Re: Don't get your GI tract in a twist...

[Dan Zuras Intervals <intervals08@xxxxxxxxxxxxxx>]

> Date: Tue, 31 Jan 2012 04:20:24 -0800 (PST)
> From: Dmitry Nadezhin <dmitry.nadezhin@xxxxxxxxxx>
> To: <intervals08@xxxxxxxxxxxxxx>
> Cc: <stds-1788@xxxxxxxxxxxxxxxxx>, <nh@xxxxxxxxxxxxxxxxx>
> Subject: Re: Don't get your GI tract in a twist...
> 
> What about one more example (again in two-digit floating-point decimal numb=
> ers) ?
> 
> [u,v] = [0.92,1.2]
> midpoint([u,v]) = roundNear((u + v)/2) = roundNear((0.92 + 1.2)/2) = =
> roundNear(1.06) = 1.1
> radius([u,v]) = roundUp((v - u)/2) = roundUp((1.2 - 0.92)/2) = roundU=
> p(0.14) = 0.14
> [roundDown(midpoint(X) - radius(X)), roundUp(midpoint(X) + radius(X))] = =
> [roundDown(1.1 - 0.14), roundUp(1.1 + 0.14)] =
> [roundDown(0.96), roundUp(1.24)] = [0.96, 1.3]
> 
> The property ([0.92,1.2] \subset [0.96, 1.3]) is false.
> 

	Now, THAT'S a good example.

	OK, we're wrong.  (I'm wrong.  I had to convince Nate. :-)

	And

		radius([0.92,1.2]) =
			max(midpoint([0.92,1.2])-0.92,1.2-midpoint([0.92,1.2]))
			= max(1.1-0.92,1.2-1.1) = max(0.18,0.10) = 0.18

	implies

		[0.92,1.2] \subset [roundDown(1.1-.18),roundUp(1.1+.18)]
				= [roundDown(0.92),roundUp(1.28)]
				= [0.92,1.3]

	clearly works.

	So I guess your definitions has the -w option (works) & ours
	does not.

	I was hoping for 2 things: (1) that the definition of radius
	reflected the definition of width & (2) that both were
	independent of the definition of midpoint.

	So before we run this play one more time let me ask you 2
	questions:

	(1) Can you think of a reason that the definition of width
	should be changed to reflect the definition of radius?

	And (2) Can you think of a definition of radius that has
	both the -w option & is independent of the definition of
	midpoint?

	Any way you answer is OK.  We will change it to work this
	time.  Thanks again for finding the flaws.


				Dan