# Difference between revisions of "Talk:TADM2E 5.13"

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The solution to 2) as I understand it ("do the same as for 1") is wrong. Take a tree where the root has 50 children, each having one child of its own. The algorithm results in the children of the root plus the root being selected for a total weight of 2*50 + 50 = 150, while an optimal cover would be the root plus all the leaves for a total weight of 50 + 50. | The solution to 2) as I understand it ("do the same as for 1") is wrong. Take a tree where the root has 50 children, each having one child of its own. The algorithm results in the children of the root plus the root being selected for a total weight of 2*50 + 50 = 150, while an optimal cover would be the root plus all the leaves for a total weight of 50 + 50. | ||

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+ | 2-coloring the tree and using either of the colors as the solution should be correct since that way, every edge is covered exactly once. |

## Revision as of 12:49, 13 September 2019

I think the answer for the third case does not work since in a weighted graph it might not be optimal to choose alternating nodes: 100 - 1 - 1 - 100 (a path with4 nodes, high weights on the leaves, low weights on the inner nodes).

## Solution to 2 is probably wrong

The solution to 2) as I understand it ("do the same as for 1") is wrong. Take a tree where the root has 50 children, each having one child of its own. The algorithm results in the children of the root plus the root being selected for a total weight of 2*50 + 50 = 150, while an optimal cover would be the root plus all the leaves for a total weight of 50 + 50.

2-coloring the tree and using either of the colors as the solution should be correct since that way, every edge is covered exactly once.