|q⟩
Bad Qubits
Play
Quest
Questions
Learn
Playground
☕
← Question Bank
Multiple choice
For Grover search over
N
=
2
10
=
1024
N=2^{10}=1024
N
=
2
10
=
1024
items, the optimal number of iterations is
k
=
⌊
(
π
/
4
)
N
⌋
k=\lfloor(\pi/4)\sqrt{N}\rfloor
k
=
⌊(
π
/4
)
N
⌋
. What is
k
k
k
?
25
16 (using
(
π
/
8
)
N
(\pi/8)\sqrt{N}
(
π
/8
)
N
)
32 (
1024
\sqrt{1024}
1024
, omitting the
π
/
4
\pi/4
π
/4
factor)
26 (rounding up instead of flooring)
Check answer