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| + | '''n = 1''' |
| | + | The single element array is already its max. Loop is not entered. Max is returned |
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| − | :[[1.1]]. Show that ''a'' + ''b'' can be less than ''min(a, b)''.
| + | Let for '''n=k''', the algorithm is true |
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| | + | For '''n = k+1''' ,two cases arise : |
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| − | :1.2. Show that ''a'' × ''b'' can be less than ''min(a, b)''.
| + | '''1)''' a[k+1] is max |
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| | + | '''2)''' a[k+1] is not max. |
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| − | :[[1.3]]. Design/draw a road network with two points ''a'' and ''b'' such that the fastest route between ''a'' and ''b'' is not the shortest route.
| + | If '''1)''' holds then at the last iteration when i = k+1, m = a[k+1] is assigned. Hence max is returned |
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| | + | Else if '''2)''' holds then we are left with the task of finding the max among '''n = k''' elements, which we already assumed that the algorithm does correctly. |
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| − | :1.4. Design/draw a road network with two points ''a'' and ''b'' such that the shortest route between ''a'' and ''b'' is not the route with the fewest turns.
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| − | | + | Back to [[Chapter 2]] |
| − | :[[1.5]]. The ''knapsack problem'' is as follows: given a set of integers ''S'' = {''s1, s2. . . , sn''}, and a target number ''T'', find a subset of ''S'' that adds up exactly to ''T''. For example, there exists a subset within ''S'' = {1, 2, 5, 9, 10} that adds up to ''T'' = 22 but not ''T'' = 23.
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| − | :Find counterexamples to each of the following algorithms for the knapsack problem. That is, give an ''S'' and ''T'' where the algorithm does not find a solution that leaves the knapsack completely full, even though a full-knapsack solution exists.
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| − | ::(a) Put the elements of ''S'' in the knapsack in left to right order if they fit, that is, the first-fit algorithm.
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| − | ::(b) Put the elements of ''S'' in the knapsack from smallest to largest, that is, the best-fit algorithm.
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| − | ::(c) Put the elements of ''S'' in the knapsack from largest to smallest.
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| − | :1.6. The ''set cover problem'' is as follows: given a set ''S'' of subsets ''S1, . . . , Sm'' of the universal set ''U'' = {1, ..., ''n''}, find the smallest subset of subsets ''T ⊆ S'' such that ''∪ti∈T ti'' = ''U''. For example, consider the subsets ''S1'' = {1, 3, 5}, ''S2'' = {2, 4}, ''S3'' = {1, 4}, and ''S4'' = {2, 5}. The set cover of {1, . . . , 5} would then be ''S1'' and ''S2''.
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| − | :Find a counterexample for the following algorithm: Select the largest subset for the cover, and then delete all its elements from the universal set. Repeat by adding the subset containing the largest number of uncovered elements until all are covered.
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| − | :[[1.7]]. The ''maximum clique problem'' in a graph ''G'' = (''V'', ''E'') asks for the largest subset ''C'' of vertices ''V'' such that there is an edge in ''E'' between every pair of vertices in ''C''. Find a counterexample for the following algorithm: Sort the vertices of ''G'' from highest to lowest degree. Considering the vertices in order of degree, for each vertex add it to the clique if it is a neighbor of all vertices currently in the clique. Repeat until all vertices have been considered.
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| − | :1.8
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| − | :[[1.9]]
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| − | :1.10
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| − | :[[1.11]]
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| − | :1.12
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| − | :[[1.13]]
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| − | :1.14
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| − | :[[1.15]]
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| − | :1.16
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| − | :[[1.17]]
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| − | :1.18
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| − | :[[1.19]]
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| − | :1.20
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| − | :[[1.21]]
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| − | :1.22
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| − | :[[1.23]]
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| − | :1.24
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| − | :[[1.25]]
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| − | :1.26
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| − | :[[1.27]]
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| − | :1.28
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| − | :[[1.29]]
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| − | :1.30
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| − | :[[1.31]]
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| − | :1.32
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| − | :[[1.33]]
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| − | :1.34
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| − | :[[1.35]]
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| − | :1.36
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| − | :[[1.37]]
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| − | :1.38
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| − | Back to [[Chapter List]] | |
n = 1
The single element array is already its max. Loop is not entered. Max is returned
Let for n=k, the algorithm is true
For n = k+1 ,two cases arise :
1) a[k+1] is max
2) a[k+1] is not max.
If 1) holds then at the last iteration when i = k+1, m = a[k+1] is assigned. Hence max is returned
Else if 2) holds then we are left with the task of finding the max among n = k elements, which we already assumed that the algorithm does correctly.
Back to Chapter 2
