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14,000x Speedup (2015)

james.hiebert.name

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Re: 14,000x Speedup (2015)

#5
I’m not sure why we can’t map a position in a big grid to a position in a small grid by doing:

X = Math.floor(bigX / bigXMax * smallXMax)

Is there some magic here I’m not seeing?

Re: 14,000x Speedup (2015)

#6
post #5

I’m not sure why we can’t map a position in a big grid to a position in a small grid by doing: X = Math.floor(bigX / bigXMax * smallXMax) Is there some magic here I’m not seeing?

Grid doesn't need to be regular

Re: 14,000x Speedup (2015)

#7
> Furthermore, there is literally no way to tell whether your program will ever actually terminate without actually executing it.

This is technically not correct, is it? Or at least phrased a bit poorly. Maybe replace "your" with "any given"?

Re: 14,000x Speedup (2015)

#8

> Furthermore, there is literally no way to tell whether your program will ever actually terminate without actually executing it. This is technically not correct, is it? Or at least phrased a bit poorly. Maybe replace "your" with "any given"?

Correct, not all programs suffer from the Halting Problem.

Re: 14,000x Speedup (2015)

#10

You could get arbitrarily large improvements by starting with even worse code ;)

Exactly. I like the enthusiasm of the author; but the self-congratulatory attitude doesnt go down very well if you should actually be ashamed of the first version. I mean it is only a few inches from:"you know in the old days we had to flip through the telephone directory from front to end to find a name. But you know what: it is actually a sorted list, so we applied a binary search algorithm and are 14000x faster. Tl;dr CS for the win"
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