Asked by qwerty

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:
Answered by anonymus
Stucklike you :(, anyone please
Answered by JLG
I have done 6A:
(1/2 5/2)
(5/2 1/2)

6C and 6D: 0

others pleasee

Answered by anonymus
6B is
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))
Answered by Anonymous
6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->
Answered by Mary
Problem 5 plz
Answered by Final ;)
Thank you guys!!
Answered by JLG
4b= 4

someone for the first??plzz
Answered by Final ;)
Any kind of help is well received ;)
Answered by JLG
3C=0
Answered by anon
1st is part c):

-z-h-
-x-h-
Answered by anon
3B is Z
Answered by anon
I meant answer to question 1 is part c.

-z-h-
-x-h-
Answered by anon
problem 5,7 please.
Answered by Anonymous
4A please
Answered by anion
problem 4 a):

the eigenvalues are 1 and -1

then the lowest eingevalue (ground energy) is : -1
Answered by quanc
4A is -1
Answered by flou
3D yes or no ?
Answered by Anonymous
3d no
Answered by e=mc^2
Problem 7a
theta = pi/4
phi = 5*pi/2
Answered by e=mc^2
Problem 7b
1/2+i/2
1/sqrt(2)
Answered by Anonymous
wrong
Answered by Gyano
7A and 7B wrong
Answered by AR
SOMEBODY KNOW ANSWER FOR: 3A? 5? 7? 2? 4C?
Answered by hikikomori
5 and 7 please
Answered by AR
hikikomori, do you have answer for the 3A? 5? 7? 2? 4c?
Answered by AR
hikikomori,Sorry, do you have answer for the 3A? 2? 4c?
Answered by Megadeth
7A) pi/2 , 5*pi/4
Answered by Megadeth
7B) 0.707 , -0.5-(0.5*i)
Answered by Anonymous
4C (001, 010, 100, 111)
Answered by Anonymous
anyone question 5?
Answered by Stud
3A: C option
Answered by s
Thanks all guys.
We now just need Q5's answers.
Answered by Stud
We need Q.2 also
Answered by Anonymous
for 2, last option is correct
Answered by JLG
Thanks all guys
well-done
Answered by Yuh
Problem 5 please?
Answered by xaad
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
Answered by INDEED Need
Anyone??? Pls Q5... guys do soething pls
Answered by Mah
PLease Problem 5!?
Answered by EdX Winner
5A: (2/K - 1), (2/K)

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

5C: (2/K - 1), (2/K)
Answered by Anonymous
EdX Winner its wrong answer
Answered by qwerty
5A: (2/K), (1 - 2/K)

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

5C: (2/K - 1), (-2/K)
Answered by EdX Winner
Wrong? It worked for me!
Answered by Anonymous
qwerty urs answer is also showing wrong dude
Answered by qwerty
All wrong?
Answered by Anonymous
YES ALL WRONG
Answered by anon
the above answers to 5th question is of assignment 6 problem 5 don't get misguided
Answered by AR
Anon.. What is the correct answer to the 5? please help us
Answered by anon
WORKING ON IT AR . DISTRACTED BY SOME PERSONAL PROBLEMS IN LIFE ... NOT ABLE TO CONCENTRATE
Answered by PhysTech
I think it should look like

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

where x = m+1
but i have failed for find oput more :-(
Answered by Stud
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.
Answered by cheers
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
Answered by xaad
problem 3b,4c and 5
Answered by cheers
well guys, i need few drinks (i'm doing my best MF) and we will be all right..cheer MF
Answered by PhysTech
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.
Answered by j
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
Answered by PhysTech
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.
Answered by Stud
@xaad

3B: Z (3rd option)
4C: Last option
5B: ay = -1; ax = 0
5A, 5C: ---Not yet solved correctly---
Answered by J
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.
Answered by J
Ok, 5C is 1/sqrt(K) for both of them. I cannot figure out what is 5A... :(
Answered by J
5A is not -1/sqrt(N) or any combinations with minus sign.
Answered by s
@J
I input 1/sqrt(K) as 5C's answer, but they are wrong.
Answered by s
@J
No they're right, I misspell the answer
Answered by cheers
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???
Answered by J
wow! Strange behaviour @cheers xD Nevertheless good news for 5C :D Now, only 5A remains behind the Fortress of Solitude...
Answered by cheers
@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
Answered by J
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...
Answered by Anonymous
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)|->
Answered by FLu
Anonymous its the first.

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

Answered by Gyr
Anyone for Problem 5 a)?
Answered by AR
Dear friends, I tried this:
5c)(1/sqrt(k),1/sqrt(k))
but doesnt work...
Anyone for 5c and 5a???
Answered by g
uppercase K
Answered by Ur
5) a)?
Answered by AR
Thank you!!
Answered by Mik
ANyone for 5)c)?
Answered by Mik
Meant 5) a)?
Thank
Answered by Nurt
5 a)?!
Answered by Andy
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)
........
Answered by Andy
guess the right answer for 5a
and don't forget to write CAPITAL K
Answered by FLu
It is this, thanks Andy!
5)a)
ay = -1/sqrt(K)
ax = -1/sqrt(K)
Answered by qwerty
thanks Andy,Flu and everyone worked for this subject!
Answered by AR
thank you!!!
Answered by Gyano
thx all!!
Answered by INDEED Need
Well done Guys... Cheers 2 all...
Answered by Arbiter
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)
Answered by Einsteinos
hey guys,i want the answer of Problem 2
Answered by helpfull
Eisteinos

2) answer is e

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Answered by ajay
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>
Answered by ss01
what is the answer for q1?
Answered by Cyber
1a. Z
1b. X
1c. X
1d. D=Z
1d. E=Z
1e. Z
1f. G=X
1f. G'=X
Answered by Anon
Please Q 7 and 8!
Answered by unknown
8a- pi/3,o
Answered by Anonymous
8D please!
Answered by Abc
Anonymous, did you get all solutions for Q 7 and 8 except 8(D).
If yes then please share......
Answered by abc
@abc did you got question 8D?
Answered by victor
thnx, all of you for your help
Answered by ME
Someone made ​​the second, please?
Answered by Any
Esta resposta para 2 esta correta? Como posso inserir esta questão?

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

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