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I am taking this puzzle from inspiration from the Four Fours to get Pi puzzle. The puzzle rules are just slightly different. Here are the rules:

You are trying to get as close to e as possible with only πs used as real numbers in the approximations, and with the following allowed operations:

  • Simple arithmetic: +, -, x, ÷
  • Exponents and logs as binary operations: $\log_a b$ and $a^b$. You can only take logs of positive numbers.
  • The root function as a binary operator $\sqrt[b]{a}$.
  • Unary operations: $(\cdot)!$ for integer arguments only, unary negation -, the square root $\sqrt{\cdot}$ for positive numbers, and floor and ceil functions: $\lfloor\cdot\rfloor$ and $\lceil \cdot\rceil$.

Any other operations are not allowed, including double factorials and decimal points. In addition, you can do the following with no penalty:

  • Parentheses (for grouping purposes only, no binomial coefficients etc.)

The score will be calculated the same way as in the four fours puzzle*.

Your score is equal to the number of digits of accuracy per operation used. That is, if you got the approximation A by using n operations, your score is

$$\frac{-\log_{10}|e - A|}{n}$$ (adjusted for e instead of π.)

*However, as a side note, you multiply your score by π/n where n is the number of πs used in your approximation.

The one with the highest score will get that sweet green checkmark on their submission.

Hint:

Try to use $$\lim_{n \to +\infty} \left( 1 + \frac{1}{n}\right)^n = e. $$

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    $\begingroup$ How long will you wait before awarding the checkmark? Also, [open-ended] puzzles (as the Four Fours one was) are no longer allowed - do you have any sort of proof for an optimal solution here? $\endgroup$ – bobble Nov 26 '20 at 17:41
  • $\begingroup$ 10 days. Maybe I could allow proofs of optimal solutions in the answers. $\endgroup$ – new QOpenGLWidget Nov 26 '20 at 17:46
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One can get

an arbitrarily high finite score

by

using the suggested formula $e\simeq\left(1+\frac1n\right)^n$ and taking $n=\pi^{\pi^{\pi^\pi}}$ with an arbitrarily large tower of $\pi$s. (Grouped so that you do the "upper" exponentiations first, of course.)

So the only question is

whether one can get an infinite score by hitting $e$ exactly. Obviously we could do it if allowed complex logarithms (with some care being needed about exactly how they're interpreted) since $e^{i\pi}=-1$, but I think it can't be done with the restrictions given in the question.

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    $\begingroup$ How about $e \approx \pi$? Not very accurate, but zero operations in the denominator of the score. $\endgroup$ – Bass Nov 26 '20 at 20:28
  • $\begingroup$ Ha! That works, but I suggest that that should simply be regarded as a bug in the definition of the score :-). $\endgroup$ – Gareth McCaughan Nov 26 '20 at 20:36
  • $\begingroup$ As $ \underset{{n\rightarrow\infty}}\lim -\frac{\log_{10}\left|e-(1+\frac{1}{\pi\uparrow\uparrow n})^{\pi\uparrow\uparrow n}\right|}{n}=\infty $, I would prefer simply consider it has infinity score. $\endgroup$ – tsh Nov 27 '20 at 9:39
  • $\begingroup$ I think the distinction between "finite but unboundedly large" and "actually infinite" is an important one. For "actually infinite" one would need (ignoring Bass's ingenious but cheaty "rubbish approximation, no operations" approach) some expression in terms of $\pi$ and the available operations that lands exactly on $e$. I think there isn't one, but it's not clear to me how to prove it. $\endgroup$ – Gareth McCaughan Nov 27 '20 at 17:00

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