Question

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⟩.
If we run for m+1 additional steps (i.e. total of 2m+1 steps from the initial state), what is the resulting superposition? Note that you can describe the superposition by specifying two numbers, αy and αx for x≠y. You may use K to denote the total number of elements. (K should be uppercase.) HINT: You may want to review Problems 2, 3, and 5 of Assignment 6.
αy :
αx for x≠y:
Now if we run for another m steps, what is the resulting superposition?
αy :
αx for x≠y:
What about after yet another m+1 steps?
αy :
αx for x≠y:

Answer 1

Stucklike you :(, anyone please

Answer 2

I have done 6A:
(1/2 5/2)
(5/2 1/2)

6C and 6D: 0

others pleasee

Answer 3

6B is
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

Answer 4

6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->

Answer 5

Problem 5 plz

Answer 6

Thank you guys!!

Answer 7

4b= 4

someone for the first??plzz

Answer 8

Any kind of help is well received ;)

Answer 9

3C=0

Answer 10

1st is part c):

-z-h-
-x-h-

Answer 11

3B is Z

Answer 12

I meant answer to question 1 is part c.

-z-h-
-x-h-

Answer 13

problem 5,7 please.

Answer 14

4A please

Answer 15

problem 4 a):

the eigenvalues are 1 and -1

then the lowest eingevalue (ground energy) is : -1

Answer 16

4A is -1

Answer 17

3D yes or no ?

Answer 18

3d no

Answer 19

Problem 7a
theta = pi/4
phi = 5*pi/2

Answer 20

Problem 7b
1/2+i/2
1/sqrt(2)

Answer 21

wrong

Answer 22

7A and 7B wrong

Answer 23

SOMEBODY KNOW ANSWER FOR: 3A? 5? 7? 2? 4C?

Answer 24

5 and 7 please

Answer 25

hikikomori, do you have answer for the 3A? 5? 7? 2? 4c?

Answer 26

hikikomori,Sorry, do you have answer for the 3A? 2? 4c?

Answer 27

7A) pi/2 , 5*pi/4

Answer 28

7B) 0.707 , -0.5-(0.5*i)

Answer 29

4C (001, 010, 100, 111)

Answer 30

anyone question 5?

Answer 31

3A: C option

Answer 32

Thanks all guys.
We now just need Q5's answers.

Answer 33

We need Q.2 also

Answer 34

for 2, last option is correct

Answer 35

Thanks all guys
well-done

Answer 36

Problem 5 please?

Answer 37

desperately need answer for question 5. can someone explain a little how we get last option as correct for problem 2. as in the original cct if we give input 1> and 0> then apply cnot then we get 11> as the target bit flips.after that once apply Z gate (which will now be 4x4 matrix)we obatain (0,-1,0,1). how last option satisfy the same with same input as 0> once pass Z gate we get same 0> later as control bit is 1> it flips it after application of cnot and we get 11> as output. how are both equivalent??plz help

Answer 38

Anyone??? Pls Q5... guys do soething pls

Answer 39

PLease Problem 5!?

Answer 40

5A: (2/K - 1), (2/K)

5B: (1- 2/K), (-2/K)

5C: (2/K - 1), (2/K)

Answer 41

EdX Winner its wrong answer

Answer 42

5A: (2/K), (1 - 2/K)

5B: (-2/K), (2/K - 1)

5C: (2/K - 1), (-2/K)

Answer 43

Wrong? It worked for me!

Answer 44

qwerty urs answer is also showing wrong dude

Answer 45

All wrong?

Answer 46

YES ALL WRONG

Answer 47

the above answers to 5th question is of assignment 6 problem 5 don't get misguided

Answer 48

Anon.. What is the correct answer to the 5? please help us

Answer 49

WORKING ON IT AR . DISTRACTED BY SOME PERSONAL PROBLEMS IN LIFE ... NOT ABLE TO CONCENTRATE

Answer 50

I think it should look like

(k-2)^x + ...
-------------
k^x

where x = m+1
but i have failed for find oput more :-(

Answer 51

I am trying my best. Cant seem to get an answer. I will keep trying. Till then, I request others to try and post solutions here as well. Thanks.

Answer 52

In the assignment 6, when you look to the posted answer it was written -answer to the part (c)- "Note that this is exactly the negation of the answer to part (a)". I think the following:
the state after " m+p steps + phase inversion" is equal to minus (-) the state after m-p steps! I checked this statement and it turns out to be true: the state after "2m step + phase inversion" must be equal to minus (-) the initial state...however when I enter the answer which becomes obvious when you apply the above observation, the grader says always "wrong"!...so guys to be honest with you: I still have only one shot for the three questions of problem 5: so or I get them all right (and for sure I will forward the right answer to all of you) or I will get them wrong...bye bye

Answer 53

problem 3b,4c and 5

Answer 54

well guys, i need few drinks (i'm doing my best MF) and we will be all right..cheer MF

Answer 55

Some ppl think there is a simple answer to the problem.
To be honest I do not beleive it. For instance let m=1 and k=171. Why not? The original state was not defined in the problem, so it just might be that way. Will the "answer" work? I am sure it won't.

Answer 56

Hi! I have just done 5B: ay=-1, ax=0, so, cheers and PhysTech, it must something about a cycle of the states from m-2, m-1, m, m+1, m+2

Answer 57

Hi J. If the grader thinks it is the proper answer it doesn't mean it is. I tried it for a several k and there is no cicle except fo k=2. Try for k=7, and goes and goes without end, never in cicle.

Answer 58

@xaad

3B: Z (3rd option)
4C: Last option
5B: ay = -1; ax = 0
5A, 5C: ---Not yet solved correctly---

Answer 59

Hi PhysTech, I am reading an article from twistedoakstudios(at)com in /blog/Post2644_grovers-quantum-search-algorithm and doing some calculations... Seems cyclic, in a geometric view.

Answer 60

Ok, 5C is 1/sqrt(K) for both of them. I cannot figure out what is 5A... :(

Answer 61

5A is not -1/sqrt(N) or any combinations with minus sign.

Answer 62

@J
I input 1/sqrt(K) as 5C's answer, but they are wrong.

Answer 63

@J
No they're right, I misspell the answer

Answer 64

dudes there is something wrong with the system: i did not answer to 5) c) and as I mentioned before I have a last shot (only one shot) and guess what? I gave the answer to 5)b) (which was -1 and 0) and automatically the grader gave me the answer for 5)c)which is (1/sqrt(k),1/sqrt(k))...!!!!! there is something wrong here...don't you think so???

Answer 65

wow! Strange behaviour @cheers xD Nevertheless good news for 5C :D Now, only 5A remains behind the Fortress of Solitude...

Answer 66

@J "sorry I did not sleep the whole night"...I apologise to everybody...I'm still working on a) please forgive my "swearing" I did not sleep for 24h...booze make you thinking...cheers

Answer 67

C'mon @cheers!! get some rest :) 5A is not (0,-1), or (0,1/sqrt(K)) or (0,-1/sqrt(N)) xD It must be something very close to it... I think the answer is translating the grover algorithm to a Bloch Sphere and see how it moves around the surface :) At least, that is what 5B,c suggest me...

Answer 68

what is the answer of 6B?
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

or

6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->

Answer 69

Anonymous its the first.

(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

Answer 70

Anyone for Problem 5 a)?

Answer 71

Dear friends, I tried this:
5c)(1/sqrt(k),1/sqrt(k))
but doesnt work...
Anyone for 5c and 5a???

Answer 72

uppercase K

Answer 73

5) a)?

Answer 74

Thank you!!

Answer 75

ANyone for 5)c)?

Answer 76

Meant 5) a)?
Thank

Answer 77

5 a)?!

Answer 78

Gyus, 5 is easier than you think!
It takes exactly m steps from starting superposition (ay = 1/sqrt(K), ax = 1/sqrt(k)) to get to the solution (ay = 1, ax = 0), then it takes exactly m+1 steps to get back to initial state, but after each cycle the sign changes!
So it looks like this:
0: ay = 1/sqrt(K), ax = 1/sqrt(k)
m: ay = 1, ax = 0
2m+1: ay = -1/sqrt(K), ax = -1/sqrt(k)
3m+1: ay = -1, ax = 0
4m+2: ay = 1/sqrt(K), ax = 1/sqrt(k)
5m+2: ay = 1, ax = 0
6m+3: ay = -1/sqrt(K), ax = -1/sqrt(k)
........

Answer 79

guess the right answer for 5a
and don't forget to write CAPITAL K

Answer 80

It is this, thanks Andy!
5)a)
ay = -1/sqrt(K)
ax = -1/sqrt(K)

Answer 81

thanks Andy,Flu and everyone worked for this subject!

Answer 82

thank you!!!

Answer 83

thx all!!

Answer 84

Well done Guys... Cheers 2 all...

Answer 85

Correct answers to 5A,5B,5C:

5A:
ay=-1/sqrt(K)
ax=-1/sqrt(K)

5B:
ay=-1
ax=0

5C:
ay=1/sqrt(K)
ax=1/sqrt(K)

Answer 86

hey guys,i want the answer of Problem 2

Answer 87

Eisteinos

2) answer is e

------*------
|
--Z---0------

Answer 88

Suppose we have a quantum circuit that takes the input |0> and outputs |+>, and also takes the input |1> and outputs −|−>. If we input √2*i/√3|+> + 1√3|−>, what does the circuit output?
In the form a│0> + b│1>

Answer 89

what is the answer for q1?

Answer 90

1a. Z
1b. X
1c. X
1d. D=Z
1d. E=Z
1e. Z
1f. G=X
1f. G'=X

Answer 91

Please Q 7 and 8!

Answer 92

8a- pi/3,o

Answer 93

8D please!

Answer 94

Anonymous, did you get all solutions for Q 7 and 8 except 8(D).
If yes then please share......

Answer 95

@abc did you got question 8D?

Answer 96

thnx, all of you for your help

Answer 97

Someone made ​​the second, please?

Answer 98

Esta resposta para 2 esta correta? Como posso inserir esta questão?

------*------
|
--Z---0------

Answer 99

The answer to 2 is correct? How do I insert this?

------*------
|
--Z---0------