Talk:Alpha max plus beta min algorithm

Is there a way to polish the solution?

Is there a way to compute a new estimate of hypot from this starting condition like is done with typical sqrt algorithms? --Jaded-view (talk) 03:29, 5 June 2008 (UTC)[reply]

Is there one wrong parameter set?

The parameter set alpha=7/8 and beta=15/16 can't be right. I tried to plot that and the graph looks completely different. My guess is that this should have been something like beta=15/32. Most beta values are close to 1/2.

TomF —Preceding unsigned comment added by 77.191.238.62 (talk) 22:56, 26 December 2008 (UTC)[reply]

This has now been corrected to alpha = 7/8 and beta = 7/16. Gaius Cornelius (talk) 12:06, 2 September 2009 (UTC)[reply]

Examples don't seem practical

This formula is extremely useful, but I have a question: Why are the examples not very good approximations of the ideal alpha and beta? For example, alpha = 31/32 and beta = 13/32 outperforms all of the other examples, and (similarly to most of the others) only requires 2 multiplies and one shift to apply those constants, with similar bit-depth requirements.

It might also be worth showing how well 16-bit approximations do (i.e. alpha = 62941/65536 and beta = 26070/65536). Many archetectures have extremely efficient and/or SIMD-able 16-bit multiplies, and those values do an excellent job.

Nathaniel bogan (talk) 18:52, 27 January 2011 (UTC)[reply]

Well just as an example (7/8, 7/16) can be done just using 2 shifts, 1 add and 1 sub:

 len = mx + (mn >> 1);
 len -= (len >> 3);

I agree that it would not hurt to add the 16bit approximation as well --92.76.235.178 (talk) 22:26, 13 June 2011 (UTC)[reply]

Polar plot

Wouldn't it be better to add a polar plot with the results of this function? as it more intuitively shows how this formula works. (a polar plot will show an octagon)213.126.27.106 (talk) 10:19, 12 August 2013 (UTC)[reply]

It wouldn't be a polygon: here is the plot for (a=1,b=1/2) for theta in (0,pi/2): Plot on wolfram alpha