a_(r+1)=t(s-1)的距离正则图

Distance-regular Graph with a_(r+1)=t(s-1)

设Γ是序为(s,t)直径为d的距离正则图,讨论了l(c,a,b)表示在交叉阵列l(Γ)中列(c,a,b)的个数,记r=r(Γ)=l(c1,a1,b1),s/=s/(Γ)=l(cr+1,ar+1,br+1),t/=t/(Γ)=l(cr+s/+1,ar+s/+1,br+s/+1).所得结论如下:设Γ=(X,E)是一个序为(s,t)的直径为d的距离正则图,如果cr+1=t,ar+1=t(s-1),则d=r+s/+1,cd=t/+1且Γ为正则拟2d边形.

Suppose F to be a distance regular graph with the diameter of d as well as an order of （s,t） , l（c,a,b） as a number of （c, a, b） in the intersection array with I （F） is discussed, which can be indicated as r =r（F） =l（c1 ,α1 ,b1 ） ,s＇ =s＇（F） -=/（c r＋1 ,α r＋1, b r＋1 ）, t＇ =t＇（F） =l（C r＋s＇＋1 ,αr＋s＇＋1 ,br＋s＇＋l ）The result will be as follows： suppose（F） = （X,E） is a distance regular graph with an order of （ s, t） and the diameter of d, given cr ＋ 1 = t, αr＋ l, = t （ s - 1 ）, then d = r ＋ s＇ ＋ 1, cd = t＇ ＋ 1 and F is a regular near 2d - polygon.