Suppose we ran m steps of Grover's algorithm on some function f (which has one marked element y) and the resulting superposition was exactly |y>.
(a) What was the state after the (m−1)th step? Note that you can describe the superposition by specifying two numbers, αy and αx for x≠y. Use K to denote the total number of elements. Please fully simply your answer.
Answer in this format: αy: αx for x≠y:
(b) Now, if we run one more step (total of m+1 steps), what is the resulting superposition?
Answer in this format: αy: αx for x≠y:
(c) What if you now apply another phase inversion?
Answer in this format: αy: αx for x≠y:
PLS HELP
help guys
Anyone please?
So the state is at |y> after m steps. So the probability of getting that is 1 and the rest is 0. So What is the mean and what if we move 1 step forward? Is the the same as moving 1 step back?
I have tried to solve it through the trigonometrics way. If in the iteration m you have probability 0 for the rest of the situations, then cos[(2k+1)g] = 0 becasuse of that k=(pi/(4*g))-(1/2) and you know that cos(g)=sqrt(k-1)/sqrt(k) so you have the value of k and the value of cos(g) so you can apply it to k-1 and try to solve the values of coefficients with the aid of the trigonometrics functions. However the marker gives me a red cross, I don´t know what I´m doing wrong! Please help.
Problem 4,5,10,11 answers guys
12 a and 13 a also pls guys
Thank you all.
And does anyone can help me with p1,p5 and p13
Problem 1
a)No
b)1
c)a; -b
Problem 13
a)
1; 0
0; 5
b)Last tick for multiple question
c)0
d)0
Anyone for Problem 5 though please?