Matriculation number: You have 90 minutes to complete the exam of InformatikIIb. The following rules apply:

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1 Departmet of Iformatics Prof. Dr. Michael Böhle Bizmühlestrasse Zurich Phoe: AlgoDat Midterm1 Sprig Name: Matriculatio umber: Advice You have 90 miutes to complete the exam of IformatikIIb. 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 (1 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. For the exam Iformatik IIb, 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. Notably 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 1 3 Total Poits Achieved Maximum Poits

2 Exercise 1 0 Poits Asymptotic Complexity ad Recurreces 1.1 [4 poits] Calculate the asymptotic tight boud for the followig fuctios ad rak them by their order of growth (lowest first). Clearly work out the calculatio steps i your solutio: f 1 () = lg 8 (3 + 8 lg ), f () = +10, f 3 () = log π log log ( 10 log 35 ), f 4 () = max((log ) 4, ), f 5 () = lg 10, f 6 () = ( + 10)! f 1 () = lg 8 (3 + 8 lg ) Θ( log ) f () = +10 Θ( ) f 3 () = log π log log ( 10 log 35 ) Θ((log ) ) f 4 () = max((log ) 4, ) Θ( ) f 5 () = lg 10 = (10 log ) 16 = (10 ) 4 Θ( 8 ) f 6 () = ( + 10)! Θ(( + 10)!) 1 Rakig: f 3, f 4, f 1, f 5, f, f 6

3 Name: Matriculatio umber: 1. [ poits] Calculate the asymptotic tight boud of the followig recurreces. If the Master Theorem ca be used, write dow a, b, f() ad which case (1-3) applies. a) T () = 7T ( 3 ) a = 7, b = 3, f() = Θ( 3 ), Case : T() = Θ( 3 log 3 ) b) T () = 9T ( ) a = 9, b = 4, f() = 3, Case 3: Θ( 3 ) Regularity coditio: if c = 9, a f(/b) = 9 ( 64 4 )3 9 T () = Θ( 3 )

4 1.3 [6 poits] Cosider the followig recurrece. { T () = a) Compute the value of T () for = 8. 4, if = 1 T (/) , if > 1 T (8) = T (4) = (T () ) + 41 = 4(T (1) ) + 83 = 159 b) Solve recurrece exactly by givig a closed form expressio (without T () o the right side of = ). T () = T ( ) = [T ( ] ) = T ( ) (1 + ) = [ T ( 3 ) ] (1 + ) = 3 T ( 3 ) (1 + + ) =... = i T ( i ) + i 5 + ( i 1 ) For i = lg, we have: T () = lg T ( lg ) + lg 5 + ( lg 1 ) = T (1) + 5lg + lg 1 = 5lg c) Fid the simplest f(), such that T () = Θ(f()). f() = lg 4

5 Name: Matriculatio umber: 1.4 [8 poits] Cosider the followig recurrece: { 1, if = 1 T () = T (/5) + T (7/10) +, if > 1 a) 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 = 9 10 = Logest brach o the left. Tree grows util ( 7 h 10) = 1 = h = log 10 7 Guess: O() b) Prove the correctess of your estimate usig iductio. Proof by iductio: T () = T (/5) + T (7/10) + (recurrece) c + c 7 + (iductive hyp.) 5 10 = c( 9) + (reformattig) 10 = ( 9c + 1) (reformattig) 10 c (for c 10, > 1) 5

6 Exercise 0 Poits Rutime ad Recursio Algorithm: alg(a,) 1 for i = 1 to / do mi = i; 3 max = i + 1; 4 if A[mi] > A[max] the 5 exchage A[mi] ad A[max]; 6 for j = i + 1 to i do 7 if A[j] < A[mi] the 8 mi = j; 9 if A[j] > A[max] the 10 max = j; 11 exchage A[i] ad A[mi]; 1 exchage A[ i + 1] ad A[max];.1 [ poits] The algorithm alg(a,) gets as iputs a array A[1...] ad the umber of itegers it cotais. Assume A=[5, 7, 6, 3,, 1, 4]. Complete the followig matrix where you eed to show the value of i ad the cotet of A after each executio of the outer for loop. The first lie of the matrix shows the iitial array A before the executio of the loop. i A[1] A[] A[3] A[4] A[5] A[6] A[7]

7 Name: Matriculatio umber:. [ poits] Describe i words (maximum three lies) what the algorithm alg(a,) does. The algorithm sorts the array. It is a variat of selectio sort. I every pass it fids both the miimum ad maximum values ad creates two sorted partitios of the array, oe with miimum values (o the left side) ad oe with maximum (o the right side). This reduces the umber of scas of the list by a factor of, elimiatig some loop overhead but ot actually decreasig the umber of comparisos or swaps. 7

8 .3 [8 poits] Aalyze the steps of the algorithm alg(a,), calculate its exact rutime ad its asymptotic complexity. lie 1 c lie c lie 3 c 3 lie 4 c 4 lie 5 c 5 α 1 lie 6 c ( ) + lie 7 c ( ) lie 8 c 8 α 1 4 ( ) lie 9 c ( ) lie 10 c 10 α3 1 4 ( ) lie 11 c 11 lie 1 c 1 * 0 α 1 1 ** For each i [1, ], it will ru ( i) (i + 1) + 1 times plus termiatio coditio check. *** ( i) = 1 4 ( ) i=1 **** 0 α 1, 0 α 3 1, α + α 3 1, Ruig time: times for 8

9 Name: Matriculatio umber: T () = c 1 + (c 1 + c + c 3 + c 4 + α 1 c 5 + c 6 + c 11 + c 1 ) ( )(c 7 + α c 8 + c 9 + α 3 c 10 ) = (c 7 + α c 8 + c 9 + α 3 c 10 ) + (c 1 + c + c 3 + c 4 + α 1 c 5 + c 6 + c 11 + c 1 c 7 α c 8 c 9 α 3 c 10 ) + c 1 O( ).4 [8 poits] Give a recursive versio of the algorithm alg(a,). Use either C or pseudocode for your solutio. Algorithm: BidirectioSelectioSortRec(A, l, r) 1 if l r the mi = l; 3 max = r; 4 if A[mi] > A[max] the 5 exchage A[mi] ad A[max]; 6 for j = l + 1 to r 1 do 7 if A[j] < A[mi] the 8 mi = j; 9 if A[j] > A[max] the 10 max = j; 11 exchage A[l] ad A[mi]; 1 exchage A[r] ad A[max]; 13 BidirectioSelectioSortRec(A, l + 1, r 1); 14 Call: BidirectioSelectioSortRec(A, 1, ); 9

10 Exercise 3 0 Poits Divide ad Coquer The mi-max search algorithm determies the miimum ad maximum elemets of a usorted array A[1...]. First, it divides the iput array ito two equal partitios: I (A[1]... A[mid]) ad II (A[mid + 1]... A[]). Afterwards, it calls itself recursively o both partitios to fid the miimum ad the maximum of each partitio, the combies them to fid the global miimum ad maximum elemets. For example, if the give array is [-, -5, 6, -, -3, 1, 5], the the miimum is 5 ad the maximum is [4 poits] Based o the above, draw a tree to illustrate the process of determiig the miimum ad the maximum of the array A = [-, -5, 6, -, -3, 1, 5]. Aotate each ode of the tree with its miimum ad maximum mi =, max = mi = 5, max = 6 mi = 3, max = mi = 1, max = mi = 5, max = 6 mi = 3, max = mi = 5, max =

11 Name: Matriculatio umber: 3. [10 poits] Desig a divide ad coquer algorithm which fids ad returs the miimum ad maximum elemets of a usorted array. Use either C or pseudocode for your solutio. Algorithm: MiMax(A,l,r) 1 if l = r the retur A[l], A[l]; 3 else 4 m = l+r ; 5 mi1, max1 = MiMax(A, l, m); 6 mi, max = MiMax(A, m+1, r); 7 if mi1 < mi the 8 mi = mi1; 9 else 10 mi = mi; 11 if max1 > max the 1 max = max1; 13 else 14 max = max; 15 retur mi, max; 3.3 [6 poits] Write the recurrece for the complexity of your algorithm i Task 3.. Solve the recurrece usig the repeated substitutio method. { T () = 0, if = 1 T ( ) + 1, if > 1 T () = T (/) + 1 = (T (/4) + 1) + 1 = 4T (/4) + 3 = 4(T (/8) + 1) + 3 = 8T (/8) + 7 T () = i T (/ i ) + i 1 = lg T (/ lg ) + lg 1 = T (1) + 1 = Θ() 11

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