"L-Shaped Positions" in Chomp In my earlier question I asked about the winning move in Ordinal Chomp played on a $3 \times 3 \times \omega$ board. So far the problem has seemed intractable. Therefore, I now consider Ordinal chomp played on an L-shaped board \begin{pmatrix}a&b&c \\ d&0&0\end{pmatrix}


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*This game is played between two players on the set $2 \times 3 \times \omega$.  A move is to pick any remaining $(i,j,k)$ and remove all  $(i^\prime,j^\prime,k^\prime)$ where $i^\prime \geq i$, $j^\prime \geq j$ and $k^\prime \geq k$. The player to take $(0,0,0)$ loses. I have represented the positions I am considering the whole remaining chomp board as a matrix to make it easier to analyse. For a more full explanation, see (https://en.wikipedia.org/wiki/Chomp)


I have been able to prove various results, including that \begin{pmatrix}\omega&\omega&c \\ d&0&0\end{pmatrix}
is a losing position for (c, d) = (0, 0), (1, 1), (2, 3), (4, 2), (3, 5) (I can post a proof if anyone is interested).
My questions are: 1) What research has been done on these positions, if any?
2) If research has been done, what other values of (c, d) are known such that the above is a losing position?
3) Is there a number n such that \begin{pmatrix}n&n&n \\ n&0&0\end{pmatrix} is a losing position?
 A: (I put this into another answer because it was too long for a comment) I made a database of losing positions. I noticed after calculating a few positions by hand that there were some positions, like \begin{pmatrix}3&3&1 \\ 3&0&0\end{pmatrix} that are losing positions, which means that there are only finitely many cases with c=1 (analogously with the other cases in my earlier comment). Therefore, I can list all the classes of solutions. Then I made a computer program, which had two functions. The first calculates for a given b, c, and d, what value of a makes \begin{pmatrix}a&b&c \\ d&0&0\end{pmatrix}
a losing position, and the second takes c and d and for all b from c to c+14, so I could find the pattern in the losing positions.
It was only then that I found that \begin{pmatrix}5&5&5 \\ 5&0&0\end{pmatrix} was a losing position. After that, I built up a database of all solutions with a fixed value of c, slowly increasing c as I covered more cases. None of the cases were too hard, except for c=4, b=10, which, because of period multiplication in the c=4 case, had period 80. For that case I had to construct a specialised program.
Actually, I just realised that the way I was programming the functions was extremely inefficient, and after correcting my program it works up to c = 200 in under a minute, and uses no cataloging of positions prior to runtime, (which I had to do for the last program). It also records positions for future use, so once you've calculated a position it will remain in the memory (until the program is reset).
A: After a bit more research (and a very helpful python program), I have been able to solve answer all of my questions. First, the values of (c, d) such that \begin{pmatrix}w&w&c \\ d&0&0\end{pmatrix} is a losing position are: (0, 0), (1, 1), (2, 3), (3, 5), (4, 2), (5, 4). Second, the position \begin{pmatrix}5&5&5 \\ 5&0&0\end{pmatrix} is losing, which means that any other position with c,d>=5 can be reduced to this, and is therefor a winning position. Also, this means that there are no other (c, d) answers to question 1. Also, I have calculated all the finite losing positions where c<=8 (going through c is much easier than d because d=0 is not even fully solved!).
Edit: I have also found that \begin{pmatrix}8&8&8 \\ 4&0&0\end{pmatrix} is a losing position.
