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Consider the following implementation of FANOUT using Toffoli gates:

enter image description here

I'm confused about the following statement: "the second bit being the input to the FANOUT and the other two bits standard ancilla states". Does it mean that the second bit is also the input to the other two ancilla states? Why do ancilla states require any inputs, aren't they pre-determined?

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    $\begingroup$ Each post should be focused on one question. Also, since the value of a question to the community depends on other people's ability to find it, please avoid using images for text. $\endgroup$ Aug 24 at 4:43
  • $\begingroup$ @user1271772 just a heads up: editing questions automatically throws them out of the review queue. That means that making trivial edits to a question can effectively be used to single-handedly override other people's close votes. For this reason, please refrain from making edits from the review queue that don't directly address the reason the question is in the queue. I'm all for correcting grammar etc, but this is a case in which it's better not to $\endgroup$
    – glS
    Aug 28 at 9:50
  • $\begingroup$ @glS that's not correct, unless something changed 2 days ago when the review queue system got changed. But my edit was longer ago. $\endgroup$ Aug 28 at 10:31
  • $\begingroup$ @user1271772 what are you saying isn't correct exactly? An edit throws a question out of the review queue. You can also see it directly in the timeline of this post, for example. See also meta.stackexchange.com/q/162034/276202 $\endgroup$
    – glS
    Aug 28 at 10:34
  • $\begingroup$ @glS I'm driving, ping me if I don't reply in 3 days. $\endgroup$ Aug 28 at 10:49
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"Does it mean that the second bit is also the input to the other two ancilla states?"

Nothing is "input to the ancilla states". The second bit in this case, is the input to the FANOUT, whereas the first and third bits are ancillas.

"Why do ancilla states require any inputs, aren't they pre-determined?"

In the diagram that you posted, the ancilla qubits are pre-determined to have inputs of 1 and 0 respectively.

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