$\tau$ is greatest! $\tau$ is all (from 1 to 20)

Question

(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.

Answer 1

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 = $\lfloor\tau^{\sqrt{\sqrt{\tau}}}\rfloor + \lfloor\sqrt{\tau}\rfloor$

Answer 2

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.

Answer 3

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.