COMP 251 Assignment 2 Solutions

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1 COMP 251 Assigmet 2 Solutios Questio 1 Exercise Treat the umbers as 2-digit umbers i radix. Each digit rages from 0 to 1. Sort these 2-digit umbers ith the RADIX-SORT algorithm preseted i Sectio 8.3. As the textboo shos, the RADIX-SORT algorithm rus i Odd time. I this case, d =2 ad =, so the total ruig time is O2 2 =O. Questio 2 Problem 8-4 (a) Compare each red jug ith each blue jug util the matchig oe is foud. I the orst case, e eed to compare the first red jug ith all blue jugs to fid its match. We the eed 1 comparisos to fid the match of the secod red jug (because e do't eed to loo at the blue jug that as already matched ith aother red jug). Similarly, e eed 2 comparisos for the third red jug. Ad so o. This gives us a orst-case ruig time of O 2. (b) To prove the algorithmic loer boud, e ill use the decisio tree cocept that as used i the miimum boud proof for compariso sortig. For this decisio tree, each ode represets the compariso of oe red jug ith a blue jug. There are three possible results to this compariso: the red jug is smaller, the blue jug is smaller, or both jugs are equal. So each compariso ode i the decisio tree has a mamum of three childre. We o eed to determie the umber of leaves for our decisio tree. Let b 1, b 2,...,b ad r 1, r 2,..., r deote the blue ad red jugs respectively. The solutio to our problem the is the set of pairs b i, r j here the b i ad r j jugs hold the same amout of ater. There are! such possible sets of pairs, so the decisio tree for our algorithm must have! leaves. Let's o loo at the miimum height for our tree. A tree of height h here each ode has at most three childre has a mamum of 3 h leaves. Therefore, e have: 3 h!/e h log 3 log 3 e log (c) Our radomized jug matchig algorithm is similar to RANDOMIZED-QUICKSORT. For this algorithm, e let R=r 1, r 2,..., r ad B=b 1, b 2,...,b be the sets of red ad blue jugs.

2 MATCH-JUGS(R,B) if( R = 0 ) retur if( R = 1 ) output (r 1, b 1 ) retur r = radomly chose jug from R compare every jug i B to r B < = all the jugs i B smaller tha r B > = all the jugs i B larger tha r b = jug that matches r compare every jug i R to b R < = all the jugs i R smaller tha b R > = all the jugs i R larger tha b output( r, b ) ed MATCH-JUGS(R <, B < ) MATCH_JUGS(R >, B > ) We ca covice ourselves of the correctess of this algorithm by remarig that the matchig jugs are alays grouped together i the B < ad R < groups, B > ad R > groups, or directly matched as r ad b. Also, each recursive call to this algorithm is made ith smaller sets of jugs, so the algorithm is guarateed to termiate. Let's o aalyze the ruig time for this algorithm. The o-recursive part of the algorithm compares all blue jugs ith r ad all red jugs ith b, so this rus i O=O. I the orst case, the radomly piced jug is alays the smallest or the largest jug. The oe of the recursive calls to MATCH-JUGS is made ith a set of 1 jugs. So the total orst case ruig time for the algorithm is: O O 1...O1=O 2. To sho that the expected ruig time for this algorithm, simply follo the procedure used to aalyze the expected ruig time for the RANDOMIZED-QUICKSORT algorithm. The same logical progressio foud i Sectio gives a expected ruig time of O log for the MATCH-JUGS algorithm. Questio 3 Problem 8-6 (a) Give 2 umbers, e ca choose of those umbers to put i the first list i sorted order. The remaiig umbers are put i the secod list i sorted order. Thus, there are 2 ays to divide 2 umbers ito 2 sorted list of umbers. (b) For this part, e eed a better appromatio of! tha /e. Goig bac to chapter 2, e fid the folloig equatio for! :!= e 2 c, here 1c e 1 12 From part (a), e o that the decisio tree must have at least 2 leaves, so the height of the tree is bouded belo by:

3 h log 2 =log 2! e e h log 2 h log h 2 log c 2 =log2! 2 log!!! 22 c 2 2 c 2 1 c 2 2 c =log22 log 1 c 2 2 c c 2 We ca easily verify that log c 2 c 2 gros sloer tha, so h 2 o. (c) Cosider to elemets a ad b from the differet sorted lists that are adjacet i the fial sorted list. The the fial list could be either: c 1, c 2...c i, a,b,c i1,...c 2 or c 1, c 2,...c i,b, a,c i1,c 2. Ca e determie hich of the to optios is the case here? If e compare a or b ith c j i c 1,... c i, e'll get that ac j ad bc j. Similarly, if e compare a or b ith c i c i1,..., c 2, e'll get that both ac ad bc. Therefore, the oly ay to o hich of a or b is smallest is to compare the to elemets to each other. (d) Let the to origial lists be A=a 1, a 2,..., a ad B=b 1,b 2,...,b. Cosider the fial sorted list S=a 1,b 1, a 2,b 2,..., a,b. Each a i elemet is adjacet to the umber b i from the other list ad each a i elemet ith i1 is adjacet to the b i 1 umber from the other list. So e have a total of 2 1 adjacet umbers from differet lists. From our part (c) e eed at least oe compariso for each of these cases so e eed a miimum of 2 1 total comparisos to merge the to lists. Questio 4 Exercise Here is the outlie of a algorithm to fid the elemets of S closest to the media: 1. Fid the media of S usig the SELECT algorithm. If is eve, fid both the upper ad loer medias of S ad tae the average of the to values. The SELECT algorithm from the textboo taes liear time, so this step rus i time. 2. Create a e set D that cotais the differece betee each elemet of S ad the media. Formally, set D i = S i M for 1 i. To do this, e have oe calculatio to mae per elemet i S, so this step taes time as ell. 3. Fid the (+1)-th value of D. Call this value d. The values of S closest to the media ill all have values less tha d aay from the media. If e use the same SELECT algorithm that e used i the first step, this also taes time. 4. Go through S, ad tae all the values of S here S i M d. This is a simple sca of the array ad so it taes time.

4 We o eed to justify that this algorithm taes time. This is simple: e have sho above ho every step rus i time, so the total ruig time for the algorithm is also. Questio 5 Problem 9-2 (a) The media value is x /2. Therefore, e have that: x i x /2 =/2 1 1 = ad x i x /2 =/ So x /2 is the eighted media of the x i 's he =1/ for i=1,2,...,. (b) This ca be doe ith this 2-step algorithm: 1. Sort the elemets by x value usig merge sort. This taes O log time. 2. Start at the begiig of the sorted array. Keepig trac of the total eight of the elemets from x 1 to the curret elemet x i, go through the array util you get to the i value here the sum of the eights is greater tha 1/2. That fial x i value is the eighted media. Sice e oly loo at each elemet oce, this taes O time i the orst-case. The total ruig time of this algorithm is O log O =O log time. Note that the algorithm ca also be ritte i pseudo-code as follos: FidWeightedMedia(X) MergeSort(X) i = 1 eightsum = 1 hile( eightsum < 1/2 ) i = i + 1 eightsum = eightsum + retur x i ed (c) The desired algorithm is very similar to the SELECT algorithm from the textboo: 1. Divide the elemets of the iput array ito groups of 5 elemets each. 2. Fid the eighted media of each of the /5 groups by isertio sortig the elemets of each group ad taig the media. 3. Use SELECT to fid the media of the /5 medias foud i the previous step. 4. Partitio the array aroud the media foud i step 3. Keep trac of the total eight of each partitio. Let l be the eight of the lo side partitio ad h =1 l be the eight of the high side partitio. 5. If l is less tha, call WEIGHT-SELECT recursively o the high partitio ith ' = l. Otherise, call WEIGHT-SELECT recursively o the lo side partitio ith ' =. The first 4 steps tae the same amout of time as the SELECT algorithm, so they ru i time. Sice our recursive calls are called o at most /2 elemets, the total ruig time for this algorithm is T =T /2. By the Master Theorem, the algorithm rus i time.

5 (d) We ill prove the result by cotradictio. To do this, e eed a fe defiitios. Let x be the eighted media of the poits. Let p be the best solutio for the 1-dimesioal post-office locatio problem. Ad let f c= i=1 c x i. The by defiitio f p is the miimum value of f. For our cotradictio, assume o that p x. We ill first cosider the case he px. Sice p is the best solutio to the problem, f p f x. No: f p f x = i=1 p x i i=1 f p f x = i=1 p x i x x i x x i There are three cases to cosider here: 1. Whe x i px, p x i x x i = p x i x x i = p x 2. Whe p x, p x i x x i = x i p x x i =2 x i p x. Sice x i p, 2 x i p2 p p= p so p x i x x i p x. 3. Whe px x i, p x i x x i = x i p x i x =x p Isertig these cases ito our equatio above gives us: f p f x x p x x x p f p f x p x x x By the defiitio of the eighted media, 1/2 so 0. Also, by our assumptio px so p x 0. Therefore, f p f x 0 hich meas that f x f p. This cotradicts the fact that p is the best solutio. I the same maer, e ca sho that px leads to the same cotradictio. Therefore, p x ca ot be the best solutio to the 1-dimesioal post-office problem ad so e have sho that the eighted mea x is ideed the best solutio. (e) With the Mahatta distace, the problem ca be split up ito idepedet problems i the x ad y directios: f p= i=1 d p,c= i=1 x p x c y p y c f p= i=1 x p x c i=1 y p y c Therefore, the x coordiate of the best solutio is the solutio to the 1-dimesioal post-office problem ith the x coordiates of the poits, ad the y coordiate of the best solutio is foud by determiig the best solutio to the 1-dimesioal post-office problem ith the y coordiates of the poits. I part (d), e have see that the solutio to the 1-dimesioal post-office problem is the eighted mea of the poits. So the best solutio to the Mahatta distace post-office problem is the poit x, y, here x is the eighted mea of the x coordiates of the poits ad y is the eighted mea of the y coordiates of the poits.

CS103X: Discrete Structures Homework 4 Solutions

CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible six-figure salaries i whole dollar amouts are there that cotai at least