Báo cáo toán học: "The Ramsey number r(K5 − P3, K5)" pptx

... bound on the number of blue edges ensures that at least one edge f ∈ F is blue (under c(x 1 , x 2 )): |F| > (1 − 2ρ ∗ ) r−2 N r−2 (r − 2)! = (1 − µ)N r−2 (r − 2)! > (1 − µ)  N r − 2  ≥ |E(T ∗ R )| where ... C r n ) is asymptotic to (2r−1)n 2r−2 . To see that this is about best, set n = (2r−2)k and consider the 2-coloring of a complete graph with (2r−1)k−2 = (...

Ngày tải lên: 07/08/2014, 21:20

... R(K m − P 3 , K n ; k) and v ∈ V (F ) then F [N F (v)] ∈ R(K m−1 −P 3 , K n ; d F (v)) and F[N F (v)] ∈ R(K m −P 3 , K n−1 ; k−1 −d F (v)). Thus ∆(F ) ≤ r(K m−1 − P 3 , K n ) − 1 and δ(F ) ≥ k − ... d(u 3 ) − d G[A i ] (u 1 ) − d G[A i ] (u 2 ) − d G[A i ] (u 3 ) ≤ d(u) + 11 + 11 − 2 − 2 − 2 = d(u) + 16, hence d(u) = 11. On the other hand, 27 = 39 −...

Ngày tải lên: 08/08/2014, 14:23

... i − 1 i − 1  = d k+1,j−1 . Also, d j,j − c j,j = d j,j − c j,j−1 = d j,j − d j,j−1 =  2j−3 j−1  −  2j−4 j−2  −  2j−4 j−3  =  2j−3 j−1  −  2j−4 j−2  −  2j−4 j−1  = 0. Thus, by induction ... 1 i  and k−4  i=0  k + i − 1 i  =  2k − 4 k − 4  so that d k,k−1 /  2k− 4 k−2  = 1−  2k− 4 k−4  /  2k− 4 k−2  = 1−( k−2)(k−3)/k...

Ngày tải lên: 07/08/2014, 07:21

... least n − (k−1)n 4(k−1) = 3n 4 vertices remain uncoloured, thus    V (G  (k−1) )    ≥ 3n/4, and for any uncoloured vertex v ∈ G  (k−1) sm  i=(s−1)m+1 |N s i (v)|≥m (k−s) |Γ G  k−1 (v)| ... G  (k−1) we have sm  i=(s−1)m+1 |N s i (v)|≥m (k−s) |Γ G  k−1 (v)| 1 ≤ s ≤ (k − 1) . Now since, |V (G  (k−1) )|≥n/k, by definition of ∆ 0 we must be able to choose a v so that |Γ G  (k...

Ngày tải lên: 07/08/2014, 06:20

... d k (v). Hence α(G) ≥  v∈G  d 1 (v) 2 k−1 α(G)  1 k−1 or α(G) k k−1 ≥ 1 2   v∈G d 1 (v) 1 k−1  or α(G) ≥ 2 −( k−1 k )   v∈G d 1 (v) 1 k−1  k−1 k as claimed. Corollary 1: Let G be regular ... d k−1 (v) 1+d 1 (v)+···+ d k (v)  /(k − 1). Since the arithmetic mean is greater than the geometric mean, we can conclude that α(G) ≥  v∈G  d 1 (v)2 −( k−2) 1+d 1 (v)+···+ d k (v)...

Ngày tải lên: 07/08/2014, 06:20

... follows: ∞  j=1 1 1 − e −cjs −1 /2 · M s  j=1 1 − e −cjs −1 /2 1 −e −cjs −1 /2 +βs −3 /2 j 2 =exp  π 2 6z + 1 2 log z 2π + O(z)      z=cs −1 /2 exp   2 M s  j=1 e βs −3 /2 j 2 −1 e cjs −1 /2 −1   . ... m =(m−k+t+3/2) log(m −k + t +1)−t −( m−k+3/2) log(m − k +1)−tlog m + o(1) =(m−k)log  1+ t m−k  −t+tlog  1 − k −t m  + o(1) =(m−k...

Ngày tải lên: 07/08/2014, 06:20

... matrix. Proof of Theorem 1.1: The square of the number of perfect matchings of G counts ordered pairs of such matchings. We claim that this is the number of spanning 2-regular subgraphs H of G ... counts the number of spanning 2-regular subgraphs H  of G, where now we allow odd cycles and cycles of length 2 as well. Here, too, each such H  is counted 2 s times, where s is the number...

Ngày tải lên: 07/08/2014, 15:22

... counted exactly  k−2 M g −1  -times since we have a free choice of selecting the remaining M g −1 cycles from C to form C  ⊇ {e, f} of size M g + 1. Hence the counting argument yields that the total number ... j  ) iff |i − i  | + |j − j  | = 2. The projective diamond grid P r of size r is obtained from D r by identifying the opposite pairs of its “boundary” vertices, th...

Ngày tải lên: 07/08/2014, 15:23

... (9),  j≥0 (−1 ) j β N j = β N 0 +  j≥1 (−1 ) j [dim(B j ) − rank(∂ j ) − rank(∂ j+1 )] = β N 0 +rank(∂ 1 )+  j≥1 (−1 ) j dim(B j ) = β N 0 +(n − β N 0 ) − m + k 2,2 +  j≥3 (−1 ) j dim(B j ) = n − m + ... −        k 1,2 −2 k 2,2 +k 3,2 −k 4,2 . . .        +        k 1,3 −k 2,3 +2k 3,3 −k 4,3 . . .        − ··· = n +  ...

Ngày tải lên: 07/08/2014, 07:21

... C n (q), where, C 2n (q)=  j  2n n − 3j  −  2n n − 3j − 1  · q n 2 −3 j 2 −j , C 2n+1 (q)=  j  2n +1 n − 3j  −  2n +1 n − 3j − 1  · q n 2 +n−3j 2 −2 j . Proof. In [K] it was shown that the ... by  m n  q := (1 − q m )(1 − q m−1 ) ···(1 − q m−n+1 ) (1 − q)(1 − q 2 ) ···(1 − q n ) , (Carl) whenever 0 ≤ n ≤ m,and0other...

Ngày tải lên: 07/08/2014, 06:20