Let Ξ£ = {1,2,3,4}. For π₯ β Ξ£β , let ππππ(π₯) be the product of symbols inβ¦
2025
Let Ξ£ = {1,2,3,4}. For π₯ β Ξ£β , let ππππ(π₯) be the product of symbols in π₯ modulo 7. We take ππππ(π) = 1, where π is the null string.
For example, ππππ(124) = (1 Γ 2 Γ 4) mod 7 = 1.
Define πΏ = {π₯ β Ξ£β | ππππ(π₯) = 2}.
The number of states in a minimum state DFA for πΏ is ___________. (Answer in integer)
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Correct answer: 6
Interpretation: the DFA state after reading a string can be taken as the product of its symbols modulo 7 (with the start state equal to 1 for the empty string).
The nonzero residues modulo 7 (1 through 6) form a multiplicative group of order 6.
The alphabet contains the symbol 3, and 3 is a generator of this multiplicative group, so by multiplying by symbols from the alphabet we can reach every nonzero residue. Hence all residues 1,2,3,4,5,6 are reachable states.
We can build a DFA with one state for each residue 1..6, with start state 1 and the accepting state being residue 2.
Distinctness argument: for two distinct residues r and s, choose a string whose product equals r^{-1}Β·2 (which exists because every nonzero residue is reachable). Multiplying r by that product gives 2 (accept), while multiplying s gives sΒ·r^{-1}Β·2 β 2. Thus r and s are distinguishable.
Conclusion: the minimal DFA requires one state for each of the six nonzero residues, so the number of states is 6.
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