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Additional info for Combinatorics, Paul Erdős is eighty, Volume 1

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The function Erase(v, U ) returns the set of half-edges obtained by erasing the preﬁx v of the words u appearing in the half-edges (u, q) ∈ U . In a second step, we build the set of states and the next state function of the resulting sequential transducer B. As for automata, we use a function Explore() which operates on the ﬂy. Explore(T , S, B) 1 T is a collection of sets of half-edges 2 S is an element of T 3 for each letter a do 4 (v, U ) ← Lcp(Next(S, a)) 5 NextB (S, a) ← (v, U ) 6 if U = ∅ and U ∈ / T then 7 T ←T ∪U 8 (T , B) ← Explore(T , U, B) 9 return (T , B) We can ﬁnally write the function realizing the determinization of a transducer into a sequential one.

The product of ρ and σ ⊂ A∗ × B ∗ is the relation ρσ = {(ur, vs) | (u, v) ∈ ρ, (r, s) ∈ σ}. Version June 23, 2004 40 Algorithms on Words The star of σ ⊂ A∗ × B ∗ is the relation σ ∗ = {(u1 u2 · · · un , v1 v2 · · · vn ) | (ui , vi ) ∈ σ, n ≥ 0}. A relation from A∗ to B ∗ is rational if it can be obtained from subsets of (A ∪ {ε}) × (B ∪ {ε}) by a ﬁnite number of operations of union, product and star. A rational relation that is a (partial) function is called a rational function. 4. g. on the alphabet {a, b} as ((a, aa) ∪ (b, bb))∗ .

The algorithm computing the composition of two transducers is easy to write. 5. 31. The right 2-shift. ComposeTransducers(S, T) 1 S and T are literal transducers 2 U ← NewTransducer() 3 for each edge (p, a, b, q) of S do 4 for each edge (r, b, c, s) of T do 5 add ((p, r), a, c, (q, s)) to the edges of U 6 for each edge (p, a, ε, q) of S do 7 for each state r of T do 8 add ((p, r), a, ε, (q, r)) to the edges of U 9 for each edge (r, ε, c, s) of T do 10 for each state p of S do 11 add ((p, r), ε, c, (p, s)) to the edges of U 12 InitialU ← InitialS × InitialT 13 TerminalU ← TerminalS × TerminalT 14 return U The composition can be used to compute an automaton that recognizes the image of a word (and more generally of a regular set) by a rational relation.