Orbit-blocking words and the average-case complexity of Whitehead’s problem in the free group of rank 2

For any element 𝑤 of F 2 , there is some g ∈ F 2 such that no cyclically reduced image of 𝑤 under an automorphism of F 2 contains 𝑔 as a subword. Such a word 𝑔 can be produced explicitly for any given 𝑤.

Theorem 2 ([3])

Suppose that some conjugate of

and some conjugate of

form a basis of F ⁢ ( a , b ) , where p ≥ 1 , q ≥ 1 , and all of the exponents are non-zero. Then, modulo the possible replacement of 𝑎 by a − 1 or 𝑏 by b − 1 throughout, there are integers t > 0 and ε = ± 1 such that either

(the latter being an equality of sets) or, symmetrically,

Every primitive pair in F 2 is conjugate to a primitive pair where both entries are cyclically reduced.

Proof

The following proof of Lemma 1 was suggested by the referee. By way of contradiction, suppose there is a primitive pair ( g − 1 ⁢ u ⁢ g , h − 1 ⁢ v ⁢ h ) of minimum total length. Then ( u , g ⁢ h − 1 ⁢ v ⁢ h ⁢ g − 1 ) is a primitive pair, too, and the total length of the latter pair is not greater than that of the original pair. If g ≠ h and there is a cancellation in g ⁢ h − 1 ⁢ v ⁢ h ⁢ g − 1 , then we get a primitive pair with smaller total length, a contradiction. If g ≠ h and there is no cancellation in g ⁢ h − 1 ⁢ v ⁢ h ⁢ g − 1 , then ( u , g ⁢ h − 1 ⁢ v ⁢ h ⁢ g − 1 ) is a Nielsen reduced pair, and therefore any non-trivial product of 𝑢 and g ⁢ h − 1 ⁢ v ⁢ h ⁢ g − 1 is either equal to u ± 1 or has length at least 2. Therefore, the subgroup of F 2 generated by 𝑢 and g ⁢ h − 1 ⁢ v ⁢ h ⁢ g − 1 cannot contain both 𝑎 and 𝑏, a contradiction. ∎

Let u i be words in F 2 such that m a ⁢ ( u i ) ≤ 1 for all i ∈ { 1 , 2 , … , n } . Then m a ⁢ ( u 1 ⁢ ⋯ ⁢ u n ) ≤ n and m a ∘ ⁢ ( u 1 ⁢ ⋯ ⁢ u n ) ≤ n + 1 .

Proof

By way of contradiction, suppose there is a tuple ( u 1 , … , u n ) of minimum total length that does not satisfy the conclusion of the lemma. Then we can assume that there are no cancellations in the products u i ⁢ u i + 1 since otherwise we can choose our u i s differently to decrease Σ ⁢ | u i | while preserving their product and conditions of the lemma. (For example, if u i ends with 𝑎 and u i + 1 starts with a − 1 , we can replace u i by u i ⁢ a − 1 and u i + 1 by a ⁢ u i + 1 ; this will decrease Σ ⁢ | u i | by 2.)

Note that, since m a ⁢ ( u 1 ) ≤ 1 , we have m a ∘ ⁢ ( u 1 ) ≤ 2 . By induction, for k ≥ 2 , if m a ⁢ ( u 1 ⁢ ⋯ ⁢ u k ) ≤ k and m a ∘ ⁢ ( u 1 ⁢ ⋯ ⁢ u k ) ≤ k + 1 , then m a ⁢ ( u 1 ⁢ ⋯ ⁢ u k ⁢ u k + 1 ) ≤ k + 1 and m a ∘ ⁢ ( u 1 ⁢ ⋯ ⁢ u k ⁢ u k + 1 ) ≤ k + 2 since u k + 1 cannot start or end with a ± 2 by the conditions of the lemma. Thus, our tuple ( u 1 , … , u n ) will satisfy the conclusion of the lemma, and this contradiction completes the proof. ∎

Proof of Theorem 1

Let 𝑤 be our given word and 𝑙 its length. Let φ ∈ Aut ⁡ ( F 2 ) send ( a , b ) to ( u , v ) . By Lemma 1, Theorem 2, and the fact that swapping 𝑎 and 𝑏 is an automorphism, we may assume m a ⁢ ( u ) , m a ⁢ ( v ) ≤ 1 . Then, by Lemma 2, m a ∘ ⁢ ( φ ⁢ ( w ) ) ≤ l + 1 . Thus, a l + 2 ⁢ b l + 2 cannot appear as a subword of the cyclic reduction of φ ⁢ ( w ) . ∎

Suppose possible inputs of the above algorithm 𝒜 are cyclically reduced words that are selected uniformly at random from the set of cyclically reduced words of length 𝑛. Then the average-case time complexity (a.k.a. expected runtime) of the algorithm 𝒜, working on a classical Turing machine, is O ⁢ ( 1 ) , a constant that does not depend on 𝑛. If one uses the “Deque” (double-ended queue) model of computing [18] instead of a classical Turing machine, then the “cyclically reduced” condition on the input can be dropped.

Proof

Suppose first that the input word 𝑢 is cyclically reduced.

(1) First we address the complexity of the algorithm 𝒯. Here we use a result of [2] saying that the number of (freely reduced) words of length 𝐿 with (any number of) forbidden subwords grows exponentially slower than the number of all freely reduced words of length 𝐿.

In our situation, we have at least one B ⁢ ( u ) as a forbidden subword. Therefore, the probability that the initial segment of length 𝑘 of the word 𝑣 does not have B ⁢ ( u ) as a subword is O ⁢ ( s k ) for some 𝑠, 0 < s < 1 . Thus, the average time complexity of the algorithm 𝒯 is bounded by

which is bounded by a constant.

(2) Now suppose that the input word 𝑣 of length 𝑛 does not have any subwords B ⁢ ( u ) , so that we have to rely on the standard Whitehead algorithm 𝒲 for an answer. The probability of this happening is O ⁢ ( s n ) for some 𝑠, 0 < s < 1 , as mentioned before.

The worst-case time complexity of the Whitehead algorithm is known to be O ⁢ ( n 2 ) in the group F 2 (see [7]).

Thus, the average-case complexity of the composite algorithm 𝒜 is

which is bounded by a constant.

(3) Now suppose the input word 𝑢 is not cyclically reduced. Then we are going to cyclically reduce it. This cannot be done in constant (or even sublinear) time on a classical Turing machine, so here we are going to use the “Deque” model of computing [18]. This allows one to move between the first and last letter of a word in constant time. We are going to show that, with this facility, one can cyclically reduce any element 𝑣 of length 𝑛, in any F r , in constant time (on average) with respect to n = | v | . In fact, this was previously shown in [16], but we reproduce the proof here to make the exposition complete.

First, recall that the number of freely reduced words of length 𝑛 in F r is

The following algorithm, that we denote by ℬ, will cyclically reduce 𝑣 on average in constant time with respect to n = | v | .

This algorithm will compare the first letter of 𝑣, call it 𝑎, to the last letter, call it 𝑧. If z ≠ a − 1 , the algorithm stops right away. If z = a − 1 , the first and last letters are deleted, and the algorithm now works with this new word.

The probability of z = a − 1 is 1 2 ⁢ r for any freely reduced word whose letters were selected uniformly at random from the set { x 1 , … , x r , x 1 − 1 , … , x r − 1 } . At the next step of the algorithm, however, the letter immediately following 𝑎 cannot be equal to a − 1 if we assume that the input is a freely reduced word, so at the next steps (if any) of the algorithm ℬ, the probability of the last letter being equal to the inverse of the first letter will be 1 2 ⁢ r − 1 . Then the expected runtime of the algorithm ℬ on an input word of length 𝑛 is

The infinite sum on the right is known to be equal to 2 ⁢ r − 1 ( 2 ⁢ r − 2 ) 2 ; in particular, it is constant with respect to 𝑛. ∎