Adele Lopez

Karma: 1,415

Adele Lopez 29 Jul 2019 3:10 UTC
6 points

in reply to: James_Miller's comment on: What supplements do you use?
Is there a reason not to take it if you’re younger than 40?

Adele Lopez 13 May 2019 4:19 UTC
2 points

in reply to: Said Achmiz's comment on: Coherent decisions imply consistent utilities
Presumably, any agent which we manage to build will be computable. So to the extent our agent is using utility functions, they will be continuous.
If an agent is only capable of computable observations, but has a discontinuous utility function, then if the universe is in a state where the utility function is discontinuous, the agent will need to spend an infinite amount of time (or as long as the universe state remains at such a point) determining the utility of the current state. I think it might be possible to use this to create a more concrete exploit.

Adele Lopez 13 May 2019 1:35 UTC
4 points

in reply to: Samuel Hapák's comment on: Coherent decisions imply consistent utilities
I think you’re right that this is part of where the intuition comes from. But it’s still irrational in a context where you actually know the probabilities accurately enough.

Adele Lopez 13 May 2019 1:27 UTC
3 points

in reply to: Said Achmiz's comment on: Coherent decisions imply consistent utilities
For continuity, it’s reasonable to assume this because all computable functions are continuous. See theorem 4.4 of https://eccc.weizmann.ac.il/resources/pdf/ica.pdf
Edit: I realized that the continuity assumption is different (though related) from assuming the utility function is continuous. My guess is that computability is still a good justification for this, but I’d have to check that that actually follows.

Adele Lopez 15 Dec 2018 18:29 UTC
6 points

on: What is abstraction?
Abstraction can be understood as a relationship between models. You can have a model of what it means to “brush your teeth”, and you can have models of what it means to “prepare toothbrush”, etc… The models of the subtasks can be composed into a model of the whole sequence, and the abstraction relationship tells you how this model is a realization of the abstract “brush your teeth” model. Similarly, we can have a model of what an “animal” is, and models for “dogs”, “humans”, “cats”. The abstraction relationship tells you how each of these models are realizations of the “animal” model. Removing properties is an easy way to create models with this relationship, but it’s not the only way. For example, you can also replace continuous time with discrete time.
You can chain this relationship to create a hierarchy, and for humans, the most concrete models we typically use are the ones that are “modeling” our raw sensory experiences. I think that adequately explains why people use it this way, but that this isn’t what abstraction is.

Adele Lopez 24 Nov 2018 7:01 UTC
3 points
AF
in reply to: lbThingrb's comment on: Topological Fixed Point Exercises
Awesome! I was hoping that there would be a way to do this!

Adele Lopez 24 Nov 2018 6:25 UTC
4 points
AF
in reply to: JacobKopczynski's comment on: Diagonalization Fixed Point Exercises
Currying doesn’t preserve surjectivity. As a simple example, you can easily find a surjective function $2 \times 2 \to 2$ , but there are no surjective functions $2 \to 2^{2}$ .

Adele Lopez 22 Nov 2018 4:17 UTC
14 points
AF
on: Iteration Fixed Point Exercises
Ex 6:
If at any point $f^{n} (b) = f^{n - 1} (x)$ , then we’re done. So assume that we get a strict increase each time up to $n = | P |$ . Since there are only $| P |$ elements in the entire poset, and $f$ is monotone, $f^{n + 1} (x)$ has to equal $f^{n} (x)$ .
Ex 7:
For a limit ordinal $α$ , define $f^{α} (x)$ as the least upper bound of $f^{n} (x)$ for all $n < α$ . If $α > | L |$ , then the set $f^{n} (x)$ for $n < α$ is a set of size $α$ that maps into a set of size $L$ by taking the value of the element. Since there are no injections between these sets, there must be two ordinals $n < m$ such that $f^{n} (x) = f^{m} (x)$ . Since $f$ is monotone, that implies that for every ordinal $l > n$ , $f^{l} (x) = f^{n} (x)$ and thus is a fixed point. Since $n < α$ this proves the exercise.
Ex 8:
Starting from $x$ , we can create a fixed point via iteration by taking $α > | L |$ , and iterating $α$ times as demonstrated in Ex 7. Call this fixed point $f_{x}$ . Suppose there was a fixed point $k$ such that $x \leq k$ and $k \leq f_{x}$ . Then at some point $f^{n} (x) \leq f^{n} (k) = k$ , but $f^{n + 1} (x) \geq f^{n + 1} (k) = k$ , which breaks the monotonicity of $f$ unless $k = f_{x}$ . So $f_{x}$ generated this way is always the smallest fixed point greater than $x$ .
Say we have fixed points $x_{i}$ . Then let $x$ be the least upper bound of $x_{i}$ , and generate a fixed point from $f_{x}$ . So $f_{x}$ will be greater than each element of $x_{i}$ since $f$ is monotone, and is the smallest such fixed point as shown in the above paragraph. So the poset of fixed points is semi-complete with upper bounds.
Now take our fixed points $x_{i}$ again. Now let $x$ be the greatest lower bound of $x_{i}$ , and generate a fixed point $f_{x}$ . Since $x \leq x_{i}$ and $f$ is monotonic, $f^{α} (x) \leq f^{α} (x_{i}) = x_{i}$ , and so $f_{x}$ is a lower bound of $x_{i}$ . It has to be the greatest such bound because $x$ itself is already the greatest such bound in our poset, and $f$ is monotonic.
Thus the lattice of fixed points has all least upper bounds and all greatest lower bounds, and is thus complete!

Adele Lopez 22 Nov 2018 2:21 UTC
13 points
AF
on: Diagonalization Fixed Point Exercises
Ex 8: (in Python, using a reversal function)
def f(s):
return s[::-1]

dlmt = ’”“”‘
code = “””def f(s):
return s[::-1]

dlmt = ’{}’
code = {}{}{}
code = code.format(dlmt, dlmt, code, dlmt)
print(f(code))”″”
code = code.format(dlmt, dlmt, code, dlmt)
print(f(code))

Adele Lopez 21 Nov 2018 18:42 UTC
13 points
AF
in reply to: Davidmanheim's comment on: Topological Fixed Point Exercises
You can use 1d-Sperner to deal with all the cases effectively.

Adele Lopez 21 Nov 2018 18:11 UTC
16 points
AF
on: Topological Fixed Point Exercises
Found a nice proof for Sperner’s lemma (#9):
First some definitions. Call a d-simplex with vertices colored in (d+1) different colored chromatic. Call the parity of the number of chromatic simplices the chromatic parity.
It’s easier to prove the following generalization: Take a complex of d-simplices that form a d-sphere: then any (d+1)-coloring of the vertices will have even chromatic parity.
Proof by induction on d:
Base d=-1: vacuously true.
Assume true for d-1: Say you have an arbitrary complex of d-simplices forming a d-sphere, with an arbitrary d+1-coloring. Choose a vertex. W.L.O.G. we will call the color of the chosen vertex blue.
Take the complex of simplices that contain this vertex. Since a sphere has no boundary or branches, this complex will be a d-ball. Delete the chosen vertex, and keep only the opposite (d-1)-simplices that are left, which will form a (d-1)-sphere, call it the shell.
We need to choose a second color, say red. We’ll call a (d-1)-simplex with vertices of all d+1 colors except red an R-chromatic face, and similarly with blue.
Now, replace all the red vertices in the shell with blue vertices, so that the shell is now R-chromatic. By induction it has an even number of R-chromatic faces. Consider what happens when we reconvert a vertex on the shell back to red: since the vertex was previously blue, any R-chromatic faces will get turned into B-chromatic faces. Let r be the number of R-chromatic faces on the shell, and b be the number of B-chromatic faces. The parity of r-b will remain even as we continue this process.
Let’s go back to the vertex in the center of the shell. All currently chromatic simplices with this vertex have opposite faces which are B-chromatic, since this vertex is blue. We’ll flip the vertex to red, which destroys chromatic simplices with opposite B-chromatic faces and creates chromatic simplices with opposite R-chromatic faces. Since r-b is even, the chromatic parity is preserved!
Since we’ve shown that arbitrary recoloring of vertices preserves the chromatic parity, it’s clear that the chromatic parity will be even for any coloring.
Corollary: Sperner’s lemma
Start with a d-simplex which has been divided into d-simplices, and where each face of the large simplex has one color which vertices on it are forbidden from taking. Take a point of each color, and match it with a face of the simplex that that color is allowed on. Then connect that vertex to each point on that face. This will create a bunch of non-chromatic simplices. Finally, create a simplex of all of the new points. This will create one chromatic simplex.
This will form a d-sphere, and thus will have an even chromatic parity. That means the original simplex must have had odd chromatic parity.

Adele Lopez 27 May 2018 6:06 UTC
2 points

on: Editor Mini-Guide
Is there a way to make drafts available to specific people, for review?