Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!


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1 Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite: put the re verties in V 1 n the lk in V 2. Not iprtite! Consier the three verties olore re. For the ske of ontrition, ssume tht it is iprtite. Pik ny one of them to e in V 1. Tht woul fore the other two to e in V 2. But they re jent, whih is ontrition. () Hyperues re iprtite.
2 (i) The following is the 4ue: She in the verties tht hve n even numer of 0 s. Explin why the 4ue is iprtite. None of the she verties re pirwise jent. None of the nonshe verties re pirwise jent. So put ll the she verties in V 1 n ll the rest in V 2 to see tht Q 4 is iprtite. (ii) Explin why Q n is iprtite in generl. [Hint: onsier the prity of the numer of 0 s in the lel of vertex.] Any two verties with n even numer of 0 s iffer in t lest two its, n so re nonjent. Similrly, ny two verties with n o numer of 0 s iffer in t lest two its, n so re nonjent. So let V 1 = { verties with n even numer of 0 s } n V 2 = { verties with n o numer of 0 s }.
3 / Grphs In Exerises fin the jeny mtrix of the given irete multigrph with respet to the verties liste in lpheti orer. Exerise () For eh of the following pirs, list their egree sequenes. Then re they isomorphi?. If not, u 1 u 3 why? If yes, give n isomorphism. 35. u 2 v 1 (i) (ii) v 4 v u 3 u 2 u 1 v 4 eg seq: 2,2,2,1,1 eg seq: 2,2,2,2,2 Isomorphi: In Exerises rw the grph represente y the given jeny mtrix. u 1 u 1 u 2 Isomorphi: u 1 v 1, u 2, u 3 v 4, u 17 v 1, u 2 vu 3, u , v 3, Is every zero one squre mtrix tht is symmetri n (iii) hs zeros on the igonl the jeny mtrix of simple grph? u 1 u 26. Use n iniene 2 mtrix to represent the grphs in Exerises 1 n 2. v Use n iniene mtrix to represent the grphs in Exerises Wht is theusum 3 of the entries in row of the jeny mtrix for n unirete grph? Forv 4 irete grph? 29. Wht is the sum of the entries in olumn of the jeny mtrix for n unirete grph? For irete grph? Wht is the sum of the entries in row of the iniene Right eg seq: 4,3,3,2,2. mtrix for n unirete grph? 31. Wht is the sum of the entries in olumn of the iniene mtrix for n unirete grph? 32. Fin n jeny mtrix for eh of these grphs. ) K n ) C n ) W n ) K m,n e) Q n 33. Fin iniene mtries for the grphs in prts () () of (v) Exerise 32. In Exerises etermine whether the given pir of grphs u is isomorphi. Exhiit 2 v n isomorphism or provie 2 rigorous rgument tht none exists. u 1 u 3 v v 1 (iv) u 3 v u 2 v 1 u 3 v 4 v v 1 u 6 u1 u u 1 u 2 u u 1 v 1 Left eg seq: 3,3,3,3,2; Deg seq: 4,4,3,3,2 40. u 6 Isomorphi: Not isomorphi: ifferent eg seq s. u 3, u 2, u 2 v 6 u u 15 v 1, 3 v 4 v 4 u 1 u 2 u 6 u 3 v 7 v 6 v 4 v 1 v 4 v 1 v 6 u 1 u 2 u 3 4 v 4 v 4 Deg seq: 3,3,2,2,2,2 Not isomorphi: Right hs 3yle; Left oesn t.
4 () How mny isomorphism lsses re there for simple grphs with 4 verties? Drw them. () How mny eges oes grph hve if its egree sequene is 4, 3, 3, 2, 2? Drw grph with this egree sequene. Cn you rw simple grph with this sequene? By the hnshke lemm, So there re 7 eges. sequene: 2 E = = 14. Here is n isomorphism lss of simple grphs tht hs tht egree () For whih vlues of n, m re these grphs regulr? Wht is the egree? (i) K n (ii) C n (iii) W n (iv) Q n (v) K m,n (i) K n : Regulr for ll n, of egree n 1. (ii) C n : Regulr for ll n, of egree 2. (iii) W n : Regulr only for n = 3, of egree 3. (iv) Q n : Regulr for ll n, of egree n. (v) K m,n : Regulr for n = m, n. (e) How mny verties oes regulr grph of egree four with 10 eges hve? By the hnshke theorem, so V = = V 4 (f) Show tht every noninresing finite sequene of nonnegtive integers whose terms sum to n even numer is the egree sequene of grph (where loops re llowe). Illustrte your proof on the egree sequene 7,7,6,4,3,2,2,1,0,0. [Hint: A loops first.] For egree sequene 1, 2,..., n, rw one vertex v i for eh egree i, n tth i /2 loops tthe to v i. Then for eh i for whih i is even, v i so fr h egree i. For the i for whih i is o, v i urrently hs egree i 1. Sine the terms sum to n even numer, there must e n even numer of i for whih i is o; pir these i s up: the first with the seon, the thir with the fourth, n so on. Now rw n ege etween ll suh pire verties. The resulting grph hs the pproprite egree sequene. (g) Show tht isomorphism of simple grphs is n equivlene reltion.
5 () Reflexive: the ientity mp on verties is n isomorphism of grph to itself. () Symmetri: If f is n isomorphism f : G 1 G 2, then f : V 1 V 2 is ijetive, n therefore hs n inverse. Sine f preserves jeny, so oes f 1. So f 1 : G 2 G 1 is n isomorphism. () Trnsitive: If f : G 1 G 2 n g : G 2 G 3 re isomorphisms, then g f : G 1 G 3 is n isomorphism, sine the omposition of ijetive n egepreserving mps is ijetive n egepreserving. Exerise 30. () Consier the grph G = (i) Give n exmple of sugrph of G tht is not inue. H = (ii) How mny inue sugrphs oes G hve? List them. There re 4 verties, so there re 2 4 inue sugrphs: n (iii) How mny sugrphs oes G hve? A grph with m eges hs extly 2 m sugrphs with the sme vertex set. So, going through the inue sugrphs (the lrgest sugrph of G with eh possile vertex set), we get sugrphs of G in totl.
6 (iv) Let e e the ege onneting n. Drw G e n G/e., G e = G/e = (v) Let e e the ege onneting n. Drw G e n G/e., G e = G/e = (vi) Let e e n ege onneting n. Drw G + e. G = (vii) Drw Ḡ. G = () Show tht G = is isomorphi to its omplement. Sine Ḡ = the mp gives n isomorphism.,,,,
7 () Fin simple grph with 5 verties tht is isomorphi to its own omplement. (Strt with: how mny eges must it hve?) Sine there re 10 possile eges, G must hve 5 eges. One exmple tht will work is C 5 : G = = Ḡ = Exerise 31. () Drw the isomorphism lsses of onnete grphs on 4 verties, n give the vertex n ege onnetivity numer for eh. κ = 1 λ = 1 κ = 1 λ = 1 κ = 1 λ = 1 κ = 2 λ = 2 κ = 2 λ = 2 κ = 3 λ = 3 () Show tht if v is vertex of o egree, then there is pth from v to nother vertex of o egree. By the Hnshke Theorem, every grph hs n even numer of o egree verties. Notie tht eh onnete omponent is n inue sugrph with the sme egrees. So eh onnete omponent lso hs n even numer of o egree verties. So if onnete omponent hs n o egree vertex, it must hve two. So those two verties re onnete y wlk. () Prove tht for every simple grph, either G is onnete, or Ḡ is onnete. Suppose G is not onnete. Let H 1, H 2,..., H k e the onnete omponents of G (i.e. the sugrphs inue y eh set of verties etermine y the onnete omponents). First, onsier two verties in ifferent onnete omponents in G: u H i, v H j, i j. Sine u n v re in ifferent onnete omponents in G, they re ertinly not jent; thus u n v re jent in Ḡ, n therefore in the sme onnete omponent. Now onsier two verties in the sme onnete omponent in G: u, v H i. Sine there is more thn one onnete omponent in G, let w H j, i j. By our previous rgument, u n v re oth in the neighorhoo of w in Ḡ, n so u, w, v is pth in (G). Thus u n v re onnete in Ḡ. Thus Ḡ is onnete. () Rell tht κ(g) is the vertex onnetivity of G n λ(g) is the ege onnetivity of G. Give exmples of grphs for whih eh of the following re stisfie. Let δ = min v V eg(v).
8 (i) κ(g) = λ(g) < min v V eg(v) κ = λ = 1, δ = 2. (ii) κ(g) < λ(g) = min v V eg(v) κ = 1, λ = 2, δ = 2. (iii) κ(g) < λ(g) < min v V eg(v) κ = 1, λ = 2, δ = 3. (iv) κ(g) = λ(g) = min v V eg(v) Cyles of length 3 or more hve κ = λ = δ = 2. (e) For the following theorem, pik ny of prts (ii) (iv) n show (refully!) tht it s equivlent to prt (i). Theorem: For simple grph with t lest 3 verties, the following re equivlent. (i) G is onnete n ontins no ut vertex. (ii) Every two verties in V re ontine in some yle. (iii) Every two eges in E re ontine in some yle, n G ontins no isolte verties. (iv) For ny three verties u, v, w V, there is pth from u to v ontining w. (ii) = (i): If every two verties u n v re ontine in some yle, then there re two internlly isjoint u v pths. Thus u n v re onnetes, n the eletion of ny thir vertex will not isonnet u from v. Thus κ(g) 2. (i) = (ii): Assume κ(g) 2. Consier u, v V, n set = (u, v). If = 1, then u n v re jent. Sine V 3, there is some w V istint from u n v. Sine κ(g) 2, there is some u w pth P 1 not ontining v; similrly there is some w v pth P 2 not ontining u. Let x e the first vertex on P 1 whih is lso on P 2 (sine w is on oth pths, suh vertex exists). Then the wlk from u on P 1 to x, then on P 2 to v, n finlly k on the ege vu, is yle ontining u n v. If > 1, suppose tht for ny w V with (u, z) <, there is some yle ontining u n w. Consier miniml length u v pth, n let w e v s neighor on this pth, so tht (u, w) = 1. By our inutive hypothesis, there is yle C ontining u n w. Either C lso ontins v, in whih we lso hve yle ontining u n v, or it oesn t. If C oes not ontin v, then sine κ(g) 2, we lso hve pth P from v to u not ontining w. Let x e the first vertex on P
9 whih is lso on C (sine u is on P n C, x exists). u C x w v P Then the wlk from v long P to x, then long C wy from w n towr u (if x = u, wlk in either iretion), through u n roun to w, n finlly long the ege wv, is yle ontining u n v. (iii) = (i): Consier u, v V. Sine there re no isolte verties in G, eg(u), eg(v) 1, so u n v re eh inient to t lest one ege. Sine V 3, there re then t lest two eges. Sine every pir of eges re ontine in ommon yle, vertex inient to n ege must e inient to t lest two eges. Thus there re istint eges e f with e inient to u n f inient to v. Sine e n f re ontine in some yle, u n v re ontine in tht sme yle. So there re two internlly isjoint u v pths. Thus u n v re onnete, n the eletion of ny thir vertex will not isonnet u from v. Thus κ(g) 2. (i) = (iii): With G = (V, E, φ) n e, f E, efine = (e, f) = min (u, v). u φ(e) v φ(f) Now ssume κ(g) 2. Fix e, f E istint n let = (e, f). Choose u φ(e) n v φ(f) with (u, v) =, n let u n v e the other verties inient to e n f, respetively. First suppose = 0 (so tht e n f re inient to ommon vertex). Sine κ(g) 2, there is pth P from u to v not ontining u = v. Then the wlk from u long P to v, then k long f then e is yle ontining e n f. If > 0, then ssume tht for ll e g E for whih (e, g) < is ontine in yle together with e. Tke minimllength pth from u to v, n let w e v s neighor on P, so tht (u, w) = 1, n so (e, wv) = 1. By our inutive hypothesis, there is yle C ontining e n wv. If v is on C, then the wlk long v v, then long C wy from v n towr u, then on roun to v, is yle ontining e n v v. Otherwise, let P e pth from v to u not ontining v (whih exists sine κ(g) 2). Let x e the first vertex on P whih is lso on C. Then the wlk from v long v, then long P to x, then long C wy from v n towr u, through on to v, is yle ontining e n f. (iv) = (i): Let u, v V. For ny w V istint from u n v, ssuming (iv) gives t lest one u w pth ontining v. The initil wlk from u to v on this pth gives u v pth not ontining w. So u n v re onnete in G n in G w. Thus κ(g) 2. (i) = (iv): Assume κ(g) 2, let u, v, w V, n let = (u, w).
10 If = 1, let P e pth from w to v not ontining u (whih exists euse κ(g) 2). Then the wlk from u to w n then long P to v is pth from u to v ontining w. If > 1, ssume tht for every x V with (u, x) <, there is u v pth through x. Tke minimllength pth from u to w, n let x e w s neighor on this pth, so tht (u, x) = 1. Then our inutive hypothesis gurntees u v pth P ontining x. If P lso ontins w, then P is u v pth ontining w s esire. Otherwise, let P e pth from w to u not ontining x. Let z e the first vertex on P whih is lso on P (whih is gurntee sine u is on oth P n P ). If z sits etween x n v on P, then the wlk from u long P to x, then long xw, then long P to z, then ontinuing long P to v, is u v pth ontining w. Otherwise, z sits etween u n x on P, so tht the wlk from u long P to z, the kwrs long P towr w, up wx, n then long P towr v is u v pth through w.
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