Alspach B., Xu M.Y.'s 1/2-Transitive Graphs of Order 3p PDF

By Alspach B., Xu M.Y.

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D^dk+i — . • . = d^dk+i = 0 . We go on until we have chosen c i , . . , dn . Let c be the shortest vector among c i , . . , cn . Then 26 LASZL6 LOVASZ To complete the argument, it suffices to note that it is easy to construct a basis in £ whose Gram-Schmidt orthogonalization is just (d n /||d n || 2 ,... 15) with a(n] = b(n)2 . Remark. 6) is not too far from A(£): there exists a basis ( & i , . . , b n ) in any lattice £ such that Let 6 be a shortest non-zero vector in the lattice £ . We may not be able to prove in polynomial time that b is shortest, but we can prove in polynomial time that 6 is "almost shortest" in the sense that no non-zero lattice vector is shorter than ||6||/n .

Lenstra and L. Lovasz (1984). 8) Corollary. A polynomial with rational coefficients can be factored into irreducible polynomials in polynomial time. Proof. Let a be any root of the given polynomial / . For simplicity, assume that a is real (else, we could apply a similar argument to the real and imaginary parts of a). 7) (a), we can design a real number box description of a . Using part (b) of this same theorem, we can determine the minimal polynomial g of a in polynomial time. Now if / = g then / is irreducible and we have nothing to prove.

S ( K , t ) — {x 6 R n : inf ye K||z - y I < e} • We let S(K, -e) = K — S(Rn — X", e) . It helps to understand the definitions below if we read y 6 S ( K , —e) as "j/ is almost in K " and y  S ( K , e] as "y is deep in K ". 8) WEAK MEMBERSHIP PROBLEM. Given a point y e Qn and a rational e > 0 , conclude with one of the following: (i) assert that y 0 , conclude with one of the following: (i) assert that y E S(K",e); (ii) find a vector c £ Qn such that \\c\\^ — 1 and CTX < cTy + c for every xeS(K,-e) .

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1/2-Transitive Graphs of Order 3p by Alspach B., Xu M.Y.


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