Latest posts on math thread topichttps://rhizzone.net/forum/topic/14538/2020-12-23T17:49:31+00:00Discussion :: Laissez's Faire :: math thread (by c_man)
2020-12-23T17:49:31+00:00405479the tiling one is interesting. the results for 2d are well enough known and there are actually often interesting physics involved in the difference between processes that produce square vs hexagonal tilings. for example, soap films will never produce rectangular tilings in 2d because four-fold vertices are unstable without external forces. however, this means that there are special cases, for example in quasi-2D tissues like skin, where external forces can stabilize rectangular tilings
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<br/>as for shannon, information theory is one of the great hopes for bridging quantum mechanics and general relativity, since both have ways of measuring information (in particular the information associated with black holes) that have to match up. this means that its not impossible that information theory ends up having a more important long term impact on the field than string theory!
Discussion :: Laissez's Faire :: math thread (by g7DKVpkTyXaZvn65)
2020-12-23T03:06:29+00:00405463studying a book-length popularization of shannon's paper, as well as tears' post about units, is what led to the power tower number system, that i've been blathering on about here. <a href="https://www.quantamagazine.org/how-claude-shannons-information-theory-invented-the-future-20201222/">https://www.quantamagazine.org/how-claude-shannons-information-theory-invented-the-future-20201222/</a>
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<br/>but <a href="https://www.quantamagazine.org/geometry-reveals-how-the-world-is-assembled-from-cubes-20201119/">this article</a> i would recommend reading over shannon's profile. much more interesting! by doing an experiment available to the ancients -- collecting rocks, and counting their sides and vertices -- the guys gave weight to plato's ancient notion that, this planet's stuff is, on average, cubic. 8 corners and 12 edges and 6 sides. of the 5 platonic solids, the cube is the one that plato linked to earth. plato was also enthralled with the number 5040. 5040 is 7!, 7*6*5*4*3*2. 5040 is highly antiprime -- many many divisors. he believed it useful to statecraft, and recommended dividing land area into 5040 parts, so as to be easily parceled into natural numbers.
Discussion :: Laissez's Faire :: math thread (by g7DKVpkTyXaZvn65)
2020-12-05T02:35:45+00:00405135</p><style type="text/css">.custom404323{color:#0B1645 !important; background-color:#BEEBCD !important; }</style><blockquote class="custom404323"><em><a href="/forum/post/404323/">toyot</a> posted:</em><br/><p class="postbody_text">i was trying to figure out, how many different numbers can you make, using such-and-so many bits? the hunch is that, maybe it's possible to make even more numbers than i first thought, if i spend bits grouping them with parentheses. </p></blockquote><p class="postbody_text">
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<br/>i looked into this. it is fewer than i thought. but i wasn't alone, all the textbooks i've read are wrong. they all say exponentiation, A^B^C^D^..., is non-associative (A^B)^C != A^(B^C), and non-communitative, B^A^C != A^B^C. but, i found many of examples of both by just making a large spreadsheet of every possible combination. i was surprised because this is something like finding an example in addition where 2+(5+6) != (2+5)+6. so these are counter-examples to a frequently-found statement, all over wikipedia for instance, that exponentiation is non-associative and non-commutative. it's non-assoc in a couple, and the default order of operations for exponentiation, going inside-out, made explicit as (A^(B^(C^D^...))), just happens to be the only non-comm way to write parentheses with exponents.
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<br/>semi-communitative for (((^)^)^)
<br/><a href="https://tinyurl.com/ycqq4h2y">(((A^B)^A)^A)</a> = (((A^A)^B)^A) = <a href="https://tinyurl.com/ybpdnud5">(((A^A)^A)^B)</a>
<br/>(((A^A)^B)^B) = (((A^B)^A)^B) = (((A^B)^B)^A)
<br/>(A^((A^B)^A)) = (A^((A^A)^B))
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<br/>semi-associative:
<br/>((A^A)^(A^A)) = ((A^(A^A))^A)
<br/>((B^B)^(A^A)) = ((B^(A^A))^B)
<br/>((A^B)^(A^A)) = ((A^(A^A))^B)
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<br/>there are lots of examples.
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<br/>associativity and communitivity are typical in exponentiation: most () combos have sister combos (for some reason, it's either 2, 3, 4, 6, or 12). all the textbooks and fuckin papers i've tried to find on this, say the power operation does not behave this way... of the 3-base (A,B,C) combinations, there are potentially 3870 unique numbers. if exponentiation were perfectly non-associative and non-communitative, you'd find 3870. but there are only 1785 uniques. for two-base, it's 548 potential, 284 actual. it becomes <em>more </em>associative/commutative, the more bases one includes. i'm hand-typing the calculations, so i'm not checking this to the infinite power tower, until i discover whatever algebraic identity is playing out here.
Discussion :: Laissez's Faire :: math thread (by second_axiom)
2020-12-02T06:17:13+00:00405057sad that joke got no upvotes. rip,,
Discussion :: Laissez's Faire :: math thread (by second_axiom)
2020-11-30T14:48:56+00:00405037tfw. wtf. ftw. whats trhe connecttion here? well, algebraically speaking, they are all moments in a cyclic permutation. you simply cant argue with these numbe
Discussion :: Laissez's Faire :: math thread (by zhaoyao)
2020-11-09T00:22:53+00:00404548particularly when the divisor is 1/2
Discussion :: Laissez's Faire :: math thread (by filler)
2020-11-08T20:56:50+00:00404537my understanding of math is that one divides into two
Discussion :: Laissez's Faire :: math thread (by sovnarkoman)
2020-11-08T09:15:16+00:00404521yeah two plus two is four minus one that s three
Discussion :: Laissez's Faire :: math thread (by drwhat)
2020-11-08T02:15:07+00:00404508did u guys figure out math yet. i was afk
Discussion :: Laissez's Faire :: math thread (by Acdtrux)
2020-11-07T05:39:34+00:00404452I have recently observed that, in nature, things don't contain each other. Most <a href="https://en.wikipedia.org/wiki/Tree_%28data_structure%29">bourgeois</a> <a href="https://en.wikipedia.org/wiki/Systems_theory">ideological</a> <a href="https://en.wikipedia.org/wiki/Actor%E2%80%93network_theory">formations</a> are fictitious.
Discussion :: Laissez's Faire :: math thread (by g7DKVpkTyXaZvn65)
2020-11-02T21:03:00+00:00404323ran into <a href="https://oeis.org/A000108">catalan numbers</a> for the first time yesterday. catalan numbers go like 1,2,5,14...
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<br/>i'll show the context. in that number system, you make numbers using 2 unique bases, like 2^ and 3^. i was trying to figure out, how many different numbers can you make, using such-and-so many bits? the hunch is that, maybe it's possible to make even more numbers than i first thought, if i spend bits grouping them with parentheses. parentheses cost bits, i wanted to see what the tradeoff was.
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<br/>people who know about catalan numbers probably already see where they popped up, but the answer is, they equal how many unique numbers can be built using this strategy.
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<br/>i'll use example numbers, a=2.0, b=2.1, c=2.2, d=2.3... and say there's no 'default' order for exponents (it's usually top to bottom), all operations must be defined w/ ( )
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<br/>1 power, 1 combo:
<br/>(a^b) = 2^2.1 = 4.287..
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<br/>2 powers, 2 combos:
<br/><a href="https://www.wolframalpha.com/input/?i=%282%5E2.1%29%5E2.2"> ((a^b)^c) </a>
<br/><a href="https://www.wolframalpha.com/input/?i=2%5E%282.1%5E2.2%29"> (a^(b^c)) </a> (default)
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<br/>3 powers, 5 combos:
<br/><a href="https://www.wolframalpha.com/input/?i=%28%28%282%5E2.1%29%5E2.2%29%5E2.3%29"> (((a^b)^c)^d) </a>
<br/><a href="https://www.wolframalpha.com/input/?i=%28%282%5E2.1%29%5E%282.2%5E2.3%29%29"> ((a^b)^(c^d)) </a>
<br/><a href="https://www.wolframalpha.com/input/?i=%28%282%5E%282.1%5E2.2%29%29%5E2.3%29"> ((a^(b^c))^d) </a>
<br/><a href="https://www.wolframalpha.com/input/?i=%282%5E%28%282.1%5E2.2%29%5E2.3%29%29"> (a^((b^c)^d)) </a>
<br/><a href="https://www.wolframalpha.com/input/?i=%282%5E%282.1%5E%282.2%5E2.3%29%29%29"> (a^(b^(c^d))) </a> (default)
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<br/>4 powers, 15 combos, which i'm not going to hassle with
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<br/>every set of ( & ) costs 10 bits, kind of a lot. i think, with no loss of information, since the start and ends of a power tower are well-defined, that i can cheat here, and use imbalanced parentheses. any imbalance is assumed to be 'made up', hanging off either end of the power tower. so like (((a^b)^c)^d) = a^b)^c)^d. can anyone see any ambiguities arising from this, if the start and end of the power tower are well-defined?
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<br/>here are 5 combinations written like this:
<br/>a^b)^c)^d
<br/>a^b)^(c^d
<br/>a^(b^c))^d
<br/>a^((b^c)^d
<br/>a^(b^(c^d
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<br/>just experimenting. here's a fractal formed by the complex power tower, z^z^z^z^z^z^z... yellow means it diverges, blue means the power series converges:
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<br/>bug playing the piano:
<br/><img class="postimg_inline" src="https://i.imgur.com/A4tEOXO.png"/>
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<br/>zoom in on bug:
<br/><img class="postimg_inline" src="https://i.imgur.com/4PWAgpY.png"/>
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<br/><img class="postimg_inline" src="https://i.imgur.com/8OrINuJ.png"/>
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<br/>bug head:
<br/><img class="postimg_inline" src="https://i.imgur.com/UX9KZ67.png"/>
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<br/>1 antenna:
<br/><img class="postimg_inline" src="https://i.imgur.com/tE6sQcy.png"/>
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<br/>hidden whirlpool:
<br/><img class="postimg_inline" src="https://i.imgur.com/ZsDJQo8.png"/>
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<br/>i put these two in the same post, because the <a href="https://maths.ucd.ie/%7Eplynch/Publications/PowerTower.pdf">author</a> only used one parentheses combination, to make that fractal. but catalan numbers get big quick. so that's only one of the possible fractals, there are a catalan #'s worth of unique fractal bugs playing piano (or whatever). all hidden this whole time in the power operation!
Discussion :: Laissez's Faire :: math thread (by second_axiom)
2020-10-15T02:53:41+00:00403583<img class="postimg_inline" src="https://i.imgur.com/0d5woWU.png"/>
Discussion :: Laissez's Faire :: math thread (by g7DKVpkTyXaZvn65)
2020-10-14T17:44:34+00:00403568oh, that's cool, you're right! <em>any </em>combination of 00 and 01 makes zero, so long as every 01 is eventually followed by 00 (the sequence ends in 00). so the total number of special characters possible in n bits is something combinatorial.
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<br/>testing for byte sequences, n=8:
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<br/>01 00 00 00 -*0+0+0+ = 0 or 1x -0 and 2x +0
<br/>01 01 00 00 -*-*0+0+ = 0 or 2x +0
<br/>01 00 01 00 -*0+-*0+ = 0 or 2x -0
<br/>01 01 01 00 -*-*-*0+ = 0 or 1x -0
<br/>00 00 00 00 (silent channel) or 4x 0+
<br/>00 01 00 00 0+-*0+0+ = 0 or 1x -0 and 2x +0
<br/>00 00 01 00 0+0+-*0+ = 0 or 1x -0 and 2x +0
<br/>00 01 01 00 0+-*-*0+ = 0 or 2x +0
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<br/>if -0 and +0 are differentiated like in programming languages, and counted as negative and positive tallys of zero, some counts of +0 and -0 are unique and some aren't. i think the count is basically equivalent to keeping a running count of 01 and 00?
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<br/>maybe obviously, it is basically binary, but with double the bits, an extra zero before each of binary's 0s and 1s. so the information content per bit, no matter what it encodes, will be exactly half of binary's then minus two end bits.
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<br/>so a 6-bit sequence has 4 listed there that end in 00, and the 8-bit has 8, and a 10-bit appends either 00 or 01 to all the entries in the table there, doubling it.
<br/>6 makes 4, 2^2
<br/>8 makes 8, 2^3
<br/>10 makes 16, 2^4
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<br/>looking at the table, then the number of possible special characters is 2^(n/2 - 1). to make the alphabet and some punctuation, 32 characters 2^(n/2-1) = 32 when n=12. this formula matches the intuitive reasoning above that it's basically half of binary, minus two bits. if 0000 is taken to mean a comma then that greatly reduces the character reserve of this system.
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<br/>my imagination is butting up against my ignorance, in terms of how an electronic computer would use this 3210 algebra. but i believe most operations would just happen in memory, applying algebraic rules to simplify long sequences of number. a computer wouldn't have to go to float, until it is asked to display numbers to a person, or talk to a regular computer. i also know that w/ quantum computers they can send an entangled bit, which is read first at the transmitter to decide what to send the receiver, something called superdense coding, if your language is in 2bit increments then maybe it's a good fit. but again i don't know what i'm talking about here.
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<br/>another interesting aspect is error correction, identifying bitflips. because numbers aren't one-to-one with the 0 and 1 sequence, the system can have canonical representations of each number (for instance, largest elements first, or using a common signpost). here are two different ways to write 9: 3^2^2^0+ and 2^3^2^0+2^0+, 3^2 = 9, and 2^3+1 = 9. if a bitflip produces the less-efficient way of storing 9, then the system could potentially detect it, by comparing it against a table. in our computers and networks, we group bits into blocks and packets and do checksums etc on the set, this would be a different strategy. a checksum in this system might be more complex, since it spans more than the integers, but it is potentially still possible.
Discussion :: Laissez's Faire :: math thread (by Acdtrux)
2020-10-14T05:38:05+00:00403565toyot:
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<br/>this is really interesting I'll have to think about the possibilities more in the future. One thought:
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<br/></p><style type="text/css">.custom403560{color:#0B1645 !important; background-color:#BEEBCD !important; }</style><blockquote class="custom403560"><em><a href="/forum/post/403560/">toyot</a> posted:</em><br/><p class="postbody_text">2 dimensions of infinity to write special characters using both the additive and multiplicative identities. but they are not very bit-efficient, like writing numbers by tally mark. </p></blockquote><p class="postbody_text">
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<br/>in this case instead of using tally marks you can intermingle the modes of encoding complicated zeros in order to represent binary codes, making the number of awkward sequences logarithmic rather than tally (linear). I.e. -1*-1*0+-1*0+-1*-1*0 is the encoding of special character 5.
Discussion :: Laissez's Faire :: math thread (by cars)
2020-10-14T05:17:52+00:00403563I wish factorials would calm down