Chapter 6. Dynamic Programming

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1 Chapter 6 Dynamic Programming CS 573: Algorithms, Fall 203 September 2, Maximum Weighted Independent Set in Trees Maximum Weight Independent Set Problem Input Graph G = (V, E) and weights w(v) 0 for each v V Goal Find maximum weight independent set in G 2 A 5 B 5 C F 2 0 E D 20 Maximum weight independent set in above graph: {B, D} Maximum Weight Independent Set in a Tree Input Tree T = (V, E) and weights w(v) 0 for each v V Goal Find maximum weight independent set in T r 0 5 a b 8 4 c d e f g h i 7 j 8 Maximum weight independent set in above tree:??

2 Towards a Recursive Solution For an arbitrary graph G: (A) Number vertices as v, v 2,..., v n (B) Find recursively optimum solutions without v n (recurse on G v n ) and with v n (recurse on G v n N(v n ) & include v n ). (C) Saw that if graph G is arbitrary there was no good ordering that resulted in a small number of subproblems. What about a tree? Natural candidate for v n is root r of T? Towards a Recursive Solution Natural candidate for v n is root r of T? Let O be an optimum solution to the whole problem. Case r O : Then O contains an optimum solution for each subtree of T hanging at a child of r. Case r O : None of the children of r can be in O. O {r} contains an optimum solution for each subtree of T hanging at a grandchild of r. Subproblems? Subtrees of T hanging at nodes in T A Recursive Solution T (u): subtree of T hanging at node u OP T (u): max weighted independent set value in T (u) v child of u OP T (v), OP T (u) = max w(u) + v grandchild of u OP T (v) Iterative Algorithm (A) Compute OP T (u) bottom up. To evaluate OP T (u) need to have computed values of all children and grandchildren of u (B) What is an ordering of nodes of a tree T to achieve above? Post-order traversal of a tree Iterative Algorithm MIS-Tree(T ): Let v, v 2,..., v n be a post-order traversal of nodes of T for i = to n do ( ) v M[v i ] = max j child of v i M[v j ], w(v i ) + v j grandchild of v i M[v j ] return M[v n ] (* Note: v n is the root of T *) Space: O(n) to store the value at each node of T Running time: (A) Naive bound: O(n 2 ) since each M[v i ] evaluation may take O(n) time and there are n evaluations. (B) Better bound: O(n). A value M[v j ] is accessed only by its parent and grand parent. 2

3 Example r 0 5 a b 8 4 c d e f g h i 7 j Dominating set Definition 6... G = (V, E). The set X V is a dominating set, if any vertex v V is either in X or is adjacent to a vertex in X. r 0 5 a b 8 4 c d e f g h i 7 j 8 Problem Given weights on vertices, compute the minimum weight dominating set in G. Dominating Set is NP-Hard! 6.2 DAGs and Dynamic Programming Recursion and DAGs Observation Let A be a recursive algorithm for problem Π. For each instance I of Π there is an associated DAG G(I). (A) Create directed graph G(I) as follows... (B) For each sub-problem in the execution of A on I create a node. (C) If sub-problem v depends on or recursively calls sub-problem u add directed edge (u, v) to graph. (D) G(I) is a DAG. Why? If G(I) has a cycle then A will not terminate on I Iterative Algorithm for Dynamic Programming and DAGs Observation An iterative algorithm B obtained from a recursive algorithm A for a problem Π does the following: For each instance I of Π, it computes a topological sort of G(I) and evaluates sub-problems according to the topological ordering. 3

4 (A) Sometimes the DAG G(I) can be obtained directly without thinking about the recursive algorithm A (B) In some cases (not all) the computation of an optimal solution reduces to a shortest/longest path in DAG G(I) (C) Topological sort based shortest/longest path computation is dynamic programming! A quick reminder A Recursive Algorithm for weighted interval scheduling Let O i be value of an optimal schedule for the first i jobs. Schedule(n): if n = 0 then return 0 if n = then return w(v ) O p(n) Schedule(p(n)) O n Schedule(n ) if (O p(n) + w(v n ) < O n ) then O n = O n else O n = O p(n) + w(v n ) return O n Weighted Interval Scheduling via Longest Path in a DAG Given intervals, create a DAG as follows: (A) Create one node for each interval, plus a dummy sink node 0 for interval 0, plus a dummy source node s. (B) For each interval i add edge (i, p(i)) of the length/weight of v i. (C) Add an edge from s to n of length 0. (D) For each interval i add edge (i, i ) of length Example p(5) = 2, p(4) =, p(3) =, p(2) = 0, p() = s 4

5 Relating Optimum Solution Given interval problem instance I let G(I) denote the DAG constructed as described. Claim Optimum solution to weighted interval scheduling instance I is given by longest path from s to 0 in G(I). Assuming claim is true, (A) If I has n intervals, DAG G(I) has n + 2 nodes and O(n) edges. Creating G(I) takes O(n log n) time: to find p(i) for each i. How? (B) Longest path can be computed in O(n) time recall O(m + n) algorithm for shortest/longest paths in DAGs DAG for Longest Increasing Sequence Given sequence a, a 2,..., a n create DAG as follows: (A) add sentinel a 0 to sequence where a 0 is less than smallest element in sequence (B) for each i there is a node v i (C) if i < j and a i < a j add an edge (v i, v j ) (D) find longest path from v 0 a a Edit Distance and Sequence Alignment Spell Checking Problem Given a string exponen that is not in the dictionary, how should a spell checker suggest a nearby string? What does nearness mean? Question: Given two strings x x 2... x n and y y 2... y m what is a distance between them? 5

6 Edit Distance: minimum number of edits to transform x into y Edit Distance Definition Edit distance between two words X and Y is the number of letter insertions, letter deletions and letter substitutions required to obtain Y from X. Example The edit distance between FOOD and MONEY is at most 4: Edit Distance: Alternate View FOOD MOOD MONOD MONED MONEY Alignment Place words one on top of the other, with gaps in the first word indicating insertions, and gaps in the second word indicating deletions. F O O D M O N E Y Formally, an alignment is a set M of pairs (i, j) such that each index appears at most once, and there is no crossing : i < i and i is matched to j implies i is matched to j > j. In the above example, this is M = {(, ), (2, 2), (3, 3), (4, 5)}. Cost of an alignment is the number of mismatched columns plus number of unmatched indices in both strings Edit Distance Problem Problem Given two words, find the edit distance between them, i.e., an alignment of smallest cost Applications (A) Spell-checkers and Dictionaries (B) Unix diff (C) DNA sequence alignment... but, we need a new metric Similarity Metric Definition For two strings X and Y, the cost of alignment M is (A) [Gap penalty] For each gap in the alignment, we incur a cost δ. (B) [Mismatch cost] For each pair p and q that have been matched in M, we incur cost α pq ; typically α pp = 0. Edit distance is special case when δ = α pq = An Example Example Alternative: o c u r r a n c e o c c u r r e n c e Cost = δ + α ae o c u r r a n c e o c c u r r e n c e Cost = 3δ Or a really stupid solution (delete string, insert other string): Cost = 9δ. o c u r r a n c e o c c u r r e n c e 6

7 Sequence Alignment Input Given two words X and Y, and gap penalty δ and mismatch costs α pq Goal Find alignment of minimum cost 6.3. Edit distance Basic observation Let X = αx and Y = βy α, β: strings. x and y single characters. Think about optimal edit distance between X and Y as alignment, and consider last column of alignment of the two strings: α x α x αx or or β y βy β y Observation Prefixes must have optimal alignment! Problem Structure Observation Let X = x x 2 x m and Y = y y 2 y n. If (m, n) are not matched then either the mth position of X remains unmatched or the nth position of Y remains unmatched. (A) Case x m and y n are matched. (A) Pay mismatch cost α xm y n plus cost of aligning strings x x m and y y n (B) Case x m is unmatched. (A) Pay gap penalty plus cost of aligning x x m and y y n (C) Case y n is unmatched. (A) Pay gap penalty plus cost of aligning x x m and y y n Subproblems and Recurrence Optimal Costs Let Opt(i, j) be optimal cost of aligning x x i and y y j. Then Base Cases: Opt(i, 0) = δ i and Opt(0, j) = δ j Dynamic Programming Solution α xi y j + Opt(i, j ), Opt(i, j) = min δ + Opt(i, j), δ + Opt(i, j ) for all i do M[i, 0] = iδ for all j do M[0, j] = jδ for i = to m do for j = to n do α xi y j + M[i, j ], M[i, j] = min δ + M[i, j], δ + M[i, j ] 7

8 0, 0 αx ix j δ... δ i, j m, n Figure 6.: Iterative algorithm in previous slide computes values in row order. shortest path from (0, 0) to (m, n) in DAG. Optimal value is a Analysis (A) Running time is O(mn). (B) Space used is O(mn) Matrix and DAG of Computation Sequence Alignment in Practice (A) Typically the DNA sequences that are aligned are about 0 5 letters long! (B) So about 0 0 operations and 0 0 bytes needed (C) The killer is the 0GB storage (D) Can we reduce space requirements? Optimizing Space (A) Recall α xi y j + M(i, j ), M(i, j) = min δ + M(i, j), δ + M(i, j ) (B) Entries in jth column only depend on (j )st column and earlier entries in jth column (C) Only store the current column and the previous column reusing space; N(i, 0) stores M(i, j ) and N(i, ) stores M(i, j) Computing in column order to save space Space Efficient Algorithm for all i do N[i, 0] = iδ for j = to n do N[0, ] = jδ (* corresponds to M(0, j) *) for i = to m do α xi y j + N[i, 0] N[i, ] = min δ + N[i, ] δ + N[i, 0] for i = to m do Copy N[i, 0] = N[i, ] 8

9 0, 0 αx ix j δ... δ i, j m, n Figure 6.2: M(i, j) only depends on previous column values. Keep only two columns and compute in column order. Analysis Running time is O(mn) and space used is O(2m) = O(m) Analyzing Space Efficiency (A) From the m n matrix M we can construct the actual alignment (exercise) (B) Matrix N computes cost of optimal alignment but no way to construct the actual alignment (C) Space efficient computation of alignment? More complicated algorithm see text book Takeaway Points (A) Dynamic programming is based on finding a recursive way to solve the problem. Need a recursion that generates a small number of subproblems. (B) Given a recursive algorithm there is a natural DAG associated with the subproblems that are generated for given instance; this is the dependency graph. An iterative algorithm simply evaluates the subproblems in some topological sort of this DAG. (C) The space required to evaluate the answer can be reduced in some cases by a careful examination of that dependency DAG of the subproblems and keeping only a subset of the DAG at any time. 6.4 All Pairs Shortest Paths Shortest Path Problems Shortest Path Problems Input A (undirected or directed) graph G = (V, E) with edge lengths (or costs). For edge e = (u, v), l(e) = l(u, v) is its length. (A) Given nodes s, t find shortest path from s to t. (B) Given node s find shortest path from s to all other nodes. (C) Find shortest paths for all pairs of nodes Single-Source Shortest Paths Single-Source Shortest Path Problems Input A (undirected or directed) graph G = (V, E) with edge lengths. For edge e = (u, v), l(e) = l(u, v) is its length. 9

10 (A) Given nodes s, t find shortest path from s to t. (B) Given node s find shortest path from s to all other nodes. Dijkstra s algorithm for non-negative edge lengths. Running time: O((m + n) log n) with heaps and O(m + n log n) with advanced priority queues. Bellman-Ford algorithm for arbitrary edge lengths. Running time: O(nm) All-Pairs Shortest Paths All-Pairs Shortest Path Problem Input A (undirected or directed) graph G = (V, E) with edge lengths. For edge e = (u, v), l(e) = l(u, v) is its length. (A) Find shortest paths for all pairs of nodes. Apply single-source algorithms n times, once for each vertex. (A) Non-negative lengths. O(nm log n) with heaps and O(nm+n 2 log n) using advanced priority queues. (B) Arbitrary edge lengths: O(n 2 m). Θ(n 4 ) if m = Ω(n 2 ). Can we do better? Shortest Paths and Recursion (A) Compute the shortest path distance from s to t recursively? (B) What are the smaller sub-problems? Lemma Let G be a directed graph with arbitrary edge lengths. If s = v 0 v v 2... v k is a shortest path from s to v k then for i < k: (A) s = v 0 v v 2... v i is a shortest path from s to v i Sub-problem idea: paths of fewer hops/edges Hop-based Recur : Single-Source Shortest Paths Single-source problem: fix source s. OP T (v, k): shortest path dist. from s to v using at most k edges. Note: dist(s, v) = OP T (v, n ). Recursion for OP T (v, k): min u V (OP T (u, k ) + c(u, v)). OP T (v, k) = min OP T (v, k ) Base case: OP T (v, ) = c(s, v) if (s, v) E otherwise Leads to Bellman-Ford algorithm see text book. OP T (v, k) values are also of independent interest: shortest paths with at most k hops 0

11 All-Pairs: Recursion on index of intermediate nodes (A) Number vertices arbitrarily as v, v 2,..., v n (B) dist(i, j, k): shortest path distance between v i and v j among all paths in which the largest index of an intermediate node is at most k 4 3 i 2 5 j i 2 5 j dist(i, j, 0) = 00 i 2 5 j 0 dist(i, j, ) = 9 2 dist(i, j, 2) = 8 00 dist(i, j, 3) = i 2 5 j i 2 5 j All-Pairs: Recursion on index of intermediate nodes dist(i, k, k ) k dist(k, j, k ) i j dist(i, j, k ) dist(i, j, k ) dist(i, j, k) = min dist(i, k, k ) + dist(k, j, k ) Base case: dist(i, j, 0) = c(i, j) if (i, j) E, otherwise Correctness: If i j shortest path goes through k then k occurs only once on the path otherwise there is a negative length cycle.

12 6.4. Floyd-Warshall Algorithm for All-Pairs Shortest Paths Check if G has a negative cycle // Bellman-Ford: O(mn) time if there is a negative cycle then return Negative cycle for i = to n do for j = to n do dist(i, j, 0) = c(i, j) (* c(i, j) = if (i, j) / E, 0 if i = j *) for k = to n do for i = to n do for j = to n do dist(i, j, k) = min { dist(i, j, k ), dist(i, k, k ) + dist(k, j, k ) Correctness: Recursion works under the assumption that all shortest paths are defined (no negative length cycle). Running Time: Θ(n 3 ), Space: Θ(n 3 ) Floyd-Warshall Algorithm for All-Pairs Shortest Paths Do we need a separate algorithm to check if there is negative cycle? for i = to n do for j = to n do dist(i, j, 0) = c(i, j) (* c(i, j) = if (i, j) / E, 0 if i = j *) not edge, 0 if i = j *) for k = to n do for i = to n do for j = to n do dist(i, j, k) = min(dist(i, j, k ), dist(i, k, k ) + dist(k, j, k )) for i = to n do if (dist(i, i, n) < 0) then Output that there is a negative length cycle in G Correctness: exercise Floyd-Warshall Algorithm: Finding the Paths Question: Can we find the paths in addition to the distances? (A) Create a n n array Next that stores the next vertex on shortest path for each pair of vertices (B) With array Next, for any pair of given vertices i, j can compute a shortest path in O(n) time. 2

13 6.4.3 Floyd-Warshall Algorithm Finding the Paths for i = to n do for j = to n do dist(i, j, 0) = c(i, j) (* c(i, j) = if (i, j) not edge, 0 if i = j *) Next(i, j) = for k = to n do for i = to n do for j = to n do if (dist(i, j, k ) > dist(i, k, k ) + dist(k, j, k )) then dist(i, j, k) = dist(i, k, k ) + dist(k, j, k ) Next(i, j) = k for i = to n do if (dist(i, i, n) < 0) then Output that there is a negative length cycle in G Exercise: Given Next array and any two vertices i, j describe an O(n) algorithm to find a i-j shortest path Summary of results on shortest paths Single vertex No negative edges Dijkstra O(n log n + m) Edges cost might be negative But no negative cycles Bellman Ford O(nm) All Pairs Shortest Paths No negative edges n * Dijkstra O(n 2 log n + nm) No negative cycles n * Bellman Ford O(n 2 m) = O(n 4 ) No negative cycles Floyd-Warshall O(n 3 ) 6.5 Knapsack Knapsack Problem Input Given a Knapsack of capacity W lbs. and n objects with ith object having weight w i and value v i ; assume W, w i, v i are all positive integers Goal Fill the Knapsack without exceeding weight limit while maximizing value. Basic problem that arises in many applications as a sub-problem Knapsack Example Item I I 2 I 3 I 4 I 5 Example Value Weight If W =, the best is {I 3, I 4 } giving value 40. Special Case When v i = w i, the Knapsack problem is called the Subset Sum Problem. 3

14 Greedy Approach (A) Pick objects with greatest value (A) Let W = 2, w = w 2 =, w 3 = 2, v = v 2 = 2 and v 3 = 3; greedy strategy will pick {3}, but the optimal is {, 2} (B) Pick objects with smallest weight (A) Let W = 2, w =, w 2 = 2, v = and v 2 = 3; greedy strategy will pick {}, but the optimal is {2} (C) Pick objects with largest v i /w i ratio (A) Let W = 4, w = w 2 = 2, w 3 = 3, v = v 2 = 3 and v 3 = 5; greedy strategy will pick {3}, but the optimal is {, 2} (B) Can show that a slight modification always gives half the optimum profit: pick the better of the output of this algorithm and the largest value item. Also, the algorithms gives better approximations when all item weights are small when compared to W Towards a Recursive Solution First guess: Opt(i) is the optimum solution value for items,..., i. Observation Consider an optimal solution O for,..., i Case item i O O is an optimal solution to items to i Case item i O Then O {i} is an optimum solution for items to n in knapsack of capacity W w i. Subproblems depend also on remaining capacity. Cannot write subproblem only in terms of Opt(),..., Opt(i ). Opt(i, w): optimum profit for items to i in knapsack of size w Goal: compute Opt(n, W ) Dynamic Programming Solution Definition Let Opt(i, w) be the optimal way of picking items from to i, with total weight not exceeding w. 0 if i = 0 Opt(i, w) if w i > w Opt(i, w) = Opt(i, w) max otherwise Opt(i, w w i ) + v i An Iterative Algorithm Running Time for w = 0 to W do M[0, w] = 0 for i = to n do for w = to W do if (w i > w) then M[i, w] = M[i, w] else M[i, w] = max(m[i, w], M[i, w w i ] + v i ) 4

15 (A) Time taken is O(nW ) (B) Input has size O(n + log W + n i= (log v i + log w i )); so running time not polynomial but pseudopolynomial! Knapsack Algorithm and Polynomial time (A) Input size for Knapsack: O(n) + log W + n i= (log w i + log v i ). (B) Running time of dynamic programming algorithm: O(nW ). (C) Not a polynomial time algorithm. (D) Example: W = 2 n and w i, v i [..2 n ]. Input size is O(n 2 ), running time is O(n2 n ) arithmetic/comparisons. (E) Algorithm is called a pseudo-polynomial time algorithm because running time is polynomial if numbers in input are of size polynomial in the combinatorial size of problem. (F) Knapsack is NP-Hard if numbers are not polynomial in n. 6.6 Traveling Salesman Problem Traveling Salesman Problem Input A graph G = (V, E) with non-negative edge costs/lengths. c(e) for edge e Goal Find a tour of minimum cost that visits each node. No polynomial time algorithm known. Problem is NP-Hard. 5

16 Drawings using TSP 6

17 Example: optimal tour for cities of a country (which one?) An Exponential Time Algorithm How many different tours are there? n! Stirling s formula: n! n(n/e) n which is Θ(2 cn log n ) for some constant c > Can we do better? Can we get a 2 O(n) time algorithm? Towards a Recursive Solution (A) Order vertices as v, v 2,..., v n (B) OP T (S): optimum TSP tour for the vertices S V in the graph restricted to S. Want OP T (V ). Can we compute OP T (S) recursively? (A) Say v S. What are the two neighbors of v in optimum tour in S? (B) If u, w are neighbors of v in an optimum tour of S then removing v gives an optimum path from u to w visiting all nodes in S {v}. Path from u to w is not a recursive subproblem! Need to find a more general problem to allow recursion A More General Problem: TSP Path Input A graph G = (V, E) with non-negative edge costs/lengths(c(e) for edge e) and two nodes s, t Goal Find a path from s to t of minimum cost that visits each node exactly once. 7

18 Can solve TSP using above. Do you see how? Recursion for optimum TSP Path problem: (A) OP T (u, v, S): optimum TSP Path from u to v in the graph restricted to S (here u, v S) A More General Problem: TSP Path Continued... What is the next node in the optimum path from u to v? Suppose it is w. Then what is OP T (u, v, S)? We do not know w! So try all possibilities for w. OP T (u, v, S) = c(u, w) + OP T (w, v, S {u}) A Recursive Solution OP T (u, v, S) = min w S,w u,v ( c(u, w) + OP T (w, v, S {u}) ) What are the subproblems for the original problem OP T (s, t, V )? OP T (u, v, S) for u, v S, S V. How many subproblems? (A) number of distinct subsets S of V is at most 2 n (B) number of pairs of nodes in a set S is at most n 2 (C) hence number of subproblems is O(n 2 2 n ) Exercise: Show that one can compute TSP using above dynamic program in O(n 3 2 n ) time and O(n 2 2 n ) space. Disadvantage of dynamic programming solution: memory! Dynamic Programming: Postscript Dynamic Programming = Smart Recursion + Memoization (A) How to come up with the recursion? (B) How to recognize that dynamic programming may apply? Some Tips (A) Problems where there is a natural linear ordering: sequences, paths, intervals, DAGs etc. Recursion based on ordering (left to right or right to left or topological sort) usually works. (B) Problems involving trees: recursion based on subtrees. (C) More generally: (A) Problem admits a natural recursive divide and conquer (B) If optimal solution for whole problem can be simply composed from optimal solution for each separate pieces then plain divide and conquer works directly (C) If optimal solution depends on all pieces then can apply dynamic programming if interface/interaction between pieces is limited. Augment recursion to not simply find an optimum solution but also an optimum solution for each possible way to interact with the other pieces Examples (A) Longest Increasing Subsequence: break sequence in the middle say. What is the interaction between the two pieces in a solution? (B) Sequence Alignment: break both sequences in two pieces each. What is the interaction between the two sets of pieces? 8

19 (C) Independent Set in a Tree: break tree at root into subtrees. What is the interaction between the sutrees? (D) Independent Set in an graph: break graph into two graphs. What is the interaction? Very high! (E) Knapsack: Split items into two sets of half each. What is the interaction? 9

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