CS:3330 (Prof. Pemmaraju ): Assignment #1 Solutions. (b) For n = 3, we will have 3 men and 3 women with preferences as follows: m 1 : w 3 > w 1 > w 2

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1 Shiyao Wag CS:3330 (Prof. Pemmaraju ): Assigmet #1 Solutios Problem 1 (a) Cosider iput with me m 1, m,..., m ad wome w 1, w,..., w with the followig prefereces: All me have the same prefereces for wome: w 1 > w > > w. All wome have the same prefereces of me: m > m 1 > > m 1. (b) For = 3, we will have 3 me ad 3 wome with prefereces as follows: m 1 : w 3 > w 1 > w m : w 1 > w 3 > w m 3 : w 3 > w 1 > w w 1 : m 1 > m > m 3 w : m 1 > m > m 3 w 3 : m > m 1 > m 3 (c) We ow show that with the above prefereces, there is a orderig of the free me that forces the Gale-Shapley algorithm to execute Ω( ) iteratios. Suppose that the me are all placed i a queue i the order m 1, m,..., m iitially. The algorithm picks me from the frot of the queue, whereas me who become free durig the course of the algorithm joi at the back of the queue. Usig this orderig o free me, we see that the algorithm executes as follows: (1) m 1 proposes to w 1 ad this is accepted. () m proposes to w 1 ad m 1 is kicked out ad m gets egaged to w 1. (3) m 3 proposes to w 1 ad m is kicked out ad m 3 gets egaged to w 1. (i) m i proposes to w 1 ad m i 1 is kicked out ad m i gets egaged to w 1. () m proposes to w 1 ad m 1 is kicked out ad m gets egaged to w 1. At the ed of this sequece of proposals, m is egaged to w 1 ad m 1, m, m 3,..., m 1 are all free. The a sequece of 1 proposals are made to w (by me i the order m 1, m,..., m 1 ) at the ed of which m 1 is egaged to w ad m 1, m, m 3,, m are free. Cotiuig this maer, we see that the total umber of proposals made by the algorithm are: + ( 1) + ( ) = ( + 1). This shows that the worst case umber of iteratios of the Gale-Shapley algorithm is Ω( ). Page 1 of 5

2 Shiyao Wag CS:3330 (Prof. Pemmaraju ): Assigmet #1 Solutios Problem 1 Problem (a) Here is a updated versio of the Gale-Shapley algorithm: Algorithm 1 Updated Gale-Shapley algorithm 1: Iitially all m M ad w W are free : while there is a ma m who is free ad has t proposed to every woma o his preferece list do 3: Choose such a ma m 4: Let w be the highest-raked woma i m s preferece list to whom m has ot yet proposed 5: if w is free ad m is o w s preferece list the 6: (m, w) become egaged 7: else if w is curretly egaged to m 8: if m is ot o w s preferece list or w prefers m to m the 9: m remais free ad (m, w ) remai egaged 10: else if m is o w s preferece list ad w prefers m to m 11: (m, w) become egaged ad m becomes free 1: ed if 13: ed if 14: ed while 15: Retur the set S of married pairs (b) The iput of this problem will be the set of preferece lists. Sice each ma/woma has at most t idividuals o his or her list, we ca claim that the largest iput size s is Θ(t). I the above algorithm, i the worst case, a ma will propose to all wome o his list. Thus the worst case ruig time is Θ(s). Sice the largest value of s is Θ(t), the worst case ruig time is Θ(t). Problem 3 Yes, we ca fid a set of preferece lists i which a switch that would improve the parter of a woma who switched her prefereces. Oe possible example is give below: The prefereces of me are as follows: m 1 : w 3 > w 1 > w m : w 1 > w 3 > w m 3 : w 3 > w 1 > w The prefereces of wome are as follows: w 1 : m 1 > m > m 3 w : m 1 > m > m 3 w 3 : m > m 1 > m 3 With this set up, Gale-Shapley algorithm will execute the followig steps: 1) m 1 proposes to w 3 ad this is accepted. ) m proposes to w 1 ad this is accepted. 3) m 3 proposes to w 3 ad gets rejected. 4) m 3 proposes to w 1 ad gets rejected. 5) m 3 proposes to w ad this is accepted. Thus the resultig marriage will be (m 1, w 3 ), (m, w 1 ) ad (m 3, w ). Now if w 3 lies ad claims her prefereces as m > m 3 > m 1 the the Gale-Shapley algorithm will execute the followig steps: 1) m 1 proposes to w 3 ad this is accepted. Problem 3 cotiued o ext page... Page of 5

3 Shiyao Wag CS:3330 (Prof. Pemmaraju ): Assigmet #1 Solutios Problem 3 (cotiued) ) m proposes to w 1 ad this is accepted. 3) m 3 proposes to w 3 ad m 1 is kicked out ad m 3 gets egaged with w 3. 4) m 1 proposes to w 1 ad m is kicked out ad m 1 gets egaged with w 1. 5) m proposes to w 3 ad m 3 is kicked out ad m gets egaged with w 3. 6) m 3 proposes to w ad this is accepted. Thus the Gale-Shapley algorithm will yield the followig marriage: (m 1, w 1 ), (m, w 3 ), (m 3, w ). The fial result is that w 3 eded up with her favorite ma m, where as whe she was truthful she eded up with a less preferred ma. Problem 4 1) Sice log =, log < log = for all 1. Thus we have: ) Sice (log ) 3 = O( 1/3 ), we have: 3) Sice 4/3 <, we have: 4) Sice 100 = O(log ), we have: log = O( (log ) 3 ). (log ) 3 = O( 4/3 ). 4/3 = O(100 ). 100 = O( log ). 5) Sice log log 3 < whe is sufficietly large, log = O(/ log 3 ). The we will have: log = O( 3 / log 3 ). 6) We kow that 3 / log 3 = O( 3 ) ad also that 3 = O( ). Thus we have: 3 / log 3 = O( ). Thus, the fial sequece of complexity from low to high is as below: log, (log ) 3, 4/3, 100, log, 3 / log 3, Problem 5 (a) The ier while-loop at each step will execute j times whe j 1 ad 0 times if j < 1, the ier loop will ot execute. The outer loop will execute times, ad after log times the ier loop will oly execute for O(1). Thus the total ruig time of the first log steps will be as below: log 1 log Problem 5 cotiued o ext page... Page 3 of 5

4 Shiyao Wag CS:3330 (Prof. Pemmaraju ): Assigmet #1 Solutios Problem 5 (cotiued) Accordig to the geomatric series, we have: log = = Θ() log The rest log steps will be liear time ad log = Θ(). Thus the total ruig time is Θ(). (b) The ier loop rus log i +1 times for each value of i. Thus the total umber of basic operatios are: log 1 + log + + log + = + Θ(log (1 3 ) = + Θ(log (!)). We ca chage the base for log fuctio usig the chage of base formula: l() = log / log e. The we ca apply Stirlig s Approximatio to the equatio above. Based o Stirlig s Approximatio, we have the fial ruig time as: + Θ(log (!)) = + Θ(l(!)) = Θ( l + + O(l )) = Θ( l ). (c) The ier loop rus /3 times always. If we cout the steps of the outer loop, the total steps will be: ( ). 3 Usig the arithmetic series formula, we have: 3 ( ) = ( + 1) = Θ( 3 ) 3 Problem 6 First we use merge sort to sort the list. This rus i Θ( log ) time. The resultig list is sorted, with idetical umbers buched together. The we go through the list i liear time, ad cout the umber of times each distict elemet appears i the list. This will take Θ() steps. Suppose that there are s distict elemets i the list with k 1, k,..., k s beig their frequecies i the list. The the total umber of pairs is: k1 + k + + ks. Note that we have computed k 1, k,..., k s durig the liear sca metioed above. The calculatio above also takes liear time sice s <. Thus the total ruig time of this algorithm is Θ( log ). Problem 7 (a) We ca see that the outer loop execute exactly times. The ier loop will execute at most times every time it is executed. Addig up items from i to j takes at most O() steps as well. Storig the result i B[i, j] takes oly costat time. Thus the ruig time of this algorithm is O( 3 ). (b) Now cosider values of i /4 ad j 3/4. The amout of work the algorithm does for these values of i ad j is a lower boud o the total amout of work doe by the algorithm. The variables i ad j take o /16 values together. Now ote that for each of these values, j i / ad hece the Problem 7 cotiued o ext page... Page 4 of 5

5 Shiyao Wag CS:3330 (Prof. Pemmaraju ): Assigmet #1 Solutios Problem 7 (cotiued) summatio will take at least / basic operatios. The total umber of basic operatios is 3 /3. This meas that the algorithm has ruig time bouded below by Ω( 3 ). (c) Cosider the followig algorithm: for i = 1,,..., B[i, i + 1] A[i] + A[i + 1] for size =, 3,..., 1 for i = 1,,..., size j i + size B[i, j] B[i, j 1] + A[j] This algorithm works sice the values B[i, j] were already computed i the previous iteratio of the for-loop. It first computes B[i, i + 1] for all i by summig A[i] with A[i + 1]. This requires O() steps. For each value of size =, 3,..., the algorithm computes B[i, j] for j = i + size by settig B[i, j] = B[i, j 1] + A[j]. For each size, the algorithm rus O() steps sice there are at most B[i, j] s of that size (i.e., such that j = i + size). There are also less tha values of size. Thus the algorithm rus i O( ) time. Page 5 of 5