Quantum Physics
posted by ss01 on .
We will carry out some steps of the quantum factoring algorithm for N=91
(a) What is the period k of the periodic superposition set up by the quantum factoring algorithm if it chooses x=8? I.e. what is the period of f(j)=8j(mod91)
(b) Using your answer to (a), find a nontrivial square root of 1(mod91). Write your answer as an integer between 0 and 91.
(c) Then, the algorithm proceeds by computing gcd(91,y) for some y. What is y? If there are multiple correct answers, provide any one of them.

anyone,
please answer this. 
yes please..
and can anyone help with problem 1e? 
In q7 and 8 does all parts need to be correct... can anyone help please

for 1eI had XZ was wrong..sorry!

does anyone know how to do q7? it seems this question requires all entries to be correct... have only 1 option left now?

I have all the solutions for ans. 5

5A) 1/sqrt(K), 1/sqrt(K)
5B) 1, 0
5C) 1/sqrt(K), 1/sqrt(K)
Make sure you enter the K as uppercase.
Please post whatever answers you get as well. 
2) pi

Any one problem 6???

@Naren thanks for Q5.
Q6(a) is 4.. 
hi again... managed to do b... 64...
can anyone help with C... my last option left... plus q7 & q8 please 
1(e)answer :z

Q6c and Q7, Q8 please!

help!!!

3b and c please!
3d is no 
3b  C
3c  0 
thx :)

6c can be calculated by answer to 6b  1

6(b):64
6(c):63 
Q1, 7 and 8 please!

1(a):z
1(b):x
1(c):x
1(d):z
1(e):z
1(f):x,x 
Thanks a lot

Q4 please!

4(a):1 4(b):4
4(c):last option 
Q:7&Q:8
PLEASE 
Q 7 and 8 Please!!!

Q7 and 8 please

8a) theta=pi/3 and phi=0. this is 100% right. thanks guys

what about others, please help rs and others..........

In this problem, assume that we are working in units such that ℏ=1. When providing Bloch sphere coordinates, please ensure that 0≤θ≤π and 0≤ϕ<2π whenever applicable.
8(a) Consider a spin qubit in the state 3√20⟩+121⟩. What are the coordinates (θ,ϕ) of this qubit on the Bloch sphere?
θ=?
ϕ=?
8b. We turn on a magnetic field in the z^ direction at time t=0. This corresponds to the Hamiltonian H=BZ, where B is a constant and Z is the usual phase flip. What is the state of our qubit at time t as a function of B and t?
8c. What are the coordinates (θ,ϕ) of this qubit on the Bloch sphere at time t, as a function of B and t?
8d. Now suppose we start with the qubit ψ(0)⟩=0⟩ and we apply a magnetic field in the x^ direction at t=0. This corresponds to the Hamiltonian H=BX, where B is a constant and X is the usual bit flip gate. What are the coordinates (θ,ϕ) of this qubit on the Bloch sphere at time t, as a function of B and t?
θ=?
ϕ=? 
Please 7 and 8b 8c 8d

anyone can help me with Q7 and 8b,c, d please.
Thanks for all who are trying to help each other. 
hi, what about 1)d) it has two parts? which one is z and what is the other one?

@rs both parts are Z...
Do you have answer for q7 or 8 
Thanks, I am working on 7 and 8. I have already posted for 8a). will post the rest as soon as it is done

on 7B answer is 2, ?, 8
Please i have the last submission to find out what insyead of "?". hope i helped you guys 
Thanks rs!

@needhelp ... how did you get 2, and 8

@Need help its 2, 9/2, 8.
do you have answer for q7a and C please..thanks 
anyone help please with q7(a)

8D please!

7A: momentum values 2, 3, 4
State for 3 is (0 1 0)
Got the seemingly correct values for 2
(1 0 i) and for 4 (1 0 i), but the system does not accept them. 
7A: momentum values 2, 3, 4
State for 3 is (0 1 0)
Got the seemingly correct states for 2
(1 0 i) and for 4 (1 0 i), but the system does not accept them. 
Great to see constructive collaboration here. I have got the same questions as most of you here. I do not have 3A; 7A, 7C; 8B, 8C, 8D. Can anyone help me out with these?
I see that someone is working on 7A. I will keep trying as well! 
3A  (1)^x  middle answer
8C  pi/3; 2*B*t 
Thanks for your help guys... what is q8(b. struggling with 7 as well.

Oy,
8b:
e^(i*B*t)*sqrt(3)/2
e^(i*B*t)/2
Do not tell anyone that I betrayed to you this great secret. 
Can anyone help with 8d and q7

thank you for the secret friend!!!

7B, again: 2, 9/2, 8.
7A: momentum values 2, 3, 4
State for 3 is (0 1 0)
Don't have anything more. 
Anyone done 8D? Please, I'm stuck there.

7c, t=pi > 1,0,0

Thanks "a" ..can you help with Q7(a)

7C complete answer:
0 0 1
1 0 0  pi rotation around x axis,
z> > z> 
Thanks...anon @4:37pm, can you also help with Q7A.. thanks

anyone can help me with Q7a and 8(d)..thanks...

I AM 90% SURE THAT THE ANSWER TO 8D IS:
theta: 2*B*t
phi: pi/2
CAN ANYONE PLEASE VERIFY THIS? I HAVE ONLY 1 CHECK LEFT... 
Yeah, that's not it the theta: 2*B*t
phi: pi/2 is wrong. 
It could be because 2*B*t is not in the [0,2*pi] range. Maybe it's theta = (2*pi  2*B*t) instead.
Anyone willing to try that? 
any other guesses?

I tried 2*pi2*B*t and pi/2, it was wrong :(

7A:
1/sqrt(2) , 0, i/sqrt(2)
2

0 , 1, 0
3

1/sqrt(2), 0, i/sqrt(2)
4

Any update on 8D? 
for 8d i tried with this but got wrong
theta=2*B*t phi= 0 or pi/2 or pi 
8(d)
I tried everything... I can't get any value for 0<=thetha<=pi 
8(d)
the answer should be
thetha = 2*B*t
phi = pi/2
I tried thetha = 2*B*t, 2*pi2*B*t (although this mayn't be always between 0 and pi), 2*B*t (i know its wrong but i tried) 
The 8D answer with theta=2*B*t and phi=pi/2 is wrong!

Yes, I was trying to say it should be the answer, but the question required 0<=thetha<=pi...
this doesn't satisfy for any answers mentioned above by me...
I feel this question is tricky... can someone find the answer? 
Shouldn't the answer for 8d be as theta = pi/2 and phi = 2*B*t ? Can you confirm?

@Anonymous. That is not the answer; That is the same one I had got when I last tried. I've got 2 attempts left now.

can anyone help with Q8d please!

theta = pi (8D)

theta = pi
phi = pi/2
is wrong Jerome! How did you get theta = pi ??
I'm down to one try, so wait for a genuine answer for 8D 
Jerome, did you check the answer ?
theta=pi
What is phi then? 2*B*t ? 
Im certain theta is correct, unsure about phi. it is not B*t or 2bt. try 2bt

hi guys... have you got answer to Q8d.. I have now one option left only

Jerome, did you try ? Both answers must be correct in order to get check marks on both.

i tried theta = pi and phi = b*t
and theta = pi and phi = 2*b*t
try it with 2bt 
I tried:
theta = pi phi=B*t
theta = pi phi=2*B*t
Both incorrect. Down to the last answer. 
Could someone try:
theta = pi phi= 2*B*t ? 
8d) Theta = pi
Phi = 2*b*t 
Jerome, is this correct? I think somebody tried it and reported as incorrect, though the answer makes sense: X converts 0> to 1>, and 2*B*t is phi for 1>

Did anyone try
8d) Theta = pi
Phi = 2*b*t ? 
wrong

jerome wrong any one got answer for 8D

theta = pi
phi = 0 
For 8d.
Theta = pi
Phi = 0 
I just tried what Jason said and it worked :)

I have helped with answers ...can someone help me with 8d...down with one option...I think both parts need to be correct...

pi and 0 work for 8d :)

I tried theta=pi and phi=0 its wrong!!!

Also tried theta = pi and phi = 0, its wrong. Anyone have any idea for 8d?

I tried what Jason said is working long time ago. It is wrong. I don't understand how it could have worked for you. Are you trolling ?

He must be trolling, lets ignore what jason and jerome say!

BTW, both parts must be correct.

What is the answer?! Someone make another question

please help with q8d please

can anyone help me with 8d please!!

It seems like some people experimenting at the expense of the others. A guy does not want to waste his trial, so he posts an answer he thinks is right as the correct answer. Then, he gets the feedback from those who unwittingly used their trials. Very mean, I say.

I think the answer is:
theta: pi/2
phi: (2*pi)(b*t) 
Only got 1 try, someone try that ^^^

Jason and Jerome please help with q8d!

@ Anonymous at 4:24pm
Your thinking is great. But did you try it? If not, why do you want the others to try? Could you give some background for your thinking ? 
Jason and Jerome are as much in darkness as we are.

theta=pi/2 and phi=2pibt is wrong!!

to those who have answers to Q8D..please can you help with explanation or your thoughts..thanks

8D Answer
theta = pi
phi = (2*pi)(B*t)
You are all welcome :) 
for pi and 0: X maps 0 to 1, meaning theta = pi. If it is at 0 or 1, then phi is irrelevant, could be 0, because 0 and 1 are along Z axis.

No the answer is:
theta = pi/2
phi = 0 
Anon @ 5:13  is it another bogus answer ?
What is the rationale for your answer ? 
@ Anonymous at 5:17:
BS, tried that long ago, wrong! 
As another anomyous @5:16 said, the answer would be theta = pi and phi= 0

help withq8d please someone out there!!!

The answer for 8d is theta = pi and phi = 0!!!!!

pi/2 and 0 are wrong!

Sorry about the mistake earlier i copied down the wrong answer, that was an answer from a previous attempt. The correct answer has already been stated on here.

Anonymous @5:44 PI is correct NOT pi/2

pi and 0 is also wrong!!

@Jerome ..what is the correct answer please!

@ Anonymous at 5:49:
He doesn't have a clue! He is just trolling... 
@Jerome..can you atleast explain how to get the answer please

Using the formula for the state psi under the act of a hamiltonian, you deduce that theta = pi and psi = bt. However the bt needs to be changed. think about why it needs to be changed

so is theta=pi and phi=2pibt

anyone please help with 8d..

theta=pi and phi=2pibt ISN'T WORKING

@Jerome thanks for the explanation so is the answer pi and 2pibt.. thank you

thanks @anon @7.09pm it also doesn't work for theta=pi and phi=2pi2bt...
HELP!! 
please help with 8d.

Jerome give the answer please!!

Just wondering has anyone got the answer for q8d..or am I the only one banging my head!!

help with q8D

did anyone already check for phi=2bt or 2b

Ive checked and its neither phi=2bt nor 2b

The answer is in both cases 0 and 0

No but almost! Pi and 0 works! Job done!!

pi and 0 does not work. Was tried long time ago. I doubt that 0, 0 would work either.

Jason, please stop trolling with your pi and 0 guano.

anyone plz try with 0 , 0
mine no chance left 
OK, here is the real answer:
theta = e*a*t
phi = s*H*i*t
Job done! 
@jason wonder did you really got the answer????????

Why not try yourself or insted give up? Only 4 points in a final exam of a whole course? Does it matter to get the Certificate? Is it so important? I assume that if you have been cheating the whole course doesn't matter not to get a 100% mark. Any explanation? If somebody gives a good explanation, I promise to give the correct answer.

ya okay maybe some people have cheated all the way through. But there may be a reason or so either to pass atleast the 80% mark because it is high cutoff and we have limited chances. Y people may do this?
To get certificate . Does certificate matters? Yes, may be they will use it in college to show off their intelligence or may be needed for some recommendation or will help them in some way....
There might be reason they can't solve all the problems infact only some get 100% mark and you wonder how much pass the 80% barrier? They might have wasted their 2 months or so........ 
We will get the certificate regardless, very well past that point. And we are not cheaters but rather the hard working overachievers. The problem is the suspense  we where studying so hard, what we have not learned so that nobody could answer that seemingly easy question ? What is the trick with that question ? Will we ever know ?
X turns 0> into 1>. So, theta is pi. Phi is the angle in the complex plane that is orthogonal to Z / Z axis where 0> and 1> lie. So, phi is not relevant. It could be 0, or anything.
Do you disagree with these conclusions ? Do you see any mistake ? 
sometimes you have to do wrong things to get into the right way may be for a cause or so

BTW, it is meaningless to talk about cheating in this course. The problems are not authentic. They are taken from a set of quantum mechanics/ quantum computing problems that is moving from a course to a course. Some "noncheaters" had no quarrel with the course because they have the answers to that set. So, we are just equalizing. Equal opportunities, that is the slogan.

@MMM you are correct!!!!!!!!

I bet "Cheaters" don't have answer to this question...

I think the answer is 0 & 0

They will both be 0 because X is perpendicular to z so the field doesn't effect it. The transformation by X doesn't either because it is still around the z axis.

@jerome are you sure?

Yes

0, 0 is wrong.

Jerome you are wrong, I think you also don't know the right answer. Stop fooling us around

Jerome is fooling you.
The actual answer is pi, pi. Jerome, go home. 
pi,pi seems to be wrong too.

Yup pi, pi is also wrong

Oh come on! pi pi/2. It is so eeeeeeeasy to see :D

Nope, pi and pi/2 also wrong.

Ha ha! Correct answers told as wrong. The good answer was already given and now the question: who is fooling who?

plz tell what is the answer?

The answer, my friend, is blowin' in the wind, the answer is blowin' in the wind.

there are various answers and only 2 chances, what you think will be the probability?

0<p<1, or something like that

P>0

come on please please please please please please please please please tell the right answer

Since phi is irrelevant, you must leave it blank. The grader is intelligent and will mark it as correct.

what nonsense is this? if you so intelligent then what is theta?
i have never seen not putting any value and getting it correct 
Well try it and you'll see.

so atleast tell what is theta?
if you r right i would worship you 
OK, theta is f*a*r*t/2

This is a special optimization question.
You just leave both fields blank and click submit. The grader will figure the correct answer for you for theta, and will simply mark the blank phi as correct.
try this! 
what are f, a & r ???

You know where it is coming from, don't you?

nope!!!!!!

It is coming from the expression:
(1/(sqrt(s*H*i*t)))*(p*o*o*p) 
you are wrong this too :D :D :D

@Jerome stop being mean, if you can help that's fine if cant please don't confuse us!!!

theta = pi
phi = 0
got 100% 
pi and 0 is wrong...I wonder how you got 100%

it is not wrong, do not fool around

0 and 0 is wrong!

I checked 0 and 0 its marked wrong!

final ans of 8D plzzzzzz!!!

@quantum...I have checked and it was wrong...you stop fooling around mate!!!thanks for yr sharing anyway!

You're welcome. Does this mistake made you don't pass the course? Come on... is this the case?

@ quantum.. actually I passed ... even if you have 100% does it matter! have some self pride mate!!

so you admit you put the wrong answer on purpose!!! don't you feel guilty quantum?

don't feed the troll ;)
back to topic.. I have two submissions left but no more ideas.. 
I am working on an idea! Anonymous @4:33pm wait for me to answer again and then try it for me please :)

what's your idea @4H34 ?

Have to change 0> to +> and > equivalents and then simplify from there on in.

@quantum I am pretty sure you must have used one of our answers at some point in this course!! you know what... no one is perfect!!

annon 234, why that ?

@4:46 because the X bit flip acts in those eigenvectors.
I think the answer is as follows:
theta = pi/2
phi = B*t
Now i am not 100% on phi; it may be 2*B*t 
Phi would be 2*B*t!!!

Wrong !!!! I try it!!

@234, havn't tried, only one try left
physics.stackexchange dot com/questions/81101/quantumcomputationhamiltonian
maybe some hints 
@ anyone... can anyone help with Q8D please!!!

pi/2 B*t and 2*B*t were tried long time ago, and failed. Anons 234 idea about conversion of 0> to +> / > basis was tried and yielded e^(2Bt)1>. The grader does not take that answer equivalent to pi 2Bt.

@ Jerome...why are you even here when you have the right answer!!1

@ Jason and Jerome does not have any clue...check new postings guys and don't be fooled!

it seems now that the answer is very simple... but what is it?

help please!!!

I am incredibly close to the answer. Will update soon!

pi/2 and Bt is wrong guys...just tried... down to my last option... help!

guys don't trust anyone..trust yr instincts...

The answer is theta = 0 and phi = 2pi !!!!
I can leave this question forever now!! 
No....
theta=0 and phi=2*pi 
Theta=2BT
Phi=pi/2 
Thankyou Anonymous @ 7:08 it worked

theta = 2bt
phi = 0 
which one is right: theta: 2Bt and pi/2 or 2bt and 0!!! please help!!

Theta=2BT
Phi=pi/2 
It works at last!

really it doesn't and did not work for me mate!!

No, it is not. All troll answers are wrong.

seriously anyone has any help to offer for q8d

help please with q8d

what works guys for 8d

For $1000, I'd give you the right answer.

done
but if it is wrong we would take 1000 from you 
So, are you ready to transfer them to my personal account at the Bank of Nigeria ?

Time is running out for you. No money means no score sufficient to get the certificate.

The answer is theta = 0 and phi = 2Bt
everyone stop being mean to the people who want the answer 
yup,
1000 USD done...... 
Theta is not 0 and phi is not 2*B*t

jason what happened?

seems you are doing a nice deal out here

Confirmation of theta = BT and Phi = pi/2

you asking us to confirm or telling the answer?

sorry theta = 0, not bt!
theta = 0 and hphi = pi/2 
are you sure???
i will give jason's deal to you 
I am fairly certain that it is

can you explain how you got theta = 0 and phi = pi/2

did you try it?

So, phi = pi/2 always, independent on time and the strength of the field ?
Hmm, maybe, maybe, because the perpendicular field separates the spin upwards/downwards. Though, I'm still skeptical. 
Citation:
"How can we create a beam of qubits in the state ψ=0? Pass a beam of spin1/2
particles with randomly oriented spins through a SternGerlach apparatus oriented along the z axis. Block the downwardpointing beam, leaving the other beam of 0 qubits. 
you try it and send us a screenshot of you r page

Guys, simply read this, pages 2 & 3
inst.eecs.berkeley.edu/~cs191/fa07/lectures/lecture15_fa07.pdf 
theta = 0 and phi = pi/2 is wrong
theta = pi/2 and phi = 0 is also wrong 
So, here is my rationale. +> and > are two eigenvectors of X, with eigenvalues of 1 and 1. So, putting them through X makes them unchanged, bar reversal of >.
0> = 1/sqrt(2)+> + 1/sqrt(2)>
That is the output:
X*0> = 1> = 1/sqrt(2)+>  1/sqrt(2)>;
Canonical form:
Psi(x) = cos (theta/2)+ e^(i*phi)sin (theta/2):
Now, how do we get from canonical form to 1> output ?
Easy:
theta/2 = pi/4, thus theta = pi/2;
e^(i*phi)= 1 iff phi = 0
So the answer should be theta = pi/2
and phi = 0.
Generally speaking, the magnetic field performs Bloch sphere rotation/transposition along the axis it is applied. Therefore, application along x axis transforms 0> into 1>
I don't remember if theta = pi/2
and phi = 0 was ever tried. But, rationally, that should be the answer.
P.S. I cannot try, down to the last submission. 
I see above that theta = pi/2
and phi = 0 were tried. Maybe, the grader is not programmed it correctly ? Did anyone really got his 8D answer accepted? 
if you so sure please risk yourself and tell us

Theta = 2bt
Phi = 3pi/2 
Found the same result but don't beleive in it. ;)

@chop are you sure???

I am quite sure

No. I'm not sure. Really. Looks like it's wrong result.

Fake @chop detected!

what is going on????

theta = 2bt
phi = pi 
what is phi tell properly

phi is pi!!

did you got it correct?

Yes i did.

could you send a screenshot of it????

Cant send screen shots on here :/

take a screen shot and upload on bit and paste the link without http if u r confident about your answer and your intelligence

i.minus dot com/i5xld8aUe08Ee.png
I CAN 
ya good screenshot is of wrong answer

anyone confident of answer please send screenshot

Theta = 2bt
Phi = pi/2
imgur dot com/eMrPvfk 
Jerome  fake! fake screen shot!

@jerome yup its fake

YEap, if you see this thread, you will see that they ask if someone has mixed phi with theta as he give an answer Theta=pi/2 a Phi=2bt. It's only matter of reading you cheaters.
physics.stackexchange dot com/questions/81101/quantumcomputationhamiltonian 
@anonymous please try and send a fake screenshot

Theta=pi
Phi=s*o*B 
@jerome
i.minus dot com/i9iaY1QUADBkO.png
You managed to fail your fake... congrats. 
please help if you can with Q8d

Hi guys... I just tried:8d
theta: 2*B*t and phi: 3*pi/2 and it worked!!
thanks to those who helped... 
@Pari. Another fake ? Where did you get theta: 2*B*t and phi: 3*pi/2 from ?

Yah fake too :(

No, it is real, though I forgot the sign:
theta: 2*B*t and phi: 3*pi/2 
fake

there is no point in faking when you can always create a new account and get an extra 3 tries(I hope you guys aren't trying these on your actual accounts).
Why not just stick to solving the problem.
I'm trying to do it like I did 8C but it's just not working.
The rotation about the xaxis changes phi to theta and theta to phi. 
My answer is correct...take it or leave it ...Q8d theta:2*B*t and phi 3pi/2..stop using my name...and confusing others...guys trust me... and stop This is my last post... anyone using pari is fake!!!I am going to report those using my name!!

thx Pari :)

Fake again...

fake Pari, or whatever it was, fake answer.
If you claim you are real, explain how you arrived to that answer, pleeeeazeee 
I tried theta = 2*B*t and phi = (3*pi)/2
AND IT WORKED
THANKYOU SO MUCH Pari!! 
@ss01, is theta= 2*B*t or 2*B*t?

Its 2*B*t

YES IT WORKED, THANK YOU GUYS
THETA= 2Bt and phi= 3pi/2 
Thanks Pari and SS01...we all know who the fake Pari may be?
so its... theta= 2*B*t and phi= (3*pi)/2
Thanks to also everyone who helped through out!! 
i dunno how m stuck with 8c. :(
i need help. 
You find the all answers somewhere in this thread:
8:
A: pi/3 0
B: e^(i*B*t)*sqrt(3)/2 e^(i*B*t)/2
C: pi/3 2*B*t
D: 2*B*t 3*pi/2
Thanks a lot for 8D 
Thanks Pari. Still, I wonder, how did you arrive to the correct answer? Seems that I missed something in the course.

"why are you faking ..."
I'm a TA. It was entertaining to observe you cheating. So, we have decided that those of you who entered the answer within 6 hours after publication on this site will not get a completion certificate, because of academic dishonesty. The dishonesty policies and the conditions for the course completion were published rather prominently on the course site. Cheating in EDX courses will not be tolerated. 
How will you find out who gave the answers by cheating and who actually solved each answer.
Some of us actually solve the answers out on paper and plug it in to the autograder.
A few of us haven't even shared a single answer and were just here for entertainment purposes.
I think you are either trolling or bluffing. Have fun with the complaints from testtakers who entered answers correctly. 
We have your IP addresses. If you visited the page for "entertainment purposes", and then entered the answer stated on this page within 6 hours after your visit, you clearly cheated. Since the "entertainment purposes" of your visit included the forbidden collaboration on the problems, you violated the terms of the course, and you will receive no certificate.

thnx frnds, all of you :)

Your method of using IP addresses would never work. Most people here have static Ip's ,unless you have a corporate connection; everyone's IP address would change the moment they restart their modem.
Obtaining the Ip address for each comment here is clearly impossible.This site does not store ip addresses and attempting to find it is illegal.
I would have to report you for performing illegal acts on behalf of EdX.
Performing such a task of filtering IP addresses on Edx's server would require admin access to the server files which you don't have and can not get because you would have to be the admin.
I don't care about the certificate; just wanted to learn. If you are confident you can find IP addresses for every person then do so and tell me which country I live in. troll. 
You don't understand how it works. You will be up to an unpleasant surprise, cheater. If you took the course just to learn, you had no business to cheat. As Mark said, the final exam was very straightforward if you studied in earnest. Since you had to resort to cheating, it appears that you have not learned much.

I discourage cheating, I haven't even given the final exam, I'm just auditing the course. (just to learn)
I'm just proving that your ideology and methodology is wrong. You could really mess up the EdX system if you decided to do it your way.
You can't see what visitors were seeing before visiting your website(EdX). Having the Ip address would mean nothing since you can't view visitors' browsing history.
I still challenge you to find what country I'm in (I don't care if you get all my ISP's details; I'll just restart my modem). If you can do it, then I'll back out and let you proceed while I make better use of my time. 
You let me to proceed ? That is most interesting. Thank you, sahib, I'm patiently waiting for your kind permission.

Love the sarcastic comment. This is the best entertainment I've received throughout the entire duration of this publication.