Fascinating. Thank you so much for that Geoff.
I've been afraid to play with ChatGPT so far - too worried abut getting
sucked in and spending way too much time ....
I did take a look at your third example (since I can never resist a
performance challenge :-)
There are a number of minor tweaks that could be made to improve performance
1. set initial value to infinity rather than calculating a distance
between the first two points.
2. "number of elements in pPoints" is unvarying within any one call - so
extract it to a variable at the start
3. use the square of the distance rather than the actual distance - save
N**2 calls to sqrt
4. use "DX * DX" rather than "DX ^ 2" (about 25% faster)
5. calculate distance in-line rather than call a function
but those all add up to maybe 10% performance improvement (or less - I
didn't test it). That's useful - but not enough.
For a modest number of points (2000 random points), this takes approx
16.5 seconds !!
We need a better algorithm. If we use a "linear scan", we can change it
from essentially Order(N**2) to approx Order(N).
Summary:
- sort the points by X coordinate
- while scanning the inner loop, as soon as the difference in Xcoord
from the 'outer' point exceeds the minDist so far, you can reject not
just this point, but all subsequent points, and hence exit the inner
loop immediately.
This brings the time down from 16500 millisecs to 25 millisecs.
BUT - I have no clue how I'd go about describing this to ChatGPT :-)
NB I changed the input parameter to be the list of points rather than
the array.
Code:
function closestPointsSQ pLines
sort pLines by item 1 of each
put pLines into pPoints
split pPoints by CR
put infinity into minDist
put the number of elements in pPoints into N
repeat with i = 1 to N-1
repeat with j = i + 1 to N
put item 1 of pPoints[j] - item 1 of pPoints[i] into t1
if t1 * t1 > minDist then exit repeat
put item 2 of pPoints[j] - item 2 of pPoints[i] into t2
put t1 * t1 + t2 * t2 into dist
if dist < minDist then
put dist into minDist
put pPoints[i] & " " & pPoints[j] into closestPoints
else if dist = minDist then
put return & pPoints[i] & " " & pPoints[j] after closestPoints
end if
end repeat
end repeat
return closestPoints
end closestPointsSQ
-- Alex.
On 20/01/2023 06:02, Geoff Canyon via use-livecode wrote:
I tested three use cases, with variations, using ChatGPT for (live)code
generation. There was a lot of back and forth. In the end, I solved all the
problems I set, but in some cases I had to hold ChatGPT's hand pretty
tightly.
That said, I learned some things as well -- about LiveCode. ChatGPT's code
for Fizz Buzz was faster than mine.
My code was faster for reversing lines. But ChatGPT, when asked to "make
the code faster" gave several suggestions, some of which were wrong or
impossible, but one of them was my method, and when I said "write that
option" it did, with only a few corrections needed.
And one of the ideas, while wrong, caused me to think of a different way to
solve the problem, and that way ended up being faster than my original
solution by over 3x on reversing 10k lines. Getting ChatGPT to write this
new method was *hard*.
In any case, I wrote it all down in a google doc
<https://docs.google.com/document/d/1W3j5WaFhYZaqSt0ceRQj8j160945gSwG_nyZsCBP6v4/edit?usp=sharing>.
If you're curious, have a read. That URL is open for comments/edit
suggestions. If you have any I'd love to hear them.
Thanks!
Geoff
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