MInimal number of faces to generate full rubik's cube state

Question

I am wondering what's the minimal number of faces we need to know in advance to be assured of the rest of the state of the cube. For a solved cube I believe just the 3 solved faces is enough to infer the entire state of the cube. But for a scrambled cube I bet 3 faces might not be enough. I am curious to know the minimal number of faces required.

Answer 1

Your "conjecture" is not true. 3 faces are not enough. Even 4 faces are not enough:

We have to show that there exists a state of the cube such that 4 faces are solved but the remaining 2 faces aren't (since you would get two different states of the cube with the same 4 faces solved but different on the remaining 2 faces, namely the solved state and the state as described above). To get such a state, apply the following permutation: U D R2 L2 U D' R2 L2 D2. Another (qualitatively different) one would be: R2 U2 B2 R U2 D2 L U D' B2 L2 U2 B2 U D'.

5 faces is obviously enough.

Edit: As Ravi Fernando correctly observed, 5 faces is enough for a solved cube but in general it is not always enough. An example of 2 states with the same 5 faces but a different 6th face are U L' F D' L2 D F' L U' L2 U' L2 U F2 U' L2 U L2 and R2 B2 L2 D' L2 B2 R2 U F' L F' L' R U R' F U2.