The name is absent

12

Proof: Consider the simplest case in which P contains a single integral point, say
w. The general case can be shown in a similar way. Choose w to be the starting
point since it is allowed. Take an arbitrary index к from N and set l(ιυ) = к
artificially. Then the algorithm will terminate with a completely labeled simplex,
say σ⅛, within a finite number of steps. If w is a vertex of σ⅛, then restore the true
label of w, i.e., Z(w) = 0. Hence σ⅛ is of type I. Otherwise, it must be of type
II. However, this can not happen according to Theorem 3.5. Hence repeat the
procedure over all indices of N. We complete the proof.                       □

It is not clear whether the Exclusion Theorem also holds for the problems in
standard form without the additional condition. In order to provide a complete
answer in that case, let us define for к ∈ ʌ a subset C⅛ of Zⁿ by

C_k = { x ∈ Zⁿ I a_j^τ x > b₃ for j ∈ N__k }.

Now we establish the following procedure.

Step (1) Set к = 1.

Step (2) Choose any starting point v^k E C_k and implement the algorithm. If an integral
point in P is found, then stop. Otherwise, к becomes к ⅛ 1.

Step (3) If к = n ⅛ 2, then stop. Otherwise, go to Step (2).

We still have to discuss how to obtain a starting point v^k ∈ C_k for each к E N.
For each к ∈ N, we define

q(^k) = -q(k∖

We adopt the following labeling rule.

Modified Labeling Rule: To x E Zⁿ, we assign x with the label

f(ʃ) = min{ j E ∖-∣- ∖ aj x — bj = min{α^s — h_k. l∣ E N__k ∖a[x — bf_l ≤ 0 } }.

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