# Linear time algorithm for the word problem on \(B_3\)

The word problem is solved if one can put \(w\) in Garside normal form. The process is trivial in \(B_3\) and can be done in linear time on a Turing machine. Remember \(\Delta = \sigma_1\sigma_2\sigma_1\).

- Rewrite all generator with negative exponents into ones with positive exponents and a \(\Delta^{-1}\).
- Move every \(\Delta^{-1}\) to the beginning of the word. First locate the last \(\Delta^{-1}\), and move it forward using the relations \(\Delta^{2n+1}\sigma_i = \sigma_{3-i} \Delta^{2n+1}\) and \(\Delta^{2n}\sigma_i = \sigma_i\Delta^{2n}\). Pick up other \(\Delta^{-1}\)'s in between.
- Move every \(\Delta\) factor in the word to the beginning of the word. Read from the end of the word, move it forward using \(\Delta\) relations, pick up other \(\Delta\) on the way. Check if moving \(\Delta\) generates another \(\Delta\). \(B_3\) don't have \(\sigma_i\sigma_j = \sigma_j\sigma_i\) relation, therefore if a new \(\Delta\) is created from moving \(\Delta\)'s around, it is a local event. To be precise, during this process, \(x\Delta^m y = x\Delta^{m+1} y'\) if and only if the first 3 letter in the current presentation of \(y\) is \(\Delta\).

Posted by Chao Xu on 2011-07-05.