ITEC 360 Data Structures and Analysis of Algorithms Spring for n 1

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1 ITEC 360 Data Structures ad Aalysis of Algorithms Sprig Prove that f () = is Θ ( ) = 66 for 1 Take C 1 = 66 f () = is O( ) Sice for 1 If we take C = 60 f () = is Ω ( ) Sice f () is O ( ) ad f () is Ω ( ), therefore f () is Θ ( ).. Prove that f() = a k k + a k-1 k a 1 + a 0 is Θ ( k ). Let C = a k + a k a 1 + a 0. The a k k + a k-1 k a 1 + a 0 a k k + a k-1 k +. + a 1 k + a 0 k = (a k + a k a 1 + a 0 ) k = C k Therefore, f() = a k k + a k-1 k a 1 + a 0 is O( k ) Sice a k k + a k-1 k a 1 + a 0 a k k f() = a k k + a k-1 k a 1 + a 0 is Ω ( k ) Thus f() = a k k + a k-1 k a 1 + a 0 is Θ ( k )

2 3. Prove that f() = 1 k + k +. + k is Θ ( k+1 ) If k is a positive iteger ad replace each iteger 1,,, by We have 1 k + k +. + k k + k + + k =. k = k+1 for 1; Hece f() = 1 k + k +. + k is O ( k+1 ) We ca obtai the lower boud, throwig away the first half of the terms 1 k + k +. + k / k + + (-1) k + k / k + + / So we ca coclude that = ( +1) / f() = 1 k + k +. + k is Ω ( k+1 ). k / k k k + / (/) (/) k = k+1 / k+1. Hece f() = 1 k + k +. + k is Θ ( k+1 ) 4. Prove that f() = lg! is Θ ( lg ). Sice! =. (-1)..,. 1 ad lg a. b = lg a + lg b lg! = lg + lg (-1) + + lg + lg 1 lg + lg (-1) + + lg + lg 1 lg + lg + + lg + lg = lg We ca coclude that lg! is O ( lg ). lg + lg (-1) + + lg + lg 1 lg + lg (-1) + + lg / lg / + lg / + + lg / ( +1) / lg / (/) lg (/)

3 Therefore, By mathematical iductio we ca show that if 4, (/) lg (/) ( lg ) / 4. I the last two iequalities, lg + lg (-1) + + lg + lg 1 ( lg ) / 4 for 4. So fially it follows that lg! is Ω ( lg ). lg! is Θ ( lg ). 5. Assumig that f 1 () is O (g 1 ()) ad f () is O (g ()), prove the followig statemets: a. f 1 () + f () is O (max (g 1 (), g ))). b. If a umber k ca be determied such that for all > k, g 1 () g (), the O (g 1 ()) + O(g ()) is O(g ()). c. f 1 () * f () is O(g 1 () * g ()) (rule of product). d. O(cg()) is O(g()). e. c is O(1). I the followig aswers, these two defiitios are used: f 1 () is O(g 1 ()) if there exist positive umbers c 1 ad N 1 such that f 1 () c 1 g 1 () for all > N 1 ; f () is O(g ()) if there exist positive umbers c ad N such that f () c g () for all > N ; (a) From the above defiitios we have f 1 () c 1. max (g 1 (), g ()) for all max ( N 1, N ) f () c. max (g 1 (), g ()) for all max ( N 1, N ) which implies that f 1 () + f () ( c1+ c ) max (g 1 (), g ()) for all max ( N1, N ) Hece for c 3 = c 1 + c ad N 3 = max (N 1, N ), f 1 ()+ f () c 3. max (g 1 (), g ()) for all N that is f 1 ()+ f () is O(max (g 1 (), g ()). 3

4 (b) If g 1 () g (), the for c = max (c 1+ c ), c g 1 () c g (), c g 1 () + c g () c g() which implies that O(g 1 ()) + O(g ()) is O(g ()). (c) The rule of product, f 1 (). f () is O(g 1 ().g ()) is true sice f 1 (). f () c 1 c g 1 (). g ()) for all max ( N1, N ) (d) O (cg()) is (O(g()) meas that ay fuctio f which is O(cg) is also O(g). Fuctio f is O(cg) if there are two costats c 1 ad N so that f () c 1 cg() for all N; i this case, for a costat c = c 1 c, f () c g(); thus by choosig properly a costat c (whose value depeds o the value of c ad c 1 ), f is O(g). (e) A costat c is O (1) if there exist positive umbers c 1 ad N such that c c 1. 1 for all N; that is, the costat fuctio c is idepedet of, ad we ca simply set c 1 = c. 6. Assumig that f 1 () is O (g 1 ()) ad f () is O (g ()), Fid couter examples refute the followig statemet: f 1 () - f () is O (g 1 () g ()). Let f 1 () = a 1 ad f () = a, the f 1 () ad f () are O (). But f 1 () - f () = (a 1 a ) is ot O( ) = O(0). Hece f 1 () - f () is ot O (g 1 () g ()). 7. Fid the complexity of the fuctio used to fid the kth smallest iteger i a uordered array of itegers it selectkth(it a[], it k, it ) { it i, j, mii, tmp; for (i = 0; i < k; i++) { mii = i; for (j = i+1; j < ; j++) if (a[j]<a[mii]) mii = j; tmp = a[i]; a[i] = a[mii]; a[mii] = tmp; retur a[k-1]; The complexity of selectkth ( ) is (-1) + (-) + + (-k) = ( k - 1) k / = O( )

5 8. Determie the complexity of the followig implemetatios of the algorithms for addig, multiplyig, ad trasposig matrices: Aswer: for (i = 0; i < ; i++) for (j = 0; j < ; j++) a[i][j] = b[i][j] + c[i][j]; for (i = 0; i < ; i++) for (j = 0; j < ; j++) for (k = a[i][j] = 0; k < ; k++) a[i][j] += b[i][k] * c[k][j]; for (i = 0; i < - 1; i++) for (j = i+1; j < ; j++) { tmp = a[i][j]; a[i][j] = a[j][i]; a[j][i] = tmp; The algorithm for addig matrices requires assigmets. Note that the couter i for the ier loop does ot deped o the couter j for the outer loop ad both of them take values 0,..., - 1. All three couters, i, j ad k, i the algorithm for matrix multiplicatio are also idepedet of each other, hece the complexity of the algorithms is 3. To traspose a matrix, i= j= i+ 1 = O( ) assigmets are required. 9. Fid the computatioal complexity for the followig four loops: a. for (ct1 = 0, i = 1; i <= ; i++) for (j = 1; j <= ; j++) ct1++; b. for (ct = 0, i = 1; i <= ; i++) for (j = 1; j <= i; j++) ct++; c. for (ct3 = 0, i = 1; i <= ; i *= ) for (j = 1; j <= ; j++) ct3++;

6 d. for (ct4 = 0, i = 1; i <= ; i *= ) for (j = 1; j <= i; j++) ct4++; Aswer: (a) The auto-icremet ct++ is executed exactly times. (b) i = 1 i = O( ); lg i = 1 (c) i = Θ( lg ) lg = (d) i = Θ() i Fid the average case complexity of sequetial search i a array if the probability of accessig the last cell equals ½ the probability of the ext to last cell equals ¼, ad the probability of locatig a umber i ay of the remaiig cells is the same ad equal to 1. 4( ) Aswer: ( ) 4( ) ( 1) = = IMPORTANT NOTE Study problems of exercises i pages of chapter.

7 Exact Aalysis Rules 1. We assume a arbitrary time uit.. Executio of oe of the followig operatios takes time 1: a) assigmet operatio b) sigle I/O operatios c) sigle Boolea operatios, umeric comparisos d) sigle arithmetic operatios e) fuctio retur 3. Ruig time of a selectio statemet (if, switch) is the time for the coditio evaluatio + the maximum of the ruig times for the idividual clauses i the selectio. 4. Loop executio time is the sum, over the umber of times the loop is executed, of the body time + time for the loop check ad update operatios, + time for the loop setup. Always assume that the loop executes the maximum umber of iteratios possible 5. Ruig time of a fuctio call is 1 for setup + the time for ay parameter calculatios + the time required for the executio of the fuctio body. Aalysis Example 1 Give: for (i = 0; i < -1; i++) { for (j = 0; j < i; j++) { aray[i][j] = 0; Rules 4 ad a: time 1 before loop Rules 4, c ad d: time 3 o each iteratio of outer loop, ad oe more test Rules 4 ad a: time 1 o each iteratio of outer loop for (i = 0; i < -1; i++) { for (j = 0; j < i; j++) { aray[i][j] = 0; So, the total time T() is give by: 1 i T ( ) = = i= 1 j= 1 3 Rules 4, c ad d: time (o each iteratio of ier loop) ad oe more test Rule a: time 1 o each pass of ier loop 7 + 3

8 Aalysis Example Rules 4 ad a: time 1 before loop Rule a: time 1 before loop Rules 4 ad a: time 1 o each iteratio of outer loop Sum = 0; for (k = 1; k <= ; k = *k) { for (j = 1; j <= ; j++) { Sum++; Rules 4, c ad d: time 3 o each iteratio of outer loop, plus oe more test Rules 4, c ad d: time (o each iteratio of ier loop) plus oe more test Rule a: time 1 o each pass of ier loop The tricky part is that the outer loop will be executed log() times assumig is a power of. So, the total time T() is give by: log T ( ) = k = 1 j= 1 Aalysis Example = 3 log + 5log + 3 How does this compare to the previous result? Rule a: time 1 before loop Sum = 0; I >> Value; while ( I ) { Rule b: time 1 before loop Rule c: time 1 at begiig of each pass, ad oce more Rule c: time 1 o each pass Rules ad: time, if doe if ( Value < 0 ) { Sum = -Sum; Sum = Sum + Value; else { Sum = Sum + Value; I >> Value; Rules ad: time, if doe Rules ad: time, if doe Rules ad: time 1 o each pass So, assumig iput values are received, the total time T() is give by: T ( ) = + k = 1 ( 3 + max( 4, )) + 1 = 7 + 3

9 Big-O Example Take the fuctio obtaied i the algorithm aalysis example earlier: 3 5 T ( ) = + 3 Ituitively, oe should expect that this fuctio grows similarly to. To show that, we will shortly prove that: 5 < 5 for all 1 ad that 3 < for all 1 Why? Because the we ca argue by substitutio (of o-equal quatities): T ( ) = = for all 1 Thus, applyig the defiitio with C = 15/ ad N = 1, T() is O( ).