April 2015 - mathfun@mailman.xmission.com

Re: [math-fun] SETI breakthrough: Project Durin Succeeds!
by Keith F. Lynch 05 Apr '15

05 Apr '15

Cris Moore <moore(a)santafe.edu> wrote: > Sorry, I should have said that there were about 1000 different sets, > with ratios measured to about 5 digits of accuracy each... That might work, but I'm not sure how much the frequency of each tuning fork might drift due to physical changes over millions of years. My idea was that the tuning forks would have some simple constant difference or constant ratio of frequencies, and that the message would be in ternary, with each trit being conveyed by the polarization of that frequency's tuning fork. (Sound in solids have three possible polarizations: Horizontal, vertical, and longitudinal. The first two are analogous to EM waves, the last one to sound waves in gases and liquids.) I'm reminded that Touch Tone frequencies were chosen so that none of them would be close to the sums or differences of frequencies (or of integer multiples of frequencies) of the others, to prevent aliasing due to nonlinerarities in telephone circuits. I don't know whether that's an issue in sound propagation in rocks, but if it is a similar approach could be used. What's the math behind such frequencies? > your point about the gravitational redshift is well taken. All sorts of subtle effects become important when discussing very high precision measurements. (I remember when I first learned about pi, and the fact that it was then known to thousands of places of precision, I was envisioning mathematicians very carefully drawing as nearly perfect a circle as possible, and measuring it very carefully with a very good ruler, with the aid of a powerful microscope.) Warren D Smith <warren.wds(a)gmail.com> wrote: > I kind of liked your idea for searching for ET by "looking down." > The particular details re tuning forks were stupid, but... Why? Is there something I'm overlooking? > I think it is an interesting idea in the abstract that the aliens > would want to "leave a message" so that they would be heard at > any time in the future, not just one time like the stupid idea of > sending a radio message once. Also, I point out that this idea > was already in Arthur Clarke's "2001" sci-fi. Yes, and in his earlier (1951) short story "The Sentinel." > His aliens left a monolith made of magic indestructible material on > the moon, etc. It was neither magic nor indestructible. It contained a superconductor with a strong current in it. This resulted in a strong magnetic field, to make it findable, and gave it power to transmit a radio signal when it was exposed to sunlight and warmed by its discoverers. > Seems to me, the Moon is a lot better place to leave a message than > Earth. It'd last longer. That depends. Clarke's message was intended to last only a few million years. In his context, that made sense. (The builders of the monolith had recently uplifted our ancestors into sapience, and reasonably expected that we would soon develop science, technology, and space travel.) But in general, time is deep, and there's no reason to expect that we and aliens are neighbors in time. Over a longer time span, if on the surface of the Moon or shallowly buried, the message would be likely to be destroyed by a meteor impact. It could be deeply buried, but that would make it harder to find. Also, it's difficult to bury things deeply. On Earth, shallow burials often spontaneously become deep burials, due to Earth's geological activity, and aliens could rely on that. And things on Earth will always be more findable by us than things elsewhere, since we will probably always study and explore the Earth more than any other planet or moon. > Perhaps a satellite in Earth orbit would be even longer lasting, and > anyhow very easy to spot. My understanding is that no Earth orbits last more than about ten million years, due to air drag in low orbits and perturbations from the Moon in high orbits. > As far as I understand they claim to be tracking everything in orbit > larger than X, where X is what? The size of a golf ball? And > nothing was there? Right. And that was strong evidence that there aren't any long-term orbits, since otherwise natural debris would have accumulated there, as it has around Jupiter and Saturn. > Certainly a golf ball sized satellite would be adequate to carry an > enormous amount of info from aliens, terabytes easily. How long > would it last? That depends on where it is. The Voyager records are expected to remain playable for at least 10 billion years, which is a respectable amount of time even by my standards. The downside is that they're as close to absolutely unfindable as you can get. Suppose our galaxy contains about a trillion randomly distributed Voyager-sized probes. Think about how any technology, no matter how advanced, could go about finding any of them, ever. I worked this problem a few years ago, and came up with absurd answers, such as converting most of the mass of the galaxy into an immense fleet of starships each equipped with such a powerful radar that not only would we have noticed if any was operating anywhere in the whole universe in our past light cone, but Marconi would have noticed too. Any science fiction novel which depicts a chance meeting in interstellar space, I will throw across the room in disgust. That will never happen, for the same reason that you'll never suffocate from all the air in the room being on the opposite site of the room from you. Physically it could happen, but it's too absurdly unlikely to think about. > Supposedly the layer of dust on the moon is only about 2 cm thick, > suggesting that if you are willing to sacrifice an outer layer a few > cm thick your satellite has decent odds of lasting a few Gyears? The Moon has significant gravity. When meteors blast debris from the Moon, almost all of it falls back to the Moon. Not so with reasonable- sized artificial satellites. Anyhow, below the few centimeters of dust isn't a pristine surface, but smashed up debris of all sizes, getting gradually larger as you dig deeper. The Moon is impacted by meteors of all sizes, and is saturation-cratered. (That's another reason why spheres with tuning forks would be more findable on Earth. To reach any large body of undisturbed bedrock on the Moon, you'd have to dig quite deep.) > And dust particles in space obviously do last for Gyears in the > sense of not evaporating, Again, it depends on where they are. The sun's light pressure exceeds the sun's gravity for dust particles, similarly with other stars, so whatever dust doesn't land on a planet is quickly pushed away from solar systems into interstellar space. > So... I doubt ET left any messages for us. Again I ask if anyone has done a search like Project Durin.

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Re: [math-fun] "Project Durin"
by Warren D Smith 04 Apr '15

04 Apr '15

I kind of liked your idea for searching for ET by "looking down." The particular details re tuning forks were stupid, but... I think it is an interesting idea in the abstract that the aliens would want to "leave a message" so that they would be heard at any time in the future, not just one time like the stupid idea of sending a radio message once. Also, I point out that this idea was already in Arthur Clarke's "2001" sci-fi. His aliens left a monolith made of magic indestructible material on the moon, etc. Having said that, the question is how should the aliens act, and how should we act? Seems to me, the Moon is a lot better place to leave a message than Earth. It'd last longer. Perhaps a satellite in Earth orbit would be even longer lasting, and anyhow very easy to spot. As far as I understand they claim to be tracking everything in orbit larger than X, where X is what? The size of a golf ball? And nothing was there? Certainly a golf ball sized satellite would be adequate to carry an enormous amount of info from aliens, terabytes easily. How long would it last? Supposedly the layer of dust on the moon is only about 2 cm thick, suggesting that if you are willing to sacrifice an outer layer a few cm thick your satellite has decent odds of lasting a few Gyears? And dust particles in space obviously do last for Gyears in the sense of not evaporating, so you're safe against that. So... I doubt ET left any messages for us.

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Re: [math-fun] SETI breakthrough: Project Durin Succeeds!
by Keith F. Lynch 04 Apr '15

04 Apr '15

Cris Moore <moore(a)santafe.edu> wrote: > The first seismic tests have succeeded in returning a signal. It > turns out that there are two sets of tuning forks deep in the > bedrock, whose resonant frequencies appear to be incommensurable. > When their ratio is expanded in binary, its first few thousand > digits reveal an image of an eldritch race, barrel-shaped creatures > with pentagonal symmetry and diaphanous wings... who knows what > secrets are buried deeper inside its binary expansion. How long would you have to listen to two tuning forks to measure their relative frequencies to a few thousand binary digits? Remember that the sphere is at most two or three billion years old (2^51 seconds), and that frequencies above about 100 MHz (2^24 Hz) won't propagate through rock. Extreme frequency stability would also require an absurdly high Q, but at least that would mean that there's no need for seismic excitation. The aliens who buried the thing set it to resonating, and it's still resonating. But for it to emit one watt of acoustic power for a billion years (10^16.5 seconds) it would have to have started with at least 10^16.5 joules of kinetic plus potential energy. That energy, by E=mc^2, has a mass of about a third of a kilogram. The gradual loss of that mass would greatly change the resonant frequency. Also, if the two tuning forks differ in altitude by even one centimeter, that would throw off the frequency ratio by about one part in 10^18, i.e. 54 binary digits, due to general relativity's gravitational time dilation. The article was of course a joke, but I tried to make it physically possible. (Similarly with my proposal to speed up the Earth's rotation by running tidal power stations in reverse; that would actually work, though you would need to build lots of large new stations.) Indeed, I'd like to see someone try something like Project Durin. I think it has just as much chance of working as radio-based SETI searches, and would be a lot less expensive. If I could find a good venue (this probably isn't it), I'd like to have a discussion of how workable the idea is. Would such a sphere be stable for hundreds of millions of years at the temperatures and pressures in the upper mantle? What would the signal strength of such a tuning fork be, and would it be lost in the background noise? How much information could be practically encoded with multiple tuning forks in such a sphere? Information could be encoded with frequency, location within the sphere, and polarization. (Sounds in solids have three possible polarizations.) Has anyone made an acoustic search for such spheres? Is the idea really original with me?

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[math-fun] John Tromp's proof of work
by Warren D Smith 03 Apr '15

03 Apr '15

Tromp's idea, I think, was something like this. (Or it may be I'm inventing something different under the delusion this was his idea, but anyhow.) >From an n-bit string x, you apply some cryptographic bijective function F1 to reach another y=F1(x). Or, you could instead have used F2, or F3. This defines a directed graph with 2^n vertices and each outvalence=3. Now, to prove you've done work, you say "starting from the n-bit string 0101101...1, apply functions 1,3,2,3,3,2,2,2,1,...,2 in succession, to return to start point." Where the whole sentence is O(n) characters long. Anybody can verify or deny the truth of your sentence very quickly. But it is not easy to find true sentences of this nature. Further, if the start point is chosen by somebody else (not you) then you will have even more difficulty finding true sentences of this nature.

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Re: [math-fun] The game called "Set"
by Warren D Smith 03 Apr '15

03 Apr '15

The goal of the problem is to determine (or merely bound) F(n), which is the maximum possible number of n-vectors over Z3, such that no three sum to 0. More precisely such that for any three, call them x,y,z, it can only happen that x+y+z=0 if x=y=z. As my first trivial remark... obviously, if it is the case that no three of the vectors x,y,z exhibit ANY linear relation a*x+b*y+c*z=0 over Z3, (a,b,c)!=(0,0,0), that suffices. For that, it in turn suffices if our set of (column) n-vectors, n>3, happens to form an "orthogonal array of strength 3." For that, it in turn suffices if the set of vectors is the dual of a hamming-distance-4 linear code over GF3. So it seems to me that by using the "duals of the extended ternary Hamming codes" we obtain a proof that F( (3^r-1)/2 + 1 ) >= 3^r. Oh. Well, that was a quite weak result, F(n) >= 2*n-1 for an infinite set of n. Supposedly it is known that F(n) > 2.21^n, a far stronger lower bound. Still, since Hamming codes are "perfect" my bound is presumably best possible using its technique. So this demonstrates that demanding all 26 linear relations be impossible causes a tremendously different effect than only demanding (what the problem calls for) it for the 7 relations (a,b,c) = (1,1,1) and perms(0,1,2). This is a very interesting problem. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Hawking black hole info paradox -- solved? Looks like it.
by Warren D Smith 03 Apr '15

03 Apr '15

> Date: Thu, 2 Apr 2015 15:43:26 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Hawking black hole info paradox -- solved? Looks > like it. > http://arxiv.org/abs/1503.01487 > Anshul Saini, Dejan Stojkovic: > Radiation from a collapsing object is manifestly unitary > Phys.Rev.Lett. 114,11 (2015) 111301 > > This claims to explicitly calculate the > full density matrix, i.e. including all off-diagonal terms, > of the Hawking radiation > from a "collapsing shell" model of a black hole. > It stays unitary at all times. > Thus the Hawking "information loss paradox" actually does not occur. > > It looks to me, at least on the surface, like they've really done it, and > they didn't need to do anything crazy. --Corrections. The "density matrix" in quantum mechanics is NOT unitary, it is (for a "pure state") of rank-1. (It is in this case of countably infinite order.) What I meant was, the density matrix M evolves via conjugation by a time-dependent unitary matrix U(t): M(t) = Uconjugatetranspose M U and they've apparently demonstrated that M remains of rank=1 and with |trace|=1, which is consistent with that.

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[math-fun] "Minsky Stock Index" correction
by Bill Gosper 03 Apr '15

03 Apr '15

rwg>But maybe not this week. Tom's crunch has so far pushed Corey's & Julian's 208T to 3.227P, Thus (unless I'm spazzing) the point density [has increased by 15/10], which predicts convergence... <rwg. --Bill

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Re: [math-fun] Computational effort
by Bill Gosper 03 Apr '15

03 Apr '15

On Fri, Mar 20, 2015 at 11:49 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > HGBaker> > Some non floating point units of computational effort: > > Bitcoin POW (SHA-256 hashes/second) > > Other POW's https://en.wikipedia.org/wiki/Proof-of-work_system > > Password cracking > > Most of these problems are "embarrassingly parallel". > > Perhaps someone here has some ideas for a non-embarrassingly-parallel > unit of computational effort. > > <hgb > > How about the Minsky Stock Index? > Oops, Neil Bickford has found a long-sought algorithm to parallelize the Minsky Circle Recurrence, and noted funster Tom Rokicki (diameter(Rubik group)=20) is excitedly throwing GPUs at it. > > Subj: What is the period of this integer sequence? > On Sun, May 4, 2014 at 10:37 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > >> 0, 1, 7, -2, -8, 3, 14, -4, -16, 5, 21, -6, -24, 7, 28, -7, -25, 7, >> 27, -7, -26, 7, 26, -6, -19, 5, 18, -4, -12, 3, 10, -2, -5, 1, 2, 0, >> 2, -1, -6, 3, 16, -5, -22, 7, 30, -8, -30, 8, 30, -7, -23, 6, 22, -5, >> -16, 4, 14, -3, -9, 2, 6, -1, -2, 1, 5, -1, -3, 1, 4, -1, ... >> >> a[0] = 0; a[1] = 1; >> a[k_?OddQ] := a[k] = a[k - 2] - Floor[9*a[k - 1]/17]; >> a[k_?EvenQ] := a[k] = a[k - 2] + Floor[15*a[k - 1]/2] >> >> This is the "Minsky Stock Index" that Corey and Julian ran fruitlessly >> from a[-10^14] to a[10^14] a few years ago. "If you haven't looked at a >> problem in the last few years, you haven't looked at it." --Ed Pegg, Jr. >> There is a slight chance that the period is infinite. >> (When it reached 18 trillion, we exclaimed AT&T!) >> >> The above recursive definition crashes Mathematica in under a million, >> even with memoizing. Instead use the iteration >> In[115]:= NestList[Function[xy, {#, xy[[2]] + Floor[15*#/2]} & >> [xy[[1]] - Floor[9*xy[[2]]/17]]], {1, 0}, 35] >> e.g., for {a[-1],a[0}, {a[1],a[2]},...{a[69],a[70]} >> Out[115]= {{1, 0}, {1, 7}, ..., {-1, -4}} >> > ~100 times faster: Clear[MinskyStepC];(*No frills*) MinskyStepC[x0_Integer, y0_Integer, d_Integer | d_Rational, e_Integer | e_Rational, n_] := minskystepc[x0, y0, Numerator[d], Denominator[d], Numerator[e], Denominator[e], n]; minskystepc = Compile[{{x0, _Integer}, {y0, _Integer}, {dn, _Integer}, {dd, _Integer}, {en, _Integer}, {ed, _Integer}, {ct, _Integer}}, Module[{x = x0, y = y0}, Do[x = x - Quotient[y*dn, dd]; y = y + Quotient[x*en, ed], {ct}]; {x, y}], RuntimeOptions -> "Speed", CompilationTarget -> "C"]; E.g., (Corey's old) In[925]:= MinskyStep[1, 0, 9/17, 15/2, 10^7] // tim 60.579526,2 Out[925]= {-11142, -46828} In[926]:= MinskyStepC[1, 0, 9/17, 15/2, 10^7] // tim 0.634458,2 Out[926]= {-11142, -46828} > >> Background: >> http://www.blurb.com/books/2172660-minskys-trinskys-3rd-edition >> Unlike Collatz or twin prime searching, this is a very specific question >> about >> very specific quantities, rather than a question about the infinitude of >> integers. >> --rwg >> > > So far, nobody(?) has figured out how to distribute this calculation. > Until now. > > I conjecture finitude for all these monsters that don't blow right up with > an obvious, > provable pattern. The bigger they get, the proportionally smaller is the > effect of the Floor > steps, and thus the closer they get to pure ellipses. I.e. their growth > rate slows down, > and the density of points they visit will increase with increasing > radius. Eventually > they will run out of new points, and thus must return to the beginning, by > reversibility. > But maybe not this week. Tom's crunch has so far pushed Corey's & Julian's 208T to 3.227P, while the circularized radius has nearly tripled. Meanwhile, as spinoff from Neil's algorithm, Tom has found a loop with these same multipliers (9/17, 15/2) of length 389T, roughly seven times the previous record for any Minsky loop. Thus, the "population" increased 30fold while the "radius" only tripled. Thus (unless I'm spazzing) the point density is increasing with time, which predicts convergence unless the rate of increase approaches zero. Which is not happening. Here is Tom's latest news, which I hope he won't mind me blabbing. The startling proximity of the current extremal points is an artifact of the search algorithm. When Neil and I first saw this, we thought Tom was April Fooling. ------------ So things have been running a bit over five days. At the moment, from (1,0), we are at forward 35432432 172100384 774871328665934 backward 35451612 172193544 -2452403557917764 so in total that's about 3.227P without looping. But that's not to say we haven't found some big loops! >From 1,003,636*(7,34) we have a closed loop of 388,651,797,405,176 steps, for instance. There are many other large closed loops. In total I've explored 1.6286e16 points. From these I have found 1842 closed loops totaling 8.71398e+14 points and 5735 yet unresolved open segments totaling 1.54146e+16 points (including our largest one from (1,0) that is about 20% of the total points). We have explored the continuations of about 5M distinguished points (the distinguished points are i*(7,34), i integer > 0). The minimum traversal from DP to DP without an intervening DP is from 189*(7,34) to 187*(7,34) in 311 steps; the largest is from 4,567,906*(7,34) to 4,553,980*(7,34) in 76,299,044,924 steps. >From a small experiment it appears that the set of all points reached from or to a distinguished point is > 99.5% of all points; this is based on a few hundred million points near the center. As we expand away from the center, points that traverse to or from a DP appear to decrease in absolute density. But most of the really short loops never get to a DP. It's quite possible some distinguished point may iterate and never come back to the set of distinguished points, but that has not happened yet from exploring more than the first five million distinguished points. If I were to start this again, I'd probably pick a different set of distinguished points (like i*(0,1)) for better coverage and to improve the efficiency of the kernel slightly. But I've decided not to restart my existing kernel. The core idea---distinguishing points and exploring their futures in parallel---I got from Neil; everything else is just a tiny bit of hacking. -tom ---------- --rwg

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[math-fun] Bird 'backpacks' help scientists discover the longest oversea migration
by Henry Baker 03 Apr '15

03 Apr '15

FYI -- This article is relevant to the recent discussion about ocean navigation by birds, etc. http://www.theguardian.com/science/2015/apr/01/bird-backpacks-help-scientis… Bird 'backpacks' help scientists discover the longest oversea migration By fitting it with tiny geolocators, scientists have proven that the blackpoll warbler completes the longest known oversea journey for any land bird Hannah Devlin, science correspondent Wednesday 1 April 2015 00.00 BST Last modified on Wednesday 1 April 2015 00.03 BST A tiny North American songbird migrates 1,500 miles non-stop over the Atlantic, scientists have discovered, in the longest known oversea journey for any land bird. The blackpoll warbler weighs only as much as a Â£2 coin and normally lives in forest environments, but once a year it embarks on a perilous three-day journey, scientists have found. By fitting the birds with tiny geolocator backpacks, they were able to track their route from North America and Canada to Puerto Rico and Cuba and then onwards to South America. Chris Rimmer, an ornithologist at the Vermont Center for Ecostudies and a co-author of the study, said: ÂThere is no longer any doubt that the blackpoll undertakes one of the most audacious migrations of any bird on earth.Â Although other migratory birds such as the albatross and arctic tern are known for travelling thousands of miles across oceans, the blackpoll warbler has the added challenge of making the journey without stopping, as a water landing would be fatal. ÂItÂs a fly-or-die journey,Â said Ryan Norris, the Canadian team leader at the University of Guelph, said. Other North American songbirds that travel south in winter take a less risky, continental route south through Mexico and Central America. However, ornithologists suspected the blackpoll warbler was taking a water route, based on observations of birds landing on boats during stormy Atlantic weather. It had also been noted that the birds put on significant amounts of weight in the autumn Â sometimes almost doubling in size Â hinting that they were preparing for a long journey ahead. ÂThey get basically so fat that they donÂt move,Â said Norris. Until now, there was no conclusive proof that they were migrating across open ocean, however. ÂSome people said thereÂs no way they could do this,Â Norris added. ÂThey donÂt seem particularly athletic birds.Â In the latest study, published in Biology Letters, scientists equipped 20 birds in Vermont and 20 more in Nova Scotia with tiny backpacks fitted with a clock and a sensor to measure ambient light levels. The number of hours of daylight gives the birdÂs latitude and the time of sunrise gives longitude. But the packs, weighing just 0.5g, were too small to include transmitters, meaning that the researchers had to search for the birds back at their nesting ground in the North American forest the following year. ÂThese birds come back every spring very close to the same place they used in the previous breeding season, so with any luck you can catch them again,Â said Norris. ÂOf course there is high mortality among migrating songbirds on such a long journey; we believe only about half return.Â The team worried that the extra 0.5g Â added to the birdÂs roughly 12g body weight Â could make an already challenging journey insurmountable. In the end, they were able to recapture three birds from the Vermont group and two from the Nova Scotia group for analyses. ÂIt was pretty thrilling to get the return birds back, because their migratory feat in itself is on the brink of impossibility,Â said Bill DeLuca, of the University of Massachusetts Amherst and the paperÂs first author. The locators showed that the blackpolls took a route directly over the Atlantic, with distances ranging from 1,410 to 1,721 miles (2,270 to 2,770 km). Grahame Madge, of the RSPB, said the findings help explain why the songbirds have occasionally been seen in Britain: records show 45 sightings, mostly in the Scilly Isles, since the 1960s. ÂItÂs quite easy to imagine how a storm system could sweep them up,Â he said. ÂThey set out thinking theyÂre heading for the Caribbean and end up in Europe.Â http://i.guim.co.uk/static/w-300/h--/q-95/sys-images/Guardian/Pix/pictures/…

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[math-fun] Is your ignition sluggish?
by Bill Gosper 03 Apr '15

03 Apr '15

Is your escargot on the cold side? Popping the hood on my repair shop loaner, gosper.org/IMG_0339.JPG --rwg

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