To put it another way, is your decision to chew gum determined by EDT our by your genes? Pick one.

It can be both. Causation is not exclusionary. I'm suggesting that you are mistaken about the aforementioned handling.

Omega's decision is based on our decision algorithm itself.

Yes, but this dependence factors through the strategy that the algorithm produces. Read Eliezer's TDT document(pdf) where he talks about 'action-determined' and 'decision-determined' problems. Whereas CDT only 'wins' at the former, TDT 'wins' at both. Note that in a decision-determined problem, Omega is allowed but a TDT-minimizer is not.

I argue that an EDT agent should integrate his own decision as evidence.

You appear to mean: When an EDT agent hypothesizes "suppose my decision were X" it's subsequently allowed to say "so in this hypothetical I'll actually do Y instead."

But that's not how EDT works - your modification amounts to a totally different algorithm, which you've conveniently named "EDT".

If EDT's decision is to one-box, then Omega's prediction is that EDT one box, and EDT should two-box.

...then Omega's prediction is that EDT will two-box and oops - goodbye prize.

What is generally meant is that having this gene induces a preference to chew gum, which is generally acted upon by whatever decision algorithm is used.

This is actually not what is meant when considering Solomon's problem. They really do mean the actual decision.

As it's been pointed out, this is not an anthropic problem, however there still is a paradox. I'm may be stating the obvious, but the root of the problem is that you're doing something fishy when you say that the other people will think the same way and that your decision will theirs.

The proper way to make a decision is to have a probability distribution on the code of the other agents (which will include their prior on your code). From this I believe (but can't prove) that you will take the correct course of action.

Newcomb like problem fall in the same category, the trick is that there is always a belief about someone's decision making hidden in the problem.

The answer is yes.

I think we have just established that the answer is no... for the definition of "approximate" that you gave...

And if you use the ArthurB definition of "approximately" (which is an excellent definition for many purposes), then a piecewise constant function would do just as well.

ok, but with this definition of "approximate", a piecewise linear function with finitely many pieces cannot approximate the Weierstrass function.

Furthermore, two nonidentical functions f and g cannot approximate each other. Just choose, for a given x, epsilon less than f(x) and g(x); then no matter how small your neighbourhood is, |f(x) - g(x)| > epsilon.

taboo "approximate" and restate.

you won't see the difference

that is because our eyes cannot see nowhere differentiable functions, so a "picture" of the Weierstrass function is some piecewise linear function that is used as a human-readable symbol for it.

Consider that when you look at a "picture" of the Weierstrass function and pick a point on it, you would swear to yourself that the curve happens to be "going up" at that point. Think about that for a second: the function isn't differentialble - it isn't "going" anywhere at that point!

Under the usual mathematical meanings of "continuous", "function" and so on, this is strictly false. See: http://en.wikipedia.org/wiki/Weierstrass_function

It might be true under some radically intuitionist interpretation (a family of philosophies I have a lot of sympathy with). For example, I believe Brouwer argued that all "functions" from "reals" to "reals" are "continuous", though he was using his own interpretation of the terms inside of quotes. However, such an interpretation should probably be explained rather than assumed. ;)