Department of Informatics Prof. Dr. Michael Böhlen Binzmühlestrasse Zurich Phone:

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1 Departmet of Iformatics Prof. Dr. Michael Böhle Bizmühlestrasse Zurich Phoe: Iformatik II Midterm1 Sprig Advice You have 90 miutes to complete the exam of Iformatik II. The followig rules apply: Aswer the questios o the exam sheets or the backside. Mark clearly which aswer belogs to which questio. Additioal sheets are provided upo request. If you use additioal sheets, put your ame ad matriculatio umber o each of them. Check the completeess of your exam (13 umbered pages). Use a pe i blue or black colour for your solutios. Pecils ad pes i other colours are ot allowed. Solutios writte i pecil will ot be corrected. Stick to the termiology ad otatios used i the lectures. Oly the followig materials are allowed for the exam: Oe A4 sheet (-sided) with your persoal hadwritte otes, writte by yourself. Sheets that do ot coform to this specificatio will be collected. A foreig laguage dictioary is allowed. The dictioary will be checked by a supervisor. No additioal items are allowed except calculators. Computers, pdas, smart-phoes, audio-devices or similar devices may ot be used. Ay cheatig attempt will result i a failed test (meaig 0 poits). Put your studet legitimatio card ( Legi ) o the desk. Sigature: Correctio slot Please do ot fill out the part below Exercise Total Poits Achieved Maximum Poits

2 Exercise 1 15 Poits Arrays ad Sortig 1.1 [8 poits] Cosider a array of itegers A[0..-1], which cotais all itegers betwee 1 ad +1 except oe. 1. Assume the array is sorted. Fid ad describe a algorithm with subliear (i.e., less tha liear) asymptotic complexity that determies the missig umber. Determie the asymptotic worst-case complexity of your algorithm.. Assume the array is ot sorted. Fid ad describe a algorithm with liear asymptotic complexity that determies the missig umber. Hit: You ca use a auxiliary array. You do ot eed to implemet the algorithm. 1. We use biary search for a sorted array: We look for the smallest idex i for which A[i] = i + ; this is the positio of our missig umber. We fid the mid elemet, mid =. If A[mid] = mid+1 we recurse o the secod half of the array A[mid+1..high], otherwise we recurse o first half of the array A[low..mid], The algorithm termiates whe the search rage cosists of oly oe elemet (i.e. A[low] == A[high]) ad A[i] = i +. The missig umber is i + 1. We get the recurrece T () = T ( ) + Θ(1), which gives a worst-case ruig time of Θ(log ). We use the auxiliary array to keep track of the umbers we have see: A[i] = 0 if we have ot see i; A[i] = 1 if we have see i. Create auxiliary array B with idices 1 to + 1 Iitialize all locatios to 0 Go through all elemets of A ad set B[A[i]] = 1 Go through all elemets of B ad retur the i for which B[i] = 0. This is the missig umber.

3 1. [7 poits] Cosider a array of itegers A[0..-1]. The array is sorted i ascedig order ad there are o duplicate elemets. Assume the array has bee rotated i.e., the elemets have bee shifted multiple times to the right with the last elemet beig moved to the first positio i each step. For istace, A = [10, 0, 5, 40, 5] becomes A = [5, 40, 5, 10, 0] after its elemets were shifted to the right three times. Implemet a algorithm that fids the positio of a elemet i a rotated array. The asymptotic ruig time complexity of your algorithm must be O(log ). Use either C or pseudocode for your solutio. it search(it arr[], it low, it high, it key) { if (low > high) retur 1; it mid = (low + high) / ; if (arr[mid] == key) retur mid; if (arr[low] arr[mid]) { if (key arr[low] && key arr[mid]) { retur search(arr, low, mid 1, key); } retur search(arr, mid+1, high, key); } if (key arr[mid] && key arr[high]) { retur search(arr, mid+1, high, key); } retur search(arr, low, mid 1, key); 3

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5 Exercise 18 Poits Asymptotic Complexity.1 [6 poits] Calculate the simplest possible asymptotic tight boud for the followig fuctios. Iclude ay calculatio steps i your solutio: a) f 1 () = log( log()) f 1 () = log( log()) = log() + log log() Θ(log()) b) f () = log( i=1 i) f () = log( (+1) i=1 i) = log( ) = log( ( + 1)) log() = log() + log( + 1) log() Θ(log ) c) f 3 () = 16 log f 3 () = 16 log = 3 log = ( log ) 3 = 3 Θ( 3 ) d) f 4 () = log + log f 4 () = log + log = 1 log + log = + log Θ( log ) e) f 5 () = log 9 f 5 () = log 9 = log 9 Θ() f) f 6 () = 3 log f 6 () = 3 log = 7 log Θ(16 ) 5

6 . [6 poits] Calculate the asymptotic tight boud of the followig recurreces. If the Master Theorem ca be used, write dow a, b, f() ad the case (1-3). a) T () = 3T ( ) + a = 3, b =, f() =, log ba = log 3 = 1.58 f() = Ω( log ba+ɛ ), ɛ > 0 Case 3: Θ( ) Regularity coditio: if c = 3 4, a f(/b) = 3 ( ) 3 4 T () = Θ( ) b) T () = 4T ( 9 ) + 7 a = 4, b = 9, f() = 7 = 7 0.5, logab = log 49 = 0.63 f() = O( logab ɛ ), ɛ = 0.13 > 0 Case 1: Θ( log 9 4 ) c) T () = 9T ( ) a = 9, b = 4, f() = 3, logba = log94 = 0.63 f() = Ω( logba+ɛ ), ɛ > 0 Case 3: Θ( 3 ) Regularity coditio: if c = 9, a f(/b) = 9 ( 64 4 )3 9 T () = Θ( 3 ) 64 3 d) T () = log + T ( ) T () = log + T ( ) = log + log + T ( ) = log + log + log + T ( ) +... = log + 1 log log + 1 log = ( ) log 4 = + i=0 ( 1 )i log = log = Θ(log ) 6

7 .3 [6 poits] Cosider the recurrece: { 1 if = 1 T () = T (/3) + T (/9) + if > 1 a) [3 poits] Draw a recursio tree ad use it to estimate the asymptotic upper boud of T (). Iclude the tree-based calculatios that led to your estimate. = h = log = 4 9 = = ( 4 9 )h h h=0 ( 4 9 )h Logest brach o the left. Tree grows util ( 1 h 3) = 1 = h = log3 To get a upper boud, we ca use the sum h h=0 ( 4 9 )h. Guess: O() b) [3 poits] Prove the correctess of your estimate usig the substitutio method. Proof by iductio: T () = T (/3) + T (/9) + c 3 + c 9 + (recurrece) (iductive hyp.) = ( 4c + 9 ) (rewritig) 9 = (c 5c ) (rewritig) 9 = c ( 5c 9 ) (rewritig) 9 c (for c 9 5 ) 7

8 Exercise 3 13 Poits Rutime Aalysis Algorithm: whatdoesitdo(a, ) 1 tmp = 0; do 3 ready = 0; 4 for i = -1 to 1 do 5 if A[i-1]>A[i] the 6 tmp = A[i-1]; 7 A[i-1] = A[i]; 8 A[i] = tmp; 9 ready =1; 10 for i = 1 to -1 do 11 if A[i-1]>A[i] the 1 tmp = A[i-1]; 13 A[i-1] = A[i]; 14 A[i] = tmp; 15 ready =1; 16 while ready=1 ; 8

9 3.1 [3 poits] The algorithm whatdoesitdo(a,) gets a array A[0...-1] of itegers as a iput. Apply the algorithm o the array A=[8, 5, 10, 1, 3] ad complete each lie of the followig matrix if the cotet of A is modified. The already-completed lie of the matrix shows the iitial cotet of A. i A[0] A[1] A[] A[3] A[4] [4 poits] Describe i words what the algorithm whatdoesitdo(a,) does (maximum three lies). What is the best ad worst case for this algorithm, ad what are the asymptotic complexities i these cases. The algorithm sorts the array A, it is a variatio of bubble sort, this variatio sorts A i both directios o each pass through the list. Best case: O(), array already sorted i ascedig order. Worst case: O( ), array sorted i descedig order. 9

10 3.3 [6 poits] Aalyze the steps of the algorithm whatdoesitdo(a,) ad calculate its ruig time (worst case). lie 1 c 1 1 lie c 1 lie 3 c 3 lie 4 c 4 ( 1) lie 5-9 c 5 ( ) lie 10 c 6 ( 1) lie c 7 ( 1) lie 16 c 8 Ruig time: c 1 + c + c 3 + c 4 ( 1) + 4 c 5 ( ) + c 6 ( 1) + 4 c 7 ( 1) + c 8 10

11 Exercise 4 14 Poits Divide ad Coquer The maximum-subarray algorithm fids the cotiguous subarray that has the largest sum withi a usorted array A[0...-1] of itegers. For example, for array A = [-, -5, 6, -, -3, 1, 5], the maximum subarray is [6, -, -3, 1, 5]. The algorithm works as follows: Firstly, it divides the iput array ito two equal partitios: I (A[0]...A[mid]) ad II (A[mid+1]...A[-1]). Afterwards, it calls itself recursively o both partitios to fid the maximum subarray of each partitio. The combiatio step decides the maximum-subarray by comparig three arrays: the maximum-subarray from the left part, the maximum-subarray from the right part, ad the maximum-subarray that overlaps the middle. The maximum-subarray that overlaps the middle is determied by cosiderig all elemets to the left ad all elemets to the right of the middle. 4.1 [4 poits] Based o the above algorithm descriptio, draw a tree that illustrates the process of determiig the maximum subarray i array A = [-1,, -3, 4, 3, -5, 1, 5] max = [ 1] max = [] max = [ 3] max = [4] max = [3] max = [ 5] max = [1] max = [5] max = [] max = [4] max = [3] max = [1, 5] max = [4] max = [1, 5] max = [4, 3, 5, 1, 5]

12 4. [10 poits] Desig a divide ad coquer algorithm that fids ad returs the maximum subarray of a usorted array. Use either C or pseudocode for your solutio. Algorithm: MaxCrossSubarray(A, low, mid, high) 1 leftsum = ; sum = 0; 3 for i = mid to low do 4 sum = sum + A[i]; 5 if sum > leftsum the 6 leftsum = sum; 7 maxleft = i; 8 rightsum = ; 9 sum = 0; 10 for j = mid + 1 to high do 11 sum = sum + A[j]; 1 if sum > rightsum the 13 rightsum = sum; 14 maxright = j; 15 retur (maxleft, maxright, leftsum+rightsum); Algorithm: FidMaxSubarray(A, low, high) 16 if high == low the 17 retur (low, high, A[low]); 18 else 19 mid = low+high ; 0 (leftlow, lefthigh, leftsum)=fidmaxsubarray(a, low, mid); 1 (rightlow, righthigh, rightsum)=fidmaxsubarray(a, mid+1, high); (crosslow, crosshigh, crosssum)=maxcrosssubarray(a, low, mid, high); 3 if leftsum rightsum ad leftsum crosssum the 4 retur (leftlow, lefthigh, leftsum); 5 else if rightsum leftsum ad rightsum crosssum the 6 retur (rightlow, righthigh, rightsum); 7 else 8 retur (crosslow, crosshigh, crosssum); 1

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