2 Apr
2014
2 Apr
'14
3:10 p.m.
True situation is more complicated. What you really need, is to prove f(x,y) = min( f(x-1,y-2), f(x+1,y-2), f(x+2,y-1), ... , f(x-1,y+2) ) with 8 arguments inside the min. That is a pretty nasty recurrence.
--except I should have added +1 to the right hand side. Duh.