2 DD2458 Popup HT Solution: Choose the activity which ends first and does not conflict with earlier chosen activities.

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1 DD2458, Problem Solvig ad Programmig Uder Pressure Lecture 1: Greedy algorithms ad dyamic programmig Date: Scribe(s: Marti Wedi ad Nilas Wagre Lecturer: Douglas Wiström This lecture cotais basic iformatio about three differet ad ofte used classes of algorithms: Divide ad Coquer, Greedy algorithms ad Dyamic programmig. 1 Divide ad Coquer Divide ad coquer algorithms ca be geeralized i three steps. 1. Divide the mai problem ito two or more subproblems. 2. Solve each subproblem recursively. 3. Combie the solutios for the subproblems to build a solutio for the mai problem. Algorithm 1: Merge Sort Iput: List L = [u 1, u 2,..., u ] Output: The elemets i L i sorted order. MergeSort(L (1 if U = 1 (2 retur L (3 L l MergeSort( [ u 1,..., u /2 ] (4 L r MergeSort( [ u /2 +1,..., u ] (5 retur Merge(L l, L r 2 Greedy algorithms 2.1 Disjoit itervals Example: Activities i lecture hall Problem: Activities are to be held i a lecture hall. The problem is that all activities do ot fit i the same hall. Each activity has a startig ad a edig time. The tas is to fit i as may activities as possible ito the schedule. Formally: Give a set of time itervals [s 1, f 1,..., [s, f, choose the maximum umber of mutually disjoit ivervals. 1

2 2 DD2458 Popup HT 2008 Solutio: Choose the activity which eds first ad does ot coflict with earlier chose activities. Algorithm 2: Maximum umber of mutually disjoit ivervals, Iput: The itervals [s 1, f 1,..., [s, f. Output: The largest subset of mutual disjoit itervals. DisjoitIterval([s 1, f 1,..., [s, f (1 Sort the itervals so that f 1 f 2... f. (2 S {1}, j 1 (3 for i = 2 to (4 if s i f j (5 S S {i} (6 j i (7 retur S Greedy choice: The iterval I mi which cotais the earliest edig time is a part of a optimal solutio. Optimal substructure: The optimal solutio to I 1,..., I I mi combied with I mi is the optimal solutio to I 1,..., I. If we tae the iterval I mi which has the earliest edig time the we ca chage the first iterval i ay optimal solutio to I mi ad still have a optimal solutio. This ca the be repeated for I I mi ad so o. Therefor greedy choice always fids a optimal solutio. 2.2 Iterval coverig Example: Lightig a old car tuel Problem: I a old car tuel there are some lamps located uevely throughout the tuel. Each lamp has a differet lightig rage. The tas is to light up the whole tuel while usig as few lamps as possible. Formally: Give a iterval [s, f], ad a set of itervals [s 1, f 1 ],..., [s, f ]. Choose a miimal subset S {1,..., } such that [s, f] i S [s i, f i ]. Solutio: Choose that lamp which covers the etrace of the tuel ad has the logest lightig rage ito the tuel. Cotiuously choose the lamp which overlap the lighted part ad has the logest rage further ito the tuel - util you reach the exit. Greedy Choise: The itervall I max which cotais s ad has the largest edig poit f max is a part of a optimal solutio.

3 Greedy algorithms ad dyamic programmig 3 Algorithm 3: Iterval coverig. Iput: Goal iterval [s, f] ad the itervals [s 1, f 1 ],.., [s, f ] such that [s, f] [s i, f i ]. Output: A smallest subset S {1,..., } such that [s, f] i S [s i, f i ]. ItervalCover([s 1, f 1 ],..., [s, f ] (1 S, t s (2 while t < f (3 f i max{f j s j t f j } (4 S S {i} (5 t f i (6 retur S Optimal Substructure. The optimal solutio to [f max, f], {I 1,..., I } I max combied with I max is the optimal solutio for {I 1,..., I }. If we tae the iterval I max which cotais the startig poit a ad has the largest edig poit the we ca chage the first iterval i ay optimal solutio to I max ad still have a optimal solutio. This ca the be repeated for I I max ad so o. There for greedy choise always fids the optimal solutio. 2.3 Cotiuous apsac problem Recall the usual apsac problem: We have items ad item i has value v i ad weight w i. Ad a bag which ca carry a maximum weight of c. The tas is to maximize the sum of the items value i the bag. However, this problem is NP-complete ad we will ot try to solve it. We will istead solve a modified versio. Problem: We ow have piles of differet metal grai. Each pile cotais a specific weight of metal grai, which has a specific value per weight. A theif carryig a bag with a weight capacity of c wats to maximize the value of his loot. The tas is to figure out how much of each pile he should steal. Formally: Give a apsac with capacity c ad a set of value/weight-pairs [(v 1, w 1,..., (v, w ]. Choose [a 1,..., a ] such that a i w i ad i=1 a i v i /w i is maximized. Solutio: Choose the grai with the highest value per weight ad tae as much as you ca ad cotiue util the bag is full or you ru out of grai. 2.4 Huffma codig Problem: Give a table of how commo each letter i a alphabet is, fid the prefixfree bit strig which gives the shortest expected strig legth. Prefix-free meas that the bit strig represetig some particular symbol is ever a prefix of the bit strig represetig ay other symbol. Formally: Give a set of letter/weight-pairs {(a 1, f 1,..., (a, f } for some alphabet A = {a 1,..., a }. Fid a prefix-free code τ : A {0, 1} that miimizes i=1 f i τ(a i, where τ(a i is the legth of the bit strig τ(a i.

4 4 DD2458 Popup HT 2008 Algorithm 4: Cotiuous apsac. Iput: The capacity W ad value/weight-pairs [(v 1, w 1,..., (v, w ]. Output: [a 1,..., a ], a i w i, such that i=1 a i v i w i is maximized. CotKapsac(W, [(v 1, w 1,..., (v, w ] (1 Sort the value/weight-pairs such that v 1.. v w. (2 for i = 1 to (3 a i mi{w, w i } (4 W W \a i (5 retur [a 1,..., a ] w 1 v 2 w 2 Figure 1: Example of Huffma codig. Solutio: We idetify a prefix-free code τ with a 0/1-labeled tree. The bit strig is represeted by the path dow the tree. How do we build the tree i the best way? We start at the leaves. We repeatedly choose the two letters with the least weight ad joi them to form a ode of the tree. I the algorithm we use the otatio a i a j, which deotes cocateatio ad is simply a way to ivet ames for iteral odes. 3 Dyamic programmig 3.1 Biomial coefficiets ad Pascals triagle is proouced choose ad is the umber of -elemet subsets from a ( -elemet set. The defitio of biomial coefficiets is (! = (!!

5 Greedy algorithms ad dyamic programmig 5 Algorithm 5: Huffma codig. Iput: Symbol/weight-paris {(a 1, f 1,..., (a, f }. Output: 0/1-brach mared tree such that i=1 f i τ(a i is miimized. Huffma({(a 1, f 1,..., (a, f } (1 S {(a 1, f 1, a 1,..., (a, f, a } (2 while S 2 (3 Fid (b i, f i, t i, (b j, f j, t j S with f i, f j as small as possible. (4 S S\{(b i, f i, t i, (b j, f j, t j } (5 S S {(a i a j, f i + f j, TreeNode((t i, t j } (6 retur S. We calculate these coefficiets ad mae a table If we study this table we ca see that the value i the th row ad the th colum is the sum of the two values i the row above, 1, i the th colum ad the colum the the left, 1. This will be eve clearer if the write the table as Pascal s triagle. Hece, we ca calculate the biomial coefficiets with this recursive formula. ( = Pascal s triagle: Memoizatio { 1 ( if = or = otherwise ( Dyamic programmig divides the problem ito smaller subproblems util we reach a simple case which ca easily be calculated. These subproblems are the combied to fid the solutio. Dyamic programmig with memoizatio wors i the same way but saves the results of the calculatio so that if we have to solve aother problem these results ca be retreived istead of recalculated. The differece ca be illustrated i these two example algorithms.

6 6 DD2458 Popup HT 2008 Algorithm 6: Biomial coefficiets with memoizatio. Iput: Itegers ad, 0. Output: ( Bi(, (1 if T [, ] = (2 if = = 0 (3 T [, ] = 1 (4 else (5 T [, ] Bi( 1, 1+Bi( 1, (6 retur T [, ] Algorithm 7: Biomial coefficiets without memoizatio. Iput: Itegers ad, 0. Output: ( Bi(, (1 for i = 0 to (2 T [i, 0] T [i, i] 1 (3 for i = 2 to (4 for j = 1 to mi{i 1}{} (5 T [i, j] T [i 1, j 1] + T [i 1, j] (6 retur T [, ] 3.2 Maximal partial sum Formally: Give a set of itegers x 1,..., x, fid max s f { f i=s x i} Algorithm 8: Fidig the maximal subarray. Iput: Itegers x 1 {,..., x. f } Output: max z f i=s x i. MaxSubarray(x 1,..., x (1 S[0] 1 (2 for i = 1 to (3 S[i] max{0, S[i 1]} + x i (4 retur max i {S[i]} 3.3 Logest icreasig subsequece Example: Curcuit plugs Two circuits has plugs each o rows mared 1, 2,..., i some order. We wat to coect the plugs parwise without ay coectios crossig each other. Formally: Give a -permutatio π, fid the size of the maximal subset S {1, 2,..., } such that π(i π(j for all i j i the set S.

7 Greedy algorithms ad dyamic programmig 7 Algorithm 9: Fidig the logest icreasig subsequece. Iput: Permutatio π. Output: The size of the maximal subset S {1, 2,..., } such that π(i < π(j for all i < j i S. LogIcSubseq(π (1 T [0] (2 for i = 1 to (3 T [i] (4 for i = 1 to (5 Fid j such that T [j 1] < π(i < π(j. (6 T [j] π(i (7 retur max{j : T [j] < }

Math 155 (Lecture 3)

Math 155 (Lecture 3) Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,