Lecture 016 - Geometry of Grover

Geometry of Grover and Approximate Sharp-SAT

Reflection operation: quantum operation that is reverse of itself

Rotation operation: quantum operation that is not reverse of itself

Geometry of Grover's SAT Algorithm

Remember Grover's SAT Algorithm: assuming unique string x \in \{0, 1\}^n satisfy f(x) = 1.

  1. We prepare uniform superposition
  2. We can convert any code into AND/OR/NOT circuit
  3. Do If C(x_1, x_2, ..., x_n) then Minus
  4. Reflect accross the mean
  5. Go to 3.

Now we aim for a geometric interpotation of Grover's algorithm.

Observations:

Geometry of Grover's Algorithm: klzzwxh:0002 is reflection accross the mean and klzzwxh:0003 is if...then...minus.

Geometry of Grover's Algorithm: V is reflection accross the mean and W is if...then...minus.

Imagine the horizontal axis denotes everything else that does not satisfy the the function f. Vertical axis is the bit we sill keep reflecting. We start from uniform position |\phi\rangle, and do "If ... then Minus" (reflection across |\alpha_{\perp}\rangle), and do "Reflection Accross Mean" (reflection across \simeq |\phi\rangle, although not exact)

If we use un-normalized representation, imagine the circle to be length \sqrt{N}

If you do too many times of V and W, for example, twice of amount, you will go pass our goal. In each combo V \cdot W, the angle \theta of the vector got changed by a constant amount. In numerical representation, we observe we get the the goal with slower and slower speed, this is because of how we measure the successfulness: we measure it by probability. At the start, we increase the probability to go to our goal a lot, but it becomes slower and slower as two angle approaches even though we change it by a constant amount every time.

The total amount of rotation is: 90 \degree - \theta/2 \simeq \pi/2 since, according to the graph above |\alpha_{\perp}\rangle \perp |\alpha\rangle.

Knowing the Probability of Success is \frac{1}{N}

Assume |v\rangle = V|\phi\rangle = [1,...,1,-1,1,...,1], |u\rangle = |\phi\rangle = [1,...,1]

We then ask: how much \theta did we rotate when we do one "combo"?

\begin{align*} \cos \theta =& \langle v' | u' \rangle\\ =& \frac{1}{\sqrt{N}} \frac{1}{\sqrt{N}} \langle v | u \rangle\\ =& \frac{1}{N} ((N - 1)(1) + 1(-1))\\ =& \frac{N - 2}{N}\\ =& 1 - \frac{2}{N}\\ =&1-2p\\ \cos^2 \theta =& (1 - \frac{2}{N})^2\\ =& 1 - \frac{4}{N} + \frac{4}{N^2}\\ \simeq& 1 - \frac{4}{N} \tag{by $\frac{4}{N^2}$ negligible}\\ 1-\sin^2 \theta =& 1 - \frac{4}{N}\\ \sin \theta =& \frac{2}{\sqrt{N}}\\ \theta \simeq& \frac{2}{\sqrt{N}}\tag{by $\theta$ small}\\ \end{align*}

If you want to do k rotations instead of 1, you will rotate by about \frac{2k}{\sqrt{N}}. Therefore, if you want to rotate \frac{\pi}{2}, you need k = \frac{\pi}{4}\sqrt{N}

Knowing the Probability of Success is \frac{m}{N}

\cos \theta = \langle v' | u' \rangle = \frac{1}{\sqrt{N}} \frac{1}{\sqrt{N}} \langle v | u \rangle = \frac{1}{N} ((N - m)(1) + m(-1)) = \frac{N - 2m}{N} = 1 - \frac{2m}{N} \implies \theta = 2\sqrt{\frac{m}{N}}

If you want to do k rotations instead of 1, you will rotate by about 2k\sqrt{\frac{m}{N}}$. Therefore, if you want to rotate \frac{\pi}{2}, you need k = \frac{\pi}{4}\sqrt{\frac{N}{m}} \in O(\frac{1}{\sqrt{p}})

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