Let f:{0,1}^3→{0,1} with f(101)=1 and f(x)=0 for x≠101. In this question, we will run through two iterations of the algorithm.

(a) What is the superposition after the initialization step? Note that you can describe the superposition ∑x∈{0,1}3αx|x> by specifying two numbers α101 and αx for x≠101.
Answer in this format: α101: αx for x≠101:

(b) After the phase inversion in iteration 1?
Answer in this format: α101: αx for x≠101:

(c) After the inversion about mean in iteration 1?
Answer in this format: α101: αx for x≠101:

(d) After the phase inversion in iteration 2?
Answer in this format: α101: αx for x≠101:

(e) After the inversion about mean in iteration 2?
Answer in this format: α101: αx for x≠101:

Anyone please!

1/sqrt(8),1/sqrt(8)

-1/sqrt(8),1/sqrt(8)
5/(2*sqrt(8)),1/(2*sqrt(8))
-5/(2*sqrt(8)),1/(2*sqrt(8))
11/(4*sqrt(8)),-1/(4*sqrt(8))

problem 6 plz

To answer these questions, we need to understand the steps involved in the algorithm.

The algorithm involves the following steps:
1. Initialization: Create the equal superposition of all possible inputs.
2. Oracle (Black Box): Apply the function f to the superposition.
3. Phase Inversion: Negate the phase of the marked element (marked by the function f).
4. Inversion about the Mean: Perform an inversion about the mean of the superposition.

Now, let's go through each step and calculate the superposition after each iteration.

(a) Initialization:
In this case, we have a function f where f(101) = 1 and f(x) = 0 for x ≠ 101.
To initialize the superposition, we create an equal superposition of all possible inputs, which is:
∑x∈{0,1}^3 (1/√8) |x>

Since f(101) = 1, α101 = 1/√8 and αx for x ≠ 101 = 1/√8.

Therefore, the superposition after initialization is:
α101: 1/√8
αx for x ≠ 101: 1/√8

(b) After the phase inversion in iteration 1:
In this step, we apply the oracle (function f) to the superposition. The function marks the element 101, which means its phase is negated.

The superposition after the phase inversion is:
α101: -1/√8
αx for x ≠ 101: 1/√8

(c) After the inversion about mean in iteration 1:
In this step, we perform an inversion about the mean of the superposition.

The mean of the superposition is (1/√8) * (|101> + ∑x≠101|x>), which is the sum of all elements (including the marked element) divided by the square root of 8.

The inversion about the mean operation flips the signs of the amplitudes about the mean.

The superposition after the inversion about mean is:
α101: -(1 - 2/√8)
αx for x ≠ 101: 1/√8

(d) After the phase inversion in iteration 2:
In this step, we apply the oracle (function f) again to the superposition. The function marks the element 101, which means its phase is negated.

The superposition after the phase inversion is:
α101: (1 - 2/√8)
αx for x ≠ 101: 1/√8

(e) After the inversion about mean in iteration 2:
In this step, we perform an inversion about the mean of the superposition.

The mean of the superposition is (1/√8) * (|101> + ∑x≠101|x>), which is the sum of all elements (including the marked element) divided by the square root of 8.

The inversion about the mean operation flips the signs of the amplitudes about the mean.

The superposition after the inversion about mean is:
α101: -(1 - 4/√8)
αx for x ≠ 101: 1/√8

Final answers:
(a) α101: 1/√8
αx for x ≠ 101: 1/√8

(b) α101: -1/√8
αx for x ≠ 101: 1/√8

(c) α101: -(1 - 2/√8)
αx for x ≠ 101: 1/√8

(d) α101: (1 - 2/√8)
αx for x ≠ 101: 1/√8

(e) α101: -(1 - 4/√8)
αx for x ≠ 101: 1/√8