pro-recursiveproblemproblem.recurrencearguments.re-nicely.proverecursi - PDF document

pro-recursiveproblemproblem.recurrencearguments.re-nicely.proverecursivecorrectrunningrecurrence Post #1 Post #2 Post #3 problemrulesare 2 Recurrences approachproblemthreeWebelow.areLargercorrespondlarger 1 2 3 ) 2 1 ) 3 1 2 ) 3 1 2 ) 3 1 2 ) 1 3 2 ) 1 2 3 ) 1 3 problemFrenchEdouardtherewerewerewereaccordingrulesAccordingTowercrumble are,sufcientimportantly,“Willbefore 1.1 Finding a Recurrence TowersHanoiproblemcanrecursivelyfromanother.=1 =3theproves  Weapproachrecursive recursively � fromthird � 1 steps. 1 2 . . . n  ) n  1 2 . . . n  � 1  3 Recurrences largestfromrequires 1 2 . . . ) n  n  � 1  n 1 2 . . . n  � 1  the � third � arerequired. 12. . . ) n  n  � 1  1 2 . . . n  requiredferent � +1Werequired T3   2
+1  = 7   2
+1   15  sufcienteffortprobablyburgerscream,over.) 1.2 Towers WePerhaps—ponderedproblem thereherewhy.fromdifferent � remaining � requires � largest � arerequired � 1smaller disks on top. Recurrences argumentrequired � +1recurrencerequiredTowerproblem =1 =2 � +1   2) Werecurrence=3;T=7;T=15;::: 1.3 Guess-and-Verify fromrecurrencerequireexpressionrequiredTowersproblempurchaseburgerscreambefore Therearedifferentrecurrences.correct,proof. n 1  1  2  3  3  7  4  15  5  31  6  63  Tn  n =2 � 1. recurrence,proofwrong.we'rewrong, let's check.) Claim. If: =1  Tn  =2+1  (for n 2) then: Tn  =2 � 1  Recurrences n Proof.proofproposition=2 � 1. 1 (1)true=1=2 � =2 n  � prove=2 n+1  � where  1. =2+1  =2(2 n  � 1)+1  =2 n+1  � 1  recurrencerelation, require 7  � 1=127Weresolveremaining= 2 64  � morebefore 2 Graduate Student Job Prospects (say,thereareInitially,therewerewereareprospectsmoreWorse,increasingprofessorsareincreasingmorerapidly.Eventually,career.problemstopparticular,“AreTAsHereareproblem. • Thereare • Congressforcedretirementretirevoluntarily.trueproblem!)therefore, • year,professorpro-year.professorsare • thereareprofessorsprofessorhired. Recurrences 2.1 Finding a Recurrence Ideally,professorsyear.areprofessorsToproblem, (0) =  0  professors; (1) =  1  professor; (2) =  1  professor; (3) =  2  prof,prof;profbusy,prof (4) =  3  prof,profs;profbusy,profs (5) =  5  profs,profs (6) =  8  profs,profs professorsprofessorshires.professors � hiresprofessors � professorsyear.recurrenceprofessors: (0) =0 (1) =1 )=  � 1) +  � 2) (  2)  recurrence.areintroducedreproduction.appear,requiregreatestarerecurrence, 2.2 Solving the Recurrence recurrencerecurrencelinear.(Well,lecture; X 7 Recurrences recurrence )=  � 1) +  � 2) + : : :  � d)  d = bif(n  � i)  i=1  where; a;: : : aarerecurrencerecurrenceordercoefcients=1recurrence.) now,recurrence;recurrenceslecture.ruleofunfamiliarrecurrence.recurrence,However,reallyrun”;proof.Rather, n )=  HereareintroducedimprovecorrectproofToimprove(0) =0 (1) =1 Verication.recurrence)=  � 1) +  � 2) gives: nn � 1  cx = cx + cx  n � 2  2  + 1  2  x  � x  � 1 =0  1   p 5  x =  2  In the rst step, we divide both sides of the equation by cx n � 2 rearrange This calculation suggests that the connstant c can be anything, but that x must be (1   Evidently,therearerecurrence: 1  � p 5  ! nn  1 +  ! )= )=  22  theoremtruerecurrences. arerecurrence)+  X X X X ! ! X �
8 Recurrences Proof.are d )=  � i)  i=1  d )=  � i)  i=1  Multiplying the rst equation by c, the second by d, and summing gives: dd )+ )= ca � )+ da � i)

i=1 i=1 d = ai cf (n  � )+  � i)  i=1  )+  phenomenon—differentialconsequently,presenttheorem 1  � p 5  ! nn  1 +  ! )=  22  torecurrenceremains(0)=0 (1)=1From 1  � p 5  ! 0  1 +  ! 0  (0) = =0  22  From 1  � p 5  ! 1  1 +  ! 1  (1) = =1  22  Wethere=1 � We'reWerecurrence 1  � p 5  ! nn  1 1 +  ! 1  )=  p52 �p52  Recurrences wrong!areexpressionsquarerootsAmazingly,however,squarerootsexpressionreally=0;:::across 2.3 Job Prospects returnare To  thereareprofessorsshufingobscure 19  years. hardexactly.However,approximateanswer.zero: 1  � p 5  ! nn  1 1 +  ! 1  )=  p52 �p52  n  1 1 +  ! (1)  p52  This is because (1  � 2=0618:::   jj tiny. Fromapproximationin n  1 1 +  ! f(n)   N p52  arewhere: log((1))  log( 1+ 2  p 5 )  = (log professorsincreasingexponentially.=10000 =20YourTAs recurrenceinterestingcorollary. 1 +  ! =  2  Recurrences Ratio”.We n  )= + (1)  p 5  We'veexpressionlargenamely,number.properties 3 General Linear Recurrences recurrencerecurrence.recurrence )=  � 1) +  � 2) + ::: � d)  )=  n recurrence, nn � d  x= a1x  n � 1  + a2x  n � 2  ::: d  x= a1x  d � 1  + a2x  d � 2  ::: � 1x+ ad  Dividing the rst equation by x n � d  recurrence.readoffcoefcientsarecoefcientsrecur-rence. recurrenceareroots • nonrepeatedroot n  recurrence. • repeatedroot nn  ; nr; nr; :::; n  k � 1  r  arerecurrence. Futhermore,Theorem 1 solutions recurrencerootsroots nnn )=  1  )=  2  )=  n  )=  3 r 3  11 Recurrences Thus, every linear combination ar  +  b r  +  cr +  d  nr  f(n)
n 1
 n 2
 n 3
 n 3  is also a solution. remainsappropriately.were(0) =0(1) =1(2) =4 (3) =9 01  02  0 d + 3  03 (0)=0 +  b  0r  0 + ar
r  cr  =  )


11  12  13  13 (1)=1 +  b  +  d 1r
1 + ar
r  cr  =  )

21  22  23  23 (2)=4 +  b  +  d 2r
4 + ar
r  cr  =  )

31  32  33  33 (3)=9 +  b  +  d 3r
9 + ar
r  cr  =  )

All the nasty r i  arerecurrence 3.1 An Example thereforever,reproducesgrow?reverseproblemwhere“reproduce” (0) =0  j  (1) =1frombeforefrombefore � year, � 1) � f(n � together,recurrence )=  � 1) + ( � 1)  � f(n  � 2)) =2 � 1)  � f(n  � 2)  2 � + 1 =0root=1 Therefore,recurrence The characteristic equation is x )= (1) n  (1) n  = c1 + c2n  The boundary conditions imply two linear equations in two unknowns: (0) =0  ) (0) =0 (1) =1  ) (1) =1  Recurrences = 0= 1Therefore,recurrence )=  = 0+(1) = n  thereareprobablyproblemmoreguess-and-verify.recurrenceeasy. 4 Inhomogeneous Recurrences Werecurrences )=  � 1)+ � 1)+::: � d)  recurrences.recurrence )=  � 1)+ � 1)+::: � )+ recurrencere-currence: )=  � 1)+ � 2)+1  recurrencesdifferentdifcult.Wethree 1. resultingrecurrencebefore.(Ignore;c;:::;crecurrence 2. restorerecurrence,Thereareguess-and-verify.wisely. 3. Add the homogeneous and particular solutions together to obtain the general solu-tion. Now use the boundary conditions to determine constants by the usual method of generating and solving a system of linear equations. differentialprobablyfamiliar.arounddifferentialfamiliar. Recurrences 4.1 An Example recurrence (1)=1 )=4 � 1)+3 n  Step 1: Solve the Homogeneous Recurrence recurrence)=4 � 1). The characteristic equation is x  � 4=0 n root=4Therefore,)= Step 2: Find a Particular Solution n recurrence)= 4 � 1)+3there n whererecurrence n d3 n  =4 n � 1  +3=4+3  � 3  Evidently,)= � 3
3 n  = � 3 n+1  is a particular solution. Step 3: Add Solutions and Find Constants We )= n  � 3 n+1  The boundary condition gives the value of the constant c: ) c4 1  � 3 1+1  =1  (1)=1  c =  ) 2  n Therefore,recurrence)=  5 4 � 3 n+1 . Piece of cake! 2 surethat= 2Fromrecurrence,(2)= 4(1)+3 2  = 13From 2 (2)=  5 4 � 3 3  =40 � 27=13 2 Recurrences 4.2 How to Guess a Particular Solution hardestrecurrenceswrong.However,rules • Generally, • )=)=)= 2  +bn +c, etc. • Moregenerally,samedegree,degreehigher,higher,)=6+5)=)= 2  +bn +c. • If g(n)is an exponential, such as 3 n )= n )= n  +c3 n  and then an 2 3 n  +bn3 n  +c3 n , etc. 15 Recurrences Short Guide to Solving Linear Recurrences recurrence )=  � 1) +  � 2) + ::: � )+ |{z} homogeneous part inhomogeneous part (0) = (1) =  1. roots x  n  = a1x  n � 1  + a2x  n � 2  ::: 2. Writeroot n nonrepeatedrootwherelater.root nn 2 nn  cr 1 r; c 2 nr; c 3 nr; :::; c k n  k � 1  r  where 1 ;:::;carelater. rk  3. recurrenceverify.degree,degreehigher,higher,)= )= )=  2 +bn+c. If g(n)is an exponential, such as 3 n )=  n  )=  n  + c3 n  and then an 2 3 n  + bn3 n  + c3 n , etc. 4. Here )=  n  + d( � 1) n  +3+ 1  |{z} homogeneous solution particular solution 5. (1) =2  2= 
2 1  + d
( � 1) 1  + 31+ 1 
) � 2 =2 � d  resulting