h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

In the following, use "~" for "equivalent", and "Ax" for "all x such that...".

To restate the parameters,

[1] c = b - a
[2] c > 0
[3] d = c / g
[4] a < x < b
[5] x ~ a when x < (a + d), and
[6] x ~ b when x > (a + d).

To restate the first case,

if Ax ~ a
then a + d = b (ref: [4] and [5])
or d = b - a
or d = c (ref: [1]).

Therefore, g = 1 (ref: [3]).

So, for the second case,

if Ax ~ b
then a + d = a (ref: [4] and [6])
or d = 0.

Therefore, g is either undefined or zero (ref: [2] and [3]).

In general, as g approaches infinity, d approaches zero.

What happens to d when g = infinity, I don't really know (maybe zero?), but I doubt it helps you. If you truly need Ax ~ b, or even mostly x ~ b with a reasonable g, then I suspect you're out of luck with the system as defined. I was slightly surprised to see the asymetry here, so maybe I'm missing something.

Hmmm. Thinking....

Well, if we juggle things a bit, as in

b2 = -(a) and a2 = -(b)

then Ax ~ a2 when g = 1

which is effective, if inelegant (and inefficient), although perhaps not suitable for your application...

...Still thinking...

Ok, if g = 42, then d = 1/42 of c, which may be enough... No?

...Thinking...

panic: towel: threadbare: transmission interrupted: hope it helped!

I can see a way of making the problem symmetrical. Instead of c/g use g2 = 1/g so c/g becomes cg. Then g is the error coefficient.

Does this run?

Thanks for tidying up the logic though

*Tries to think about this one....*
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*Sometime later decides he forgot logic sometime ago, round about when he started working...... So gives up...*

Hmm, sorry! I'm sure I would have quite helpful about 2 years ago... but I'm just too rusty these days

Maths student? If so, could you help me get ready for STEP?

Actually, the inabillity to determine a situation of always rounding up could be important in itself. It sounds quite a deep area.