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[math-fun] Orthodiagonal quadrilaterals
by Colin Wright 26 Aug '20

26 Aug '20

I've been sent a paper by an amateur mathematician showing a link between a perfect Euler brick, and orthodiagonal quadrilaterals. Is a connection between these well-known? Obvious? Or is this of at least some interest, even if it's just a simple observation? As always, if it's new or at all significant, the mathematician in question is keen to have their name attached to it, but can't get anyone interested, so I'd like to know people's reactions to the idea that these things are related. To save you looking them up: Perfect Euler brick: a cuboid with integer sides, face diagonals, and long diagonal. Orthodiagonal Quadrilateral: A quadrilateral whose diagonals cross at right angles. Best regards, Colin -- The power of accurate observations is commonly called cynicism by those who haven't got it. -- George Bernard Shaw

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Re: [math-fun] MoMath's Mindbenders for the Quarantined
by Dan Asimov 26 Aug '20

26 Aug '20

Thanks! I've felt allergic to getting myself on too many e-mail lists since they can generate large volumes of e-mail, but maybe this one is safe. And thanks for posting several of the top-voted puzzles, and the link to the factorial-square discussion on StackExchange. —Dan Mike Beeler wrote: ----- Yes, they, like all nonprofits, want donations. But they aren’t pushy. I find them a worthy cause. There were “chat” requests on the Zoom meeting suggesting they put the puzzles online. At least one was from Winkler’s books. He is working on another book, that will include (some? all?) of these MoMath ones. Many of the puzzles are from other sources. ... ... -----

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Re: [math-fun] MoMath's Mindbenders for the Quarantined
by Dan Asimov 26 Aug '20

26 Aug '20

I don't know, this seems a bit dogmatic. Maybe it's true on average for random things posted on random websites, but for interesting math problems, that's not so obvious. E.g., Stan Wagon's POW (Problem of the Week) on the Macalester College website was well-known to problem enthusiasts. The word spreads fast among math fans. —Dan ----- Dan said: > From: Dan Asimov <dasimov(a)earthlink.net> > Subject: Re: MoMath's Mindbenders for the Quarantined > > There's no reason anyone should have to > "sign up" when MoMath could easily just > post the problems on their website. There's a huge difference. If you post stuff on your website, you are relying on people coming to visit, and there's a clear drop-off of people doing that. On the other hand, if you deliver things to people's inbox, that reminds them of your existence, and potentially prompts them to come and visit the website, as well as providing the material in question. Simply posting stuff on a website does not work. Email lists do work. It's as simple as that. -----

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Re: [math-fun] MoMath's Mindbenders for the Quarantined
by Colin Wright 26 Aug '20

26 Aug '20

Dan said: > From: Dan Asimov <dasimov(a)earthlink.net> > Subject: Re: MoMath's Mindbenders for the Quarantined > > There's no reason anyone should have to > "sign up" when MoMath could easily just > post the problems on their website. There's a huge difference. If you post stuff on your website, you are relying on people coming to visit, and there's a clear drop-off of people doing that. On the other hand, if you deliver things to people's inbox, that reminds them of your existence, and potentially prompts them to come and visit the website, as well as providing the material in question. Simply posting stuff on a website does not work. Email lists do work. It's as simple as that. Best regards, Colin -- The power of accurate observations is commonly called cynicism by those who haven't got it. -- George Bernard Shaw

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[math-fun] [SOLUTION] Combinatorics of shell substitution
by Robin Houston 26 Aug '20

26 Aug '20

Since Tomas has now found the answer, perhaps it’s time for me to reveal my method. If you’re still thinking about it, and you don’t want a spoiler, don’t read on! . . . . SPOILER BREAK . . . . The reason this problem is not amenable to naive brute force is that the strings explode in length very rapidly. So the problem is really to find a more manageable representation of these strings. What I did – which I believe is the natural representation in a sense – is to represent them as **elements of the group ring of the infinite dihedral group**. Don’t worry, I shall explain… Notation: I shall write ‘$x’ to refer to the two-character string consisting of a dollar sign followed by the letter x, and I shall write $x to refer to the value of the shell variable x. If there were no quotation marks, and we were to start, more simply, with: x=\$x\$x\$x then each time we execute eval x=$x the number of occurrences of ‘$x’ would be squared, and so the sequence would be 3, 9, 81, ... with the (n+1)’st term equal to 3^2^n. This is therefore an upper bound on the true sequence. That is not a great deal of use by itself, but it does suggest that the growth rate of the true sequence could be doubly-exponential – and certainly that it is no greater than that. The complication we need to take account of in the real sequence is that occurrences of ‘$x’ that appear in single quotes inside $x will not be substituted. If $x contains α occurrences of ‘$x’ that are in single quotes and β that are not, then the next value, obtained after executing eval x=$x will contain α + β(α + β) occurrences of ‘$x’. For example, the initial value has α=1 and β=2, hence the second value of the sequence is 7. In order to do this repeatedly, however, we need to know how many of the occurrences of ‘$x’ in the *new* value of $x occur within single quotes. And we must retain enough information that we can do this as many times as we need. It would suffice to know, for each occurrence of ‘$x’ in $x, the sequence of ' and " characters to the left of that occurrence. Strings over a two-character alphabet are, of course, elements of the free monoid on two generators, which I’ll denote 2*. So it would be sufficient to represent each value of $x as a set of elements of 2*. This is not a useful representation, yet, since it’s larger than just storing the whole string; but we shall improve it. To determine whether an occurrence of ‘$x’ should be subject to substitution, we need to know whether it is inside single quotes. There is a (surjective) homomorphism of monoids from 2* to the symmetric group S_3 that maps ' to the permutation (0 1) and " to the permutation (0 2). I shall refer to the image under this homomorphism of some string of quotation marks as the “quotational effect” of that string. To determine whether an occurrence of ‘$x’ is inside single quotes, we may take the string of quote marks to the left of that occurrence and compute the element of S_3 that it maps to under this homomorphism. Consider this permutation as a bijection π: {0,1,2} → {0,1,2}. If π(0) = 1 then the occurrence is inside single quotes; if π(0) = 2 then it is inside double quotes; and if π(0) = 0 then it is not inside quotes at all. So, to determine whether an occurrence of ‘$x’ should be substituted we need only check whether or not π(0) = 1. The element of 2* does, however, contain information that is irrelevant to us. For example, the strings $x ''$x ""$x "''"$x (and so forth) are all equivalent for our purposes, since the quotational action is all fully resolved before the occurrence of ‘$x’ has been reached. On the other hand, we cannot go all the way to elements of S_3, for these do not contain enough information. For example, the occurences of ‘$x’ in the strings $x '"'"'"$x"'"'"' have the same quotational effect – the identity element of S_3 – but after one round of evaluation this is no longer the case: the second one has an additional "' preceding the substituted value of $x, which has a non-trivial quotational effect (0 2 1). Therefore a suitable quotient is to ignore paired quotes, i.e. to take the quotient monoid 〈 a, b | a^2 = 0, b^2 = 0 〉 – where the generators a and b represent the characters ' and " respectively. This quotient monoid is in fact a group: the infinite dihedral group D_∞. Our “quotational effect” homomorphism 2* → S_3 factors through this quotient, so we have 2* → D_∞ → S_3. Therefore a suitable representation for our strings is as multisets of elements of D_∞, which we can more conveniently think of as elements of the group ring of D_∞ (over the integers, say). Since D_∞ is the group of isometries of the integers, we may represent an element of D_∞ as a pair (t, r), where t ∈ ℤ and r ∈ {+1, -1}. The action of such a pair on the integers is z ↦ r(z + t): “t” stands for “translate”, and “r” for “reflect”. Our generators a and b will be represented by (0, -1) and (-1, -1), i.e. z ↦ -z and z ↦ 1 - z, respectively. Composition is: (t1, r1) ; (t2, r2) = (t1 + r1 × t2, r1 × r2). Now we need only work out how to compute, for each element of D_∞: * which element of D_∞ it becomes after evaluation; and * whether the occurrences of ‘$x’ it represents should be substituted. It is not too hard to see that (t, r) reduces to (⌈t/3⌉, r'), where r' depends only on r and the value of t modulo 3. In fact, reducing t modulo 3 is essentially the “quotational effect” homomorphism, so we can also decide whether or not to substitute these occurrences of ‘$x’ from the values of r and (t mod 3). The mapping from (r, t mod 3) to (r', substitute?) is: (0, +1) ↦ (+1, True) # () (1, +1) ↦ (-1, False) # (0 1 2) (2, +1) ↦ (-1, True) # (0 2 1) (0, -1) ↦ (+1, False) # (0 1) (1, -1) ↦ (-1, True) # (1 2) (2, -1) ↦ (+1, True) # (0 2) where I have also annotated each line with the relevant quotational effect. All of this means that the following Python (3) program should print out the values of the sequence: import math TABLE = { (0, +1): (+1, True), # () (1, +1): (-1, False), # (0 1 2) (2, +1): (-1, True), # (0 2 1) (0, -1): (+1, False), # (0 1) (1, -1): (-1, True), # (1 2) (2, -1): (+1, True) # (0 2) } def compose(tr1, tr2): (t1, r1), (t2, r2) = tr1, tr2 return (t1 + r1*t2, r1*r2) def subst(x, y): z = {} for ((t0, r0), n0) in x.items(): # Let tr1 be the dequoted version of (t0, r0) t1 = -math.ceil(t0 / 3) r1, should_substitute = TABLE[t0 % 3, r0] tr1 = (t1, r1) # and substitute where appropriate if should_substitute: for (tr2, n2) in y.items(): tr3 = compose(tr1, tr2) z[tr3] = z.get(tr3, 0) + n0 * n2 else: z[tr1] = z.get(tr1, 0) + n0 return z x = { (-1, +1): 1, (0, +1): 1, (+1, +1): 1 } print(sum(x.values())) for i in range(10): x = subst(x, x) print(sum(x.values())) Indeed, running the program produces: 3 7 37 973 642493 274981436065 50075673362453978752897 1660424958340272612902157434339071057416705025 1828659082532127294015555219377679316426731045207698280944194119017937814697416523736342529 2221932973262746587938621327316508263261099068744811444411319257981535885894143217671769171584514785091039842436855683166325319643148964213893272271303784306333684832769323249729537 3284840218479775743047790632317572154909424991433212164298865481805700811352338269615571535863546774083036854200912305419781334801792798527519528263346227635678885261969168464305471249372299734845553556695985956938868825596778725419448920773764852977163215319470541619936785213643936937876887597557533802625700859018364222540571399561867801071781407682811002881 where the last line is the answer to the puzzle, and matches what Tomas found with a different method. Cheers, Robin

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Re: [math-fun] MoMath's Mindbenders for the Quarantined
by Dan Asimov 26 Aug '20

26 Aug '20

There's no reason anyone should have to "sign up" when MoMath could easily just post the problems on their website. I suspect they just want a bunch of people they can hit up for money. Peter Winkler has written at least two books of mathy puzzles, which I own. Can it be that these puzzles are from those? Could you (Mike) post a few of the best ones to math-fun? —Dan ----- I think MoMath’s weekly puzzle series hasn’t been mentioned on math-fun. In case you want a nice weekly puzzle, consider subscribing (it’s free). They range from numerical, to logical, to somewhat geometric. None require any math beyond algebra, and they each involve some kind of “aha!” insight, explained in the solution sent 6 days later. -----

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Re: [math-fun] Another Lex. Earliest Sequence: Scott Shannon's "Enots Wolley sequence A336957
by Neil Sloane 26 Aug '20

26 Aug '20

if a(4) = 9 then a(5) does not exist. therefore a(4) > 9 Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com On Tue, Aug 18, 2020 at 2:47 AM Frank Stevenson < frankstevensonmobile(a)gmail.com> wrote: > Hi Niel, > > I was trying to compute this sequence, but I am having problems > understanding the concept of lexicographical earliest. Why is not 9 the 4th > number in the sequence, but 15 is ? Why not 105, 1005, 100005 etc ? What am > I missing ? > > Regards, > Frank > > > On Sun, Aug 16, 2020 at 5:56 AM Neil Sloane <njasloane(a)gmail.com> wrote: > >> Obviously this is an inverted version of the Yellowstone sequence A098550 >> ! >> The name Enots Wolley is for personal use only, it must not be mentioned >> in >> the OEIS! We frown on such made-up names. >> >> Definition: Lexicographically earliest sequence {a(n)} of distinct >> positive >> numbers such that, for n>2, a(n) has a common factor with a(n-1) but not >> with a(n-2). >> 1, 2, 6, 15, 35, 14, 12, 33, 55, 10, 18, 21, 77, 22, 20, 45, 39, 26, 28, >> 63, ... >> The original idea was due to Scott, with a different sequence, but this is >> my (canonical!) version. >> >> Could someone please prove the conjecture that this is a permutation of >> the >> set {1, all numbers with at least two distinct prime factors} ? >> >> I can't even prove that every number 2*p (p prime) appears, or that there >> are infinitely many even terms (although I've found a dozen false proofs). >> It's a slippery problem. >> >> Neil >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >

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[math-fun] MoMath's Mindbenders for the Quarantined
by Mike Beeler 26 Aug '20

26 Aug '20

I think MoMath’s weekly puzzle series hasn’t been mentioned on math-fun. In case you want a nice weekly puzzle, consider subscribing (it’s free). They range from numerical, to logical, to somewhat geometric. None require any math beyond algebra, and they each involve some kind of “aha!” insight, explained in the solution sent 6 days later. Last night they had a “Best of Mindbenders” meeting via Zoom. People voted for their most favorite, least favorite, hardest, and most poorly defined puzzle. There have been 23 puzzles so far, and Peter Winkler thinks he has enough more to last until the pandemic is over. The MoMath lady who introduced the meeting said they have 9500 subscribers in many different countries. These puzzles are probably back-of-the-envelope (if not do-in-your-sleep) for math-funsters, but many have generalizations that are non trivial. — Mike > Begin forwarded message: > > From: "National Museum of Mathematics" <functions(a)momath.org <mailto:functions@momath.org>> > Subject: Mind-Benders for the Quarantined! > Date: March 20, 2020 at 12:03:01 PM EDT > To: "Mr. Mike Beeler" <mikebeeler2(a)gmail.com <mailto:mikebeeler2@gmail.com>> > Reply-To: "National Museum of Mathematics" <functions(a)momath.org <mailto:functions@momath.org>> > > Dear MoMath friends, > MoMath is pleased to announce a new program to keep your minds fresh! Sign up now for Mind-Benders for the Quarantined! at mindbenders.momath.org <https://momath.org/wp-content/plugins/civicrm/civicrm/extern/url.php?u=2427…> and each Sunday, MoMath will send you a challenging mathematical puzzle from the collection of our own puzzle master, Dr. Peter Winkler. On Tuesday, you'll receive a subtle hint; on Thursday, a serious push; on Saturday, the solution. And the next day, of course, a new puzzle. > These puzzles are for your edification, enjoyment, and personal satisfaction. If you wish, however, you may send your answers and/or complaints to mindbenders(a)momath.org <https://momath.org/wp-content/plugins/civicrm/civicrm/extern/url.php?u=2428…> at any time. Dr. Winkler won't be able to reply, but he will be happy to benefit from your comments, and will announce a winner (for the most right answers with the least help) when the epidemic is over. > Regards, > National Museum of Mathematics > > Support MoMath at momath.org/contribute <https://momath.org/wp-content/plugins/civicrm/civicrm/extern/url.php?u=2429…> > momath.org <https://momath.org/wp-content/plugins/civicrm/civicrm/extern/url.php?u=2430…> > 212-542-0566 > 11 E 26th St > New York, NY 10010 > United States

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[math-fun] Combinatorics of shell substitution
by Robin Houston 25 Aug '20

25 Aug '20

I posted the below puzzle to Twitter, where no one has solved it. I think it’s too hard. Although it’s a silly question, the method of solution is quite interesting IMHO. (The computation would be doable by hand, if one were patient and careful, but can be done more easily and reliably by a computer.) I know some of you are aware of a related problem I solved a few years ago. The method of solution of this one is related, but not identical. Cheers, Robin A Unix shell puzzle. If you were to run this script – which I don’t recommend doing – how many occurrences of $x would appear in the output? x=\"\'\$x\'\"\$x\'\"\$x\"\' eval x=$x eval x=$x eval x=$x eval x=$x eval x=$x eval x=$x eval x=$x eval x=$x eval x=$x eval x=$x echo "$x"

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Re: [math-fun] Legendary Chalk
by Dan Asimov 25 Aug '20

25 Aug '20

The article, with lines like ----- How did a brand of chalk develop a cult-like following? ----- seems to be fact-free. (I always used to buy An-Du-Septic dustless chalk, which worked just fine. <https://shop.crayola.com/color-and-draw/an-du-septic-anti-dust-chalk-sticks…>.) I'm happy to use chalk on a blackboard or markers on a whiteboard, both of which seem to have gone out of style in favor of overhead or PowerPoint-type presentations. —Dan Hilarie Orman wrote: ----- Does anyone really use chalk anymore? If so, is Hagoromo really this good? https://www.cnn.com/2020/08/22/us/hagoromo-chalk-great-big-story-trnd/index… -----

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