21
Example 9. We are given
|
-1 |
-1 |
-1 | |
|
3 |
1 |
1 | |
|
A = |
1 |
3 |
1 |
|
1 |
1 |
3 | |
|
Then |
0 |
1 |
0 |
|
U = |
0 |
0 |
1 |
|
-1 |
-1 |
-1 | |
|
such that |
1 |
0 |
0 |
|
-1 |
2 |
0 | |
|
À = |
-1 |
0 |
2 |
|
-3 |
—2 |
—2 |
Now let us give a proof of Theorem 2.2. We are given a polytope which has the
standard form. Since P contains no integral point, the algorithm terminates with
a completely labeled simplex of type 77, say, σ1(⅛1,π), within a finite number of
steps. Suppose to the contrary that there is another completely labeled simplex of
tpye 77, say, σ2(y1, p). Without loss of generality we may assume that π is equal to
(1,..., n), l(xt^) = i for any i ∈ .V. and хг = yl for all i ∈ ʌ except for some index k,
1 < к < n -∖- 1. It implies that l(xt^) = l(yt) = i for all i ∈ N. We have to consider
the following cases:
(1). If к = 1, then /∕l = xn+1 ⅛ ¢(1). Since l(xn+1~) = n ⅛ 1, we have
<P+∖x" + ' - ^n+ι > alχn+1 - bt,l = 1, ■ ■ ■ ,n.