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I received this problem and unsure how to figure it out. Apparently it is possible to decode the value correctly 99% of the time.

An example: Suppose the target is 20, and there is a function (let's call it noisy_encode) that will return the number of 1's of the XOR of 20, X, and u. You can specify X, and u will range between 0 to 20, with a bias to 2 with 20% probability (remaining are uniform). You can call noisy_encode with different values as many times as you would like.

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  • $\begingroup$ Do I get to observe noisy_encode for many X's? Or do I have to fix a single X? $\endgroup$ – rikhavshah Oct 5 '18 at 1:59
  • $\begingroup$ Many X's are allowed $\endgroup$ – Imran Q Oct 5 '18 at 19:50
  • $\begingroup$ In that case, my answer below works. $\endgroup$ – rikhavshah Oct 5 '18 at 21:46
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Simply set $X=0$. Let $t$ be the target.

noisy_encode returns $t\oplus u$ which will be $t\oplus 2$ about 20% of the time, and look random the rest of the time. The mode output of noisy_encode will be $t\oplus 2$, and so $\text{mode}\oplus 2$ extracts $t$.

If instead of XOR, noisy_encode uses AND, we can assume $u$ is completely uniform instead:

First pick $X = 1$. Then the first bit of $X \wedge u$ will be the same as the first bit of $u$, then the rest of its bits will be $0$. Thus $X \wedge u$ is either equal to $0$ or to $X$. There is a nonzero probability of each of these.

Now consider $X\wedge u\wedge t$. It will either be $0\wedge t$ or $X\wedge t$. In the first case, the output will be $0$. In the second case, the output will depend on $t$. Specifically, the first bit will be the same as the first bit of $t$, then the rest of its bits will be $0$. Thus, if the first bit of $t$ is $0$, then $X\wedge u\wedge t=0$ always. On the other hand, if this bit is $1$, then $X\wedge u\wedge t$ can be either $0$ or $X$.

If you observe the function many times over and over, you will be able to determine which case you are highly likely to be in.

Repeat this process with $X=2,4,8,16,\cdots$ to get all the other bits of $t$.

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  • $\begingroup$ If X=1, wouldn't the first digit of XOR(X,u) be the opposite of the first digit of u? $\endgroup$ – Imran Q Oct 12 '18 at 16:59
  • $\begingroup$ You are right, I read XOR as AND. $\endgroup$ – rikhavshah Oct 17 '18 at 17:45
  • $\begingroup$ I fixed my answer. $\endgroup$ – rikhavshah Oct 17 '18 at 17:50

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