]=vert; 
      } 
    } 
} 

function enqueue (vert, t){ 
    var tmp = new elem();    //Enqueues labels for scanning. 
    tmp.vertex=vert;  //No label that's dequeued during the execution 
    tmp.type=t;    //of augmentMatching() goes back to the queue. 
    queue[queueBack++]=tmp; //Thus, circular arrays are unnecessary. 
} 

function newBlossom (a, b){  //newBlossom() will be called after the paths are evaluated. 
    var i,base,innerBlossom,innerBase; 
    for (i=0; path[0][i]==path[1][i]; i++); //Find the lowest common ancestor of a and b 
    i--;          //it will be used to represent the blossom. 
    base=blossom[path[0][i]];     //Unless it's already contained in another... 
               //In this case, all will be put in the older one. 
    for (var j=i; j<endPath[0]; j++) blossom[path[0][j]]=base; 
    for (var j=i+1; j<endPath[1]; j++) blossom[path[1][j]]=base; //Set all nodes to this 
    for (var p=0; p<2; p++){          //new blossom. 
     for (var j=i+1; j<endPath[p]-1; j++){ 
      if (label[path[p][j]].even==undef){  //Now, new labels will be applied 
       label[path[p][j]].even=path[p][j+1]; //to indicate the existence of even 
       enqueue(path[p][j],0);     //and odd length paths. 
      } 
      else if (label[path[p][j]].odd[0]==undef && label[path[p][j+1]].even==undef){ 
       label[path[p][j]].odd[0]=path[p][j+1]; 
       enqueue(path[p][j],1);     //Labels will only be put if the vertex 
      }           //doesn't have one. 

      else if (label[path[p][j]].odd[0]==undef && label[path[p][j+1]].even!=undef){ 
       /* 
        If a vertex doesn't have an odd label, but the next one in the path 
        has an even label, it means that the current vertex is the base node 
        of a previous blossom and the next one is contained within it. 
        The standard labeling procedure will fail in this case. This is fixed 
        by going to the last node in the path inside this inner blossom and using 
        it to apply the dual label. 
        Refer to backtrace() to know how the path will be built. 
       */ 
       innerBlossom=blossom[path[p][j]]; 
       innerBase=j; 
       for (; blossom[j]==innerBlossom && j<endPath[p]-1; j++); 
       j--; 
       label[path[p][innerBase]].odd[0]=path[p][j+1]; 
       label[path[p][innerBase]].odd[1]=path[p][j]; 
       enqueue(path[p][innerBase],1); 
      } 
     } 
    } 
    if (g[a][b]==unmatched){   //All nodes have received labels, except 
     if (label[a].odd[0]==undef){ //the ones that called the function in 
      label[a].odd[0]=b;   //the first place. It's possible to 
      enqueue(a,1);    //find out how to label them by 
     }        //analyzing if they're in the matching. 
     if (label[b].odd[0]==undef){ 
      label[b].odd[0]=a; 
      enqueue(b,1); 
     }        
    } 
    else if (g[a][b]==matched){ 
      if (label[a].even==undef){ 
      label[a].even=b; 
      enqueue(a,0); 
      } 
      if (label[b].even==undef){ 
      label[b].even=a; 
      enqueue(b,0); 
      } 
    } 
} 

function augmentPath(){   //An augmenting path has been found in the matching 
    var a,b;     //and is contained in the path arrays. 
    for (var p=0; p<2; p++){ 
     for (var i=0; i<endPath[p]-1; i++){ 
      a=path[p][i];    //Because of labeling, this path is already 
      b=path[p][i+1];   //lifted and can be augmented by simple 
      if (g[a][b]==unmatched) //changing of the matching status. 
       g[a][b]=g[b][a]=matched; 
      else if (g[a][b]==matched) 
        g[a][b]=g[b][a]=unmatched; 
     } 
    } 
    a=path[0][endPath[0]-1]; 
    b=path[1][endPath[1]-1]; 
    if (g[a][b]==unmatched) g[a][b]=g[b][a]=matched; 
    else if (g[a][b]==matched) g[a][b]=g[b][a]=unmatched; 
    //After this, a and b are included in the matching. 
    match[path[0][0]]=match[path[1][0]]=true; 
} 

function augmentMatching (n){ //The main analyzing function, with the 
    var node,nodeLabel;  //goal of finding augmenting paths or 
    initAlg(n);    //concluding that the matching is maximum. 
    for (var i=0; i<n; i++) if (!match[i]){ 
     label[i].even=empty; 
     enqueue(i,0);   //Initialize the queue with the exposed vertices, 
    }       //making them the roots in the forest. 

    while (queueFront<queueBack){ 
     node=queue[queueFront].vertex; 
     nodeLabel=queue[queueFront].type; 
     if (nodeLabel==0){ 
      for (var i=0; i<n; i++) if (g[node][i]==unmatched){ 
       if (blossom[node]==blossom[i]); 
       //Do nothing. Edges inside the same blossom have no meaning. 
       else if (label[i].even!=undef){ 
        /* 
         The tree has reached a vertex with a label. 
         The parity of this label indicates that an odd length 
         alternating path has been found. If this path is between 
         roots, we have an augmenting path, else there's an 
         alternating cycle, a blossom. 
        */ 
        endPath[0]=endPath[1]=0; 
        backtrace(node,0,empty,0,reverse); 
        backtrace(i,1,empty,0,reverse); 
        //Call the backtracing function to find out. 
        if (path[0][0]==path[1][0]) newBlossom(node,i); 
        /* 
         If the same root node is reached, a blossom was found. 
         Start the labelling procedure to create pseudo-contraction. 
        */ 
        else { 
          augmentPath(); 
          return true; 
          /* 
          If the roots are different, we have an augmenting path. 
          Improve the matching by augmenting this path. 
          Now some labels might make no sense, stop the function, 
          returning that it was successful in improving. 
          */ 
        } 
       } 
       else if (label[i].even==undef && label[i].odd[0]==undef){ 
        //If an unseen vertex is found, report the existing path 
        //by labeling it accordingly. 
        label[i].odd[0]=node; 
        enqueue(i,1); 
       } 
      } 
     } 
     else if (nodeLabel==1){ //Similar to above. 
      for (var i=0; i<n; i++) if (g[node][i]==matched){ 
       if (blossom[node]==blossom[i]); 
       else if (label[i].odd[0]!=undef){ 
        endPath[0]=endPath[1]=0; 
        backtrace(node,0,empty,1,reverse); 
        backtrace(i,1,empty,1,reverse); 
        if (path[0][0]==path[1][0]) newBlossom(node,i); 
        else { 
          augmentPath(); 
          return true; 
        } 
       } 
       else if (label[i].even==undef && label[i].odd[0]==undef){ 
        label[i].even=node; 
        enqueue(i,0); 
       } 
      } 
     } 
     /* 
      The scanning of this label is complete, dequeue it and 
      keep going to the next one. 
     */ 
     queueFront++; 
    } 
    /* 
     If the function reaches this point, the queue is empty, all 
     labels have been scanned. The algorithm couldn't find an augmenting 
     path. Therefore, it concludes the matching is maximum. 
    */ 
    return false; 
} 

function findMaximumMatching (n){ 
    //Initialize it with the empty matching. 
    for (var i=0; i<n; i++) match[i]=false; 
    //Run augmentMatching(), it'll keep improving the matching. 
    //Eventually, it will no longer find a path and break the loop, 
    //at this point, the current matching is maximum. 
    while (augmentMatching(n)); 
} 

function main(){ 
    var n; 
    n=readGraph(); 
    findMaximumMatching(n); 
    for (var i=0; i<n; i++){ 
     for (var j=i+1; j<n; j++) if (g[i][j]==matched) 
      console.log(i+1,j+1); 
    } 
    return 0; 
} 

Выход:

var graph = [[1,2] 
      ,[1,3] 
      ,[2,4]]; 

graph["n"] = 4; 
graph["e"] = 3; 

main() 

1 3 
2 4 

Answer 2

Просто для удовольствия - FIDDLE.

Он читает содержимое divs, а затем создает массив всех возможных перестановок, «пары» (3 'переменные, взятые по 2 за раз).

Затем отдельные буквы, составляющие содержимое разделов, разбираются.

Затем каждая буква сопоставляется с парами, чтобы увидеть, сколько раз это происходит - если это всего лишь один раз для каждой пары, эта конкретная «пара» добавляется к массиву решений.

После анализа всех комбинаций представлен массив решений.

Не очень элегантный, но, похоже, работает.

JS

var totaldivs = $('div').length;     //count total divs or objects 
    console.log("Total divs = " + totaldivs); 

var elements = [];         //Read objects from the divs and assign to array elements 
for(var i = 0; i < totaldivs; i++) 
    { 
    elements[i] = $('div:eq(' + i + ')').text(); 
    elements[i] = elements[i].replace(/\s+/g, ''); 
    console.log(elements[i]); 
    } 

var objects = [];         //make array of all individual objects 
var loopvar = 0; 
for(var n = 0; n < totaldivs; n++) 
    { 
    poshyphen = elements[n].indexOf('-'); 
    objects[loopvar] = elements[n].substring(0, poshyphen); 
    objects[loopvar+1] = elements[n].substring(poshyphen+1, elements[n].length); 
    loopvar = loopvar + 2; 
    } 
$('.putmehere2').html(objects); 

var pair = []; 
var pairindex = 0;         //make and array of all combinations of objects - pair 
for(var r = 0; r < totaldivs; r++) 
    { 
    for(var s = 0; s <totaldivs; s++) 
     { 
     if(elements[r] != elements[s]) 
      { 
      pair[pairindex] = elements[r] + '-' + elements[s]; 
      pairindex++; 
      } 
     } 
    } 

$('.putmehere').html(pair); 
var solution = []; 
var sol = 0; 
var count = []; 
var reg = new RegExp("regex","g"); 

for(var q = 0; q < pair.length; q++) 
    { 
    for(var r = 0; r < 4; r++) 
     { 
     var regex = new RegExp(objects[r], 'g'); 
     count[r] = (pair[q].match(regex)||[]).length; 
     console.log("Pair= " + pair[q] + " Objects = " + objects[r] + " Count= " + count[r]); 
     } 
    if(count[0] == 1 && count[1] == 1 && count[2] == 1 && count[3] == 1) 
      { 
       solution.push(pair[q]); 
      } 
    } 

$('.putmehere2').html(solution);