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| courses:cs211:winter2012:journals:paul:home [2012/03/26 21:50] – [Chapter 6] nguyenp | courses:cs211:winter2012:journals:paul:home [2012/04/06 02:17] (current) – [Chapter 7] nguyenp | ||
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| + | * Section 6.2 - Principles of Dynamic Programming: | ||
| + | * The book first goes over what makes the previous algorithm dynamic and efficient | ||
| + | * If I was the author, I would have put that in the last chapter. It has no business being here. | ||
| + | * It says nothing else useful. | ||
| + | * Section 6.3 - Segmented Least Squares: Multi-way Choices | ||
| + | * The problem we are focusing in this section is the one where we have a bunch of dots and want to know what line (if it even is a line) that is most similar to the data (you know what I mean) | ||
| + | * If it is not a line, then we do not try and form a curve, instead we form a sequence of lines. | ||
| + | * How do we do this? | ||
| + | * Consider the first and last points. | ||
| + | * For every point p_n, it can only belong to one line segment | ||
| + | * We define cost for each point as cost: c (segment cost) + error of segment | ||
| + | * Let OPT(j) = minimum cost for points p_1, p_i+1 , … , p_j (the min cost of all the points) | ||
| + | * Let e(i, j) = minimum sum of squares for points p_i, p_i+1 , … , p_j | ||
| + | * How to compute OPT(j)? | ||
| + | * Cost = e(i, j) + c + OPT(i-1). | ||
| + | * OPT(J): | ||
| + | * if j=0 | ||
| + | * return 0 | ||
| + | * else | ||
| + | * min {e(i, | ||
| + | * Algorithm: page 265 | ||
| + | * INPUT: n, p_1,…,p_N , c | ||
| + | * Segmented-Least-Squares() | ||
| + | * M[0] = 0 | ||
| + | * e[0][0] = 0 | ||
| + | * for j = 1 to n | ||
| + | * for i = 1 to j | ||
| + | * e[i][j] = least square error for the segment p_i, …, p_j | ||
| + | * | ||
| + | * for j = 1 to n | ||
| + | * M[j] = min 1 ≤ i ≤ j (e[i][j] + c + M[i-1]) | ||
| + | * | ||
| + | * return M[n] | ||
| + | * The magic really lies in the math behind the least square error. That's how we determine which segment something belongs to. | ||
| + | * Efficiency: | ||
| + | * The first group of nested for loops is O(n^3). | ||
| + | * This can be reduced to quadratic if we pre-compute and write down the values for p_whatever | ||
| + | * YAY MEMOIZING | ||
| + | * the last for loop is actually quadratic even though it initially looks linear because it's just a single for loop, but remember that min is a linear time function | ||
| + | * For some reason the book likes to do things in this very not straightforward way of finding all the data we need using one algorithm and then later using that data to find the information we really need to get our results. | ||
| + | * So, since we really need to know how to split up the points into a sequence of segments, we will go through all the segments we just assigned | ||
| + | * Algorithm: page 266 | ||
| + | * FindSegments(j): | ||
| + | * if j = 0: | ||
| + | * output nothing | ||
| + | * else: | ||
| + | * Find an i that minimizes ei,j + c + M[i-1] | ||
| + | * Output the segment {pi, …, pj} | ||
| + | * FindSegments(i-1) | ||
| + | * Section 6.4 - Subset Sums and Knapsacks: Adding a variable | ||
| + | * We're focusing on the knapsack problem | ||
| + | * We have n things and a knapsack | ||
| + | * item i has weight w_i (always positive) and also has a value v_i | ||
| + | * We can only carry so much weight | ||
| + | * We want to maximize the value (more bang for the buck type of deal) | ||
| + | * So, for each item, we either pick it or we don't pick it | ||
| + | * OPT(i) = max profit subset of items 1,…, i | ||
| + | * How to compute OPT(i)? | ||
| + | * if it does not select object i | ||
| + | * OPT selects best of { 1, 2, …, i-1 } | ||
| + | * if it does | ||
| + | * Does not mean we must not pick everything lower than i | ||
| + | * So, we don't know what to do. | ||
| + | * MORE SUBPROBLEMS!!! w0000ooooooo... | ||
| + | * So, we set a new weight limit, our current available weight minus what we jsut added, i.e. w – w_i | ||
| + | * We let OPT select the best of these | ||
| + | * That's it! That's the algorithm | ||
| + | * Algoriddim be on Page 269 mahn | ||
| + | * Input: N, w_1, | ||
| + | * for w = 0 to W | ||
| + | * M[0, w] = 0 | ||
| + | * for i = 1 to N | ||
| + | * for w = 1 to W | ||
| + | * if wi > w : | ||
| + | * M[i, w] = M[i-1, w] | ||
| + | * else | ||
| + | * M[i, w] = max{ M[i-1, w], vi + M[i-1, w-wi] } | ||
| + | * return M[n, W] | ||
| + | * The reason for why this works is suprisingly simple once we realize the important of just changing the amount of weight left over and over again | ||
| + | * Efficiency: | ||
| + | * This is a strange one. | ||
| + | * It's Θ(n W) | ||
| + | * So it both depends on the knap sack capacity and the number of objects | ||
| + | * they call it Pseudo-polynomial. | ||
| + | * Weird stuff. Never thought algorithms like this existed. | ||
| + | * | ||
| + | |||
| + | ====== Chapter 7 ====== | ||
| + | * Section 7.1 - The Maximum-Flow Problem and the Ford-Fulkerson Algorithm | ||
| + | * Problem | ||
| + | * You have a directed graph | ||
| + | * Each edge has a capacity | ||
| + | * Have a start node s and end node t | ||
| + | * Greedy | ||
| + | * Find an s-t path with highest capacity | ||
| + | * do it! | ||
| + | * keep doing this until you cant do it any more | ||
| + | * Locally optimal, but not globally optimal (it's very easy to think of a counter example, counter example is on page 339 in book) | ||
| + | * Awesome Solution: Residual Graphs | ||
| + | * Edges have a capacity, but also a flow | ||
| + | * We don't have to use all of it at once | ||
| + | * Residual edge is one where you have some going one way and some going the other | ||
| + | * The Ford-Fulkerson Algorithm uses residual edges and works awesomely (optimal) | ||
| + | * page 342 - 344 | ||
| + | * Informal explanation | ||
| + | * graphs | ||
| + | * keep updating the residual edges using the following formula: | ||
| + | * If edge is the same as it was originally, set the forward rate to the bottle neck + original rate (which is the edge with the smallest flow rate) | ||
| + | * Otherwise, set the backwards residual rate to what it originally was minus the bottle neck | ||
| + | * Section 7.2 - Maximum Flows and Minimmum Cuts in a Network | ||
| + | * Cut: s-t cut is a partition (A,B) where s is in A and t is in B | ||
| + | * Problem: Find an s-t cut of minimum capacity | ||
| + | * The capacity of a cut is the sum of all the possible stuff that can go through the edge at once | ||
| + | * Flow Value Lemma: Let f be any flow, and let (A, B) be any s-t cut. Then, the value of the flow is = f_out(A) – f_in(A). | ||
| + | * Weak Duality: let f be any flow and let (A, B) be any s-t cut. Then the value of the flow is at most the cut’s capacity. | ||
| + | * Corollary: Let f be any flow, and let (A, B) be any cut. If v(f) = cap(A, B), then f is a max flow and (A, B) is a min cut | ||
| + | * Augmenting path theorem: Flow f is a max flow iff there are no augmenting paths | ||
| + | * Max-flow min-cut theorem: The value of the max flow is equal to the value of the min cut. | ||
| + | * More explanation on page 350 | ||
| + | * Boom and we are done. Just use the same stuff as in the past section and we have all we need. | ||
| + | * Section 7.5 - A First Application: | ||
| + | * Given an undirected bipartite graph | ||
| + | * We must match so each node appears at most once on each edge | ||
| + | * How do we find one with the max number of nodes? This is our problem. | ||
| + | * Here's how we do it. We got a bunch of red and blue nodes. Add nodes s and t, which will be the start and end nodes. Have the start node s have a unit edge (edge of size 1) connect it to all the red nodes. Have all the blue nodes have a unit edge connect them to the end node t. Still have same edges in the middle. | ||
| + | * Now, use same algorithm as in the first section. BOOOM!!! Right? Yep. It's pretty intuitive, these later sections are more like practical uses rather than actual algorithms. I guess this is kind of good since pretty much all computer science problems go back to graph theory. | ||
| + | * Extensions: The structore of bipartite graphs with no perfect matching | ||
| + | * Dicusses how to tell if it is impossible to find a perfect matching in a bipartite graph. | ||
| + | * Can easily find an algorithm based on the following theorem: | ||
| + | * Assume that the bipartite graph G=(V,E) has two sides X and Y such that |X| and |Y|. Then the graph G either has a perfect matching or these is a subset A such that |\G(A)|< | ||
| + | * Section 7.7 - Extensions to Max-Flow Problem | ||
| + | * Circulations with Demands | ||
| + | * We have a directed graph | ||
| + | * edges have capacities | ||
| + | * nodes have supply and demands | ||
| + | * A circulation is a function that satisfies the following | ||
| + | * the flows for a particular edge are always less than the capacity (makes sense and is very intuitive) | ||
| + | * the demand for a node is equal to the stuff coming in minnus the stuff going out (the the demand that his node has? I guess that makes enough sense...) | ||
| + | * To keep the world form exploding, the sum of supplies == sum of demands | ||
| + | * Algorithm: more details on pg 381 and 382 | ||
| + | * Add a new source s and sink t | ||
| + | * For each v with d(v) < 0, add edge (s, v) with capacity -d(v) | ||
| + | * For each v with d(v) > 0, add edge (v, t) with capacity d(v) | ||
| + | * MAGIC: G has circulation iff G' has max flow of value D | ||
| + | * MORE MAGIC : Given (V, E, c, d), there does not exist a circulation iff there exists a node partition (A, B) such that the sum of all the demands is greater than cap(A,B) | ||
| + | * These two pretty much solve the problem for us. more details are on page 382 if you get confused rereading this sometime in the future | ||
| + | * Circulation with Demands and Lower Bounds | ||
| + | * Same givens as last problem | ||
| + | * except!!!! we have lower bounds on each edge | ||
| + | * algorithm on page 384 | ||
| + | * Here's how we do | ||
| + | * we set teh flow along an edge to be the lower bound and then we update the demand on both nodes attached to it. | ||
| + | * the rest is self explanatory | ||
