Posted by qwerty on Tuesday, April 16, 2013 at 12:32am.
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):
zh
xh
3B is Z
I meant answer to question 1 is part c.
zh
xh
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)
5?
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
welldone
5?
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 dont 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
(k2)^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.
kdhfksdfh
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 mp 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 m2, m1, 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_groversquantumsearchalgorithm 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???
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???
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???
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???
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
hey guys,i want the answer of Problem 2
Eisteinos
2) answer is e
*

Z0
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?
Someone made the second, please?
Someone made the second, please?
Esta resposta para 2 esta correta? Como posso inserir esta questão?
*

Z0
The answer to 2 is correct? How do I insert this?
*

Z0
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