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

described by the following join-table:∨ a b ma a m mb m b mm m m mand ÊA 3is its endomorphismsemiring. From Proposition 2.1 the elements of ÊA 3are: ≀aaa≀, ≀bbb≀, the identity≀abm≀, ≀bam≀, ≀amm≀, ≀bmm≀, ≀mam≀, ≀mbm≀ and the absorbing element ≀mmm≀.We have the following addition table:+ ≀aaa≀ ≀bbb≀ ≀abm≀ ≀bam≀ ≀amm≀ ≀bmm≀ ≀mam≀ ≀mbm≀ ≀mmm≀≀aaa≀ ≀aaa≀ ≀mmm≀ ≀amm≀ ≀mam≀ ≀amm≀ ≀mmm≀ ≀mam≀ ≀mmm≀ ≀mmm≀≀bbb≀ ≀mmm≀ ≀bbb≀ ≀mbm≀ ≀bmm≀ ≀mmm≀ ≀bmm≀ ≀mmm≀ ≀mbm≀ ≀mmm≀≀abm≀ ≀amm≀ ≀mbm≀ ≀abm≀ ≀mmm≀ ≀amm≀ ≀mmm≀ ≀mmm≀ ≀mbm≀ ≀mmm≀≀bam≀ ≀mam≀ ≀bmm≀ ≀mmm≀ ≀bam≀ ≀mmm≀ ≀bmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀≀amm≀ ≀amm≀ ≀mmm≀ ≀amm≀ ≀mmm≀ ≀amm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀≀bmm≀ ≀mmm≀ ≀bmm≀ ≀mmm≀ ≀bmm≀ ≀mmm≀ ≀bmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀≀mam≀ ≀mam≀ ≀mmm≀ ≀mmm≀ ≀mam≀ ≀mmm≀ ≀mmm≀ ≀mam≀ ≀mmm≀ ≀mmm≀≀mbm≀ ≀mmm≀ ≀mbm≀ ≀mbm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mbm≀ ≀mmm≀≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀)So in the semigroup(ÊA3 , + there is no neutral element and the endomorphism≀mmm≀ is the absorbing element. The multiplication table is:.· ≀aaa≀ ≀bbb≀ ≀abm≀ ≀bam≀ ≀amm≀ ≀bmm≀ ≀mam≀ ≀mbm≀ ≀mmm≀≀aaa≀ ≀aaa≀ ≀bbb≀ ≀aaa≀ ≀bbb≀ ≀aaa≀ ≀bbb≀ ≀mmm≀ ≀mmm≀ ≀mmm≀≀bbb≀ ≀aaa≀ ≀bbb≀ ≀bbb≀ ≀aaa≀ ≀mmm≀ ≀mmm≀ ≀aaa≀ ≀bbb≀ ≀mmm≀≀abm≀ ≀aaa≀ ≀bbb≀ ≀abm≀ ≀bam≀ ≀amm≀ ≀bmm≀ ≀mam≀ ≀mbm≀ ≀mmm≀≀bam≀ ≀aaa≀ ≀bbb≀ ≀bam≀ ≀abm≀ ≀mam≀ ≀mbm≀ ≀amm≀ ≀bmm≀ ≀mmm≀≀amm≀ ≀aaa≀ ≀bbb≀ ≀amm≀ ≀bmm≀ ≀amm≀ ≀bmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀≀bmm≀ ≀aaa≀ ≀bbb≀ ≀bmm≀ ≀amm≀ ≀mmm≀ ≀mmm≀ ≀amm≀ ≀bmm≀ ≀mmm≀≀mam≀ ≀aaa≀ ≀bbb≀ ≀mam≀ ≀mbm≀ ≀mam≀ ≀mbm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀≀mbm≀ ≀aaa≀ ≀bbb≀ ≀mbm≀ ≀mam≀ ≀mmm≀ ≀mmm≀ ≀mam≀ ≀mbm≀ ≀mmm≀≀mmm≀ ≀aaa≀ ≀bbb≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀ ≀mmm≀)The noncommutative semigroup(ÊA3 , · has a neutral element ≀abm≀. Note thatthe subset(ÊA3 , ·)\{≀aaa≀, ≀bbb≀} is a noncommutative semigroup with absorbingelement ≀mmm≀.In the semiring ÊA 3the element ≀abm≀ is identity and ≀mmm≀ is an aditivelyabsorbing element. The set {≀ a a a ≀, ≀ b b b ≀, ≀ m m m ≀ } is an ideal of semiring ÊA 3.Since in the semiring ÊA nthere are elements which are roots of the additivelyabsorbing element ≀ m m . . . m ≀, namely≀ m a 1 m . . . m ≀ 2 = ≀ m m . . . m ≀.it follows that ÊA nis not a Viterbi semiring.22