The name is absent



24

for all к, I E N. Now it immediately follows from (4.2) that

λkh3k = 0

for all h and k. Hence xk = x0 and l(xk) = 0. It is a contradiction. We complete
the proof.                                                                       □

Theorem 2.4 is a straightforward result of the above results.

5 Extension to general п-dimensional polytopes

Let M denote the set of integers { 1, ∙ ∙ ∙ , m }. The problem is to test the integral
property of a general n-dimensional polytope
P given by

P = { x Rn I Ax ≤ b },

where ¾τ = (аг1, ...,c⅛n) is the г-th row of the m by n matrix A for i = 1, ..., m,
and b = (δι, ∙ ∙ ∙ , δm)τ is a vector of Rm. It is clear that m ≥ n ⅛ 1. As usual we
may assume that ɑɪ,...,
am are integral vectors of Rn, and b = (b1,   , bm)τ is an

integral vector of Rm. Finally we assume that none of the constraints ɑʃæ ≤ bi,
i
M ,is redundant, and that there is a subset J, with cardinality n ⅛ 1, of M such
that

{ x Rn I ɑɪx ≤ bi, i ,J }

is an п-dimensional simplex. In the sequel we take J = N for simplicity of notation.
Compared with the labeling rule in Section 2, we have the following generalized
labeling rule.

Generalized Labeling Rule: If there is an index i E M for which ajx — bi > 0,
we assign
x E Zn with the label

/(æ) = min{ j E N aj x — bj = max{ ɑɪæ — ∕y} }.



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