When is "do-almost-nothing" a good idea in CHOMP?

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The proof by strategy-stealing that CHOMP on a rectangular board is a first-player win involves player 1 taking the top-right square on their first move. Of course given the proof-by-contradiction nature of the argument, this does not mean that the top-right square is in general a good opening move, and there are plenty of rectangular boards on which it is in fact a losing move for the first player.

Note: different versions of CHOMP place the "poison square" at different locations. For simplicity, I'm sticking with the convention from the above-linked notes.

I'd like to know how often it is in fact a good move. (Note: by a strategy stealing argument, if it's winning in a given board it is in fact the only winning move in that board. I missed this at first.)

There are a couple ways to phrase this problem:

Question 1: For fixed $n$, let $c_n$ be the upper asymptotic density of the set of $m$ such that moving in the top-right square is the winning move for the first player on the $m$-by-$n$ board. What is $c_3$?

(It's easy to see that $c_1=0$ and $c_2=1$, so this is the first interesting case. Of course the higher $c_i$s are also interesting, but I worry that they are intractable.)

Question 2: Is it the case that, for every $k$, there are $m,n>k$ such that the top-right square is the winning move for the first player in the $m$-by-$n$ board? (Is this even true for $k=2$?)

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Partial answer too big for a comment:

For a $3×n$ board, the "do almost nothing" move is never the right move. Using the recurrence from this page:

Let f(q,r) be the unique p such that (p,q,r) is a P-position. (Here (p,q,r) stands for (p,min(p,q),min(p,q,r)).) There is a simple recurrence for f(q,r):

If r > q, then f(q,r) = f(q,q). (That is, f is constant on verticals above the diagonal.) Otherwise, if f(q-1,r) < q, then f(q,r) = f(q-1,r). (That is, if f(q,r) = q then f remains constant on the rest of this row.) If neither case occurs, then f(q,r) is the smallest positive integer not among f(a,r) for a < q or among f(q,b) for b < r.

The author of that page draws a table:

24|                                                                       42
23|                                                                    40 41
22|                                                                 39 38 40
21|                                                              37 38 36 39
20|                                                           35 36 37 33 38
19|                                                        34 33 35 36 31 37
18|                                                     32 33 31 29 35 34 36
17|                                                  30 31 29 32 33 34 26 35
16|                                               28 29 26 31 30 32 33 23 23
15|                                            27 26 28 29 30 23 31 22 22 22
14|                                         25 26 23 27 28 22 29 30 31 32 33
13|                                      24 23 22 25 26 27 19 19 19 19 19 19
12|                                   21 23 22 24 19 17 17 17 17 17 17 17 17
11|                                20 19 22 17 23 24 25 21 26 27 28 29 30 31
10|                             18 19 20 21 14 14 14 14 14 14 14 14 14 14 14
 9|                          16 17 14 18 19 20 21 22 23 24 25 26 27 28 29 30
 8|                       15 14 16 17 12 12 12 12 12 12 12 12 12 12 12 12 12
 7|                    13 14 12 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
 6|                 11 12 13  9  9  9  9  9  9  9  9  9  9  9  9  9  9  9  9
 5|              10  9 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
 4|            8  9 10  7  7  7  7  7  7  7  7  7  7  7  7  7  7  7  7  7  7
 3|         6  7  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5
 2|      4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
 1|   3  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2
 0|1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
----------------------------------------------------------------------------
q: 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

He then notes that elements on the diagonal are always the largest in their column:

The column (q,*) contains q+1 different numbers f(q,r), all of them positive integers, so f(q,r) > q. It follows that f(q,r) is the smallest element that does not occur earlier in the same row or column. Since the numbers f(q,s) are not at the (q,r) position, it follows that they all occur earlier in row (*,r), but not at (t,r) with t <= r since above the diagonal verticals are constant. So, they all occur at (t,r) with r+1 <= t <= q-1. But there are more numbers s than positions t, contradiction.

This is easily extended to show that $f(q,q-1)$ is at most third greatest. But there are $q-2$ numbers below $f(q,q-1)$ - and neither $1$ nor $3$ are in the column, so $f(q,q-1)$ is at least $q+1$.

Finite for any $m>2$ on boards $m×n$

A similar table can be created for any $m$ - just much more complicated because it would have more dimensions. And the same argument would apply to the diagonal of the table - $f(k,k,k)$ would be the greatest value in its column, and so on. $f(k,k,k-1)$ would be at least the fifth greatest, and so on.

Thus, for every $m>2$ there exists an $N$ such that "do almost nothing" is a losing move on all boards $m×n$, where $n>N$.

Given that $N$ increases with $m$, I think it's safe to say that there are no solutions for $m=4,5$ (I have with computer search checked boards up to $6×65$).