Here’s the deal. I’m going to give you a square grid, with some of the cells colored and others possibly left blank. We’ll call this a template. Perhaps the grid will be one of these 3×3 templates:

You have a supply of transparent plastic overlays that match the grid in size and shape and that also bear patterns of black dots:

Note that each of these patterns has exactly three dots, with one dot in each row and each column. The six overlays shown are the only 3×3 grids that have this property.

Your task is to assemble a subset of the overlays and lay them on the template in such a way that dots cover all the colored squares but none of the blank squares. You are welcome to superimpose multiple dots on any colored square, but overall you want to use as few overlays as possible. To make things interesting, I’ll suggest a wager. I’ll pay you $3 for a correct covering of a 3×3 template, but you have to pay me $1 for each overlay you use. Is this a good bet?

Before going further, I should mention that not every conceivable template can be covered under these rules. To take an obvious example, no 3×3 template with fewer than three colored squares can possibly be covered by any combination of the six overlays. But I promise to submit only templates that can be covered by some combination of the given dot patterns; if I err about this, I forfeit the bet.

How does the game play out? If I give you the template marked “1″ above, you can easily win; just choose permutations a and b, which together cover all the colored squares and no others. You pay $2 and get $3. Template 2, with all nine squares colored, looks like it might be the toughest challenge. Clearly, it cannot be covered with fewer than three overlays, since we need a total of nine dots; and it turns out that exactly three overlays are required. Indeed, there are two ways of covering the template with three overlays: a + d + e and b + c + f. Thus this template is a breakeven proposition: You earn $3 and pay $3.

Now we come to template 3, which has eight colored squares and one blank. Surely if you can cover the full nine squares with just three overlays, then you should also be able to cover eight squares—no? I invite you to try it. In fact the only covering that works requires four overlays: b + d + e + f. Thus you shouldn’t take my bet, since I can always give you a template with just one blank, and you’ll have a net loss of $1.

Some background. I’ll return to the gaming table momentarily, but first let me explain what this is all about and where it came from. A few weeks ago, I was writing about “ranges of rankings,” which led me into the topic of permutation matrices. To recapitulate:

• A permutation matrix is a square matrix with a single 1 in each column and each row, and all the rest of the elements 0.
• An ormat is a superposition of permutation matrices, formed by applying the Boolean OR function to corresponding elements of the permutation matrices. For example:
• Not all square matrices with (0,1) entries can be formed by OR-ing permutation matrices, but there’s an efficient algorithm for deciding whether or not a given matrix is an ormat. (I thank some helpful commenters for enlightening me on this point.)
• Given an ormat, the total number of distinct permutation paths that can be threaded through the 1 entries of the matrix is equal to the permanent of the matrix. Calculating the permanent is known to be a hard computational problem.

In a comment, Barry Cipra posed the following query:

The permanent tells us the maximum number of different permutations that can be OR-summed to produce a given ormat, but what is the corresponding minimum number? Also, in how many different ways can the minimum be achieved?

The connection between ormats and my little game is probably apparent by now. The template of colored and blank squares is an ormat; the dotted overlays represent permutation matrices; to maximize your payoff in the game (or to minimize your loss), you need to answer Barry’s first question, finding the minimum number of permutations that can be combined to yield the given ormat.

For 3×3 matrices, we can solve this problem by exhaustive search, calculating the OR-sums of all possible combinations of the six 3×3 permutation matrices taken 1, 2, 3, …, 6 at a time. I did this with pencil and paper on a recent airplane trip. Here is a summary of the results:

Some of the numbers on this card are easy to explain. The six ormats with just three 1 entries are the permutation matrices themselves. There are six of them because there are 3! = 6 permutations of three things. There are no ormats with four 1 entries for a reason that bears thinking about: There can be no permutations that differ from one another in just one element. When you superimpose any of the six overlays shown above, you can wind up with three, five or six dots, but never four.

At the other end of the scale, it’s no surprise that there’s exactly one ormat with nine 1 entries, and that it takes three permutations to produce it. And then there are the nine ormats with eight 1 entries, which each require four permutations to be OR-ed. These are the single-blank patterns like template 3 above.

Based on these results, I began speculating about what I would see in a tabulation of all 4×4 ormats.

There would have to be 4! = 24 patterns with four 1 entries, and just one pattern with all 1s, generated by OR-ing four permutations. And there should be 16 ormats that require five permutations, namely the 16 matrices with a single 0 element. This last prediction seemed a little less self-evident than the others.

Pocket change and Cheerios. My thoughts about the single-zero (or single-blank) case went something like this. To cover 15 squares with sets of four dots each, we need at least four sets, or else we simply won’t have enough dots. So a useful starting point is one of the optimal arrangements that cover all 16 squares without gaps or overlaps. By this time I had grown tired of drawing zillions of dots, and so I started working with sets of coins.

In this arrangement each coin denomination forms a permutation, with no two pennies, nickels, dimes or quarters in the same row or the same column. We have successfully covered all the colored squares, but unfortunately we’ve also covered the blank at the lower right. Thus this pattern of coins is not an acceptable solution, but maybe we can fix it up somehow?

Moving the penny from the blank square to another square in the same column solves one problem but creates another: Now the arrangement of pennies is no longer a permutation. There are two pennies in the third row.

So now we have to shift another penny to restore the one-per-row-and-column property. Inevitably, this leaves a colored square uncovered. The only way we can cover that exposed square is to introduce a fifth permutation. Since I had run out of coin denominations, I chose a popular brand of breakfast toroids. Voila:

There’s nothing special about the particular moves I chose in this sequence. If you try some alternatives, you should be able to persuade yourself that moving the penny that covers the blank to any other square in the fourth column (or in the fourth row) would lead to essentially the same situation. Likewise the game would come out the same if the single blank square were placed anywhere else in the grid. And you could also start with a different set of initial permutations (provided they cover all the squares).

This coin-shuffling exercise demonstrates that we can cover any 4×4 template that has a single blank by combining no more than five permutations, but how do we know that five are actually needed? Maybe there’s some totally different arrangement that would do the job with just four permutations? Well, think about what such an arrangement would look like. It would have to differ at exactly one position from some other layout of permutations that covers the full 16-square grid. But no two permutations can differ at one and only one place. Thus the reason there can be no four-permutation cover of 15 squares is essentially the same as the reason no 4×4 ormat pattern can cover just five squares.

This argument generalizes to k×k matrices: For any integer k, there must be at least k ormat patterns that cannot be covered with fewer than k+1 permutations. But then comes the bigger speculative leap: Perhaps k+1 is an upper bound. Perhaps part of the answer to Barry’s question is that no k×k ormat pattern requires more than k+1 permutations. At one point I even had a “proof” of this conjecture. Then I wrote a program to check it, doing much the same thing I did with the dots on the airplane.

Out of bounds. My program found the expected 16 ormat patterns that require five permutations—and it found many more as well. In all it identified 2,032 4×4 ormats that can’t be composed from fewer than five permutations. And then came a bigger surprise: The program also found 480 patterns that require six permutations. So much for my proposed upper bound.

One of those problematic 480 ormats takes this form:

Looking over this pattern, I thought I understood where my earlier reasoning had gone awry. This matrix is just like the single-zero pattern, but with two zeros! (I do mean for that statement to make sense. Bear with me.) Suppose we start again with a set of four permutations that completely cover the grid, including the two blanks.

Then we can uncover each blank just as we did in the coin-shuffling procedure above, although we have to be careful the two sets of movements don’t interfere with each other. (Not much point in removing the penny from a blank square, then putting the nickel there.) Here is a strategy for clearing both blank squares while maintaining the one-per-column-and-row permutation property:

Inevitably, when we uncover the two blank squares, we also remove coins from two colored squares, which now have to be filled in again. The key point is that no single permutation can repair that damage, because the two open colored squares are in the same row. To cover both of those squares we need two additional permutations.

Other ways of reshuffling the coins avoid putting the two open squares in the same row or column, but they still foil all attempts to complete the covering with just five permutations. Try adding four Cheerios to the diagram below. If you cover both of the open blue squares, then either you also cover one of the blank squares or you wind up with two Cheerios in the same row.

So now it’s clear we need as many as six permutations to cover a 4×4 ormat. Does that suggest that the general upper bound might be k+2 rather than k+1? Or perhaps the appropriate formula is 2k–2? In support of this latter possibility I offer these two ormats, which require 8 and 10 permutations respectively:

Another wager. Having fooled myself several times about the upper bound on minimal ormat coverings, I feel I should build in a little margin for error before I invite you to make a further wager. We already have direct evidence that covering a k×k ormat can take as many as 2k–2 permutations. So I’ll be generous and offer a full $2k for a proper covering, while charging $1 per permutation. If k=3 or k=4, you can definitely make money on this deal. But is it a good bet for larger k? (Hint: I’d be willing to play the game on these terms for real money.)

•     •     •

Update 2010-08-19: No takers for my bet, eh? Too bad; I had already spent my winnings.

Barry Cipra, who raised the question about minimal ormat covers in the first place, sends this illuminating letter:

I’m going to tiptoe a short ways out on a long long limb and conjecture (really just guess) that the “worst case” behavior, in terms of the minimum number of permutations it takes to produce a given ormat, occurs for ormats of the following form, shown here for k=7:

So as not to abuse existing matrix terminology, I’ll call any (square) matrix of this type—i.e., whose entries below the main subdiagonal are all 0—”uppity triangular.” I can (and will!) show that this uppity triangular ormat for k=7 requires (at least) 16 permutations—and the number appears to grow concavely upwards from that, so I, for one, will definitely not take you up on your $2k wager.

The trick, I realized, is to view each ormat as the “shadow” of what I’ll call an “addmat.” If you let P1, P2, …, Pr be k×k permutation matrices, their addmat is simply the ordinary result of addition: S = P1 + P2 + … + Pr, whose entries are positive integers wherever one or more of the constituent permutations has a 1 and otherwise 0. The associated ormat is obtained by changing each of these entries to a 1, while leaving the 0’s alone. In this sense, the ormat’s 1’s are the “shadows” of the addmat’s positive entries.

What’s crucial is that addmats have a lovely little property not shared with their shadows: the row and columns sums of the entries of an addmat all equal the number of permutations that produce them, r.

Come now, let us reason together…. The uppity triangular ormat example above (for k=7) must come from an addmat of the form

where a, b, and all the *’s are positive integers. In particular, each * is at least 1. Since all row and column sums must be equal, the sum a+b must equal the sum of b and all 6 *’s above it. Hence a is at least 6. Likewise a+b must equal the sum of a and all 6 *’s above it, so b is also at least 6. Hence a+b is at least 12, which means the OR-sum that produced the given ormat involves at least 12 permutations.

This clearly generalizes to arbitrary k, which is more than “direct evidence” that covering a k×k ormat can take as many as 2k–2 permutations, it’s rigorous proof! But we can immediately do better, at least on a case-by-case basis. If we try to get by with just 12 permutations for this uppity triangular ormat, we quickly run into trouble. We obviously must have a = b = 6, and it follows that all the *’s above them are 1’s (to make those column sums 12). That is, we have the addmat

where I now wish to call your attention to the entry labeled “@”. To make its row-sum equal 12, we need @ = 10. But that means its column sum (with the 5 *’s above it) is at least 15, which cannot be! So we are forced to try larger values of a and/or b—which is to say, we need more permutation matrices to produce this addmat.

It turns out you can’t satisfy the row and column sum condition until you get to a = b = 8. I won’t take you through all the steps, but just give you a taste with the penultimate possibility, a = 7, b = 8. The best you can hope for in this case is

Note that I put as much of the “weight” in the last two columns as close to the 7 and 8 as possible, so that I could use the smallest possible value (10) as the entry with 5 positive entries above it. This makes the last three rows, and the right three columns all have the same sum, 15, but now we see a problem in the 12’s column: Its sum is at least 16. So once again, we’re screwed. It’s only with the next attempt that we avoid contradiction:

This matrix finally has all its row and column sums equal. Please note, this may or may not be an actual addmat of a set of 16 permutation matrices—I suspect it probably is, but I haven’t bothered to check. All we know is that it satisfies a necessary condition of being an addmat, namely that its row and column sums are all equal. (It’d be nice if that were also a sufficient condition, but something tells me it isn’t.)

This example, which can clearly be played out for larger values of k, suggests that not only are you safe with a $2k wager, but with a $(2k+2) wager and higher I’ve played around with this a bit, and persuaded myself that the number of permutation matrices will go to 2k + a lot—for k=10, if I did things correctly, you need 24 permutations (or possibly more, if the uppity triangular matrix the analysis leads to is not an actual addmat). I am entirely convinced that some additional careful thought can streamline the analysis into a nice, slick proof. I’m just not sure I haven’t already made a mistake, and built an elaborate house of cards….

Does any of this jibe with what you’ve already found to be the case?

It does indeed jibe.

First of all, to answer a small question Barry left open, here is a set of 16 permutations that will successfully cover his 7×7 “uppity triangular” matrix:

{1,2,3,4,5,6,7} {2,1,4,3,6,7,5} {1,3,2,5,4,7,6} {1,3,4,2,6,5,7}
{2,3,1,5,6,4,7} {1,2,3,5,6,7,4} {1,2,4,5,3,6,7} {1,2,4,5,6,3,7}
{1,2,4,5,6,7,3} {1,3,4,5,2,6,7} {1,3,4,5,6,2,7} {1,3,4,5,6,7,2}
{2,3,4,1,5,6,7} {2,3,4,5,1,6,7} {2,3,4,5,6,1,7} {2,3,4,5,6,7,1}

This was found with a simple greedy search.

My own attempts to find an upper bound have focused not on uppity triangular matrices but on matrices I’ve been calling “flags,” like this 7×7 case:

This matrix also requires 16 permutations for a proper covering. To see why, try threading permutations through the columns of the matrix, starting at the left edge and in each column choosing a 1 element (never a 0) from a different row. Because of the block of zeros at the upper left, the first three elements of every permutation must lie in rows 4 through 7. Thus each permutation “uses up” three of the last four rows in the first three columns, and the rest of the permutation can revisit this range of rows only once. It follows that each permutation can touch only one element in the 4×4 block of 1s in the lower right corner of the matrix, and at least 16 permutations are needed to cover all the 1s in the matrix. Showing that 16 are sufficient is not hard.

This kind of analysis works for any odd k, and thus we know that such matrices can require as many as

permutations. (For even k the situation is a little less symmetrical, and I haven’t worked out the exact details.)

These results give us a lower bound on the upper bound on the number of permutations that may be needed to cover a k×k ormat. But we haven’t proved it’s the true upper bound. Are there other ormats that require even more permutations? My guess is no, but keep in mind that almost all my conjectures along these lines have turned out to be wrong.

The National Research Council is getting ready to release a new assessment of graduate-education programs in the U.S. The previous study, published in 1995, gave each Ph.D.-granting department a numerical score between 0 and 5, then listed all the programs in each discipline in rank order. For example, here’s the top-10 list for doctoral programs in mathematics (as presented by H. J. Newton of Texas A&M University):

 rank    school                 score   
    1    Princeton               4.94
    2    Cal Berkeley            4.94
    3    MIT                     4.92
    4    Harvard                 4.90
    5    Chicago                 4.69
    6    Stanford                4.68
    7    Yale                    4.55
    8    NYU                     4.49
    9    Michigan                4.23
   10    Columbia                4.23

Note that the scores of the first two schools are identical (to two decimal places), and the first four scores differ by less than 1 percent. Given the uncertainties in the data, it seems reasonable to suppose that the ranking could have turned out differently. If the whole survey had been repeated, the first few schools might have appeared in a different order. Doctoral candidates in mathematics are presumably sophisticated enough to understand this point. Nevertheless, the spot at top of the list still carries undeniable prestige, even when you know that the distinction could be merely an artifact of statistical noise.

The committee appointed by the NRC to conduct the new graduate-school study wants to avoid this “spurious precision problem.” They’ve adopted some jazzy statistical methods—mainly a technique called resampling—to model the uncertainty in the data, and they’ve also decreed that the results will be presented differently. There will be no sorted master list showing overall ranks in descending order. Instead the programs in each discipline will be listed alphabetically, and each program will be given a range of possible ranks. For example, a program might be estimated to rank between fifth place and ninth place. Let’s call such a range of ranks a rank-interval, and denote it {5, 6, 7, 8, 9} or {5–9}.

For a hypothetical set of 10 institutions, A through J, here’s what a set of rank-intervals might look like.

Acknowledging the uncertainty in your findings is commendable. But let’s be realistic. If you actually want to make use of these results—for example, if you’re a student choosing a grad-school program—the first thing you’re going to do is sort those bars into some sort of rank order, trying to figure out which school is best and how they all stack up against one another. In other words, you’re going to undo all the elaborate efforts the NRC committee has put into obscuring that information.

Below is one possible ordering of the bars. I have sorted first on the top of the rank-intervals, then, if two columns have the same top rank, I’ve sorted on the bottom rank. Other sorting rules give similar but not identical results. For example, sorting on the midpoints of the intervals would interchange columns B and F.

Question 1. Does sorting a set of rank-intervals by one of these simple rules yield a consistent and meaningful total ordering of the data? To put it another way, can you trust this attempt to reconstruct a ranking?

I hasten to add that this is not really a practical question about finding the best grad school. If you’re facing such a choice in real life, the NRC rank-intervals are not the only available source of information. But, for the sake of the mathematical puzzle, let’s pretend that all we know about schools A through J is embodied in those ranges of rankings.

It turns out that rank-intervals have some fairly peculiar behavior. Ranges of ratings are not a problem. If the NRC merely gave each school a fuzzy rating on the 0-to-5 scale, no one would have much trouble interpreting the results. But when you turn fuzzy ratings into fuzzy rankings, there are hidden constraints. For example, not all sets of rank-intervals are well-formed.

The set at left is impossible because there’s no one in last place. (We can’t all be above average.) The example at right is also nonsensical because D has no ranking at all. For a set of rank-intervals to be valid, there has to be at least one entry in each row and each column.

That’s a necessary condition, but not a sufficient one, as the two graphs below illustrate.

Do you see the problem with the example at left? Column B has a rank-interval of {1–2}, but in fact B can never rank first because A has no alternative to being first. The case at right is conceptually similar but a little subtler: If B is ranked third, then either first place or second place will have to remain vacant.

The underlying issue here is the presence of constraints or linkages within a set of rankings. Suppose you have calculated ratings and rankings of several schools, and then some new information turns up about one school. You can change the rating of that school without any need to adjust other ratings, but not so the ranking. If a school goes from third place to fourth place, the old fourth-place school has to move to some other rung of the ladder, and somebody has to fill the vacancy in third place. These interdependencies are obvious in a non-fuzzy ranking, but they also exist in the fuzzy case. You can’t just assign arbitrary rank-intervals to the items in a set and assume they’ll all fit together. This observation leads to a second question:

Question 2. What are the admissible sets of rank-intervals? How do we characterize them?

I have a partial answer to this question. It goes like this. Any ranking of k things must be a permutation of the integers from 1 through k. A permutation can be embodied in a permutation matrix—a square k × k matrix in which every row has a single 1, every column has a single 1, and all the other entries are 0. For example, here are the six possible 3 × 3 permutation matrices:

They correspond to the rankings (1, 2, 3), (1, 3, 2), (2, 1, 3), (3, 1, 2), (2, 3, 1) and (3, 2, 1).

Since a permutation matrix represents a specific (non-fuzzy) ranking, we can build up a set of rank-intervals by taking the OR-sum of two or more permutation matrices. What do I mean by an OR-sum? It’s just the element-by-element sum of the matrices using the boolean OR operator, ∨, instead of ordinary addition. OR has the following addition table:

                      0 ∨ 0 = 0
                      0 ∨ 1 = 1
                      1 ∨ 0 = 1
                      1 ∨ 1 = 1

For the first two 3 × 3 matrices shown above the arithmetic sum is:

whereas the OR-sum looks like this:

Every valid set of rank-intervals must correspond to an OR-sum of permutation matrices, simply because a set of rank-intervals is in fact a collection of permutations. The converse also holds: Any OR-sum of permutation matrices yields an admissible set of rank-intervals. Thus the OR-sums of permutation matrices—let’s call them ormats for brevity—are in one-to-one correspondence with the admissible sets of rank-intervals. (There’s just one catch when applying this idea to the NRC study. The columns of an ormat may well have “gaps,” as in the column pattern (0 1 1 0 0 1 1), which corresponds to the rank-interval {2–3, 6–7}. Will the NRC allow such discontinuous ranges in their grad-school assessments? Perhaps the issue will never come up in practice. In any case, I’m ignoring it here.)

Arithmetic sums of permutation matrices form an open-ended, infinite series; in contrast, there are only finitely many distinguishable OR-sums. The reason is easy to see: Ormats have k² entries, each of which can take on only two possible values, and so there can’t be more than \(2^{k^{2}}\) distinct matrices. Because of the various constraints on the arrangement of the entries, the actual number of ormats is smaller. For example, at k = 3 the \(2^{k^{2}}\) upper bound allows for 512 ormats, but there are only 49:

Thus we come to the next question.

Question 3. For each k ≥ 1, how many distinct ormats can we build by OR-ing subsets of k × k permutation matrices? Is there a closed-form expression for this number?

I have answers only for puny values of k.

   k       upper bound        # of ormats  
   1                 1                  1
   2                16                  3
   3               512                 49
   4            65,536              7,443
   5        33,554,432          6,092,721
   6    68,719,476,736                  ?

The tallies of ormats were calculated by direct enumeration, which is not a promising approach for larger k. (I note—to spare folks the bother of looking—that the sequence 1, 3, 49, 7443, 6092721 does not yet appear in the OEIS.)

To extend this series, we might try to exploit the internal structure and symmetries of the ormats. By sorting the columns and rows of the matrices, we can reduce the 49 3×3 ormats to just six equivalence classes, with the following exemplars:

Enumerating just these reduced sets of matrices should make it possible to reach larger values of k, but I have not pursued this idea. (Furthermore, the two-dimensional sorting of matrices looks to be a curiously challenging task in itself.)

By the way, I think the number of ormats will approach the \(2^{k^{2}}\) upper bound asymptotically as k increases. Many of the features that disqualify a matrix from ormathood—such as all-zero rows or columns—become rarer when k is large. I have tested this conjecture by generating random (0,1) matrices and then counting how many of them turn out to be ormats.

For k = 1 through 5 the results are in close agreement with the actual counts of ormats, and up to k = 10 the trend is clearly upward. But continuing this inquiry to larger values of k will depend on a positive answer to the next question.

Question 4. Given a square matrix with (0,1) entries, is there an efficient algorithm for deciding whether or not it is an OR-sum of permutation matrices, and thus an admissible set of rank-intervals?

The question asks for a recognition predicate—a procedure that will return true if a matrix is an ormat and otherwise false. If efficiency doesn’t matter, there’s no question such an algorithm exists. At worst, we can generate all the k × k ormats and see if a given matrix is among them. But that’s like saying we can factor integers by producing a complete multiplication table. It just won’t do in practice. Isn’t there a quick and easy shortcut, some distinctive property of ormats that will let us recognize them at a glance?

If we could replace the OR-sum with the ordinary arithmetic sum, the answer would be yes. Permutation matrices have the handy property that all rows and columns sum to 1. An arithmetic sum of r permutation matrices has rows and columns that all sum to r. (It is a semi-magic square.) The converse is also true (though harder to prove): If a matrix of nonnegative integers has rows and columns that all sum to r, it is a sum of r permutation matrices. This fact yields a simple test: Sum the rows and the columns and check for equality.

Unfortunately, the trick won’t work for ormats, because the boolean OR operation throws away even more information than summing does. Because 0 ∨ 1 = 1 ∨ 0 = 1 ∨ 1, infinitely many sets of operands map into the same result, and there’s no obvious way to recover the operands or even to determine how many permutation matrices entered into the OR-sum.

Maybe there’s some other clever trick for recognizing ormats, but I haven’t found it. Let me make the question more concrete. Below are three (0,1) square matrices. Two of them are ormats but the third is not. Can you tell the difference?

If it’s so hard to recognize an ormat, how did I count the ormats among a bunch of randomly generated (o,1) matrices? By hard work: I reconstructed the set of permutations allowed by each matrix. Visualize a permutation as a path threading its way through the matrix from left to right, connecting only non-zero elements and touching each column and each row just once. When you have drawn all possible permutation paths, check to see if every non-zero element is included in at least one path; if so, then the matrix is an ormat. Note that this is not an efficient recognition procedure. In the worst case (namely, an all-ones matrix), there are k! permutations, so this method has exponential running time. But k! is better than \(2^{k^2}\); and, besides, for sparse matrices the number of permutations is much smaller than k!. The 10 × 10 matrix presented as an example at the start of this post gives rise to 580 permutations, a manageable number. Here’s what they look like, plotted as a spider web of red paths across the bar chart.

Every nonzero site is visited by at least one permutation path, so this set of rank-intervals is indeed valid.

This process of lacing permutations through a matrix finally brings me back to Question 1, about how to make sense of the NRC’s fuzzy ranking scheme. Let’s take a small example:

Examining the graph above shows that A must rank either first or second—but which is more likely? In the absence of more-detailed information, it seems reasonable to assume the two cases are equally likely; we assign them each a probability of 1/2. Similarly, B has the rank-interval {1–3}, and so we might suppose that each of these three cases has probability 1/3. Continuing in the same way, we assign probabilities to every element of the matrix.

But wait! This can’t be right; our probabilities have sprung a leak. Any proper set of probabilities has to sum to 1. Our procedure assures that each column obeys this rule, but there is no such guarantee for the rows. In row 1, we’re missing one-sixth of our probability, and in row 2 we have an excess of 1/2; row 4 comes up short by 1/3.

Is there any self-consistent assignment of probabilities for the elements of this matrix? Sure. As a matter of fact, there are infinitely many such assignments, including this one:

I’ll return in a moment to the question of how I plucked those particular numbers out of the air, but note first what they imply about the ranking of items A through D. For item A, with the rank-interval {1–2}, the odds are two-to-one that it ranks first rather than second. B has the behavior we expected from the outset, with probability uniformly distributed over the three cases. But if you pick either C or D, each with the rank-interval {2–4}, your chance of getting second place is only 1/6, and half the time you’ll be in last place.

Where do these numbers come from? Instead of starting with the assumption that probability is uniformly distributed over each rank-interval, assume that each possible permutation of the ranks is equiprobable. For this matrix there are six allowed permutations: (1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (2, 1, 3, 4) and (2, 1, 4, 3). Observe that four of the six ordering put A first, and only two permutations place A second. We can also tally up such “occupation numbers” for all the other matrix elements:

Dividing these numbers by the total number of permutations, 6, yields the probabilities given above.

We can do the same computation for the 10 × 10 example matrix, which turns out to allow 580 permutations:

If you care to check, you’ll find that each column and each row sums to 580; dividing all the entries by this number yields a probability matrix with columns and rows that sum to 1 (also known as a doubly stochastic matrix).

This process of tabulating permutation paths recovers some of the information we would have gotten from the arithmetic sum of the permutation matrices—information that was lost in the OR-ing operation. But we get back only some of the information because we have to assume that each permutation included in the OR-sum appears only once. (This is just another way of saying that the allowed permutations are equiprobable.) There’s no particularly good reason to make this assumption, but at least it leads to a feasible probability matrix.

Is there any way of calculating the entries in the doubly stochastic matrix without explicitly tracing out all the permutation paths? I’m sure there is. I think the construction of the matrix can be approached as an integer-programming problem, and perhaps through other kinds of optimization technology. What seems less likely is that there’s some simple and efficient shortcut algorithm. But I could be wrong about that; there’s a lot of mathematics connected with this subject that I don’t understand well enough to write about (e.g., the Birkhoff polytope). I hope others will fill in the gaps.

Getting back to the assessment of grad schools—have we finally found the right way to understand those rank-intervals that the NRC promises to publish any day now? My sense is that a semi-magic square (or, equivalently, a doubly stochastic matrix) will give a less-misleading impression than a simple eyeball sorting on the spans or midpoints of the rank-intervals. But what a lot of bother to get to that point! How many prospective grad students are going to repeat this analysis?

Acknowledgment: Thanks to Geoff Davis of PhDs.org for introducing me to this story. PhDs.org will have the new ratings as soon as the NRC releases them, and may even find a way to make them intelligible! Disclaimer: I’ve done paid work for the PhDs.org web site (but this is not a paid endorsement).

Update 2010-07-27: If you’ve gotten this far, please read the comments as well. A number of commenters have provided important insights and context, which have helped me understand what’s going on in the matrices I’ve been calling ormats. But I’m still a bit murky about the best way to recognize and count them. I’m not sure that publishing my still-murky thoughts is terribly helpful, but maybe someone else will read what follows and give us a dazzling, gemlike synthesis.

For the ormat-recognition problem (Question 4 above), three basic approaches have been mentioned: enumerating the permutation paths through the matrix, examining matrix minors, and looking for perfect matchings in a bipartite graph defined by the matrix. It seems to me that all of these methods are doing the same thing.

Start with Barry Cipra’s method of minors. The basic operation is to choose a nonzero matrix element, then delete the row and the column in which that element occurs. You then apply the same operation to the remaining, smaller matrix.

In tracing permutation paths, we’re looking for sequences of nonzero elements, drawing one element from each column and each row. A way of organizing this search is to choose a nonzero element and then, after recording its location, delete the corresponding column and row, so that no other elements can be chosen from that column or row.

In the method based on Hall’s theorem, as explained by John R., we view the ormat as the adjacency matrix of a bipartite graph, where every nonzero element designates an edge connecting a row vertex to a column vertex. To find a matching, we delete an edge, along with the two vertices it connects (and also all the other edges incident on those vertices). Then we recurse on the smaller remaining graph. (See further update below.) If you translate this operation on the graph back into the language of matrices, deleting an edge and its endpoints amounts to deleting a row and a column of the adjacency matrix.

I am not asserting that these three algorithms are all identical, but they all rely on the same underlying operation. To say more, we would need to consider the control structure of the algorithms—how the basic operations are organized, how the recursion works, all the details of the bookkeeping. I don’t trust myself to make those comparisons without trying to implement the three methods, which I have not yet done. However, at this point I just don’t see how any method can guarantee correct results without something resembling backtracking (or else exhaustive search through an exponential space). After all, we’re not looking for just one matching in the graph, or one decomposition into matrix minors, or one permutation path; we have to examine them all.

Here’s a further hand-wavy argument for the essential difficulty of the task. For a (0,1) matrix, the number of permutation paths that avoid all zero entries is equal to the permanent of the matrix. Computing the permanent of such a matrix is known to be #P-complete.

Update 2010-07-31: With lots of help from my friends, I think I finally get it. Although there could be as many as k! permutation paths in a k × k matrix, you don’t need to examine all of the paths to decide whether or not the matrix is an ormat. It’s enough to establish that one such path passes through each nonzero element. This is what the algorithm based on Hall’s theorem does. As Frans points out in a comment below, I misunderstood the essential nature of that algorithm (in spite of having it explained to me several times). There is no recursive deconstruction into progressively smaller matrix minors; instead, we just loop over all the nonzero elements of the matrix, find the minor associated with each such element, then check for a perfect matching in the minor. (Still more refinements are possible—but already we have a polynomial algorithm.)

With this efficient recognizer predicate, it’s easy to measure the proportion of ormats in random matrices at larger values of k:

As expected, the fraction of ormats approaches 1 beyond about k = 20.

So much for identifying ormats. I am still unable to extend the series of exact counts beyond k = 5. The tabulations for random (0,1) matrices suggest that for k = 6 there should be about 20 billion ormats, and counting that high is just too painful. I need to work out the symmetries of the problem.

As far as I can tell, assigning exact probabilities to the nonzero matrix elements requires a full enumeration of all the permutation paths, and thus a calculation equivalent to the permanent. There may be a useful approximation.

Barry Cipra asks a really good question: The permanent tells us the maximum number of permutations that could possibly be included in a given ormat, but what is the minimum number? A naive upper bound is the number of 1s in the matrix, but I don’t see an easy path to an exact count. But enough for now.

Martin Gardner died over the weekend. He was 95 and living in Norman, Oklahoma, not too far from his birthplace in Tulsa.

Like many others, I grew up on Martin’s “Mathematical Games” column in Scientific American. Later I joined the staff of that magazine—but don’t imagine that Martin and I became office buddies. As a matter of fact, I never once saw him in the office. He worked at home. He attended none of our editorial meetings. We never had lunch together. Indeed, I never would have met him at all except for the coincidence that we lived a few blocks apart, and every now and then I would be called upon to deliver a package of urgent proofs. (By the way, his address in those days was on Euclid Avenue!)

Someone on the magazine staff described Martin as “a shy woodland creature,” and the tag stuck. Looking back, however, I think it tells only half the story. Yes, Martin was no schmoozer, and he preferred to stay out of the spotlight, both in person and in print. Most of his best-known columns reported on someone else’s discoveries—Conway’s game of life, RSA’s cryptosystem, Penrose’s tilings. He delighted in annotating other people’s work, as in his celebrated edition of Lewis Carroll. Yet he was anything but timid or retiring. In an argument, the shy woodland creature was a grizzly bear. He had strongly held opinions and philosophical convictions, and he knew the worth of his own work. He was a man of ideas, to be taken seriously, yet also a man who had fun with his ideas.

As a celebration of Martin’s peculiar genius, I would like to revive the little puzzle that formed the basis of his very first column, in January of 1957. (He had published a few articles in Scientific American earlier, but this was the first column to appear under the “Mathematical Games” title.)

On the bingo card above, choose any number, circle it, and cross out all the other numbers in the same column or row. Now select a second number from among those that remain unmarked, and again circle your choice and then cross out the rest of the column and row. Continue in this way until there are no unmarked numbers left to choose.

The sum of the circled numbers is 57. How did I know that? How did Martin construct the matrix?

Update 2010-05-29: Here is Martin’s own explanation of the 1957 puzzle:

Like most tricks, this one is absurdly simple when explained. The square is nothing more than an old-fashioned addition table, arranged in a tricky way. The table is generated by two sets of numbers: 12, 1, 4, 18, 0 and 7, 0, 4, 9, 2. The sum of these numbers is 57. If you write the first set of numbers horizontally above the top row of the square, and the second set vertically beside the first column [see Fig. 9], you can see at once how the numbers in the cells are determined. The number in the first cell (top row, first column) is the sum of 12 and 7, and so on through the square.

You can construct a magic square of this kind as large as you like and with any combination of numbers you choose. It does not matter in the least how many cells the square contains or what numbers are used for generating it. They may be positive or negative, integers or fractions, rationals or irrationals. The resulting table will always possess the magic property of forcing a number by the procedure described, and this number will always be the sum of the two sets of numbers that generate the table.

It’s a cruel irony: As the citizens of Zimbabwe sink into bitter poverty, they are becoming millionaires and billionaires. Inflation is eroding the value of the Zimbabwean dollar so rapidly that everyday transactions turn into lessons in the arithmetic of large numbers. When the photo above was made on July 17, the largest currency denomination in circulation was a note for ZW$50,000,000,000. Last week the nation’s central bank issued a ZW$100,000,000,000 bill. (I’ll spare you the trouble of counting zeroes: That’s 10¹¹, or 100 billion by American reckoning.)

The Zimbabwean inflation is the worst in the world at the moment, but it is not (yet) setting all-time records. Probably the most famous episode of extreme inflation was that of the German Weimar Republic (a story told vividly in Erich Maria Remarque’s novel The Black Obelisk.) In 1921, German marks traded at about 60 to the U.S. dollar; two years later, in December of 1923, the exchange rate was 4.2×10¹² per dollar. The Hungarian inflation following World War II reached even greater numerical heights. In a single year the exchange rate for the Hungarian pengo went from 100 per U.S. dollar to 4×10²⁹. As Feynman said, astronomical numbers are dwarfed by economical ones.

Takayuki Mizuno, Misako Takayasu and Hideki Takayasu have analyzed the German and Hungarian episodes of “hyperinflation.” (Citation: Physica A 308 (2002) 411; there’s also an arXiv preprint.) Inflation at its worst, they find, proceeds at a doubly exponential rate. In other words, prices rise not just as an exponential function of time—exp(t)—but as an exponentiated exponential—exp(exp(t))—or:

This growth law has a simple meaning in terms of everyday experience. With “ordinary,” single-exponential inflation, prices have a constant doubling time. If bus fare was 1 million last month and 2 million this month, it will be 4 million next month. Under double-exponential growth, the doubling time itself decreases exponentially. In the last months of the Hungarian inflation the doubling time fell from about 20 days to 15 hours.

On a logarithmic scale, a simple exponential function yields a straight-line graph. Here is the Mizuno-Takayasu evidence that the final phase of the Hungarian inflation was superexponential:

And here are the data for the final six months plotted as log(log(p(t))), showing a simple linear trend:

How does the Zimbabwean economy look when submitted to this kind of scrutiny? I don’t know of a reliable source of data on prices in Zimbabwe, but foreign exchange rates can serve as a rough proxy. Until three months ago, the official ZW$ rate was pegged at roughly 30,000 per US$, but on May 10 the currency was allowed to float free, and the rate immediately jumped to 190,000,000 ZW$ per US$. By July 31 the rate had reached 57,381,544,140. Thus the 50 billion ZW$ note in the photo above was worth a little less than a 1 US$ by the end of last month. And that’s at the official rate of exchange; the street value is reportedly about a tenth of the official quote.

Here’s how the official exchange rate has varied in the 84 days between May 10 and August 1, as plotted on a linear scale:

And here’s the same data after a logarithmic transformation:

Although there’s more bumpiness here than in the Mizuno-Takayasu data, the trend looks reasonably linear to me. The fitted line has slope 0.03358, which yields a doubling time of about nine days. I see no hint of superexponential growth. I’d like to think this is an encouraging sign, a glimmer of hope that Zimbabwe will be spared an even more pernicious phase, when even inflation has inflation.

Runaway inflation is usually blamed on the incompetence or malevolence of governments and the central banks that implement their policies. In the case of Zimbabwe, the government of Robert Mugabe certainly has a lot to answer for. The country was once the shining success story of southern Africa—I have friends who migrated across the continent to go to school there—but the nation is now a basket case, and inflation is only one of many urgent crises. (The unemployment rate is reported to be 80 percent.) The Mugabe regime can’t escape blame for this situation. Still, it seems that hyperinflation is not to be explained purely in terms of fundamental economic imbalances—too many dollars and not enough goods. Sometimes it seems there is also a psychological component. When you believe that prices will double next week, you raise your own prices in anticipation. It’s a self-reinforcing process.

One sign of such a feedback loop in the inflationary spiral is that inflation sometimes stops even though the underlying economic situation hasn’t really changed. The Weimar hyperinflation ended with the introduction of the Rentenmark, which was set equal to 10¹² old marks but really had no firmer backing than the earlier Papiermark. The change in currency did nothing to solve Germany’s problems of debt and unemployment, but the inflation ended anyway. Evidently, people chose to believe that the value of the Rentenmark would remain stable, and it did.

The central bank of Zimbabwe has just announced a similar effort at currency reform, devaluing the ZW$ by a factor of 10¹⁰. In other words, the ZW$100,000,000,000 note introduced a week ago is equal in value to a new ZW$10 bill. According to press reports, the main motive for the change was simply logistical convenience:

Gideon Gono, the Central Bank governor, … acted because the high rate of inflation was hampering the country’s computer systems. Computers, electronic calculators and automated teller machines at Zimbabwe’s banks cannot handle basic transactions in billions and trillions of dollars. (AP/Baltimore Sun)

But perhaps one can hope that the newly denominated currency will bring more than numerical benefits. Over the weekend, the official exchange rate has held at 6.569 new Zimbabwe dollars to the U.S. dollar. We’ll have to wait a few more days to see if the curve has really flattened out.

Update 2008-09-04: With another month of exchange-rate data, here’s what the situation looks like:

The blue line in the semilog graph is the same as the one in the corresponding earlier graph—that is to say, it is fitted to the first 80 days of data. It appears that the inflation rate has diminished slightly since the revaluation at the end of July. But that slightly lower rate is still formidable; in a little more than a month the value of the new Zimbabwe dollar has fallen from about 15 cents (U.S.) to about 2 cents.

Update 2008-10-02: After another month, what passes for good news is that the rate of exponential growth does not seem to be growing:

On the other hand, news reports suggest that the situation in Harare is bleaker than ever. Money is scarce as well as nearly worthless; people stand in line all night for the privilege of withdrawing the equivalent of a dollar or two from their own bank accounts. (Note that the equivalent of $1 U.S. is $ZW137 in the devalued currency issued in August. In pre-devaluation Zimbabwe dollars, it comes to $ZW1.37 trillion.)

Isn’t it curious that both here in the U.S. and in Zimbabwe, the financial pages are filled with such enormous numbers.

Update 2008-11-02: One more month of data:

Still no sign of “hyperinflation”—if that term is taken to mean doubly exponential growth—but that can’t be much solace to the Zimbabweans whose currency has yet again lost three-fourths of its value over the course of a month. Adjusting for the August devaluation, one U.S. dollar now buys 5.6 trillion Zimbabwean dollars.

I’ve been Scrabbling by email lately. In today’s game my partner started out by playing

                    H
                    E
                    X

and I responded with

                   A H
                   W E
                   E X

At this point my opponent might well have continued with another three-letter word to make a tidy square block such as:

                  Y A H
                  E W E
                  S E X

In actuality she did something quite different. She played a seven-letter “bingo,” using all her letters to earn a 50-point bonus; as a result I’m hopelessly far behind in the scoring. But let us say no more about the tawdry details of winning and losing; there’s a puzzle here. Looking at that three-by-three block of letters and words, it occurs to me there must surely be legally reachable configurations of a Scrabble board that have no legal continuation. Scrabble rules say that, except for the first move, letters can be added to the board only on squares adjacent to existing letters, and all sequences of two or more letters (both vertically and horizontally) must be dictionary words. The rules say nothing about the situation where continued play is impossible.

I’m sure there must be many stymied positions, where no words can be formed, regardless of what letters you have on your rack. Or so I assert; but, the fact is, I haven’t been able to find even one such configuration. A cursory examination of the list of all allowed two-letter words argues that no two-by-two block of letters can be stymied. What about two-by-three or three-by-three blocks? Somebody must have settled these questions, but my Googling has failed to find the answer. What is the smallest stymied position? (I don’t require that a solution be a rectangular block of letters, but having stray letters dangling off the edges of a block makes it easier rather than harder to form words.)

A few weeks ago I playfully suggested that the Democratic nominee for POTUS might be selected this year by a game of hex played on the map of the lower 48 states. I emphasize the word “playfully.” This was not a serious suggestion. But life has a way of overtaking us, even in our most frivolous moments. As the primary season now spirals down to the last bitter dregs, the nomination remains undecided, and so does the game of electoral hex.

The goal in this version of hex is to assemble a continuous chain of states that spans the country either east to west or north to south (or both). As the map below shows, both candidates are tantalizingly close to this goal, but neither has actually achieved it. The recent round of voting in North Carolina and Indiana, and then yesterday’s result in West Virginia, failed to clinch it.

But it will all be over soon. Kentucky is the key. The nomination may or may not be settled after that state’s primary election next Tuesday, but the game of hex will surely be decided, one way or the other. Whichever candidate wins Kentucky will form both east-west and north-south chains, and his or her opponent will be shut out from creating either kind of chain.

Update 2008-05-21: Game over. As the map from today’s Times shows, Hillary Clinton percolates. She can drive coast to coast or border and border and never stray outside of states that supported her candidacy. As far as I know, she has not yet cited this fact in support of her decision to stay in the race; maybe she’s saving that argument for the convention.

The interesting thread of comments on Wednesday’s post about “electoral hex” led me to look more closely at the graph for the Lower 48 states:

I’ve constructed the graph—and it wasn’t easy, by the way—on the rule that two states are linked by an edge if they share a common boundary of more than one point. National boundaries are determined in the same way. Nodes half-shaded orange are east coast or west coast states; those half-shaded blue are north coast or south coast states. Underwater boundaries count in this scheme. Thus, for example, Illinois is adjacent to Michigan because there’s a section of boundary line between them in the middle of Lake Michigan. Similarly, Wisconsin is not a north coast state, even though it abuts Lake Superior, because it has no shared boundary with Canada.

Some properties of the graph:

• It has 48 vertices (no surprise!), 107 edges, and 60 faces (excluding the outer, embedding, face).
• It’s a planar graph (again no surprise).
• All but one of the faces are triangles. The exception is a quadrilateral formed by the Four Corners states.
• The degree sequence runs: 8 8 7 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 2 2 2 2 1. (The two octopus states are Missouri and Tennessee. The singleton is Maine.)
• The diameter is 11, and the characteristic path length (the length of the shortest path between two vertices, averaged over all pairs of vertices) is 4.1.

What about playing hex on this graph? In standard hex, one player tries to connect east to west, and the other player has to find a path from north to south. Those rules would hardly be fair with such an asymmetrical graph. So let’s adopt a modified version of the rules that I suggested for the political primaries: The players take turns, and the first to create a pathway either from north to south or from east to west is the winner. (Barry Cipra suggested that we call this the game of pox.)

Is the game a sure win for either the first or the second player (assuming correct play on both sides)? If so, what’s the right opening move?

It’s easy to find some really bad moves. As Paul Zorn noted, the New England states are not worth bothering with, because they’re sealed off by New York. But it turns out New York is not a very smart choice either. (If you take New York, I take Pennsylvania; then, if you want to get anywhere, you have to win all three of New Jersey, Delaware and Maryland, which gives me three chances to block you.)

The shortest bridging paths are along the I-5 and I-15 corridors, out west. In particular, there are five ways to construct a path of length 3 from north-to-south with the states WA, OR, CA, ID, NV, UT, AZ. But I don’t believe there’s any way a player can force a win using these states alone.

In hex, the best opening moves are generally near the middle of the board. The analogous pox strategy would choose one of the central and highly connected states such as MO, TN or KY. This seems sensible, but the games are too complicated to work out by hand.

On a regular 7 × 7 board, hex has been fully solved: The entire game tree is known for every possible opening move, so that each cell of the board is classified as a win or a loss for the first player. (This impressive feat of analysis and computation was done by R. Hayward, Y. Björnsson, M. Johanson, M. Kan, N. Po and J. van Rijswijck of the University of Alberta.) The Lower 48 graph is roughly the same size as the 7 × 7 hex board, so a similar method would presumably be feasible. Feasible but not easy.

Another question is whether the game can end without a winner. Jim Ward asked about this in the comments to the previous post, but I didn’t really understand the full force of the question. In the game of presidential primaries—where elections can occur simultaneously in multiple states—two or more winning paths could be created at the same time. Enforcing a strict alternating-turns policy, as I’m assuming here, eliminates that problem. But there’s another way the game might possibly fail to yield a winner, namely if all the states are chosen but there’s still no path along either axis.

In hex it’s easy to show that no such drawn game is possible. You can block the progress of a north-south path only by completing an east-west path. Hex is played on a regular and symmetrical lattice whose structure is equivalent to this graph:

The graph of the Lower 48 is not quite so uniform. The diagonals don’t all go the same way, and there’s that quadrilateral with no diagonal at the Four Corners. Does that feature (or some other oddity of the graph) allow all the states to be divided into two categories without creating either a north-south or an east-west path? I don’t think so; I haven’t been able to find such an assignment. But I’m not quite sure it can’t exist.

Update 2008-03-17: By popular request, I’ve posted a PDF version of the lower-48 graph and a text file with the vertex list and edge list. The vertex and edge lists actually take the form of a Scheme definition, but the semantics should be obvious and translation to other formats should be straightforward.

Democrats here in the U.S. are having quite a primary season. With high hopes of winning the general election in November, we are deadlocked over which candidate to nominate. Should we just wait to let the delegates slug it out at the convention in August? Should the trailing candidate withdraw in the interest of party unity? Could the two contenders flip a coin to decide who gets to head the ticket and who gets to drink the bucket of warm piss?

I have a proposal. Here’s the geography of the current situation, following Obama’s victory in Mississippi yesterday (map brazenly swiped from the New York Times):

The solution is obvious, no? We follow the rules of the game of hex: The nomination goes to the first candidate to form a continuous chain of states either east to west (Atlantic to Pacific) or north to south (from the Canadian border to Mexico or the Gulf Coast). Thus the three key states are Pennsylvania (which votes April 22), North Carolina (May 6) and Kentucky (May 20). Clinton could create an east-west chain by taking North Carolina, or a north-south path by winning Kentucky. For Obama, Kentucky creates an east-west link; to construct a north-south chain, Obama would have to win both North Carolina and either Kentucky or Pennsylvania.

I get to cast my ballot on May 6 in North Carolina. All I’m saying for now is that I’m not voting for Ralph Nader.

The answer to the question in the headline is: Too many. After I wrote about Sudoku a couple a years ago, I thought I had cured my addiction; but I’ve been a shameless backslider.

I return to the subject now, in this public confessional mode, to correct an error in my 2006 column. Citing a paper by Bertram Felgenhauer and Fraser Jarvis (pdf), I stated that there are 6,670,903,752,021,072,936,960 Sudoku solutions. This is simply the number of ways of filling in the 81 squares without violating the no-two-alike constraints in rows, columns and blocks. I then went on to discuss the symmetries of the Sudoku grid, and noted:

When all these symmetries are taken into account, the number of essentially different Sudoku patterns is reduced substantially…. although [the reduction] still leaves an impressive number of genuinely different solutions: 3,546,146,300,288, or 4×10¹².

The latter number also comes from Felgenhauer and Jarvis, but I misinterpreted it. Contrary to my statement, it takes into account only a few of the simplest symmetries. The true number of essentially distinct Sudoku solutions is reported in a slightly later publication by Ed Russell and Jarvis. The correct tally is smaller by three orders of magnitude: 5,472,730,538.

Thanks to Richard E. Dickerson for pointing out my error.

It needs to be emphasized that both of these calculations count the number of solutions, not the number of puzzles. For any given solution, there are many possible puzzles, each created by erasing a different subset of the entries in the solution. How many puzzles per solution? Does anyone know? An exact count seems beyond reckoning, but random sampling ought to yield a rough estimate. For example, consider just those puzzles that have 17 initial clues (the apparent minimum). For any given solution grid, there are (81 choose 17) ways of forming a 17-clue puzzle; this works out to 128,447,994,798,305,325. Most of those puzzles will not have a unique solution, and so they are disqualified as proper Sudokus. The random sampling is needed to estimate how many of the potential puzzles are admissible.

I’d try doing the calculation myself, but I’m too busy solving Sudokus.

Update 2008-04-05: Richard E. Dickerson discovered the error mentioned above in the course of preparing his own discourse on Sudoku-solving strategies. That work now has its own web site, The Sudokerson Matrix, where you can download the 61-page PDF file and an accompanying spreadsheet. Dickerson also mentions some other resources on Sudoku strategy, including a recent book by Andrew C. Stuart that I haven’t yet seen. Dickerson comments:

Few books on Sudoku go into strategies in any depth. Stuart’s The Logic of Sudoku (Michael Mepham, 2007) is a wonderful exception. But his book and mine represent the two opposite poles of Sudoku-solving strategy. I regard many of his methods as superfluous, and I am sure that he regards most of my strategies as heretical. You need to have a close look at both.

What I find intriguing in all this is the very idea that we could have contending doctrines of Sudoku strategy. But before the talk of “heresy” gets out of hand and leads to Sudoku holy wars, I’d like to remind everyone that there’s another view of Sudoku in which all strategies become superfluous. Give me a computer and a program for solving constraint-satisfaction problems, and every 9×9 Sudoku yields in milliseconds.

Don Elgee, a retired teacher of mathematics and computer science from Ottawa, sends the following inquiry:

In hockey, when a team is down by a goal with about one minute to go, the goalie is pulled in favor of another offensive player. This illustrates a key point in the strategy of most games.

The object is not to maximize points for, and minimize points against; rather, it is to win or tie the game. As the game nears its conclusion the winning team plays more conservatively and the losing team more radically. But when—and to what extent—should a team change its style of play? 

To clarify the problem I tried to conceive the simplest possible example. It turned out to be much more complex than I expected.

Consider the following simple game: Two players toss a die on alternate turns. The first one to reach 100 is the winner. Players have a choice of two dice—one is normal (6, 5, 4, 3, 2, 1) and the other has faces with 6, 6, 5, 1, 1, 1. When should the losing player switch to the second die?

Is there a mathematical solution? Could you determine this by experimental computer simulation? Does the best strategy depend on the strategy the opponent is likely to choose next?

This simple problem is difficult enough, but the sports situation could better be simulated by adding a third die with 4, 4, 3, 3, 3, 3.

Has this issue been subject to mathematical study? With all the statistics used in professional sport has anyone tried to include this factor? Has it ever been incorporated in computer simulations?

Hockey is not my sport, but these are interesting questions. Inspired by Elgee’s letter, I’ve run a few of the obvious simulations. For the time being, however, I’m not going to say much about the results. What I’ve learned so far does not lead to any simple, pithy rule of thumb that a hockey coach could use to decide when to pull the goalie. Nor do I have a clear grasp of how best to play the dice game. I’m hoping someone else will offer a sharper analysis. Toward that end I have some further notes and observations.

Under the stated rules—alternating turns, with victory going to the first player who reaches 100—the privilege of moving first brings a substantial advantage. In a 100-point game with standard dice, the first player can expect to win about 55 percent of the time. If two players are tied at 99, the first to roll is definitely going to win, regardless of which die is chosen. Because of this bias, we probably need to consider different strategies for the first and second players (like white and black in a chess game). An alternative is to add a further element of randomness to the game: In each round, the player to move is determined by the toss of a fair coin.

Elgee’s two nonstandard dice have faces that sum to 20, less than the standard die’s 21. As a model of hockey tactics, this bias makes sense: It reflects the risk of leaving the goal untended. But if we ignore hockey and just consider the dice game on its own merits, it might be more interesting to explore a symmetrical contest with dice marked 6, 6, 6, 1, 1, 1 and 4, 4, 4, 3, 3, 3. Then we’d be looking at three distributions that have the same mean but different shapes.

Generalizing further, we could allow dice marked with any six non-negative integers that sum to 21. What would happen to the game if we allowed dice such as 100, -16, -16, -16, -16, -15? Or, we could allow the players to choose continuous probability density functions.

It’s easy to imagine circumstances where a “high-risk” die should be advantageous. If the score is 99 to 94, then the trailing player should be willing to accept any trade-off that improves the odds of rolling a 6; even a 6, 6, 0, 0, 0, 0 die is preferable to the standard die. Conversely, if you’re ahead near the end of the game, the “low-risk” die seems favorable. All this suggests there ought to be some fairly simple guidelines for choosing the best die in any situation. But, as Elgee noted, it’s more complicated than you might expect.

In my simulations I found that with the score tied at 94, switching to the 6, 6, 5, 1, 1, 1 die (while your opponent continues to play the standard die) yields an advantage of 2.9 percent. But when the score is 96 to 96, making the same switch produces a deficit of 3.7 percent. Interestingly, choosing the 4, 4, 3, 3, 3, 3 die brings similar results: a gain at 94 to 94 but a loss at 96 to 96. (The simulations used the coin-flip protocol, but the same kind of anomalies appear under other rules.)

It gets even more complicated. The results cited above are for players who always choose the same die, no matter what the circumstances. But Elgee’s rules allow a player to choose a different die on each roll. Presumably, players who adapt or learn will do better than any player with a fixed strategy. I find that a very simple adaptive strategy—use 6, 6, 5, 1, 1, 1 when behind, 4, 4, 3, 3, 3, 3 when ahead, and 6, 5, 4, 3, 2, 1 when the score is even—gives mostly but uniformly good results in end-game situations.

I echo Elgee’s questions about whether this problem has been analyzed previously. As it happens, I’ve been able to find a bit of literature on the original hockey issue. Tom Benjamin’s NHL Weblog has pointers to three papers published in the 1980s in Interfaces, an operations research journal. In those papers several authors—fans of rival teams—derive a formula for the optimal time to pull the goalie. (They find that coaches almost always wait too long.)

I would have thought that Elgee’s nonstandard dice would also have been investigated before. Some years ago Bradley Efron invented a wonderful set of nontransitive dice, which Martin Gardner wrote about (Scientific American, December 1970, pp. 110–114). Gardner also had a column on various kinds of rigged and trick dice (Scientific American, November 1968, pp. 140–146). A game called Piggy has certain elements reminiscent of Elgee’s game, but those elements don’t include dice that differ in variance.

Perhaps I’ve just failed to find the right search term on Google or MathSciNet, and a reader will supply a reference.

Update: Minutes after posting the above, I discovered an article by B. De Schuymera, H. De Meyera and B. De Baetsb, “Optimal strategies for equal-sum dice games” (Discrete Applied Mathematics 154 (2006) 2565–2576), that covers some of the territory. They analyze a game in which players can choose any die whose n faces have the same sum σ. But they consider only games in which players bet independently on each throw of the dice. Under these conditions, they show that with six-sided dice summing to 21, the optimal die is the standard one, with faces marked 6, 5, 4, 3, 2, 1. Again, however, if I understand correctly, this result applies only to individual throws of the die.