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Union of disjoint sets


Union of disjoint sets

By : user2954483
Date : November 22 2020, 10:31 AM
help you fix your problem First, when you get keys, you need to find them in the hash table.
Hash table contains entries: (key, pointer-to-node).
code :
Algorithm(k1,k2){
   int i=0,j=0;
   int i1,i2;
   while (i<max and A[i1 = h1(k1)+i*h2(k1) mod size(A)]->key!=k1){
          i++;
   }
   while (j<max and A[i2 = h1(k2)+j*h2(k2) mod size(A)]->key!=k2){
          j++;
   }
   if (A[i1]->key!=k1) return;
   if (A[i2]->key!=k2) return;

   pointer node1,node2,root1,root2;
   node1=A[i1]->node;
   node2=A[i2]->node;
   root1=UpTreeFind(node1);
   root2=UpTreeFind(node2);
   if (root1==root2){
      printf("The nodes belong to the same up tree");
      return;
   }

   // path compression
   pointer tmp,tmpParent;

   tmp = node1;
   while (tmp->parent!=root1) {
       tmpParent=tmp->parent;
       tmp->parent=root1;
       tmp=tmpParent;
   }

   tmp = node2;
   while (tmp->parent!=root2) {
       tmpParent=tmp->parent;
       tmp->parent=root2;
       tmp=tmpParent;
   }

   UpTreeUnion(root1,root2);
}


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dynamic-set operation UNION takes two disjoint sets S1 and S2 as input

dynamic-set operation UNION takes two disjoint sets S1 and S2 as input


By : Jevi
Date : March 29 2020, 07:55 AM
I wish did fix the issue. This is my homework question i have tried to solve it just need someone to look and tell me if i am doing it right or worng.. , Yea, that's the same approach I'd take.
code :
S1:
A1->A2->A3

S2:
B1->B2->B3

Tail node of S1 (A3) linked to head node of S2 (B1)

S1US2:
A1->A2->A3*->*B1->B2->B3
Why is my Union Find Disjoint Sets algo for this problem not passing all test cases?

Why is my Union Find Disjoint Sets algo for this problem not passing all test cases?


By : Vyshak
Date : March 29 2020, 07:55 AM
like below fixes the issue The for loop you put in union ruins the performance. Just take it out.
In the histogram loop, you need int root = find(j,parent);
O(1) Make, Find, Union in Disjoint Sets Data Structure

O(1) Make, Find, Union in Disjoint Sets Data Structure


By : user3910535
Date : March 29 2020, 07:55 AM
seems to work fine My intuition agrees with your colleague. You say:
u.set.list.append(v.set.list); // hypothetical method, append a list in O(1)
Implementing Disjoint Sets (Union Find) in C++

Implementing Disjoint Sets (Union Find) in C++


By : Saif
Date : March 29 2020, 07:55 AM
it should still fix some issue Not a perfect implementation by any means (I did write it after all!), but does this help?
Detect if a graph is bipartite using union find (aka disjoint sets)

Detect if a graph is bipartite using union find (aka disjoint sets)


By : yamini
Date : March 29 2020, 07:55 AM
I wish this help you Given a graph represented as an adjacency-list (i.e. a list of edges), you can determine if it's bipartite as follows:
Initialize a disjoint-set data structure SETS, with a singleton set for each vertex. (If there is an even-length path between two vertices, then we will ultimately unify those two vertices into the same set, unless we return ꜰᴀʟꜱᴇ first.) Initialize a mapping MAP from each vertex to ɴɪʟ. (As we examine edges, we will populate MAP with a mapping from each vertex to one of its neighbors.) For each edge {u, v}: If u and v belong to the same set in SETS, then return ꜰᴀʟꜱᴇ. If MAP[u] = ɴɪʟ, set MAP[u] := v.
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