Consider the following undirected graph G: Choose a value for x that will…

Answer: set x = 5. The number of minimum weight spanning trees is 4.

Reasoning (concise):

• The edges of weights 1 and 3 must be in every minimum spanning tree (they are strictly the smallest), so those two endpoints (top, bottom-left, bottom-right) are already connected.

• To attach the top-left vertex we need one of the two edges of weight 4 (top–top-left or top-left–bottom-left). That gives 2 independent choices for top-left.

• To attach the top-right vertex we use the cheaper of the two edges to it: top–top-right (weight x) or bottom-right–top-right (weight 5). To maximize the number of minimum trees we want a tie here, so choose x = 5. Then there are 2 equal choices for attaching top-right.

• The choices for attaching top-left (2 choices) and top-right (2 choices) are independent, so the total number of MWSTs is 2 × 2 = 4.

• If x ≠ 5 there is no tie for the top-right edge: if x < 5 the top–top-right edge is strictly preferred; if x > 5 the bottom-right–top-right edge is strictly preferred. In both cases there is only 1 choice for top-right, giving fewer total MWSTs (2).