September 2006 - mathfun@mailman.xmission.com

[math-fun] continued subscripts
by Marc LeBrun 10 Nov '06

10 Nov '06

Recent correspondence has inspired me that, analogous to stacking superscripts for iterated exponentiation, we may stack the subscripts that are vulgarly used as "base" notation. For example, we may apprehend, among 239's many sterling properties, the following gratuity: 10 11 12 13 14 15 16 = 239. 17 18 19 20 21 22 23 That is, working up from the bottom: ... ... ... 21 = 21 = 97 and so on = 239. 22 48 23 Question: If stacking superscripts produces a "tower", what are stacked subscripts? A "dungeon"? The sequence following this incrementing pattern of subscripts, namely, 10, 10, 10, 10, ... 11 11 11 12 12 13 has (if I've computed them correctly) the values 10 11 13 16 20 25 31 38 46 55 65 87 135 239 463 943 1967 4143 8751 18479 38959 103471 306223 942127 2932783 9153583 28562479 89028655 277145647 861652015 2675637295 10173443119 41132125231 168836688943 695134284847 2862242032687 11778342480943 48433422101551 199016451894319 817199416306735 3353334654921775 16351027752823855 82924577766468655 423717750455364655 2167310727003204655 11083411175259204655 56654591244123204655 289463880726363204655 1478277273827163204655 7546178967779163204655 38504861079779163204655 227971995605219163204655 1387065053878499163204655 8475364910242019163204655 51807613088766179163204655 ... (which I will submit to the OEIS as soon as I can craft a concise explanation!). Puzzle: the least-significant digits seem to be stabilizing ... why or why not? Of course we can write the numerals in the stack in other bases, such as binary: 10, 10, 10, 10, ... 11 11 11 100 100 101 and these have as their first few values 2 3 5 26 1370 9840770 851566070376026 ... but it grows rapidly--the next two terms have ~144 and ~450 digits respectively (and seem beginning to show a distinct tendency to end with the digits ...26 ?!) Question: What is a good approximate formula for a(n)? And of course other continued subscript patterns besides simply incrementing the terms are possible.

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[math-fun] Decimal question
by David Wilson 30 Sep '06

30 Sep '06

Is there a finite decimal that uses the same digits as its reduced fractional representation?

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RE: [math-fun] A.I. question re ongoing chess championship match
by Daniel Asimov 29 Sep '06

29 Sep '06

Andy writes: << [I wrote]: > A.I. QUESTION: Is there a reasonable way, based on Kramnik's play > in previous games, to distinguish moves that he conceived solely on > his own from ones that are so brilliant that they were probably > computer-aided ? I don't think so. There could be evidence that Kramnik is *not* using computer assistance, if he's making fatal mistakes that a computer would not make. But if Kramnik were using a computer, and he knows there's a chance of getting caught, and he's smart, he won't just do whatever the computer recommends, because that might risk detection. Instead, he could just play his own game except in situations in which he felt that he had a close decision between two alternatives. In such a case, he could let the computer decide which play was better, and make that play. This would result in play indistinguishable to any human from Kramnik's play, that would nonetheless gain an advantage from use of the computer. >> Someone as smart as a top chessplayer would realize that even with illicit access to a good chess computer for use during a match, it can't be too obvious, and this alone may explain numerous suboptimal moves Kramnik has made. I don't want to believe he's cheating, but I must admit that even for fairly acute gastronomic distress or an overload of diuretics, 50 bathroom visits per game, if true, is suspicious. Even more so now that he's said he will not continue after today's game (which he lost, leaving the score at Kramnik 3, Topalov 2), because he has been barred from continued use of the players' private rest room, which is unmonitored. Seems to me Kramnik should be going out of his way to demonstrate that he is not using any kind of chess computer or access to one. --Dan

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RE: [math-fun] A.I. question re ongoing chess championship match
by Andy Latto 29 Sep '06

29 Sep '06

> -----Original Message----- > From: math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com]On > Behalf Of Daniel Asimov > Sent: Thursday, September 28, 2006 10:23 PM > To: math-fun > Subject: [math-fun] A.I. question re ongoing chess championship match > > > Right now there is an ongoing chess championship, which (as followers > of the sport will know) has been accepted by both major camps > as the determiner of the next "world champion". > > (The match is to be 12 games, and is played in Elista, located close > to the middle of nowhere. After 4 games, Topalov has 1 point and > Kramnik has 3 points -- 2 wins & 2 draws.) > > Topalov has just signed a complaint saying that Kramnik is averaging > 50 bathroom breaks per game -- and the private, players' bathroom > is not monitored in any way. Topalov wants monitoring. (Cf. > < http://www.nytimes.com/2006/09/29/crosswords/chess/29chess.html >.) > Evidently Topalov is concerned Kramnik may be accessing a chess > computer for advice. > > A.I. QUESTION: Is there a reasonable way, based on Kramnik's play > in previous games, to distinguish moves that he conceived solely on > his own from ones that are so brilliant that they were probably > computer-aided ? I don't think so. There could be evidence that Kramnik is *not* using computer assistance, if he's making fatal mistakes that a computer would not make. But if Kramnik were using a computer, and he knows there's a chance of getting caught, and he's smart, he won't just do whatever the computer recommends, because that might risk detection. Instead, he could just play his own game except in situations in which he felt that he had a close decision between two alternatives. In such a case, he could let the computer decide which play was better, and make that play. This would result in play indistinguishable to any human from Kramnik's play, that would nonetheless gain an advantage from use of the computer. Andy Latto andy.latto(a)pobox.com

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[math-fun] Re: inconvenient directed graphs
by James Propp 29 Sep '06

29 Sep '06

I also posted my question to the domino forum (see jamespropp.org/about-domino) and got (1) a solution by Hugh Thomas and (2) a reference to the literature: H. Wielandt: Unzerlegbare, nicht negative Matrizen, (Engl.: Irreducible non-negative matrices) Math. Z. 52 (1950) 642-648. The optimal (or pessimal, depending on your point of view) matrices look like [0 + 0 0] [0 0 + 0] [0 0 0 +] [+ + 0 0] (as Dan Asimov observed). Thanks to all who responded (on- or off-line)! Jim Propp

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[math-fun] Tetrahedron puzzle
by Daniel Asimov 29 Sep '06

29 Sep '06

Let 0 < p < 2 and consider the graph of y = x^p in the xy-plane as well as the graph of z = y^q in the yz-plane for any q. In particular, consider the normal line to y = x^p at x=c, and the triangle T(p,c) that it makes with the nonnegative x- and y-axes. At the vertex (0,y0) of T(p,c) on the positive y axis, evaluate the function z = y^q and contruct a z-altitude A of height (y0)^q that's points parallel to the positive z-axis. Finally, let V(p,q,c) be the volume of the tetrahedron with T(p,c) as base and altitude A. Assume q = q(p) is chosen so that lim as c -> 0+ of V(p,q,c) exists and is positive. Call this limit L(p). PROBLEM: Find the maximum value of L(p) over all p in (0,2) to three decimal places. --Dan

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[math-fun] Q-factorials
by Schroeppel, Richard 29 Sep '06

29 Sep '06

I've been playing with Q-factorials: QF(0) == 1 QF(N+1) == QF(N) * (1 + Q + Q^2 + ... + Q^N). When Q=1, this gives QF(N) = N!. The degree of QF(N) is Triangle(N-1) = N(N-1)/2. Q-binomials seem to be polynomials: QB(A,B) = QF(A+B) / (QF(A) QF(B)). The degree of QB(A,B) is AB. The degree of QB(A,B,C) = QF(A+B+C) / (QF(A)QF(B)QF(C)) is AB+AC+BC. Etc. Perhaps we can use some of our standard bag of tricks for defining QF(N) for non-integer N. Put QF(Z) ~= Q ^ [ Z(Z-1)/2 ] for RealPart(Z) large, and divide by (Q^Z-1)/(Q-1) to go from Q(Z) to Q(Z-1). There's a set of functional equations similar to X! (X+1/2)! = (2X+1)! * stuff. Presumably there's also a reflection equation with QF(Z) QF(-Z) = more-stuff that looks like pi Z / sin(pi Z). Rich

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[math-fun] A.I. question re ongoing chess championship match
by Daniel Asimov 29 Sep '06

29 Sep '06

Right now there is an ongoing chess championship, which (as followers of the sport will know) has been accepted by both major camps as the determiner of the next "world champion". (The match is to be 12 games, and is played in Elista, located close to the middle of nowhere. After 4 games, Topalov has 1 point and Kramnik has 3 points -- 2 wins & 2 draws.) Topalov has just signed a complaint saying that Kramnik is averaging 50 bathroom breaks per game -- and the private, players' bathroom is not monitored in any way. Topalov wants monitoring. (Cf. < http://www.nytimes.com/2006/09/29/crosswords/chess/29chess.html >.) Evidently Topalov is concerned Kramnik may be accessing a chess computer for advice. A.I. QUESTION: Is there a reasonable way, based on Kramnik's play in previous games, to distinguish moves that he conceived solely on his own from ones that are so brilliant that they were probably computer-aided ? --Dan

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Re: [math-fun] Graph Question
by Michael Reid 29 Sep '06

29 Sep '06

> Is there a name for the operation which takes one graph to another, > such that the points of the new graph are the edges of the original > graph, and there is an edge between two such points if their > originating edges shared a point? I.e., each edge becomes a point, and > a point with valence n becomes C(n,2) edges. i believe the resulting graph is called the _line_graph_ of the original graph. i do not know if there's a specific term for the *operation* involved. mike

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[math-fun] Faro shuffling with jokers
by Marc LeBrun 29 Sep '06

29 Sep '06

It's fairly well-known that perfectly Faro shuffling a standard deck of playing cards 8 times will return them to their original order. For some reason I happened to compute Sloane A123320, which gives a table for different size decks cut at different places. I noticed that if you add two jokers to the deck and cut evenly the number is 52. If however you cut the regular deck evenly and place the jokers on top of the pile whose top card winds up on top it's still 8. On the other hand, if you do the same but put them under the pile whose bottom card winds up on the bottom it's 380. I can't think of any use for this factoid--other than maybe one magician hacking another (it'd require marks that can Faro shuffle!)--but perhaps it's still fun enough to share here...

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