<p><a href="http://dx.doi.org/10.1016/0020-0190(75)90036-8">de Groote showed</a> that one could compute both $uv$ and $vu$ using only a total of 10 non-scalar multiplications.</p>

<p>Howell and Lafon, in a <a href="http://www.theworld.com/~sweetser/quaternions/ps/cornellcstr75-245.pdf">tech report</a> (that I could not find a refereed version of and have not checked myself, but is probably correct) showed that one can do quaternion product with only 8 multiplications. </p>

<p>7 is certainly a lower bound, as noted in both of the above papers.</p>

<p>(While this was an interesting question and fun to look up, I found the answer in about 5 minutes of Googling. Google is your friend.)</p>

CC BY-SA 3.0
2015-04-24T03:14:36.033Z
2015-04-24T03:14:36.033Z
Joshua Grochow
<p><em>Karatsuba multiplication</em> 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?</p>

CC BY-SA 3.0
2015-04-24T05:26:54.527Z
Cross-posted from Stackoverflow at http://mathoverflow.net/q/203759/5340 (which gives the right answer in a comment), and Math Stackexchange at http://math.stackexchange.com/q/1222820/24908
Zsbán Ambrus
CC BY-SA 3.0
2015-04-24T18:53:32.277Z
@Kaveh: Does this mean I should post the answer at those other sites as well? Or just on math.SE, as that seems to be the only version of this question that hasn't been put on hold. What's the procedure here?
Joshua Grochow
<ds.algorithms>
Minimum number of real multiplications to multiply two quaternions
Adhvaitha
# Minimum number of real multiplications to multiply two quaternions

(Asked by: username_0 on 2015-04-24)

*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?

**Tags:** ds.algorithms

> **Comments:**
> **username_1**: Cross-posted from Stackoverflow at http://mathoverflow.net/q/203759/5340 (which gives the right answer in a comment), and Math Stackexchange at http://math.stackexchange.com/q/1222820/24908  
> **username_2**: @Kaveh: Does this mean I should post the answer at those other sites as well? Or just on math.SE, as that seems to be the only version of this question that hasn't been put on hold. What's the procedure here?  

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## 1 answers

### Answer by username_2 on 2015-04-24. Score: 6

[de Groote showed](http://dx.doi.org/10.1016/0020-0190(75)90036-8) that one could compute both $uv$ and $vu$ using only a total of 10 non-scalar multiplications.

Howell and Lafon, in a [tech report](http://www.theworld.com/~sweetser/quaternions/ps/cornellcstr75-245.pdf) (that I could not find a refereed version of and have not checked myself, but is probably correct) showed that one can do quaternion product with only 8 multiplications.

7 is certainly a lower bound, as noted in both of the above papers.

(While this was an interesting question and fun to look up, I found the answer in about 5 minutes of Googling. Google is your friend.)


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