Talk:Deutsch–Jozsa algorithm
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I found a small mistake in the description of the algorithm, and I corrected it. To me, the wording of the article is still a bit sloppy; but I'm not going to try to fix it at the moment. At least now the algorithm works. :) Karadoc** 05:27, 3 July 2006 (UTC)
what we measuring?
As wrot there "Definition: A quantum algorithm to determine whether a function is constant or balanced, that is, returns 1 for half the ___domain and 0 for the other half. For a function taking n input qubits, first, do Hadamards on n 0's, forming all possible inputs, and a single 1, which will be the answer qubit. Next, run the function once; this exclusive or's the result with the answer qubit. Finally, do Hadamards on the n inputs again, and measure the answer qubit. If it is 0, the function is constant, otherwise the function is balanced. - http://www.nist.gov/dads/HTML/deutschJozsaAlgo.html" - need measure only one the answer qubit, but how wrote at wikipedia "The algorithm is as follows. First, do a Hadamard transform on a quantum register of n 0s, forming all possible inputs, and a single 1, which will be the answer qubit. Next, run the function once. This is done by using the n input qubits as input of the function of a Function-Controlled NOT gate that works on the answer qubit. Finally, do Hadamards on the n inputs again, and measure them". So what we have measure answer qubit or over all except answer qubit. And if we must measure answer qubit, then why we must do hdamard transform (after then qubits pass through CNOT gate) on qubits that not be measured? CNotGate
- By "answer qubit" they are refering to the target of f. That is |x>|y> becomes |x>|f(x)+y> and the "|y>" qubit is the "answer" (of f). The behaviour of f should have been explicitly stated as it has in my recent change.Skippydo 16:28, 27 June 2007 (UTC)
If i correct understand, on output measuring is the answer qubit (it is just mathematical model that measuring must be all first (top) qubits)? Or maybe becouse all firsts qubits is entangled, they gives answer if we measure one of those?..
Deutsch algorithm with n qubits remake
I don't understand why so stupid and dificul is calculation of such simple thing...
Supose we have n qubits on input, for example |0>|1>|0>|1>...|n>=|0,1,0,1,...,n>. Let's mark all qubits like this: . Then summ last qubits with all firsts qubits except last by MOD 2: If all firsts qubits is zeros, then function is constant, if all first qubits doesn't zeros, then function is balanced. Example. We have on input |0>|1>|0>|0>|1>|1>=|0,1,0,0,1,1>. As we see, last qubit is 1. Then plus last qubit by Mod 2: |0+1,1+1,0+1,0+1,1+1,1>=|1,0,1,1,1,1>, all firsts qubits is not zeros, so function is balanced. Another example. We has on input |1,1,1,1,1,1>. Last qubit is 1. So by Mod 2 plus last qubit to all firsts: |1+1,1+1,1+1,1+1,1+1,1>=|0,0,0,0,0,1>. As we can see all firsts qubits is zeros, so function is constant. One more example. We has on input |0,0,0,0,0,0,0,0>. Sum up firsts all qubits with last one by Mod 2. |0+0,0+0,0+0,0+0,0+0,0+0,0+0,0>=|0,0,0,0,0,0,0,0>. As we can see all firsts qubits is zeros, so function is constant.
sharlatans
I think, that people who gave formula how calculate Deutsch-Jzosa algorithm, himselfs don't understood what they do. Becouse imposible to calculate something concrete with this selfinvent trough dreams formula .
![{\displaystyle {\frac {1}{2^{n}}}\sum _{x=0}^{2^{n}-1}(-1)^{f(x)}\sum _{y=0}^{2^{n}-1}(-1)^{x\cdot y}|y\rangle =\sum _{y=0}^{2^{n}-1}[\sum _{x=0}^{2^{n}-1}(-1)^{f(x)}(-1)^{x\cdot y}]|y\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/20e9b5b91d6fe30d5a6e51db8bd36b583a39d8df)
- I was the one who rewrote this section. I do not understand your concern. I have calculated something concreate as a demonstrate in line after that which you have stated. Perhaps I do not understand your question. I disagree with your recent additions, they do not add clearity and may be interpreted incorrectly. For instance,
- is only the first term of one of the summations, it does not equal the previous equation. Tell me what you think. Skippydo 07:05, 9 July 2007 (UTC)
- I tray to understand that is exactly is |x>, and i know that this equation is incorect in contrast with first equations, but i try to calculate somthing concrete, but don't successfully. So maybe you can give answer, what exactly is ? -cnotgate
- That is |y>? It is very confused, muddy.-cnotgate
- Another note, what I had added is remarkably similar to what is located in one of the references.
- http://beige.ovpit.indiana.edu/M743-talk-2/node13.html
- I also now realise that the header of this post is an insult directed at me. This is not something I appreciate. Skippydo 07:15, 9 July 2007 (UTC)
If your equation is realy good, then you proof it, and calculate something concrete. Becouse it seems like sharaltanism. For example I with my Formula can easy calculate somthing concrete. For example: on input is . On output will be . Becouse all qubits except last not zeros, function is balanced. Or if your more like this answer: .
It insult not for you one, but for all sharlatans, who gives some formulas, but can't show, how calculate somthing concrete.
Maybe sombody can explain,what exactly means say or say ?
That is your formula: : . And how you can explain for example this ?
