Given an Undirected graph G with with weight on its edges and 2 different minimal spanning trees: T, T'
Then I want to prove the following:
For every edge e in T that's not in T', there is an edge e' in T'
that's not in T such that if we replace e with e' in T (let's call
it T_new) then it's still a minimal spanning tree of G.
I think I am too close for finding the right algorithm but stuck a little:
I have proved that weight(e) must be exactly equal to weight(e').
Since T is a tree, deleting e will result in 2 separated components, then for T_new to be a tree it must use one of the edges connecting two vertices from those different components.
But, I wasn't able to know which edge e' exactly will work. Plus I wasn't able to prove that always there is such an edge (I just found some requirements for e' that is must satisfy).
Some notes: I know Kruskal algorithm, and familiar with an algorithm in which we can paint some edges in yellow and request it to generate minimal spanning trees with maximum yellow edges (In other words from all found minimal spanning trees return the one with maximum number of yellow edges)
Let T1 and T2 be the two connected components of T \ {e}. Consider the path P joining the endpoints of e in T'. Since e connects T1 and T2, so does P, and therefore there exists an edge e' in P that connects T1 and T2. The edge e' cannot be lighter than e, or else T would not be minimum (T \ {e} U {e'}). The edge e' cannot be heavier than e, or else T' would not be minimum (T' \ {e'} U {e}).
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I am trying to design an algorithm where, given a connected weighted graph G = (V, E) and a subset of vertices U that is in V, will construct a minimum spanning tree such that all vertices in U are leaves (other vertices may also be leaves), or returns that no such tree exists (False).
This is all I got, adapting Prim's algorithm (fair warning, its really bad; don't even know if it works/is efficient or what data structures to use, I will accept literally any other correct algorithm instead):
Let x be an arbitrary node in G
Set S = {x}
While S != V:
Let (u,v) be the cheapest edge with u in S and v not in S
Add (u,v) to tree T if u is not in U, add v to S
If all u in U is in the tree T:
return T
Else:
return False
I also have a picture of what I think it would do to this graph I drew:
pic here
A proof that the algorithm is correct would also give me some peace of mind.
If all vertices u ∈ U are to be leaves in a solution, no u can be used in that solution to connect two other vertices. All vertices not in U must be connected by edges not incident to any u.
Remove U and all edges incident to U. Find the minimum spanning tree, then connect each u to the tree by the smallest-weighted edge available from those we removed.
For a connected, weighed, undirected graph G:
G has a unique MST, if for every cut of G there is a unique minimum weight edge crossing the cut.
Is this statement true?
I think false because for the following graph in the given link there can be multiple MSTs.
https://drive.google.com/file/d/1yDK3juPxeDBdS-aEOx0aAsphy4odZ55l/view?usp=drivesdk
If you mean a connected graph G, with edge costs that are all distinct. Then G has a unique minimum spanning tree.
proof:
Suppose there are two different MSTs, call them T1 and T2, who have different sets of edges--{t11, t12, … t1n-1} for T1 and {t21, t22, … t2n-1} for T2. So, let ti be the edge with the smallest weight only in T1 (not in T2). Since it is the smallest, ti must be included in "every choice" of MST. That is, both MST of T1 and T2 would have it. But this contradicts the definition of ti.
I have a connected graph G=(V,E) V={1,2,...,n} and a cost function c:E->R
and a second partial graph G'=(V,T) where T={ for every vertex v∈ V find the neighbor with the minimum cost and add the new edge to T}
If G' graph has at least 2 connected components with the set of vertices we consider the graph H where
iff the set of edges (from the initial graph G) is not null.We define over the edges of H a cost function.
Let's say I choose V(H)={a,e,f} and E(H)={ae,af,fe} and
E12={ab,bc,bd,ed}
E23={eg,ef} E31={fc,fd}
c'(ae)=min{c(ab),c(bc),c(bd),c(ed)}=4
c'(af)=min{c(fc),c(fd)}=9
c'(fe)=min{c(eg),c(ef)}=8
Now for every edge e ∈ E(H) we note with e' the edge (from the original graph G)
for which this minimum is attained.
So e'={bc,df,eg} because bc=4 , df=9 and eg=8 and are the min edges that connect my components.
And I have a minimum spanning tree in H relative to the cost function c' and A' is the set of edges for this tree.
So A'={ae,fe} (I deleted the edge with the maximum cost=af from my graph H to create a min spanning tree)
and I have another set of edges A'={e'|e∈A'} and
is a min spanning tree in G relative to the function cost c.
But none of my edges from A' are the same with the ones from e'.
What I'm I doing wrong?
Looks like you're implementing Boruvka's algorithm. If you look at the notation, it says there's an edge from one new node vC1 to a new node vC2 if there are a pair of nodes x ∈ C1 and y ∈ C2 that are adjacentnin the original graph G. In other words, there's an edge between two new nodes if the connected components they correspond to in G' are adjacent in G. The cost of the edge running between them is then the lowest of the costs of any of the edges running between those CC's in the original graph G.
I am studying algorithms, and I have seen an exercise like this
I can overcome this problem with exponential time but. I don't know how to prove this linear time O(E+V)
I will appreciate any help.
Let G be the graph where the minimum spanning tree T is embedded; let A and B be the two trees remaining after (u,v) is removed from T.
Premise P: Select minimum weight edge (x,y) from G - (u,v) that reconnects A and B. Then T' = A + B + (x,y) is a MST of G - (u,v).
Proof of P: It's obvious that T' is a tree. Suppose it were not minimum. Then there would be a MST - call it M - of smaller weight. And either M contains (x,y), or it doesn't.
If M contains (x,y), then it must have the form A' + B' + (x,y) where A' and B' are minimum weight trees that span the same vertices as A and B. These can't have weight smaller than A and B, otherwise T would not have been an MST. So M is not smaller than T' after all, a contradiction; M can't exist.
If M does not contain (x,y), then there is some other path P from x to y in M. One or more edges of P pass from a vertex in A to another in B. Call such an edge c. Now, c has weight at least that of (x,y), else we would have picked it instead of (x,y) to form T'. Note P+(x,y) is a cycle. Consequently, M - c + (x,y) is also a spanning tree. If c were of greater weight than (x,y) then this new tree would have smaller weight than M. This contradicts the assumption that M is a MST. Again M can't exist.
Since in either case, M can't exist, T' must be a MST. QED
Algorithm
Traverse A and color all its vertices Red. Similarly label B's vertices Blue. Now traverse the edge list of G - (u,v) to find a minimum weight edge connecting a Red vertex with a Blue. The new MST is this edge plus A and B.
When you remove one of the edges then the MST breaks into two parts, lets call them a and b, so what you can do is iterate over all vertices from the part a and look for all adjacent edges, if any of the edges forms a link between the part a and part b you have found the new MST.
Pseudocode :
for(all vertices in part a){
u = current vertex;
for(all adjacent edges of u){
v = adjacent vertex of u for the current edge
if(u and v belong to different part of the MST) found new MST;
}
}
Complexity is O(V + E)
Note : You can keep a simple array to check if vertex is in part a of the MST or part b.
Also note that in order to get the O(V + E) complexity, you need to have an adjacency list representation of the graph.
Let's say you have graph G' after removing the edge. G' consists have two connected components.
Let each node in the graph have a componentID. Set the componentID for all the nodes based on which component they belong to. This can be done with a simple BFS for example on G'. This is an O(V) operation as G' only has V nodes and V-2 edges.
Once all the nodes have been flagged, iterate over all unused edges and find the one with the least weight that connects the two components (componentIDs of the two nodes will be different). This is an O(E) operation.
Thus the total runtime is O(V+E).
Let G = (V, E) be a weighted, connected and undirected graph and let T be a minimum spanning tree. Let e be any edge not in E (and has a weight W(e)).
Prove or disprove:
T U {e} is an edge set that contains a minimum spanning tree of G' = (V, E U {e}).
Well, it sounds true to me, so I decided to prove it but I just get stuck every time...
For example, if e is the new edge with minimum weight, who can promise us that the edges in T weren't chosen in a bad way that would prevent us from obtaining a new minimum weight without the 'help' of other edges in E - T ?
I would appreciate any help,
Thanks in advance.
Let [a(1), a(2), ..., a(n-1)] be a sequence of edges selected from E to construct MST of G by Kruskal's algorithm (in the order they were selected - weight(a(i)) <= weight(a(i + 1))).
Let's now consider how Kruskal's Algorithm behaves being given as input E' = E U {e}.
Let i = min{i: weight(e) < weight(a(i))}. Firstly algorithm decides to choose edges [a(1), ..., a(i - 1)] (e hasn't been processed yet, so it behaves the same). Then it need to decide on e - if e is dropped, solution for E' will be the same as for E. So let's suppose that first i edges selected by algorithm are [a(1), ..., a(i - 1), e] - I will call this new sequence a'. Algorithm continues - as long as its following selections (for j > i) satisfy a'(j) = a(j - 1) we are cool. There are two scenarios that break such great streak (let's say streak breaks at index k + 1):
1) Algorithm selects some edge e' that is not in T, and weight(e') < weight(a(k+1)). By now a' sequence is:
[a(1), ..., a(i-1), e, a(i), a(i+1), ..., a(k-1), a(k), e']
But if it was possible to append e' to this list it would be also possible to append it to [a(1), ..., a(k-1), a(k)]. But Kruskal's algorithm didn't do it when looking for MST for G. That leads to contradiction.
2) Algorithm politely selected:
[a(1), ..., a(i-1), e, a(i), a(i+1), ..., a(k-1), a(k)]
but decided to drop edge a(k+1). But if e was not present in the list algorithm would decide to append a(k+1). That means that in graph (V, {a(1), ..., a(k)}) edge a(k+1) would connect the same components as edge e. And that means that after considering by algorithm edge a(k + 1) in case of both G and G' the division into connected components (determined by set of selected edges) is the same. So after processing a(k+1) algorithm will proceed in the same way in both cases.
When ever a edge is add to a graph without adding a node , then that edge creates a cycle in minimum spanning tree of graph, cycle length may vary from 2 to n where n= no of nodes in graph.
T = Minimum spanning tree of G
Now to find the MST for (T + added edge) , we have to just remove one edge from that cycle .. so remove that edge which has maximum weight.
So T' always comes from T U {e}.
And if you are thinking that this doesn't prove that new MST will be an edge set of T U {e} then analyse Kruskal algorithim for for new graph. i.e. if e is of minimum weight it must have been selected for MST acc to Kruskal algorithim and same here if it is minimum it can not be removed from cycle.