Improving The Problem Not the Code

Transcription

1 mprovig Te Probem Not te Code

2 Fidig Te Max ad Mi of a Number Comparig two adjacet umbers ad fidig te maximum ad miimum e.g Normay T()= N- Compariso wit a orace, i tis case a touramet teory N-1 matces are ecessary & sufficiet to fid a wier i a touramet of payers Miimum as to be amog tose wo as ost first matc, so we ca use osers of first compariso ad try to miimize tis set of umbers. Te oser set wi at east cosist of N/ umbers if we pay pair wise, so i tat case tota wi be N/+(N/-1)+(N/-1)=3N/- stead of improvig te code we wet ito te costructio of te probem. t is to be soved by dividig te umbers ito two sets we takig iput of two umbers eac.

3 Fidig max & extmax Secod argest must ecessariy oose from te argest. So secod argest must b foud from tose wo were compared & ost from te argest, ad te set of suc umber soud be miimized. Te egt of te pat of te wier wi give umber of compared eemets wit te wier. Storig umbers from to

4 Towers of Haoi At ay time smaer disk soud ot be beow te arger disk Oy disk 1 disk case 3 disk case <a,b> <a,c>,<a,b><c,b> <a,c><a,b><c,a><b,c><a,c><a,b><c,a><c,b><a,b> <a,b><a,c><b,c><a,b><c,a><c,b><a,b> 3(a,b,c) (a,c,b,) (1,a,b,c) (,c,b,a) Base: (==1) L=<a,b> ductio: (>1) 1=towers(-1,a,c,b) =towers(1,a,b,c) 3=towers(-1,c,b,a)

5 Fidig Greatest Commo Divisor put: Two iteger s m a Output: Greatest factor tat divides bot 1. Factorize m: fid primes m1, m, m3 suc tat m= m1*m*m3. Factorize : fid primes 1,, 3 suc tat = 1**3 3. Fid commo factors i bot mutipy ad prit resut Additioa Beefits: we ad foud prime umbers aso o te sides

6 Eucid Agoritm t is usig a cever tecique wic was expored after studyig te deeper properties of te umbers ivoved. 1 divide m by ad et r be te remaider if r=0, agoritm termiates; is te aswer 3 set m=; =r; ad go to step 1

7 Agoritm witout swappig 1 divide m by ad et r be te remaider if r=0, agoritm termiates; is te aswer 3 divide by r ad et m be te remaider 4 if m=0, agoritm termiates; r is te aswer 5 divide r by m ad et be te remaider 6 if =0, agoritm termiates; m is te aswer 7. go to step 1

8 Agoritm witout additioa variabe 1. Divide m by ad et m be te remaider. f m=0, te agoritm termiates wit aswer 3. Divide by m ad et be te remaider 4. if =0 te agoritm termiates wit aswer m; oterwise go to step 1

9 f we go deeper te et m=ag ad =bg aso m=x*+r T(a, b) = 1 + T(b, r 0 ) = + T(r 0, r 1 ) = = N + T(r N, r N 1 ) = N + 1 f te Eucidea agoritm requires N steps for a pair of atura umbers a > b > 0, te smaest vaues of a ad b for wic tis is true are te Fiboacci umbers F N+ ad F N+1, respectivey. Worst-case umber of steps By iductio. f N = 1, b divides a wit o remaider; te smaest atura umbers for wic tis is true is b = 1 ad a =, wic are F ad F 3, respectivey.

10 Now assume tat te resut ods for a vaues of N up to M 1. Te first step of te M-step agoritm is a = q 0 b + r 0, ad te secod step is b = q 1 r 0 + r 1. Sice te agoritm is recursive, it required M 1 steps to fid GCD(b, r 0 ) ad teir smaest vaues are F M+1 ad F M. Te smaest vaue of a is terefore we q 0 = 1, wic gives a = b + r 0 = F M+1 + F M = F M+. For N steps, te b is greater ta or equa to F N+1 wic i tur is greater ta or equa to φ N 1, were φ is te gode ratio. Sice b φ N 1, te N 1 og φ b. Sice og 10 φ > 1/5, (N 1)/5 < og 10 φ og φ b = og 10 b. Tus, N 5 og 10 b. Tus, te Eucidea agoritm aways eeds ess ta O() divisios, were is te umber of digits i te smaer umber b.

11 Recursio Discussio o differet pases of ay recursio agoritm maiy Base coditio, Decompositio ad recompositio. Aso ow recursio may be usefu i makig te code sorter ad simper, but i some cases it may prove very costy. Exampe of fiboacci sequece ike fib(6) breakig ito fib(5),fib(4) ad so o ad makig te umber of fuctio cas of te expoetia order. Wie te iterative fiboacci sequece wi be of te order of O(). Fiboacci sequece ca aso be soved i og time by recursive squarig metod F+1 F = F F = x x is to fid te t fiboacci umber t is θ(og) F F-1 F-1 F- 1 0

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15 Substitutio Metod 1. Guess te form of te soutio. Verify by iductio ad sow tat te soutio works for a set of costats T()=T(/) + Wic meas tat ow te probem as bee divided ito two sub probems ad size of te sub probem is / We guess tat it works for T() = O( g ) To prove tat T() c g for a appropriate c. Assumig tat te guess works we wi ave T() = ( c (/) g (/))+ = c g(/) + = c g c g + = c g (c-1) cg tis is true for a c 1

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17 Tota = + (3/16) + (3/16) k /16 k +. Tis is a geometric series (1/1-(3/16)) θ( )

18 Master Metod T()= at(/b) + f() : Were a 1 ad b>1 are costats ad f() is a asymptoticay positive fuctio We are dividig a probem of size ito a sub probems, eac of size /b, were a ad b are positive costats. a sub probems are soved recursivey eac i time T(/b) were a ad b are positive costats. Te cost of dividig te probem ad combiig te resuts of te sub probems is described by te fuctio f(). 1. f F() = O( og b a -ε )for some costat ε > 0, te T() = θ ( og b a ). f, f() = θ ( og b a og k )te. T()= θ( og b a og k+1 ) 3. f f()=ω( og b a+ ε ) for some costat ε > 0, ad if a f (/b) cf () for some costat c < 1 ad a sufficiety arge, te T () = Θ(f ())

19 Tree cases do ot cover a te possibiities for f (). Tere is a gap betwee cases 1 ad we f () is smaer ta but ot poyomiay smaer. Simiary, tere is a gap betwee cases ad 3 we f () is arger ta but ot poyomiay arger. f te fuctio f () fas ito oe of tese gaps, or if te reguarity coditio i case 3 fais to od, te master metod caot be used T() = 4T(/) + og b a = wic is poyomiay arger te f() so we are i case 1 ad aswer is θ( ) T() = 4T(/) + og b a = wic is same simiar to f() so we are i case ad aswer is θ( og) T() = 4T(/) + 3 og b a = wic is poyomiay arger from f() so we are i case 3 ad aswer is θ( 3 ) T()=T(/)+og case O(og ) T() = 4T(/) + / og Master metod does ot appy Biary Searc T() = 1.T(/) + O(1)= θ(og) Fidig Powers of a umber T()= T(/)+ θ(1) = θ(og)

20 Case1: Te Weigt icreases Geometricay from te root to te eaves. Te eaves od a costat fractio of te tota weigt. Last term is domiatig. Case : for k=0 weigt is approximatey same o a eves. t is poyomia or sowy decreasig fuctio. Case 3: Te Weigt decreases Geometricay from te root to te eaves. Te root od a costat fractio of te tota weigt. First term is domiatig.

21 teger mutipicatio How fast we ca mutipy Normay we require compexity to mutipy two umbers Mutipy two -bit itegers ad. Divide step: Spit ad ito ig-order ad ow-order bits / / We ca te defie * by mutipyig te parts ad addig: / / / / ) )*( ( * So, T() = 4T(/) +, wic impies T() is O( ). But tat is o better

22 / / / / ) )*( ( * / / / ) ( ] ) [( ] ) )( [( * So, T() = 3T(/) +, wic impies T() is O( og 3 ), by te Master Teorem. Tus, T() is O( ).

23 Matrix Mutipicatio Suppose we wat to mutipy two matrices of size N x N: for exampe A x B = C. C 11 = a 11 b 11 + a 1 b 1 C 1 = a 11 b 1 + a 1 b C 1 = a 1 b 11 + a b 1 C = a 1 b 1 + a b x matrix mutipicatio ca be accompised i 8 mutipicatio. T()= 8T(/) + = ( og 8 = 3 ) Strasse sowed tat x matrix mutipicatio ca be accompised i 7 mutipicatio ad 18 additios or subtractios.

24 T() = 7T(/) + = ( og 7 =.807 ) Tis reductio ca be doe by Divide ad Coquer Approac Divide matrices ito sub-matrices: A 0, A 1, A etc Use bocked matrix mutipy equatios Recursivey mutipy sub-matrices Termiate recursio wit a simpe base case P1 = (A11+ A)(B11+B) P = (A1 + A) * B11 P3 = A11 * (B1 - B) P4 = A * (B1 - B11) P5 = (A11 + A1) * B P6 = (A1 - A11) * (B11 + B1) P7 = (A1 - A) * (B1 + B) C11 = P1 + P4 - P5 + P7 C1 = P3 + P5 C1 = P + P4 C = P1 + P3 - P + P6 C 11 = P 1 + P 4 - P 5 + P 7 = (A 11 + A )(B 11 +B ) + A * (B 1 - B 11 ) - (A 11 + A 1 ) * B +(A 1 - A ) * (B 1 + B ) = A 11 B 11 + A 11 B + A B 11 + A B + A B 1 A B 11 -A 11 B -A 1 B +A 1 B 1 + A 1 B A B 1 A B = A 11 B 11 + A 1 B 1 Strasse beats orma matrix mutipicatio for 5 approx.

25 fiite Wa Probem * You ave a ifiite wa o bot sides were you are stadig ad it as a gate somewere i oe directio, you ave to fid out te gate, o desig-ifiite time icremeta desig- you go oe step i oe directio, come back go to oter directio oe step, come back ad te go steps i oter directio : : -1+.(-1) i+3 4(-1)/+3= +

26 Usig Divide ad Coquer You go 0 step i oe directio, come back go 1 step i oter te 1 directio i oe way te come back ad te up to k : K-1 +. K-1 + K-1 3. K : 4( K-1 + K- + 1)+ 3. K 4( K -1)+ 3. K 7. K -4 7N-4