1 Disjoint-set data structure.

Transcription

1 CS 124 Setion #4 Union-Fin, Greey Algorithms 2/20/17 1 Disjoint-set ata struture. 1.1 Operations Disjoint-set ata struture enale us to effiiently perform operations suh as plaing elements into sets, querying whether two elements are in the same set, an merging two sets together. (Thus they re very useful for simulating graph onnetivity.) Must implement the following operations: makeset(): reate a new set ontaining the single element. union(, y): replae sets ontaining an y y their union. fin(): return name of set ontaining. We a for onveniene the funtion link(, y) where, y are roots: link hanges the parent pointer of one of the roots to e the other root. In partiular, union(, y) = link(fin(), fin(y)), so the main prolem is to make the fin operations effiient. 1.2 Optimization Heuristis We have two main methos of optimization for isjoint-set ata strutures: Union y rank. When performing a union operation, we prefer to merge the shallower tree into the eeper tree. Path ompression. After performing a fin operation, we an simply attah all the noes touhe iretly onto the root of the tree. Eerise 1. Draw how the isjoint set ata struture hanges after eah of the following operations using a partiular heuristi: (a) union(, y) with union y rank. a y () fin() with path ompression. 1

2 a e f g a y a e f g Eerise 2. When using the union y rank optimization only, what is the asymptoti runtime of the operation fin()? When we use the union y rank heuristi, we have the property that the rank of a tree only inreases when we union together two trees of the same rank. (Reall that rank = height). If you union a tree with rank 3 with one of rank 4, you get a tree of rank 4. Only when you union two trees of rank 4 an you get a tree of rank 5. 2

3 With this property, we an show that a tree of rank h has at least 2 h noes. This an e prove using inution. Base Case: When the rank of a tree is 0, there is always 1 noe, whih is inee at least 2 0. Inutive Case: Suppose it is true that every tree of rank h has at least 2 h noes. If we ha a tree of rank h + 1, we know that it must have ome aout through the union of two trees of rank h + 1 at some point in the history of this tree. By the inutive hypothesis, those two trees of rank h must have ha 2 h noes eah, an thus the numer of noes in this tree of rank h + 1 must e at least 2 h + 2 h = 2 h+1. Therefore, we have that the run-time of fin() when using only the union y rank heuristi is O(log n) eause if there are n noes, then the maimum rank of any tree is log n. 2 Greey Algorithms A greey algorithm is an algorithm whih attempts to fin the gloally optimal solution to a prolem y making loally optimal eisions. Eamples from Leture: Kruskal s: We greeily hoose the lightest ege in the graph that oesn t form a yle. Prim s: We greeily hoose the lightest ege ajaent to the spanning tree we ve forme so far. Horn Formula: We start y assigning all variales to false an only set variales to true when an impliation fores you to. Huffman Enoing: We greeily onstrut the enoing y taking the two least use haraters an merging them into one new harater. Set Cover (O(k log n) approimation): We greeily hoose the set that overs the most numer of the remaining unovere elements at the given iteration. In orer to prove that a greey algorithm is orret, we nee to show that inee hoosing the loally optimal solution keeps us on trak to fining the gloally optimal solution. The ut property with MST s guarantees that the greey MST algorithms are orret. Prolems with orret greey algorithms are rare. We will soon e introue to Dynami Programming, a tehnique that lets you solve a muh wier range of optimization prolems. It is argualy the most useful tehnique you will learn in CS 124. Eerise 3. Suppose that we have a set S = {a 1,..., a n } of propose ativities. Eah ativity a i has a start time s i an a finish time f i. We an only run one ativity at a time. Your jo is to fin a maimal set of ompatile ativities. Whih of the following greey algorithms is orret? (a) Sort all the ativities y their uration an greeily piking the shortest ativity that oes not onflit with any of the alreay hosen ativities. () Pik the ativity that onflits with the fewer numer of remaining ativities. Remove the ativities that the hosen ativity onflits with. Break ties aritrarily. 3

4 () Sort all the ativities y their en time an greeily pik the ativity with the earliest en time that oes not onflit with any of the alreay hosen ativities. (a) Inorret. It s not always est to hoose the shortest length ativities Counter-eample: () Inorret. This one is triky, ut it s not always est to hoose the ativity with the least a mount of onflits. Counter-eample: () Corret. Proof: Let g 1 e the first ativity hosen y the greey solution an let t 1 e the first ativity y the optimal solution. We know that ase on our seletion tehnique, g 1 is has the earliest en-time out of all the ativities, an thus the en time of t 1 must e equal to or after that of g 1. A similar argument an e mae for t 2 an g 2. We know that t 2 oes not overlap with g 1 eause g 1 ene efore t 1 ene. Therefore t 2 must en at the same time or later than g 2. This argument an e mae inutively to show that the en time of g i is always earlier than the en time of t i. Why oes this imply that the greey algorithm is optimal? Imagine a situation where we have one more ativity in the optimal solution than in greey. Let t l e the last event hosen y the optimal solution, ut suppose that greey only goes up to g l 1. Well we know y the informal inution we i that the en time of g l 1 is no later than the en time of t l 1. This implies that t l oes not onflit with g l 1. So why i the greey algorithm not hoose t l after hoosing g l 1? Contraition. Eerise 4. Let s go ak to greey. from the first CS 50 prolem set. The question was to etermine the fewest numer of US oins neessary to make hange for a given amount of money. (a) Desrie a greey algorithm to make hange onsisting of quarters, imes, nikels, an pennies. Prove that your algorithm yiels an optimal solution. () Suppose that the availale oins are in the enominations that are powers of, i.e., the enominations are 0, 1,..., k for some integers > 1 an k 1. Show that the greey algorithm always yiels an optimal solution. () Give a set of oin enominations for whih the greey algorithm oes not yiel an optimal solution. Your set shoul inlue a penny so that there is a solution for every value of n. (a) Our greey algorithm will e to use the largest oin at eah iteration. Suppose we are making 4

5 hange for n ents. If n < 5, then greey will use n pennies whih is optimal. If 6 n < 10, then greey will use one nikle an then n 5 pennies, whih is also optimal. If 11 n < 25, then greey will use a ime first. Can the optimal solution not use any imes? Notie that the optimal solution will never use more than 4 pennies or 1 nikel (eause otherwise you oul just always replae some of them with a nikle or a ime). 4 pennies an a nikle is only 9 ents, so it is efinitely su-optimal to not use a ime first when n is in this range. If n 25, y a similar reasoning as efore, we will never use more than 2 imes eause if you use 3 imes, you might as well replae it with a nikle an a quarter. Therefore, if n > 29, then ertainly we will use a quarter first. You an easily hek that if n is etween 25 an 29, then taking the quarter an then n 25 pennies is optimal. () We will generalize the argument use in the eample aove. The important realization was that you will never use more than 1 opies of a partiular oin eause otherwise you shoul replae opies of that oin with 1 opy of the net higher oin. The eeption is of ourse for the largest oin. Therefore, for any enomination n, let m e the size of the largest enomination less than n. If I only use oins of enomination m 1 an less, while using only at most 1 opies of eah, the maimum amount of money I oul make hange for is: m 1 ( 1) i = ( 1) m 1 1 = m 1 i=0 Given our assumption that n m, we know that it is impossile to have an optimal solution that oesn t use the oin with value m. () Suppose we only ha pennies, imes an quarters. Then, the optimal solution for n = 31 woul e to take 3 imes an a penny (4 oins total), ut the greey algorithm woul take 1 quarter an 6 pennies (7 oins total). Note that in this ase, it was est to take a oin with smaller value first (the penny) rather than the one with the highest value (the quarter). 3 Divie an Conquer Divie an Conquer algorithms work y reursively reaking the prolem into smaller piees an solving the suprolems. You nee to think aout how to reursively reak own the prolem as well as how to omine the various piees together. Eamples from Leture: Mergesort an StoogeSort: We reursively sorte a fration of the list an then i some work to omine those frations of the list together. 5

6 Integer Multipliation: We split two integers in half eah, forming 4 piees an then performe only 3 multipliations on half-size numers. Strassen s Algorithm: We perform matri multipliation on two n n matries y performing 7 multipliations on n/2 n/2 matries. Eerise 5. Given an array of n elements, you want to etermine whether there eists a majority element (that is an element whih ours at least n+1 2 times) an if so, output this element. Show how to o this in time O(n log n) using Divie an Conquer. Split the list into 2 halves an reursively fin the majority element of eah half. If oth sies return "no majority", then return "no majority". This is true eause if there eists a majority element, then it also has to e the majority on at least one of the two. If oth sies return an element, hek to see if the elements are the same an if so return it. Otherwise, for eah element returne y the reursive all, o a linear time traversal through the entire list to see if it is the majority element. Return it if it is an "no majority" if you on t etet a majority element this way. It is not har to onvine yourself that this algorithm is orret, ut what is its run-time? We make 2 reursive alls on lists of size n/2 an then perform up to 2 linear traversals through the list. Thus: T (n) = 2T (n/2) + O(n) whih solves to O(n log n) just like mergesort. Eerise 6. (2015 Prolem Set 4) You are given an n-igit positive integer written in ase-2. Give an effiient algorithm to return its representation in ase-10 an analyze its running time. Assume you have lak-o aess to an integer multipliation algorithm whih an multiply two n-igit numers in time M(n) for some M(n) whih is Ω(n) an O(n 2 ). Consier the n-igit ase-2 integer as a string of ase-2 igits, [0...n]. Assume n is a power of 2. If not, we an pa the left sie of the input integer with zeros. This will inrease the numer of igits y less than a fator of 2, whih will not affet the ig-o runtime. We offer a reursive algorithm. In the ase ase, for an input of size 0, output 0. Divie the integer into two halves, [0... n 2 ] an [ n 2...n]. Then, reursively onvert eah half into a ase-10 integer. Call these (onverte) ase-10 integers y l an y r. Then, the esire integer an e foun y omputing y l 2 n/2 + y r. Note that, p = 2 n/2 must e in ase-10 also. One way we oul get the ase-10 representation of p is y onverting it reursively, too, along with the halves of the input integer. Then we make three reursive alls to our proeure, eah with a ase-2 integer whih is roughly half the size of the prolem input. After we otain these ase-10 integers, we have to perform an integer multipliation (y l p) an an integer aition (y l p) + (yr). The run-time is: T (n) = 3T (n/2) + M(n/2) + O(n) Suppose M(n) = O(n k ) for some 1 k 2. Then, M(n/2) + O(n) = O(M(n/2)) = O(n k ). Thus, we an apply the master theorem on what k is: 6

7 If k > log 2 3, then T (n) = O(n k ). If k = log 2 3, then T (n) = O(n k log n) an if k < log 2 3, then T (n) = O(n log 2 3 ). 7