August 2020 - mathfun@mailman.xmission.com

Re: [math-fun] MoMath's Mindbenders for the Quarantined
by Keith F. Lynch 28 Aug '20

28 Aug '20

Mike Beeler <mikebeeler2(a)gmail.com> wrote: > Voted most favorite puzzle (24% of votes) was ?Factorials and > Squares?: from the product 1! * 2! * 3! * ? * 100!, remove one > ?term? K!, to make the result a perfect square. Attributed to > Jeremy Kun. I just brute-force solved it, but the unique solution gives me no insight. I suspect that there's a symmetry argument which I'm missing. I don't know how many decimal digits the product of all the factorials through 100 has, but the last 1124 of them are zeros. And the last 1112 digits of the perfect square are zeros. (Of course the latter must be an even number.) I reinvented A248663 in the process. Note that in the digest option of this list, all non-ASCII characters are turned into question marks.

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[math-fun] D120 claim
by Bill Gosper 28 Aug '20

28 Aug '20

The roundest polyhedron in PolyhedronData: In[86]:= MaximalBy[# | (# | N@# &@Min@PolyhedronData[#, "DihedralAngles"]) & /@ PolyhedronData[], #[[2, 2]] &] // tim During evaluation of In[86]:= 0.537811 (* seconds *),1 (* winner *) Out[86]= {"DisdyakisTriacontahedron" | (ArcCos[1/241 (-179 - 24 Sqrt[5])] | 2.87783661046122)} {. . . | Sharpest dihedral} In[87]:= Labeled[PolyhedronData@%[[1, 1]], %[[1, 1]]] Out[87]=DisdyakisTriacontahedron <http://gosper.org/D120.png> In[91]:= PolyhedronData@120 Out[91]= {"DisdyakisTriacontahedron", "IcosahedronSixCompound", {"IcosahedronStellation", 3}, {"IcosidodecahedronStellation", 1}} In[92]:= Tally[PolyhedronData[%[[1]], "DihedralAngles"]] Out[92]= {{ArcCos[1/241 (-179 - 24 Sqrt[5])], 180}} claims that all 180 edges (60 short, 60 medium, 60 long) have this same angle. Is this obvious? In[95]:= PolyhedronData[%91[[1]], "AlternateNames"] Out[95]= {"28\[Hyphen]uniform dual polyhedron", "hexakis icosahedron"} —rwg

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Re: [math-fun] Orthodiagonal quadrilaterals
by Dan Asimov 26 Aug '20

26 Aug '20

Yeah, that became clear as soon as I actually thought about it instead of shooting my mouth off. —Dan Gene Salamin wrote: ----- Start by drawing the diagonals, two arbitrary orthogonal line segments. -----

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Re: [math-fun] Orthodiagonal quadrilaterals
by Dan Asimov 26 Aug '20

26 Aug '20

Okay, second guess: Aren't those just the rhombi and kite-shapes? ----- Orthodiagonal Quadrilateral: A quadrilateral whose diagonals cross at right angles. -----

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Re: [math-fun] Orthodiagonal quadrilaterals
by Dan Asimov 26 Aug '20

26 Aug '20

Y'mean those aren't just the rhombi? Rheally? —Dan Colin Wright wrote ----- ... ... Orthodiagonal Quadrilateral: A quadrilateral whose diagonals cross at right angles. -----

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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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