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Word problem for symmetric group is linear on RAM

1 Linear time algorithm for symmetric group

Problem1The word problem for symmetric groups

Input:* \(w\) is a word of length \(l\) from the presentation \(S_n = \langle x_1,x_2,\ldots,x_n \mid x_i^2 = 1, x_{i+1}x_ix_{i+1} = x_ix_{i+1}x_i, x_ix_j = x_jx_i \rangle\) where \(|i-j|\neq 1\).

Output: Return true if \(w\) is the identity, else return false.

The representation was crucial for coming up with a linear time algorithm respect to \(n\) and \(l\). This is not a word problem on one group, but on a set of group.


The following is a \(O(n+l)\) algorithm for the word problem for symmetric groups on RAM.

  1. Produce an array a of size \(n\), Such that a[i] = i. (Array start with index 1)
  2. Reading the word letter by letter. If one encounters \(x_i\), swap(a[i],a[i+1]).
  3. Test if the a[i] == i for all \(i\). If true, return true, else return false.

The algorithm takes \(O(n+l)\) time is obvious. The correctness needs to be justified.

\(x_i\) can be represented as the transposition \((i~i+1)\). Define \((n~n+1) = (n~1)\).

Represent a element of the group as a permutation \(\pi\) in the 2 line notation. wlog, assume \(\pi(j) = i\) and \(\pi(k) = i+1\).

\[ \begin{pmatrix} 1 & \cdots &j & \cdots & k& \cdots & n \\ \pi(1) & \cdots & i & \cdots & i+1 & \cdots & \pi(n)\end{pmatrix}(i~i+1) = \begin{pmatrix} 1 & \cdots &j & \cdots & k& \cdots & n \\ \pi(1) & \cdots & i+1 & \cdots & i & \cdots & \pi(n)\end{pmatrix} \]

If we call \(j\) the index of \(i\) if \(\pi(j) = i\). Then each transposition is a swap of indices.

The value of a[i] in the algorithm stores the index of \(i\). Array a represent the identity iff a[i] = i for all \(i\).

This proves the the correctness of the algorithm.

The algorithm can be modified so it runs in \(O(l n!)\) time for a Turing machine. For each fixed \(n\), a Turing machine \(M\) can construct another Turing machine \(M_n\), such that it store the state of the array as the state of the Turing machine.

This proves every symmetric group is automatic. For any fixed \(S_n\), the Turing machine \(M_n\) can solve the problem in \(O(l)\) time without writing anything to the tape and can only move to the right, which is equivalent to a finite state automata.


Automatic is a property of a group, not a set of groups. That's why \(n\) is ignored in the \(O(ln!)\), because it's fixed for each \(S_n\). I was confused for a while before I read a concrete definition.

2 Algorithms on reduce the word to normal form

The normal form of a word \(w\in S_n\) is \(w = u_1u_2\ldots u_n\), such that \(u_i\in U_i\), and \(U_i = \{1, x_n, x_nx_{n-1}, \ldots, x_n\ldots x_1\}\).

One can construct a purely symbolic algorithm that apply only the group relations. We measure the time by the amount of group relations used.

Siegel proposed an \(O(l^2)\) algorithm to solve this problem. It is shown in these slides.

If there exist an algorithm \(A(w)\) that write the word \(w\) of length \(l\) in normal form in \(O(f(l,n))\) time., then one can always make it into an algorithm taking \(O(l f(n^2,n))\) time.

Observe that \(w = w'yz\) where \(y\) and \(z\) are words in normal form, and the length of \(|z|\) is maximized. \(A(w) =A(w'A(yz))\). Here \(y\) can be worst case, one single letter, it doesn't change the complexity. Let's introduce two algorithms. A' and A''.

\(A'(w)\) first find the \(z\) in the description, then returns the value of \(A''(w',A(x_iz))\), where \(w = w'x_iz\).

Recursively calculate \(A''(w,z)\) with the following definition. \(A''(1,z) = z\). \(A''(w,z) = A''(w', A(x_iz))\), where \(w = w'x_i\).


\(A(w) = A'(w)\) and runs in \(O(l f(n^2,n))\) time.


\(A''(w,z)\) can ran at most \(l\) times, each time it makes a call to \(A(w)\), contribute the factor \(O(f(n^2,n))\).

In particular, Siegel's algorithm can be modified to run in \(O(l n^4)\) time.

Posted by Chao Xu on 2011-06-21.
Tags: BSU REU, computational complexity, group theory.