Karatsuba multiplication of two complex numbers can be performed with just three real multiplications (instead of four) as follows: $$(a+bi)(c+di) = (ac-bd) + i ((a+b)(c+d) - ac-bd)$$ We only need the products $ac$, $bd$ and $(a+b)(c+d)$. Does there exists a similar trick for quaternion multiplication? A naive multiplication would need $16$. This can probably be reduce to $9$ using the same trick as above (and taking into account that quaternion multiplication is non-commutative). My question is it known that $9$ is the lower bound? If not, does there exist an algorithm, which does quaternion multiplications in less than $9$ real multiplications?
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closed as off-topic by Will Jagy, darij grinberg, Joonas Ilmavirta, Alex Degtyarev, Dima Pasechnik Apr 24 '15 at 5:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – darij grinberg, Joonas Ilmavirta, Alex Degtyarev, Dima Pasechnik
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3$\begingroup$ Duplicate of math.stackexchange.com/questions/1222820/… $\endgroup$ – Igor Rivin Apr 24 '15 at 2:24
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1$\begingroup$ Indeed, word-for-word duplicate. 17762, you ought to have included a link to the other question at each site. But I do note that no answers were given at the other site. $\endgroup$ – Gerry Myerson Apr 24 '15 at 2:32
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2$\begingroup$ ...though an apparently useful comment was given by a certain Gerry Myerson... $\endgroup$ – Noam D. Elkies Apr 24 '15 at 3:11
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1$\begingroup$ theworld.com/~sweetser/quaternions/ps/cornellcstr75-245.pdf claims magic number $8$. $\endgroup$ – Brout Apr 24 '15 at 4:19
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4$\begingroup$ Also cross-posted to Theoretical Computer Science: cstheory.stackexchange.com/q/31251/8067 $\endgroup$ – Zsbán Ambrus Apr 24 '15 at 5:26
