Problem Set 1 Solutions

Transcription

1 V R N N N R f ] R S Itroductio to Algorithms September 12, 2003 Massachusetts Istitute of echology 6046J/18410J rofessors Shafi Goldwasser ad Silvio Micali Hadout 7 roblem Set 1 Solutios roblem 1-1 Recurrece Relatios Solve the followig recurreces Give a boud for each problem If you are uable to fid a boud, provide as tight upper ( or ) ad lower ( or ) bouds as you ca fid Justify your aswers You may assume that (a)! #"%$'&() - By part 1 of the Master Method (b) *,+-!!/1032 /4032* - By part 3 of the Master Method (c) 6, 6 87: ; < Substitute = for he resultig relatio is =>?@ BA C;= Clearly, this is => ad thus 6 (d) *D EF G H ( ) ( ) 2 NN I G %J K M M M M M M M M M M M MM R R R R 16 2 R S S S S S S S x S x16 S x2 S S S S S S 16x 16x16 16x2 2x 2x16 2x2 he figure above shows the recursio tree that helps us i guessig a solutio We have that U #"%$XW BY \^] a3bdc e \^] a3bdc g ïhj ( H gk l hus we guess that lm * ad we prove it by substitutio Assume that =>UDop = for a appropriate costat o ad for all =rqs he we have that

2 v ( \ 2 Hadout 7: roblem Set 1 Solutios t v ou sou a sou 3w a3b ox, c {z o yh a3b o h c o a-b ~} for o {z Hece, * the other had by ispectio we have g ƒ % * Hece l % (e) D hk{ * ˆ 87:? j (f) *D /1032s lš * (g) /40323 :* - By ispectio ad substitutio G k*œ & Œ /4032* - By part 2 of the Master Method roblem 1-2 Asymptotic Notatio Rak the followig fuctios by order of growth; that is, fid a arragemet }' }pžpžpž }' of the H fuctios satisfyig,,,, E k H artitio your list ito equivalece classes such that x ad are i the same class if ad oly if x* *' (he fuctio /40-2 is discussed o pages -6 of CRS) G 4 # & z ]%]%]%] z /4032! #"%$8 #"%$ /4032* #"%$ š/4032,{b /1032dC œ he followig are ordered asymptotically from smallest to largest, are as follows (two fuctios, ad are o the same lie if x *X ): ]%]%]%] z

3 Hadout 7: roblem Set 1 Solutios 3 /4032 š/4032 /4032d ž G 1 # & /40-2 #"%$ #"%$3 #"%$ z roblem 1-3 Sieve of Eratosthees DB he Sieve of Eratosthees, iveted circa 200 BC, is a algorithm to fid all prime umbers betwee 2 ad a iput umber Ÿ he algorithm works as follows: we begi with a list of all itegers from 2 to Ÿ For each = Ÿ, we cross out (ie mark as composite) each multiple of = (> = ) that is less tha or equal to Ÿ Whe this process termiates, oly the prime umbers betwee 2 ad Ÿ are umarked

4 4 Hadout 7: roblem Set 1 Solutios Below, we give pseudocode for this algorithm At the begiig of the algorithm, every etry i the array is iitialized to true, ie is true for all } ˆŸ At the ed of the algorithm, is true iff is prime ERASHENES-SIEVE Ÿ : 1 et for all from 2 to Ÿ!ª «2 for = to Ÿ : 3 4 while y p= kÿ : l # {= `±6²! 6 D 7 edwhile 8 edfor he table below shows the values of the array elemets ³ŽpŽpŽXŸ (lie 7) durig a executio of the algorithm ru o iput Ÿ µ{z at the ed of each while loop i #¹ º #» º ¼ ½pº #¾ º # º ¼ ÁÀBº # º #à º ¼ œä)åæº œä{ä'º œä¹ º œä» º 1 2 F F F F F 3 F F F F F F 4 F F F F F F F F F F F F 13 F F F F F F rove that the ERASHENES-SIEVE algorithm is correct; that is, prove that upo termiatio, is true iff is prime Hit: You ca use the followig pre-coditio ad post-coditio ad you ca prove the suggested loop ivariat for the for loop (lie 2) et Ç is true iff is prime Ç Ç re-coditio: ŸÈ, ost-coditio: }ŠÉ such that ~ oop Ivariat: } É such that DŸ D= represet the statemet: l 6 We will use the give loop-ivariat for the for loop (lie 2) ad the relevat pre- ad postcoditios to prove the correctess of the algorithm We will ow prove the give loop ivariat Namely, we will prove by iductio that whe = is assiged the value (lie 2), Ç holds for all } Whe = is assiged the value 2, Ê is true sice it was iitialized to be true hus, we have proved that before we execute the for loop the first time, the base case of the loop ivariat holds

5 Hadout 7: roblem Set 1 Solutios Now we will assume that whe = use this iductive hypothesis to prove that whe = })~, is assiged the value, Ç holds for all }g is assiged the value µ, Ç We will holds for all et s cosider a executio of the while loop with =r Sice goes from to ËŸ =Ì, y = is always at least Sice oly elemets with ˆy {= are (re)set to false, o elemet l 6 6 with will chage value hus, Ç for }, will hold at the ed of the executio of the while loop It remais to show that Ç holds at the ed of the executio of the while loop If Í is true, the it caot have ay divisor Î such that ÏÎ the it does ot have a divisor ÎÐ ivariat is correct So k must be prime If k is prime, hus, it must be true hus, we have proved that the loop Whe =mƒÿ, o the fial executio of the for loop, we have that Ç due to the correctess of the loop ivariat holds for all, ƒÿ,