A Triad of Pies

Question

Make the 41 integers between -20 and 20 (-20, -19, ..., 0, 1, ...20) using only the four basic arithmetical operations, square root, the floor function, and, Surprise Surprise, exactly 3 π’s. No more and no less. No other digits or symbols are allowed. You can use as many square roots and floor functions as you like. Parenthesis are allowed. Unary minus sign is not allowed. Exponentiation is not allowed. Anything not explicitly allowed is disallowed.

The floor function for x, written $\lfloor x \rfloor$, is equal to the greatest integer smaller or equal to x. For example, $\lfloor \pi \rfloor = 3$, $\lfloor \pi \times \pi \rfloor = 9$ and $\lfloor -\pi \rfloor = −4$.

Shorter is better (i.e. try to find expressions with the minimal number of +, -, $\times$, $\div$, $\lfloor\rfloor$ and $\sqrt{}$'s.)

Note: 20 and -11 to -20 are very challenging. (0 to 10 are fun)

Answer 1

Thanks to Retudin for $14,15,16,20$ and the inspiration necessary to find some others.

• $-20=\left\lfloor\lfloor\pi\rfloor\div\left(\lfloor\sqrt{\pi}\rfloor-\sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

$-19$ not found yet

• $-18=\left\lfloor\pi\div\left(\sqrt{\sqrt{\pi}}-\sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

• $-17=\left\lfloor\sqrt{\sqrt{\pi}}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

$-16,-15,-14$ not found yet

• $-13=\left\lfloor\pi\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\pi}}\right)\right\rfloor$

• $-12=\left\lfloor\sqrt{\sqrt{\pi}}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} - \sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

• $-11=\left\lfloor\pi\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} - \sqrt{\sqrt{\pi}}\right)\right\rfloor$

• $-10=\left\lfloor\pi\div\left(\lfloor\sqrt{\pi}\rfloor-\sqrt{\sqrt{\pi}}\right)\right\rfloor$

$-9$ not found yet

• $-8=\lfloor\sqrt{\pi}\rfloor-\lfloor\pi\times\pi\rfloor$

• $-7=\left\lfloor\sqrt{\pi}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\pi}}\right)\right\rfloor$

• $-6=\lfloor\pi\rfloor-\lfloor\pi\times\pi\rfloor$

• $-5=\lfloor\sqrt{\pi}\rfloor-\lfloor\pi+\pi\rfloor$

• $-4=\lfloor\pi-\pi-\pi\rfloor$

• $-3=\pi-\pi-\lfloor\pi\rfloor$

• $-2=(\pi\div\pi)-\lfloor\pi\rfloor$

• $-1=\lfloor\sqrt{\pi}\rfloor-\lfloor\sqrt{\pi}\rfloor-\lfloor\sqrt{\pi}\rfloor$

• $0=\lfloor\pi\div(\pi\times\pi)\rfloor$

• $1=\lfloor\sqrt{\pi}\times\pi\div\pi\rfloor$

• $2=\lfloor\pi\rfloor-(\pi\div\pi)$

• $3=\lfloor\pi\times\pi\div\pi\rfloor$

• $4=\lfloor\pi\rfloor+(\pi\div\pi)$

• $5=\lfloor\sqrt{\pi\times\pi\times\pi}\rfloor$

• $6=\lfloor\pi+\pi\rfloor\times\lfloor\sqrt{\pi}\rfloor$

• $7=\lfloor\pi+\pi\rfloor+\lfloor\sqrt{\pi}\rfloor$

• $8=\lfloor\sqrt{\sqrt{\pi}}\times(\pi+\pi)\rfloor$

• $9=\lfloor\pi+\pi+\pi\rfloor$

• $10=\lfloor\pi\times\pi\rfloor+\lfloor\sqrt{\pi}\rfloor$

• $11=\lfloor\sqrt{\pi}\times(\pi+\pi)\rfloor$

• $12=\lfloor\pi\times\pi\rfloor+\lfloor\pi\rfloor$

• $13=\lfloor\sqrt{\sqrt{\pi}}\times\pi\times\pi\rfloor$

• $14=\lfloor(\pi+\sqrt\pi)\times\lfloor\pi\rfloor\rfloor$

• $15=\lfloor\lfloor\pi\rfloor\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

• $16=\lfloor\pi\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

• $17=\lfloor\pi\times\pi\times\sqrt{\pi}\rfloor$

• $18=\lfloor\pi\rfloor\times\lfloor\pi+\pi\rfloor$

• $19=\lfloor\pi\times(\pi+\pi)\rfloor$

• $20=\left\lfloor\sqrt{\sqrt{\sqrt{\pi}}}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}}\right)\right\rfloor$

Answer 2

As OP moved the goal posts half-way through I can't be bothered to fix this ...

$0 = (\pi-\pi)\times\pi$
$\pm 1 = \pm (\pi^{\pi-\pi})$
$\pm 2 = \pm \frac{\pi+\pi}{\pi}$
$\pm 3 = \pm \lfloor \pi+\pi-\pi \rfloor$
$\pm 4 = \mp \lfloor \pi-\pi-\pi \rfloor$
$\pm 5 = \mp \lfloor -\frac {\pi}{\pi}-\pi \rfloor$
$\pm 6 = \pm \lfloor \pi \times \pi - \pi \rfloor$
$\pm 7 = \mp \lfloor \pi - \pi \times \pi \rfloor$
$\pm 8 = \pm \lfloor \frac {\lfloor \pi \rfloor^{\lfloor \pi \rfloor}} {\pi} \rfloor$
$\pm 9 = \pm \lfloor \frac {\pi^{\lfloor \pi \rfloor}} {\pi} \rfloor$
$\pm 10 = \mp \lfloor -\frac {\pi^{\lfloor \pi \rfloor}} {\pi} \rfloor$
$\pm 11 = \pm \lfloor \frac {\pi^{\pi}} {\pi} \rfloor$
$\pm 12 = \mp \lfloor -\frac {\pi^{\pi}} {\pi} \rfloor$
$\pm 13 = \pm \lfloor -\frac {\lfloor \pi \rfloor^{\lfloor \pi \rfloor}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 14 = \mp \lfloor \frac {\lfloor \pi \rfloor^{\lfloor \pi \rfloor}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 15 = \pm \lfloor \frac {\lfloor \pi \rfloor^{\pi}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 16 = \mp \lfloor \frac {\lfloor \pi \rfloor^{\pi}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 17 = \pm \lfloor \pi\pi \sqrt{\pi} \rfloor$
$\pm 18 = \mp \lfloor -\pi\pi \sqrt{\pi} \rfloor$
$\pm 19 = \pm \lfloor \pi(\pi+\pi) \rfloor$
$\pm 20 = \mp \lfloor -\pi(\pi+\pi) \rfloor$

Answer 3

'missing' positives
14-16

$14=\lfloor(\pi+\sqrt\pi)\times\lfloor\pi\rfloor\rfloor$

$15=\lfloor\lfloor\pi\rfloor\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

$16=\lfloor\pi\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

20

$20 = \lfloor\sqrt(\sqrt(\sqrt\pi)) / (\sqrt(\sqrt(\sqrt(\sqrt\pi))) - \sqrt(\sqrt(\sqrt(\sqrt(\sqrt(\sqrt\pi))))) \rfloor) $

Answer 4

The missing numbers (including some Rand also solved in the meantime)

$-9 = \lfloor \lfloor \sqrt{\pi} \rfloor - \pi \rfloor \times \lfloor \pi \rfloor$
$-10 = \lfloor \lfloor \lfloor \sqrt{\pi} \rfloor - \pi \rfloor \times \pi \rfloor$
$-12 = \lfloor\frac{\sqrt{\pi}}{\lfloor \sqrt{\pi} \rfloor - \sqrt{\sqrt{\sqrt{\pi}}}} \rfloor$
$-13 = \lfloor \frac{\sqrt{\pi}}{\lfloor \pi \rfloor - \pi}\rfloor$
$-14 = \lfloor\frac{\lfloor \sqrt{\pi} \rfloor}{\lfloor \sqrt{\pi} \rfloor - \sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} \rfloor$
$-15 = \lfloor \frac{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}}}}}} - \sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}\rfloor$
$-16 = \lfloor \frac{ \sqrt{\pi} }{ \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} - \sqrt{\sqrt{\sqrt{\pi}}}}\rfloor$
$-17 = \lfloor \frac{\lfloor \pi \rfloor}{ \sqrt{\sqrt{\sqrt{\pi}}} - \sqrt{\sqrt{\pi}}}\rfloor$
$-18 = \lfloor \frac{\pi}{ \sqrt{\sqrt{\sqrt{\pi}}} - \sqrt{\sqrt{\pi}}}\rfloor$
$-19 = \lfloor \frac{\sqrt{\sqrt{\pi}}}{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}}}}}}- \sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} \rfloor$
$-20 = \lfloor\frac{\lfloor \pi \rfloor}{\lfloor \sqrt{\pi} \rfloor - \sqrt{\sqrt{\sqrt{\pi}}}} \rfloor$

Answer 5

Those are "minimal" expressions. (some of you posted alternate forms with equal number of operations, but others were not minimal).