MAT3707. Tutorial letter 201/1/2017 DISCRETE MATHEMATICS: COMBINATORICS. Semester 1. Department of Mathematical Sciences MAT3707/201/1/2017

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1 MAT3707/201/1/2017 Tutoril lttr 201/1/2017 DISCRETE MATHEMATICS: COMBINATORICS MAT3707 Smstr 1 Dprtmnt o Mtmtil Sins SOLUTIONS TO ASSIGNMENT 01 BARCODE Din tomorrow. univrsity o sout ri

2 SOLUTIONS TO ASSIGNMENT 01 (SEMESTER 1) CLOSING DATE: 10 Mr 2017 UNIQUE NR.: T nswrs to only som o t qustions will mrk. 1. Drw () tr non-isomorpi rps wit 4 vrtis n 3 s. (3) T tr rps r non-isomorpi us t irst rp s two vrtis o r on, t son rp s no vrtis o r on n t tir rp s tr vrtis o r on. () two non-isomorpi rps wit 5 vrtis n 6 s. (2) T two rps r non-isomorpi us t son rp s vrtx o r our wil t irst rp os not v su vrtx. Explin in s wy your rps r non-isomorpi. 2. T list o rs o t vrtis o rp, rrn in non-rsin orr, is ll t r squn o t rp. () Drw rp wit r squn 1, 1, 1, 5 or sy wy no su rp xists. (2) Su rp os not xist us t vrtx o r iv s to v iv niours wil t rp only s our vrtis. () Drw rp wit r squn 1, 1, 1, 1, 1, 5 or sy wy no su rp xists. (2) 2

3 MAT3707/201/1/2017 () Drw two non-isomorpi rps wit r squn 1, 1, 1, 2, 2, 3. (4) n () Prov tt up to isomorpism, tr is only on rp wit r squn 2, 3, 3, 4, 5, 5. (5) W iv onstrutiv proo s ollows. T rp s six vrtis, two o wi v r iv. Hn ts two vrtis v to jnt to otr n lso to t otr our rminin vrtis: I w lt ts two vrtis o r 5 rom t rp (n tir inint s), wt rmins is rp on our vrtis wit r squn 0, 1, 1, 2. (T rminin our vrtis in t oriinl rp rs 2, 3, 3, n 4 n i w lt t two s twn o ts vrtis n t two vrtis o r 5, t rs o vrtx rss y two.) Tr is only on su rp, nmly rp onsistin o n isolt vrtx totr wit pt o orr 3: Tis mns tt t only possil wy to s to t irst rp is y in two s su tt ty sr niour: 3. () Lt G rp wit vrtx st V (G) n st E(G). Din t omplmnt G o G. (2) Lt t vrtx st n st o G ivn y V (G) n E(G) rsptivly. Tn V (G) = V (G) n E(G) i n only i / E(G). () Giv n xmpl o rp n its omplmnt on our vrtis. (2) n 3

4 () Drw sl-omplmntry rp (i.. rp wi is isomorpi to its omplmnt) wit i. our vrtis (3) n T two rps r isomorpi sin ty r ot pts o orr our. ii. iv vrtis (3) n T two rps r isomorpi sin ty r ot iruits on iv vrtis. Explin in s wy your rp is sl-omplmntry. 4. Prov tt i rp is not onnt, tn its omplmnt must onnt. (6) Suppos tt G is isonnt rp. Lt x n y ny two vrtis in G. W will now prov tt tr xists n x y pt in G, t omplmnt o G. W n to onsir t ollowin two ss. First suppos tt x n y li in t sm omponnt o G. Sin G is isonnt, tr xists notr omponnt tt ontins t lst on vrtx, sy v. Sin xv / E(G) n yv / E(G) it ollows y t inition o G tt xv E(G) n yv E(G). Hn xvy is n x y pt in G. Now suppos tt x n y li in two irnt omponnts o G. Tn xy / E(G) n in y t inition o o G, xy E(G). Hn xy is n x y pt in G. Sin x n y wr osn ritrrily w v sown tt tr is pt twn ny two vrtis o G n tror G is onnt. 5. Lt G rp wit itn s, n wit ll vrtis o t sm r. Dtrmin ll t possil vlus or t numr o vrtis o G, n iv n xmpl o su rp or possil vlu. Suppos tr r n vrtis, o r r. Lt (v) not t r o t vrtx v n m t numr o s o G. Sin (v) = 2m n m = 15, it ollows tt n tus v V (G) rn = 30 n = 30 r. (5) 4

5 MAT3707/201/1/2017 So, t ivisors o 30 r {1, 2, 3, 5, 6, 10, 15, 30}. I r = 1, tn tr r 30 vrtis, r = 2, tn tr r 15 vrtis, r = 3, tn tr r 10 vrtis, r = 5, tn tr r 6 vrtis. T nxt ivisor o 30 is 6 - ut oviously tr is no rp wit iv vrtis, o r 6. T sm is tru i r {10, 15, 30}. Nxt w iv n xmpl o su rp or o r {1, 2, 3, 5}. I r = 1 n n = 30, tn 15K 2 is su rp: I r = 2 n n = 15, t rp onsistin o tr iruits o orr 5 is su rp (3C 5 ): For r = 3 n n = 10 t ollowin is n xmpl: Finlly, i r = 5 n n = 6, t only possiility is t omplt rp on 6 vrtis, K 6 : 5

6 6. () Wt is t smllst numr o vrtis tt iprtit plnr rp wit 18 s n v? Giv n xmpl o iprtit plnr rp ttinin tis numr o vrtis. (4) Lt v t numr o vrtis o su rp. From Eulr s Corollry in t stuy ui or plnr iprtit rps it ollows tt 18 2v 4. Hn v 11. So w n t lst 11 vrtis. To k tt 11 vrtis is possil w onstrut t ollowin iprtit plnr rp wit 11 vrtis n 18 s: () I onnt plnr rp wit n vrtis, ll o r 3 s 6 rions, trmin n. Skt n xmpl o su onnt plnr rp. (4) Lt not t numr o s. Eulr s ormul ivs T sum o t rs is 3n, so w lso v Sustitut k in t irst qution: 6 = n n = 2 i.. = 3n/2. 6 = 3n/2 n + 2 i.. n = 8. Hr is n xmpl o onnt plnr rp wit 8 vrtis, ll o r 3 wit 6 rions. T rions r numr rom 1 to 6. Not tt t outsi is lso rion!

7 MAT3707/201/1/ Dtrmin wtr t ollowin rp s i j k () n Eulr tril (2) : Sin t rp s xtly two vrtis o o r (vrtis n o r 3) t rp ontins n Eulr tril. () n Eulr yl (2) : Sin not vry vrtx is o vn r, t rp os not ontin n Eulr yl. () Hmilton pt (2) : Tr r mny Hmilton pts; or xmpl ijk. () Hmilton iruit (2) : Tr r mny Hmilton iruits; or xmpl ijk. 8. For o t ollowin pirs o rps, trmin wtr ty r isomorpi or not. I ty r, iv vrtx orrsponn. I ty r not, iv rson wy. () (5) 7

8 () : T two rps r not isomorpi. T irst rp ontins 5-iruit nmly wil t son rp is iprtit : (5) T two rps v two vrtis o r 2 n o tm r jnt to two vrtis o r our. Witout loss o nrlity w my ssum tt 1 n 6. T our niours o ts two vrtis in rp inu K 4. Hn t two rps r isomorpi. T ollowin is vrtx orrsponn twn t vrtis o t two rps: Prov tt no Hmilton iruit in t rp low xists. i j k l m (6) 8

9 MAT3707/201/1/2017 : By rul 1 o Tukr w v to us t two s n inint wit n t two s k n kl inint wit k. Now onsir t vrtx. W nnot us ot t s n inint wit otrwis w will t propr suiruit violtin rul 2. By t symmtry o t rp w n lt n tn us t s n (rul 1). At w tn v to lt t s n (rul 3). Now w r or to us t s n inint wit (rul 1). By rul 3 w v to lt t s n i inint wit. But now vrtx is o r 1 n tror nnot li on ny Hmilton iruit. Hn t rp os not ontin Hmilton iruit. 10. Us t irl-or mto to trmin wtr t ollowin rp is plnr. I it is, iv plnr rwin. I it is not, in K 3,3 - or K 5 - oniurtion. : (8) W us t Hmilton iruit. By insi-outsi symmtry w oos t to rwn insi. Tn t s to rwn outsi. Now t nnot rwn insi or outsi. Hn t rp is not plnr: 9

10 T ollowin is K 3,3 oniurtion o t rp, 11. Dtrmin t romti numr o o t ollowin rps. Giv miniml olourin n proo tt smllr numr o olours nnot us. () () (5) : Sin t rp s s, χ(g) 2. Sin t rp is iprtit, χ(g) 2. Hn χ(g) = 2. T ollowin is 2-olourin o t vrtis o G lu r,,,,,, j i (5) : Sin t rp ontins n o iruit (.. ), χ(g) 3. T ollowin is 3-olourin o t vrtis o G: lu r rn,, i, j,,,, 10

11 MAT3707/201/1/2017 () i j k (5) : Sin t rp ontins trinls (.. k), χ(g) 3. W will now try to or 3-olourin o t vrtis o G. So lt us olour k r, rn n lu. Tn j is jnt to rn n r vrtx n s to olour lu. Now i s to rn. T vrtx is jnt to lu n rn vrtx n is or to olour r. Now s to lu n s to olour rn. Tis ors to r. But now is jnt to r (), lu () n rn () vrtx n nnot olour. Hn tr olours r not nou to olour t vrtis o t rp. I w olour vrtx wit ourt olour, sy yllow, tn w n olour t vrtis o t rp in our olours: lu r rn yllow,,, j,, k,, i Tis sows tt χ(g) 4 n n χ(g) = Lt G onnt rp wit 30 s. Dtrmin t mximum possil numr o vrtis o G. Lt v t numr o vrtis o G. Sin G is onnt, G s spnnin tr T. Lt t numr o s o T. Tn 30. Also, = v 1 (Torm 2 in Stion 3.1 o Tukr). Tror v 1 30 n so v 31. Tis mns tt G nnot v mor tn 31 vrtis. W still v to sow tt 31 vrtis is possil. For tis w just lt G ny tr wit 31 vrtis (n tror 30 s). Tis sows tt t mximum quls Lt T tr (not root) on 10 vrtis wit vrtx o r itr 1 or 3. () How mny vrtis o T v r 3? (3) Consir ny tr T wit 10 vrtis, o r 1 or 3. Lt t t numr o vrtis o r 3. Tn t sum o t rs qul 3t + 1 (10 t) = 2t (4) 11

12 T numr o s o T is 10 1 = 9, n it ollows y t ountin torm tt t sum o t rs lso qul 2 9 = 18, i.. 2t + 10 = 18, n t = 4. So w v tt tr r 4 vrtis o r 3, n 6 o r 1. () I w rmov ll t vrtis o r 1, will t rp tt rmins still tr? Explin. (2) I w rmov ll t vrtis o r 1 n tir inint s, t rminin rp will still onnt. So wt is lt is still tr, ut now wit 4 vrtis. () List ll su non-isomorpi trs (not root) on 10 vrtis wit vrtx o r itr 1 or 3 n xplin wy t trs in your list r non-isomorpi. (3) Tr r two: n Consir t surps o t ov two rps orm y t vrtis o r 3. In t irst rp w otin n in t son rp Ty r lrly not isomorpi. Hn nitr r t oriinl rps. () Us () n () ov to prov tt your list ontins ll su trs. (3) From () w know tt i w rmov ll t vrtis o r 1 n tir inint s, w r still lt wit tr. From () w know tt tis tr s 4 vrtis. It is sy to list ll trs wit 4 vrtis: n 12

13 MAT3707/201/1/2017 In s tr is only on wy o puttin k t vrtis o r 1: n So tr r only ts two, n t list in () is omplt. 14. Dtrmin t numr o s o () inry tr on 67 vrtis. (2) A inry tr on 67 vrtis s 66 s y Torm 2 o Stion 3.1 o Cptr 3. () trnry tr on 67 vrtis. (2) Ain t nswr is 66, or t sm rson. 15. For o t ollowin, trmin i it xists. I it os, iv n xmpl. I it osn t iv rson wy. () A 4-ry tr wit 11 lvs. (2) In t nottion o Stion 3.1 o Tukr w v i = l 1 m 1 = = 10 3, wi is not n intr. Tror, su tr os not xist. () A 4-ry tr wit 13 lvs. (2) Now w v i = l 1 m 1 = = 12 = 4 intrnl vrtis. 3 Tis osn t yt tll us tt it xists; w v to try n onstrut on: Tror su tr xists. 13

14 16. In t rp low, trmin i j () pt-irst spnnin tr wit s root. (4) i j () rt-irst spnnin tr wit s root. (4) i j 14

15 MAT3707/201/1/ Fin miniml spnnin tr o t wit rp low, usin 2 1 i j 1 k m n o p l () Kruskl s loritm (4) () Prim s loritm. (4) In spnnin tr sow t orr in wi t s v n slt y numrin tm orinly (i.. o not iv t wits). () Kruskl s loritm: W strt wit n mpty T, n t stp to T sortst tt os not orm iruit wit s lry in T. (Not tt tr n mor tn on su sortst. You n oos ny on o tm. So it is possil to otin irnt miniml spnnin trs pnin on your oi, ut ty will ll v t sm totl wit.) In t spnnin tr low w sow t orr in wi t s v n slt (not t wits) j i k l m n o p 13 T totl ost is

16 () Prim s loritm: W strt wit T onsistin o ny sortst, n t stp, to T sortst twn T n vrtx not in T. (Ain tr is mor tn on oi. Coos nyon. ) In t spnnin tr low w sow t orr in wi t s v n slt (not t wits) j i k l m n o p T totl ost is