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reviewed Approve books on very large scale linear optimization
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revised Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
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answered Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
Sep
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comment Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
HJRW - this (= going over a ball in the Cayley graph) is actually how I am producing drafts of posters right now. I still think it would be nice to know a priori how far one needs to go.
Sep
23
comment Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
Well, one can take care of the region $\Re(y)>1$ in the obvious way, and then apply a Monte Carlo method like this one to the rest. Still, this isn't really that satisfying.
Sep
22
comment Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
This is a good idea, but I'd still like to have an explicit bound. $2 A(S)/\epsilon$ (where $A(S)$ is presumably the Euclidean area of $S$) can be pretty large.
Sep
22
comment Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
In the case of a box $\lbrack 0,N\rbrack \times \lbrack 0,N\rbrack$ (a good example, actually), the hyperbolic area of $S$ is $\infty$, yet the question still makes sense. So, you really have to know what $S$ is. (And as I said, you may assume it is a box.)
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asked Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large
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comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
The point is that I should have been a bit more precise about the wording - I think most have understood the "right meaning" by now, but it is worth repeating. Sean Eberhard's reply below contains a clean formulation.
Jul
8
comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
... but if anybody can get $< (1-\epsilon) p$ for quadruple exponentiation, Kate and I will be very happy.
Jul
8
comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Yes, we know that - that gives that there are at most $(1-\epsilon) p$ solutions. The answer below gives $<\epsilon p$ (without attempting to follow closely the group-theoretical ideas that give $(1-\epsilon) p$). Notice that the supposedly non-group-theoretical proof for two iterations (and possibly that for three iterations) is soft enough that it can be adapted even if there are $o(p)$ breaking points.
Jul
8
comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
...and actually, you mean either "if one disproves a slightly weaker statement than the above", or, what is the same, "if one proves a slightly stronger statement than the negative of the statement above". I'm trying to keep "positive" and "negative" answers straight...
Jul
8
comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Kate, do you have to give everything away? :)
Jul
8
comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
On your second comment: which answer is positive and which answer is negative? I take you mean to say that, if $2^{2^{2^{2^x}}} = x \mod p$ had $> (1-\epsilon) p$ fixed points, then the discrete logarithm would be easy to compute? (Agreed.)
Jul
8
comment Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Actually, Andreas, your example (the group $H$) strengthens the case for an algebraic approach. Its soficity corresponds to $2 (x+1) = 2x + 2$, which is, um, true. Or do you have a different soficity-algebra dictionary in mind? Please explain.