Biggest number of creatures in forest In crazy forest there are 6 werewolf's,17 unicorns and 55 spiders.


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*Werewolf can eat unicorn and spider,but can't eat another
werewolf.

*Spider can eat unicorn,but can't eat werewolf or another spider

*Unicorn can't eat spider,werewolf or another unicorn


Whenever a werewolf eats a spider he turns into a unicorn,and whenever he eats a unicorn he turns into a spider.Whenever a spider eats an unicorn he turns into a werewolf.
At the time creatures start to eat until no other creature can be eaten.
What's the biggest number of creatures in the crazy forest left after the creatures ate?
My attempt
We're looking so that there is only 1 kind of creatures in forest,first try is to maximize the number of werewolf's.
We have that 17 spiders ate 17 unicorns, now there are 23 werewolf's and 38 spiders left if 19 werewolf eat 19 spiders,than there are 4 werewolf's, 19 unicorns and 19 spiders and now 19 spiders eat 19 unicorns we get back 23 werewolf's.
If there is even number of unicorns and $n$ spiders we can get that 1 spider can eat a unicorn and than that unicorn turns into werewolf and that werewolf can eat a unicorn to turn back into a spider,so we have that we can always get back to $n$ of spiders so we can consider case with only 1 unicorn,so if there are $n$ spiders and 1 unicorn we have that spider eats unicorn and it turn into a werewolf and that werewolf eats a spider and turns into an unicorn so there are $n-2$ spiders and 1 unicorn.It doesn't matter whether the 6 werewolf's eat spider or unicorn since there would still be odd number of unicorns.
So we're left with maximizing the number of unicorns it's clear that the number of unicorns can be 23 at most so the answer is 23.
 A: Your argument that $23$ is optimal is not convincing. In particular, at the end a certain number of wolves survives,  not unicorns.
Denote the momentaneous state by $(w,u,s)$. There are three kinds of moves, namely
$$\eqalign{
T_1:\quad &(w,u,s)\to (w+1,u-1,s-1),\cr
T_2:\quad &(w,u,s)\to (w-1,u+1,s-1),\cr
T_3:\quad &(w,u,s)\to (w-1,u-1,s+1)\cr}$$
(note the symmetry!). Each move decreases the sum $t:=w+u+s$ by $1$; furthermore the parity of the differences $w-u$, $u-s$, $s-w$ remains unchanged.
The game ends when two of the three variables are $0$. Given the initial state $(w_0,u_0,s_0)=(6,17,55)$ the only possibility to reach such a state is when $w$ is odd and $u=s=0$. Since at least $55$ moves are needed to bring $u$ down to $0$ the final population $t_{\rm end}$ is at most $t_0-55=23$.
The following scenario shows that $t_{\rm end}=23$ can in fact be realized: Begin with $17$ moves $T_1$. These lead from $(6,17,55)$ to $(23,0,38)$. Then perform $19$ double steps
$$T_1\circ T_2:\quad (23,0, s)\mapsto(22,1,s-1)\mapsto(23,0,s-2)\ ,$$
ending in $(23,0,0)$. These are $55$ moves in all, and $23$ wolves remain.
