N-Queens on 50g (RPL language)

[Claudio L.]

Easy solution: replace n=10*k and work with the integers divided by 10, then compensate with the numeric factor, and no more integers anywhere:

Code:


<< 0. 1000. 0.1 FOR n
   0.02 n DUP 0.1 + * / +
   -0.1 STEP
>>

The exact result is:
1 followed by 'x' 9's, then an 8, followed by 'x' 0's, then repeats the sequence with a 1 followed by...
'x' is the exponent of the end of the loop. If the loop starts from 1e3, there's 3 9's and 3 zeroes:
1.999 8 000 1 999 8 000 1 999 8 000 1 999 8 000 ...

So it's easy to check for rounding errors even at higher precisions (which I want to benchmark too).