11
for i = 2, ∙ ∙ ∙, n + 1. Notice that
kt, = max kt-
JEN 1
for any i ∈ N.
The following cases need to be addressed:
(1). If 1 ≤ Z ≤ n, we have that for any i ∈ N, it holds
ajxl-bι = ɑɪx0 - bι ⅛ E⅛=ι klhafq(hf
≤ — Цац — ∑^nh=ι,h≠ι k^lhaih + ⅛+ι E⅛=ι aih
≤ -Цац - ∑nh=1,h≠l k}aιh + ⅛+ι ∑⅛=ι α⅛
≤ (⅛+1-⅛)Σ∑1¾
≤ 0.
It implies that /(жг) Ц I for z = l, ∙∙∙,n⅛l.
(2). If I = n + 1, we have that for any i ∈ N, it holds
an+ιχt - bn+1
= af+1x0 - bn+1 + E⅛=∣ ⅛⅛(⅛)
≤ EL⅜⅛(⅛)
≤ - E⅛=l ^⅛β(ra+l)⅛ + kn+1 E⅛=l a(n+l~)h
— ~ ∑⅛=l kn+ia(n+l)h + kn+1 E⅛=l a(n+l)h
— (⅛ + l - ⅛÷1) ∑⅛=l a(n+l)h
= 0.
It implies that /(жг) Ц n ⅛ 1 for all i = 1, ∙ ∙ ∙, n ⅛ 1. We conclude from (1) and
(2) with a contradiction. We are done.
Moreover, it is easy to derive the following lemma.
Lemma 3.6 Let a simplex P be given in standard form with the additional
condition that ∑j≠i ∣¾∣ < <⅛ holds for all i = 1, ∙ ∙ -, n. If P contains an integral
point, then the Labeling Rule results in at least n ⅛ 1 completely labeled simplices of
type I and no completely labeled simplex of type II.