CS 5150/6150: Assignment 1 Due: Sep 23, 2010

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1 CS 5150/6150: Assigmet 1 Due: Sep 23, 2010 Wei Liu September 24, 2010 Q1: (1) Usig master theorem: a = 7, b = 4, f() = O(). Because f() = log b a ε holds whe ε = log b a = log 4 7, we ca apply the first case of master theorem ad get T () = Θ( log b a ) = Θ( 1.4 ). (2)a = 2, b = 2, f() = O( log 2 ) = log 2. If we check the coditio f() = O(log b a log k ), we fid it holds whe k = 2. So T () = Θ( log b a log k+1 = Θ( log 3 ). (3) We ca make a guess that T () = O( log ), hece T (/ log ) = log log log = log log log log + = log log + = log log log + = O( log ) (4). T () = π T (1) = π 2 T (2) 2a = Θ(c). Q2: We ca do a merge sort usig devide ad coquer approach, which is O( log )-time, ad get a sorted array y first. The to fid two umber y i ad y j satisfyig y i + y j = x, we do searchig from both sides of the array y, that is, searchig begi with i = 1 ad j =, here is the array size. Durig the searchig, if y i + y j < x, we update i = i + 1, else if y i + y j > x, we update j = j 1, the algorithm will stop whe y i +y j = x. This searchig algorithm yields a O() time. So the total computatio complexity of the procedure is O( log ) + O(). Sice O( log ) is domiated, the algorithm 1

2 is O( log )-time. Q3: We eed to look at the poits where two lies meet, because these poits seperate visible segmet ad ivisible segmets. The algorithm takes a few steps: Fid all possible poits where two lies meet. There is at most ( ) 2 umber of poits, so this step takes O( 2 ). For each lie equatio, plug i all poits i previous step, to see which side the poits are, so we kow if each poits are uppermost. This is O( 2 ) poits cross with lies, so this step takes O( 3 ). Fid the lies that meet at these uppermost poits. These lies are what we eed. Q4: (a).oe disadvatage of usig a eve umber is there is o media for each subgroup, so we may eed to compute the mea ad use the mea istead. Aother disadvage lies i that if usig eve umber, the size of the subproblem will be bigger tha that of usig odd umber. For example,for eve umber 2k, the subproblem size is T ( 4k k) = T ( 3 ), while for 4k 4 odd umber 2k + 1, the subproblem size is T ( 4k+2 (k+1) ) = T ( 3k+1) = 4k+2 4k+2 T (( k + 1 k )). It is easy to see that T (( + 1)) is smaller tha T ( 3). 4k+2 2 4k (b). Usig 5 yields: T () 5 = O() + T (/5) + T (7/10) O() is the time to compute medias, T (/5) is the time to compute pivot, ad T (7/10) is the time for subproblem. While usig 3, we have: T () 3 = O() + T (/3) + T (2/3) O() time to compute medias, T (/3) time to compute pivot, T (2/3) is the time for subproblem. Sice T () 3 > T () 5, we use 5 istead of 3. (c). For usig a odd umber, eg, m = 2k +1, as we discussed before, the size of subproblem is T (( k + 1 )). If icrease the umber from 5 to 7, 9, 4k or bigger, we ca see that the size of subproblem icreases, whe m = 5, size of subproblem is T ( 7 ) = T (0.7), whe m = 7, size of subproblem is 10 2

3 T ( ) T (0.714), whe m = 9, size of subproblem is T ( ) T (0.722) So usig larger odd ca t give us better result. O the other had, as the icreasig of the odd umber, the time of computig medias will also icrease. let s take 5 ad 7 as examples: the time of computig medias are 5 log 5 = (log 5) ad 7 log 7 = (log 7), we observe a icrease of time 5 7 whe switch from 5 to 7. So i a sese, 5 if optimal. Q5: Here is a example that the algorithm gives wrog aswer. week 1 week 2 week 3 week 4 l h The algorithm will pick l 1 = 6 at week 1, because l 1 + l 2 > h 2.Ad actually choosig h 1 = 40 is a better solutio i this case. So the problem of the algorithm is that it does ot take h 1 ito accout, so eve if i the followig steps it chooses the optimal values, it ca t get a global maximum i the ed. (b). The algorithm below modified the first step of the origial oe, which make sure that h 1 is take ito accout: 3

4 StartIdex = 1; If h 2 l 1 + l 2 the Edif if h 1 > l 1 the Output Choose a high-stress job h 1 i week 1 Output Choose a low-stress job l 1 i week 1 Set StartIdex = StartIdex + 1; If h 1 > h 2 the Output Choose a high-stress job h 1 i week 1 Set StartIdex = StartIdex +1; Output Choose o job i week 1 Output Choose a high-stress job h 2 i week 2 Set StartIdex = StartIdex+2; Do the origial algorithm begi with i = StartIdex. Q6: Segmeted-Least_Squares() Array M[0...] Set M[0] = 0 For all pairs i<=j Compute the least squares errors e_i,e_j for the segmet p_i,...,p_j Edfor For j=1,2,..., Use the recurrece (6.7) to compute M[j] Edfor Retur M[] 4

5 Fid-Segmets(j) If j=0 the Output othig Fid a i that miimizeds e_ij + C + M[i-1] Output the segmet {p_i,...,p_j} ad the result of Fid-Segmets(i-1) Edif This homework is a joit effort with Bo wag. 5