Homework 1 Solutions CSE 101 Summer 2017

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1 Homewok 1 Soutions CSE 101 Summe Waming Up 1.1 Pobem and Pobem Instance Find the smaest numbe in an aay of n integes a 1, a 2,..., a n. What is the input? What is the output? Is this a pobem o a pobem instance? If you beieve this is a pobem, give an instance of this pobem. Soution: So: Input: the aay a 1, a 2,..., a n. Output: the smaest numbe 1.2 Lowe Bound vs. Uppe Bound Anayse owe bound and uppe bound of an agoithm Pobem: Does thee exist an eement 0 in an intege aay a 1, a 2,..., a n. Agoithm: Sweep fom the eft to ight unti the end of the aay. If it sees 0, etun yes; othewise, etun no. Lowe bound: What is the best case fo this agoithm? How ong does the best case take? Uppe bound: What is the wost case fo this agoithm? How ong does the wost case take? So: Best case: a 1 is 0. Constant time. O(n). Wost case: no eement is 0 in the aay. O(n) Big O and its eatives Show that og(n!) = Θ(n og n). Hint: Show the uppe bound by compaing with n n, owe bound by compaing with (( n 2 ) n 2 ). 1.3 Effective vs. Efficient Ae seection sot, bubbe sot and insetion sot effective agoithms fo Minsot pobem? In tems of time, is any agoithm above significanty moe efficient to sot a vey age unsoted ist than the othes? Besides time, what ese can you impove to make the agoithm moe efficient? Give an exampe. So: Yes, they ae effective because they ae coect and teminite. No, they a take O(n 2 ) time. Efficiency = Output / Input Impove Efficiency: 1. Output + / Same Input, e.g. soting moe numbes in the same time (if this is the Input measue) o in the same space o with the same enegy consumption 2. Output the same / Input -, e.g. sot the same numbe of integes in ess time with ess memoy equiements, with ess enegy consumption Time is not the ony measue,eg. Space and Enegy efficient agoithms. 1

2 2 Anaysing Agoithms 2.1 Iteative agoithm Hee is an agoithm that, given two soted ists A[1] < A[2] <...A[n] and B[1] < B[2] <.. < B[n], decides whethe thee is a pai of indices i, j with 1 i n and 1 j n so that A[i] = B[j]. Intesect(A[1..n], B[1..n]: soted ist of integes) 1. I 1, J 1, F ound F ase. 2. Whie I n and J n and F ound = F ase do: 3. IF A[I] = B[J] THEN F ound T ue. 4. IF A[I] > B[J] THEN J + + ELSE I Retun F ound. a) Suppose A is {1,4,6,8} and B is {2,3,5,6}, show that how the agoithm woks with this pobem instance. b) Pove that this agoithm is coect, i.e., that it etuns tue if and ony if such a pai i, j exists. c) Give a time anaysis, up to ode, fo this agoithm. Be sue to expain you answe. So: b) We need to pove two things: Fist, that if the agoithm etuns Tue, the ists intesect, both containing some vaue X; and second, that if the ists intesect, the agoithm etuns Tue. The fist is simpe. The ony ine whee Found gets set to Tue is ine 3, and it ony gets set to tue if A[I] = B[J]. Setting X = A[I], X is in both ists, and so the ists intesect. The othe diection is a bit moe difficut. Assume X is in both ists, X = A[i] and X = B[j] fo some 1 i, j n. We show as a oop invaiant that, if the agoithm has not etuned Tue aeady, that I i and J j. This is tue at the stat of the agoithm, since I = 1 i and J = 1 j. Assume at the stat of an iteation I i and J j. If I = i and J = j, A[I] = A[i] = X = B[j] = B[J]. Thus, the If in ine 3 is tue, and the agoithm hats etuning Tue. If I = i and J < j, then B[J] < B[j] = X = A[I] and the agoithm incements J, not changing I. Since the new vaue of J is at most j, we sti have I i and J j. The case I < i and J = j is identica. Finay, if I < i and J < j, eithe one we incement, we wi sti have I i and J j. Thus, by induction on the numbe of iteations, eithe we have etuned Tue o I i and J j. Now, afte 2n iteations, the agoithm must teminate, since each iteation that we don t teminate, we incement I o J and we stop when eithe eaches n + 1. When this happens, by the oop invaiant I i n and J j n, so we cannot teminate because eithe I o J exceed n. Thus, we must teminate because F ound is T ue, at which point we etun T ue. Atenative fomat: Say X = A[i] = B[j]. Assume we didn t etun Tue, to get a contadiction. At the end I o J must exceed n, and since we ony incement them by at most one each iteation, they must take on a possibe vaues in between. So thee must be a point whee I = i o a point whee J = j. Conside the fist time when eithe event happens. Without oss of geneaity, assume its when I = i (since the J = j case is symmetica). When this happens, J j, since eithe it happens in the beginning when J = 1 j, o J hasn t yet eached j since the I = i case happened fist. Then, whie J < j, B[J] < B[j] = X = A[i] = A[I], so we wi incement J and not change I unti J = j. At this point, B[J] = B[j] = X = A[i] = A[I], so the agoithm wi set Found to Tue and teminate. This contadicts ou assumption that the agoithm did not etun Tue. c) Time anaysis: Each opeation within the whie oop takes constant time, as does each opeation befoe it. As mentioned above, the whie oop can execute at most a tota of 2n times befoe the agoithm teminates. Thus, T (n) O(n). If the ists do not intesect, at east one counte needs to get incemented n times, so the wost-case time is aso O(n). 2

3 2.2 Recusive Mege RMege(A[1..k], B[1..]: soted ist of integes) 1. IF k = 0 etun B[1,..., ] 2. IF = 0 etun A[1,..., k] 3. IF A[1] B[1] THEN etun A[1] RMege(A[2,..., k], B[1,..., ]) 4. ELSE etun B[1] RMege(A[1,..., k], B[2,..., ]) a) Suppose A is {1,4,6} and B is {2,3,5}, show that how the agoithm woks with this pobem instance. b) Pove RMege(A[1,..., k], B[1,..., ]) is a soted aay containing a eements fom both soted aay A and B by induction on n = k +. c) Give a time anaysis fo this agoithm. So: b) Pove by induction on n, the tota input size. (1) Base case: When n=0. Then both ists ae empty. RMege etuns an empty ist, which is coect. (2) Induction: Assume n >= 1 and RMege(A[1,..., k], B[1,..., ]) etuns a soted ist containing a eements fom eithe ist wheneve k + = n 1. We want to pove:rm ege(a[1,..., k], B[1,..., ]) etuns a soted ist containing a eements fom eithe ist wheneve k + = n. Case 1: One of the ists is empty. The fist o second ine of RMege etuns the othe ist, which is coect. Case 2: Both ists ae nonempty. If A[1] B[1], then A[1] is the smaest eement among a the eements in both ists, because both A, B ists ae soted. Accoding to the thid ine RMege etuns A[1] RMege(A[2,..., k], B[1,..., ]). Obviousy, A[2,..., k], B[1,..., ] have ony (k 1) + = n 1 eements, so by assumption, RMege(A[2,..., k], B[1,..., ]) etuns a soted ist with eements fom A[2,..., k], B[1,..., ]. So RMege(A[1,..., k], B[1,..., ]) woks when k + = n if A[1] B[1]. Simiay, RMege(A[1,..., k], B[1,..., ]) aso woks when k + = n if A[1] B[1]. Concusion, RMege(A[1,..., k], B[1,..., ]) woks when k + = n. c) Time anaysis: Accoding to the ecuence eation, T (n) = T (n 1) + c. Soving it we have T (n) O(n). 2.3 MegeSot MegeSot(A[1,..., n]) 1. IF n = 1 Retun A 2. B[1,..., n/2] MegeSot(A[1,..., n/2]) 3. C[1,..., n/2] MegeSot(A[n/2 + 1,..., n]) 4. Retun Mege(B[1,..., n/2], C[1,..., n/2]) a) Suppose A is {4,3,2,1},show that how the agoithm woks with this pobem instance. b) Pove MegeSot(A[1,..., n]) etun a soted aay with exacty the eements in A[1,..., n]. c) Give a time anaysis fo this agoithm. So: b) (1) Base case: When n=1, the ist has ony one eement, it is soted aeady, the MegeSot woks; (2) Induction: Assume that MegeSot sots n = 1, 2, 3,..., k eements. We want to show MegeSot sots n = k +1 eements. This is tue because MegeSot(A[1,..., (k +1)/2]) etuns the soted fist n/2 eements, and MegeSot(A[(k + 1)/2 + 1,..., (k + 1)) etuns the soted est haf 3

4 eements. Finay M ege etuns a soted ist with eements exacty in eithe aay. So M egesot wi sot n = k + 1 eements. c) Time anaysis: Accoding to the ecuence eation, we have By soving it we have. T (n) = 2T ( n 2 ) + co(n). T (n) O(nogn) Note: Maste theoem is used to sove ecuence eation that takes the fom T (n) = at ( n b ) + f(n). See moe about Maste theoem in wiki o discussion section. 4

5 3 Two pointes Pobem: You ae given a sequence of n positive integes - a 1, a 2,..., a n. Find a continuous subsequence a, a +1,..., a, such that the sum of a numbes in it is ess than K and its ength is as age as possibe. 1. Deveop an agoithm that soves this pobem. 2. Pove its coectness. 3. Show (pove) its time compexity. Soution: 1. Agoithm: We wi sove this pobem with a popua technique caed two pointes. Let s have two vaiabes = 0 and = 0. Let s scan ou sequence fom eft to ight. Vaiabes and wi denote the beginning and the end of the cuent subsequence espectivey and et S(, ) be its sum. At each step, if S(, ) is ess than K, we have found the subsequence of the ength + 1. Othewise, whie S(, ) K and, we wi incease the vaue of vaiabe. At the end of each step, we incease the vaue of. Duing the pocess, we aso have vaiabe ans = 0 which we wi update evey time we find a new vaid sequence: ans = max(ans, + 1). The agoithm stops when = n. 2. Coectness: Let s pove that the agoithm finds the agest possibe subsequence. Let a, a +1,..., a be the optima answe fo the pobem. At some time of ou agoithm, wi be moe than o equa to. We wi have and and. At any time whie, wi not exceed. Indeed, if at some point wee bigge than, it woud mean that thee was, such that S(, ) K. Howeve, this is not possibe as a numbes ae positive. Theefoe, S(, ) S(, ) < K. Consequenty, thee wi be a moment duing ou agoithm, when and =. As sequence a, a +1,..., a is optima, it means that =. So ou agoithm finds the optima answe fo this pobem. 3. Time compexity: Time compexity of the agoithm is θ(n). Indeed, at each step, we incease eithe (in a oop) o. Both and can not exceed n. Each step woks in θ(1) time. Thee ae no moe than 2n steps. Theefoe, the compexity is θ(n). 5

6 4 Counting invesions Pobem: You ae given an aay of n eements. Find the numbe of invesions in this aay. The invesion is a pai (a i, a k ), such that i < k and a i > a k. 1. Deveop an agoithm that soves this pobem. 2. Pove its coectness. 3. Show (pove) its time compexity. Soution: 1. Agoithm: Thee is a simpe bute foce soution, whee we just ook at a possibe pais. 2. Coectness: As we enumeate a possibe pais, we wi be abe to find a invesions. 3. Time compexity: As the numbe of pais equas n(n 1) 2, the time compexity wi be θ(n 2 ). 1. Agoithm: Anothe appoach is to use divide-and-conque. We wi beak ou aay into two pats, such that thei sizes do not diffe by moe than one: B = A[0,..., [ n 2 ] 1] and C = A[[ n 2 ],..., n 1]. We wi ecusivey find the numbe of invesions in aays B and C and then find a invesions that have one eement in B and the othe one in C (the ecusion stops if thee is ony one eement in the aay; the esut wi be zeo fo this case). Fo this pupose, et s assume that eements in B and C ae soted. We wi scan both aays fom eft to ight. Initiay, i = k = 0. Then, if we compae eements B i and C k : (a) if B i C k, then B i doesn t fom invesions with any eements C j, such that k j. Incease i by one. (b) if B i > C k, then C k fom invesions with a eements B j, such that i j. Incease k by one. 2. Coectness: Let s pove the coectness by the induction: The base case is when thee is ony one eement in the aay. The esut is zeo, which is tue. Hypothesis: Let s assume that the agoithm finds coecty the numbe of invesions fo a aays of the size ess than n. Induction step: Evey time we have B i > C k, B i wi be the fist eement in B that satisfies this condition (the smaest eement in B that is bigge than C k ). So, if we take any pai (B j, C k ), such that B j > C k, then i j and B i B j. So fo evey C k, we find a possibe numbes in B, with which C k foms invesions. Accoding to the hypothesis, we wi coecty find a invesions in aays B and C, as thei sizes ae ess than n (max([ n 2 ], n [ n 2 ]) < n fo n > 1). Then, if we add a these numbes, we wi get the tota numbe of invesions in the aay. 3. Time compexity: As we can see, at each step we need both aays B and C to be soted. Fo this pupose we can use a mege sot agoithm. As you can see, ou method is just a sight modification of the mege sot, which aso aows us to find the numbe of a the invesions. By adding a few opeations that wok in constant time to the meging pocess of the two aays in the mege sot, we can sove ou pobem. So the time compexity of this appoach equas the time compexity of the mege sot, i.e. θ(n og n). 6

7 5 Quicksot Pobem: Quicksot is a vey famous and simpe soting agoithm that is used in many appications. The pobem is the foowing we have an aay of n eements: A 0, A 1,..., A n 1 that we need to eaange in a way that it becomes soted in an inceasing ode, i.e. A 0 A 1... A n 1. To accompish this, we wi do the foowing: et s choose a andom eement fom the aay. We wi ca this eement a pivot. Now, et s beak, o patition, the emainde of the aay into two goups (maybe empty): the fist pat contains a eements that ae ess than o equa to the pivot, the second pat contains a eements that ae sticty age than the pivot. We wi ca this pocess patitioning. Let s ook at the exampe: The aay contains 10 eements. Let s choose a andom eement. Let it be the thid eement of the aay. Its vaue equas 5. Now et s beak the aay into two pats as was descibed above: Red-coo eements beong to the fist goup, whie bue-coo ones beong to the second goup. The pivot eement s coo is geen. The eements in both goups can be aanged in any way inside thei goup. As ong as the conditions that a eements in the eft goup ae ess than o equa to the pivot and that a eements in the goup ae sticty age that the pivot ae satisfied, any patition wi be vaid. Fo exampe, anothe vaid patition is: As we can see, if a the eements in both goups wee soted in an inceasing ode, the esuting aay afte the patitioning woud be soted too. We wi use this obsevation to sot the aay. We wi ca the soting method quicksot. Fist, we choose a andom eement in the aay and pefom pationing. Then, we appy the quicksot ecusivey to each of the esuting goups. The ecusion stops when thee is ony one eement in the aay. Appaenty, as a esut, we wi get a soted aay. 1. Deveop an agoithm that pefoms patitioning. Ty to make it as efficient as possibe. 2. Pove the coectness of the patitioning. 3. Show (pove) the patition agoithm s time compexity. 4. Pove the coectness of the quicksot. 5. Show (pove) the wost case unning time of the quicksot. 6. Show (pove) the aveage unning time of the quicksot. (the aveage unning time of an agoithm is the numbe of opeations aveaged ove a possibe inputs) Soution: 1. Agoithm: Let s descibe how patitioning woks. Let s ook at the same exampe with the same pivot eement:

8 Fist et s swap the fist eement of the aay and the pivot eement. We wi get: Now, et s have two pointes and, which wi initiay point to the beginning and to the end of the aay espectivey ( = 0, = n 1). And et s denote the pivot eement as p Fist, we ty to incease as much as possibe. We incement pointe, if A[] p Then, we ty to decease as much as we can. We decement pointe, if p < A[] Then, if <, we swap eements A[] and A[] If <, we aso epeat a these thee opeations again (inceasing, deceasing and swapping and again, if < ). As soon as >, we swap eements A[] and p and stop patitioning

9 2. Coectness (patitioning): Let s pove that patitioning is coect. Indeed, afte we incease and decease, a eements to the eft of A[] ae ess than o equa to p, and a eements to the ight of A[] ae sticty age than p. Afte this pocess, we get that A[] > p and A[] p. If <, we swap A[] and A[] and get that A[] p, A[] > p. If >, we swap A[] and p, which ae both ess than o equa to p and stop patitioning. As a esut, we get that a eements to the eft of p ae ess than o equa to p, and a the eements to the ight of p ae sticty age than p. 3. Time compexity (patitioning): The time compexity fo patitioning wi be θ(n). Indeed, at each step we incease o decease. Each step takes O(1) time. Both and can t be inceased/deceased moe than n times, so thee ae not moe than 2n steps. On the othe hand, the numbe of iteations wi be at east equa to n, as we continue patitioning ti > 4. Coectness (quicksot): Now, et s pove that the quicksot is coect. The base case is when we have ony one eement. One eement is an odeed aay, so the base case is coect. The hypothesis is that the quicksot sots coecty a aays of the size ess than n. Now, thee is the inductive step. Fo an aay of size n, we, fist, choose a pivot eement p. Then, we pefom patitioning bases on this eement. As a esut, the aay wi be divided into two goups: the eft goup in which a eements ae ess than o equa to p, and the ight goup in which a eements ae sticty age than p. The sizes of these goups ae ess than n (because they at east do not contain the pivot eement). Then, we use the quicksot on them ecusivey. Accoding to the hypothesis, both these goups of eements wi be soted coecty. As a esut, the initia aay wi consist of the eft pat which is soted coecty, the pivot eement and the ight pat which is aso soted coecty. Theefoe, the whoe aay wi be soted coecty. 5. Time compexity (quicksot): Let T (n) be the numbe of opeations the quicksot makes fo the aay of size n. Then, whee A[i] is a pivot eement. T (0) = θ(1) T (n) = T (i) + T (n i 1) + θ(n) Let s say, that at each iteation, the pivot eement is the smaest eement in the aay. Then, we have: Theefoe, T (n) Ω(n 2 ) T (n) = T (0) + T (n 1) + θ(n) T (n) T (n 1) + n 1 T (n) T (n 2) + n 2 + n 1... T (n) n 2 + n 1 T (n) (n 1)n 2 So in the wost case, the quicksot wi take quadatic time to sot the aay. But this case happens aey. Let s see how fast the quicksot woks on aveage (the expected time compexity). 6. Time compexity (quicksot): Let patitioning take cn opeations and T (n) be the aveage unning time. Then, we wi have the foowing ecuent equation: T (n) = 1 n 1 (T (i) + T (n i 1) + cn) n i=0 9

10 T (n) = 2 n 1 T (i) + cn n i=0 n 1 nt (n) = 2 T (i) + cn 2 i=0 n 2 (n 1)T (n 1) = 2 T (i) + c(n 1) 2 i=0 nt (n) (n 1)T (n 1) 2T (n 1) + 2cn n i=1 T (n) T (n 1) + 2c n + 1 n n + 1 T (i) i + 1 n i=1 ( T (i 1) i + 2c i + 1 ) T (n) n + 1 T n c( 1 i 1) i=1 T (n) 2c(n n 1) n + 1 T (n) c n n n T (n) θ(n og n) 10