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# Tag Info

## Hot answers tagged deutsch-jozsa-algorithm

4

If you look at the formula you want to prove term-by-term, you'll notice that the sum and the $(-1)^f(x)$ part is the same in both formulas; you just need to show that $$H^{\otimes n} |x\rangle = \frac{1}{\sqrt{2^n}} \left( \sum_{y=0}^{2^n-1} (-1)^{x \cdot y} |y\rangle \right )$$ You can either show this strictly by induction (similar to this question but ...

1

I don't think you can factor. There are many cases where the result of the gate will produce a combined CNOT output (4 numbers) that cannot be factored. I think most states between two qbits is entangled from our perspective. So you cannot trust in factoring. Instead, the four products remain in memory while your program writes the data to both of your ...

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