Raul, I see the result of `location` as being the number of steps, but not a grid location (as you suggest) because its result is simply 696 in my data's case, for example. Of course, that's all that AoC 11 required.
My approach using Jimmy's algorithm (I think) does perhaps not tell the exact final location because that final location can apparently be reached with several different combinations of directions including the following. But it at least tells the general direction of the end result. Right? Steps NE N 696 0 698 1 697 2 ... 586 110 N NE N 110 586 0 109 585 1 etc. (others) On the other hand, I may be wrong, and the algorithm Jimmy suggest DOES provide a way of telling exactly the unique location reached. I just cannot work out what it tells. Can anyone? On Sun, Dec 17, 2017 at 12:27 PM, Raul Miller <[email protected]> wrote: > Here's how I would interpret the final stage's result: > > dss=:-:<:3 #.inv 15 21 19 11 5 7 > nms=: ;:'n ne se s sw nw' > location=:3 :0 > {:+/||:+/\dss{~nms i.y > ) > > This translates a symbolic sequence into a grid location. So as a > basic integrity check, I would run it against both your source > sequence and your result sequence. If the location of both is the > same, you are probably on the right track. > > But I think this problem requires at least two passes. If you do not > figure out where you are going before you try to find the shortest way > to get there, you wind up trying to find the shortest distance to > "everywhere". (And, ok, that can be a quicky description of a viable > algorithm, but you still wind up with multiple passes of a sort in > your implementation.) > > Thanks, > > -- > Raul > > > > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
