Dynamic Programming

divide and conquer - use sub problems and build up to correct sultion.

Designing a Recursive Algorithm

• need a base case
• then a step that makes it smaller, then you call the algorithm again.

Weighted interval scheduling

• choose if we do the job or not
• If we choose the job we run the algorithm on the compatiable jobs
• if we don't choose the job j we look at the problem j-1
• key is we need a memoization array - to make the problem more efficient. (THIS COMES IN 255-257

• 6.1 OPT(J) = max(Vj+opt(p(j), opt(j-1))
• 6.2 Request j belons to the optimal solution on the set if and only if vj+opt(p(j)) >= opt(j-1)
• algorithm on p. 254
• NOW that we have the value of the optimal solution need to do post work to find the actual solution.

Computing the optimal solution

• 6.5 given the array M of the optimal values of the sub-problems Find Solution returns an optial solution in O(n)

• Memoization - array details. 259
• can also do iterative algorithms

A basic outline of Dynamic Programming

• There are only a polynoimal number of sub problems
• the solution to the original problem can be easily computed from the solutions to the subproblems
• there is a natural ordering on subproblems from smallest to largest together with an easy to compute recurrence and that allows one to determine the solution to a subproblem from the solution to some number of smaller subproblems.

The problem

• minimize the sum of the distances the points are from the line, by using more than one line. (Find the lines of best fit)
• instead of having a binary choice we have a multi way choice. do we do one line from point i to j or a line from i-2 to j. Each line has a cost also though so it can't just be number lines.

Designing the alogirthm

• see the algorithm on page 265
• 6.6 if the last segment of the optimal partition is pi …. pn then the value of the optimal solution is OPT(N) = ei,n + c + opt(i-1)
• 6.7 for the sub problem on the points pi…pj opt(j) = min(ei,j + C + opt(i-1)), and the segment pi…pj is used in the optimal solution for the subproblem if and only if the minimum is obtained using index i.

The problem

• n items
• knapsack can hold W total weight
• each item i has Wi
• each item i also has a value to the person
• want to max value which staying under W
• key is using memoization array
• second key is many many sub problems

Designing an algorithm

• algorithm on 6.4
• 6.8 if W < Wi then opt(i, W) = opt(i-1, W) Otherwise Opt(i,W) = max(opt(i-1, W), Wi + opt(i-1, w-wi))
• 6.9 the subset-sum (n,W) algorithm correctly computes the value of the problem and runs in O(nW) time.
• 6.10 given a table M of the optimal values of the subproblems, the optimal set S can be found in O(n) time.
• (AGAIn need to do post back work to get the solution.)

There is an expansion section on the knapsack problem that looks at a different restriction on weight.

Problem: suppose I type provlem in google as opposed to problem. google looks for problem as the match. So basically matching sequences of letters to find workds. - we can either have a match, or a gap. If it is a match it can be correct or a mismatch or a gap can be on the x word or the y word. - we give weights to each option then we have figure out the optimal sequence from that. - see 6.16 - see page 282 or the algorithm - the graph image on 283 makes it really easy for me to understand.

• changing sequence alignment using divide and conquer
• algorithm on 285
• split the graph into two different sections and then go forward from your start point and then backwards from your final point to the midel.
• find the path that minimizes the weight

• weighted graph with negative weights
• have to adjust to use dijkstras
• note if there is negative cycles then we will never have a shortest path
• algorithm on page 294
• implementation O(mn)