Weighted Intervals Problem EXAMPLE:Suppose we have the following weighted intervals. start finish weight 11 14 4 10 13 5 9 11 5 11 14 3 8 9 2 To solve the problem, we start out by sorting the intervals by finish time. This gives us the following. start finish weight 8 9 2 9 11 5 10 13 5 11 14 4 11 14 3 Next we calculate the p-values. P(j) = k where interval k is the last interval prior to interval j in the list below which does not conflict with interval j, or P(j) = 0 if all intervals prior to interval j conflict with it. # start finish weight P 1 8 9 2 0 2 9 11 5 1 3 10 13 5 1 4 11 14 3 2 5 11 14 4 2 Finally, we use the formula OPT(j) = max{OPT(j-1), weight(j)+ OPT(P(j))} to fill in the OPT value for each j. (The base case where j=1 is that OPT(1)=weight(1)). # start finish weight P OPT 1 8 9 2 0 2 2 9 11 5 1 7 3 10 13 5 1 7 4 11 14 3 2 10 5 11 14 4 2 11 BACKTRACKING:Since OPT(5)=11>10=OPT(4), we know that interval 5 is part of an optimal solution. Since the P value for 5 is 2, we know that the rest of the intervals are chosen from among intervals 1 and 2.Since OPT(2)=7>2=Opt(1), we know that interval 2 is part of an optimal solution that uses interval 5. Since the P value for 2 is 1, we know that the rest of the intervals are chosen from "among" interval 1. As a base case of the backtracking, we see that we can either take interval 1, and get its weight 2, or not take it, and get nothing, so we conclude that interval 1 is a part of the solution.Thus, by backtracking, we have found that intervals 5, 2, and 1 form an optimal solution, with total weight 11. The example above shows how to work this kind of problem. Your assignment is to work the following problem, and to produce the corresponding table and backtracking results, according to the technique illustrated in the example.YOUR ASSIGNED PROBLEM: start finish weight 13 15 2 11 13 4 13 16 2 14 17 6 10 13 6 Let me know in class if you have questions about how such problems are solved. The algorithm is covered in sections 6.1 and 6.2.
Weighted Intervals Problem
EXAMPLE:
Suppose we have the following weighted intervals.
start finish weight
11 14 4
10 13 5
9 11 5
11 14 3
8 9 2
To solve the problem, we start out by sorting the intervals by finish time. This gives us the following.
start finish weight
8 9 2
9 11 5
10 13 5
11 14 4
11 14 3
Next we calculate the p-values. P(j) = k where interval k is the last interval prior to interval j in the list below which does not conflict with interval j, or P(j) = 0 if all intervals prior to interval j conflict with it.
# start finish weight P
1 8 9 2 0
2 9 11 5 1
3 10 13 5 1
4 11 14 3 2
5 11 14 4 2
Finally, we use the formula OPT(j) = max{OPT(j-1), weight(j)+ OPT(P(j))} to fill in the OPT value for each j. (The base case where j=1 is that OPT(1)=weight(1)).
# start finish weight P OPT
1 8 9 2 0 2
2 9 11 5 1 7
3 10 13 5 1 7
4 11 14 3 2 10
5 11 14 4 2 11
BACKTRACKING:
Since OPT(5)=11>10=OPT(4), we know that interval 5 is part of an optimal solution. Since the P value for 5 is 2, we know that the rest of the intervals are chosen from among intervals 1 and 2.
Since OPT(2)=7>2=Opt(1), we know that interval 2 is part of an optimal solution that uses interval 5. Since the P value for 2 is 1, we know that the rest of the intervals are chosen from "among" interval 1.
As a base case of the backtracking, we see that we can either take interval 1, and get its weight 2, or not take it, and get nothing, so we conclude that interval 1 is a part of the solution.
Thus, by backtracking, we have found that intervals 5, 2, and 1 form an optimal solution, with total weight 11.
The example above shows how to work this kind of problem. Your assignment is to work the following problem, and to produce the corresponding table and backtracking results, according to the technique illustrated in the example.
YOUR ASSIGNED PROBLEM:
start finish weight
13 15 2
11 13 4
13 16 2
14 17 6
10 13 6
Let me know in class if you have questions about how such problems are solved. The
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