1995
DOI: 10.1002/jgt.3190190113
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The crossing number of c4 × c4

Abstract: We prove t h a t t h e crossing number of C4 X Ca is 8. 0 1995 John Wiley & Sons, Inc.It has long been conjectured that the crossing number of C, X C,, denoted u(C, X C n ) , is ( m -2)n for m I n . While ( m -2)n is easily seen to be an upper bound (see Figure l), equality has been established only for m = 3,4. Harary, Kainen, and Schwenk [3] proved that u(C3 X C,) = 3, and Beineke and Ringeisen [4] used this to prove that u(C3 X C,) = n for n 2 3 . Beineke and Ringeisen [ l ] also proved that v(C4 X C,) = 2… Show more

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Cited by 40 publications
(41 citation statements)
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“…In the opposite case by deleting the vertices of C f 4 and the edges of F ∪I from the graph H ×C 4 we obtain the drawing of the subgraph isomorphic to C 4 × C 4 with at most seven crossings. This is in contradiction with cr(C 4 × C 4 ) = 8 (see [5]). …”
Section: Lemmamentioning
confidence: 57%
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“…In the opposite case by deleting the vertices of C f 4 and the edges of F ∪I from the graph H ×C 4 we obtain the drawing of the subgraph isomorphic to C 4 × C 4 with at most seven crossings. This is in contradiction with cr(C 4 × C 4 ) = 8 (see [5]). …”
Section: Lemmamentioning
confidence: 57%
“…Harary et al [8] conjectured that the crossing number of C m × C n is (m − 2)n, for all m, n satisfying 3 m n. Progress toward a possible proof of the conjecture has come slowly. The conjecture has been proved only for m, n satisfying n m, m 7 [1][2][3][4][5]8,15,16,18]. It was recently proved by Glebsky and Salazar [6] that the crossing number of C m × C n equals its long-conjectured value at least for n m(m + 1).…”
Section: Introductionmentioning
confidence: 98%
“…Harary et al [4] showed that cr(C 3 × C 3 ) = 3, and Ringeisen and Beineke [8] showed that cr(C 3 × C n ) = n. Dean and Richter [2] proved that cr(C 4 × C 4 ) = 8, and Beineke and Ringeisen [1] proved that cr(C 4 × C n ) = 2n and cr(K 4 × C n ) = 3n. Let S n−1 and P n be the star and path with n vertices, respectively.…”
Section: Introductionmentioning
confidence: 96%
“…. , 7, the base case n = m was published separately [2,3,9,11,25]. The recent state-of-the-art on the problem is the result of Glebsky and Salazar from [10]: cr(C m C n ) = (m − 2)n for m ≥ 3 and n ≥ 1 2 (m + 1)(m + 2).…”
Section: Introductionmentioning
confidence: 98%