Need to let loose a primal scream without collecting footnotes first? Have a sneer percolating in your system but not enough time/energy to make a whole post about it? Go forth and be mid: Welcome to the Stubsack, your first port of call for learning fresh Awful youāll near-instantly regret.
Any awful.systems sub may be subsneered in this subthread, techtakes or no.
If your sneer seems higher quality than you thought, feel free to cutānāpaste it into its own post ā thereās no quota for posting and the bar really isnāt that high.
The post Xitter web has spawned soo many āesotericā right wing freaks, but thereās no appropriate sneer-space for them. Iām talking redscare-ish, reality challenged āculture criticsā who write about everything but understand nothing. Iām talking about reply-guys who make the same 6 tweets about the same 3 subjects. Theyāre inescapable at this point, yet I donāt see them mocked (as much as they should be)
Like, there was one dude a while back who insisted that women couldnāt be surgeons because they didnāt believe in the moon or in stars? I think each and every one of these guys is uniquely fucked up and if I canāt escape them, I would love to sneer at them.
(Credit and/or blame to David Gerard for starting this.)
One thing that I couldnāt easily figure out is what is the constant factor. If the constant factor is significantly worse than for Strassen, then it would be much slower than Strassen except for very large matrices.
Letās say the constant factor is k.
N should be large enough that N^((log(49)-log(48))/log(4)) > k where k is the constant factor. Letās say the difference in exponents is x, then
N^x > k
log(N)*x > log(k)
N > exp(log(k)/x)
N > k^(1/x)
So lets say x is 0.01487367169 , then weāre talking [constant factor]^67 for how big the matrix has to be?
So, 2^67 sized matrix (2^134 entries in it) if Googleās is 2x greater constant than Strassen.
That donāt even sound right, but I double checked, (k^67) ^ 0.01487367169 is approximately k.
edit: Iām not sure what the cross over points would be if you use Googleās then Strassenās then O( n^3 )
Also, Strassenās algorithm works on reals (and of course, on complex numbers), while the new āimprovementā reduces by 1 the number of real multiplications required for a product of two 4x4 complex-valued matrices.