The Universe of Discourse

Sun, 16 Sep 2007

Thank you very much for that bulletin
I'm about to move house, and so I'm going through a lot of old stuff and throwing it away. I just unearthed the decorations from my office door circa 1994. I want to record one of these here before I throw it away and forget about it. It's a clipping from the front page of the New York Times from 11 April, 1992. It is noteworthy for its headline, which only one column wide, but at the very top of page A1, above the fold. It says:


Sometimes good articles get bad headlines. Often the headlines are tacked on just before press time by careless editors. Was this a good article afflicted with a banal headline? Perhaps they meant there was internecine squabbling among the diplomats charged with the negotiations?

No. If you read the article it turned out that it was about how darn hard it was to end the war when folks kept shooting at each other, dad gum it.

I hear that the headline the following week was DOG BITES MAN, but I don't have a clipping of that.

[Other articles in category /lang] permanent link

Sat, 15 Sep 2007

The Wilkins pendulum mystery resolved
Last March, I pointed out that:

  • John Wilkins had defined a natural, decimal system of measurements,
  • that he had done this in 1668, about 110 years before the Metric System, and
  • that the basic unit of length, which he called the "standard", was almost exactly the same length as the length that was eventually adopted as the meter
("John Wilkins invents the meter", 3 March 2006.)

This article got some attention back in July, when a lot of people were Google-searching for "john wilkins metric system", because the UK Metric Association had put out a press release making the same points, this time discovered by an Australian, Pat Naughtin.

For example, the BBC Video News says:

According to Pat Naughtin, the Metric System was invented in England in 1668, one hundred and twenty years before the French adopted the system. He discovered this in an ancient and rare book...
Actually, though, he did not discover it in Wilkins' ancient and rare book. He discovered it by reading The Universe of Discourse, and then went to the ancient and rare book I cited, to confirm that it said what I had said it said. Remember, folks, you heard it here first.

Anyway, that is not what I planned to write about. In the earlier article, I discussed Wilkins' original definition of the Standard, which was based on the length of a pendulum with a period of exactly one second. Then:

Let d be the distance from the point of suspension to the center of the bob, and r be the radius of the bob, and let x be such that d/r = r/x. Then d+(0.4)x is the standard unit of measurement.
(This is my translation of Wilkins' Baroque language.)

But this was a big puzzle to me:

Huh? Why 0.4? Why does r come into it? Why not just use d? Huh?

Soon after the press release came out, I got email from a gentleman named Bill Hooper, a retired professor of physics of the University of Virginia's College at Wise, in which he explained this puzzle completely, and in some detail.

According to Professor Hooper, you cannot just use d here, because if you do, the length will depend on the size, shape, and orientation of the bob. I did not know this; I would have supposed that you can assume that the mass of the bob is concentrated at its center of mass, but apparently you cannot.

The usual Physics I calculation that derives the period of a pendulum in terms of the distance from the fulcrum to the center of the bob assumes that the bob is infinitesimal. But in real life the bob is not infinitesimal, and this makes a difference. (And Wilkins specified that one should use the most massive possible bob, for reasons that should be clear.)

No, instead you have to adjust the distance d in the formula by adding I/md, where m is the mass of the bob and I is the moment of intertia of the bob, a property which depends on the shape, size, and mass of the bob. Wilkins specified a spherical bob, so we need only calculate (or look up) the formula for the moment of inertia of a sphere. It turns out that for a solid sphere, I = 2mr2/5. That is, the distance needed is not d, but d + 2r2/5d. Or, as I put it above, d + (0.4)x, where d/r = r/x.

Well, that answers that question. My very grateful thanks to Professor Hooper for the explanantion. I think I might have figured it out myself eventually, but I am not willing to put a bound of less than two hundred years on how long it would have taken me to do so.

One lesson to learn from all this is that those early Royal Society guys were very smart, and when they say something has a mysterious (0.4)x in it, you should assume they know what they are doing. Another lesson is that mechanics was pretty well-understood by 1668.

[Other articles in category /physics] permanent link

Fri, 14 Sep 2007

Why spiders hang with their heads down
Katara asked me last week why spiders hang in their webs with their heads downwards, and I said I would try to find out. After a cursory Google search, I was none the wiser, so I tried asking the Wikipedia "reference desk" page. I did not learn anything useful about the spiders, but I did learn that the reference desk page is full of people who know even less about spiders than I do who are nevertheless willing to post idle speculations.

Fortunately, I was at a meeting this week in Durham that was also attended by three of the world's foremost spider experts. I put the question to Jonathan A. Coddington, curator of arachnids for the Smithsonian Institution.

Professor Coddington told me that it was because the spider prefers (for obvious mechanical and dynamic reasons) to attack its prey from above, and so it waits the upper part of the web and constructs the web so that the principal prey-catching portion is below. When prey is caught in the web, the spider charges down and attacks it.

I had mistakenly thought that spiders in orb webs (which are the circular webs you imagine when you try to think of the canonical spiderweb) perched in the center. But it is only the topological center, and geometrically it is above the midline, as the adjacent picture should make clear. Note that more of the radial threads are below the center than are above it.

[Other articles in category /bio] permanent link

Thu, 13 Sep 2007

Girls of the SEC
I'm in the Raleigh-Durham airport, and I just got back from the newsstand, where I learned that the pictorial in this month's Playboy magazine this month is "Girls of the SEC". On seeing this, I found myself shaking my head in sad puzzlement.

This isn't the first time I've had this reaction on learning about a Playboy pictorial; last time was probably in August 2002 when I saw the "Women of Enron" cover. (I am not making this up.) I wasn't aware of The December 2002 feature, "Women of Worldcom" (I swear I'm not making this up), but I would have had the same reaction if I had been.

I know that in recent years the Playboy franchise has fallen from its former heights of glory: circulation is way down, the Playboy Clubs have all closed, few people still carry Playboy keychains. But I didn't remember that they had fallen quite so far. They seem to have exhausted all the plausible topics for pictorial features, and are now well into the scraping-the-bottom-of-the-barrel stage. The June 1968 feature was "Girls of Scandinavia". July 1999, "Girls of Hawaiian Tropic". Then "Women of Enron" and now "Girls of the SEC".

How many men have ever had a fantasy about sexy SEC employees, anyway? How can you even tell? Sexy flight attendants, sure; they wear recognizable uniforms. But what characterizes an SEC employee? A rumpled flannel suit? An interest in cost accounting? A tendency to talk about the new Basel II banking regulations? I tried to think of a category that would be less sexually inspiring than "SEC employees". It's difficult. My first thought was "Girls of Wal-Mart." But no, Wal-Mart employees wear uniforms.

If you go too far in that direction you end up in the realm of fetish. For example, Playboy is unlikely to do a feature on "girls of the infectious disease wards". But if they did, there is someone (probably on /b/), who would be extremely interested. It is hard to imagine anyone with a similarly intense interest in SEC employees.

So what's next for Playboy? Girls of the hospital gift shops? Girls of State Farm Insurance telephone customer service division? Girls of the beet canneries? Girls of Acadia University Grounds and Facilities Services? Girls of the DMV?

[ Pre-publication addendum: After a little more research, I figured out that SEC refers here to "Southeastern Conference" and that Playboy has done at least two other features with the same title, most recently in October 2001. I decided to run the article anyway, since I think I wouldn't have made the mistake if I hadn't been prepared ahead of time by "Women of Enron". ]

[ Addendum 20070913: This article is now on the first page of Google searches for "girls of the sec playboy". ]

[ Addendum 20070915: The article has moved up from tenth to third place. Truly, Google works in mysterious ways. ]

[Other articles in category /misc] permanent link

Wed, 12 Sep 2007

The loophole in the U.S. Constitution: the answer
In the previous article, I wondered what "inconsistency in the Constitution" Gödel might have found that would permit the United States to become a dictatorship.

Several people wrote in to tell me that Peter Suber addresses this in his book The Paradox of Self-Amendment, which is available online. (Suber also provides a provenance for the Gödel story.)

Apparently, the "inconsistency" noted by Gödel is simply that the Constitution provides for its own amendment. Suber says: "He noticed that the AC had procedural limitations but no substantive limitations; hence it could be used to overturn the democratic institutions described in the rest of the constitution." I am gravely disappointed. I had been hoping for something brilliant and subtle that only Gödel would have noticed.

Thanks to Greg Padgett, Julian Orbach, Simon Cozens, and Neil Kandalgaonkar for bringing this to my attention.

M. Padgett also pointed out that the scheme I proposed for amending the constitution, which I claimed would require only the cooperation of a majority of both houses of Congress, 218 + 51 = 269 people in all, would actually require a filibuster-proof majority in the Senate. He says that to be safe you would want all 100 senators to conspire; I'm not sure why 60 would not be sufficient. (Under current Senate rules, 60 senators can halt a filibuster.) This would bring the total required to 218 + 60 = 278 conspirators.

He also pointed out that the complaisance of five Supreme Court justices would give the President essentially dictatorial powers, since any legal challenge to Presidential authority could be rejected by the court. But this train of thought seems to have led both of us down the same path, ending in the idea that this situation is not really within the scope of the original question.

As a final note, I will point out what I think is a much more serious loophole in the Constitution: if the Vice President is impeached and tried by the Senate, then, as President of the Senate, he presides over his own trial. Article I, section 3 contains an exception for the trial of the President, where the Chief Justice presides instead. But the framers inexplicably forgot to extend this exception to the trial of the Vice President.

[ Addendum 20090121: Jeffrey Kegler has discovered Oskar Morgenstern's lost eyewitness account of Gödel's citizenship hearing. Read about it here. ]

[ Addendum 20110525: As far as I know, there is no particular reason to believe that Peter Suber's theory is correct. Morgenstern knew, but did not include it in his account. ]

[ Addendum 20160315: I thought of another interesting loophole in the Constitution: The Vice-President can murder the President, and then immediately pardon himself. ]

[Other articles in category /law] permanent link

Sun, 09 Sep 2007

The loophole in the U.S. Constitution

Gödel took the matter of citizenship with great solemnity, preparing for the exam by making a close study of the United States Constitution. On the eve of the hearing, he called [Oskar] Morgenstern in an agitated state, saying he had found an "inconsistency" in the Constitution, one that could allow a dictatorship to arise.
(Holt, Jim. Time Bandits, The New Yorker, 29 February 2005.)

I've wondered for years what "inconsistency" was.

I suppose the Attorney General could bring some sort of suit in the Supreme Court that resulted in the Court "interpreting" the Constitution to find that the President had the power to, say, arbitrarily replace congresspersons with his own stooges. This would require only six conspirators: five justices and the President. (The A.G. is a mere appendage of the President and is not required for the scheme anyway.)

But this seems outside the rules. I'm not sure what the rules are, but having the Supreme Court radically and arbitrarily "re-interpret" the Constitution isn't an "inconsistency in the Constitution". The solution above is more like a coup d'etat. The Joint Chiefs of Staff could stage a military takeover and institute a dictatorship, but that isn't an "inconsistency in the Constitution" either. To qualify, the Supreme Court would have to find a plausible interpretation of the Constitution that resulted in a dictatorship.

The best solution I have found so far is this: Under Article IV, Congress has the power to admit new states. A congressional majority could agree to admit 150 trivial new states, and then propose arbitrary constitutional amendments, to be ratified by the trivial legislatures of the new states.

This would require a congressional majority in both houses. So Gödel's constant, the smallest number of conspirators required to legally transform the United States into a dictatorship, is at most 269. (This upper bound would have been 267 in 1948 when Gödel became a citizen.) I would like to reduce this number, because I can't see Gödel getting excited over a "loophole" that required so many conspirators.

[ Addendum 20070912: The answer. ]

[ Addendum 20090121: Jeffrey Kegler has discovered Oskar Morgenstern's lost eyewitness account of Gödel's citizenship hearing. Read about it here. ]

[ Addendum 20160129: F.E. Guerra-Pujol has written an article speculating on this topic, “Gödel’s Loophole”. Guerra-Pujol specifically rejects my Article IV proposal for requiring too many conspirators. ]

[Other articles in category /law] permanent link

Sat, 08 Sep 2007

The missing deltahedron
I recently wrote about the convex deltahedra, which are the eight polyhedra whose faces are all congruent equilateral triangles:

Tetrahedron 464
Triangular dipyramid 695
Octahedron 8126
Pentagonal dipyramid 10157
Snub disphenoid 12188
Triaugmented triangular prism 14219
Gyroelongated square dipyramid 162410
Icosahedron 203012
The names are rather horrible, so I think that from now on I'll just refer to them as D4, D6, D8, D10, D12, D14, D16, and D20.

The number of edges that meet at a vertex is its valence. Vertices in convex deltahedra have valences of 3, 4, or 5. The valence can't be larger than 5 because only six equilateral triangles will fit, and if you fit 6 then they lie flat and the polyhedron is not properly convex.

Let V3, V4, and V5 be the number of vertices of valences 3, 4, and 5, respectively. Then:

D8 6 
D10 52
D12 44
D14 36
D16 28
D20  12
There's a clear pattern here, with V3s turning into V4s two at a time until you reach the octahedron (D8) and then V4s turning into V5s one at a time until you reach the icosahedron (D20). But where is V4=1, V5=10? There's a missing deltahedron. I don't mean it's missing from the table; I mean it's missing from the universe.

Well, this is all oversubtle, I realized later, because you don't need to do the V3V4V5 analysis to see that something is missing. There are convex deltahedra with 4, 6, 8, 10, 12, 14, and 20 faces; what happened to 18?

Still, I did a little work on a more careful analysis that might shed some light on the 18-hedron situation. I'm still in the middle of it, but I'm trying to continue my policy of posting more frequent, partial articles.

Let V be the number of vertices in a convex deltahedron, E be the number of edges, and F be the number of faces.

We then have V = V3 + V4 + V5. We also have E = ½(3V3 + 4V4 + 5V5). And since each face has exactly 3 edges, we have 3F = 2E.

By Euler's formula, F + V = E + 2. Plugging in the stuff from the previous paragraph, we get 3V3 + 2V4 + V5 = 12.

It is very easy to enumerate all possible solutions of this equation. There are 19:

Solutions in green correspond to convex deltahedra. What goes wrong with the other 11 items?

(3,1,1) fails completely because to have V5 > 0 you need V ≥ 6. There isn't even a graph with (V3, V4, V5) = (3,1,1), much less a polyhedron.

There is a graph with (3,0,3), but it is decidedly nonplanar: it contains K3,3, plus an additional triangle. But the graph of any polyhedron must be planar, because you can make a little hole in one of the faces of the polyhedron and flatten it out without the edges crossing.

Another way to think about (3,0,3) is to consider it as a sort of triangular tripyramid. Each of the V5s shares an edge with each of the other five vertices, so the three V5s are all pairwise connected by edges and form a triangle. Each of the three V3s must be connected to each of the three vertices of this triangle. You can add two of the required V3s, by erecting a triangular pyramid on the top and the bottom of the triangle. But then you have nowhere to put the third pyramid.

On Thursday I didn't know what went wrong with (2,2,2); it seemed fine. (I found it a little challenging to embed it in the plane, but I'm not sure if it would still be challenging if it hadn't been the middle of the night.) I decided that when I got into the office on Friday I would try making a model of it with my magnet toy and see what happened.

It turned out that nothing goes wrong with (2,2,2). It makes a perfectly good non-convex deltahedron. It's what you get when you glue together three tetrahedra, face-to-face-to-face. The concavity is on the underside in the picture.

(2,0,6) was a planar graph too, and so the problem had to be geometric, not topological. When I got to the office, I put it together. It also worked fine, but the result is not a polyhedron. The thing you get could be described as a gyroelongated triangular dipyramid. That is, you take an octahedron and glue tetrahedra to two of its opposite faces. But then the faces of the tetrahedra are coplanar with the faces of the octahedron to which they abut, and this is forbidden in polyhedra. When that happens you're supposed to eliminate the intervening edge and consider the two faces to be a single face, a rhombus in this case. The resulting thing is not a polyhedron with 12 triangular faces, but one with six rhombic faces (a rhombohedron), essentially a squashed cube. In fact, it's exactly what you get if you make a cube from the magnet toy and then try to insert another unit-length rod into the diagonal of each of the six faces. You have to squash the cube to do this, of course, since the diagonals had length √2 before and length 1 after.

So there are several ways in which the triples (V3,V4,V5) can fail to determine a convex deltahedron: There is an utter topological failure, as with (3,1,1).

There is a planarity failure, which is also topological, but less severe, as with (3,0,3). (3,0,3) also fails because you can't embed it into R3. (I mean that you cannot embed its 3-skeleton. Of course you can embed its 1-skeleton in R3, but that is not sufficient for the thing to be a polyhedron.) I'm not sure if this is really different from the previous failure; I need to consider more examples. And (3,0,3) fails in yet another way: you can't even embed its 1-skeleton in R3 without violating the constraint that says that the edges must all have unit length. The V5s must lie at the vertices of an equilateral triangle, and then the three unit spheres centered at the V5s intersect at exactly two points of R3. You can put two of the V3s at these points, but this leaves nowhere for the third V3. Again, I'm not sure that this is a fundamentally different failure mode than the other two.

Another failure mode is that the graph might be embeddable into R3, and might satisfy the unit-edge constraint, but in doing so it might determine a concave polyhedron, like (2,2,2) does, or a non-polyhedron, like (2,0,6) does.

I still have six (V3,V4,V5) triples to look into. I wonder if there are any other failure modes?

I should probably think about (0,1,10) first, since the whole point of all this was to figure out what happened to D18. But I'm trying to work up from the simple cases to the harder ones.

I suppose the next step is to look up the proof that there are only eight convex deltahedra and see how it goes.

I suspect that (2,1,4) turns out to be nonplanar, but I haven't looked at it carefully enough to actually find a forbidden minor.

One thing that did occur to me today was that a triple (V3, V4, V5) doesn't necessarily determine a unique graph, and I need to look into that in more detail. I'll be taking a plane trip on Sunday and I plan to take the magnet toy with me and continue my investigations on the plane.

In other news, Katara and I went to my office this evening to drop off some books and pick up some stuff for the trip, including the magnet toy. Katara was very excited when she saw the collection of convex deltahedron models on my desk, each in a different color, and wanted to build models just like them. We got through all of them, except D10, because we ran out of ball bearings. By the end Katara was getting pretty good at building the models, although I think she probably wouldn't be able to do it without directions yet. I thought it was good work, especially for someone who always skips from 14 to 16 when she counts.

On the way home in the car, we were talking about how she was getting older and I rhapsodized about how she was learning to do more things, learning to do the old things better, learning to count higher, and so on. Katara then suggested that when she is older she might remember to include 15.

[Other articles in category /math] permanent link

Fri, 07 Sep 2007

Families of scalars
I'm supposedly in the midst of writing a book about fixing common errors in Perl programs, and my canonical example is the family of scalar variables. For instance, code like this:

     if ($FORM{'h01'}) {$checked01 = " CHECKED "}
     if ($FORM{'h02'}) {$checked02 = " CHECKED "}
     if ($FORM{'h03'}) {$checked03 = " CHECKED "}
     if ($FORM{'h04'}) {$checked04 = " CHECKED "}
     if ($FORM{'h05'}) {$checked05 = " CHECKED "}
     if ($FORM{'h06'}) {$checked06 = " CHECKED "}
(I did not make this up; I got it from here.) The flag here is the family $checked01, $checked02, etc. Such code is almost always improved by replacing the family with an array, and the repeated code with a loop:

        $checked[$_] = $FORM{"h$_"} for "01" .. "06";
Actually in this particular case a better solution was to eliminated the checked variables entirely, but that is not what I was planning to discuss. Rather, I planned to discuss a recent instance in which I wrote some code with a family of variables myself, and the fix was somewhat different.

The program I was working on was a digester for the qmail logs, translating them into a semblance of human-readable format. (This is not a criticism; log files need not be human-readable; they need to be easy to translate, scan, and digest.) The program scans the log, gathering information about each message and all the attempts to deliver it to each of its recipient addresses. Each delivery can be local or remote.

Normally the program prints information about each message and all its deliveries. I was adding options to the program to allow the user to specify that only local deliveries or only remote deliveries were of interest.

The first thing I did was to add the option-processing code:

  } elsif ($arg eq "--local-only" || $arg eq '-L') {
    $local_only = 1;
  } elsif ($arg eq "--remote-only" || $arg eq '-R') {
    $remote_only = 1;
As you see, this is where I made my mistake, and introduced a (two-member) family of variables. The conventional fix says that this should have been something like $do_only{local} and $do_only{remote}. But I didn't notice my mistake right away.

Later on, when processing a message, I wanted to the program to scan its deliveries, and skip all processing and display of the message unless some of its deliveries were of the interesting type:

  if ($local_only || $remote_only) {
I had vague misgivings at this point about the test, which seemed redundant, but I pressed on anyway, and found myself in minor trouble. Counting the number of local or remote deliveries was complicated:
  if ($local_only || $remote_only) {
    my $n_local_deliveries = 
      grep $msg->{del}{$_}{lr} eq "local", keys %{$msg->{del}};
    my $n_remote_deliveries = 
      grep $msg->{del}{$_}{lr} eq "remote", keys %{$msg->{del}};
There is a duplication of code here. Also, there is a waste of CPU time, since the program never needs to have both numbers available. This latter waste could be avoided at the expense of complicating the code, by using something like $n_remote_deliveries = keys(%{$msg->{del}}) - $n_local_deliveries, but that is not a good solution.

Also, the complete logic for skipping the report was excessively complicated:

  if ($local_only || $remote_only) {
    my $n_local_deliveries = 
      grep $msg->{del}{$_}{lr} eq "local", keys %{$msg->{del}};
    my $n_remote_deliveries = 
      grep $msg->{del}{$_}{lr} eq "remote", keys %{$msg->{del}};

    return if $local_only  && $local_deliveries == 0
           || $remote_only && $remote_deliveries == 0;

I could have saved the wasted CPU time (and the repeated tests of the flags) by rewriting the code like this:

  if ($local_only) {
    return unless
      grep $msg->{del}{$_}{lr} eq "local", keys %{$msg->{del}};
  } elsif ($remote_only) {
    return unless
      grep $msg->{del}{$_}{lr} eq "remote", keys %{$msg->{del}};
but that is not addressing the real problem, which was the family of variables, $local_only and $remote_only, which inevitably lead to duplicated code, as they did here.

Such variables are related by a convention in the programmer's mind, and nowhere else. The language itself is as unaware of the relationship as if the variables had been named $number_of_nosehairs_on_typical_goat and $fusion_point_of_platinum. A cardinal rule of programming is to make such conventional relationships explicit, because then the programming system can give you some assistance in dealing with them. (Also because then they are apparent to the maintenance programmer, who does not have to understand the convention.) Here, the program was unable to associate $local_only with the string "local" and $remote_only with "remote", and I had to make up the lack by writing additional code.

For families of variables, the remedy is often to make the relationship explicit by using an aggregate variable, such as an array or a hash, something like this:

  if (%use_only) {
    my ($only_these) = keys %use_only;
    return unless
      grep $msg->{del}{$_}{lr} eq $only_these, keys %{$msg->{del}};
Here the relationship is explicit because $use_only{"remote"} indicates an interest in remote deliveries and $use_only{"local"} indicates an interest in local deliveries, and the program can examine the key in the hash to determine what to look for in the {lr} data.

But in this case the alternatives are disjoint, so the %use_only hash will never contain more than one element. The tipoff is the bizarre ($only_these) = keys ... line. Since the hash is really storing a single scalar, it can be replaced with a scalar variable:

  } elsif ($arg eq "--local-only" || $arg eq '-L') {
    $only_these = "local";
  } elsif ($arg eq "--remote-only" || $arg eq '-R') {
    $only_these = "remote";
Then the logic for skipping uninteresting messages becomes:

  if ($only_these) {
    return unless
      grep $msg->{del}{$_}{lr} eq $only_these, keys %{$msg->{del}};
Ahh, better.

A long time ago I started to suspect that flag variables themselves are a generally bad practice, and are best avoided, and I think this example is evidence in favor of that theory. I had a conversation about this yesterday with Aristotle Pagaltzis, who is very thoughtful about this sort of thing. One of our conclusions was that although the flag variable can be useful to avoid computing the same boolean value more than once, if it is worth having, it is because your program uses it repeatedly, and so it is probably testing the same boolean value more than once, and so it is likely that the program logic would be simplified if one could merge the blocks that would have been controlled by those multiple tests into one place, thus keeping related code together, and eliminating the repeated tests.

[Other articles in category /prs] permanent link

Thu, 06 Sep 2007

Followup notes about dice and polyhedra
I got a lot of commentary about these geometric articles, and started writing up some followup notes. But halfway through I got stuck in the middle of making certain illustrations, and then I got sick, and then I went to a conference in Vienna. So I decided I'd better publish what I have, and maybe I'll get to the other fascinating points later.

  • Regarding a die whose sides appear with probabilities 1/21 ... 6/21

    • Several people wrote in to cast doubt on my assertion that the probability of an irregular die showing a certain face is proportional to the solid angle subtended by that face from the die's center of gravity. But nobody made the point more clearly than Robert Young, who pointed out that if I were right, a coin would have a 7% chance of landing on its edge. I hereby recant this claim.

    • John Berthels suggested that my analysis might be correct if the die was dropped into an inelastic medium like mud that would prevent it from bouncing.

    • Jack Vickeridge referred me to this web site, which has a fairly extensive discussion of seven-sided dice. The conclusion: if you want a fair die, you have no choice but to use something barrel-shaped.

    • Michael Lugo wrote a detailed followup in which he discusses this and related problems. He says "What makes Mark's problem difficult is the lack of symmetry; each face has to be different." Quite so.

  • Regarding alternate labelings for standard dice

    • Aaron Crane says that these dice (with faces {1,2,2,3,3,4} and {1,3,4,5,6,8}) are sometimes known as "Sicherman dice", after the person who first brought them to the attention of Martin Gardner. Can anyone confirm that this was Col. G.L. Sicherman? I have no reason to believe that it was, except that it would be so very unsurprising if it were true.

    • Addendum 20070905: I now see that the Wikipedia article attributes the dice to "Colonel George Sicherman," which is sufficiently clear that I would feel embarrassed to write to the Colonel to ask if it is indeed he. I also discovered that the Colonel has a Perl program on his web site that will calculate "all pairs of n-sided dice that give the same sums as standard n-sided dice".

    • M. Crane also says that it is an interesting question which set of dice is better for backgammon. Both sets have advantages: the standard set rolls doubles 1/6 of the time, whereas the Sicherman dice only roll doubles 1/9 of the time. (In backgammon, doubles count double, so that whereas a player who rolls ab can move the pieces a total of a+b points, a player who rolls aa can move pieces a total of 4a points.) The standard dice permit movement of 296/36 points per roll, and the Sicherman dice only 274/36 points per roll.

      Ofsetting this disadvantage is the advantage that the Sicherman dice can roll an 8. In backgammon, one's own pieces may not land on a point occupied by more than one opposing piece. If your opponent occupies six conscutive points with two pieces each, they form an impassable barrier. Such a barrier is passable to a player using the Sicherman dice, because of the 8.

    • Doug Orleans points out that in some contexts one might prefer to use a Sicherman variant dice {2,3,3,4,4,5} and {0,2,3,4,5,7}, which retain the property that opposite faces sum to 7, and so that each die shows 3.5 pips on average. Such dice roll doubles as frequently as do standard dice.

    • The Wikipedia article on dice asserts that the {2, 3, 3, 4, 4, 5} die is used in some wargames to express the strength of "regular" troops, and the standard {1, 2, 3, 4, 5, 6} die to express the strength of "irregular" troops. This makes the outcome of battles involving regular forces more predictable than those involving irregular forces.

  • Regarding deltahedra and the snub disphenoid

    • Several people proposed alternative constructions for the snub disphenoid.

      1. Brooks Moses suggested the following construction: Take a square antiprism, squash the top square into a rhombus, and insert a strut along the short diagonal of the rhombus. Then squash and strut the bottom square similarly.

        It seems, when you think about this, that there are two ways to do the squashing. Suppose you squash the bottom square horizontally in all cases. The top square is turned 45° relative to the bottom (because it's an antiprism) and so you can squash it along the -45° diagonal or along the +45° diagonal, obtaining a left- and a right-handed version of the final solid. But if you do this, you find that the two solids are the same, under a 90° rotation.

        This construction, incidentally, is equivalent to the one I described in the previous article: I said you should take two rhombuses and connect corresponding vertices. I had a paragraph that read:

        But this is where I started to get it wrong. The two wings have between them eight edges, and I had imagined that you could glue a rhombic antiprism in between them. . . .

        But no, I was right; you can do exactly this, and you get a snub disphenoid. What fooled me was that when you are looking at the snub disphenoid, it is very difficult to see where the belt of eight triangles from the antiprism got to. It winds around the polyhedron in a strange way. There is a much more obvious belt of triangles around the middle, which is not suitable for an antiprism, being shaped not like a straight line but more like the letter W, if the letter W were written on a cylinder and had its two ends identified. I was focusing on this belt, but the other one is there, if you know how to see it.

        The snub disphenoid has four vertices with valence 4 and four with valence 5. Of its 12 triangular faces, four have two valence-4 vertices and one valence-5 vertex, and eight have one valence-4 vertex and two valence-5 vertices. These latter eight form the belt of the antiprism.

      2. M. Moses also suggested taking a triaugmented triangular prism, which you will recall is a triangular prism with a square pyramid erected on each of its three square faces, removing one of the three pyramids, and then squashing the exposed square face into a rhombus shape, adding a new strut on the diagonal. This one gives me even less intuition about what is going on, and it seems even more strongly that it shou,ld matter whether you put in the extra strut from upper-left to lower-right, or from upper-right to lower-left. But it doesn't matter; you get the same thing either way.

      3. Jacob Fugal pointed out that you can make a snub disphenoid as follows: take a pentagonal dipyramid, and replace one of the equatorial *----*----* figures with a rhombus. This is simple, but unfortunately gives very little intuition for what the disphenoid is like. It is obvious from the construction that there must be pentagons on the front and back, left over from the dipyramid. But it is not at all clear that there are now two new upside-down pentagons on the left and right sides, or that the disphenoid has a vertical symmetry.

    • A few people asked me where John Batzel got they magnet toy that I was using to construct the models. It costs only $5! John gave me his set, and I bought three more, and I now have a beautiful set of convex deltahedra and a stellated dodecahedron on my desk. (Actually, it is not precisely a stellated dodecahedron, since the star faces are not quite planar, but it is very close. If anyone knows the name of this thing, which has 32 vertices, 90 edges, and 60 equilateral triangular faces, I would be pleased to hear about it.) Also I brought my daughter Katara into my office a few weekends ago to show her the stella octangula ("I wanna see the stella octangula, Daddy! Show me the stella octangula!") which she enjoyed; she then stomped on it, and then we built another one together.

    • [ Addendum 20070908: More about deltahedra. ]

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