For strings $w$ and $t$, if they have the same length and comprise the same characters (namely, they are two permutations of these characters), then $w\sim t$. For a string $w$, define an operator $\operatorname {SCRAMBLE}$ such that $$\operatorname {SCRAMBLE}(w):=\{t\mid t\sim w\}, $$ and for a language $A$, $$\operatorname {SCRAMBLE}(A):=\{t\mid t\sim w, \exists w\in A\}.$$ Prove that: given an alphabet $\Sigma:=\{0,1\}$, for any regular language $L$, $\operatorname {SCRAMBLE}(L)$ is a context-free language.

I have tried to use the pumping lemma of regular language to show that it satisfies the pumping lemma of context-free language after the $\operatorname {SCRAMBLE}$ operation, but I failed to find the pumping substring of $\operatorname {SCRAMBLE}(L)$. I have also tried PDA construction but to no avail. Can anyone help me with it? Any answer or hint will be appreciated. Thanks in advance.

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    $\begingroup$ You can not use the Pumping lemma in this way; non-context-free languages can fulfill its conditions. You need to show that there is a CFG or NPDA (or...) for that language. $\endgroup$
    – Raphael
    Apr 5, 2016 at 17:05
  • $\begingroup$ In the definition of $\operatorname {SCRAMBLE}(A)$, the quantifier $\exists w$ should probably go before the statement $t \sim w$. $\endgroup$
    – DylanSp
    Apr 5, 2016 at 17:15
  • $\begingroup$ @Raphael but it seems hard for me to construct a CFG or NPDA... So do you have any suggestion about it? $\endgroup$
    – Vim
    Apr 5, 2016 at 23:40
  • $\begingroup$ Thanks, @YuvalFilmus -- this has been re-opened. Vim, may I encourage you to edit the question based on the feedback and comments you've received? $\endgroup$
    – D.W.
    Apr 6, 2016 at 19:21
  • $\begingroup$ This is also proved in a note by Dexter Kozen: cs.cornell.edu/~kozen/papers/2and3.pdf. $\endgroup$ May 15, 2016 at 19:56

1 Answer 1


First, two remarks. When $L = (01)^*$, we have $\newcommand{\scramble}{\mathrm{SCRAMBLE}}\scramble(L) \cap 0^*1^* = \{ 0^n 1^n : n \geq 0 \}$, and this shows that we can't expect $\scramble(L)$ to be regular.

Second, when $L=(012)^*$, we have $\scramble(L) \cap 0^*1^*2^* = \{ 0^n 1^n 2^n : n \geq 0 \}$. This shows that it's important that the alphabet have only two symbols.

Now for the proof. It is actually enough to assume that $L$ is context-free (and not necessarily regular). For a word $w$, let $p(w) = (\#_0(w),\#_1(w))$, and define $P(L) = \{ p(w) : w \in L \}$. Parikh's theorem states that $P(L)$ is a semilinear set, that is, a finite union of sets of the form $$ x + \sum_{i=1}^m \mathbb{N} y_i, $$ where $x,y_1,\ldots,y_m \in \mathbb{N}^2$. It is thus enough to show that for every linear set $S$, the language $L(S)$ of words $w$ such that $p(w)$ belongs to $S$ is context-free.

A pushdown automaton accepting $L(S)$ proceeds as follows. We will assume for simplicity that $x = (0,0)$; for general $x = (x_0,x_1)$, the PDA will ignore the first $x_0$ zeroes and $x_1$ ones, but will only accept if these actually appear.

When the PDA starts running, it guesses a mode $i \in \{1,\ldots,m\}$. When it reads $0$, it pushes it to the stack. Meanwhile, it keeps track of the number of $1$s it has read. Once it has read $(y_i)_1$ of them, it removes $(y_i)_0$ zeroes from the stack (pushing, if necessary, "negative" $0$s which will get cancelled later with actual $0$s). Then it guesses another mode $i \in \{1,\ldots,m\}$ (possibly the same one).

Upon reaching the end of input, the automaton verifies that it hasn't read any $1$s in its current phase, and that the stack is empty. This completes the construction. We leave it to the reader to prove that it actually works.

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    $\begingroup$ Correct me if I am wrong, but it seems that you proved that if $L$ is context-free, then scramble($L$) is accepted by a non-deterministic one-counter automata. $\endgroup$ Apr 10, 2016 at 19:22
  • $\begingroup$ That's right. So scrambling could make a language easier. $\endgroup$ Apr 10, 2016 at 19:34
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    $\begingroup$ It says for any semiliner set, the language of words whose parikh vectors are in the set is regular. However, the language of stings with an equal number of 0's and 1's is not regular. $\endgroup$ Apr 10, 2016 at 20:19
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    $\begingroup$ Whoever wrote the article probably meant that every semilinear set can be "realized" by a regular language – in your case $(01)^*$. If you find a mistake in Wikipedia, go ahead and correct it – that's part of the features of Wikipedia. $\endgroup$ Apr 10, 2016 at 20:23
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    $\begingroup$ Great! Thanks for the clarification. Next, how large are the y_i's in terms of the state complexity of the context-free language? 2^n or 2^n^2 maybe? $\endgroup$ Apr 10, 2016 at 20:26

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