Proof Since the graph between X and Y is not c-regular,there must be two subsets X and Yi of X and Y,respectively,with Xi>X,Yil >eYl and ld(X1,Yi)-al>e.Let X2=X\X1,Y2 =Y\Yi and u(Xi,Yi)=d(Xi,Yi)-a.Then 点器r -品w-品+盆 品器 isiJs2XlY = Note that the second line holds since 多-xn-n The result therefore follows. 3Proof Since the graph between X and Y is not -regular, there must be two subsets X1 and Y1 of X and Y , respectively, with |X1| ≥ |X|, |Y1| ≥ |Y | and |d(X1, Y1)−α| > . Let X2 = X\X1, Y2 = Y \Y1 and u(Xi , Yj ) = d(Xi , Yj ) − α. Then  4 ≤ X 1≤i,j≤2 |Xi ||Yj | |X||Y | u(Xi , Yj ) 2 = X 1≤i,j≤2 |Xi ||Yj | |X||Y | d(Xi , Yj ) 2 − 2α X 1≤i,j≤2 |Xi ||Yj | |X||Y | d(Xi , Yj ) + α 2 X 1≤i,j≤2 |Xi ||Yj | |X||Y | = X 1≤i,j≤2 |Xi ||Yj | |X||Y | d(Xi , Yj ) 2 − α 2 . Note that the second line holds since X 1≤i,j≤2 |Xi ||Yj | |X||Y | d(Xi , Yj ) = d(X, Y ) = α. The result therefore follows. ✷ 3