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Time Complexity Analysis - Grover

Time Complexity Analysis - Grover

Let’s learn why Grover's Algorithm runs in O(sqrt(N)) time.

This is going to be a geometric proof to show that the number of Grover iterations we must do to achieve a high probability of measuring the winner state is bounded by O(N)O(\sqrt{N}).

Equal superposition

We will carry forward what we’ve learned from the past lessons and start with the situation where some arbitrary state x|x\rangle could be measured from the superposition, and it lies in between w|w\rangle and E|E\rangle in the state space. The x|x'\rangle state is orthogonal to w|w\rangle and the angle between x|x'\rangle and E|E\rangle is Δ\Delta

Reflecting about the w|w\rangle state

The state x|x\rangle makes an angle θ\theta with the w|w\rangle, upon reflection about the w|w\rangle state, our new state x1|x_1\rangle would end up with the same angle but on the opposite side.

Reflecting about the E|E\rangle state

If we denote the angle between E|E\rangle and x|x\rangle as ϕ\phi then after this reflection, our new state x2|x_2\rangle would have an angle of 2θ+ϕ2\theta+\phi with E|E\rangle. Using all the information we have gathered so far, we need to focus on finding the angle between our final state and the original state.

Take a look at the upper half (first and second quadrants) of the diagram given below, the sum of the angles should be equal to π\pi radians.

2ϕ+2θ+2Δ=π2\phi + 2\theta + 2 \Delta = \pi

Let’s have a look at x2-|x_2\rangle because that’s essentially the same state (ignoring global phase, the 1-1 sign), which is now closer to w|w\rangle. Remember that this state is simply in the opposite direction. Geometrically, the angle between x2|x_2\rangle and x2-|x_2\rangle ...

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