Math 210 Exam 4 - Practice Problem Solutions. 1. Answer the following questions based on the rooted tree shown below:

Transcription

1 Mt 0 Exm 4 - Prctce Proem Soutons. Answer te foowng questons se on te roote tree sown eow: c m n o p q r s t () Lst te cren of vertex. n,o,p () Lst te ncestors of vertex s m,,,, (c) Lst te sngs of vertex q r,s,t () Fn te numer of eves n ts roote tree = (e) Lst eve vertces n ts roote tree.,,,, (f) Fn te est m for wc ts tree s roote m-ry tree. m = 4 snce m s 4 cren n no oter vertex s more tn 4 cren. (g) Fn te egt of ts roote tree. = 5. () Fn te orer tt you wou vst te vertces of ts tree f you use postorer trvers to vst te vertces.,q,r,s,t,m,,,e,,n,o,p,,f,,,g,c, () Fn te orer tt you wou vst te vertces of ts tree f you use preorer trvers to vst te vertces.,,,,,m,q,r,s,t,e,c,f,,n,o,p,g,, () Fn te orer tt you wou vst te vertces of ts tree f you use norer trvers to vst te vertces.,,q,m,r,s,t,,,e,,n,,o,p,f,c,,g,. A cn etter strts wen person sens etter to 5 peope. Ec person wo sens te etter to 5 oter peope wo ve never receve t or oes not sen t to nyone. Suppose tt 0,000 peope sen out te etter efore te cn ens n tt no one receves more tn one etter. How mny peope receve te etter? How mny peope o not sen t out? Notce tt snce every person wo sens out te etter sens t to excty 5 oter peope, n no two peope receve te etter twce, ts stuton cn e moee usng fu roote 5-ry tree. Te root represents te person wo frst sens out te etter, n te cren of ny vertex represent te 5 peope tt te rete person sent etters. From ts, we see tt = 0,000. Te tot numer of peope wo receve te etter cn e foun y computng te numer tot numer of vertces n te roote tree. Usng Teorem 4, ts s n = m+ = 5(0,000)+ = 50,00. Ts counts te root, wo strte te etter ut not receve t, so 50,000 peope receve te etter. Notce tt eves represent peope wo not m te etter to 5 oter peope. Tus, gn usng Teorem 4, = (m )+ = 4(0,000)+ = 40,00, so 40,00 peope not sen t out fter tey receve t.

2 . Prove tt every tree on n vertces s n eges. We w procee y nucton on te numer of vertces n te grp. Bse Cse: Conser tree wt n = vertces. Notce tt suc tree must ve no eges, snce tree s smpe grp n ence s no oop eges. Ten te teorem s true wen n =. Inuctve Step: Suppose tt tree wt vertces s eges. Conser tree T wt + vertces. We cm tt tere must e t est one ege n T tt s pennt ege. To see ts, conser mxm pt n T. Snce tree s no crcuts n te grp T s fnte, te fn ege n ts pt must e pennt ege. Let v e te fn vertex n te egree sequence of ts pt. Let u e te unque vertex cent to v. If we remove remove vertex v ong wt te pennt ege {u,v}, we otn grp T. We cm ts T s tree. Notce tt snce te ony ege we remove ws pennt ege, n we so remove te vertex v, ten T s connecte. Aso note tt snce eges were remove n no eges new were e, tere re no smpe crcuts n T. Ten T s tree wt vertces. Hence, ppyng te nucton ypotess to T, ts tree must ve eges. But ten te orgn tree T s eges. Ts proves te teorem.. 4. Prove tt n m-ry tree of egt s t most m eves. We w procee usng strong nucton on te egt of te tree. Bse Cse: Suppose T s roote m-ry tree wt egt. Ten T conssts of root vertex n cren of tt root vertex. Snce T s m-ry, te root s t most m cren, so te grp s t most m eves. Inuctve Step: Suppose tt ny roote m-ry tree of egt < s t most m eves. Conser n m-ry tree of egt. As ove, te root vertes of T s t most m cren. Let T,T,...,T e te sutrees of T roote t te eve cren of te root vertex. Ten m, n ec T, m cn e tougt of s roote tree wt egt t most. Usng te nuctve ypotess on ec T, ec of tese sutrees s t most m eves. However, ec ef of T s ef of one of tese sutrees, so L, te numer of eves of T stsfes te nequty L m m m = m. Ts proves te teorem.. 5. () Drw nry serc tree for te sentence Now s te tme for goo men to come to te of ter country. now s te for men goo of tme to ter come country () How mny comprsons re neee to octe te wor tme n ts tree? (c) How mny comprsons re neee to te wor wffe to ts tree? 4 6. How mny wegngs re neee to fn counterfet cons mong 8 tot cons f te counterfet cou e eter ever or gter tn norm con. Gve n gortm tt proves your nswer. Te counterfet con cn e foun n tree wegngs. Frst, use te sce to compre cons n on one se gnst cons n 4 on te oter se. Tere re two cses: Cse : Suppose te two ses o not nce. Ten te counterfet must e one of tese four cons, n te remnng four re genune. Terefore, to fn te counterfet, we weg cons n on one se gnst cons 5 n 6 on te oter se.

3 Sucse : If cons n n cons 5 n 6 nce, ten te counterfet s eter or 4. To fns, we weg con gnst con 5 (wc we now s genune). If n 5 weg te sme, ten te counterfet s con 4. Oterwse, must e te counterfet. Sucse : If cons n n cons 5 n 6 o not nce, ten te counterfet s eter or. To fns, we weg con gnst con 5 (wc we now s genune). If n 5 weg te sme, ten te counterfet s con. Oterwse, must e te counterfet. Cse : Suppose te two ses o nce. Ten te counterfet must e one of te oter four cons, n te frst four re genune. Terefore, to fn te counterfet, we weg cons n on one se gnst cons 5 n 6 on te oter se. Sucse : If cons n n cons 5 n 6 nce, ten te counterfet s eter or 8. To fns, we weg con (wc we now s genune) gnst con. If n weg te sme, ten te counterfet s con 8. Oterwse, must e te counterfet. Sucse : If cons n n cons 5 n 6 o not nce, ten te counterfet s eter 5 or 6. To fns, we weg con (wc we now s genune) gnst con 5. If n 5 weg te sme, ten te counterfet s con 6. Oterwse, 5 must e te counterfet.. How mny wegngs re neee to fn two counterfet cons mong 5 tot cons, one ever tn norm con n te oter gter tn norm con. Gve n gortm tt proves your nswer. Note tt t s posse (tougt not necessry) tt te evy n gt con to comne to weg te sme s norm cons. *** I te gortm n ecson tree for ts ter f I ve tme *** 8. Conser te foowng tree: c m n o p q () Lst te vertces n te orer tt you wou vst tem f you trverse te tree n preorer.,,,e,,,n,o,f,c,g,,,m,p,q, () Lst te vertces n te orer tt you wou vst tem f you trverse te tree n norer.,,,e,n,,o,f,,g,c,,,p,m,q, (c) Lst te vertces n te orer tt you wou vst tem f you trverse te tree n postorer.,,n,o,,e,f,,g,,p,q,m,,,c,. Conser te expresson: ( 4 ) (5+(5 ) ) () Drw te roote tree representng ts computton.

4 _ + _ 5 * _ 4 5 () Wrte ts expresson n prefx notton (c) Wrte ts expresson n postfx notton Drw posse spnnng trees for te foowng grp:

5 [I tn tt ts s compete st f you see one I msse, et me now. Te frst person to pont out ec mssng spnnng treee gets n extr cret pont]

6 . Gven te foowng grp: () Use ept frst serc to fn spnnng tree for ts grp strtng t root vertex. e c g f () Use ret frst serc to fn spnnng tree for ts grp strtng t root vertex e e c f g. For te wegte grps gven eow: 8 0 c c e f 5 g e f 5 g () Use Prm s Agortm to fn mnmum spnnng tree for ec grp. 8 0 c c 5 0 e f 5 g 4 e f 5 g 4 4 For, te eges re e n te foowng orer: {f,g},{e,f},{e,},{,},{,},{,f},{,},{,c},{c,},{g,},{,} For, te eges re e n te foowng orer: {,g},{f,g},{,},{e,},{e,},{,},{g,},{,c},{c,},{c,},{,}

7 () Use Krus s Agortm to fn mnmum spnnng tree for ec grp. 8 0 c c 5 0 e f 5 g 4 e f 5 g 4 4 For, te eges re e n te foowng orer: {f,g},{c,},{e,f},{e,},{,},{,},{g,},{,f},{,c},{g,},{,} For, te eges re e n te foowng orer: {,g},{c,},{,e},{f,g},{c,},{e,f},{e,},{,},{g,},{,c},{,}. Descre n gortm to etermne weter or not smpe connecte grp s Hmton Crcut usng ept frst serces. Strt y puttng tot orer on te vertces of te smpe connecte grp ; {v,v,...,v n }. Next, egn crryng out ept frst serc for spnnng tree strtng t v. Wenever you ve coce of wc vertex to next, coose te vertex tt ppers erest on te orere st of vertces we cose ove. Wen you cn no onger ny more vertces to your nt pt, f you ve vste every vertex, cec to see f you cn return to v usng n ege tt s not rey een use. If you cn, te pt you foun s Hmton Pt, n te ege tng you c to v cn e e to te pt to form Hmton Crcut. Oterwse, f tere s vertex tt s not yet een vste or f te en vertex s not cent to v v n unuse ege, ten te current pt cnnot e extene to Hmton Crcut. Next, go c n crry out new ept frst serc strtng t v gn, ut te frst tme you ve coce etween more tn one vertex to to te pt, coose te erest vertex t tt step not cosen n prevous pont n te constructon. Repet unt you eter construct Hmton Crcut or unt posse wys of crryng out ept frst serc ve een exuste. Note: we cn e certn tt ts gortm w prouce Hmton Crcut wenever one exsts, snce f s Hmton Crcut, f we eete one of te eges n te crcut tt re cent to v, te resut s pt grp spnnng tree of egnnng t v.