The non-abelian squares are not context-free

. Clearly z ∈ (Σ 2 ) * ∩ R. Suppose m is the total number of 1's in z. Then m = 3n! + 4n + 2. However, the number of 1's in the second half of z is 4n! 3 + 2n + 1, which is not equal to m 2 . Hence z is not an abelian square. Thus z ∈ L. Definition 4. Let w = 0 s 0 10 s 1 • • • 10 s k be a word over the alphabet Σ. We define alt(w) = |{1 ≤ i < k : s i = s i+1 }|.

Define alt(•) over a language K as follows: alt(K) = max w∈K alt(w). Definition 5. A sequence of non-negative integers (a k ) n k=1 is called uneven if n > 1 and ∃i ∈ [1, n] such that a i = a i+1 . Here a n+1 = a 1 . Otherwise, it is called even. Definition 6. Suppose w is a word that contains a 1. Then w is called uneven if the sequence (s 1 + s k+1 , s 2 , . . . , s k ) is uneven, where w = 0 s 1 10 s 2 1 • • • 0 s k 10 s k+1 . Otherwise, it is called even.

Example 7. Suppose w is an even word and of the form 0 s 1 10 s 2 1 • • • 0 s k 10 s k+1 . Now we consider what w looks like. Since w is even, we get that

Proof. (By Ogden’s Lemma) For any n > 4, let z = w n 4 w 3 w n!+n 2 w 3 w 2(n!+n) 3

. By Lemma 3 we see that z ∈ T . Mark the first 4n bits of z, that is, the bits corresponding to w n 4 . Let m(s) denote the number of bits marked in s. Now we show by contradiction that no decomposition z = u 0 v 0 w 0 x 0 y 0 satisfies all the following three conditions:

Before further consideration into all possible decompositions, we first mark z with different colors. Mark the bits corresponding to w n 4 red. Mark the next 3 bits corresponding to w 3 blue. Mark the bits corresponding to w n!+n 2 green. Mark the bits corresponding to w 3 w 2(n!+n) 3 black. Define a new function m(color, x) as the number of bits in x colored color. Note that m(x) in our former definition is the same as m(red, x). Here is a picture of how z is colored:

Now we list all possible cases.

(i) Either v or x is the empty word. Without loss of generality, suppose x is empty.

(ii) v contains a 1 and v is uneven.

(iii) v contains a 1 and v is even.

(ii) Both v and x are non-empty words.

(i) v ∈ 0 + or x ∈ 0 + .

(ii) Both v and x contain a 1; v is uneven or x is uneven.

(iii) Both v and x contain a 1 and are even.

• m(blue, x) = 0 and m(green, x) > 1.

• m(blue, x) = 0 and m(green, x) = 1. (iii) m(red, x) = m(green, x) = 0 and m(black, x) > 0 Suppose there exists a decomposition z = uvwxy satisfying the above three conditions simultaneously.

Case i: First we consider the case when either v or x is empty. Without loss of generality, suppose x is empty. Then v cannot be empty, since vx is non-empty.

Case i.i: Suppose v = 0 k for some k ∈ N + . Then we select i = 4. Since there are more than 3 successive 0’s in v 4 , this is also true for uv 4 wx 4 y. However, no word in T contains more than 3 successive 0’s. Hence we get a contradiction. Case i.ii: Suppose v contains a 1 and is uneven. We pick i = 6. Then alt(v 6 ) ≥ 5 > alt(R) = 4 by Lemma 8. So uv 6 wx 6 y ∈ T , which violates condition C. Case i.iii: Now we consider when v is even. In this case v can be written in the form 0 k 1(0 k+s 1) p 0 s for some k, s, p ∈ N (as we mentioned in Example 7). Then it follows that m(green, v) = 0 and k + s = 3 by the following argument. Suppose m(green, v) > 0. Then m(blue, v) = 3. That is to say, the w 3 between w 4 ’s and w 2 ’s lies in v. Then v must be of the form r 1 01001r 2 for some words r 1 and r 2 . It follows that k + s = 2, since v is even. Now we select i = 2. Then uv 2 wx 2 y = w n 4 w l 3 w n!+n 2 w 3 w 2(n!+n) 3

for some l > 1, which violates condition C. Now suppose k + s = 3. Then we pick i = 2. It follows that uv 2 wx 2 y is of the form w l 4 w 2+2p k+s+1 w j 4 w 3 w n!+n

) is an abelian square, a contradiction.

Case ii: Both v and x are non-empty. In this case, we first show that both v and x contain a 1. Then, we show v and x are even. Finally we rule out all subcases under the condition that v and x are even.

Case ii.i: Suppose v = 0 k or x = 0 l for some k, l ∈ N + . By a similar analysis in Case i.i, we get that this case violates condition C.

Case ii.ii: Suppose v is uneven. By a similar analysis in Case i.ii, we see that this case violates condition C. The same applies to the case when x is uneven.

Case ii.iii: Now it remains to consider when v and x are even. Suppose v = 0 k 1(0 k+s 1) p 0 s for some k, s, p ∈ N, and x = 0 c 1(0 c+d 1) e 0 d for some c, d, e ∈ N.

Case ii.iii.i: First of all we consider the case when m(red, v) = 0. Then m(red, x) = 0 since x precedes v in z. It follows that m(red, vx) = 0, which violates condition A.

Case ii.iii.ii: Now we turn to the case when m(red, v) > 0. By the same argument in Case i.iii, we get that m(green, v) = 0 and k + s = 3. Note that p < n, for otherwise the condition m(green, v) = 0 cannot be satisfied. Now we consider the following subcases:

Case ii.iii.ii.i: If m(red, x) > 0, then m(green, x) = 0 and c

) is an abelian square, which violates condition C again.

Case ii.iii.ii.ii: If m(red, x) = 0 an

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