We received lots and lots of very
well-described solutions to this Teddy Bear problem - I'm sorry
that we can't mention everyone. It was interesting to see what you
called a 'move'. Jessica decided that a move was a 'swap':
Start: BBBB RRRR YYYY GGGG
Step 1, Switch the second blue and the second red:
BRBB RBRR YYYY GGGG
Step 2, Switch the fourth blue and the fourth yellow:
BRBY RBRR YYYB GGGG
The fourth red is not used because that would result in two reds
being side by side. Switching fourth with fourth makes it easier to
keep track of what has been done.
Step 3, Switch the fourth red and the fourth green:
BRBY RBRG YYYB GGGR
Step 4, Switch the second yellow and the second green:
BRBY RBRG YGYB GYGR
The bears are now fully mixed-up. This is not a unique
Jessica says this is not a unique
solution, meaning that it can be done in four moves, but in
different ways. Will, Milly and Emily from Swiss Gardens Primary
School agreed with Jessica and sent in this picture of the four
moves they made, which are slightly different from
Gemma and Maggie, also from Swiss
Gardens, looked at simpler cases with smaller numbers of bears.
We did $2\times2$ which was 1 move also $3\times3$ which was 2
$4\times4$ was 4 moves.
It can be useful to do this to see if
you can identify and explain the pattern. I wonder whether you
could predict the least number of moves for five lots of five
Jodie thought about a 'move' in a different
way and here is what she did:
If you call the blue bears B, the red bears R, the yellow bears Y,
and the green bears G, you get:
1. Move one of the green bears to separate the first two blue
2. Move one of the yellow bears to separate the second two blue
3. Move one of the red bears to separate the last two blue
4. Move one of the green bears to separate the second two red
5. Move one of the yellow bears to separate the final two red
6. Move one of the green bears to separate the final two yellow
Elijah also used this idea and, like Gemma and
Maggie, he investigated simpler cases. He even tried five lots of
five bears as well:
I found that the smallest number of moves to mix up the four
lots of four colours was $6$.
Call the bears a1, a2, a3, a4, b1, b2, b3, b4, c1, c2, c3, c4,
d1, d2, d3, d4 (the letters mean different colours).
Move a1 between b1 and b2
Move a2 between b2 and b3
Move a3 between b3 and b4
Now you have a4, b1, a1, b2, a2, b3, a3, b4
You can't do it in less than 3 moves because in 4 bears
coloured 'b' there are 3 pairs that need to be split up.
Now do the same for the c's and d's.
That is $6$ moves.
You can do other moves and get a different mixture at the
I tried it for two lots of two colours and could do that in
Then I tried it for three lots of three colours. This was a
bit harder because it doesn't split into two halves and you have to
mix the colours up. I did it like this:
Start with a1, a2, a3, b1, b2, b3, c1, c2, c3
Move a1 between b1 and b2
Move c1 between b2 and b3
Move a2 between c2 and c3
Now you have a3, b1, a1, b2, c1, b3, c2, a2, c3
I noticed that the sequence $1, 3, 6 ... $ is the start of the
So then I thought that for five lots of five colours the
answer would be $10$. I think this is right - I couldn't do it in
less than $10$ moves.
I think these are right for the smallest numbers of moves
because you can imagine the bears as pairs of two colours (like the
two lots of two colours problem) and see how many moves it must
take to mix them up.
Fantastic reasoning, Elijah, thank you for
your detailed solution. We can really understand how you were
We also received the following message from a
mum which I couldn't resist including!
My daughter was trying this as we came back from Science Festival
today when her little brother found the solution: he picked all the
sets of buttons she was using (for counters) and threw them in the
air ... as they fell they all scattered ..."Mum! It's just ONE move
when HE does it!" Not sure this is what you intended,