Counting Problem
I do seem to have made some cases, but some of them do not work out to be easily computable.
30 Replies
@Apu
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I'm hella rusty at combi, so yeah. 🥲
Just a moment, are we considering rotationally symmetric colourings as identical?
Or are taking labelled squares?
Labelled squares.
Oh that makes it easier
Okay, I think I messed up my case-working ig.
I got 96 but that's too low right?
I don't have the numerical answer, but this seems feasible I think.
What I did was considering each 2×2
There are sixteen such.
We choose two from the four squares to be one colour. By default the other two need to be the other colour.
So 16×(4C2)
If we had to consider rotational symmetry, it would be tougher
But for unique cells, I think this is feasible
But consider a square a2,a3,b2,b3; by 16.4C2, we also count cases where the two red squares on square a1,a2,b1,b2 and a3,a4,b3,b4 are towards the sides of intersection, which defies the condition that each 2x2 square must have 2 red and yellow squares right? [Where a,b... is column and 1,2... is the row.]
Oh nvm
Okay, I think I have a solution...
There would be 2 cases one where 2 reds (or yellows) are adjacent and one where no 2 reds are adjacent?
shocking
Ok, bit of thinking later, I think there are only three types of solutions
Because if 2 reds are adjacent, that forces the entire column (if they are side-by-side) to be alternating and similarly the column next to it has to be alternating either opposite or similar to the scheme. (2 choices, but again, leads to over counting ig)?
Yeah.
So we have alternating rows of colours, alternating columns of colours, and checkerboard
That's 6?
Consider a column color scheme: R,R,Y,R,Y,R,R,Y where R and Y denote the color of the top most cell and colors are alternating, that works too doesn't it?
Ahh. So the problem translates to "Number of binary strings of length 8 such that there are atleast 2 similar digits together".
Also consider this.
RRYYRRYY
YYRRYYRR
And so on
Which would be 2^8-2 for the columns, 2^8-2 for the rows and the 2 chessboard colorings ig.
RRRRYYYY
YYYYRRRR also works
Yup
Yeah, this seems correct.
But you could also have nonalternating in one direction
Ie,
RRRRRRRR
YYYYYYYY
RRRRRRRR
YYYYYYYY
And so on
Did we account for that?
Yes those are accounted for.
We discounted the alternating patterns then?
Yeah, because that would be overcounting
Yup, I got your solution
Because 2^8-2 represents the binary strings where atleast 2 similar digits are together (Call red 0 and yellow 1)
Yup, I got it.
+solved @Opt
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