Volume 61 Issue 2 (2011) - Годишник на ТУ - София - Технически ...

( ) n − 1(k = 1, . . . , n) from the all a i by ways. So the number of all possibilitiesk − 1isn∑( ) ( ) ( ) ( )n − 1 n − 1 n − 1n − 1= + + · · · + = 2 n−1k − 1 0 1n − 1k=1and this completes the proof.From the proof of the last theorem immediately followsCorollary 3.5 In the semiring ID (E An ) the addition and the multiplicationtables coincides and the identity i is both an additively neutral and a multiplicativelyneutral element.Using that all elements of ID (E An ) are both additively and multiplicativelyidempotent followsCorollary 3.6 The semiring ID (E An ) is a Viterbi semirimg.Let us consider for any i = 1, . . . , n − 1 the endomorphisms f i ∈ ID (E An )defined by{ai , if i = kf i (a k ) =m, if i ≠ k .But these maps are almost absorbing endomorphisms. They are idempotentsand f i · f j = f j · f i = ∞. The set of these endomorphisms is denoted byAA (ID (E An )). Immediately from Proposition 2.5 followsCorollary 3.7semiring ID (E An ).For every n ≥ 2 the set AA (ID (E An )) is an ideal of theLet the endomorphism f ∈ ID (E An ) has no more than k fixed elements fromthe set {a 1 , . . . , a n−1 }, where k < n − 1, and transforms all other elements inm. We denote the subset of ID (E An ) of these endomorphisms by ID k (E An ).So, ID 1 (E An ) = AA (N I (E An )).Let f, g ∈ ID k (E An ), the fixed points of f are a i1 , . . . , a is , where s ≤ k, andthe fixed points of g are a j1 , . . . , a jr , where r ≤ k. If{a k1 , . . . , a km } = {a i1 , . . . , a is } ∩ {a j1 , . . . , a jr }then the fixed points of f + g = f · g are a k1 , . . . , a km and m ≤ k. So, f + g =f · g ∈ ID k (E An ).Let φ ∈ ID (E An ). Then the fixed elements of the endomorphism f + φ =f · φ = φ · f are part of the fixed elements a i1 , . . . , a is of the endomorphism f. Sof · φ = φ · f ∈ ID k (E An ).Let us denote ID 0 = {∞}, ID 1 = ID 1 (E An ), . . . , ID n−2 = ID n−2 (E An ),ID n−1 = ID (E An ). Thus we prove27