tag:blogger.com,1999:blog-67578052019-02-22T03:02:02.186ZKen's blogmostly on computers and mathematicsKennoreply@blogger.comBlogger7681125tag:blogger.com,1999:blog-6757805.post-29132154320890692712019-02-22T03:02:00.000Z2019-02-22T03:02:02.105Z[dqqoaxhc] Games on a 3x3 board<ol style="list-style-type: decimal"><li>Tic-tac-toe</li>
<li><a href="/2009/06/hbskhzci-anti-mirroring-misere-tic-tac.html">Misere tic-tac-toe</a></li>
<li>Wild tic-tac-toe: X or O.</li>
<li>Notakto: all Xs.</li>
<li>Three-men's morris</li><li>Dodgem</li>
<li>3x3 (16 initial points) Dots and Boxes (but with 3 additional free boxes for the second player to counterbalance the advantage of the first player who can force a 6-3 win with perfect play)</li><li>16-point Sprouts (normal or misere)</li>
</ol><p>More: <a href="https://boardgamegeek.com/geeklist/16886/games-played-3x3-grid">https://boardgamegeek.com/geeklist/16886/games-played-3x3-grid</a></p>Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-39488782468160636492019-02-20T17:55:00.000Z2019-02-20T17:55:37.918Z[wnpwrynz] Single-elimination free-throw contest<p>Conduct a basketball free-throw contest in the format of a spelling bee: contestants take turns taking one shot each per round. One miss and you are out, unless everyone in the round also misses.</p> <p>Will the winners typically be NBA professionals? Or could free-throw specialists from outside the NBA compete successfully against the pros?</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-58148993136858971542019-02-14T17:42:00.000Z2019-02-14T17:44:42.393Z[rgujafxj] EE things to make<p>Some important electronic components. Consider making them from scratch. How difficult would it be to return to making electronics after an apocalyptic destruction of all technology?</p><ol><li>1-bit register</li>
<li>laser</li>
<li>LED</li>
<li>LCD</li>
<li>electron gun (CRT)</li>
<li>electro-optical sensor</li>
<li>quartz crystal oscillator</li>
<li>supercapacitor</li>
<li>solid state storage</li>
</ol>Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-55465415132266879272019-02-12T17:30:00.000Z2019-02-13T16:19:45.902Z[xrmrjvqa] Nth annual 10th anniversary<p>In any base N > 1, 10 in base N encodes the value of N.</p><p>10_{N}^{th}<br />
10<sub>N</sub><sup>th</sup><br />
ten<sub>n</sub><sup>th</sup><br/>te<sub>n</sub><sup>th</sup><br/>te<sub>n</sub>th</p>Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-52400022321981055592019-02-05T04:32:00.001Z2019-02-10T03:48:12.949Z[ghrlvjid] High pressure ice<p>Freeze water in a very sturdy container that does not deform or break due to the expansion of normal ice when it freezes. Maybe a diamond anvil, though that might be overkill. Alumina (e.g., synthetic ruby or sapphire) or silicon carbide is cheaper.</p> <p>We expect the result to be some exotic form of high pressure ice with density greater than liquid water. Examining <a href="http://www1.lsbu.ac.uk/water/water_phase_diagram.html">this phase diagram of water</a>, ice II, IV, and VI seem possible, as well as refusing to freeze. Which happens? Surely this has already been done. What kind of equipment is needed?</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-74725268449174965842019-02-05T04:32:00.000Z2019-02-05T04:32:06.120Z[xxobyyfq] Manchester cheese / Muenster United<p>If Worcester is pronounced Wuster (similarly Gloucester, Leicester), Manchester ought to be pronounced Muenster like the cheese.</p><p>Start an urban legend that that's where the cheese originated, but the spelling was changed to get people to pronounce the name of the city correctly.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-63106136033319460962019-02-03T14:04:00.000Z2019-02-03T14:16:36.822Z[igpmbijs] Pairing wines with Pringles aromas<p>When <a href="https://www.foodandwine.com/news/drink-wine-pringles-can">drinking wine out of an empty Pringles can</a>, which wines go well with which Pringles flavors?</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-21123417900376549222019-01-25T00:02:00.000Z2019-01-25T13:17:09.859Z[xarlozhw] VR sphere double cover<p>You see a rotating sphere, perhaps manipulating it in virtual reality to control its rotation. For simplicity, orthographic projection. The sphere is paradoxical: it has much greater surface area than a regular sphere of that size. Rotating it any direction 360 degrees does not put you back where you started; rather, somewhere less.</p> <p><a href="https://math.stackexchange.com/questions/1697915/is-it-possible-to-have-a-connected-manifold-that-is-a-double-cover-of-a-2-sphere">It seems impossible to nicely put the area of 2 spheres on 1</a>, and we assume the result also holds for any larger integer. Therefore, we consider imperfect solutions.</p> <p>An infinite manifold of spherical curvature seems possible.</p> <p>Double cover (or more) with discontinuities. What are some elegant places to put the discontinuities?</p> <p>If we limit rotation to one axis only, then one can pack multiple covers of a cylinder. This is equivalent in 2D to multiple covers of a circle on a circle. We need to prevent seeing the entire neighborhood around a pole.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-32320931282196589742019-01-23T04:37:00.000Z2019-01-23T04:37:07.330Z[azeevreu] Tetrahedron octahedron building blocks<p>Create toy building blocks based on the tetrahedral-octahedral honeycomb. In this honeycomb, the polyhedra attach in a checkerboard pattern: no two tetrahedra share a face, nor do two octahedra. Therefore, we can have asymmetric connectors between them.</p> <p>Three possible simple asymmetric connectors are magnets, Lego, and velcro. Asymmetric connectors seem generally mechanically simpler than symmetric connectors that allow anything to attach to anything. For many connector types, we will need some flexibility in the pieces, because we may want to simultaneously connect to multiple pieces at once when building a structure, e.g., fit a tetrahedron point first into a tetrahedral hole. Magnetic connectors avoid this problem.</p> <p>We could also do <a href="/2017/12/vyffujal-beams-and-connectors-in.html">beams and connectors</a>.</p> <p>Pointy tetrahedra left on the floor are of course even more hazardous to step on with bare feet than regular Lego. One could round off vertices quite a bit, even all the way to spheres (though that makes <a href="/2019/01/kudfmfyp-octahedron-tetrahedron-volume.html">the volume difference</a> even greater, assuming the octahedra keep their shape, because they naturally lay flat).</p><p>Fun things to build: an <a href="/2015/10/yjbljbsv-quarter-octahedron.html">octahedron twice the size</a>, a tetrahedron twice the size.</p><p>One could also apply the idea of asymmetric connectors joining pieces based on an underlying checkerboard honeycomb to the cubical honeycomb: color the cubes in a checkerboard pattern.</p><p>We can have larger pieces, e.g., beams and plates spanning multiple cells of the honeycomb. Such pieces will have both kinds of connectors.</p> <p>Many others have thought of this, including <a href="http://clowder.net/hop/deltablks/deltablks.html">Hop's Delta Blocks, Patent Number 6,152,797. Date of Patent: Nov. 28, 2000 (so probably expired?)</a>, which references Escher's <i>Flatworms</i>. Note that its tetrahedron face tile and octahedron face tile are subtly different (in beveling) so ought to be colored differently to avoid confusion unlike in the virtual mock-ups. Also, unless the pieces are flexible, these pieces will have the simultaneous attachment problem mentioned above.</p>Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-15728832580301361422019-01-23T04:32:00.000Z2019-01-23T04:32:51.824Z[kudfmfyp] Octahedron tetrahedron volume ratio<p>Assuming congruent edges, the regular octahedron has volume 4 times a regular tetrahedron. Any easy way to illustrate this is to consider two tetrahedra, a big one with edge length 2 and a small one with edge length 1. The big tetrahedron has volume 8 times the small one, because volume scales by the cube of the linear dimension.</p> <p>Place 4 copies of the small one inside the big one, one inside each vertex. (Incidentally, this draws a Zelda Triforce on each face. The whole arrangement could be considered a 3-dimensional analogue of the Triforce, maybe a Tetraforce.)</p> <p>The remaining space in the middle has the volume of 8-4 = 4 small tetrahedra. <a href="/2015/10/yjbljbsv-quarter-octahedron.html">The remaining space in the middle is a regular octahedron</a> with the same edge length as the small tetrahedra. Therefore, the octahedron has the volume of 4 tetrahedra.</p> <p>Motivating question: how uneven in size are the components of the tetrahedral-octahedral honeycomb? Answer: they differ in volume by a ratio of 4, which is a lot compared the cubical honeycomb (or any parallelopiped) whose components are of course all the same size. However, such a volume ratio would correspond to a ratio of linear dimensions of cbrt 4 = 1.6, which might make it seem not so much. Incidentally, that cube root can be approximated by 16^3=2^12=4096.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-67199383426470569022019-01-23T02:51:00.000Z2019-01-23T02:51:01.322Z[crxqvylh] Composites are more interesting than primes<p>Primes get all the attention, but composite numbers are individually more interesting, having varied and unique factorizations. The main way primes become individually interesting is through <a href="http://web.mit.edu/kenta/www/three/prime/prime-arithmetic.html">the factorization of the composite p-1, which induces the structure of its multiplicative group</a>.</p> <p>What are some ways to bring out the beauty of the composites?</p> <p>Color them so that smooth numbers are easy to spot: color by the largest prime factor. On the other end of the spectrum, primes and semiprimes could look practically alike.</p><p>Color by the magnitude of the smallest prime factor. There will be stripes caused by multiples of small primes: maybe do some processing to eliminate some of the stripes to see more subtle structure?</p> <p><a href="/2012/10/wsinccxo-factorization-diagrams.html">Factorization diagrams.</a></p> <p>Investigate twin smooths: smooth numbers which are unusually near each other, inspired by twin primes.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-61021963994621314862019-01-22T00:57:00.000Z2019-01-22T00:57:39.996Z[vavwxmxv] Kilton's English-only cryptocurrency<p>Kilton's oblique reference to Bitcoin in Zelda Breath of the Wild (BOTW) seems to be only in the English translation (localization):</p> <p>Kilton: Mon is a currency I invented to destabilize the market and fight the establishment!</p> <p>Kilton: Just kidding, there is no establishment in Hyrule! I just love monsters so much that I turned them into money!</p> <p>Here is the dialogue in Japanese, transcribed here for easy input into tools like Google Translate:</p> <p>キルトン: マモは私が発明した<br> 私の店でしか使えない通貨でございます <a href="https://youtu.be/M3RiA5TpTPg?t=3062">Video</a></p> <p>Originally from here: <a href="https://boards.fireden.net/v/thread/401634591/" class="uri">https://boards.fireden.net/v/thread/401634591/</a></p> <p>Is Bitcoin a popular fad only in English-speaking countries? Are English-speaking countries more tolerant and appreciative of jokes about unstable currency and upsetting the status quo?</p> <p>"Grinding for monster parts" as proof of work. Monsters are called cryptids in cryptozoology, so cryptocurrency.</p> <p>Incidentally, the Japanese name of his shop, マモノショップ Mon Shop, is more boring than the English name, Fang and Bone. マモ is a coined word, probably from 魔物 ma+mono, similar to how Mon is the word Monster shortened.</p> <p><a href="https://youtu.be/uXloOSGBR6Y?t=394">Some video in Spanish, which is a mostly a straight translation from Japanese, so includes none of the colorful English dialogue.</a> The name of the shop is Monstruoteca.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-32236057581717422172019-01-19T00:36:00.000Z2019-02-18T19:02:24.350Z[byopuoco] AI manager<p>Groups of people can change the world through politics, but only if they are organized. To be organized traditionally requires a manager, someone hired by the group of people to coordinate organization.</p> <p>Can AI significantly decrease the cost of management? If so, such technology might profoundly change the world, because it would give poor people more access to management, so it would allow the poor to organize amongst themselves to improve their lot in life.</p> <p>One of the traditional techniques for those in power to stay in power has been to prevent the opposition from organizing, while they themselves are organized.</p> <p>Therefore, this technology, AI managers, will be asymmetric in who it affects. Those with more money were already employing human managers (to stay in power). This technology will decrease their costs, but it won't increase the amount of organization they already have. In contrast, the poor didn't have organization in the first place, so access to this technology will affect them much more.</p> <p>To be effective, people need to trust their manager. People likely will be more willing to trust a machine as their manager than a human: one can, for example, audit the source code of the management software to verify that it is working in your and your group's best interests. (If your best interests don't align with your group's best interests, you shouldn't be part of the organization.) Cryptographic techniques (e.g., homomorphic encryption) can credibly assure that private information you share with the manager stays private and won't be used against you.</p><p>Of course, if the poor become more politically effective, then the rich will probably move on to other techniques -- violence -- to stay in power.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-70632670105938730162019-01-19T00:11:00.000Z2019-01-19T00:11:06.569Z[lreqztkn] History of the size of an atom<p>When, and how, did we discover the rough size of an atom? Assuming atoms are hard spheres, this is equivalent to measuring the spacing between atoms in a solid, e.g., a crystal lattice. So the answer came from X-ray crystallography, though we need to know the wavelength of an X-ray to calibrate.</p> <p>Knowing what we know now, we can do the following to measure the wavelength of X-rays. Set up an X-ray tube of a known high voltage. We assume we can measure voltages accurately. We know the charge of electron (Millikan oil drop), so can deduce the maximum energy of X-rays being emitted (Charles Glover Barkla). Make a leap of faith that the X-rays actually are near those energies. Divide by h (measured by Planck) to go from energy to frequency, then apply the speed of light (Fizeau and Foucault, later Michelson) to get wavelength. Demonstrate diffraction (Max von Laue, also Arnold Sommerfeld?) to show that inter-atomic distances are on the same scale as the X-rays.</p> <p>(There's a pile of important experiments and Nobel prizes there.)</p> <p>How did the determination atomic scale actually historically happen? Was it shocking when we first learned how small atoms are?</p><p>On the opposite end of scale: history of the cosmic distance ladder. I think it was shocking when we first learned how far stars, then galaxies, are.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-22338395036915324562019-01-17T07:51:00.000Z2019-02-11T01:04:27.325Z[zewqywan] Electrostatic repulsion in the nucleus<p><a href="/2015/08/zcdpxksx-speed-of-weaponizing-nuclear.html">In 1911, Rutherford discovered the atom has a nucleus.</a> Let's assume that from his discovery he deduced a model for the atom consisting of electrons somehow wandering around the entire volume of the atom, and a positively charged nucleus occupying a tiny space in the center. The linear scale of that tiny space is approximately 100000 times smaller than the diameter of the atom. (We'll mention an alternative possible model at the end.)</p><p>A fun mathematical exercise -- did Rutherford try this? -- then is to <a href="/2016/04/dmuwqriz-keep-protons-together.html">calculate the potential energy in the electrostatic repulsion of that much charge crammed into that tiny volume of the nucleus</a>. There are two possible simple models: all the positive electric charge is uniformly spread out over the surface of a sphere of that volume, or all the charge is uniformly spread out within the interior of a sphere. These calculations are probably easy, though I haven't done them.</p><p>The answer, the total electrostatic potential energy locked into the nucleus of an atom, is likely shockingly huge, especially when multiplied up to a macroscopic quantity of material. How did scientists react to such huge numbers? It was probably an exciting time, as the universe had revealed itself to be profoundly weird.</p><p>Scientists then of course realized that there must be a very strong force, now called the nuclear force or the residual strong force, that cancels out (and them some) electrostatic repulsion within the nucleus. If the nuclear force were to disappear, how much energy would it take to assemble an atomic nucleus? (This energy is equivalent to the electrostatic potential energy.)</p><p>How does this energy compare to the mass-energy of the atom itself? The "mass defect" is not quite what we want: it sums the potential energies of attractive nuclear force and the repulsive electrostatic force, so there's some canceling out of signs going on. We want just the electrostatic component. </p><p>If the nuclear force were to disappear, how much force would it take to hold together one atomic nucleus against electrostatic repulsion? This might be a human-scale number.</p><p>* * *</p><p>It might have been that Rutherford's gold foil alpha particle experiment only yielded a ratio between the size of the nucleus and the space between, while the absolute atomic scale remained unknown <a href="/2019/01/lreqztkn-history-of-size-of-atom.html">until X-ray crystallography later</a>. When did we first realize how weird the atomic nucleus is?</p><p>* * *</p><p>It's also possible that scientists considered models in which the electrons and the nuclear positive charge were both contained in the nucleus, eliminating the difficulty of huge electrostatic forces within the nucleus. However, this would require some sort of magic to permeate the empty space between atoms, keeping atoms separated by typical atomic distances. Though in retrospect, such magic subjectively seems less preposterous than the nuclear force.</p>Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-67471205795779747682019-01-12T21:24:00.000Z2019-01-12T22:05:45.707Z[fzrukklc] Towers of Hanoi clock<p>Create an automated demonstration of the Towers of Hanoi that transfers a stack discs from one peg to another <a href="/2015/05/qswkpfzs-24n-clock.html">in exactly 8 hours</a>. It completes a full cycle through the three pegs, returning to the initial configuration, in 24 hours.</p> <p>14 discs would have a disc move every 1.76 seconds. 15 discs, 0.88 seconds.</p> <p>It's a clock: tell time by the position of the discs. 8 is conveniently a power of 2, so each hour corresponds to a certain large disc configuration.</p> <p>Easiest would be simulated in video, which could also incorporate other time cues, e.g., lighting. Highly reliable, constantly running mechanical would be difficult.</p><p>There's the legend of the universe ending after a 64-disk transfer is completed.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-73142000907322986872019-01-12T21:22:00.000Z2019-01-14T23:35:15.851Z[nmrqcxez] Finite sequences of primes<p>Fermat numbers grow as O(2^2^n). <a href="/2018/12/arhynpov-statistical-infinitude-of-twin.html">Apply the Prime Number Theorem as a heuristic -- a number n is prime with probability 1/log(n) -- and then replace the sum with an integral,</a> then conclude that there are only a finite expected number of Fermat primes. A similar approximation argument works for any sequence growing as (1+alpha)^(1+beta)^n, for positive alpha and beta.</p> <p>Here are some more slowly growing asymptotic sequence growth rates that would also have a finite expected number of primes. Fermat numbers grow so fast that they are overkill.</p> <p>(1+alpha)^n^(1+beta)</p> <p>(1+alpha)^(n*(log n)^(1+beta))</p> <p>(1+alpha)^(n*(log n)*(log log n)^(1+beta))</p> <p>These follow from the convergence of the integrals of 1/x^(1+beta), 1/(x*(log x)^(1+beta)), and 1/(x*(log x)*(log log x)^(1+beta)).</p> <p>Incidentally, <a href="/2018/12/vsogakpo-nn-and-mersenne-numbers.html">Mersenne numbers grow as 2^(n*log n)</a>, so we expect an infinite number of them to be prime, but only barely. The integral of 1/(x*log x) is log log x, which diverges extremely slowly.</p> <p>Are the any sequences that grow faster than (1+alpha)^(n*(log n)*(log log n)*(log log log n)*...) that have an infinite number of primes? We probably need to limit the type of sequences somehow, or else it would be easy to construct something along the lines of f(n)=nextprime(2^2^n).</p> <p>We explored 1.5^n^1.5 (n <= 3820) and found the following primes:</p> <p>ceil(1.5^1^1.5) = 2<br> floor(1.5^2^1.5) = 3<br> floor(1.5^8^1.5) = 9649<br> ceil(1.5^42^1.5) = 852069747811116972109875222464252771091039570931<br> ceil(1.5^60^1.5) = 6915469149559541382089000664679518689857578042557766452578156297444346096339139131<br> floor(1.5^61^1.5) = 784004647503044231367089868577092132418449967431905553965609437027199675493821852237<br> ceil(1.5^163^1.5) = 2843017533260985010694548776892309855603589981873575072997240935668178294887672889892483420546588216491406511265136751706516520216010851901076208798125523476834397113088228375446152254181894670397711424402046795371831348295827170539437281033397169434713977200649666292247119320731085463593586047121077004471593138233344780787933699241119783143640475074753926994619789</p> <p>We explored 1.5^(n*log(n)^1.5) (n <= 6846) and found the following primes:</p> <p>ceil(1.5^(2*log(2)^1.5)) = 2<br> ceil(1.5^(3*log(3)^1.5)) = 5<br> ceil(1.5^(8*log(8)^1.5)) = 16759<br> floor(1.5^(50*log(50)^1.5)) = 133515421876799711956512714908896295020000818230293796389265675254297<br> floor(1.5^(73*log(73)^1.5)) = 1735871086055272518034305068724289808613087895373465352838436980978395419910575340052199712423367433395137993242363</p> <p>Next, we explore sequences which much closer to the boundary between expecting a finite versus infinite number of primes. (Possibly previously related: <a href="/2012/09/nvzmrton-derivative-of-super-logarithm.html">What is the derivative of iterated logarithm?</a>) We define the product of iterated natural logarithm function in Pari/GP as follows:</p> <p>productiteratedlog(x)=local(y);y=log(x);if(y>1,x*productiteratedlog(y),x)</p><p>The integral of the reciprocal of this function is a piecewise thing that behaves like log x + C for a while, then log log x + C for a while, then log log log x + C, and so on. It therefore diverges (so these sequences are expected to contain an infinite number of primes). Can we define a function whose integral diverges even more slowly? (What's the derivative of the inverse Ackerman function?)</p> <p>We explored sequences of the form exp(productiteratedlog(n)) up to n=4811 and found the following primes. Note that there are no primes of the form exp(n*log(n)) because that is equivalent to the n^n.</p> <p>floor(exp(1)) = 2<br> ceil(exp(1)) = 3<br> floor(exp(2)) = 7<br> floor(exp(30*log(30)*log(log(30)))) = 1760128056820064244010305004494635245456824791132541637</p> <p>We explored sequences of the form 1.5^productiteratedlog(n) up to n=8587 and found the following primes. Note that "log" is still the natural logarithm function, not log base 1.5.</p> <p>ceil(1.5^1) = 2<br> floor(1.5^2) = 2<br> ceil(1.5^2) = 3<br> floor(1.5^(3*log(3))) = 3<br> ceil(1.5^(6*log(6))) = 79<br> ceil(1.5^(7*log(7))) = 251<br> floor(1.5^(11*log(11))) = 44129<br> floor(1.5^(13*log(13))) = 744127<br> floor(1.5^(18*log(18)*log(log(18)))) = 5294466019<br> ceil(1.5^(49*log(49)*log(log(49)))) = 4282685512139315751672739845547275333810677903<br> ceil(1.5^(139*log(139)*log(log(139)))) = 6221632666571163256101566337246863650786499780605768974125245025330883899622185682413811743744661868380530596471794351854330573314331143314872956807522819830556618344028551855268029363025956851<br> ceil(1.5^(211*log(211)*log(log(211)))) = 3623494910954830585594313737287881530765752272229835760527651812393582429644929820469389670214667905333199180860386175763265704984813233955603461933338053203761636195958993503806976854699767461110674905487021598284019232000497521739654465437636807465852318560446512975205883072019435693838691994972399554208850816983281868149450551863<br> floor(1.5^(352*log(352)*log(log(352)))) = 7289878734250178733557869371226896257197695691444233340924792985960426450348980980173587882815055284500903729584130609072757886746227645416034122373248464805270371112957337295293685646178665534636450775554372470252121103912396741477323903733624670105182142642422668301774110539909227413055340142789999395433643294917957997726155270670994619845935016902355835220563425700321313729883919366920336703580184344433746670937276952428746193471967580404410209867212289698162796403647146180978879155890648375724608924682947343951640209056626802452338726872019682305806205921186649908142725637212012518872063615446327545327031106057072327429927943759967<br> ceil(1.5^(2704*log(2704)*log(log(2704)))) ~= 2.113 * 10^7788</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-45914202603934069182019-01-12T21:20:00.000Z2019-01-12T21:20:13.649Z[ugwprsop] Gaussian composites<p>What does the pattern of Gaussian integer multiples of a given complex number look like? This should be easy.</p> <p>Potentially nice demo: hover over a lattice of integers in the complex plane, and the multiples of the lattice point under the mouse cursor light up. It will probably make nice symmetric patterns. This would be useful for understanding the Gaussian primes, which aren't multiples of anything.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-70277235404897184592019-01-12T21:05:00.000Z2019-01-12T21:05:22.266Z[zdzrnlzi] 12 choose 6 = 924<p>The central binonial coefficient 12!/6!^2 seems a nice number to use in a game. 12 and 6 are nice numbers, and 924 seems a good number of possible outcomes of a game or stage of a game. Factorization is 2*2 * 3 * 7 * 11.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-36069590827773836002019-01-12T20:57:00.000Z2019-01-12T20:57:04.225Z[vphyqpen] Concealing the incentive to lie<p>Sometimes a lie gets uncovered as a consequence of it first becoming discovered that the liar had incentive to lie. The investigators then know to look for the lie.</p> <p>Therefore, to <a href="/2018/12/uohbrdbd-keep-people-believing-in-lie.html">maintain a lie</a>, you also need to conceal the incentives to lie. How is this being done by those successful at maintaining a lie?</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-42353863934620936092019-01-12T20:47:00.000Z2019-01-12T20:47:10.175Z[aqnpuilb] Robot baseball batter<p>Create a robot that can hit baseballs pitched by a skilled pitcher. Has this already been done? Some ways to make it even more difficult:</p> <ol style="list-style-type: decimal"> <li>Instead of just hitting home runs, hit specific targets.</li> <li>Moving targets.</li> <li>Taller pitcher's mound.</li> <li>Closer pitcher's mound.</li> <li>More textured ball.</li> <li>Thinner bat.</li> <li>Limit bat speed to how fast a human can swing.</li> </ol> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-30831838960914888142019-01-12T20:43:00.000Z2019-01-12T20:43:56.931Z[xropqqon] Logically assuming based on dress<p>What can you conclude from how someone is dressed? Logically, you can conclude that they are not part of the group who would never dress like that, the contrapositive.</p> <p>The types of clothing you would never wear is often something deeply psychological: you could imagine (or more likely you would rather not even try to imagine) the huge discomfort you would feel or huge resistance you would put up if someone forced you to dress that way.</p> <p>Famous examples: men in women's clothing, women in sexy clothing, nudity.</p> <p>That huge discomfort is probably a phobia. It might also be <a href="/2015/06/oiumqler-changing-yourself.html">identity</a>.</p> <p>What mechanism shapes a person's psychological profile of the clothing they would never wear? Does the mechanism also cause other things, in which case one can probabilistically conclude (through correlation) those other things based on dress?</p> <p>It seems like people with high self-qi would not have any strong clothing objections.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-32714360412007217992019-01-10T18:39:00.000Z2019-01-10T18:39:04.749Z[jndsrmas] Things to recharge every night<ol style="list-style-type: decimal"> <li>Phone</li> <li>Tablet</li> <li>Smart watch</li> <li>External battery charger</li> <li>Wireless headphones</li> <li>Laptop</li> <li>Bike light</li> <li>Camera</li> <li>Mobile gaming device</li> </ol> <p>That's a lot of items. Is there a better way? Some sort of wireless charging that permeates an entire room or house would be nice, but seems scary.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-52293599796365833232019-01-06T20:14:00.000Z2019-01-10T17:46:06.769Z[wetcelbr] Statistics of permutation inversions<p>The number of inversions in a random permutation of n items is normally distributed with parameters</p> <p>mean = binomial(n,2)/2<br> variance = n(n-1)(2n+5)/72</p> <p>This is from "Normal Approximations for Descents and Inversions of Permutations of Multisets" by Mark Conger and D. Viswanath, though these results are not the main subject of the paper.</p> <p>Normal is only an asymptotic approximation. The bizarre paper "Permutations with Inversions" by Barbara Margolius discusses some aspects of how much the actual distribution differs from the normal approximation. The paper is bizarre because, even though its main subject is this normal approximation, it does not explicitly give the expressions for the parameters above. Maybe Margolius was trying to conceal the answers to a homework problem?</p> <p>How computationally difficult is it to exactly compute the distribution? Typically we want the p-value of a given number of inversions x. Naively this would require summing from x to n! which is at least O(n!) work, too much.</p> <p>What are some two-tailed confidence intervals for permutations of 52 items, at 90% 95% 99%?</p> <p>Obviously, this can be used to statistically test how good your card shuffling is: are there large subsets of cards which remain in order (or reversed)? Many common shuffling methods obviously have this problem when done too little.</p> <p>What are some simple permutations which are obviously not random but don't have an unusual number of inversions? Perhaps they exactly hit the expected number of inversions given above. I don't think the following works: cut a deck exactly in half, reverse the order of one half, then put the halves together.</p> Kennoreply@blogger.com0tag:blogger.com,1999:blog-6757805.post-25908366229857234312019-01-06T19:55:00.000Z2019-01-06T19:55:49.371Z[efcdzagl] Random pipes on a 3D lattice<p>Animate a random walk within a cubical lattice, keeping marked traversed edges. Inspired by the "pipes" screensaver.</p> <p>Avoid going over the same edge twice. More restrictively, avoid visiting the same node twice. There's some lookahead required to avoid running into dead ends. Somewhat reminiscent of Paterson's Worm.</p><p>Incidentally, consider ways of avoiding dead ends on 2D lattices. This is similar to heuristics for solving mazes.</p> <p>Instead of the cubical lattice, consider the lattice of the tetrahedral-octahedral honeycomb (alternated cubic honeycomb). 12 edges exit each node, twice that of the cubical lattice, though this is not too useful if a node can be visited at most once.</p> <p>Prefer paths that stay within the viewport (frustrum) of the fixed camera. Prefer paths that stay close to the camera.</p><p>What angle should the camera be pointed to avoid near nodes obscuring farther nodes? If it is an orthographic projection, then some irrational slope would do (which one is best?). What about perspective projection?</p> Kennoreply@blogger.com0