# Re: Almost perfect nonlinear permutations

I thought I'd posted a reply earlier, but it's not showing up and I
think I pressed Discard by mistake.

Firstly, the function mapping 6-> 9 (not 7 -> 9) wasn't an example of
what I'm trying to achieve here. It was just the closest I'd come
while still mapping all inputs with Hamming weight 1 to themselves.

The key stumbling block seems to be this statement.

" the upper n-m bits will be a linear mapping of the input,
specifically the identity function."

I think you're stating that all 00*** map to 00***, all 11*** to
11***, all 01*** to 01***, and all 10*** to 10***. Equivalently, that
the upper two bits have ANFs y_0 = x_0 and y_1 = x_1.

Now, while I agree that (assuming the first eight truth table entries
were a bijection) all 00*** map to 00***, 01000 maps to 01000, and
10000 maps to 10000, I do not see that anything in my statement of the
problem implies that all other 01*** and 10*** must map to outputs
with the same two upper bits.

Plus, my posting dated June 16th contains a description of the
algebraic normal forms of y_0 and y_1 that shows how they could map
all 00*** to 00***, and 01000/10000 to themselves, without being
linear; hence suggesting that the upper two bits are not necessarily
the identity mapping.

James.
.

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