WRITTEN ASSIGNMENT 1 ANSWER KEY

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1 CISC 65 Itrodutio Desig ad Aalysis of Algorithms WRITTEN ASSIGNMENT ANSWER KEY. Problem -) I geeral, this problem requires f() = some time period be solve for a value. This a be doe for all ase expet lg ad!, whih will be evaluated umerially. For example, F() = lg mirose = se => lg = E6 => = E6 seod miute hour day moth year etury lg E6 6E7 3.6E9 8.64E0.59E 3.5E3 3.5E E 3.60E5.30E9 7.46E 6.7E4 9.95E6 9.95E30 E6 6E7 3.6E9 8.64E0.59E 3.5E3 3.5E5 lg E6.33E8.75E9 7.8E0 7.97E 6.86E3.00E E E04.94E05.6E06 5.6E06 5.6E ! I solvig for problems lie! ad lg, umerially solvig i Exel is a aeptable approah. For example, let ell A equal ad ell A equal =FACT(A) Next, we reogize seod equals x0 6 miroseods. So, irease ell A util ell A equals x0 6. Rather tha maually hagig ell A, you ould use Exel s Goal See futio.. Exerise.-) Cost Time legth[a] for j to - do smallest j 3 - for i j + to 4 ( + )! t j =! j = " do if A[i] < A[smallest] 5 ( t! ) = the smallest i 6 " " j! (! ) ( j! ) = " j = exhage A[j] A[smallest] 7 - " ( t j! )

2 CISC 65 Itrodutio Desig ad Aalysis of Algorithms The algorithm maitais the loop ivariat that at the start of eah iteratio of the outer for loop, the subarray A[,, j-] osists of the j- smallest elemets i the array A[,, ], ad this subarray is the sorted order. After the first - elemets, the subarray A[,, -] otais the smallest - elemets, sorted, ad therefore elemet A[] must be the largest elemet. Best Case) Best-ase will our whe the array is already sorted, thus 6 will ever be exeuted. Hee the ruig time is: T() = () + () + 3 (-) + 4 ( +-)/ + 5 ( -)/ + 7 (-) T() = ( 4 / + 5 /) + ( / 5 / + 7 ) + ( ) T() = a + b + T() = Θ( ) Worse Case) Worst-ase will our whe the array is usorted i reverse order (i.e. <5, 4, 3,, >), thus 6 will our (-)/ time. Hee the ruig time is: T() = () + () + 3 (-) + 4 ( +-)/ + 5 ( -)/ + 6 ( -)/ + 7 (-) T() = ( 4 / + 5 / + 6 /) + ( / 5 / - 6 / + 7 ) + ( ) T() = a + b + T() = Θ( ) The ruig time for the algorithm is Θ( ) for all ases. 3. Exerise.-4) Modify the algorithm so it tests whether the iput satisfies some speial-ase oditio ad, if it does, output a pre-omputed aswer. The best-ase ruig time is geerally ot a good measure of a algorithm. 4. Exerise.3-)

3 CISC 65 Itrodutio Desig ad Aalysis of Algorithms Exerise.3-3) Prove by idutio that # T ) = "! T ( / ) + is ( T ( ) = lg if = if =, for > To prove by idutio, we must ) show it is true for some base ase, ) hypothesize for some value of, ad 3) show that it holds for +. Base Case: whe = => = T() = lg = lg = x =. Hypothesis Step: T ( ) = ( )lg( ) T ( ) = ( )lg( Idutive Step +: T ( ) = T ( T ( T ( ) = ) = T ( ) = Q.E.D. ) [ lg ] 6. Exerise.3-5) + (lg lg / ) ) = T ( ) + = lg + 3

4 CISC 65 Itrodutio Desig ad Aalysis of Algorithms The biary searh proedures taes a sorted array A, a value v, ad a rage [low high] of the array A, i whih to searh for the value v. ITERATIVE-BINARY-SEARCH(A, v, low, high) while low high ( low + high) / do mid! " If v = A[mid] the retur mid if v > A[mid] the low mid + else high mid - retur NIL RECURSIVE-BINARY-SEARCH(A, v, low, high) if low > high the retur NIL ( low + high) / mid! " if v = A[mid] the retur mid if v > A[mid] the retur RECURSIVE-BINARY- SEARCH(A, v, mid+, high) else retur RECURSIVE-BINARY- SEARCH(A, v, low, mid-) Based o the ompariso of v to the middle elemet i the searh rage, the searh otiues with the rage halved. 4

5 CISC 65 Itrodutio Desig ad Aalysis of Algorithms =0 = = lg T(/) T(/) T(/4) lg... T() Total: lg The depth of the tree is omputed by otig that T() ours whe T(/ )=T(), or / j =. This meas = j, or j = lg. 7. Exerise 3.-4) (a) Is + = O( )? f() = O(g()) 0 f() g() where > 0 ad > 0 Hee, 0 + where > / where > 0 0 where > 0 ad > 0 Therefore, + = O( ) Q.E.D. (b) Is = O( )? f() = O(g()) 0 f() g() where > 0 5

6 CISC 65 Itrodutio Desig ad Aalysis of Algorithms Hee, 0 where > 0 0 / where > 0 0 where > 0 0 o Now, 0 o where > 0 0 +o where > for > 0 0 for > 0 Is a otraditio so, O( ) Q.E.D. 8. Show that (-)/ is O( ) f() = O(g()) 0 f() g() where > 0 ad > 0 Hee, 0 (-)/ where > 0 ad > 0 0 /-/ where > 0 ad > 0 0 /-/4 where = /4 Hee, 0 (-)/ /4 where Therefore, (-)/ = O( ) 9. Usig "big-o" otatio, give the worst ase ruig times of the followig proedures as a futio of. proedure mystery (: iteger); var i,j,: iteger; begi for i := to - do for j := i + to do 6

7 CISC 65 Itrodutio Desig ad Aalysis of Algorithms ed for := to j do... a statemet requirig O() time Lie Pseudo ode Cost Times for i := to - do for j := i + to do j = 3 for := to j do 3! + " ( + 5)( j = 6 + +! "" = 3 ) 4... a statemet requirig O() time 4 "" = j = ( + )(! ) 3 T() = + ( + )/ + 3 ()(+5)(-)/6 + 4 ()(+)(-)/3 T() = ( ) 3 /3 + ( + 3 ) / + ( + / 5 3 /6 4 /3) T() a 3 + b + + d T() = O( 3 ) 7