Fair warning: This will be surprisingly simple.

Since you understand the 2-qubit Grover’s Algorithm now, it’s quite easy to scale up to 3-qubits. Firstly, regarding the oracle, there is a CCZ gate which has a matrix \(\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}\) which already marks \(|111\rangle\) with a negative just like that. And using the same idea from last time, we can manipulate what it marks using X gates before and after the CCZ gate. By the way, in Jupyter, a CCZ gate is created with two Hadamard gates, one on each side of the target qubit, and a Toffoli gate (CCNOT).

Also, for reference, the columns and rows essentially just increase in binary all the way to \(111\). They start at \(000\), then \(001\), then \(010\), \(011\)…

So now that the oracle is completed, we just need to be able to reflect the amplitude which is also surprisingly easy as we already have a general formula \(H^{\otimes n}(2|0\rangle\langle 0|-I)H^{\otimes n}\) which we can accomplish with this circuit:

Jupyter has no formatting method for circuit drawing so here is a neater visual that I drew:

Don’t forget that we need to calculate how many times to run Grover’s iteration though. Using the formula $$sin(\theta) = \frac{2\sqrt{M(N-M)}}{N}$$ with \(M = 1\) and \(N = 2^3 = 8\), we can easily calculate \(\theta\) to be \(41.41^\circ\) and thus the starting angle to be \(20.7^\circ\). Clearly, with two rotations, we get the closest to \(\frac{\pi}{2} radians = 90^\circ\) even though we rotate slightly past it. This also falls in line with the other formula $$R \leq \lceil\frac{\pi}{4}\sqrt{\frac{N}{M}}\rceil$$ in which we can calculate R to be 3. Thus, at most 3 rotations are required to get close to \(|\beta\rangle\) and we calculated the exact number to be 2.

You can try this out for yourself in Jupyter. I included my code and my drawn circuit down below:

Simulating this, I got

Keep in mind, this 3-qubit form of Grover’s algorithm is so easy precisely because a CCZ gate exists. For four qubits, there is no CCCZ gate and thus is considerably more complex.