Chameleons with a possible ending

1. Don't follow specific one-on-one encounters; look for a magnitude that changes in a controlled way.
2. Differences between quantities of colors behave in a particularly rigid manner.
3. If an ending is possible, you will have to respect that restriction from the beginning.

Answer: Yes it is possible; The minimum is 7 meetings. Initial state:

$$ (R,G,B)=(4,7,10),\quad R+G+B=21. $$ Let: - $x$ = Red-Green encounters, - $y$ = Red-Blue matches, - $z$ = Green-Blue encounters. To finish all blue, we want $R'=0$ and $G'=0$ with: $$

R'=4-x-y+2z,

$$ $$

G'=7-x-z+2y.
$$ Solving:

$$ y=z-1,\quad x=z+5. $$ Total number of meetings: $$

x+y+z=(z+5)+(z-1)+z=3z+4.
$$ Minimum with $y\ge 0$ is $z=1$, then:

$$ (x,y,z)=(6,0,1),\quad x+y+z=7. $$ Concrete sequence (7 steps): 1. A Green-Blue match: $(4,7,10)\to(6,6,9)$. 2. Six Red-Green matches in a row: $(6,6,9)\to(5,5,11)\to(4,4,13)\to(3,3,15)\to(2,2,17)\to(1,1,19)\to(0,0,21)$. They are all blue. **Reusable idea:** in population dynamics, combine modular invariant with linear system to construct an optimal sequence. $$