Let s = startups, b = bigcorps, z = zombies 1) In any of the transformations [(-,-,+),(-,+,-),(+,-,-)], the delta b/w numbers (s-b, s-z, b-z) changes by 0 or 2

Let s = startups, b = bigcorps, z = zombies 1) In any of the transformations [(-,-,+),(-,+,-),(+,-,-)], the delta b/w numbers (s-b, s-z, b-z) changes by 0 or 2. What this tells us is that if the delta is odd to start, it will always be odd, if even always even. 2) Since the final state will involve only one type left standing (or else the stronger could eat the weaker), we know that the final counts will be two at 0, one at something >0. Since our losers end with an even delta (0), they must always have had an even delta. In the scenario you proposed, that's s and z (2055 - 2017 = 38). So b, bigcorps, is going to be our winner. 3) Now that we know who will win, we need to find the maximum possible number of bs remaining. Since each move reduces the total population by one, the maximum bs will be produced by the shortest sequence of moves that gets rid of all zs and ss. 4) As long as z>0 and s>0, our best move is to have z eat s, as it drops the population of z and s by 2. So our first sequence is to have all zs eat ss, leaving us with s = 38, b = 4023, z = 0. 5) When z = 0, our only possible move is b eats s, creating a z, which can then be eaten by an s to produce a b, thereby restoring b to its original count and lowering s by 2. After repeating this sequence 19 times to get rid of the 38 ss, we'll have s = 0, b = 4023, z = 0. Ergo, maximal final state is 4023 bigcorps.