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How do i solve this? Thank you Question 5 To see how bad the (greedy) Nearest Neighbour algorithm can be consider this deliberately pathological example:

image text in transcribedHow do i solve this? Thank you

Question 5 To see how bad the (greedy) Nearest Neighbour algorithm can be consider this deliberately pathological example: The complete graph Kio has vertices labelled 1, , 10 and each edge {a,b} is given weight ab. Do not attempt to draw this graph. (a) Starting at the vertex determined by the last digit of your ANU ID (use vertex 10 if that digit is 0), apply the Nearest Neighbour algorithm to find a Hamilton circuit of supposedly low total weight. Give your answer as a list of vertices, and calculate the total weight (b) By avoiding large products such as 9 x 10, find (without the help of any algorithm) a Hamilton circuit of much lower total weight. Again give your answer as a list of vertices, and calculate the total weight. ) How many diffits would you need to consider in order to prove your answer to (b) has the lowest possible total weight? Circuits are different if and only if there is at least one edge on one that is not on the other. Question 5 To see how bad the (greedy) Nearest Neighbour algorithm can be consider this deliberately pathological example: The complete graph Kio has vertices labelled 1, , 10 and each edge {a,b} is given weight ab. Do not attempt to draw this graph. (a) Starting at the vertex determined by the last digit of your ANU ID (use vertex 10 if that digit is 0), apply the Nearest Neighbour algorithm to find a Hamilton circuit of supposedly low total weight. Give your answer as a list of vertices, and calculate the total weight (b) By avoiding large products such as 9 x 10, find (without the help of any algorithm) a Hamilton circuit of much lower total weight. Again give your answer as a list of vertices, and calculate the total weight. ) How many diffits would you need to consider in order to prove your answer to (b) has the lowest possible total weight? Circuits are different if and only if there is at least one edge on one that is not on the other

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