#math - Fri 11 May 2007 between 02:01 and 02:13

gkrZ_n = Z/nZ, right?
TRWBWgkr: yes
sloof3gkr: of course
actiongkr thought he was all wrong
gkrthought he was all wrong
TRWBWgkr: you mean mod n, yes?
Capsosloof3: No, the point is: no matter WHAT input, you get the SAME output, for the SAME function definition.
sloof3Capso: why would we get the same output
Capsosloof3: That's the whole thing, isn't it?
sloof3: You know your output, you know what the system is doing, you just don't know what you gave the system.
sloof3I was thinking in terms of a system with an observable output and changing input
CapsoYour 'observability' defines the probabilities of what you MIGHT have given the system.
sloof3Capso: yes
CapsoSo, the OUTPUT needs to stay the same.
sloof3If the output always stayed the same that wouldn't help us would it?
CapsoYou can only CONSIDER one output.
All the input values which would get that output.
sloof3We don't know the inputs though
CapsoThe thing is: you're not DETERMINING anything, you're simply getting a rating of a definition FROM an observation.
Right, we don't know input.
sloof3Originally I was only considering a system that had changing inputs.
We would need at least n outputs for n inputs
CapsoSorry, had some disruption here.
No, your GOAL is to define 'observability' by knowing ONLY the following:
(1) The number of inputs a function might take
(2) The operation (definition) of the function
(3) The output(s?) of the function.
And the 'observability' will simply be you *ability to define the inputs to obtain the output(s?)*.
sloof3define the inputs given the outputs
CapsoThat's vague.
What I've stated is clearer.
The reason I include the '(s?)' in 'output(s?)' is that I'm still considering whether we should take into account > 1 output.
You need certain inputs to GET those outputs.
Okay, suppose:
f(x,y) = x + y; -- Definition
f(x,y) = 5
f(x,y) = 6

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