Ngày tải lên :
08/08/2014, 12:23
... W ∈ A( C2 ⊕ C2n ) and W | ( A) A We have |W | = 2a +2 and thus by Lemma 4.9 |A| −|W | +2 = 2n +2 ( 2a +2) +2 ∈ L(B) So, we have 2( n a +1) ∈ {2, 3, 2n, 2n +1, 2n +2} and consequently n a +1 ∈ {1, n, n +1} Yet, ... which by the minimality of A3 implies that A1 = ghf1 f2 Since |A1 | + |A2 | = 4n and the lengths of A1 and A2 is odd, we may assume that |A2 | ≥ 2n + Again, let g1 g2 h1 h2 | A2 such that π(gi ... [0, 2n − 1] and bI , cI ∈ {0, 1} Since by F2 g {2, 3} = g2 + g3 = g1 + g {1, 2, 3} , it follows that a2 + a3 ≡ (mod 2n), b2 = b3 , and c2 = c3 ; as well as a1 + a {1, 2, 3} ≡ (mod 2n), b1 = b {1, 2, 3} , and...