# Membership problem for context sensitive languages PSPACE-complete

I have read that the membership problem for CSL is PSPACE-complete but I couldn't find the proof anywhere. So I tried it myself.

Let's mark the membership problem for CSL as MEM. First I have to proof that $MEM \in PSPACE$. This should be easy, just take Turing Machine that generate words from L in lexicographic order and check if any is the same as the input word w. We can stop with the Turing machine when we reach length w+1.

Seccond, make a reduction from some PSPACE language. Quantified Boolean formula problem (QBF) seems to be suitable for this reduction. I have seen how to make a reduction from MEM to QBF but here I need the opposite. If I had a word w, I could make a formula based on the configuration I must go through to get w and all those configurations would mean true for the QBF. The representation could be just going from the binary code to some formulas.

But when going from the opposite direction, I don't know how to make CSL work from any given formula.

• Consult a good textbook. Apr 18 '16 at 10:24
• @YuvalFilmus Of which topic? I don't think this particular theorem is usually included in formal languages textbooks, and probably not in many (undergraduate?) complexity theory texts.
– Raphael
Apr 18 '16 at 10:29
• @Raphael If it's not in undergraduate texts, perhaps the OP should consult graduate texts. Another suggestion is to locate the original paper and read the proof. Apr 18 '16 at 10:30
• That said, Google gives rich results and even the Wikipedia article on PSPACE seems to point to the material you need.
– Raphael
Apr 18 '16 at 10:30

I'm assuming that the context-sensitive language is given to you as a context-sensitive grammar. Summarizing the Wikipedia article, here is how to show that the word problem for context-sensitive grammars is $\mathsf{PSPACE}$-complete.
First, we show that the word problem is in $\mathsf{PSPACE}$. The productions in a context-sensitive grammar cannot decrease the size of the word. Therefore, given a context-sensitive grammar $G$ and a word $w$, we can repeatedly apply productions in reverse, in the hope of eventually reaching the starting symbol $S$. We also keep a counter, and give up after $|\Sigma|^{|w|}$ steps (this ensures that the machine always halts), where $\Sigma$ contains both terminals and non-terminals; the size of this counter is $|w| \log |\Sigma| = O(n\log n)$ (here $n$ is the input size). This puts the word problem in $\mathsf{NSPACE}(O(n\log n)) \subseteq \mathsf{NPSPACE} = \mathsf{PSPACE}$, using Savitch's theorem.
In the other direction, Kuroda showed how to convert a linear bounded automaton (that is, a machine in $\mathsf{NSPACE}(n)$) to a context-sensitive grammar in such a way that the LBA accepts a word iff the CSG accepts it. Given a language $L \in \mathsf{SPACE}(p(n))$, we consider the language $L' = \{x\#^{p(|x|)} : x \in L \}$ obtained by padding $L$, and note that $L' \in \mathsf{NSPACE}(n)$. Kuroda's theorem gives an equivalent context-sensitive grammar $G'$. In order to check whether $x \in L$, we first compute $x' = x\#^{p(x)}$, and then check whether $G'$ accepts $x'$. This gives a polynomial time reduction from any language in $\mathsf{PSPACE}$ to the word problem for context-sensitive grammars, and so the latter is $\mathsf{PSPACE}$-complete.
• Does not alter your conclusion, but in the 2nd paragraph, should it not mean to give up after $|\Sigma|^{|w|+1}$ steps, if your argument is to not repeat a "configuration"-string in the "reverse parsing", whose number is upper bounded by $|\Sigma|^{|w|+1}$? Sep 7 '20 at 19:56