Harry has to redo his computations

(This is on integer complexity so notation as usual.) OK. Let's let D(n)=f(n)-L(n). Suppose m is a canopy number (i.e. D(m)=0). If m=g(k) for some k, then m+1 is also a canopy number, and f(m+1)=f(m)+1, f((m+1)3^k)=3k+f(m)+1. So say m≠g(k) for any k. If m+1 is also a canopy number, then since L(m)=L(m+1), f((m+1)3^k)=3k+f(m+1)=3k+f(m). And if m+1 is not such, then clearly f(m+1)≤f(m)+1, and also as D(m+1)≥1, f(m+1)≥L(m+1)+1=L(m)+1=f(m)+1, f(m+1)=f(m)+1; furthermore f((m+1)3^k)=3k+f(m)+1, as if D(n)=1, then D(n*3^k)=1 as D(n*3^k)≤D(n), but D(n*3^k)≠0 as D(n)≠0.

So if m is a canopy number, f((m+1)3^k) is usually 3k+f(m)+1... but if m+1 is also a canopy number, and m is not g(k) for any k, then it's just 3k+f(m).

So I have to go recompute a bunch of stuff, because I was doing a bunch of computations while ignoring that the latter was a possibility. Then I got to 5 and noticed something was wrong. So I had to go and figure out all canopy numbers m (other than 3^k, 2*3^k, and 4*3^k) such that m+1 is also a canopy number.

Here I just sat down to do some computation, not some tedious... whatever you call this. Regardless, the complete list of exceptions is as follows: For k≥1, 2*3^k+1, 4*3^k+2, 2*3^k+2, 4*3^k+2, 4*3^k+3, as well as the numbers 5, 63, and 512. (Yes, 5=4*3^0+1, but writing it that way would have made things more complicated.)

Now I have to actually go recompute stuff, accounting for this. But not now, that was tiring.

EDIT next day: Silly me, 7 is redundant anyway as it's 2*3+1.

-Harry

Incidentally, last time I played against it, Vanessa wasn't in the lead

Thought: Playing against that deck-attackers deck is really annoying because it violates your assumption that a low-health character is still far better than a dead character.