Computer Science Asked on November 5, 2021
Let ||L|| be the sum of all lengths of words in L und N(L) the number of equivalence claesses for the Relation $equiv_L$ from Myhill–Nerode theorem. Proof, which values d can have with $d:=||L||-N(L),dinmathbb{Z}$
Some conclusions I came up with so far:
So d can be anything from ${ 0,…,infty }$
Might this be enough for an answer or am I missing any cases/details? Where could I be proofing more formally?
In order to make the question nontrivial, you have to assume that $L$ is finite. You haven't specified what the alphabet is, so I will assume that the alphabet is arbitrary.
First, let us show that $N(L) leq |L|+2$. We can construct a DFA for $L$ whose states consist of all prefixes of words in $L$, together with a failure state. This DFA contains an initial state (corresponding to the empty prefix), a failure state, and one state for every non-empty prefix. In particular, $$ N(L) leq 2 + sum_{w in L} |w| = 2 + |L|. $$ In other words, $d geq -2$.
Second, let us consider the language $L_Sigma = {sigma : sigma in Sigma}$ over an alphabet $Sigma$. The minimal DFA for this language contains three states, and so the value of $d$ for $L_Sigma$ is $$ |L_Sigma| - N(L_Sigma) = |Sigma| - 3. $$ As $Sigma$ ranges over all possible alphabets, we obtain all values which are at least $-2$.
Answered by Yuval Filmus on November 5, 2021
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