New formula has overflow error

[aadler]

Even after fixing the exponentiation, I get the following error in the test case:

F:\Elite\TD>trade.py run -vvv --progress --summary --avoid imperial,slaves --insurance 1.51m --pad L --ly 15.06 --cap 264 --no-planet --credits 24m --from "tollan/gord" --to "LHS 2094/Patt" --hops 3
* Hop   1: .........1 origins
NOTE: Pruned 23 origins too far from any end stations
* Hop   2: ........63 origins .. 33,528-382,536cr gain, 127-1,449cr/ton
[=====================    ] Traceback (most recent call last):
  File "F:\Elite\TD\trade.py", line 104, in <module>
    main(sys.argv)
  File "F:\Elite\TD\trade.py", line 77, in main
    results = cmdenv.run(tdb)
  File "F:\Elite\TD\commands\commandenv.py", line 81, in run
    return self._cmd.run(results, self, tdb)
  File "F:\Elite\TD\commands\run_cmd.py", line 1220, in run
    newRoutes = calc.getBestHops(routes, restrictTo=restrictTo)
  File "F:\Elite\TD\tradecalc.py", line 979, in getBestHops
    penalty = (-1 + 1 / (cruiseKls + 1) ** ((cruiseKls + 1) / 4)) / 2
OverflowError: (34, 'Result too large')

Possibly solved by going to logs. Will look into it if not fixed by this evening.

[eyeonus]

That's odd, because that should always result in a number between 0 and -0.5.

Hmmm. Maybe it's the "x^(x/4)' bit that's overflowing?

I'll look into it.

[eyeonus]

So, I did this:

                    try:
                        penalty = (-1 + 1 / (cruiseKls + 1) ** ((cruiseKls + 1) / 4)) / 2
                    except OverflowError as e:
                        print(str(e) + "cKls: " + str(cruiseKls))

and got this:

[                         ] (34, 'Result too large')cKls: 4218.0
[==                       ] (34, 'Result too large')cKls: 4218.0
[====                     ] (34, 'Result too large')cKls: 4218.0
[=============            ] (34, 'Result too large')cKls: 4218.0
[=========================] (34, 'Result too large')cKls: 514.4

First of all, what damned system has a station 4.2 MILLION ls away from the entry point?

Secondly, now I need to decide which of two options to use, those being:

1. Use "sys.float_info.max" instead of "(cruiseKls + 1) ** ((cruiseKls + 1) / 4)" when an overflow occurs.
2. Use the decimal module to do this math, which doesn't overflow until 9E+999999999999999999.
   (Which might still overflow if any stations are far enough away, but in that case I can easily get away with assuming "1/(cruiseKls + 1) ** ((cruiseKls + 1) / 4)" == 0.)

Thoughts?

[aadler]

4.2m ls, pshaw, that's nothing. Hutton Orbital in Proxima Centaurii is 6,784,404 ls! Over a 40 minute trip (I've yet to do it).

Converting the penalty into logs would look like:

penalty = (1 / exp(0.25 * (cruiseKls + 1) * log(cruiseKls + 1)) - 1) * 0.5

However, in this case, using logs doesn't really help, since log(Max(Double) is a hair over 709.78 and Hutton orbital would require exp(0.25 * 6201 * log(6201)) which becomes exp(13537.51) which overflows like mad. Actually, you have a form of Lambert W function here since you have a*Xe^X whew a is 0.25 and X is cruiseKls + 1. The largest Lambert W able to be calculated using double precision is a bit over 703.22. So, I'm not at home so I cannot try it, but perhaps the following will work:

cappeddist = min(0.25 * (cruiseKls + 1) * log(cruiseKls + 1), 709.78)
penalty = (1/exp(cappeddist) - 1) * 0.5

This should work and return the -0.5 for large numbers like Hutton Orbital without overflowing, If you haven't tried it by tonight, I will. Thanks!

[aadler]

Changed cap to 709.78 since we're not actually calculating Lambert W_0 but the exponentiation of capped dist. Not tghat it really matters, that's all going to be -0.5 anyway, we just need to make sure Python doesn't see a number it doesn't like.

if you wanted to use try, you'd probably want:

 try:
    penalty = (-1 + 1 / (cruiseKls + 1) ** ((cruiseKls + 1) / 4)) / 2
 except OverflowError:
    penalty = -0.5

Which is faster, try:except or the min?

[eyeonus]

Your math is off.

https://goo.gl/nPgbth

(-1 + 1 / (x + 1) ** ((x + 1) / 4)) / 2 != (1 / exp(0.25 * (x + 1) * log(x + 1)) - 1) * 0.5

[aadler]

My math is correct when you take into account that the FooPlot website has an ln operator so log is base 10, where R and Python both assume log to be natural :)

http://bit.ly/2MpyRDt