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--- ws1s_expressive_power_and_quantifier_elimination [2007/06/29 11:08]
vaibhav.rajan
+++ ws1s_expressive_power_and_quantifier_elimination [2007/06/29 16:39]
vaibhav.rajan
@@ Line 1: / Line 1: @@
+We know that Weak Monadic Second-order theory of 1 Successor (WS1S) is decideable through automata, that is, the set of satisfying interpretations of a subformula is represented by a finite-state automaton [Buc60b, Elg61]. The decision procedures (using the automata technique) have been implemented efficiently (for WS1S with certain restrictions) in MONA.
+The main problem with this approach is exponential blowup in the states of the automaton due to nested quantifiers [Nil01].
+We were interested in finding another way of deciding WS1S without using the automata. We began by investigating whether we can characterize relations in WS1S in a way that helps us eliminate quantifiers. Since we couldn't get any substantial results, we began looking at a subset of WS1S. This paper describes the key results that we obtained.
 * why the problem is interesting
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-**The binary case for the W-problem** Consider a binary relation $R(x,y)$ in $L_R$ and the corresponding formula $F(x,y)$ with two free variables $x$ and $y$. $F(x,y)$ defines a set $Q=\{(x,y)|\ F(x,y)\}$. Elements of the set $Q$ are recognized by an automaton $A$ with parallel inputs and input alphabet $\Sigma = \{\binom{0}{0},\binom{0}{1},\binom{1}{0},\binom{1}{1}\}$. The language L of A is a subset of the regular language corresponding to $\binom{0}{0}^*\binom{1}{0}\binom{0}{0}^*\binom{0}{1}\binom{0}{0}* + \binom{0}{0}^*\binom{0}{1}\binom{0}{0}^*\binom{1}{0}\binom{0}{0}^* + \binom{0}{0}^*\binom{1}{1}\binom{0}{0}^*$. Thus the input string to the automaton is of the form $\binom{0}{0}^{k_1}\binom{1}{0}\binom{0}{0}^{k_2}\binom{0}{1}\binom{0}{0}^{k_3}$ or $\binom{0}{0}^{k_4}\binom{0}{1}\binom{0}{0}^{k_5}\binom{1}{0}\binom{0}{0}^{k_6}$ or $\binom{0}{0}^{k_7}\binom{1}{1}\binom{0}{0}^{k_8}$.\\
+**The binary case for the W-problem** Consider a binary relation $R(x,y)$ in $L_R$ and the corresponding formula $F(x,y)$ with two free variables $x$ and $y$. $F(x,y)$ defines a set $Q=\{(x,y)|\ F(x,y)\}$. Elements of the set $Q$ are recognized by an automaton $A$ with parallel inputs and input alphabet $\Sigma = \{\binom{0}{0},\binom{0}{1},\binom{1}{0},\binom{1}{1}\}$. The language L of A is a subset of the regular language corresponding to $\binom{0}{0}^*\binom{1}{0}\binom{0}{0}^*\binom{0}{1}\binom{0}{0}^* + \binom{0}{0}^*\binom{0}{1}\binom{0}{0}^*\binom{1}{0}\binom{0}{0}^* + \binom{0}{0}^*\binom{1}{1}\binom{0}{0}^*$. Thus the input string to the automaton is of the form $\binom{0}{0}^{k_1}\binom{1}{0}\binom{0}{0}^{k_2}\binom{0}{1}\binom{0}{0}^{k_3}$ or $\binom{0}{0}^{k_4}\binom{0}{1}\binom{0}{0}^{k_5}\binom{1}{0}\binom{0}{0}^{k_6}$ or $\binom{0}{0}^{k_7}\binom{1}{1}\binom{0}{0}^{k_8}$.\\
 **Claim** A pair $(x,y)$ belongs to some set $Q=\{(x,y)|\F(x,y)\}$ for some $F$ iff the sets $\{k_i\}_i$ of possible lengths of the exponents $k_i$ are ultimately periodic.\\
 **Proof**
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 (a)$\binom{0}{0}^{k_1}\binom{1}{0}\binom{0}{0}^{k_2}\binom{0}{1}\binom{0}{0}^{k_3}$ \\
 (b)$\binom{0}{0}^{k_7}\binom{1}{1}\binom{0}{0}^{k_8}$\\
-case (a): We want to show that the sets {k_1}, {k_2} and {k_3} are ultimately periodic. Due to the little problem, it is sufficient to show that the language generated by the unary alphabet $\binom{0}{0}$: $\binom{0}{0}^{k_i}, i = 1,2,3$ is regular.
+case (a): We want to show that the sets $\{k_1\}$, $\{k_2\}$ and $\{k_3\}$ are ultimately periodic. Due to the little problem, it is sufficient to show that the language generated by the unary alphabet $\binom{0}{0}$: $\binom{0}{0}^{k_i}, i = 1,2,3$ is regular.
-We show that $\binom{0}{0}^{k_2}$ is regular by constructing an accepting automaton, $A = (\Sigma, \delta, s, f, S)$. Let $M = (\Sigma^{'}, \delta^{'}, s^{'}, f^{'}, S^{'})$ be the automaton that accepts the original input string (a). Let $q$ be the accepting run in $M$. In $q$, there exists a unique state $s$ such that $\delta^{'}(m, \binom{1}{0}) = n$ where $m$,$n$ are states in $q$. Define $s = n$. Similarly there exists $\delta^{'}(g, \binom{0}{1}) = h$. Define $f = g$. $S$ contains all the states of $S^{'}$ that are in $q$ between $s$ and $f$. And, $\delta$ is the restriction of $\delta^{'}$ to the transitions in $q$ between $s$ and $f$. $\Sigma = \{\binom{0}{0}\}$. Thus the automaton is defined unambiguously.
+We show that $\binom{0}{0}^{k_2}$ is regular by constructing an accepting automaton, $A = (\Sigma, \delta, s, f, S)$. Let $M = (\Sigma^{'}, \delta^{'}, s^{'}, f^{'}, S^{'})$ be the automaton that accepts the original input string (a). Let $q$ be the accepting run in $M$. In $q$, there exist unique states $m$ and $n$ such that $\delta^{'}(m, \binom{1}{0}) = n$. Define $s = n$. Similarly there exist unique $g, h$ such that $\delta^{'}(g, \binom{0}{1}) = h$. Define $f = g$. $S$ contains all the states of $S^{'}$ that are in $q$ between $s$ and $f$. And, $\delta$ is the restriction of $\delta^{'}$ to the transitions in $q$ between $s$ and $f$. $\Sigma = \{\binom{0}{0}\}$. Thus the automaton is defined unambiguously.
 We can similarly construct automata for the languages $\binom{0}{0}^{k_1}$ and $\binom{0}{0}^{k_3}$. In the former case, the start state corresponds to $s^{'}$ and the final state is the state (in $q$) from which there is a transition on consuming $\binom{1}{0}$. In the latter case, the final state corresponds to $f^{'}$ and the start state is that state in $q$ that is reached after consuming $\binom{0}{1}$.
-The construction of the automata for case (c) are along the same lines.
+The construction of the automata for case (c) is along the same lines.
 **If**: If $\{k_i\}$ is an ultimately periodic set. then the expression $p^k$ where $p$ belongs to a unary alphabet and $k \in \{k_i\}$ is a regular expression. This is because for each set $S(a,b)$ one can construct a regular expression (ppp...a times) concatenated with the kleene closure of (ppp...b times). Since $\{k_i\}$ is the finite union of sets of the form $S(a,b)$, one can construct a regular expression which is the union of all the regular expressions formed as described above. Thus, $\binom{0}{0}^{k}$ is a regular expression for $k \in \{k_i\}$.
-We now extend our alphabet to include $\binom{1}{0}$, $\binom{0}{1}$ and $\binom{1}{1}$. The above argument for $\binom{0}{0}^{k}$ still holds. (a), (b) and (c) are all regular expressions formed by concatenation. Thus, the input strings are accepted by finite automata and are satisfying interpretations of some formula.
+We now extend our alphabet to include $\binom{1}{0}$, $\binom{0}{1}$ and $\binom{1}{1}$. The above argument for $\binom{0}{0}^{k}$ still holds. (a), (b) and (c) are all regular expressions formed by concatenation. Thus, the input strings are accepted by finite automata. They are satisfying interpretations of some formula in $L^R$ because (a) can be written as $\exists k_1, k_2, x = k_1 \wedge y = x + k_2$, and (b) is simply $x = y$. (c) is symmetric to (a).
 The above reasoning can be extended to the n-ary case. The input string has n non-zero elements and n+1 expressions of the form $(\{(0...0)^{-1}\})^k$. A tuple $(x_1, x_2,...x_n)$ belongs to an n-ary relation iff the possible values of each exponent $k$  forms an ultimately periodic set.