9
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(This is basically an extension of $\pi$ Day puzzle one to twenty)

$\tau$ is greater than $\pi$ and $\tau>\pi$.

Create the numbers from $1$ to $20$ using only:

  • Tau ($\tau$, equivalent to $2\pi$)
  • Basic arithmetic operations ($+-\times\div$)
  • Square roots ($\sqrt{x}$ or $\sqrt[2]{x}$)
  • Exponentiation ($x^y$)
  • Negative tau ($-\tau$)
  • Floor functions ($\lfloor x\rfloor$)

Anything not in this list is forbidden. You are not allowed to have negative signs outside of $\tau$ or not as an operation (E.g $-\lfloor\tau\rfloor$ is forbidden, but $\tau-\lfloor\tau\rfloor$ is allowed.) You are also not allowed to use parentheses, although $\lfloor x\rfloor$ can make a good substitute.

Some basic MathJaX syntax:

$\tau, +, -, \times, \div, \sqrt{\tau^{\tau}}, \lfloor\tau\rfloor, \sqrt[2]{\tau}$

$\tau, +, -, \times, \div, \sqrt{\tau^{\tau}}, \lfloor\tau\rfloor, \sqrt[2]{\tau}$

Some more
Remember the order of operations.

1 = $\tau\div\tau$ (Uses 2 $\tau$s, worse score)
1 = $\lfloor\sqrt{\sqrt\tau}\rfloor$ (Uses 1 $\tau$, better score)

2 = $\tau\div\tau+\tau\div\tau$ (Uses 4 $\tau$s, worse score)
2 = $\lfloor\sqrt\tau\rfloor$ (Uses 1 $\tau$, better score)

Try and use the least $\tau$s possible.

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  • $\begingroup$ Curiously, I was taught (back in the day) that $\tau$ was the symbol for the golden ratio. $\endgroup$ – Ian MacDonald Mar 26 '15 at 17:49
  • 3
    $\begingroup$ I've seen that in some textbooks too; nowadays we use $\varphi$. $\endgroup$ – Joe Z. Mar 26 '15 at 17:57
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I used 43 $\tau$ to get to 20.
(44 before GOTO 0's suggestion)

1 = $\lfloor\sqrt{\sqrt{\tau}}\rfloor$
2 = $\lfloor\sqrt{\tau}\rfloor$
3 = $\lfloor\sqrt{\tau}\rfloor + \lfloor\sqrt{\sqrt{\tau}}\rfloor$
4 = $\lfloor\sqrt{\tau}\rfloor \times \lfloor\sqrt{\tau}\rfloor$
5 = $\lfloor\tau\rfloor - \lfloor\sqrt{\sqrt{\tau}}\rfloor$
6 = $\lfloor\tau\rfloor$
7 = $\lfloor\tau\rfloor + \lfloor\sqrt{\sqrt{\tau}}\rfloor$
8 = $\lfloor\tau\rfloor + \lfloor\sqrt{\tau}\rfloor$
9 = $\lfloor\sqrt{\sqrt{\tau}} \times \tau\rfloor$
10 = $\lfloor\sqrt{\tau}^{\sqrt{\tau}}\rfloor$
11 = $\lfloor\tau\rfloor + \lfloor\tau\rfloor - \lfloor\sqrt{\sqrt{\tau}}\rfloor$
12 = $\lfloor\tau\rfloor + \lfloor\tau\rfloor$
13 = $\lfloor\tau\rfloor - \lfloor-\tau\rfloor$
14 = $\lfloor\tau\rfloor - \lfloor-\tau\rfloor + \lfloor\sqrt{\sqrt{\tau}}\rfloor$
15 = $\lfloor\tau\rfloor - \lfloor-\tau\rfloor + \lfloor\sqrt{\tau}\rfloor$
16 = $\lfloor\sqrt{\sqrt{\tau}}^{\tau}\rfloor - \lfloor\sqrt{\sqrt{\tau}}\rfloor$
17 = $\lfloor\sqrt{\sqrt{\tau}}^{\tau}\rfloor$
18 = $\lfloor\tau^{\sqrt{\sqrt{\tau}}}\rfloor$
19 = $\lfloor\tau \times \tau \div \lfloor\sqrt{\tau}\rfloor\rfloor$
20 = $\lfloor\tau^{\sqrt{\sqrt{\tau}}}\rfloor + \lfloor\sqrt{\tau}\rfloor$

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  • 3
    $\begingroup$ You can write 10 as $\lfloor\sqrt\tau^\sqrt\tau\rfloor$ and save one $\tau$. $\endgroup$ – GOTO 0 Mar 26 '15 at 19:36
  • $\begingroup$ Interesting observation. $\endgroup$ – Ian MacDonald Mar 26 '15 at 19:38
  • $\begingroup$ I think there's a mistake on 19. Are you missing a 'floor' around $\tau\div\lfloor\sqrt{\tau}\rfloor$? $\endgroup$ – KSmarts Mar 26 '15 at 20:18
  • $\begingroup$ @KSmarts: No. Only the floors around the final result and the inner square root are needed. In fact, if you were to put another floor in there, it would resolve to 18, not 19. $\endgroup$ – Ian MacDonald Mar 26 '15 at 20:21
  • $\begingroup$ My bad. I only have a "regular" calculator, so I probably messed up my button-pressing. $\endgroup$ – KSmarts Mar 26 '15 at 20:51
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I've got some different answers. For brevity, I didn't include the ones that were the same as (or very similar to) Ian's answers:

$$\begin{align}4& =\lfloor\sqrt{\tau}^\sqrt{\sqrt{\tau}}\rfloor\\5& =\left\lfloor\sqrt{\tau}\times\lfloor\sqrt{\tau}\rfloor\right\rfloor\\8&=\left\lfloor\sqrt{\lfloor\sqrt{\tau}\rfloor^{\tau}}\right\rfloor\\10&=\left\lfloor\tau^{\sqrt{\sqrt{\sqrt{\tau}}}}\right\rfloor\\15&=\lfloor\tau\times\sqrt{\tau}\rfloor\\19&=\left\lfloor\sqrt{\sqrt{\tau}}^{\;\tau}+\sqrt{\sqrt{\tau}}\right\rfloor\\20&=\left\lfloor\lfloor\tau+\sqrt{\tau}\rfloor\times\sqrt{\tau}\right\rfloor\end{align}$$

My solution for $15$ saves one $\tau$ over Ian's answer. The rest use the same number.

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1
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For sake of completeness, I have done a computer search and have found that 40 $\tau$s is the absolute minimum one can achieve. This can be done using Ian's expressions for 1–13 and 17–20, KSmarts' expression for 15, and two new expressions for 14 and 16:

$$14=\left\lfloor\sqrt{\lfloor\tau\rfloor}\times\lfloor\tau\rfloor\right\rfloor$$ $$16=\left\lfloor\sqrt{\sqrt{\lfloor\tau\rfloor}}^\tau\right\rfloor$$

The only expressions requiring 3 $\tau$s are those for 11, 19, and 20, and exhaustive search has shown that no expression with 2 $\tau$s and the operators given can produce any of those numbers.

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