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Babak Farhang

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Babak Farhang's posts

Babak Farhang

Computer Science, Seriously Computer Science

14w

Wrote a short the other day on big-O space requirements for reversible computers. Because of my physics background, I've always been interested in these--their theoretical energy efficiency, that is. Thought I'd share. Perhaps of some pedagogic value, not sure..

http://babaksjournal.blogspot.com/2017/04/a-quick-argument-on-linearity-of-space.html

Babak Farhang

Mathematics Research

33w

An old post about a delayed differential equation governing the distribution of primes using probability heuristics alone

Q'(x) = - Q(x) Q( √x ) / x

http://babaksjournal.blogspot.com/2008/07/differential-equation-estimating.html

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Babak Farhang

Philosophy of Mind Soapbox

34w

Thoughts on sentience and how it relates to AI http://babaksjournal.blogspot.com/2016/12/the-sentient-social.html

The Sentient Social

babaksjournal.blogspot.com

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Babak Farhang

Mathematics Math Questions

41w

A little geometry problem I'd like to share..

I'm modeling a tetrahedron (for a computer program) by specifying the lengths of its 6 edges. Now not all 6-tuples of such positive values can represent a valid tetrahedron, so I need a way for my program to validate the given lengths. One easy-to-check constraint (a necessary condition), is that for each of the faces of the tetrahedron, the given lengths must satisfy the triangle inequality. Is that condition sufficient? (I'm thinking it is, but can't prove it.)

Andres Soolo: +Babak Farhang: Come to think of it, there must be a quadratic equation in calculating the height coordinates of the top point, so you can determine whether it's on the same plane as the base triangle by calculating the quadratic equation's discriminant and checking whether it's zero. ~~However, you'll probably have to do this four times, once for each base triangle.~~ EDIT: no, because if all the four points must lie on a plane, it doesn't matter which triangle you use as the base.

Andres Soolo: +Babak Farhang: Actually, I'm fairly sure the calculation is equivalent to the determinant. (You'll note, of course, that this particular determinant's matrix has a shape that allows many of the calculation steps to be discarded because they're trivial or zeroes.) It's only the organisation of the calculation steps that differs.

The proof should be pretty simple. We assume we've got six edges, of which we can build four triangles. Now, let's use one triangle as the basis of the prospective tetrahedron. Adding the second triangle restricts the top of the tetrahedron with the precision of a circle; adding the third one reduces the set of possible top points to one or two, never zero. Adding the fourth triangle is redundant, because the fourth triangle's sides have already been added — each of them is one side of one of the first three triangles. So, starting from edges as specified, we can always assemble them into a triangle and at least one valid top. (The other possible top point is a mirror of the first one across the plane specified by the base triangle.) The only way for this to not form a proper tetrahedron is for the top to lie on the same plane as the base triangle, and this can happen iff one of the four triangles has an area equal to the sum of the areas of the three other triangles.

Bruce Mincks: It isn't sufficient because the triangle is a plane figure and the tetrahedron is solid. If, by "the triangle inequality" you refer to the composition of angles in a straight line, the problem with the tetrahedron is that there can be no center of the volume which relates the edges, faces, and vertices equally. The equilateral faces of the regular tetrahedron are not directly in perspective with each other, and while they bear a 2/3 relation to the edges of a cube, the triangle at the base of the tetrahedron and the square at the basis of a cube imply different heights inside the same sphere.

https://plus.google.com/108657187448883149300/posts/W8gXvSUTfED

Solid proportions are based on heights from equivalent bases, not the scale of parallels intercepted by transversals. That gives you a pencil projected onto the base of a tetrahedron from its vertex and a square determining six faces by reflection in a cube.

Of course, "regular" polyhedra will differ from irregular ones, but the combinations are determinate for different systems based on irrational numbers in either case, pyramid or cube, where the diameter of the inscribing sphere is considered rational in order to admit the side of an equilateral triangle as something greater than the edge of a cube.

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Babak Farhang

Public

Aug 9, 2016

Babak Farhang commented on a post on Blogger.

I speculate sentience is an emergent property of living in numbers. A kind of social phenomenon. The idea of identity and individuation is one of the first things a living instance learns as its born into its social environment. An infant born to a barren world (but still fed) would likely never learn of its own existence.

Consciousness

languageofair.blogspot.com

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Babak Farhang

Public

22w

Spoiler: yes

Is Verlinde’s Emergent Gravity compatible with General Relativity?

backreaction.blogspot.com

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Babak Farhang

Artificial Intelligence General

34w

When will our AI become self aware? Sooner than we think, perhaps. http://babaksjournal.blogspot.com/2016/12/the-sentient-social.html

The Sentient Social

babaksjournal.blogspot.com

Babak Farhang

Mathematics Puzzles/Riddles

41w

Once at a party, I heard that this ridde was originally presented at the opening lecture of an information theory course. I had originally heard a variation on it as an interview question. I like the party version better. Hope you enjoy..

You will be dealt 5 cards at random from a full deck (of 52). You and your friend are to devise a protocol with which on your showing only 4 of those 5 cards (in some order) she can correctly infer what the 5th card is. What's your protocol?

(Needless to say, at the party, no one could solve it on spot.)

Bonus question: Generalize this. If instead you were dealt 6 cards of a deck of x unique cards (and you're to show 5 of those cards in some order), what's the maximum value of x?

Patrick Hurley: Using the pigeon hole principle, any hand of 5 cards has at least two cards of the same suit. Pick one of these cards and lay it down first. Now the other person (the psychic) knows the suit. There are 13 cards of each suit. Organise them as a circle in your head. Then the two cards of the same suit must be within 6 spaces of each other. Choose two lower of these two cards as the first card to lay down. Now organise the other three cards in one of six ways. They are Lowest. Middle. Highest. So we have LMH LHM MLH MHL HLM HML as the six ways to lay the next three cards. Now we know how many places above the first card the hidden card is. So we know it's suit and number. Trick done.

Brundizzle B (verified by your mother): the only hunch i have so far is that the 5 cards will have at least one face double (four faces, 5 cards).. so the first card you show could indicate the face.. three cards left... i suposse you pick either the highest or lowest number of the face to be guessed.. so the first card also tells the card to be identified is higher than... worst case scenario is that you still have 12 possible cards (assuming ace=1 and you had to use your ace to show the face).. you only have 3 cards left.. so just the order could only identify 6 ... and that is where i get stuck :)..

Bruce Mincks: 5/4 turns out to be the ratio of radii to squares inscribed and excribed by a circle. {5, 4, 2} presents the subcontrary (a/k/a subaltern) ratio to the proportional mean.

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Babak Farhang

Public

49w

On the Conversion of Matter to Gravitational Waves

babaksjournal.blogspot.com

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Babak Farhang

Public

Aug 7, 2016

Recording Computer Generated Proofs Using Blockchain Technologies

babaksjournal.blogspot.com

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