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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 <math>\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 </math>. [[Cnotgate]]
: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,
::<math>=\frac{1}{2^n} ((-1)^{f(0)}+(-1)^{f(0)}+...+(-1)^{f(0)}) ((-1)^{(0\cdot y_0)\oplus(0\cdot y_1)\oplus...\oplus(0\cdot y_{n-1})} |y\rang)=</math>
:is only the first term of one of the summations, it does not equal the previous equation. Tell me what you think. [[User:Skippydo|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 <math>|x\rang</math>?
::That is |y>? It is very confused, muddy.
::: |x> disappears after the the (second) haddamard transform is applied but a phase involving x remains. Perhaps we should be clearified what the haddamard transform does to an arbitrary qubit? [[User:Skippydo|Skippydo]] 12:17, 9 July 2007 (UTC)
::Another note, what I had added is remarkably similar to what is located in one of the references.
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That is your formula: <math>|\frac{1}{2^n}\sum_{x=0}^{2^n-1}(-1)^{f(x)}|^2</math>. And how you can explain for example this <math>|\frac{1}{2^2}\sum_{x=0}^{2^2-1}(-1)^{f(x)}|^2=|\frac{1}{2^2}((-1)^{f(00)}+(-1)^{f(01)}+(-1)^{f(10)}+(-1)^{f(11)}|^2</math>?
Ah, but you say, x is 0, but 0's can't be so much ( <math>2^n</math> ). [[Cnotgate]]
:Again, there is no need to continue to insult me. It will not win you anyone's agreement.
:I do not understand what you mean by concrete. But as for your direct question. f is a boolean function it outputs 1 or 0 so (-1)^f(x) = -1 when f(x) is 1 and (-1)^f(x)=1 when f(x) is 0. If you're prone to the notation of the basis states |1>, |-1> then the coeficent (-1)^f(x) could be replaced with f(x).
:On to your next question. x is not 0. We are measuring 0. Before the haddamard there are 2^n |x>'s. After the haddamard each |x> is repalced with 2^n |y>'s. The summation is rearranged so that equal |y>'s appear together, ie 2^n |y>'s each with it's own summation as a coefficent. To determine the probability of measuring all qubits zero we plug in y=0 to the |y>=|0> coefficent and calculate the square magnitude. If this argument convinces you please suggest what parts of it we should add to the article. [[User:Skippydo|Skippydo]] 12:17, 9 July 2007 (UTC)
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