1. Peter Černo

2.  Suppose we have a sample computation for (delta) clearing restarting automata.  Suppose that the inferred automaton accepts some wrong words.  There are two ways how to restrict the resulting inferred language:  We can increase k – the length of contexts.  We can change the sample computation by incorporating some delta rules.

3.  Consider the following sample computation: ¢ abababababababab $ ⊢ M ¢ abababababababb $ ⊢ M ¢ abababababbabb $ ⊢ M ¢ abababbabbabb $ ⊢ M ¢ abbabbabbabb $ ⊢ M ¢ abbabbabbab $ ⊢ M ¢ abbabbabab $ ⊢ M ¢ abbababab $ ⊢ M ¢ abababab $ ⊢ M ¢ abababb $ ⊢ M ¢ abbabb $ ⊢ M ¢ abbab $ ⊢ M ¢ abab $ ⊢ M ¢ abb $ ⊢ M ¢ ab $ ⊢ M ¢ λ $ accept.  We can collect 15 reductions.

4.  For k = 4 we get the following instructions:  ({¢ab, abab}, a, {b$, babb}),  ({¢a, abba}, b, {b$, bab$, baba}),  (¢, ab, $).  For the resulting 4-cl-RA -automaton M the following holds: L(M) ∩ {(ab) n | n>0} = {(ab) 2 m | m≥0}.  However, this does not work for smaller k.

5.  For k = 3 we get the following instructions:  ({¢ab, bab}, a, {b$, bab}),  ({¢a, bba}, b, {b$, bab}),  (¢, ab, $).  Now the resulting 3-cl-RA -automaton M accepts the wrong word ababab :  ababab ⊢ M abbab ⊢ M abab ⊢ M abb ⊢ M ab ⊢ M λ.  We blame the instruction (¢ab, a, bab) which caused the reduction: ababab ⊢ M abbab.

6.  Why? The instruction reduced the wrong word ababab to the correct word abbab.  abbab is ok – it is in the sample computation.  Idea: Let us replace the instruction:  ({¢ab, bab}, a, {b$, bab}) by the delta instruction:  ({¢ab, bab}, a → Δ, {b$, bab}).  How does the sample computation change?

7.  The modified sample computation is now: ¢ abababababababab $ ⊢ M ¢ abababababababΔb $ ⊢ M ¢ abababababΔbabΔb $ ⊢ M ¢ abababΔbabΔbabΔb $ ⊢ M ¢ abΔbabΔbabΔbabΔb $ ⊢ M ¢ abΔbabΔbabΔbab $ ⊢ M ¢ abΔbabΔbabab $ ⊢ M ¢ abΔbababab $ ⊢ M ¢ abababab $ ⊢ M ¢ abababΔb $ ⊢ M ¢ abΔbabΔb $ ⊢ M ¢ abΔbab $ ⊢ M ¢ abab $ ⊢ M ¢ abΔb $ ⊢ M ¢ ab $ ⊢ M ¢ λ $ accept.  Unfortunately, it does not help us if k = 3.

8.  Consider the following sample reduction:  abababΔb ⊢ M abΔbabΔb  As you can see, this reduction also works with the word ababab :  ababab ⊢ M abΔbab because the length of contexts is k = 3.  Is there any other way how to modify the instruction ({¢ab, bab}, a, {b$, bab}) ?  Yes, it is.

9.  It is not enough to replace a single letter a :  ({¢ab, bab}, a → Δ, {b$, bab}).  We have two simple choices for two letters:  ({¢ab, bab}, ab → Δ, {$, ab}),  ({¢a, ba}, ba → Δ, {b$, bab}).  We move one letter from the context to the rule and thus extend the context horizon.  In the following we show how does the sample computation change in both cases.

10.  If we use ({¢ab, bab}, ab → Δ, {$, ab}) we get the following new sample computation: ¢ abababababababab $ ⊢ M ¢ abababababababΔ $ ⊢ M ¢ abababababΔabΔ $ ⊢ M ¢ abababΔabΔabΔ $ ⊢ M ¢ abΔabΔabΔabΔ $ ⊢ M ¢ abΔabΔabΔab $ ⊢ M ¢ abΔabΔabab $ ⊢ M ¢ abΔababab $ ⊢ M ¢ abababab $ ⊢ M ¢ abababΔ $ ⊢ M ¢ abΔabΔ $ ⊢ M ¢ abΔab $ ⊢ M ¢ abab $ ⊢ M ¢ abΔ $ ⊢ M ¢ ab $ ⊢ M ¢ λ $ accept.

11.  For k = 3 we get the following instructions:  ({¢ab, Δab}, Δ, {$, ab$, aba}),  ({¢ab, bab}, ab → Δ, {$, abΔ}),  (¢, ab, $).  For the resulting 3-cl-RA -automaton M the following holds: L(M) = {(ab) 2 m | m≥0} ∪ {λ}.  Note that the symbol Δ codes the letter b from the original sample computation.

12.  If we use ({¢a, ba}, ba → Δ, {b$, bab}) we get the following new sample computation: ¢ abababababababab $ ⊢ M ¢ ababababababaΔb $ ⊢ M ¢ ababababaΔbaΔb $ ⊢ M ¢ ababaΔbaΔbaΔb $ ⊢ M ¢ aΔbaΔbaΔbaΔb $ ⊢ M ¢ aΔbaΔbaΔbab $ ⊢ M ¢ aΔbaΔbabab $ ⊢ M ¢ aΔbababab $ ⊢ M ¢ abababab $ ⊢ M ¢ ababaΔb $ ⊢ M ¢ aΔbaΔb $ ⊢ M ¢ aΔbab $ ⊢ M ¢ abab $ ⊢ M ¢ aΔb $ ⊢ M ¢ ab $ ⊢ M ¢ λ $ accept.

13.  For k = 3 we get the following instructions:  ({¢a, Δba}, Δ, {b$, bab}),  ({¢a, aba}, ba → Δ, {b$, baΔ}),  (¢, ab, $).  For the resulting 3-cl-RA -automaton M the same formula holds: L(M) = {(ab) 2 m | m≥0} ∪ {λ}.  Again the symbol Δ codes the letter b from the original sample computation.

14.  Note that the for k < 3 the previous algorithm does not work.  It is also not obvious how to replace more letters or even more instructions without disturbing the sample computation.  The question is whether we can do these replacements in some automated way?  Is this idea applicable to other examples?