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May 10, 2017 at 2:26 history edited user440858 CC BY-SA 3.0
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May 10, 2017 at 2:25 comment added user440858 Yes, I'll modify the statement a little.
May 9, 2017 at 22:46 comment added user44191 This direction that has a unique output would have to be in the sub space, right? Otherwise, my answer below still does work.
May 7, 2017 at 22:48 history edited user440858 CC BY-SA 3.0
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May 7, 2017 at 22:15 vote accept user440858
May 4, 2017 at 9:20 answer added user44191 timeline score: 1
Apr 27, 2017 at 8:34 comment added user440858 Yes, I know that $f$ is a linearized polynomial. I do not know how this helps to answer my question, unfortunately.
Apr 27, 2017 at 6:01 comment added user44191 This may help: a $\mathbb{F}_q$-linear function $f: \mathbb{F}_{q^n} \rightarrow \mathbb{F}_{q^n}$ has a unique expression of the form $\sum_{i = 0}^{n - 1} a_i x^{q^i}, a_i \in \mathbb{F}_{q^n}$.
Apr 27, 2017 at 4:44 comment added user44191 Ah, I misunderstood; I thought you were talking about restricting the output, not the input. My apologies.
Apr 27, 2017 at 4:42 comment added user440858 I do not really see why that is impossible. If $x$ doesn't run through the whole field, you have less points (of the form (x,f(x)), so this smaller set of points will determine less directions. No?
Apr 27, 2017 at 3:43 comment added user44191 Your second question implies that the lower bound becomes lower when adding an extra constraint; shouldn't that be impossible?
Apr 27, 2017 at 0:54 review First posts
Apr 27, 2017 at 1:00
Apr 27, 2017 at 0:54 history asked user440858 CC BY-SA 3.0