Patrick Dehornoy's handle reduction [1] technique that solves the word problem in the braid group can be modified so that it seems to solve the word problem in the groups of the form $B_{n}(d)$. In practice, for all the braid words $w$ that I have tested that are in the normal subgroup generated by $\sigma_{i}^{d}$, the algorithm is able to quickly show that $w$ is equivalent in $B_{n}(d)$ to the identity $e$ by applying the braid group identities together with the identity $\sigma_{i}^{d}=e$. If the algorithm gets stuck and is unable to reduce $w$ to $e$ after enough time, then we can be confident that $w$ is not equivalent to the identity $e$.

We say that a braid word is a $\sigma_{i}$-handle if it is of the form
$\sigma_{i}^{c}u\sigma_{i}^{-c}$ where $u$ is a braid word that does not contain any instance of $\sigma_{j}$ or $\sigma_{j}^{-1}$ whenever $j\geq i$. A good $\sigma_{i}$-handle is a $\sigma_{i}$-handle that does not contain a subword that is a $\sigma_{i-1}$-handle; equivalently, a good $\sigma_{i}$-handle is a $\sigma_{i}$-handle that either has no instance of $\sigma_{i-1}$ or has no instance of $\sigma_{i-1}^{-1}$.

Consider the following moves that relate braid words to other braid words.

Move 1: (handle reduction) Suppose that $z=u\sigma_{i}^{r}v\sigma_{i}^{-r}w$, $r\in\{-1,1\}$, and
$\sigma_{i}^{r}v\sigma_{i}^{-r}$ is a good $\sigma_{i}$-handle where $r\in\{-1,1\}$. Then perform the replacement $z\mapsto uv'w$ where $v'$ is the word obtained from $v$ by replacing each instance of $\sigma_{i-1}^{s}$ where $s\in\{-1,1\}$ with
$\sigma_{i-1}^{-r}\sigma_{i}^{s}\sigma_{i-1}^{r}$. Observe that the resulting braid word obtained after applying this move represents the same braid.

Move 2: Replace each $\sigma_{i}^{r}$ where $r\in\{1,-1\}$ with $\sigma_{n-i-1}^{r}$. This move conjugates the braid with the half-twist $\Delta_{n}$. This move together with Move 1 allows us to reduce right handles as well as left handles.

Move 3a: $u\sigma_{i}^{r}v\mapsto u\sigma_{i}^{d-r}v$ whenever $|d-r|\leq|r|$

Move 3b: $u\sigma_{i}^{r}v\mapsto u\sigma_{i}^{d+r}v$ whenever $|d+r|\leq|r|$

Observe that the resulting braid word obtained after applying Move 3 represents the same element in $B_{n}(d)$. We observe that if we obtain the empty word from $w$ by applying Moves 1-3, then $w$ represents the identity in $B_{n}(d)$.

Patrick Dehornoy has proven that the process of simply applying Move 1 (handle reduction) always terminates, and a braid word $w$ represents the trivial braid in $B_{n}$ if and only if this process terminates with the identity braid. In practice, handle reduction solves the word problem for braid groups very quickly. Moves 1-2 are both applied in order to heuristically minimize the length of a braid word representing a braid, and this braid word minimization technique has been used to attack braid based cryptosystems [2].

In all the words $w$ that I have shown to be equivalent in $B_{n}(d)$ to the identity, I was able to reduce $w$ to the identity simply by haphazardly applying Moves 1-3. Unfortunately, I do not know how to formally prove that one can solve the word problem in $B_{n}(d)$ by demonstrating that Moves 1-3 always simplifies a word $w$ to the empty word in a reasonable amount of time whenever $w$ represents the identity in $B_{n}(d)$.

[1] https://www.lmno.cnrs.fr/archives/dehornoy/Surveys/Dhn.pdf

[2] A practical attack on a braid group based cryptographic protocol. Alexei Myasnikov, Vladimir Shpilrain, and Alexander Ushakov. 2005
https://iacr.org/archive/crypto2005/36210085/36210085.pdf

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