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AlgoPARCICS 491: Competitive ProgrammingProf. Nodari SitchinavaLecture 5: Dynamic ProgrammingICS 491: Competitve Programming – Lecture 5: Dynamic Programmingwww.algoparc.ics.hawaii.edu
Practice, practice, practice.ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Practice, practice, practice.How many of you solved last week’s contest problems?ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Weekly Mini-Contest – 75 minComplete SearchICS 491: Competitve Programming – Lecture 5: Dynamic Programming
SolutionsICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Dynamic ProgrammingAt each ICPC:1-2 (simple) ad-hoc problems1-2 complete search problems1-2 dynamic programming problems2-3 Graph problemsICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Dynamic Programming (DP)Recursive Complete Search,Pruned with a Lookup (Memo) TableICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Dynamic Programming (DP)Recursive Complete Search,Pruned with a Lookup (Memo) Table(Faster if there are overlapping subproblems)ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Example: Subset SumProblem (Subset Sum). Given a set S of 1 n 20 integersand a positive integer x, is there a subset of S that sums to x?S {17, 5, 7, 15, 3, 8}ICS 491: Competitve Programming – Lecture 5: Dynamic Programmingx 16
Example: Subset SumProblem (Subset Sum). Given a set S of 1 n 20 integersand a positive integer x, is there a subset of S that sums to x?S {17, 5, 7, 15, 3, 8}x 16Solution: Generate all possible subsets, add up the elements ofeach subset and check if the sum equals xICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Example: Subset SumProblem (Subset Sum). Given a set S of 1 n 20 integersand a positive integer x, is there a subset of S that sums to x?S {17, 5, 7, 15, 3, 8}0 1 0 0 1 1x 16Solution: Generate all possible subsets, add up the elements ofeach subset and check if the sum equals xICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Example: Subset SumProblem (Subset Sum). Given a set S of 1 n 20 integersand a positive integer x, is there a subset of S that sums to x?S {17, 5, 7, 15, 3, 8}0 1 0 0 1 1x 16Solution: Generate all possible subsets, add up the elements ofeach subset and check if the sum equals xfor (i 0; i (1 n); i ) { // i-th subset out of 2 nsum 0;for (int j 0; j n; j )if (i & (1 j))// is j-th bit set in i?sum S[j]// j is part of the subsetif (sum x) return true;// or could return i}return false;ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Example: Subset SumProblem (Subset Sum). Given a set S of 1 n 20 integersand a positive integer x, is there a subset of S that sums to x?S {17, 5, 7, 15, 3, 8}0 1 0 0 1 1x 16i 19Solution: Generate all possible subsets, add up the elements ofeach subset and check if the sum equals xfor (i 0; i (1 n); i ) { // i-th subset out of 2 nsum 0;for (int j 0; j n; j )if (i & (1 j))// is j-th bit set in i?sum S[j]// j is part of the subsetif (sum x) return true;// or could return i}return false;ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive SolutionICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive SolutionTop-down Solution: Prune search space as you generatepartial solutionsS {17, 5, 7, 15, 3, 8}0 1 0 0 1 1ICS 491: Competitve Programming – Lecture 5: Dynamic Programmingx 16
Subset Sum: Recursive SolutionTop-down Solution: Prune search space as you generatepartial solutionsS {17, 5, 7, 15, 3, 8}0 1 0 0 1 100x 1611 010100011000000 000001001000010 00001111111100 111101ICS 491: Competitve Programming – Lecture 5: Dynamic Programming01111110 111111
Subset Sum: Recursive SolutionICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive Solutionint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(S[i:n], x):elsebool Si notSelected SubsetSum(S[i 1:n], x);bool Si selected SubsetSum(S[i 1:n], x-S[i]);return (Si notSelected or Si selected);ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive Solutionint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(S[i:n], x):if (x 0 or i n) return false;else if (x 0)return true;elsebool Si notSelected SubsetSum(S[i 1:n], x);bool Si selected SubsetSum(S[i 1:n], x-S[i]);return (Si notSelected or Si selected);ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive Solutionint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(S[i:n], x):if (x 0 or i n) return false;else if (x 0)return true;elsebool Si notSelected SubsetSum(S[i 1:n], x);bool Si selected SubsetSum(S[i 1:n], x-S[i]);return (Si notSelected or Si selected);main() {return SubsetSum(S[1:n], x);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive Solutionint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(S[i:n], x):if (x 0 or i n) return false;else if (x 0)return true;elsereturn SubsetSum(S[i 1:n], x) or// S[i] not selectedSubsetSum(S[i 1:n], x-S[i]); // S[i] selectedmain() {return SubsetSum(S[1:n], x);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive Solutionint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(i, x):if (x 0 or i n) return false;else if (x 0)return true;elsereturn SubsetSum(i 1, x) or// S[i] not selectedSubsetSum(i 1, x-S[i]);// S[i] selectedmain() {return SubsetSum(1, x);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Recursive Solutionint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSS[i][x]SubsetSum(i, x):if (x 0 or i n) return false;else if (x 0)return true;elsereturn SubsetSum(i 1, x) or// S[i] not selectedSubsetSum(i 1, x-S[i]);// S[i] selectedmain() {return SubsetSum(1, x);}ICS 491: Competitve Programming – Lecture 5: Dynamic ProgrammingSS[i 1][x] orSS[i 1][x-S[i]]
Subset Sum: Dynamic Programmingint S[20] {.}; int x .;// initialize S & x// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(i, x):if (x 0 or i n) return false;else if (x 0)return SS[i][x] true;else if (SS[i][x] ! UNDEFINED) return SS[i][x];elsereturn SS[i 1][x] SubsetSum(i 1, x) orSS[i 1][x-S[i]] SubsetSum(i 1, x-S[i]);main() {return SubsetSum(1, x);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Subset Sum: Dynamic Programmingint S[20] {.}; int x .;// initialize S & xint SS[20][MAX X]; memset(SS, UNDEFINED, sizeof(SS));// returns true iff exists subset within S[i:n]// that adds up to xSubsetSum(i, x):if (x 0 or i n) return false;else if (x 0)return SS[i][x] true;else if (SS[i][x] ! UNDEFINED) return SS[i][x];elsereturn SS[i 1][x] SubsetSum(i 1, x) orSS[i 1][x-S[i]] SubsetSum(i 1, x-S[i]);main() {return SubsetSum(1, x);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 KnapsackProblem (0-1 Knapsack). Given a set S of n items, each with itsown value Vi and weight Wi for all 1 i n and a maximumknapsack capacity C, compute the maximum value of the itemsthat you can carry. You cannot take fractions of items.ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 KnapsackProblem (0-1 Knapsack). Given a set S of n items, each with itsown value Vi and weight Wi for all 1 i n and a maximumknapsack capacity C, compute the maximum value of the itemsthat you can carry. You cannot take fractions of items.Optimization version of Subset SumICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 KnapsackProblem (0-1 Knapsack). Given a set S of n items, each with itsown value Vi and weight Wi for all 1 i n and a maximumknapsack capacity C, compute the maximum value of the itemsthat you can carry. You cannot take fractions of items.Optimization version of Subset SumExample:{(Vi , Wi )} {(10, 17), (5, 7), (3, 8), (9, 15)}ICS 491: Competitve Programming – Lecture 5: Dynamic ProgrammingC 16
0-1 Knapsack: Recursive Solution{(Vi , Wi )} {(10, 17), (5, 7), (3, 8), (9, 15)}ICS 491: Competitve Programming – Lecture 5: Dynamic ProgrammingC 16
0-1 Knapsack: Recursive Solution{(Vi , Wi )} {(10, 17), (5, 7), (3, 8), (9, 15)}C 16maxV (i, C) returns the maximum value among items S[i : n] withremaining knapsack capacity of SICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: Recursive Solution{(Vi , Wi )} {(10, 17), (5, 7), (3, 8), (9, 15)}C 16maxV (i, C) returns the maximum value among items S[i : n] withremaining knapsack capacity of SmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: Recursive SolutionmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: Recursive SolutionmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )maxV(i,C) {if (i n C 0) return 0;if (W[i] C)//can’t take i-th itemreturn maxV(i 1, C);return max(maxV(i 1, C),//don’t take i-th itemV[i] maxV(i 1, C-W[i]));// take i-th item}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: Recursive SolutionmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )maxV(i,C) {if (i n C 0) return 0;if (W[i] C)//can’t take i-th itemreturn maxV(i 1, C);return max(maxV(i 1, C),//don’t take i-th itemV[i] maxV(i 1, C-W[i]));// take i-th item}main() {return maxV(1, C);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: DP solutionmaxV (i, C) M[i][C] 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )maxV(i,C) {if (i n C 0) return 0;M[i 1][C]if (W[i] C)//can’t take i-th itemreturn maxV(i 1, C);M[i 1][C]return max(maxV(i 1, C),//don’t take i-th itemV[i] maxV(i 1, C-W[i]));// take i-th item}main() {return maxV(1, C);}ICS 491: Competitve Programming – Lecture 5: Dynamic ProgrammingM[i 1][C-W[i]]
0-1 Knapsack: DP solutionmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )maxV(i,C) {if (i n C 0) return 0;if (W[i] C)//can’t take i-th itemreturn maxV(i 1, C);return max(maxV(i 1, C),//don’t take i-th itemV[i] maxV(i 1, C-W[i]));// take i-th item}main() {return maxV(1, C);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: DP solutionmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )maxV(i,C) {if (i n C 0) return 0;if (M[i][C] ! UNDEFINED) return M[i, C];if (W[i] C)//can’t take i-th itemreturn M[i][C] maxV(i 1, C);return M[i][C] max(maxV(i 1, C),//don’t take i-th itemV[i] maxV(i 1, C-W[i]));// take i-th item}main() {return maxV(1, C);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
0-1 Knapsack: DP solutionmaxV (i, C) 0 0if i nif C 0if Wi C if Wi CmaxV (i 1, C) maxV (i 1, C )maxVi maxV (i 1, C Wi )int M[maxN 1][maxC];maxV(i,C) {if (i n C 0) return 0;if (M[i][C] ! UNDEFINED) return M[i, C];if (W[i] C)//can’t take i-th itemreturn M[i][C] maxV(i 1, C);return M[i][C] max(maxV(i 1, C),//don’t take i-th itemV[i] maxV(i 1, C-W[i]));// take i-th item}main() { memset(M, UNDEFINED, sizeof(M));return maxV(1, C);}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Longest Increasing Subsequence (LIS)Problem. Given a sequence A[1.n], determine the length of itslongest subsequence (not necessarily contiguous) that is inincreasing order.ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Longest Increasing Subsequence (LIS)Problem. Given a sequence A[1.n], determine the length of itslongest subsequence (not necessarily contiguous) that is inincreasing order.Example: A { 7, 10, 9, 2, 3, 8, 8, 1}.Solution: LIS 4: { 7, 2, 3, 8}ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Longest Increasing Subsequence (LIS)Problem. Given a sequence A[1.n], determine the length of itslongest subsequence (not necessarily contiguous) that is inincreasing order.Example: A { 7, 10, 9, 2, 3, 8, 8, 1}.Solution: LIS 4: { 7, 2, 3, 8}Hint 1: LIS(i) returns the size of the longest increasingsubsequence that terminates on A[i].ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Longest Increasing Subsequence (LIS)Problem. Given a sequence A[1.n], determine the length of itslongest subsequence (not necessarily contiguous) that is inincreasing order.Example: A { 7, 10, 9, 2, 3, 8, 8, 1}.Solution: LIS 4: { 7, 2, 3, 8}Hint 1: LIS(i) returns the size of the longest increasingsubsequence that terminates on A[i].Hint 2: When deciding whether to add A[i], find the longestsubsequence in A[1 : i 1] that allows adding A[i].ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
LIS: Recursive SolutionICS 491: Competitve Programming – Lecture 5: Dynamic Programming
A { 7, 10, 9, 2, 3, 8, 8, 1}LIS: Recursive SolutionLIS(i) 1 if i 1 1 LIS(1) 1 max LIS(2) 1LIS(3) 1.LIS(i 1) 1if A[i]if A[i]if A[i].if A[i]ICS 491: Competitve Programming – Lecture 5: Dynamic Programming A[1] A[2] A[3] A[i 1] if i 1
A { 7, 10, 9, 2, 3, 8, 8, 1}LIS: Recursive SolutionLIS(i) LIS(i) 1 if i 1 1 LIS(1) 1 max LIS(2) 1LIS(3) 1.LIS(i 1) 1if A[i]if A[i]if A[i].if A[i]if (i 1) return 1;best 1; // LIS { A[i] }for (int j 1; j i; j )if (A[i] A[j]) {current LIS(j) 1;if (best current)best current;}return best;ICS 491: Competitve Programming – Lecture 5: Dynamic Programming A[1] A[2] A[3] A[i 1] if i 1
A { 7, 10, 9, 2, 3, 8, 8, 1}LIS: Recursive SolutionLIS(i) LIS(i) 1 if i 1 1 LIS(1) 1 max LIS(2) 1LIS(3) 1.LIS(i 1) 1if A[i]if A[i]if A[i].if A[i]if (i 1) return 1;best 1; // LIS { A[i] }for (int j 1; j i; j )if (A[i] A[j]) {current LIS(j) 1;if (best current)best current;}return best;ICS 491: Competitve Programming – Lecture 5: Dynamic Programming A[1] A[2] A[3] A[i 1] if i 1 main()best 0;for (i 1; i n; i ){current LIS(i);if (best current)best current;}return best;
A { 7, 10, 9, 2, 3, 8, 8, 1}LIS: Recursive SolutionLIS(i) 1 if i 1 1 LIS(1) 1 max LIS(2) 1LIS(3) 1.LIS(i 1) 1if A[i]if A[i]if A[i].if A[i] A[1] A[2] A[3] if i 1 A[i 1]LIS(i)if (L[i]! UNDEFINED) return L[i];if (i 1) return L[i] 1;best 1; // LIS { A[i] }main()for (int j 1; j i; j )best 0;if (A[i] A[j]) {for (i 1; i n; i ){current LIS(j) 1;current LIS(i);if (best current)if (best current)best current;best current;}}return L[i] best;return best;ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Traveling Salesperson Problem (TSP)Problem. Given n cities and their pairwise distance in the formof a n n matrix D, compute the minimum cost of making a tourthat starts from any city s, goes through all the other n 1 citiesexactly once, and finally returns to the starting city s.ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Traveling Salesperson Problem (TSP)Problem. Given n cities and their pairwise distance in the formof a n n matrix D, compute the minimum cost of making a tourthat starts from any city s, goes through all the other n 1 citiesexactly once, and finally returns to the starting city s.Solution (Iterative Complete Search): Try each of (n 1)!permutations of the cities that define possible tours and computethe cost of each tourICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Traveling Salesperson Problem (TSP)Problem. Given n cities and their pairwise distance in the formof a n n matrix D, compute the minimum cost of making a tourthat starts from any city s, goes through all the other n 1 citiesexactly once, and finally returns to the starting city s.Solution (Iterative Complete Search): Try each of (n 1)!permutations of the cities that define possible tours and computethe cost of each tourSolution (Recursive Complete Search):Consider a tour starting at city posHave a choice of the next city to go to.Try each city i that hasn’t been visited yet, and find theminimum cost tour of the remaining cities starting at i.ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
TSP: Recursive solutionICS 491: Competitve Programming – Lecture 5: Dynamic Programming
TSP: Recursive solutionTSP(pos, visited)// Base casebest INFINITY;for (i 0; i n; i ) {// for each of n citiesif (! visited[i]) {visited[i] true// mark i as visitedcurrent D[pos, i] TSP(i, visited)if (best current)best currentvisited[i] false// backtrack}return bestICS 491: Competitve Programming – Lecture 5: Dynamic Programming
TSP: Recursive solutionTSP(pos, visited)allVisited true;for (i 0; i n; i )// check if visited all citiesallVisited allVisited && visited[i];if (allVisited) return dist[pos][s];// return to startbest INFINITY;for (i 0; i n; i ) {// for each of n citiesif (! visited[i]) {visited[i] true// mark i as visitedcurrent D[pos, i] TSP(i, visited)if (best current)best currentvisited[i] false// backtrack}return bestICS 491: Competitve Programming – Lecture 5: Dynamic Programming
TSP: Recursive solutionTSP(pos, visMask)if (visMask (1 n)-1)return D[pos, s]// Are all n bits set?best INFINITY;for (i 0; i n; i ) {// for each of n citiesif (visMask & (1 i) 0) {visMask (1 i)// mark i as visitedcurrent D[pos, i] TSP(i, visMask)if (best current)best currentvisMask & (1 i)// backtrack (unmark i)}return bestICS 491: Competitve Programming – Lecture 5: Dynamic Programming
TSP: Recursive solutionT[pos, visMask]TSP(pos, visMask)if (visMask (1 n)-1)return D[pos, s]// Are all n bits set?best INFINITY;for (i 0; i n; i ) {// for each of n citiesif (visMask & (1 i) 0) {visMask (1 i)// mark i as visitedcurrent D[pos, i] TSP(i, visMask)if (best current)best currentvisMask & (1 i)// backtrack (unmark i)}return bestICS 491: Competitve Programming – Lecture 5: Dynamic Programming
TSP: Recursive solutionT[pos, visMask]TSP(pos, visMask)if (visMask (1 n)-1)// Are all n bits set?return T[pos,visMask] D[pos, s]if (T[pos,visMask] ! UNDEFINED)return T[pos,visMask]best INFINITY;for (i 0; i n; i ) {// for each of n citiesif (visMask & (1 i) 0) {visMask (1 i)// mark i as visitedcurrent D[pos, i] TSP(i, visMask)if (best current)best currentvisMask & (1 i)// backtrack (unmark i)}return T[pos,visMask] bestICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Other Classic Problems (review ICS 311)Rod cuttingCoin changingMatrix-chain multiplicationLongest Common Subsequence (LCS)Optimal binary search treeProblems on Directed Acyclic Graphs (DAGs)ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
Questions?ICS 491: Competitve Programming – Lecture 5: Dynamic Programming
ICS 491: Competitve Programming Lecture 5: Dynamic Programming Example: Subset Sum Problem (Subset Sum) . Given a set S of 1 n 20 integers and a positive integer x , is there a subset of S that sums to x ? Solution: Generate all possible subsets, add up the elements of each subset and check if the sum equals
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Dorothy Inouye, Marilyn Kobata, Drs. Margaret and William Won, and Dr. Belinda Aquino enjoy a behind-the-scenes look at the making of . Nā Mele. PBS Hawai‘i Supporters . Meet and Greet Nā Mele Performer Josh Tatofi. PBS Hawai‘i supporters enjoy a behind-the-scenes look at the m
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