Is $g(x_1||x_2) = f(x_1 \wedge x_2)$ a one way function assuming f is a one way function

CHTM

2/10/24, 12:33 AM

Intuitively I think not because assuming the bit string $x_1,x_2 \sim \{0,1\}^{n/2}$, $x_1 \wedge x_2$ is not uniformly random so if $g$ were still a one-way function then the fact that the definition of one way function requires the input string $x$ to be uniformly random seems unneeded.

But I'm not sure how to construct the $f$ required. I tried the usual $f(x) = 0^{n/2}||f(x_{[1:n/2]})$ but got stuck.

0 + 2

one-way-function

complexity

fgrieu

2/10/24, 8:03 AM

Welcome to crypto-SE. I suggest you [edit](https://crypto.stackexchange.com/posts/104154/edit) the question, adding the definition of One Way Function assumed in the context. This may help you, and will show the level of formalism expected.

Daniel S

2/10/24, 8:15 AM

HINT: Suppose that you could invert $g$ with non-negligible probability, could you use this ability to create pseudo-inversions of $f$?

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Is $g(x_1||x_2) = f(x_1 \wedge x_2)$ a one way function assuming f is a one way function

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