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:
Stucklike you :(, anyone please
I have done 6A:
(1/2 5/2)
(5/2 1/2)
6C and 6D: 0
others pleasee
6B is
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))
6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->
Problem 5 plz
Thank you guys!!
4b= 4
someone for the first??plzz
Any kind of help is well received ;)
3C=0
1st is part c):
-z-h-
-x-h-
3B is Z
I meant answer to question 1 is part c.
-z-h-
-x-h-
problem 5,7 please.
4A please
problem 4 a):
the eigenvalues are 1 and -1
then the lowest eingevalue (ground energy) is : -1
4A is -1
3D yes or no ?
3d no
Problem 7a
theta = pi/4
phi = 5*pi/2
Problem 7b
1/2+i/2
1/sqrt(2)
wrong
7A and 7B wrong
SOMEBODY KNOW ANSWER FOR: 3A? 5? 7? 2? 4C?
5 and 7 please
hikikomori, do you have answer for the 3A? 5? 7? 2? 4c?
hikikomori,Sorry, do you have answer for the 3A? 2? 4c?
7A) pi/2 , 5*pi/4
7B) 0.707 , -0.5-(0.5*i)
4C (001, 010, 100, 111)
anyone question 5?
3A: C option
Thanks all guys.
We now just need Q5's answers.
We need Q.2 also
for 2, last option is correct
Thanks all guys
well-done
Problem 5 please?
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
Anyone??? Pls Q5... guys do soething pls
PLease Problem 5!?
5A: (2/K - 1), (2/K)
5B: (1- 2/K), (-2/K)
5C: (2/K - 1), (2/K)
EdX Winner its wrong answer
5A: (2/K), (1 - 2/K)
5B: (-2/K), (2/K - 1)
5C: (2/K - 1), (-2/K)
Wrong? It worked for me!
qwerty urs answer is also showing wrong dude
All wrong?
YES ALL WRONG
the above answers to 5th question is of assignment 6 problem 5 don't get misguided
Anon.. What is the correct answer to the 5? please help us
WORKING ON IT AR . DISTRACTED BY SOME PERSONAL PROBLEMS IN LIFE ... NOT ABLE TO CONCENTRATE
I think it should look like
(k-2)^x + ...
-------------
k^x
where x = m+1
but i have failed for find oput more :-(
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.
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
problem 3b,4c and 5
well guys, i need few drinks (i'm doing my best MF) and we will be all right..cheer MF
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.
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
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.
@xaad
3B: Z (3rd option)
4C: Last option
5B: ay = -1; ax = 0
5A, 5C: ---Not yet solved correctly---
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.
Ok, 5C is 1/sqrt(K) for both of them. I cannot figure out what is 5A... :(
5A is not -1/sqrt(N) or any combinations with minus sign.
@J
I input 1/sqrt(K) as 5C's answer, but they are wrong.
@J
No they're right, I misspell the answer
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???
wow! Strange behaviour @cheers xD Nevertheless good news for 5C :D Now, only 5A remains behind the Fortress of Solitude...
@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
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...
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)|->
Anonymous its the first.
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))
Anyone for Problem 5 a)?
Dear friends, I tried this:
5c)(1/sqrt(k),1/sqrt(k))
but doesnt work...
Anyone for 5c and 5a???
uppercase K
5) a)?
Thank you!!
ANyone for 5)c)?
Meant 5) a)?
Thank
5 a)?!
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)
........
guess the right answer for 5a
and don't forget to write CAPITAL K
It is this, thanks Andy!
5)a)
ay = -1/sqrt(K)
ax = -1/sqrt(K)
thanks Andy,Flu and everyone worked for this subject!
thank you!!!
thx all!!
Well done Guys... Cheers 2 all...
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)
hey guys,i want the answer of Problem 2
Eisteinos
2) answer is e
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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>
what is the answer for q1?
1a. Z
1b. X
1c. X
1d. D=Z
1d. E=Z
1e. Z
1f. G=X
1f. G'=X
Please Q 7 and 8!
8a- pi/3,o
8D please!
Anonymous, did you get all solutions for Q 7 and 8 except 8(D).
If yes then please share......
@abc did you got question 8D?
thnx, all of you for your help
Someone made the second, please?
Esta resposta para 2 esta correta? Como posso inserir esta questão?
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The answer to 2 is correct? How do I insert this?
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