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Re: [math-fun] Air traffic control
by Henry Baker 05 Dec '17

05 Dec '17

With modern "Graphics Processing Units" ("GPU's") having vast parallel floating point processing capabilities, it's easy to forget how hard it was to do real-time graphics programming -- even of the minimal line-drawing variety -- back in the early 1980's. I can well imagine that the simple "(x'-x)^2+(y'-y)^2+(z'-z)^2<1 mile" calculation would be too much for C(N,2) distances to compute for N airplanes. Many graphics operations first did *bounding box checks* prior to doing more expensive double precision multiplications. I can easily imagine doing such rectangular box checks in the early 1980's. Bounding using octagonal boxes (probably using 45 degree angles) instead of rectangular boxes (at least in the latitude/longitude dimensions, but probably not in the altitude dimension, where airplanes are *never* simultaneously allowed unless they're in a proper direction) could be done, and might still be slightly less expensive than a true ball radius calculation, but the advantage would seem to be minimal. At 06:44 PM 12/5/2017, James Propp wrote: >An air traffic controller in the 1980s told me that the antiquated software >that was then in use nationally approximated circles by polygons >(dodecagons I think, though maybe octagons). There was a rule saying that >two planes shouldn't be within a mile of each other, and the rule was >supposed to enforced by software (so that some sort of alarm would go off >if one plane was in the circular one-mile neighborhood of the other), but >the software used a polygonal approximation to those circles, more >precisely INSCRIBED circles, so it was possible for two planes to be >slightly less than a mile apart without setting off the alarm. He told me >that the controllers sometimes would arrange for this to happen for sport. > >Anyone know anything about this? > >My memory is unreliable, so I may have major details of the story wrong.

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[math-fun] Air traffic control
by James Propp 05 Dec '17

05 Dec '17

An air traffic controller in the 1980s told me that the antiquated software that was then in use nationally approximated circles by polygons (dodecagons I think, though maybe octagons). There was a rule saying that two planes shouldn't be within a mile of each other, and the rule was supposed to enforced by software (so that some sort of alarm would go off if one plane was in the circular one-mile neighborhood of the other), but the software used a polygonal approximation to those circles, more precisely INSCRIBED circles, so it was possible for two planes to be slightly less than a mile apart without setting off the alarm. He told me that the controllers sometimes would arrange for this to happen for sport. Anyone know anything about this? My memory is unreliable, so I may have major details of the story wrong. Jim Propp

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[math-fun] Rubik's cube: the base case
by Cris Moore 05 Dec '17

05 Dec '17

https://www.youtube.com/watch?v=LyjGhtba_zs <https://www.youtube.com/watch?v=LyjGhtba_zs> - Cris

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[math-fun] SET / Rubik mashup
by James Propp 05 Dec '17

05 Dec '17

Is this sort of hybrid between the games SET and the Rubik's cube a new thing? https://twitter.com/nomad_penguin/status/937810975025086464 Jim Propp

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Re: [math-fun] How do you ask a question in mathematical notation?
by Dan Asimov 03 Dec '17

03 Dec '17

When scribbling a math question to myself on a pad of paper, I'll often use notation like ∃? x in X ∋— [condition] [* In case it's illegible in your mail reader: i) the first symbol is a backwards capital E, a "there exists" symbol] ii) the symbol ∋— after the X looks like a pitchfork with the handle to the right. This means "such that". to mean "Does there exist a member of the set X such that [condition] holds?" where the such-that symbol ∋— is something I picked up somewhere from my professors, but haven't seen many other people using it. (To simulate it here I've used the "contains as a member" backwards epsilon special character followed by an em-dash (option-shift-plus on a Mac keyboard).) —Dan

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Re: [math-fun] Parenthesizing
by Dan Asimov 02 Dec '17

02 Dec '17

Suppose ^ means exponentiation and we want to count the *different integers* that 2^2^...^2 is equal to (N 2's in all) using various parenthesizations. E.g. #(2^2) = #{4} = 1 #(2^2^2) = #{16} = 1 #(2^2^2^2) = #{64, 256, 65536} = 3 #(2^2^2^2^2) = ??? Of course, any 3 twos grouped together make 4^2 = 2^4 = 16. Is there an asymptotic formula? Or better, an exact one? —Dan

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Re: [math-fun] Parenthesizing
by Dan Asimov 02 Dec '17

02 Dec '17

If the arrows are taken as implication, as in the predicate calculus, then Jim's question seems to make good sense to me. (But the question can of course be asked in general with other meanings assigned to the arrows.) —Dan ----- > Do all the Catalan-many parenthesizations of p_1 --> p_2 --> ... --> p_n > yield distinct Boolean functions? That doesn't quite parse. ... -----

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[math-fun] golden rectangle to ⃞
by Bill Gosper 01 Dec '17

01 Dec '17

"Annotated" dissection of Golden (or any) Rectangle to square: gosper.org/gold2square.gif Seven of the "swimming triangle" dissections (two very similar) of the half-rectangle to half-square: gosper.org/gold2halfsquare.gif Each triplet demonstrates the area formula A = ½bh for the intermediate (non-right) △ for two different choices of (b,h). (The middle school "proof" via a half-parallelogram fails if it's too skew to dissect into a rectangle in one cut.) --rwg Oops! Target #2 isn't Right! Ignore that one for now. If you're adventuresome, the notebook that made these, gosper.org/easyarea.nb was saved while running a ListAnimate of gold2halfsquare. If I load this nb and fail to pause the ListAnimate (near the end), it crashes Mathematica in a minute or two!

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[math-fun] muffin problem -- clusters and bounds
by Huddleston, Scott 01 Dec '17

01 Dec '17

I enclose examples of some of the bounds in my 28 Feb. post, and a derivation of the most involved example. S(12/7) = 3/7 = r/4 = (3-r)/3 Muffins: 1/7 * { [3,4]*12 } Persons: 1/7 * { [4*3]*4, [3*4]*3 } S(7/4) = 5/12 = (3-r)/3 = (2r-1)/6 Muffins: 1/12 * { [5,7]*6, [6*2] } Persons: 1/12 * { [5*3,6]*2, [7*3]*2 } S(40/23) = 19/46 = (2r-1)/6 = (12-5r)/8 Muffins: 1/46 * { [19,27]*24, [20,26]*12, [23*2]*4 } Persons: 1/46 * { [19*3,23]*8, [20*4]*3, [27*2,26]*12 } S(228/131) = 54/131 = (12-5r)/8 = (7r-6)/15 Muffins: 1/131 * { [54,77]*120, [55,76]*24, [57,74]*60, [65,66]*24 } Persons: 1/131 * { [54*3,66]*24, [54*2,55,65]*24, [57*4]*15, [77*2,74]*60, [76*3]*8 } S(449/258) = 319/774 = (7r-6)/15 = (73-32r)/42 Muffins: 1/774 * { [319,455]*210, [320,454]*30, [325,449]*42, [336,438]*30, [337,437]*90, [384,398]*42, [387*2]*5 } Persons: 1/774 * { [319*3,390]*42, [319*2,325,384]*42, [320*3,387]*10, [336,337*3]*30, [455*2,437]*90, [455,454,438]*30, [449*3]*14 } Here's a derivation of 449/258 example. 449 muffins -> 898 pieces, used in 124 4-parts with 496 pieces and 134 3-parts with 402 pieces. We split muffins between a 3-part and a 4-part when possible, but 4-parts contain 94 more pieces than 3-parts. We divide these 94 pieces (47 muffins) as evenly as possible among the 124 4-parts. 47 pairs of 4-parts contain both halves of a muffin, which I'll call an A^1 cluster and write A^1 = [a1*3,m]*2 where m denotes the two muffin halves and the other pieces are denoted a1, indicating a set of pieces of possibly different sizes that occure in A^1 clusters. 34 4-parts contain only muffin pieces whose mates appear in 3-parts. These I call A^0 clusters and write A^0 = [a0*4] The 449/258 example uses 30 A^0 clusters ([a0*4])*30 (120 a0 pieces, 30 4-parts) 47 A^1 clusters ([a1*3,m]*2)*47 (282 a1 pieces, 94 4-parts) Note that the A^1 cluster by itself sets an upper bound on S -- the cluster is 2 persons (size 2r), it cntains 1 complete muffin (size 1, pieces m*2), and it has 6 unmatched muffin pieces (a1*6), giving S(449/258) <= avg(a1) = (2r-1)/6. A^0 clusters also bound S by S(449/258) <= avg(a0) = r/4, but the A^1 cluster bound is tighter. Note that this (2r-1)/6 bound holds for examples 7/4 and 40/23. The muffin mates of a0 pieces (resp. a1 pieces) are z0 pieces (resp. z1 pieces). We distribute z0 and z1 pieces as evenly as possible across the 134 3-parts (402 pieces). With 282 z1 pieces, at least 14 3-parts must consist of only z1 pieces. These are Z^0 clusters, Z^0 = [z1*3] = [z10*3] where I refine piece labels so that z1 pieces appearing in Z^0 clusters are labelled z10. I create general Z^k clusters by finding minimal sets of persons that contain complete a0/z0 muffins. Maximally even distribution of the 120 z0 pieces places one per 3-part. Then a Z^1 cluster, closed under a0/z0 muffins, is Z^1 = [a0*4],[z0,z1*2]*4 = A^0,[z0,z11*2]*4 where z1 pieces appearing in Z^1 clusters are relabelled z11. Z^0 clusters bound S by S(449/258) <= avg(a10) = 1-avg(z10) = 1-r/3 = (3-r)/3 and Z^1 clusters provide an even tighter bound S(449/258) <= avg(a11) = 1-avg(z11) = 1-(5r-4)/8 = (12-5r)/8. The 449/258 example uses 14 Z^0 clusters = ([z10*3])*14 ( 42 z10 pieces, 14 3-parts) 30 Z^1 clusters = (A^0,[z0,z11*2]*4)*30 (240 z11 pieces, 120 3-parts) As before, we refine the labelling of a1 pieces into a10 and a11 pieces, according to z10 or z11 complements. The next level of clustering is constructed by taking closure under a10/z10 muffins. The 282 a1 pieces in the 47 A^1 clusters are divided into 240 a11 pieces and 42 a10 pieces. Thus at least 5 A^1 clusters must contain only a11 pieces, and with maximally even distribution we have 42 A^1 clusters with 1 a10 piece and 5 a11 pieces. Continuing to refine pieces according to their appearance in A^10 or A^11 clusters, we have A^10 = A^1 = {a110*6, ...} (6 a110 pieces and some other pieces) A^11 = Z^0, (A^1)*3 = (a111*15, ...) with 449/258 consisting of 5 A^10 clusters ( 5 A^1 clusters, 30 a110 pieces) 14 A^11 clusters (42 A^1 clusters, 252 a111 pieces) A^11 clusters provide bound S(449/258) <= avg(a111) = (7r-6)/15 The 449/258 example lies on the intersection of this bound and the next level cluster bound. The next cluster level takes a110/z110 muffin closure and induces clusters Z^10 = Z^1 Z^11 = A^10, (Z^1)*6 with 449/258 consisting of 0 Z^10 clusters 5 Z^11 clusters (i.e., only Z^11 clusters) and bound S(449/258) <= avg(a1111) = 1-avg(z1111) = (73-32r)/42. I conclude with some computer generated additional information on these bounds. If anyone sort of understands this post and would like to use my program as is (i.e., no manual and documentation is mostly my posts), drop me a note and I'll send the C++ source (300 lines). B<4>(r) = (1r-0)/4; S'(r) <= B<4>(r) on (3/2,2/1] tight bound on (3/2,8/5] and at r = (4+8k)/(2+5k) (3/2,8/5] contains mediant(3/2,8/5) = 11/7 S'( (3/2,8/5] ) = (3/8,2/5]; S'(11/7) = 11/28 B<>(r) = (1-0r)/2 B<4>(r) = B<>(r) at r = 2/1 = 2.0000000 S'(2/1) = 1/2 = 0.50000000 B<4,1>(r) = (3-1r)/3; S'(r) <= B<4,1>(r) on [12/7,2/1) tight bound on [9/5,2/1) and at r = (12+9k)/(7+5k) [9/5,2/1) contains mediant(9/5,2/1) = 11/6 S'( [9/5,2/1) ) = [2/5,1/3); S'(11/6) = 7/18 B<4>(r) = (1r-0)/4 B<4,1>(r) = B<4>(r) at r = 12/7 = 1.7142857 S'(12/7) = 3/7 = 0.42857143 B<4,1,1>(r) = (2r-1)/6; S'(r) <= B<4,1,1>(r) on (12/7,7/4] tight bound on (12/7,59/34] and at r = (21+59k)/(12+34k) (12/7,59/34] contains mediant(12/7,7/4) = 19/11 S'( (12/7,59/34] ) = (17/42,7/17]; S'(19/11) = 9/22 B<4,1>(r) = (3-1r)/3 B<4,1,1>(r) = B<4,1>(r) at r = 7/4 = 1.7500000 S'(7/4) = 5/12 = 0.41666667 B<4,1,1,1>(r) = (12-5r)/8; S'(r) <= B<4,1,1,1>(r) on [40/23,7/4) tight bound on [148/85,7/4) and at r = (80+148k)/(46+85k) [148/85,7/4) contains mediant(47/27,7/4) = 54/31 S'( [148/85,7/4) ) = [7/17,13/32); S'(54/31) = 51/124 B<4,1,1>(r) = (2r-1)/6 B<4,1,1,1>(r) = B<4,1,1>(r) at r = 40/23 = 1.7391304 S'(40/23) = 19/46 = 0.41304348 B<4,1,1,1,1>(r) = (7r-6)/15; S'(r) <= B<4,1,1,1,1>(r) on (40/23,228/131] tight bound on (40/23,1119/643] and at r = (228+1119k)/(131+643k) (40/23,1119/643] contains mediant(40/23,47/27) = 87/50 S'( (40/23,1119/643] ) = (142/345,265/643]; S'(87/50) = 103/250 B<4,1,1,1>(r) = (12-5r)/8 B<4,1,1,1,1>(r) = B<4,1,1,1>(r) at r = 228/131 = 1.7404580 S'(228/131) = 54/131 = 0.41221374 B<4,1,1,1,1,1>(r) = (73-32r)/42; S'(r) <= B<4,1,1,1,1,1>(r) on [449/258,228/131) tight bound on [9349/5372,228/131) and at r = (1347+9349k)/(774+5372k) [9349/5372,228/131) contains mediant(134/77,47/27) = 181/104 S'( [9349/5372,228/131) ) = [1107/2686,2267/5502); S'(181/104) = 75/182 B<4,1,1,1,1>(r) = (7r-6)/15 B<4,1,1,1,1,1>(r) = B<4,1,1,1,1>(r) at r = 449/258 = 1.7403101 S'(449/258) = 319/774 = 0.41214470 B<4,1,1,1,1,1,1>(r) = (61r-60)/112; S'(r) <= B<4,1,1,1,1,1,1>(r) on (449/258,764/439] tight bound on (449/258,149516/85913] and at r = (10696+149516k)/(6146+85913k) (449/258,149516/85913] contains mediant(449/258,764/439) = 1213/697 S'( (449/258,149516/85913] ) = (11909/28896,35408/85913]; S'(1213/697) = 32173/78064 B<4,1,1,1,1,1>(r) = (73-32r)/42 B<4,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1>(r) at r = 764/439 = 1.7403189 S'(764/439) = 2533/6146 = 0.41213798 B<4,1,1,1,1,1,1,1>(r) = (1101-487r)/615; S'(r) <= B<4,1,1,1,1,1,1,1>(r) on [160212/92059,764/439) tight bound on [1313469/754729,764/439) and at r = (160212+6567345k)/(92059+3773645k) [1313469/754729,764/439) contains mediant(19549/11233,764/439) = 20313/11672 S'( [1313469/754729,764/439) ) = [1555262/3773645,111271/269985); S'(20313/11672) = 986147/2392760 B<4,1,1,1,1,1,1>(r) = (61r-60)/112 B<4,1,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1,1>(r) at r = 160212/92059 = 1.7403187 S'(160212/92059) = 37941/92059 = 0.41213787

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Re: [math-fun] Parenthesizing
by Allan Wechsler 01 Dec '17

01 Dec '17

Rich politely shielded me from the jeers of the rest of the list in pointing out an "algebraic" identity that applies to 3 as it does to any number, namely: (a^a)^(a^a) = (a^(a^a))^a In general, if K and L are two different exponentiation-towers, then (a^K)^L = a^(KL) = (a^L)^K. This inspires two new questions: (1) Is this "KL law" the only algebraic law that exponentiation-towers obey in general, or are there other such general relations not inferrable from the KL law? (2) In particular for a=3, are there any equivalences that aren't true for general a? (Probably a careful reading of Guy and Selfridge [1973], cited by Neil earlier, would answer both questions.) On Wed, Nov 29, 2017 at 2:30 PM, <rcs(a)xmission.com> wrote: > (3^3)^(3^3) = (3^(3^3))^3 = 3^(3^4) = 3^81 . > The 3 and 3^3 in the exponent are multiplied, which commutes. > > Rich > > ----------- > > Quoting Allan Wechsler <acwacw(a)gmail.com>: > >> I got my head in a little twist over this whole thread. Reassure me: do >> all >> the different parethesizations of 3^3^3^...^3, for n 3's, give different >> values? The original problem (using principle values of a^b) is >> interesting >> because i is special, right? >> >> >> On Wed, Nov 29, 2017 at 9:46 AM, James Propp <jamespropp(a)gmail.com> >> wrote: >> >> Apropos of parenthesizing exponential towers, here's an easier question >>> (probably not new): Look at all the (Catalan-many) ways to parenthesize >>> a_1 >>> - a_2 - ... - a_n. Not all are distinct as linear functions of a_1, a_2, >>> ..., a_n; e.g., >>> >>> (a - (b - c)) - d = a - (b - (c - d)). >>> >>> How many different functions can be obtained? I don't know the answer. Is >>> it 2^(n-2)? >>> >>> >>> Jim Propp >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >>> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> > >

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