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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>.
: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)
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