Sleep math: red clay blue clay

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 50

The 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, 1/2

R1 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

 

Edit: As others have pointed out, this is not the best strategy.

___________________________
 

Nice problem! The best strategy seems to be to mix the red clay with the blue clay in small infinitesimal steps. Every bit of red clay then becomes as cold as possible, meaning that as much energy as possible is transferred to the blue clay.

Here's what we get with two steps:

• Mix 1/2 red at 100 degrees + 1 blue at 0 degrees => 33.33 degrees.
• Now remove the red clay, and add the other half.
• 1/2 red at 100 degrees + 1 blue at 33.33 degrees => 55.55 degrees.

Now let's compute this with $n$ steps. So at each step we add a $1/n$ fraction of the red clay to the blue clay, then remove it. Let the temperature of the blue clay after $k$ steps be $T_k$.

Initial: $T_0=0$

Adding a $1/n$ piece of red clay to the blue clay:

$T_{k+1}=(T_k+100\ast \frac{1}{n})/(1+\frac{1}{n})=\frac{n{T}_k+100}{n+1}$

To simplify our expression, let $r$ be the ratio: $\frac{n}{n+1}$.

$T_{k+1}=r(T_k+\frac{100}{n})$

Unrolling our definition gives a geometric series:

$T_n=\frac{100r}{n}(1+r+...+r^n)$

Replacing the geometric series with its sum:

$T_n=\frac{100r}{n}\cdot \frac{1-r^{n+1}}{1-r}$

Simplifying, $\frac{r}{n(1-r)}=\frac{\frac{n}{n+1}}{n(1-\frac{1}{n+1})}=\frac{\frac{n}{n+1}}{\frac{n}{n+1}}=1$

So $T_n=100r(1-r^{n+1})$.

Now $r^{n+1}={\left(\frac{n}{n+1}\right)}^{n+1}={(1-\frac{1}{n+1})}^{n+1}$. As $n$ approaches infinity, it's well-known that this approaches $1/e$.

Final temperature: 

$T_n=100(1-1/e)$ = 63.2 degrees.

As a final note, the LaTeX support for LessWrong is the best I've seen anywhere. My thanks to the team!

Do you know Lewis Carroll's Pillow Problems?

Lbh pna trg O neovgenevyl pybfr gb 100.

Cebbs ol vaqhpgvba. Fhccbfr fbzr cebprqher pna trg O gb 100-K, naq N gb K. Gura gurer vf n cebprqher gung jvyy trg N gb K-(K/20)^2. Ercrng rabhtu gvzrf gb trg O neovgenevyl pybfr gb 100.

Gur cebprqher: gnxr unys bs O naq unys bs N naq qb gur bevtvany cebprqher. Gura gnxr guvf unys bs N naq pbzovar jvgu gur bgure unys bs O, gura gnxr gur bgure unys bs N (ng 100) naq pbzovar jvgu gur 2aq unys bs O ol hfvat gur bevtvany cebprqher, gura zvk gur Nf naq Of gbtrgure frcnengryl.

Erfhygf ng rnpu fgrc:

1. Unys bs O: 100-K. Unys bs N: K
2. 2aq unys bs O: K/2. 1fg unys bs N: K/2
3. 2aq unys bs O: K/2 + (100-K)*(100-K/2)/100 2aq unys bs N: K/2 + K(100-K/2)/100
4. Nirentr bs Nf: K/2 + K(100-K/2)/200 K-(K/20)^2

Jr pna gura cyht va K=50 naq frr jurer jr trg: K=43.75 K=38.96 K=35.17 K=32.08

Guvf shapgvba znl gnxr n ybat gvzr gb trg gb mreb ohg vs lbh vgrengr ybat rabhtu vg unf gb trg gurer, gurer'f ab bgure cbvagf nybat gur jnl vg pbhyq fgbc ng.

Haven't fully verified this.

Treat the blue clay (mass 1, temperature T) as a single lump. Feed it infinitesimal lumps of red clay (mass $dX$, temperature $T_R$). After each infinitesimal feeding, the temperature of the blue clay changes by

$dT=\frac{T+T_{R}dX}{1+dX}-T=-TdX+T_{R}dX$

(Final equality comes from series expansion).

Then you can integrate: ${\int }_0^{T_{final}}\frac{dT}{T_R-T}={\int }_0^{1}dX.$

i.e. $\log (\frac{T_R}{T_R-T_{final}})=1$ 

The final solution is:

$T_{final}=T_R\cdot (1-1/e)$.

It's worth saying that this is the most efficient way to transfer energy from red to blue because each feeding step is thermodynamically reversible.