It isn’t really notation so much as a recording of the 3 states each of the pieces goes through (each piece is equilibrated n times for when the other block is split into n pieces, so I record the state of the piece after each of it’s equilibrations), expressed as how much of the maximum temperature the piece has (I suppose it would have been cleaner if I’d included the implicit initial states of 0 for the blue pieces, and 1 for the red pieces)
Stevie Lantalia Metke
Divide both clay blocks into n pieces
For each piece of blue, equalize it with each piece of red in turn, from coldest to warmest
The first red piece will be 100*(1/2^n)
The last red piece will be 50The best I’ve done in my head is to get red down to a bit below 25, at 5x5, but a spreadsheet I just rolled up says blue passes above 90 at 32x32
I may have made an error of course, but breaking down the 3x3 case (and expressing T and fractions of 100)
B1 1⁄2, 3⁄4, 7⁄8
B2 1⁄4, 1⁄2, 11⁄16
B3 1⁄8, 5⁄16, 1⁄2R1 1⁄2, 1⁄4, 1⁄8
R2 3⁄4, 1⁄2, 5⁄16
R3 7⁄8, 11⁄16, 1⁄2
And recombining gets us R 5⁄16, B 11⁄16
As far as I can tell, you can push the temperature of blue arbitrarily close to 100, and the temperature of red arbitrarily close to 0. Am I missing something, as this seems substantially better than the other answers
Can’t you do better by splitting both blocks up, and doing equilibrating each piece of blue with each piece of red, in turn? By not dividing blue, you are leaving the red pieces a lot warmer than necessary
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