CS6515 Exam 2 with correct answers
If graph G has more than |V | − 1 edges, and there is a unique heaviest edge, then this
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edge cannot be part of a minimum spanning tree - correct answer ✔False, because the
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unique heaviest edge may not be part of a cycle
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If G has a cycle with a unique heaviest edge e, then e cannot be part of any MST. -
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correct answer ✔True, if the unique heaviest edge is part of a cycle then it will be
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removed first. |!|
Let e be any edge of minimum weight in G. Then e must be part of some MST. - correct
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answer ✔True, in order create the MST we use the edges with minimum weight.
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If the lightest edge in a graph is unique, then it must be part of every MST. - correct
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answer ✔True, we always choose the lightest edge when building the MST.
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If e is part of some MST of G, then it must be a lightest edge across some cut of G. -
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correct answer ✔True, due to cut property
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If G has a cycle with a unique lightest edge e, then e must part of every MST. - correct
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answer ✔False, lightest edge in a cycle may not be necessary to create an MST because
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the remaining edges may be necessary to connect to the other vertices
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The shortest-path tree computed by Dijkstra's algorithm is necessarily an MST - correct
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answer ✔False, the shortest path may not visit all the nodes in the MST tree
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The shortest path between two nodes is necessarily an MST - correct answer ✔False, the
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shortest path between two nodes may not visit all the nodes in that make up an MST
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If G contains an r-path from node s to t, then every MST of G must also contain an r-
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path from node s to node t.
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If graph G has more than |V | − 1 edges, and there is a unique heaviest edge, then this
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edge cannot be part of a minimum spanning tree - correct answer ✔False, because the
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unique heaviest edge may not be part of a cycle
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If G has a cycle with a unique heaviest edge e, then e cannot be part of any MST. -
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correct answer ✔True, if the unique heaviest edge is part of a cycle then it will be
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removed first. |!|
Let e be any edge of minimum weight in G. Then e must be part of some MST. - correct
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answer ✔True, in order create the MST we use the edges with minimum weight.
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If the lightest edge in a graph is unique, then it must be part of every MST. - correct
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answer ✔True, we always choose the lightest edge when building the MST.
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If e is part of some MST of G, then it must be a lightest edge across some cut of G. -
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correct answer ✔True, due to cut property
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If G has a cycle with a unique lightest edge e, then e must part of every MST. - correct
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answer ✔False, lightest edge in a cycle may not be necessary to create an MST because
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the remaining edges may be necessary to connect to the other vertices
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The shortest-path tree computed by Dijkstra's algorithm is necessarily an MST - correct
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answer ✔False, the shortest path may not visit all the nodes in the MST tree
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The shortest path between two nodes is necessarily an MST - correct answer ✔False, the
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shortest path between two nodes may not visit all the nodes in that make up an MST
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If G contains an r-path from node s to t, then every MST of G must also contain an r-
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path from node s to node t.
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