Test: Minimum Spanning Trees

To get the minimum spanning tree with vertex 0 as leaf, first remove 0th row and 0th column and then get the minimum spanning tree (MST) of the remaining graph. Once we have MST of the remaining graph, connect the MST to vertex 0 with the edge with minimum weight (we have two options as there are two 1s in 0th row).

Path: 1 -> 0 -> 4 -> 2 Weight: 1 + 4 + 3

The edge (d-e) cannot be considered before (d-c) in Kruskal's minimum spanning tree algorithm because Kruskal’s algorithm picks the edge with minimum weight from the current set of edges at each step.

(a) and (b) are always true. (c) is false because (b) is true. (d) is true because all edge weights are distinct for G.

Minimum spanning tree of such a graph is

Weight of the minimum spanning tree = 2|2 - 1| + 2|3 - 2| + 2|4 - 3| + 2|5 - 4| .... + 2| n - (n-1) | = 2n - 2

Squaring the weights of the edges in a weighted graph will not change the minimum spanning tree. Assume the opposite to obtain a contradiction. If the minimum spanning tree changes then at least one edge from the old graph G in the old minimum spanning tree T must be replaced by a new edge in tree T' from the graph G' with squared edge weights. The new edge from G' must have a lower weight than the edge from G. This implies that there exists some weights C1 and C2 such that C1 < C2 and C12 >= C22. This is a contradiction. Sums of squares of two or more numbers is always smaller than square of sum. Example: 2^2 + 2^2 < (4)^2 But there is one counter example when the graph has only one edge. In that case, the two values are same.

In the sequence (b, e) (e, f) (b, c) (a, c) (f, g) (c, d) given option D, the edge (a, c) of weight 4 comes after (b, c) of weight 3. In Kruskal's Minimum Spanning Tree Algorithm, we first sort all edges, then consider edges in sorted order, so a higher weight edge cannot come before a lower weight edge.

Below diagram shows a minimum spanning tree. Highlighted (in green) are the edges picked to make the MST.

In the right side of MST, we could either pick edge ‘a’ or ‘b’. In the left side, we could either pick ‘c’ or ‘d’ or ‘e’ in MST. There are 2 options for one edge to be picked and 3 options for another edge to be picked. Therefore, total 2*3 possible MSTs.

The minimum weight edge on any s-t cut is always part of MST. This is called Cut Property. This is the idea used in Prim's algorithm. The minimum weight cut edge is always a minimum spanning tree edge. Why B (the weighted shortest path from s to t) is not an answer? See below example, edge 4 (lightest in highlighted red cut from s to t) is not part of shortest path. 

(a,c), (a,d), (d,b), (b,g), (g,h), (h,f), (h,i), (i,j), (i,e) = 31   Background required - Minimum Spanning Tree (Prims / Kruskal) In these type of questions, always go for kruskal’s algorithm to find out the the minimum spanning tree as it is easy and there are less chances of doing silly mistakes. Algorithm: Always pick the minimum edge weight and try to add to current forest (Collection of Trees) if no cycle is formed else discard. As soon as u have added n-1 edges to the forest, stop and you have got your minimum spanning tree. See the below image for construction of MST of this question.

Weight of minimum spanning tree = Sum of all the edges in Minimum Spanning tree = 31

In every cycle, the weight of an edge that is not part of MST must by greater than or equal to weights of other edges which are part of MST. Since all edge weights are distinct, the weight must be greater. So the minimum possible weight of ED is 7, minimum possible weight of CD is 16 and minimum possible weight of AB is 10. Therefore minimum possible sum of weights is 69.

Since there are 100 vertices, there must be 99 edges in Minimum Spanning Tree (MST). When weight of every edge is increased by 5, the increment in weight of MST is = 99 * 5 = 495 So new weight of MST is 500 + 495 which is 995

The shortest path may change. The reason is, there may be different number of edges in different paths from s to t. For example, let shortest path be of weight 15 and has 5 edges. Let there be another path with 2 edges and total weight 25. The weight of the shortest path is increased by 5*10 and becomes 15 + 50. Weight of the other path is increased by 2*10 and becomes 25 + 20. So the shortest path changes to the other path with weight as 45. The Minimum Spanning Tree doesn't change. Remember the Kruskal's algorithm where we sort the edges first. IF we increase all weights, then order of edges won't change.

One graph that has maximum possible weight of spanning tree

I is NOT true. Let G=(V, E) be a rectangular graph where V = {a, b, c, d} and E = {ab, bc, cd, da, ac}. Let the edges have weights: ab = 1, bc = 2, cd = 4, da = 5, ac = 3. Then, clearly, ac is the lightest edge of the cycle cdac, however, the MST abcd with cost 7 (= ab + bc + cd) does not include it. Let the edges have weights: ab = 6, bc - 7, cd = 4, da = 5, ac = 3. Then, again, ac is the lightest edge of the cycle cdac, and, the MST bacd with cost 13 (= ba + ac + cd) includes it. So, the MSTs of G may or may not include the lightest edge.

II is true Let the heavies edge be e. Suppose the minimum spanning tree which contains e. If we add one more edge to the spanning tree we will create a cycle. Suppose we add edge e' to the spanning tree which generated cycle C. We can reduce the cost of the minimum spanning tree if we choose an edge other than e from C for removal which implies that e must not be in minimum spanning tree and we get a contradiction. 

A graph is connected iff all nodes can be traversed from each node. For a graph with n nodes, there will be n-1 minimum number of edges. 
Given that there are n edges, that means a cycle is there in the graph.
The simplex graph with these conditions may be:

Now we can make a different spanning tree by removing one edge from the cycle, one at a time.
Minimum cycle length can be 3, So, there must be atleast 3 spanning trees in any such Graph.