# BTC price hovers above \$61K amid fresh concerns over fate of physical Bitcoin ETF

Suspicions arise over U.S. regulators’ acceptance of physical Bitcoin ETF products later this year, amid reports than investors are voting with their feet.

Can a series of attached tetrahedrons be used for cryptography?

(based on a prior question with triangle)

Given a regular tetrahedron T1 (with his 4 vertices A,B,C,D lying in a finite Field $$mathbb F_N^3$$ with $$N$$ a prime) another attached regular tetrahedron can be constructed by reflection 1 vertex at the plane defined by the 3 other vertices. E.g. vertex $$A$$ can be reflected to $$A’$$ with:
$$A’ equiv (B+C+D)cdot 2 cdot(3)^{-1} – A mod N$$
This can be repeated multiple times (with alternating vertices).
The edge length in between all vertices of a tetrahedron remains constant during this (square euclidean norm e.g. $$const = |A-B|^2 mod N$$ ).

Question:
Given just two random tetrahedron T1 and T2 (and $$mathbb F_N^3$$) can an adversary compute how to construct T2 out T1?

(Assuming there is a way and modulo $$N$$ chosen high enough that testing all tetrahedron would take to long)
Or can he do it (much) faster than just applying it step by step?
(like for example at EC a generator $$g$$ don’t need to be applied $$m$$ times if we want to compute $$g^m$$. I’m looking for a more-than-one-dimensional technique which can not be reduced to a single dimension problem and need to be computed (to some amount) step-by-step.)

Some tests:
In euclidean space a regular tetrahedron can not fill up all space. If the attachment progress is in one direction it has an equal structure as the Boerdijk–Coxeter Helix with vertices coordinates $$(rcos ntheta,rsin ntheta,n h)$$. Two vertices can not have the same $$x$$ and $$y$$ coordinates there.
Doing the same inside a finite Field $$mathbb F_N^3$$ they will match up again at some point.

In tests this took up to $$N^2+N$$ steps.

If we attach the next tetrahedrons around an edge it won’t fill up in euclidean space (left side) but in tests it did match up again with the starting tetrahedron after $$N-1$$ steps in $$mathbb F_N^3$$(right side)

If they just share one common point it also won’t work in euclidean space (left side) but it does in $$mathbb F_N^3$$ :

The number needed for $$mathbb F_N^3$$ depends on $$N$$. Except for $$N=11$$ ($$66$$ there) the max number seems to be $$3N pm 3$$
(could not find any rule yet)
some examples:

N8389971011031071091131271311371397192017100374001350021
number42264241021025433048384664084202154504301083078150060

The prior related question with triangle instead of tetrahedrons could be reduces to a much simpler problem because a large subset sharing the same orientation.
In tests so far tetrahedrons have more than $$N^2$$ orientations if they also need $$N^2+N$$ steps in one direction to match up again (for $$N>11$$). It looks like they even get bigger with larger $$N$$ ($$11: 110$$, $$13: 182$$, $$17: 408$$, $$31:2480$$).

Given a tetrahedron with a certain edge length more than $$N^5$$ different tetrahedron could be produced (for $$N$$ where they also need $$N^2+N$$ steps in one direction to match up again (for $$N>11$$)). It also looks like the amount gets even bigger with larger $$N$$ ($$11: 11^4cdot 10$$, $$13: 13^4cdot 14$$, $$17: 17^4cdot 21$$).

Finding $$1$$ tetrahedron in $$N^5$$ would be very hard even $$1$$ in $$N^2$$ would be hard. Does the problem scale linear with the number of orientations or can it simplified to an easier problem? Or is it even harder than the number of orientations?

#### BTC price hovers above \$61K amid fresh concerns over fate of physical Bitcoin ETF

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