CS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations


 Bennett Newton
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1 CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad hoc techique. However, oe very importat class of recurrece relatios ca be explicitly solved i a systematic way. These are the recurrece relatios that express the terms of a sequece as a liear combiatio of previous terms. Defiitio: A liear homogeeous recurrece relatio of degree k with costat coefficiets is a recurrece relatio of the form: a = c 1 a 1 + c a +L+ c k a k where c 1,c,c 3,K,c k are real umbers, ad c k 0. The recurrece relatio i the defiitio is liear sice the righthad side is a sum of multiple of the previous terms of the sequece. The recurrece relatio is homogeeous sice o terms of the recurrece relatio fail to ivolve a previous term of the sequece i some way. The coefficiets of the terms of the sequece are all costats, rather tha fuctios that deped o. The degree is k because a is expressed i terms of the previous k terms of the sequece. A cosequece of strog iductio is that a sequece satisfyig the recurrece relatio i the defiitio is uiquely determied by this recurrece relatio ad the k iitial coditio: a = c 1 a 1 + c a +L+ c k a k a 0 = C 0 a 1 = C 1 KK a k 1 = C k 1 Solvig the Little Mosters The basic approach for solvig liear homogeeous recurrece relatios is to look for solutios of the form a = r, where r is a costat. Note that a = r is a solutio to the recurrece if ad oly if: r = c 1 r 1 + c r + c 3 r 3 +L+ c k r k Whe both sides of the equatio are divided by r k ad the righthad side is subtracted from the left, we obtai the equivalet equatio: r k c 1 r k 1 c r k c 3 r k 3 L c k 1 r c k = 0
2 Cosequetly, the sequece with a = r is a solutio if ad oly if r is a solutio to this last equatio, which is called the characteristic equatio of the recurrece relatio. The solutios of this equatio are called characteristic roots of the recurrece relatio. As we ll see, these characteristic roots ca be used to give a explicit formula for all the solutios of the recurrece relatio. First, we should develop results that deal with liear homogeeous recurrece relatios with costat coefficiets of degree two. The results for secod order relatios ca be exteded to solve higherorder equatios. The mathematics ivolved i provig everythig is really messy, so eve though I ll refer to it i lecture, I do t wat to place it i here, because you might icorrectly thik you re resposible for kowig the proof, ad you re ot. Let s tur our udivided attetio to liear homogeeous recurrece relatios of degree two. First, cosider the case where there are two distict characteristic roots. A Fact Without Proof: Let c 1 ad c be real umbers. Suppose that r rc 1 c = 0 has two distict roots r 1 ad r. The the sequece { a } is a solutio of the recurrece relatio a = c 1 a 1 + c a if ad oly if a = b 1 r 1 + b r for = 0,1,,3,4,5,K where b 1 ad b are costats determied by the iitial coditios of the recurrece relatio. Problem: Fid a explicit solutio to the tilig recurrece relatio we developed last time. You may recall that it wet asomethig like this: T = T 1 + T T 0 = 1 T 1 = 1 Well, we surmise that the solutio will be a liear combiatio or terms havig the form T = r. Followig the rule from above, we just plug i T = r ito out recurrece relatio ad see what costraits are placed o r. Let s liste i: T = T 1 + T guess that T = r r = r 1 + r r = r + 1 r r 1= 0 r 1 = 1+ 5 ;r = 1 5 Well, so the guess that T = r is a good oe provided that r be equal to oe of the two roots above. I fact, ay liear combiatio of 1+ 5 ad iitial coditios for a momet that is, T = b 1 will also satisfy the recurrece if you igore the b for ay costat coefficiets as factors. It s oly whe you supply the iitial coditios that b 1 ad b are required to adopt a specific value. I fact, the iitial coditios here dictate that:
3 3 b b 1 + b = 1 + b 1 5 = 1 Ad after solvig for b 1 ad b do we fially arrive at the uique solutio to our recurrece problem: T = How very satisfyig. Problem: Fid a closed form solutio to the bit strig problem modeled i our last hadout. Welllllll, the recurrece relatio, save the iitial coditios, is exactly the same as it that for the tilig problem; that traslates to a solutio of the same basic form. But the fact that the iitial coditios are differet hits that the costats multiplyig the idividual terms: 1+ 5 B = b b ote same form, but possibly differet costats The differet iitial coditios place differet costrait o the values of b 1 ad b do we. Now the followig system of equatios must be solved: b b 1 + b = 1 + b 1 5 = We ed up with somethig like this as our solutio i this case: B = Problem: Solve the recurrece relatio give below J = J J 0 = 3J 1 = 5 You may be askig, Jerry, where i the world is the J 1 term? Relax it s i there it s just that its multiplyig coefficiet is zero, so I did t bother writig it i. It s still a homogeeous equatio of degree two, so I solve it like ay other recurrece of this type.
4 4 J = J J 0 = 3J 1 = 5 J = J r = 1 r 1 = 0 r 1 = 1;r = 1 J = b 1 ( 1) +b ( 1) =b 1 +b ( 1) Fuy little twist droppig the ( 1), sice it s trasparet to us ayway. b 1 + b = 5 b 1 b = 5 3 b 1 = 3,b = 5 3 J = ( 1) Solvig Coupled Recurrece Relatios The first recurreces hadout icluded examples where solutios ivolved coupled recurrece relatios. The circular Tower of Haoi Problem, the 3 tilig problem, the pillar problem, ad the Martia DNA problem were all complex eough to require (or at least beefit from) the ivetio of a secod coutig problem ad a secod recurrece variable. Remember the pillar recurrece? If ot, here it is oce more: 1 S = S 1 + S + 4T 1 = 0 = 1 0 T = 1 S 1 + T 1 = 0 = 1 Because S ad T are defied i terms of oe aother, it s possible to use the secod recurrece of the two to elimiate all occurreces or T from the first. It takes a algebra ad a little igeuity, but whe we rid of the secod variable, we are ofte left with a sigle recurrece relatio which ca be solved like other recurreces we ve see before. The above system of recurreces reduces to a sigle liear, homogeeous, costatcoefficiet equatio oce we get rid of T. Do t believe me? Read o, Thomas. Problem: Solve the above system of recurreces for S by elimiatig T ad solvig the liear equatio that results. I wat to completely elimiate all traces of T ad arrive at (what will tur out to be) a (cubic) recurrece relatio for just S, ad the solve it. First thigs first: We wat to get rid of the T 1 term from the recurrece relatio for S, ad to do that, I make the eat little observatio that the secod equality below follows from the first by replacig by 1: S = S 1 + S + 4T 1 S 1 = S + S 3 + 4T Subtractig the secod equatio from the first, we get:
5 5 ( ) ( S + S 3 + 4T ) S S 1 = S 1 + S + 4T 1 = S 1 S S 3 + 4T 1 4T S = 3S 1 S S 3 + 4T 1 4T ( ) = 3S 1 S S T 1 T Believe it or ot, this is progress, because I have somethig very iterestig to say about T 1 T it s always equal to S. Just look at the crafty maipulatios I work up from the T recurrece relatio: T = S 1 + T 1 T 1 = S + T T 1 T = S + T T = S How crafty! Now I ca eradicate ay metio of T from the S recurrece relatio, ad I do so likea this: S = 3S 1 S S 3 + 4( T 1 T ) = 3S 1 S S 3 + 4S = 3S 1 + 3S S 3 Therefore, a ucoupled recurrece relatio for S ca be expressed as follows: 1 S = S 1 + S 0 + 4T 1 = 9 3S 1 + 3S S 3 = 0 = 1 = 3 Admittedly, that was a lot of work, but this is pretty excitig, because we re o the verge of takig what is clearly a homogeeous, costatcoefficiet equatio ad comig up with a closed form solutio. As always, guess a solutio of the form S = a ad substitute to see what values of a make everythig work out regardless of the iitial coditios. More algebra: S S = a = 3S 1 + 3S S 3 S =a a = 3a 1 + 3a a 3 a 3 = 3a + 3a 1 0 = a 3 3a 3a + 1 ( ) ( )( a 3) 0 = ( a + 1) a 4a = ( a + 1) a + 3 a = 1, ± 3 That meas that the geeral solutio to our recurrece (igorig the iitial coditios) is ( ) + y( 3) + z 1 S = x + 3 require that: ( ), where x, y, ad z ca be ay real umbers. Boudary coditios
6 6 S 0 = x + y + z = 1 S 1 = ( + 3 )x + ( 3)y z = S = ( )x + ( 1 4 3)y + z = 9 This is a system of three liear equatios for three ukows, ad it admits exactly oe solutio. Solvig for x, y, ad z is a matter of algebra, ad after all that algebra is over, we coverge o a closed formula of S = ( ), which is ( ) ( 6 3 ) ( ) +1 rouded to the earest iteger. Neat! Problem: Revisit the Martia DNA problem, ad show that the umber of valid Martia DNA strads of legth is give as a + b = F 3 + (yes, the Fiboacci umber!) Let s brig back the recurrece that defied a ad b. 0 = 0 a = = 1 a 1 + b 1 1 = 0 b = 3 = 1 a 1 + 3b 1 Elimiate oe of the recurrece terms i order to get a closed solutio (though this time we ll ultimately eed to solve for both variables, because you re iterested i their sum.) a = a 1 + b 1 b = a 1 + 3b 1 Notice that the first oe tells us somethig about a a 1, so let s subtract a shifted versio of the secod equatio from the origial to get at somethig that s all b ad o a. b = a 1 + 3b 1 b 1 = a + 3b b b 1 = a 1 a ( ) + 3b 1 3b ( ) + 3b 1 3b b = b 1 + a 1 a The first equatio tells us that a 1 a = b. Substitutio yields b = b 1 + 4b + 3b 1 3b = 4b 1 + b The characteristic equatio here is r 4r 1 = 0 ; b = c 1 ( + 5 ) + c ( 5). Because b 0 = 1 ad b 1 = 3, we determie c 1, c by solvig the followig two equatios:
7 7 c 1 + c = 1 ( + 5)c 1 + ( 5)c = 3 c 1 = 5 + 5, c = 5 5. b = ( + 5) ( 5). You might thik you have to do the same thig for a, ad i theory you re right i that you eed to solve it, but we more or less have. Because b = a 1 + 3b 1 we actually kow that b = a 1 + 3b 1 a 1 = b 3b 1 a 1 = 1 b 3 b 1 a = 1 b +1 3 b a + b = 1 b +1 3 b + b = 1 b +1 1 b Recall that we re really just iterested i a + b, ad sice it ca be defied just i terms of the b, we get: a + b = 1 b +1 3 b + b = 1 b +1 1 b = 1 = 1 = 1 = = ( + 5) (( + 5) 1) ( + 5) ( 1 + 5) ( + 5) ( + 5) ( + 5) ( 5) (( 5) 1) 5 ( 5) ( 1 5) 5 ( 5) ( ) ( + 5) ( ) ( 5) That s typically the closedform you d leave it i, but we also wat to show that a + b = F 3 +. Here goes:
8 8 F 3 + = 1 5 = 1 5 = 1 5 = = = = ( + 5) ( + 5) ( + 5) ( + 5) ( + 5) Therefore, a + b = F 3 +, ad we rejoice. ( + 5) ( + 5) ( + 5) Problem: What about the Circular Tower of Haoi recurrece? Why ca t we solve that oe as easily? Well, you probably already oticed that both the pillar ad the DNA recurreces did t have ay ihomogeeous terms aywhere. That s ot the case with the Circular Tower of Haoi recurrece, so while we are certaily ivited to elimiate oe of the variables from the set of recurreces, there s ot much hope for solvig it like we did for Martia DNA. Q = 0; R if = 0 if > 0 R = 0; Q + Q if = 0 if > 0 Clearly, it s a piece of cake to defie Q i terms of previous Qs, but these 1s just are t goig to go away. so that Q = R ( ) + 1 = Q 1 + Q + 1 = Q 1 + Q Q = 1 Q + Q if = 0 if = 1 if > 1 We re sorta bummig because of that ihomogeeous 3. There is a trick to solvig this oe, but it s ot somethig I m formally goig to require you to kow, sice there s little pedagogical value i memorizig tricks. For those of you with othig to do o a Friday
9 9 ight, try substitutig Q = A 1 1 ito the above system (makig sure to adjust those base cases as well.) Solve for A, the take whatever you got there ad subtract 1 from it to get Q. Geeral Approach to Solvig all Liear FirstOrder Recurreces There is a geeral techique that ca reduce virtually ay recurrece of the form a T = b T 1 + c to a sum. The idea is to multiply both sides by a summatio factor, s, to arrive at s a T = s b T 1 + s c. This factor s is chose to make s b = s 1 a 1. The if we write S = s a T we have a sumrecurrece for S as: S = S 1 + s c Hece S = s 0 a 0 T 0 + s k c k = s 1b 1 T 0 + s k c k 1 k 1 T = s s a 1 b 1 T 0 + s k c k. 1 k 1 k ad the solutio to the origial recurrece becomes So, you ask: How ca we be clever eough to fid the perfect s? Well, the relatio s = s 1 a 1 be ufolded by repeated substitutio for the s i to tell us that the fractio: b ca s = a 1 a a 3 Ka 1 b b 1 b Kb (or ay coveiet costat multiple of this value) will be a suitable summatio factor. Problem: Solve V = 5 =0 V 1 + 3! 1 factor. by fidig the appropriate summatio Derivatio of the solutio just follows protocol. You're specifically told to use summatio factors here, so we should do that. Notice here that a = 1 ad b =, so that the summatio factor should be chose as: s = 1 1! = 1! V 0 = 5 for sure, but the recurrece formula V = V 1 + 3! becomes 1! V = 1 ( 1)! V Let T = 1! V for the time beig, just so we ca solve a easier recurrece relatio. We tackle T = T 1 + 3, ad would't you kow it: T = 3 + T 0 = ! V 0 = T = 1! V. We eed V =!T, so therefore we have that V =! ( 3 + 5). Woo!
10 0 =0 Problem: Solve C = summatio factor. C k 0 k 1 by fidig the appropriate Let s write out the recurreces for C ad C 1 a little more explicitly. C = C k 0 k 1 C 1 = + 1 C k 0 k Subtractig the secod oe from the first oe is a good idea, but oly after the multiply through by a factor that ll make each of the summatios equal to each other: C = C k 0 k 1 1 C 1 = = 1 + C k 0 k C k 0 k Subtractig the secod oe from the first, we arrive at: C 1 C 1 = C k 1+ 0 k 1 = + C 1 C = + 1 C 1 + C = ( + 1)C 1 + C k 0 k If we choose a summatio factor of s = a 1 a a 3 Ka 1 = ( 1) ( ) ( 3)K1 = b b 1 b Kb ( + 1)( 1)K3 through, we arrive at: + 1 ( ) C = ( + 1) ( ) C = C 1 + ( )C ( ) C + 1 = C ( + 1) + 1 ( ) ad multiply
11 If we let ( + 1)D = C, the we fially arrive at a equatio that looks reasoable: D = D Repeated substitutio yields the followig: D = D = D = D = D M 1 = D 0 + k k ( ) = H +1 1 = H = H +1 Recall that C = ( + 1)D, so that C = + 1 ( ) H = ( +1)H
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