Documents de Travail du Centre d Economie de la Sorbonne
|
|
- Dominic Parks
- 6 years ago
- Views:
Transcription
1 Documents de Traval du Centre d Econome de la Sorbonne Detectng nduced subgraphs Benamn LEVEQUE, Davd Y. LIN, Frédérc MAFFRAY, Ncolas TROTIGNON Mason des Scences Économques, boulevard de L'Hôptal, Pars Cedex 13 ISSN : X
2 Detectng nduced subgraphs Benamn Lévêque, Davd Y. Ln, Frédérc Maffray, Ncolas Trotgnon October 9, 27 Abstract An s-graph s a graph wth two nds of edges: subdvsble edges and real edges. A realsaton of an s-graph B s any graph obtaned by subdvdng subdvsble edges of B nto paths of arbtrary length (at least one). Gven an s-graph B, we study the decson problem Π B whose nstance s a graph G and queston s Does G contan a realsaton of B as an nduced subgraph?. For several B s, the complexty of Π B s nown and here we gve the complexty for several more. Our NP-completeness proofs for Π B s rely on the NP-completeness proof of the followng problem. Let S be a set of graphs and d be an nteger. Let Γ d S be the problem whose nstance s (G, x, y) where G s a graph whose maxmum degree s at most d, wth no nduced subgraph n S and x, y V (G) are two non-adacent vertces of degree 2. The queston s Does G contan an nduced cycle passng through x, y?. Among several results, we prove that Γ 3 s NPcomplete. We gve a smple crteron on a connected graph H to decde whether Γ + {H} s polynomal or NP-complete. On the other hand, the polynomal cases rely on the algorthm three-n-a-tree, due to Chudnovsy and Seymour. AMS Mathematcs Subect Classfcaton: 5C85, 68R1, 68W5, 9C35 Key words: detectng, nduced, subgraphs 1 Introducton In ths paper graphs are smple and fnte. A subdvsble graph (s-graph for short) s a trple B = (V, D, F) such that (V, D F) s a graph and D F =. The edges n D are sad to be real edges of B whle the edges n F are sad to be subdvsble edges of B. A realsaton of B s a graph obtaned from B by subdvsng edges of F nto paths of arbtrary length (at least one). The problem Laboratore G-SCOP, 46 Avenue Félx Vallet, 3831 Grenoble Cedex, France. (benamn.leveque@g-scop.npg.fr, Frederc.Maffray@g-scop.npg.fr) Prnceton Unversty, Prnceton, NJ, dyln@prnceton.edu Unversté Pars I, Centre d Économe de la Sorbonne, boulevard de l Hôptal, Pars cedex 13, France. ncolas.trotgnon@unv-pars1.fr Ths wor has been partally supported by ADONET networ, a Mare Cure tranng networ of the European Communty. 1
3 T H Fgure 1: s-graphs yeldng trvally polynomal problems B 1 B 2 B 3 Fgure 2: Pyramds, prsms and thetas Π B s the decson problem whose nput s a graph G and whose queston s Does G contan a realsaton of B as an nduced subgraph?. On fgures, we depct real edges of an s-graph wth straght lnes, and subdvsble edges wth dashed lnes. Several nterestng nstance of Π B are studed n the ltterature. For some of them, the exstence of a polynomal tme algorthm s trval, but efforts are devoted toward optmzed algorthms. For example, Alon, Yuster and Zwc [2] solve Π T n tme O(m 1.41 ) (nstead of the obvous O(n 3 ) algorthm), where T s the s-graph depcted on Fgure 1. Ths problem s nown as trangle detecton. Taran and Yannaas [1] solve Π H n tme O(n + m) where H s the s-graph depcted on Fgure 1. For some Π B s, the exstence of a polynomal tme algorthm s non-trval. A pyramd (resp. prsm, theta) s any realsaton of the s-graph B 1 (resp. B 2, B 3 ) depcted on fgure 2. Chudnovsy and Seymour [5] gave an O(n 9 )-tme algorthm for Π B1 (or equvalently, for detectng a pyramd). As far as we now, that s the frst example of a soluton to a Π B whose complexty s non-trval to settle. In contrast, Maffray and Trotgnon [8] proved that Π B2 (or detectng a prsm) s NP-complete. Chudnovsy and Seymour [4] gave an O(n 11 )-tme algorthm for P B3 (or detectng a theta). Ther algorthm reles on the soluton of a problem called three-n-a-tree, that we wll defne precsely and use n Secton 2. The three-n-tree algorthm s qute general snce t can be used to solve a lot of Π B problems, ncludng the detecton of pyramds. These facts are a motvaton for a systematc study of Π B. A further motvaton s that very smlar s-graphs can lead to a drastcally dfferent complexty. The followng example may be more strng than pyramd/prsm/theta : Π B4, Π B6 are polynomal and Π B5, Π B7 are NP-complete, where B 4,...,B 7 are the s-graphs depcted on fgure 3. Ths wll be proved n secton 3.1. Notaton and remars By C ( 3) we denote the cycle on vertces, by K l (l 1) the clque on l vertces. We denote by I l (l 1) the tree on l + 5 vertces obtaned by 2
4 B 4 B 5 B 6 B 7 Fgure 3: Some s-graphs wth pendng edges Fgure 4: I 1 tang a path of length l wth ends a, b, and addng four vertces, two of them adacent to a, the other two to b; see Fgure 4. When a graph G contans a graph somorphc to H as an nduced subgraph, we wll often say G contan an H. Let (V, D, F) be an s-graph. Suppose that (V, D F) has a vertex of degree one ncdent to an edge e. Then Π (V,D {e},f \{e}) and Π (V,D\{e},F {e}) have the same complexty, because a graph G contans a realsaton of (V, D {e}, F \ {e}) f and only f t contans a realsaton of (V, D \ {e}, F {e}). For the same reason, f (V, D F) has a vertex of degree two ncdent to the edges e f then Π (V,D\{e} {f},f \{f} {e}), Π (V,D\{f} {e},f \{e} {f}) and Π (V,D\{e,f},F {e,f}) have the same complexty. If F 1 then Π (V,D,F) s clearly polynomal. Thus n the rest of the paper, we wll consder only s-graphs (V, D, F) such that: F 2; no vertex of degree one s ncdent to an edge of F; every nduced path of (V, D F) wth all nteror vertces of degree 2 and whose ends have degree 2 has at most one edge n F. Moreover, ths edge s ncdent to an end of the path; every nduced cycle wth at most one vertex v of degree at least 3 n (V, D F) has at most one edge n F and ths edge s ncdent to v f v exsts (f t does not then the cycle s a component of (V, D F)). 2 Detecton of holes wth prescrbed vertces Let (G) be the maxmum degree of G. Let S be a set of graphs and d be an nteger. Let Γ d S be the problem whose nstance s (G, x, y) where G s a graph such that (G) d, wth no nduced subgraph n S and x, y V (G) are two non-adacent vertces of degree 2. The queston s Does G contan a hole passng through x, y?. For smplcty, we wrte Γ S nstead of Γ + S (so, the graph n the nstance of Γ S has unbounded degree). Also we wrte Γ d nstead 3
5 of Γ d (so the graph n the nstance of Γd has no restrcton on ts nduced subgraphs). Benstoc [3] proved that Γ = Γ s NP-complete. For S = {K 3 } and S = {K 1,4 }, Γ S can be shown to be NP-complete, and a consequence s the NP-completeness of several problems of nterest: see [8] and [9]. In ths secton, we try to settle Γ d S for as many S s and d s as we can. In partcular, we gve the complexty of Γ S when S contans only one connected graph and of Γ d for all d. We also settle Γ d S for some cases when S s a set of cycles. The polynomal cases are ether trval, or are a drect consequence of an algorthm of Chudnovsy and Seymour. The NP-complete cases follow from several extensons of Benstoc s constructon. Polynomal cases Chudnovsy and Seymour [4] proved that the problem whose nstance s a graph and three vertces a, b, c, whose queston s Does the graph contans a tree passng through a, b, c as an nduced subgraph? can be solved n O(n 4 ). We call ths algorthm three-n-a-tree. Three-n-a-tree can be used drectly to solve Γ S for several S s. Let us call subdvded claw any tree wth one vertex u of degree 3, three vertces v 1, v 2, v 3 of degree 1 and all the other vertces of degree 2. Theorem 2.1 Let H be a graph on vertces that s ether a path or a subdvded claw. There s an O(n )-tme algorthm for Γ {H}. proof Here s an algorthm for Γ {H}. Let (G, x, y) be an nstance of Γ H. If H s a path on vertces then every hole n G s on at most vertces. Hence, by a brute-force search on every -tuple, we wll fnd a hole through x, y f there s any. Now we suppose that H s a subdvded claw. So 4. For convenence, we put x 1 = x, y 1 = y. Let x, x 2 (resp. y, y 2 ) be the two neghbors of x 1 (resp. y 1 ). Frst chec whether there s n G a hole C through x 1, y 1 such that the dstance between x 1 and y 1 n C s at most 2. If = 4 or = 5 then the vertex-set of any such hole must be ncluded n {x, x 1, x 2, y, y 1, y 2 }, so t can be found n constant tme. Now suppose 6. For every l-tuple (x 3,...,x l+2 ) of vertces of G, wth l 5, test whether P = x x 1 x l+2 y 2 y 1 y s an nduced path, and f so delete the nteror vertces of P and ther neghbors except x, y, and loo for a shortest path from x to y. Ths wll fnd the desred hole f there s one, after possbly swappng x, x 2 and dong the wor agan. Ths taes tme O(n 3 ). Now we assume that n every hole through x 1, y 1, the dstance between x 1, y 1 s at least 1. Let be the length of the unque path of H from u to v, = 1, 2, 3. Note that = Let us chec every ( 4)-tuple z = (x 3,...,x 1+1, y 3,..., y 2+ 3 ) of vertces of G. For such a ( 4)-tuple, test whether x x 1 x 1+1 and P = y y 1 y 2+ 3 are nduced paths of G wth no edge between them except possbly x 1+1y If not, go to the next ( 4)-tuple, but f yes, delete the nteror vertces of P and ther neghbors except y, y Also delete the neghbors of x 2,...,x 1, except x 1, x 2,..., x 1, x 1+1. Call G z the resultng graph and run three-n-a-tree n 4
6 G z for the vertces x 1, y 2+ 3, y. We clam that the answer to three-n-a-tree s YES for some ( 4)-tuple f and only f G contans a hole through x 1, y 1 (after possbly swappng x, x 2 and dong the wor agan). If G contans a hole C through x 1, y 1 then up to a symmetry ths hole vsts x, x 1, x 2, y 2, y 1, y n ths order. Let us name x 3,...,x 1+1 the vertces of C that follow after x 1, x 2, and let us name y 3,..., y 2+ 3 those that follow after y 1, y 2. Note that all these vertces exst and are parwse dstnct snce n every hole through x 1, y 1 the dstance between x 1, y 1 s at least 1. So the path from y to y 2+ 3 n C \ y 1 s a tree of G z passng through x 1, y 2+ 3, y, where z s the ( 4)-tuple (x 3,..., x 1+1, y 3,..., y 2+ 3 ). Conversly, suppose that G z contans a tree T passng through x 1, y 2+ 3, y, for some ( 4)-tuple z. We suppose that T s vertex-ncluson-wse mnmal. If T s a path vstng y, x 1, y 2+ 3 n ths order, then we obtan the desred hole of G by addng y 1, y 2,..., y to T. If T s a path vstng x 1, y, y 2+ 3 n ths order, then we denote by y the neghbor of y 2+ 3 along T. Note that T contans ether x or x 2. If T contans x, then there are three paths n G: y T x x 1 x 1, y T y y 3+2 and y y 1 y 3. These three paths form a subdvded claw centered at y that s long enough to contan an nduced subgraph somorphc to H, a contradcton. If T contans x 2 then the proof wors smlarly wth y T x 1+1 x 1 x 1 nstead of y T x x 1 x 1. If T s a path vstng x 1, y 2+ 3, y n ths order, the proof s smlar, except that we fnd a subdvded claw centered at y If T s not a path, then t s a subdvded claw centered at a vertex u of G. We obtan agan an nduced subgraph of G somorphc to H by addng to T suffcently many vertces of {x,...x 1+1, y,..., y 2+ 3 }. NP-complete cases (unbounded degree) Many NP-completeness results can be proved by adaptng Benstoc s constructon. We gve here several polynomal reductons from the problem 3- Satsfablty of Boolean functons. These results are gven n a framewor that nvolves a few parameters, so that our result can possbly be used for other problems of the same type. Recall that a Boolean functon wth n varables s a mappng f from {, 1} n to {, 1}. A Boolean vector ξ {, 1} n s a truth assgnment for f f f(ξ) = 1. For any Boolean varable z on {, 1}, we wrte z := 1 z, and each of z, z s called a lteral. An nstance of 3-Satsfablty s a Boolean functon f gven as a product of clauses, each clause beng the Boolean sum of three lterals; the queston s whether f admts a truth assgnment. The NP-completeness of 3-Satsfablty s a fundamental result n complexty theory, see [6]. Let f be an nstance of 3-Satsfablty, consstng of m clauses C 1,..., C m on n varables z 1,..., z n. For every nteger 3 and paramaters α {1, 2}, β {, 1}, γ {, 1}, δ {, 1, 2, 3}, ε {, 1}, ζ {, 1} such that f α = 2 then ε = β = γ, let us buld a graph G f (, α, β, γ, δ, ε, ζ) wth two specalzed vertces x, y of degree 2. There wll be a hole contanng x and y n G f (, α, β, γ, δ, ε, ζ) f and only f there exsts a truth assgnment for f. In G f (, α, β, γ, δ, ε, ζ) (we wll sometmes wrte G f for short), there wll be two 5
7 nds of edges: blue and red. The reason for ths dstncton wll appear later. For each varable z ( = 1,...,n), prepare a graph G(z ) wth 4 vertces a,r, b,r, a,r, b,r, r {1,..., } and 4(m + 2)2 vertces t,2p+r, f,2p+r, t,2p+r, f,2p+r, p {,..., m + 1}, r {,..., 2 1}. Add blue edges so that the four sets {a,1,... a,, t,,..., t,2(m+2) 1, b,1,...,b, }, {a,1,...a,, f,,...,f,2(m+2) 1, b,1,...,b, }, {a,1,... a,, t,,..., t,2(m+2) 1, b,1,...,b, }, {a,1,...a,, f,,..., f,2(m+2) 1, b,1,..., b, } all nduce paths (and the vertces appear n ths order along these paths). See Fgure 5. Add red edges accordng to the value of α, β, γ, as follows. If α = 1 then, for every p = 1,..., m + 1, add all edges between {t,2p, t,2p+β } and {f,2p, f,2p+γ }, between {f,2p, f,2p+γ } and {t,2p, t,2p+β }, between {t,2p, t,2p+β } and {f,2p, f,2p+γ }, between {f,2p, f,2p+γ } and {t,2p, t,2p+β }. If α = 2 then, for every p = 1,..., m, add all edges between {t,2p+ 1, t,2p+ 1+β } and {f,2p+ 1, f,2p+ 1+γ } ; for every p = 1,...,m + 1, add all edges between {f,2p+ 1, f,2p+ 1+γ } and {t,2p, t,2p+β }, between {t,2p, t,2p+β } and {f,2p, f,2p+γ }, between {f,2p, f,2p+γ } and {t,2(p 1)+ 1, t,2(p 1)+ 1+β }. See Fgures 6, 7. For each clause C ( = 1,...,m), wth C = y 1 y2 y3, where each y q (q = 1, 2, 3) s a lteral from {z 1,..., z n, z 1,..., z n }, prepare a graph G(C ) wth 2 vertces c,p, d,p, p {1,..., } and 6 vertces u q,p, q {1, 2, 3}, p {1,...,2}. Add blue edges so that the three sets {c,1,..., c,, u q,1,..., uq,2, d,1,...d, }, q {1, 2, 3} all nduce paths (and the vertces appear n ths order along these paths). Add red edges accordng to the value of δ. If δ =, add no edge. If δ = 1, add u 1,1 u2,1, u1,2 u2,2. If δ = 2, add u 1,1 u2,1, u1,2 u2,2, u1,1 u3,1, u1,2 u3,2. If δ = 3, add u1,1 u2,1, u 1,2 u2,2, u1,1 u3,1, u1,2 u3,2, u2,1 u3,1, u2,2 u3,2. See Fgure 8. The graph G f (, α, β, γ, δ, ε, ζ) s obtaned from the dsont unon of the G(z ) s and the G(C ) s as follows. For = 1,...,n 1, add blue edges b, a +1,1 and b, a +1,1. Add a blue edge b n, c 1,1. For = 1,...,m 1, add a blue edge d, c +1,1. Introduce the two specal vertces x, y and add blue edges xa 1,1, xa 1,1 and yd m,, yb n,. See Fgure 9. Add red edges accordng to f, ε, ζ. For q = 1, 2, 3, f y q = z, then add all possble edges between {f,2+ 1, f,2+ 1+ε } and {u q,, uq,+ζ } and between {f,2+ 1, f,2+ 1+ε } and {uq,, uq,+ζ }; whle f yq = z then add all possble edges between {t,2+ 1, t,2+ 1+ε } and {u q,, uq,+ζ } and between {t,2+ 1, t,2+ 1+ε } and {uq,, uq,+ζ }. See Fgure 1. Clearly the sze of G f (, α, β, γ, δ, ε, ζ) s polynomal (actually quadratc) n the sze n + m of f, and x, y are non-adacent and both have degree two. Lemma 2.2 f admts a truth assgnment f and only f G f (, α, β, γ, δ, ε, ζ) contans a hole passng through x, y. proof Recall that f α = 2 then ε = β = γ. We wll prove the lemma for β =, γ =, ε =, ζ = because the proof s essentally the same for the other possble values. 6
8 p 1 2 m 1 m m + 1 r t,2p+r a,1 a, b, f,2p+r b,1 t,2p+r a,1 b, f,2p+r a, b,1 Fgure 5: The graph G(z ) (only blue edges are depcted) p 1 2 m 1 m m + 1 r t,2p+r a,1 a, b, f,2p+r b,1 t,2p+r a,1 b, f,2p+r a, b,1 Fgure 6: The graph G(z ) when α = 1, β =, γ = p 1 2 m 1 m m + 1 r t,2p+r a,1 a, b, f,2p+r b,1 t,2p+r a,1 b, f,2p+r a, b,1 Fgure 7: The graph G(z ) when α = 2, β =, γ = 7
9 1p 2 1 u 1,p u 2,p c,1 c, d,1 d, u 3,p Fgure 8: The graph G(c ) when δ = 3 x y Fgure 9: The whole graph G f Fgure 1: Red edges between G(z ) and G(c ) when ε = ζ = 8
10 Suppose that f admts a truth assgnment ξ {, 1} n. We can buld a hole n G by selectng vertces as follows. Select x, y. For = 1,..., n, select a,p, b,p, a,p, b,p for all p {1,...,}. For = 1,...,n, select c,p, d,p for all p {1,...,}. If ξ = 1 select t,p, t,p for all p {,..., 2(m + 2) 1}. If ξ = select f,p, f,p for all p {,...,2(m + 2) 1}. For = 1,...,m, snce ξ s a truth assgnment for f, at least one of the three lterals of C s equal to 1, say y q = 1 for some q {1, 2, 3}. Then select uq,p for all p {1,..., 2}. Now t s a routne matter to chec that the selected vertces nduce a cycle Z that contans x, y, and that Z s chordless, so t s a hole. The man pont s that there s no chord n Z between some subgraph G(C ) and some subgraph G(z ), for that would be ether an edge t,p u q,r wth yq = z and ξ = 1, or, symmetrcally, an edge f,p u q,r wth yq = z and ξ =, and n ether case ths would contradct the way the vertces of Z were selected. Conversely, suppose that G f (, α, β, γ, δ, ε, ζ) admts a hole Z that contans x, y. (1) For = 1,..., n, Z contans at least 4 +4(m+2) vertces of G(z ): 4 of these are a,p, a,p, b,p, b,p where p {1,..., }, and the others are ether the t,p, t,p s or the f,p, f,p s where p {,...,2(m + 2) 1}. Let us frst deal wth the case = 1. Snce x Z has degree 2, Z contans a 1,1,...,a 1, and a 1,1,..., a 1,. Hence exactly one of t 1,, f 1, s n Z. Lewse exactly one of t 1,, f 1, s n Z. If t 1,, f 1, are both n Z then there s a contradcton: ndeed, f α = 1 then, t 1,,..., t 1,2 and f 1,,...,f 1,2 must all be n Z, and snce t 1,2 sees f 1,2, Z cannot go through y; and f α = 2 the proof s smlar. Smlarly, t 1,, f 1, cannot both be n Z. So, there exsts a largest nteger p 2(m + 2) 1 such that ether t 1,,..., t 1,p and t 1,,...,t 1,p are all n Z or f 1,,..., f 1,p and f 1,,..., f 1,p are all n Z. We clam that p = 2(m+2) 1. For otherwse, some vertex w n {t 1,p, t 1,p, f 1,p, f 1,p } s ncdent to a red edge e of Z. If α = 1 then, up to a symmetry, we assume that t 1,,...,t 1,p and t 1,,..., t 1,p are all n Z. Let w be the vertex of e that s not w. Then w (whch s ether an f 1,, an f 1, or a u, ) s a neghbor of both t 1,p, t 1,p. Hence, Z cannot go through y, a contradcton. Ths proves our clam when α = 1. If α = 2, we dstngush between the followng sx cases. Case 1: p = 1. Then e = t 1, 1 f 1,2. Clearly t 1,,..., t 1, 1 must all be n Z. If t 1,,...,t 1,2 are n Z, there s a contradcton because of t 1,2 f 1,2, and f f 1,,...,f 1,2 are n Z, there s a contradcton because of e. Case 2: p = 2l where 1 l m + 1 and w = t 1,2l. Then e IS t 1,2l f 1,2l+ 1 or t 1,2l f 1,2l. In ether case, t 1,2l,..., t 1,2l+ 1 are all n Z, and there s a contradcton because of the red edge f 1,2l+ 1 t 1,2l+ 1 or t 1,2(l 1)+ 1 f 1,2l, or when l = m + 1 because of b 1,1. Case 3: p = 2l where 1 l m + 1 and w = f 1,2l. Then e s f 1,2l t 1,2(l 1)+ 1 or t 1,2l f 1,2l. In ether case, f 1,2l,..., f 1,2l+ 1 are all n Z, and there s a contradcton because of the red edge t 1,2(l 1)+ 1 f 1,2(l 1)+ 1 or t 1,2l f 1,2l+ 1, or when l = 1 because of a 1,. Case 4: p = 2l + 1 where 1 l m and w = t 1,2l+ 1. Then e s t 1,2l+ 1 f 1,2l+ 1, t 1,2l+ 1 f 1,2(l+1), or t 1,2l+ 1u q, for some, q. In the last case, there s a contradcton snce t 1,2l+ 1 Z also sees uq,. For the same reason, t 1,2l+ 1 uq, s not an edge of Z and t 1,2l+ 1,..., t 1,2(l+1) are 9
11 all n Z. So there s a contradcton because of the red edge t 1,2l f 1,2l+ 1 or t 1,2(l+1) f 1,2(l+1). Case 5: p = 2l + 1 where 2 l m and w = f 1,2l+ 1. Then e s f 1,2l+ 1 t 1,2l+ 1, f 1,2l+ 1 t 1,2l, or f 1,2l+ 1u q, for some, q. In the last case, there s a contradcton snce f 1,2l+ 1 Z also sees uq,. For the same reason, f 1,2l+ 1 uq, s not an edge of Z and f 1,2l+ 1,..., f 1,2(l+1) are all n Z. So there s a contradcton because of the red edge t 1,2l f 1,2l or t 1,2l+ 1 f 1,2(l+1). Case 6: p = 2(m + 1) + 1 and w = f 1,2(m+1)+ 1. Then there s a contradcton because of the red edge t 1,2(m+1) f 1,2(m+1). Ths proves our clam. Snce p = 2(m + 2) 1, b 1,1 s n Z. We clam that b 1,2 s n Z. For otherwse, the two neghbors of b 1,1 n Z are t 1,2(m+2) 1 and f 1,2(m+2) 1. Ths s a contradcton because of the red edges t 1,2m+ 1 f 1,2(m+1), t 1,2(m+1) f 1,2(m+1)+ 1 (f α = 2) or t 1,2(m+1) f 1,2(m+1), t 1,2(m+1) f 1,2(m+1) (f α = 1). Smlarly, b 1,1, b 1,2 are n Z. So b 1,1,..., b 1, and b 1,1,...,b 1, are all n Z. Ths proves (1) for = 1. The proof for = 2 s essentally the same as for = 1, and by nducton the clam holds up to = n. Ths proves (1). (2) For = 1,...,m, Z contans c,1,..., c,, d,1,..., d, and exactly one of {u 1,1,..., u1,2 }, {u2,1,...,u2,2 }, {u3,1,...,u3,2 }. Let us frst deal wth the case = 1. By (1), b n, s n Z and so c 1,1,..., c 1, are all n Z. Consequently exactly one of u 1 1,1, u 2 1,1, u 3 1,1 s n Z, say u 1 1,1 up to a symmetry. Note that the neghbour of u 1 1 n Z \ c 1 cannot be a vertex among u 2 1,1, u3 1,1 for ths would mply that Z contans a trangle. Hence u1 1,2,...,u1 1, are all n Z. The neghbour of u 1 1, n Z \ u1 1, 1 cannot be n some G(z ) (1 n). Else, up to a symmetry we assume that ths neghbor s t 1,p, p {,...,2(m + 2) 1}. If t 1,p Z, there s a contradcton because then t 1,p s also n Z by (1) and t 1,p would be a thrd neghbour of u 1 1, n Z. If t 1,p / Z, there s a contradcton because then the neghbor of t 1,p n Z \ u 1 1, must be t 1,p+1 (or symmetrcally t 1,p 1 ) for otherwse Z contans a trangle. So, t 1,p+1, t 1,p+2,... must be n Z, tll reachng a vertex havng a neghbor f 1,p or f 1,p n Z (whatever α). Thus the neghbour of u1 1, n Z \ u1 1, 1 s u1 1,+1. Smlarly, we prove that u 1,+2,...,u 1,2 are n Z, that d 1,1,...,d 1, are n Z, and so the clam holds for = 1. The proof of the clam for = 2 s essentally the same as for = 1, and by nducton the clam holds up to = m. Ths proves (2). Together wth x, y, the vertces of Z found n (1) and (2) actually nduce a cycle. So, snce Z s a hole, they are the members of Z and we can replace at least by exactly n (1). We can now mae a Boolean vector ξ as follows. For = 1,...,n, f Z contans t,, t, set ξ = 1; f Z contans f,, f, set ξ =. By (1) ths s consstent. Consder any clause C (1 m). By (2) and up to symmetry we may assume that u 1, s n Z. If y1 = z for some {1,.., n}, then the constructon of G mples that f,2+ 1, f,2+ 1 are not n Z, so t,2+ 1, t,2+ 1 are n Z, so ξ = 1, so clause C s satsfed by x. If y 1 = z for some {1,...,n}, then the constructon of G f mples that 1
12 t,2+ 1, t,2+ 1 are not n Z, so f,2+ 1, f,2+ 1 are n Z, so ξ =, so clause C s satsfed by z. Thus ξ s a truth assgnment for f. Theorem 2.3 Let 5 be an nteger. Then Γ {C3,...,C,K 1,6} and Γ {I1,...,I,C 5,...,C,K 1,4} are NP-complete. proof It s a routne matter to chec that the graph G f (, 2,,,,, ) contans no C l (3 l ) and no K 1,6 (n fact t has no vertex of degree at least 6). So Lemma 2.2 mples that Γ {C3,...,C,K 1,6} s NP-complete. It s a routne matter to chec that the graph G f (, 1, 1, 1, 3, 1, 1) contans no K 1,4, no C l (5 l ) and no I l (3 l ). So Lemma 2.2 mples that Γ {K1,4,C 5,...,C,I 3,...,I } s NP-complete. Complexty of Γ {H} when H s a connected graph Theorem 2.4 Let H be a connected graph. Then ether : H s a path or a subdvded claw and Γ {H} s polynomal. H contans one of K 1,4, I for some 1, or C l for some l 3 as an nduced subgraph and Γ {H} s NP-complete. proof If H contans one of K 1,4, I for some 1, or C l for some l 3 as an nduced subgraph then Γ {H} s NP-complete by Theorem 2.3. Else, H s a tree snce t contans no C l, l 3. If H has no vertex of degree at least 3, then H s a path and Γ {H} s polynomal by Theorem 2.1. If H has a sngle vertex of degree at least 3, then ths vertex has degree 3 because H contans no K 1,4. So, H s a subdvded claw and Γ {H} s polynomal by Theorem 2.1. If H has at least two vertces of degree at least 3 then H contans an I l, where l s the length of the unque path of H onnng these two vertces. Ths s a contradcton. Interestngly, a smlar theorem was proved by Aleseev: Theorem 2.5 (Aleseev, [1]) Let H be a connected graph that s not a path nor a subdvded claw. Then the problem of fndng a maxmum stable set n H-free graphs s NP-hard. But the complexty of the maxmum stable set problem s not nown n general for H-free graphs when H s a path or a subdvded claw. See [7] for a survey. NP-complete cases (bounded degree) Here, we wll show that Γ d s NP-complete when d 3 and polynomal when d = 2. If S s any fnte lst of cycles C 1, C 2,..., C m, then we wll also show that Γ 3 S s NP-complete as long as C 6 / S. Let f be an nstance of 3-Satsfablty, consstng of m clauses C 1,..., C m on n varables z 1,..., z n. For each clause C ( = 1,...,m), wth C = y 3 2 y 3 1 y 3, then y ( = 1,...,3m) s a lteral from {z 1,..., z n, z 1,..., z n }. 11
13 α 1+ α 2+ α 3+ α 4+ α α 1 α α 2 α 3 α 4 β β 1+ β 2+ β β 3+ β 4+ β 1 β 2 β 3 β 4 Fgure 11: The graph G(y ) Let us buld a graph G f wth two specalzed vertces x and y of degree 2 such that (G f ) = 3. There wll be a hole contanng x and y n G f f and only f there exsts a truth assgnment for f. For each lteral y ( = 1,...,3m), prepare a graph G(y ) on 2 vertces α, α, α 1+,..., α 4+, α 1,..., α 4, β, β, β 1+,..., β 4+, β 1,..., β 4. (We drop the subscrpt n the labels of the vertces for clarty). For = 1, 2, 3 add the edges α + α (+1)+, β + β (+1)+, α α (+1), β β (+1). Also add the edges α 1+ β 1, α 1 β 1+, α 4+ β 4, α 4 β 4+, αα 1+, αα 1, α 4+ α, α 4 α, ββ 1+, ββ 1, β 4+ β, β 4 β. For each clause C ( = 1,...,m), prepare a graph G(C ) wth 1 vertces c 1+, c 2+, c 3+, c 1, c 2, c 3, c +, c 12+, c, c 12. (We drop the subscrpt n the labels of the vertces for clarty). Add the edges c 12+ c 1+, c 12+ c 2+, c 12 c 1, c 12 c 2, c + c 12+, c + c 3+, c c 12, c c 3. c 1+ c 1 c 12+ c 12 c + c 2+ c 2 c c 3+ c 3 Fgure 12: The graph G(C ) 12
14 For each varable z ( = 1,...,n), prepare a graph G(z ) wth 2z + 2z + vertces, where z s the number of tmes z appears n clauses C 1,..., C m and z + s the number of tmes z appears n clauses C 1,..., C m. Let G(z ) consst of two nternally dsont paths P + and P wth common endponts d + and d and lengths 1 + 2z and 1 + 2z + respectvely. Label the vertces of P + as d +, p+,1,..., p+,2f, d and label the vertces of P as d +, p,1,..., p,2g, d. The fnal graph G f (see fgure 2) wll be constructed from the dsont unon of all the graphs G(y ), G(C ), and G(x ) wth the followng modfcatons: For = 1,...,3m 1, add the edges α α +1 and β β +1. Ths creates one connected chan of the graphs G(y ). For = 1,...,m 1, add the edge c c Ths creates one connected chan of the graphs G(C ). For = 1,..., n 1, add the edge d d+ +1. Ths creates one connected chan of the graphs G(x ). For = 1,..., n, let y n1,..., y nz be the occurrences of x over all lterals. For = 1,...,z, delete the edge p+,2 1 p+,2 and add the four edges p +,2 1 α2+ n, p +,2 1 β2+ n, p +,2 α3+ n, p +,2 β3+ n. For = 1,...,n, let y n1,..., y nz + be the occurrences of x over all lterals. For = 1, 2,...,z +, delete the edge p,2 1 p,2 and add the four edges p,2 1 α2+ n, p,2 1 β2+ n, p,2 α3+ n, p,2 β3+ n. For = 1,...,m and = 1, 2, 3, add the edges α 2 β 2 3( 1)+ c+, β 3 3( 1)+ c. Add the edges α 3md + 1 and β 3mc + 1 Add the vertex x and add the edges xα 1 and xβ 1. Add the vertex y and add the edges yc m and yd n. 3( 1)+ c+, α 3 3( 1)+ c, It s easy to verfy that (G f ) = 3, that G f s polynomal (actually lnear) n the sze n + m of f, and that x, y are non-adacent and both have degree two. Lemma 2.6 f admts a truth assgnment f and only f G f contans a hole passng through x and y. proof Frst assume that f admts a truth assgnment ξ {, 1} n. We wll pc a set of vertces that nduce a hole contanng x and y. P + d + P d Fgure 13: The graph G(z ) 13
15 x y Fgure 14: The fnal graph G f 1. Pc vertces x and y. 2. For = 1,...,3m, pc the vertces α, α, β, β. 3. For = 1,...,3m, f y s satsfed by ξ, then pc the vertces α 1+, α 2+, α 3+, α 4+, β 1+, β 2+, β 3+, and β 4+. Otherwse, pc the vertces α 1, α 2, α 3, α 4, β 1, β 2, β 3, and β For = 1,...,n, f ξ = 1, then pc all the vertces of the path P + and all the neghbors of the vertces n P + of the form α 2+ or α 3+ for any. 5. For = 1,..., n, f ξ =, then pc all the vertces of the path P and all the neghbors of the vertces n P of the form α 2+ or α 3+ for any. 6. For = 1,...,m, pc the vertces c + and c. Choose any {3 2, 3 1, 3} such that ξ satsfes y. Pc vertces α 2, and α 3. If = 3 2, then pc the vertces c 12+, c 1+, c 1, c 12. If = 3 1, then pc the vertces c 12+, c 2+, c 2, c 12. If = 3, then pc the vertces c 3+ and c 3. It suffces to show that the chosen vertces nduce a hole contanng x and y. The only potental problem s that for some, one of the vertces α 2+, α3+, α 2, or α3 was chosen multple tmes. If α 2+ and α 3+ were pced n Step 3, then y s satsfed by ξ. Therefore, α 2+ and α 3+ were not chosen n Step 4 or Step 5. Smlarly, f α 2 and α 3 were pced n Step 6, then y s satsfed by ξ and α 2 and α 3 were not pced n Step 3. Thus, the chosen vertces nduce a hole n G contanng vertces x and y. Now assume G f contans a hole H passng through x and y. The hole H must contan α 1 and β 1 snce they are the only two neghbors of x. Next, ether both α 1+ 1 and β1 1+ are n H, or both α 1 1 and β1 1 are n H. Wthout loss of generalty, let α 1+ 1 and β1 1+ be n H (the same reasonng that follows wll hold true for the other case). Snce β1 1 and α 1 1 are both neghbors of two members n H, they cannot be n H. Thus, α 2+ 1 and β1 2+ must be n H. Snce α 2+ 1 and β1 2+ have the same neghbor outsde G(y 1 ), t follows that H must contan α 3+ 1 and β1 3+. Also, H must contan α4+ 1 and β1 4+. Suppose that α 4 1 and β1 4 are n H. Because α 1 has the same neghbor as β1 outsde G(y 1 ) for = 2, 3, t follows that H must contan α 3 1, α2 1, and α 1 1. But then H s not a hole contanng b, a contradcton. Therefore, α4 1 and β1 4 cannot both be n H, so H must contan α 1, β 1, α 2, and β 2. By nducton, we see for = 1, 2,...,3m that H must contan α, α, β, β. Also, for each, ether H contans α 1+, α 2+, α 3+, α 4+, β 1+, β 2+, β 3+, β 4+ or H contans α 1, α 2, α 3, α 4, β 1, β 2, β 3, β 4. As a result, H must also contan d + 1 and c+ 1. By symmetry, we may assume H contans p + 1,1 and α2+ for some. Snce α 1+ s adacent to two vertces 14
16 n H, H must contan α 3+. Smlarly, H cannot contan α4+, so H contans p + 1,2 and p+ 1,3. By nducton, we see that H contans p+ 1, for = 1, 2,...,z+ and d 1. If H contans p, then H must contan p 1,z 1, for = z,..., 1, a contradcton. Thus, H must contan d + 2. By nducton, for = 1, 2,...,n, we see that H contans all the vertces of the path P + or P and by symmetry, we may assume H contans all the neghbors of the vertces n P + or P of the form α 2+ or α 3+ for any. Smlarly, for = 1, 2,..., m, t follows that H must contan c + and c. Also, H contans one of the followng: and ether α 2 and α 3 or β 2 and β 3 (where α 2 c 12+, c 1+, c 1, c 12 s adacent to c 1+ ). c 12+, c 2+, c 2, c 12 and ether α 2 and α 3 or β 2 and β 3 (where α 2 s adacent to c 2+ ). c 3+ and c 3 and ether α 2 and α 3 or β 2 and β 3 (where α 2 s adacent to c 3+ ). We can recover the satsfyng assgnment ξ as follows. For = 1, 2,..., n, set ξ = 1 f the vertces of P + are n H and set ξ = f the vertces of P are n H. By constructon, t s easy to verfy that at least one lteral n every clause s satsfed, so ξ s ndeed a satsfyng assgnment. Theorem 2.7 The followng statements hold: For any d Z wth d 2, the problem Γ d s NP-complete when d 3 and polynomal when d = 2. If H s any fnte lst of cycles C 1, C 2,...,C m such that C 6 / H, then Γ 3 H s NP-complete. proof In the above reducton, (G f ) = 3 so Γ d s NP-complete for d 3. When d = 2, there s a smple O(n) algorthm. Any hole contanng x and y must be a component of G so pc the vertex x and consder the component C of G that contans x. It taes O(n) tme to verfy whether C s a hole contanng x and y or not. To show the second statement, let K be the length of the longest cycle n H. In the above reducton, do the followng modfcatons. For = 1, 2, 3 and = 1, 2,...,3m, replace the edges α + α(+1)+, α α(+1), β + β (+1)+, and β β (+1) by paths of length K. For = 1, 2,..., 3m 1, replace the edges α α +1 and β β +1 by paths of length K. Replace the edges xα 1 and xβ 1 by paths of length K. Ths new reducton s polynomal n n and m and s H-free. The proof of Lemma 2.6 stll holds for ths new reducton so therefore, Γ 3 H s NP-complete. 15
17 3 Π B for some specal s-graphs 3.1 Holes wth pendng edges and trees Here, we study Π B4,..., Π B7 where B 4,...,B 7 are the s-graphs depcted on Fgure 3. Our motvaton s smply to gve a strng example and to pont out that surprsngly, pendng edges of s-graphs matter and that even an s-graph wth no cycle can lead to NP-complete problems. Theorem 3.1 There s an O(n 13 )-tme algorthm for Π B4 but Π B5 s NPcomplete. proof A realsaton of B 4 has exactly one vertex of degree 3 and one vertex of degree 4. Let us say that the realsaton H s short f the dstance between these two vertces n H s at most 3. Detectng short realsatons of B 4 can be done n tme n 9 as follows: for every 6-tuple F = (a, b, x 1, x 2, x 3, x 4 ) such that G[F] has edge-set {x 1 a, ax 2, x 2 b, bx 3, bx 4 } and for every 7-tuple F = (a, b, x 1, x 2, x 3, x 4, x 5 ) such that G[F] has edge-set {x 1 a, ax 2, x 2 x 3, x 3 b, bx 4, bx 5 }, delete x 1,...,x 5 and ther neghbors except a, b. In the resultng graph, chec whether a and b are n the same component. The answer s YES for at least one 7-or-6-tuple f and only f G contans at least one short realsaton of B 4. Here s an algorthm for Π B4, assumng that the entry graph G has no short realsaton of B 4. For every 9-tuple F = (a, b, c, x 1,..., x 6 ) such that G[F] has edge-set {x 1 a, bx 2, x 2 x 3, x 3 x 4, cx 5, x 5 x 3, x 3 x 6 } delete x 1,..., x 6 and ther neghbors except a, b, c. In the resultng graph, run three-n-a-tree for a, b, c. It s easly checed that the answer s YES for some 7-tuple f and only f G contans a realsaton of B 4. Let us prove that Π B5 s NP-complete by a reducton of Γ 3 to Π B5. Snce by Theorem 2.7, Γ 3 s NP-complete, ths wll complete the proof. Let (G, x, y) an nstance of Γ 3. Prepare a new graph G : add four vertces x, x, y, y to G and add four edges xx, xx, yy, yy. Snce (G) 3, t s easly seen that G contan a hole passng through x, y f and only f G contans a realsaton of B 5. The proof of the theorem below s omtted snce t s smlar to the proof of Theorem 3.1. Theorem 3.2 There s an O(n 14 )-tme algorthm for Π B6 but Π B7 s NPcomplete. 3.2 Induced subdvsons of K 5 Here, we study the problem of decdng whether a graph contan an nduced subdvson of K 5. More precsely, we put : sk 5 = ({a, b, c, d, e},, ( ) {a,b,c,d,e} 2 ). Theorem 3.3 Π sk5 s NP-complete. proof We consder an nstance (G, x, y) of Γ 3. Let us denote by x, x the two neghbors of x and by y, y the two neghbors of y. 16
18 a b a b e e x x x x x x G G y y y y y y c d c d Fgure 15: Graphs G and G Let us buld a graph G by addng fve vertces a, b, c, d, e. We add the edges ab, bd, dc, ca, ea, eb, ec, ed, ax, bx, cy, dy. We delete the edges xx, xx, yy, yy. We defne a very smlar graph G, the only change beng that we do not add edges cy, dy but edges cy, dy nstead. See fgure 15. Now n G (and smlarly G ) every vertex has degree at most 3, except for a, b, c, d, e. We clam that G contans a hole gong through x and y f and only f at least one of G, G contans an nduced subdvson of K 5. Indeed, f G contans a hole passng through x, x, y, y, y, x n that order then G obvously contans an nduced subdvson of K 5, and the hole passes n order through x, x, y, y, y, x then G contans such a subgraph. Conversely, f G (or symmetrcally G ) contans an nduced subdvson of K 5 then a, b, c, d, e must be the vertces of the underlyng K 5, because they are the only vertces wth degree at least 4. Hence there s a path from x to y n G \ {x, y} and a path from x to y n G \ {x, y}, and consequently a hole gong through x, y n G. 3.3 Π B for small B s Here, we survey the complexty Π B when B has at most four vertces. By the remars n the ntroducton, f V 3 then Π (V,D,F) s polynomal. Up to symmetres, we are left wth twelve s-graphs on four vertces. 17
19 For the followng two s-graphs, there s a polynomal algorthm usng threen-a-tree: The next two s-graphs yeld an NP-complete problem: (by Π {C4}) (by Π {K3}) For the remanng eght ones, we do not now the answer: As a concluson, we would le to pont out that every detecton problem assocated to an s-graph for whch a polynomal tme algorthm s nown can be solved by usng three-n-a-tree or by some easy brute-force enumeraton. References [1] V.E. Aleseev. On the local restrctons effect on the complexty of fndng the graph ndependence number. Combnatoral-algebrac methods n appled mathematcs, 132:3 13, Gory Unversty Press, Gory, n Russan. [2] N. Alon, R. Yuster, and U. Zwc. Fndng and countng gven length cycles. In Proceedngs of the 2nd European Symposum on Algorthms. Utrecht, The Netherlands, pages , [3] D. Benstoc. On the complexty of testng for odd holes and nduced odd paths. Dscrete Math., 9:85 92, See also Corrgendum by B. Reed, Dscrete Math., 12, (1992), p. 19. [4] M. Chudnovsy and P. Seymour. The three-n-a-tree problem. Manuscrpt. [5] M. Chudnovsy, G. Cornuéols, X. Lu, P. Seymour, and K. Vušovć. Recognzng Berge graphs. Combnatorca, 25: , 25. [6] M.R. Garey and D.S. Johnson. Computer and Intractablty : A Gude to the Theory of NP-completeness. W.H. Freeman, San Franssco, [7] A. Hertz and V. V. Lozn. The maxmum ndependent set problem and augmentng graphs. Manuscrpt. [8] F. Maffray and N. Trotgnon. Algorthms for perfectly contractle graphs. SIAM Jour. of Dscrete Math., 19(3): , 25. [9] F. Maffray, N. Trotgnon, and K. Vušovć. Algorthms for square- 3P C(, )-free Berge graphs. SIAM Journal on Dscrete Mathematcs. To appear. [1] R.E. Taran and M. Yannaas. Smple lnear-tme algorthms to test chordalty of graphs, test acyclcty of hypergraphs, and selectvely reduce acyclc hypergraphs. SIAM Journal on Computng, 13: ,
Problem Set 9 Solutions
Desgn and Analyss of Algorthms May 4, 2015 Massachusetts Insttute of Technology 6.046J/18.410J Profs. Erk Demane, Srn Devadas, and Nancy Lynch Problem Set 9 Solutons Problem Set 9 Solutons Ths problem
More informationNP-Completeness : Proofs
NP-Completeness : Proofs Proof Methods A method to show a decson problem Π NP-complete s as follows. (1) Show Π NP. (2) Choose an NP-complete problem Π. (3) Show Π Π. A method to show an optmzaton problem
More informationGraph Reconstruction by Permutations
Graph Reconstructon by Permutatons Perre Ille and Wllam Kocay* Insttut de Mathémathques de Lumny CNRS UMR 6206 163 avenue de Lumny, Case 907 13288 Marselle Cedex 9, France e-mal: lle@ml.unv-mrs.fr Computer
More informationCalculation of time complexity (3%)
Problem 1. (30%) Calculaton of tme complexty (3%) Gven n ctes, usng exhaust search to see every result takes O(n!). Calculaton of tme needed to solve the problem (2%) 40 ctes:40! dfferent tours 40 add
More informationMaximizing the number of nonnegative subsets
Maxmzng the number of nonnegatve subsets Noga Alon Hao Huang December 1, 213 Abstract Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what s the maxmum
More informationFoundations of Arithmetic
Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a 3 + + a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an
More informationComplete subgraphs in multipartite graphs
Complete subgraphs n multpartte graphs FLORIAN PFENDER Unverstät Rostock, Insttut für Mathematk D-18057 Rostock, Germany Floran.Pfender@un-rostock.de Abstract Turán s Theorem states that every graph G
More informationAffine transformations and convexity
Affne transformatons and convexty The purpose of ths document s to prove some basc propertes of affne transformatons nvolvng convex sets. Here are a few onlne references for background nformaton: http://math.ucr.edu/
More informationEvery planar graph is 4-colourable a proof without computer
Peter Dörre Department of Informatcs and Natural Scences Fachhochschule Südwestfalen (Unversty of Appled Scences) Frauenstuhlweg 31, D-58644 Iserlohn, Germany Emal: doerre(at)fh-swf.de Mathematcs Subject
More informationVolume 18 Figure 1. Notation 1. Notation 2. Observation 1. Remark 1. Remark 2. Remark 3. Remark 4. Remark 5. Remark 6. Theorem A [2]. Theorem B [2].
Bulletn of Mathematcal Scences and Applcatons Submtted: 016-04-07 ISSN: 78-9634, Vol. 18, pp 1-10 Revsed: 016-09-08 do:10.1805/www.scpress.com/bmsa.18.1 Accepted: 016-10-13 017 ScPress Ltd., Swtzerland
More informationDifference Equations
Dfference Equatons c Jan Vrbk 1 Bascs Suppose a sequence of numbers, say a 0,a 1,a,a 3,... s defned by a certan general relatonshp between, say, three consecutve values of the sequence, e.g. a + +3a +1
More informationFinding Dense Subgraphs in G(n, 1/2)
Fndng Dense Subgraphs n Gn, 1/ Atsh Das Sarma 1, Amt Deshpande, and Rav Kannan 1 Georga Insttute of Technology,atsh@cc.gatech.edu Mcrosoft Research-Bangalore,amtdesh,annan@mcrosoft.com Abstract. Fndng
More informationMath 217 Fall 2013 Homework 2 Solutions
Math 17 Fall 013 Homework Solutons Due Thursday Sept. 6, 013 5pm Ths homework conssts of 6 problems of 5 ponts each. The total s 30. You need to fully justfy your answer prove that your functon ndeed has
More informationCollege of Computer & Information Science Fall 2009 Northeastern University 20 October 2009
College of Computer & Informaton Scence Fall 2009 Northeastern Unversty 20 October 2009 CS7880: Algorthmc Power Tools Scrbe: Jan Wen and Laura Poplawsk Lecture Outlne: Prmal-dual schema Network Desgn:
More information= z 20 z n. (k 20) + 4 z k = 4
Problem Set #7 solutons 7.2.. (a Fnd the coeffcent of z k n (z + z 5 + z 6 + z 7 + 5, k 20. We use the known seres expanson ( n+l ( z l l z n below: (z + z 5 + z 6 + z 7 + 5 (z 5 ( + z + z 2 + z + 5 5
More informationMin Cut, Fast Cut, Polynomial Identities
Randomzed Algorthms, Summer 016 Mn Cut, Fast Cut, Polynomal Identtes Instructor: Thomas Kesselhem and Kurt Mehlhorn 1 Mn Cuts n Graphs Lecture (5 pages) Throughout ths secton, G = (V, E) s a mult-graph.
More informationTHE CHVÁTAL-ERDŐS CONDITION AND 2-FACTORS WITH A SPECIFIED NUMBER OF COMPONENTS
Dscussones Mathematcae Graph Theory 27 (2007) 401 407 THE CHVÁTAL-ERDŐS CONDITION AND 2-FACTORS WITH A SPECIFIED NUMBER OF COMPONENTS Guantao Chen Department of Mathematcs and Statstcs Georga State Unversty,
More informationThe L(2, 1)-Labeling on -Product of Graphs
Annals of Pure and Appled Mathematcs Vol 0, No, 05, 9-39 ISSN: 79-087X (P, 79-0888(onlne Publshed on 7 Aprl 05 wwwresearchmathscorg Annals of The L(, -Labelng on -Product of Graphs P Pradhan and Kamesh
More informationMATH 241B FUNCTIONAL ANALYSIS - NOTES EXAMPLES OF C ALGEBRAS
MATH 241B FUNCTIONAL ANALYSIS - NOTES EXAMPLES OF C ALGEBRAS These are nformal notes whch cover some of the materal whch s not n the course book. The man purpose s to gve a number of nontrval examples
More informationk(k 1)(k 2)(p 2) 6(p d.
BLOCK-TRANSITIVE 3-DESIGNS WITH AFFINE AUTOMORPHISM GROUP Greg Gamble Let X = (Z p d where p s an odd prme and d N, and let B X, B = k. Then t was shown by Praeger that the set B = {B g g AGL d (p} s the
More information3.1 Expectation of Functions of Several Random Variables. )' be a k-dimensional discrete or continuous random vector, with joint PMF p (, E X E X1 E X
Statstcs 1: Probablty Theory II 37 3 EPECTATION OF SEVERAL RANDOM VARIABLES As n Probablty Theory I, the nterest n most stuatons les not on the actual dstrbuton of a random vector, but rather on a number
More informationAnti-van der Waerden numbers of 3-term arithmetic progressions.
Ant-van der Waerden numbers of 3-term arthmetc progressons. Zhanar Berkkyzy, Alex Schulte, and Mchael Young Aprl 24, 2016 Abstract The ant-van der Waerden number, denoted by aw([n], k), s the smallest
More informationFACTORIZATION IN KRULL MONOIDS WITH INFINITE CLASS GROUP
C O L L O Q U I U M M A T H E M A T I C U M VOL. 80 1999 NO. 1 FACTORIZATION IN KRULL MONOIDS WITH INFINITE CLASS GROUP BY FLORIAN K A I N R A T H (GRAZ) Abstract. Let H be a Krull monod wth nfnte class
More informationBeyond Zudilin s Conjectured q-analog of Schmidt s problem
Beyond Zudln s Conectured q-analog of Schmdt s problem Thotsaporn Ae Thanatpanonda thotsaporn@gmalcom Mathematcs Subect Classfcaton: 11B65 33B99 Abstract Usng the methodology of (rgorous expermental mathematcs
More informationn ). This is tight for all admissible values of t, k and n. k t + + n t
MAXIMIZING THE NUMBER OF NONNEGATIVE SUBSETS NOGA ALON, HAROUT AYDINIAN, AND HAO HUANG Abstract. Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what
More informationSelf-complementing permutations of k-uniform hypergraphs
Dscrete Mathematcs Theoretcal Computer Scence DMTCS vol. 11:1, 2009, 117 124 Self-complementng permutatons of k-unform hypergraphs Artur Szymańsk A. Paweł Wojda Faculty of Appled Mathematcs, AGH Unversty
More informationCase A. P k = Ni ( 2L i k 1 ) + (# big cells) 10d 2 P k.
THE CELLULAR METHOD In ths lecture, we ntroduce the cellular method as an approach to ncdence geometry theorems lke the Szemeréd-Trotter theorem. The method was ntroduced n the paper Combnatoral complexty
More informationTHE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens
THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of
More information2.3 Nilpotent endomorphisms
s a block dagonal matrx, wth A Mat dm U (C) In fact, we can assume that B = B 1 B k, wth B an ordered bass of U, and that A = [f U ] B, where f U : U U s the restrcton of f to U 40 23 Nlpotent endomorphsms
More informationMATH 5707 HOMEWORK 4 SOLUTIONS 2. 2 i 2p i E(X i ) + E(Xi 2 ) ä i=1. i=1
MATH 5707 HOMEWORK 4 SOLUTIONS CİHAN BAHRAN 1. Let v 1,..., v n R m, all lengths v are not larger than 1. Let p 1,..., p n [0, 1] be arbtrary and set w = p 1 v 1 + + p n v n. Then there exst ε 1,..., ε
More informationarxiv: v1 [math.co] 1 Mar 2014
Unon-ntersectng set systems Gyula O.H. Katona and Dánel T. Nagy March 4, 014 arxv:1403.0088v1 [math.co] 1 Mar 014 Abstract Three ntersecton theorems are proved. Frst, we determne the sze of the largest
More informationarxiv: v2 [cs.ds] 1 Feb 2017
Polynomal-tme Algorthms for the Subset Feedback Vertex Set Problem on Interval Graphs and Permutaton Graphs Chars Papadopoulos Spyrdon Tzmas arxv:170104634v2 [csds] 1 Feb 2017 Abstract Gven a vertex-weghted
More informationMore metrics on cartesian products
More metrcs on cartesan products If (X, d ) are metrc spaces for 1 n, then n Secton II4 of the lecture notes we defned three metrcs on X whose underlyng topologes are the product topology The purpose of
More informationSection 3.6 Complex Zeros
04 Chapter Secton 6 Comple Zeros When fndng the zeros of polynomals, at some pont you're faced wth the problem Whle there are clearly no real numbers that are solutons to ths equaton, leavng thngs there
More informationLinear, affine, and convex sets and hulls In the sequel, unless otherwise specified, X will denote a real vector space.
Lnear, affne, and convex sets and hulls In the sequel, unless otherwse specfed, X wll denote a real vector space. Lnes and segments. Gven two ponts x, y X, we defne xy = {x + t(y x) : t R} = {(1 t)x +
More information20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The first idea is connectedness.
20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The frst dea s connectedness. Essentally, we want to say that a space cannot be decomposed
More informationA Simple Research of Divisor Graphs
The 29th Workshop on Combnatoral Mathematcs and Computaton Theory A Smple Research o Dvsor Graphs Yu-png Tsao General Educaton Center Chna Unversty o Technology Tape Tawan yp-tsao@cuteedutw Tape Tawan
More informationHMMT February 2016 February 20, 2016
HMMT February 016 February 0, 016 Combnatorcs 1. For postve ntegers n, let S n be the set of ntegers x such that n dstnct lnes, no three concurrent, can dvde a plane nto x regons (for example, S = {3,
More informationThe Minimum Universal Cost Flow in an Infeasible Flow Network
Journal of Scences, Islamc Republc of Iran 17(2): 175-180 (2006) Unversty of Tehran, ISSN 1016-1104 http://jscencesutacr The Mnmum Unversal Cost Flow n an Infeasble Flow Network H Saleh Fathabad * M Bagheran
More informationSociété de Calcul Mathématique SA
Socété de Calcul Mathématque SA Outls d'ade à la décson Tools for decson help Probablstc Studes: Normalzng the Hstograms Bernard Beauzamy December, 202 I. General constructon of the hstogram Any probablstc
More informationFormulas for the Determinant
page 224 224 CHAPTER 3 Determnants e t te t e 2t 38 A = e t 2te t e 2t e t te t 2e 2t 39 If 123 A = 345, 456 compute the matrx product A adj(a) What can you conclude about det(a)? For Problems 40 43, use
More informationDistance Three Labelings of Trees
Dstance Three Labelngs of Trees Jří Fala Petr A. Golovach Jan Kratochvíl Bernard Ldcký Danël Paulusma Abstract An L(2, 1, 1)-labelng of a graph G assgns nonnegatve ntegers to the vertces of G n such a
More informationModule 9. Lecture 6. Duality in Assignment Problems
Module 9 1 Lecture 6 Dualty n Assgnment Problems In ths lecture we attempt to answer few other mportant questons posed n earler lecture for (AP) and see how some of them can be explaned through the concept
More informationprinceton univ. F 17 cos 521: Advanced Algorithm Design Lecture 7: LP Duality Lecturer: Matt Weinberg
prnceton unv. F 17 cos 521: Advanced Algorthm Desgn Lecture 7: LP Dualty Lecturer: Matt Wenberg Scrbe: LP Dualty s an extremely useful tool for analyzng structural propertes of lnear programs. Whle there
More informationarxiv: v1 [cs.gt] 14 Mar 2019
Stable Roommates wth Narcssstc, Sngle-Peaked, and Sngle-Crossng Preferences Robert Bredereck 1, Jehua Chen 2, Ugo Paavo Fnnendahl 1, and Rolf Nedermeer 1 arxv:1903.05975v1 [cs.gt] 14 Mar 2019 1 TU Berln,
More informationFINITELY-GENERATED MODULES OVER A PRINCIPAL IDEAL DOMAIN
FINITELY-GENERTED MODULES OVER PRINCIPL IDEL DOMIN EMMNUEL KOWLSKI Throughout ths note, s a prncpal deal doman. We recall the classfcaton theorem: Theorem 1. Let M be a fntely-generated -module. (1) There
More informationn α j x j = 0 j=1 has a nontrivial solution. Here A is the n k matrix whose jth column is the vector for all t j=0
MODULE 2 Topcs: Lnear ndependence, bass and dmenson We have seen that f n a set of vectors one vector s a lnear combnaton of the remanng vectors n the set then the span of the set s unchanged f that vector
More informationU.C. Berkeley CS294: Spectral Methods and Expanders Handout 8 Luca Trevisan February 17, 2016
U.C. Berkeley CS94: Spectral Methods and Expanders Handout 8 Luca Trevsan February 7, 06 Lecture 8: Spectral Algorthms Wrap-up In whch we talk about even more generalzatons of Cheeger s nequaltes, and
More informationIntroductory Cardinality Theory Alan Kaylor Cline
Introductory Cardnalty Theory lan Kaylor Clne lthough by name the theory of set cardnalty may seem to be an offshoot of combnatorcs, the central nterest s actually nfnte sets. Combnatorcs deals wth fnte
More informationModulo Magic Labeling in Digraphs
Gen. Math. Notes, Vol. 7, No., August, 03, pp. 5- ISSN 9-784; Copyrght ICSRS Publcaton, 03 www.-csrs.org Avalable free onlne at http://www.geman.n Modulo Magc Labelng n Dgraphs L. Shobana and J. Baskar
More informationA new construction of 3-separable matrices via an improved decoding of Macula s construction
Dscrete Optmzaton 5 008 700 704 Contents lsts avalable at ScenceDrect Dscrete Optmzaton journal homepage: wwwelsevercom/locate/dsopt A new constructon of 3-separable matrces va an mproved decodng of Macula
More informationApproximate Smallest Enclosing Balls
Chapter 5 Approxmate Smallest Enclosng Balls 5. Boundng Volumes A boundng volume for a set S R d s a superset of S wth a smple shape, for example a box, a ball, or an ellpsod. Fgure 5.: Boundng boxes Q(P
More informationGeometric drawings of K n with few crossings
Geometrc drawngs of K n wth few crossngs Bernardo M. Ábrego, Slva Fernández-Merchant Calforna State Unversty Northrdge {bernardo.abrego,slva.fernandez}@csun.edu ver 9 Abstract We gve a new upper bound
More informationFor now, let us focus on a specific model of neurons. These are simplified from reality but can achieve remarkable results.
Neural Networks : Dervaton compled by Alvn Wan from Professor Jtendra Malk s lecture Ths type of computaton s called deep learnng and s the most popular method for many problems, such as computer vson
More informationEdge Isoperimetric Inequalities
November 7, 2005 Ross M. Rchardson Edge Isopermetrc Inequaltes 1 Four Questons Recall that n the last lecture we looked at the problem of sopermetrc nequaltes n the hypercube, Q n. Our noton of boundary
More informationYong Joon Ryang. 1. Introduction Consider the multicommodity transportation problem with convex quadratic cost function. 1 2 (x x0 ) T Q(x x 0 )
Kangweon-Kyungk Math. Jour. 4 1996), No. 1, pp. 7 16 AN ITERATIVE ROW-ACTION METHOD FOR MULTICOMMODITY TRANSPORTATION PROBLEMS Yong Joon Ryang Abstract. The optmzaton problems wth quadratc constrants often
More informationa b a In case b 0, a being divisible by b is the same as to say that
Secton 6.2 Dvsblty among the ntegers An nteger a ε s dvsble by b ε f there s an nteger c ε such that a = bc. Note that s dvsble by any nteger b, snce = b. On the other hand, a s dvsble by only f a = :
More informationStanford University CS254: Computational Complexity Notes 7 Luca Trevisan January 29, Notes for Lecture 7
Stanford Unversty CS54: Computatonal Complexty Notes 7 Luca Trevsan January 9, 014 Notes for Lecture 7 1 Approxmate Countng wt an N oracle We complete te proof of te followng result: Teorem 1 For every
More informationAPPENDIX A Some Linear Algebra
APPENDIX A Some Lnear Algebra The collecton of m, n matrces A.1 Matrces a 1,1,..., a 1,n A = a m,1,..., a m,n wth real elements a,j s denoted by R m,n. If n = 1 then A s called a column vector. Smlarly,
More informationSupplement: Proofs and Technical Details for The Solution Path of the Generalized Lasso
Supplement: Proofs and Techncal Detals for The Soluton Path of the Generalzed Lasso Ryan J. Tbshran Jonathan Taylor In ths document we gve supplementary detals to the paper The Soluton Path of the Generalzed
More informationLecture 10: May 6, 2013
TTIC/CMSC 31150 Mathematcal Toolkt Sprng 013 Madhur Tulsan Lecture 10: May 6, 013 Scrbe: Wenje Luo In today s lecture, we manly talked about random walk on graphs and ntroduce the concept of graph expander,
More information18.1 Introduction and Recap
CS787: Advanced Algorthms Scrbe: Pryananda Shenoy and Shjn Kong Lecturer: Shuch Chawla Topc: Streamng Algorthmscontnued) Date: 0/26/2007 We contnue talng about streamng algorthms n ths lecture, ncludng
More informationALGEBRA HW 7 CLAY SHONKWILER
ALGEBRA HW 7 CLAY SHONKWILER 1 Whch of the followng rngs R are dscrete valuaton rngs? For those that are, fnd the fracton feld K = frac R, the resdue feld k = R/m (where m) s the maxmal deal), and a unformzer
More informationarxiv: v3 [cs.dm] 7 Jul 2012
Perfect matchng n -unform hypergraphs wth large vertex degree arxv:1101.580v [cs.dm] 7 Jul 01 Imdadullah Khan Department of Computer Scence College of Computng and Informaton Systems Umm Al-Qura Unversty
More informationComputing Correlated Equilibria in Multi-Player Games
Computng Correlated Equlbra n Mult-Player Games Chrstos H. Papadmtrou Presented by Zhanxang Huang December 7th, 2005 1 The Author Dr. Chrstos H. Papadmtrou CS professor at UC Berkley (taught at Harvard,
More informationSpectral Graph Theory and its Applications September 16, Lecture 5
Spectral Graph Theory and ts Applcatons September 16, 2004 Lecturer: Danel A. Spelman Lecture 5 5.1 Introducton In ths lecture, we wll prove the followng theorem: Theorem 5.1.1. Let G be a planar graph
More informationPerron Vectors of an Irreducible Nonnegative Interval Matrix
Perron Vectors of an Irreducble Nonnegatve Interval Matrx Jr Rohn August 4 2005 Abstract As s well known an rreducble nonnegatve matrx possesses a unquely determned Perron vector. As the man result of
More informationMath 261 Exercise sheet 2
Math 261 Exercse sheet 2 http://staff.aub.edu.lb/~nm116/teachng/2017/math261/ndex.html Verson: September 25, 2017 Answers are due for Monday 25 September, 11AM. The use of calculators s allowed. Exercse
More informationISSN: ISO 9001:2008 Certified International Journal of Engineering and Innovative Technology (IJEIT) Volume 3, Issue 1, July 2013
ISSN: 2277-375 Constructon of Trend Free Run Orders for Orthogonal rrays Usng Codes bstract: Sometmes when the expermental runs are carred out n a tme order sequence, the response can depend on the run
More information8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS
SECTION 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS 493 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS All the vector spaces you have studed thus far n the text are real vector spaces because the scalars
More informationEconomics 101. Lecture 4 - Equilibrium and Efficiency
Economcs 0 Lecture 4 - Equlbrum and Effcency Intro As dscussed n the prevous lecture, we wll now move from an envronment where we looed at consumers mang decsons n solaton to analyzng economes full of
More informationProblem Do any of the following determine homomorphisms from GL n (C) to GL n (C)?
Homework 8 solutons. Problem 16.1. Whch of the followng defne homomomorphsms from C\{0} to C\{0}? Answer. a) f 1 : z z Yes, f 1 s a homomorphsm. We have that z s the complex conjugate of z. If z 1,z 2
More informationCaps and Colouring Steiner Triple Systems
Desgns, Codes and Cryptography, 13, 51 55 (1998) c 1998 Kluwer Academc Publshers, Boston. Manufactured n The Netherlands. Caps and Colourng Stener Trple Systems AIDEN BRUEN* Department of Mathematcs, Unversty
More informationCOMPLEX NUMBERS AND QUADRATIC EQUATIONS
COMPLEX NUMBERS AND QUADRATIC EQUATIONS INTRODUCTION We know that x 0 for all x R e the square of a real number (whether postve, negatve or ero) s non-negatve Hence the equatons x, x, x + 7 0 etc are not
More informationLecture 12: Discrete Laplacian
Lecture 12: Dscrete Laplacan Scrbe: Tanye Lu Our goal s to come up wth a dscrete verson of Laplacan operator for trangulated surfaces, so that we can use t n practce to solve related problems We are mostly
More informationNUMERICAL DIFFERENTIATION
NUMERICAL DIFFERENTIATION 1 Introducton Dfferentaton s a method to compute the rate at whch a dependent output y changes wth respect to the change n the ndependent nput x. Ths rate of change s called the
More informationSUCCESSIVE MINIMA AND LATTICE POINTS (AFTER HENK, GILLET AND SOULÉ) M(B) := # ( B Z N)
SUCCESSIVE MINIMA AND LATTICE POINTS (AFTER HENK, GILLET AND SOULÉ) S.BOUCKSOM Abstract. The goal of ths note s to present a remarably smple proof, due to Hen, of a result prevously obtaned by Gllet-Soulé,
More informationOn the Multicriteria Integer Network Flow Problem
BULGARIAN ACADEMY OF SCIENCES CYBERNETICS AND INFORMATION TECHNOLOGIES Volume 5, No 2 Sofa 2005 On the Multcrtera Integer Network Flow Problem Vassl Vasslev, Marana Nkolova, Maryana Vassleva Insttute of
More informationWeek 2. This week, we covered operations on sets and cardinality.
Week 2 Ths week, we covered operatons on sets and cardnalty. Defnton 0.1 (Correspondence). A correspondence between two sets A and B s a set S contaned n A B = {(a, b) a A, b B}. A correspondence from
More informationAn Introduction to Morita Theory
An Introducton to Morta Theory Matt Booth October 2015 Nov. 2017: made a few revsons. Thanks to Nng Shan for catchng a typo. My man reference for these notes was Chapter II of Bass s book Algebrac K-Theory
More informationSolutions to the 71st William Lowell Putnam Mathematical Competition Saturday, December 4, 2010
Solutons to the 7st Wllam Lowell Putnam Mathematcal Competton Saturday, December 4, 2 Kran Kedlaya and Lenny Ng A The largest such k s n+ 2 n 2. For n even, ths value s acheved by the partton {,n},{2,n
More informationThe Second Eigenvalue of Planar Graphs
Spectral Graph Theory Lecture 20 The Second Egenvalue of Planar Graphs Danel A. Spelman November 11, 2015 Dsclamer These notes are not necessarly an accurate representaton of what happened n class. The
More information10-701/ Machine Learning, Fall 2005 Homework 3
10-701/15-781 Machne Learnng, Fall 2005 Homework 3 Out: 10/20/05 Due: begnnng of the class 11/01/05 Instructons Contact questons-10701@autonlaborg for queston Problem 1 Regresson and Cross-valdaton [40
More informationLectures - Week 4 Matrix norms, Conditioning, Vector Spaces, Linear Independence, Spanning sets and Basis, Null space and Range of a Matrix
Lectures - Week 4 Matrx norms, Condtonng, Vector Spaces, Lnear Independence, Spannng sets and Bass, Null space and Range of a Matrx Matrx Norms Now we turn to assocatng a number to each matrx. We could
More information12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA. 4. Tensor product
12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA Here s an outlne of what I dd: (1) categorcal defnton (2) constructon (3) lst of basc propertes (4) dstrbutve property (5) rght exactness (6) localzaton
More informationLecture 4: November 17, Part 1 Single Buffer Management
Lecturer: Ad Rosén Algorthms for the anagement of Networs Fall 2003-2004 Lecture 4: November 7, 2003 Scrbe: Guy Grebla Part Sngle Buffer anagement In the prevous lecture we taled about the Combned Input
More informationKernel Methods and SVMs Extension
Kernel Methods and SVMs Extenson The purpose of ths document s to revew materal covered n Machne Learnng 1 Supervsed Learnng regardng support vector machnes (SVMs). Ths document also provdes a general
More informationThe computational complexity of the parallel knock-out problem
Theoretcal Computer Scence 393 (2008) 182 195 www.elsever.com/locate/tcs The computatonal complexty of the parallel knock-out problem Hajo Broersma, Matthew Johnson, Danël Paulusma, Ian A. Stewart Department
More informationAssortment Optimization under MNL
Assortment Optmzaton under MNL Haotan Song Aprl 30, 2017 1 Introducton The assortment optmzaton problem ams to fnd the revenue-maxmzng assortment of products to offer when the prces of products are fxed.
More information(k,?)-sandwich Problems: why not ask for special kinds of bread?
Couto-Fara-Klen-Prott-Noguera MC 2014/4/9 19:04 page 17 #1 Matemátca Contemporânea, Vol. 42, 17 26 c?2014, Socedade Braslera de Matemátca (k,?)-sandwch Problems: why not ask for specal knds of bread? F.
More informationBallot Paths Avoiding Depth Zero Patterns
Ballot Paths Avodng Depth Zero Patterns Henrch Nederhausen and Shaun Sullvan Florda Atlantc Unversty, Boca Raton, Florda nederha@fauedu, ssull21@fauedu 1 Introducton In a paper by Sapounaks, Tasoulas,
More informationThe Multiple Classical Linear Regression Model (CLRM): Specification and Assumptions. 1. Introduction
ECONOMICS 5* -- NOTE (Summary) ECON 5* -- NOTE The Multple Classcal Lnear Regresson Model (CLRM): Specfcaton and Assumptons. Introducton CLRM stands for the Classcal Lnear Regresson Model. The CLRM s also
More informationRandom Walks on Digraphs
Random Walks on Dgraphs J. J. P. Veerman October 23, 27 Introducton Let V = {, n} be a vertex set and S a non-negatve row-stochastc matrx (.e. rows sum to ). V and S defne a dgraph G = G(V, S) and a drected
More informationMatrix Approximation via Sampling, Subspace Embedding. 1 Solving Linear Systems Using SVD
Matrx Approxmaton va Samplng, Subspace Embeddng Lecturer: Anup Rao Scrbe: Rashth Sharma, Peng Zhang 0/01/016 1 Solvng Lnear Systems Usng SVD Two applcatons of SVD have been covered so far. Today we loo
More informationA combinatorial problem associated with nonograms
A combnatoral problem assocated wth nonograms Jessca Benton Ron Snow Nolan Wallach March 21, 2005 1 Introducton. Ths work was motvated by a queston posed by the second named author to the frst named author
More information5 The Rational Canonical Form
5 The Ratonal Canoncal Form Here p s a monc rreducble factor of the mnmum polynomal m T and s not necessarly of degree one Let F p denote the feld constructed earler n the course, consstng of all matrces
More informationA permuted random walk exits faster
A permuted random wal exts faster Rchard Pymar Perla Sous Aprl 4, 013 Abstract Let σ be a permutaton of {0,..., n}. We consder the Marov chan X whch jumps from 0, n to σ( + 1) or σ( 1), equally lely. When
More informationOn the Repeating Group Finding Problem
The 9th Workshop on Combnatoral Mathematcs and Computaton Theory On the Repeatng Group Fndng Problem Bo-Ren Kung, Wen-Hsen Chen, R.C.T Lee Graduate Insttute of Informaton Technology and Management Takmng
More informationfind (x): given element x, return the canonical element of the set containing x;
COS 43 Sprng, 009 Dsjont Set Unon Problem: Mantan a collecton of dsjont sets. Two operatons: fnd the set contanng a gven element; unte two sets nto one (destructvely). Approach: Canoncal element method:
More informationThe Order Relation and Trace Inequalities for. Hermitian Operators
Internatonal Mathematcal Forum, Vol 3, 08, no, 507-57 HIKARI Ltd, wwwm-hkarcom https://doorg/0988/mf088055 The Order Relaton and Trace Inequaltes for Hermtan Operators Y Huang School of Informaton Scence
More information