Chameleons with a possible ending

1. Reusable idea: In population dynamics, combine modular invariant with linear system to construct an optimal sequence.
2. To end up all blue, we want R'=0 and G'=0 with: R'=4-x-y+2z, G'=7-x-z+2y.
3. Total number of encounters: x+y+z=(z+5)+(z-1)+z=3z+4.

Back to problem
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 encounter: $(4,7,10)\to(6,6,9)$.
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.