I stumbled on a Twitter account yesterday called Wolfram|Alpha Can’t. It posts bizarre queries that Wolfram Alpha can’t answer. Here’s one that caught my eye.

result of extracting the i’s, j’s, and k’s in order from Finnegans Wake and interpreting as a quaternion product

— Wolfram|Alpha Can’t (@wacnt) May 17, 2017

Suppose you did extract all the *i*‘s, *j*‘s, and *k*‘s from James Joyce’s novel Finnegans Wake. How would you answer the question above?

You could initialize an accumulator to 1 and then march through the list, updating the accumulator by multiplying it by the next element. But is is there a more efficient way?

Quaternion multiplication is not commutative, i.e. the order in which you multiply things matters. So it would not be enough to have a count of how many times each letter appears. Is there any sort of useful summary of the data short of carrying out the whole multiplication? In other words, could you scan the list while doing something other than quaternion multiplication, something faster to compute? Something analogous to sufficient statistics.

We’re carrying out multiplications in the group Q of unit quaternions, a group with eight elements: ±1, ±*i*, ±*j*, ±*k*. But the input to the question about Finnegans Wake only involves three of these elements. Could that be exploited for some slight efficiency?

How would you best implement quaternion multiplication? Of course the answer depends on your environment and what you mean by “best.”

Note that we don’t actually need to implement quaternion multiplication in general, though that would be sufficient. All we really need is multiplication in the group Q.

You could implement multiplication by a table lookup. You could use an 8 × 3 table; the left side of our multiplication could be anything in Q, but the right side can only be *i*, *j*, or *k*. You could represent quaternions as a list of four numbers—coefficients of 1, *i*, *j*, and *k*—and write rules for multiplying these. You could also represent quaternions as real 4 × 4 matrices or as complex 2 × 2 matrices.

If you have an interesting solution, please share it in a comment below. It could be interesting by any one of several criteria: fast, short, cryptic, amusing, etc.

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