The fabled carefree residents of the Carryless Islands in the remote South Pacific have very few possessions, which is just as well, since their notion of arithmetic is ill-suited to accurate record-keeping. When they add or multiply numbers, they follow similar rules to ours, except that there are no carries into other digit positions. Addition and multiplication of single-digit numbers are performed by a process that we would call “reduction mod 10.” Any carry digits are simply ignored. So 9 + 4 = 3, 5 + 5 = 0, 9 × 4 = 6, 5 × 4 = 0, and so on.

With this fable, David Applegate, Marc LeBrun and N. J. A. Sloane introduce a new scheme of arithmetic in a paper newly posted on the arXiv.

And if you think the mathematics sounds trivial, try explaining the structure of this sequence:

21, 23, 25, 27, 29, 41, 43, 45, 47, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 63,

which lists the first 20 “carryless primes.”

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.

The Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables (better known as Abramowitz and Stegun) is a much-storied book. Not that it’s a book full of stories; truth is, there’s not much of a narrative thread running through those formulas, graphs and mathematical tables. But it’s a book with a story behind it.

The story began in the late 1930s, when the Mathematical Tables Project was launched in a factory building on the West Side of Manhattan. Supported by the Works Progress Administration, the Tables Project had dual aims: first, preparing high-quality tables of trigonometric functions, logarithms, and the like; and, second, providing work for unemployed New Yorkers. The 450 human computers hired for the project were chosen more on the basis of need than skill, and the work was done on a kind of numerical assembly line. According to David Alan Grier:

Each group was taught to perform a single arithmetic operation. One group knew how to add positive numbers, a second to subtract, and the third to multiply single digits. The last and most sophisticated group did long division.

I have to admit there’s a certain nightmare aspect to this scene. Working in a numbers factory sounds no more appealing than stamping sheet metal all day, although there was less danger of getting a finger crushed in the machinery.

A few years later, the Tables Project was swept up in war work; then, afterward, many of the key personnel moved to the National Bureau of Standards (now NIST, the National Institute of Standards and Technology). There they conceived the Handbook. Apparently the initial plan was a greatest-hits album of tables, but by the early 1950s the future of table-making was looking pretty dim. And so the project changed course and put more emphasis on the mathematical functions that lay behind the tables—familiar functions such logarithms, more specialized ones such as Bessel functions and some more recondite topics such as Mathieu functions and orthogonal polynomials. There were still tables listing numeric values of functions, but the Handbook also presented the mathematics you would need to evaluate (or approximate) the function for yourself.

The editors in charge of the Handbook were Milton Abramowitz and Irene Stegun, both veterans of the New York table office. They recruited about 30 young mathematicians to write chapters. In 1958 Abramowitz died suddenly; Stegun saw the project through to publication in 1964.

The Handbook seems an unlikely best-seller, but the U.S. government has distributed more than 150,000 copies, and editions from other publishers are estimated to bring the total copies in print to something near a million. (As a product of government work, the Handbook is not covered by copyright; there are scanned versions on the web.)

Announced last week is a new Handbook, officially retitled the NIST Handbook of Mathematical Functions. The ink-and-paper version, which I have not yet seen, is published by Cambridge University Press. Perhaps even more interesting is the web edition, called the NIST Digital Library of Mathematical Functions (DLMF), which I have just begun to explore. It is recognizably the same book as Abramowitz and Stegun, with the same terse style of presentation. But much has changed. The hundreds of pages of tables are finally gone; this is not the place to look up the sine of 23 degrees. But there are handsome color graphics now, and a new emphasis on methods of computation, including pointers to recommended software. And the selection of topics has expanded somewhat. For example, there are new chapters on the Painlevé equations and on functions whose argument is a matrix. Elsewhere, the Lambert W function (a personal favorite of mine) is a newcomer to the chapter on elementary functions.

Apart from the content, the DLMF is interesting as an experiment in presenting mathematics on the web. It’s the most ambitious project I’ve seen based on MathML, and it seems to work well, at least when viewed in recent versions of Firefox. (In other browsers I’ve tried, MathML gets garbled, but the equations can be displayed as images and are still quite readable in that format—even with an ancient version of Internet Explorer.) Here’s part of a page as seen in Firefox:

Mousing over the “i” icon at the right margin provides access to encodings of the equations in MathML or TeX and as PNG images, as well as definitions, cross-references and such. Very slick.

The editor in chief of the new Handbook is Frank W. J. Olver of the University of Maryland College Park and NIST, whom I have mentioned before both here at bit-player and in American Scientist. As a young mathematician, half a century ago, Olver wrote one of the Handbook chapters on Bessel functions. (As an even younger mathematician, in the 1940s, he worked with Alan Turing at the National Physical Laboratory in Britain.) There are three more principal editors: Daniel W. Lozier, Ronald F. Boisvert and Charles W. Clark, all of NIST, as well as a long roster of associate editors and domain experts. It’s too soon to say whether some combination of these names will eventually replace the moniker “Abramowitz and Stegun.”

Notes: The quotation above from David Alan Grier appears in “The Math Tables Project of the Work Projects Administration: The Reluctant Start of the Computing Era,” IEEE Annals of the History of Computing, Vol. 20, No. 3, 1998. Grier has also written a profile of Stegun in “Irene Stegun, the Handbook of Mathematical Functions, and the Lingering Influence of the New Deal,” American Mathematical Monthly, August-September 2006. Boisvert and Lozier have written a brief account of the history of the Handbook. (Oddly, the images in this PDF file are negatives.)

A few weeks ago I committed an act of deception. I needed a graph showing a few sets of data points, with curves fitted to them. This was strictly for show; neither the points nor the curves would mean anything; it just had to look good. So, out of laziness, I did it backwards: I drew a few arbitrary curves, chose some points on the curves, and then perturbed those points by small random displacements. The result is Figure 2 in Home Baked Graphics.

Apparently I have gotten away with this fakery, since no one has complained. But it nags at my conscience. The curves shown are not fitted to the given data points, but the other way around. And barring an extraordinary coincidence, we can’t expect the curves to give the best possible trend through the points.

I began to wonder: What if you iterate this process? In other words, you start out by generating a curve, say a simple quadratic:

You select a few points on the curve, in this case the five points with x coordinates equal to {1, 2, 3, 4, 5}. You perturb the y coordinates of these points by choosing random variates from a normal distribution with unit standard deviation:

Now you do a least-squares fit to the five selected points, generating a new quadratic:

At this stage you can start over, selecting five points on the new curve and randomly displacing them:

And a least-squares fit to those five purple points yields yet another quadratic curve:

So what happens if we continue in this way, repeatedly jiggling a few points on a curve and then fitting a new curve to the jiggled points? There’s a seeming hint of convergence in the three curves we’ve seen so far. They look like they might be approaching some fixed trajectory. Could that possibly be true?

I’ll post an update in a few days to say what more I’ve been able to learn about this question. If you just can’t wait, click here for a sneak peek.

The promised update (2010-04-02): You can’t fool anybody in this crowd, not even circa April 1. Of course there is no convergence to a limiting curve.

The butterfly above shows 100 quadratic curves, selected from 100,000 iterations of the least-squares fitting process, with that process applied at points whose x values are in the set {–5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5}. Clearly enough, the coefficients of the quadratic curves are wandering widely; in particular note that the x² term can be either positive or negative, yielding curves that are concave upward or downward.

On the other hand, these are not just any old quadratic curves. They all have their apex (the point where f′(x) = 0) fairly close to the origin. Here’s a closer view:

(When Ros glanced at this graph on the screen, her comment was: “Bad hair day.”) And here’s a tracing of the wandering of the apex (taken from a different, longer, run of the program):

The path has the look of a classic random walk, but note that it’s confined to a narrow range of x values. What attractive force holds the apex in this region? It’s the decision to fit the curves to points in the interval –5 ≤ x ≤ 5. Here’s what happens if we instead fit to points in the range 100 ≤ x ≤ 110:

And note that without the symmetry-breaking x¹ term, the curve would always be a parabola with its apex on the y axis; only the curve’s shape (or scale) and the height of the y intercept could vary.

This whole process of repeatedly fitting curves to data and then deriving new data from the fitted curves is weird and useless enough that it seems unlikely to appear anywhere in the universe outside my own idle mind. And yet…. Think of the Dow-Jones index, which is calculated as a weighted average of the prices of 30 selected stocks. They are not always the same 30 stocks. Indeed, recent years have seen quite a lot of turnover in the components of the Dow: AIG was replaced by Kraft Foods; General Motors was dropped in favor of Cisco Systems; Honeywell lost out to Chevron. Would it be too utterly cynical of me to suggest that what’s going on here resembles a process of fitting a curve to randomly chosen points, then choosing more points that lie near the curve?

All my closest friends know about my strange obsession with the mathematics of mattress flipping. A few thousand other people also know my secret, since I have written about it in an American Scientist column (HTML, PDF), which later became a chapter in a book.

To recapitulate: Fussy housekeepers rotate their mattress twice a year to ensure even wear. But a mattress has three axes of twofold symmetry (roll, pitch and yaw). Rotating around the same axis over and over will not cycle through all four possible states of the mattress, so you need to vary the procedure. Perhaps you do a roll one time and a yaw the next. But in the spring you may have forgotten which way you turned last fall. Thus the quest for a “golden rule” of mattress flipping:

A golden rule of mattress flipping would be some set of geometric maneuvers that you could perform in the same way every time in order to cycle through all the configurations of the mattress. Following this algorithm might entail extra physical labor on each occasion–perhaps making multiple flips or turns–but at least it would eliminate the mental effort of remembering.

The rules of this peculiar game forbid putting any marks on the mattress (which would break the symmetry). If we abide by this constraint, the multiplication table for the group of flip operations looks like this:

R, P and Y indicate half-turns around the roll, pitch and yaw axes; I is the identity or do-nothing operation. This table is bad news for golden rules. We already know that no single operation will cycle through all four states of the system, and the table shows that every combination of two operations is equivalent to some single operation. Hence there is no golden rule.

There the matter stood until a few days ago, when I received a letter from Tim Knoll:

I just completed reading your collection of columns entitled “Group Theory in the Bedroom” and I wanted to offer up an alternative solution to the mattress flipping problem. I believe if you add one more operation to the mattresses you can achieve your “golden rule.” What you need to add is a second bed to allow a shift operation in addition to the rotations. If you have the second bed oriented in the opposite manner as the first bed (with the first having the headboard facing north, the second facing south) you can simply have an operation of a non-rotating shift from the first (north-facing) bed to the second (south-facing) and then do a shift in addition to a rotate about the pitch axis from the south-facing bed to the north-facing bed. This mattress juggling should also achieve the desired results using the roll axis in the second step. I unfortunately can’t implement this method in my own apartment since it is equipped with one queen-sized bed and one full-sized bed, but my tests using an index card seemed accurate.

Ingenious, no? But is it truly a golden rule? I’m going to leave that question for readers to decide.

Knoll also points out that even if the new “shift” operation doesn’t yield a golden rule, it does succeed in getting two mattresses flipped with the same mental effort that would otherwise be needed for one mattress. This raises a further question: Why stop at two beds? What about mattress flipping in a barracks, where many beds are lined up in a row? And then there’s the job of flipping mattresses in the Hilbert Hotel, which has infinitely many beds. Does the mental effort per mattress go to zero in this limit?

Writing about Pólya’s recurrence theorem led me to pick up Gerry Alexanderson’s book The Random Walks of George Pólya. He tells a sweet anecdote about the origin of the theorem.

The random walk is sometimes called the drunkard’s walk, but Pólya’s musings on the subject were not inspired by a disorienting pub crawl. On the contrary, the idea came to him while he was living at the Kurhaus Zürichberg, which Alexanderson notes was founded as a temperance hotel. (It seems to be a somewhat different establishment now, and probably beyond the means of a Privatdozent.)

Alexanderson describes the incident that led Pólya to the recurrence theorem:

There was a particular wooded area near the hotel where he enjoyed taking extended walks while thinking about mathematics. He would carry pencil and paper so he could jot down ideas as they came to him. Some students also lived at the Kurhaus and Pólya got to know some of them. One day while out on his walk he encountered one of these students strolling with his fiancée. Somewhat later their paths crossed again and even later he encountered them once again. He was embarrassed and worried that the couple would conclude that he was somehow arranging these encounters. This caused him to wonder how likely it was that walking randomly through paths in the woods, one would encounter others similarly engaged. This led to one of his most famous discoveries, his 1921 paper on random walk, a phrase used for the first time by Pólya.

It’s interesting that the question raised by those embarrassing woodland encounters differs from the usual statement of the recurrence theorem, which talks about a single walker’s return to the point of origin. The answer is the same, however. In an appendix to the Alexanderson book, K. L. Chung mentions an easy proof of this equivalence: Suppose two walkers begin together at A and run into each other again at B. Nothing about the statistics of a random walk changes if you reverse the direction of all the steps. Thus the probability of the rencontre at B is the same as that of one walker going from A to B and then back to A again.

The illustration below shows the paths of two random walkers that meet three times (green circles) after leaving the origin, supporting the plausibility of Pólya’s claim that he was not stalking those young lovers in the woods.

(What’s not so plausible is that any non-inebriated walker would follow a truly random path.)

Fred Gruenberger may well have been the first blogger on computational topics. When he was writing, back in the 1970s, there was no RSS, and so he distributed his musings in a monthly newsletter called Popular Computing. A typical issue was 16 or 20 typewritten pages–stapled, folded, stamped and delivered by mail. It was always worth reading.

Gruenberger had been working and playing with computers since the 1940s. For a long stretch he was at the RAND Corporation, the famous think tank in Santa Monica. Later he taught at Cal State Northridge. In addition to Popular Computing he was involved in the startup of Datamation magazine and published at least a dozen books. I haven’t been able to learn much about his later years; he died in 1998.

A slogan that appeared in some issues of Popular Computing proclaimed: “The way to learn computing is to compute.” I took this advice to heart, although I was hampered by a total lack of hardware. Later on I acquired a programmable calculator, which helped on some of the problems and exercises.

The problem reproduced above appeared in the December 1976 issue of Popular Computing (Vol. 4, No. 12). At the time, I made no attempt to work this one out, but evidently the problem seemed interesting enough to be worth filing away. When I came upon the old clipping recently, I gave it a closer look and realized I have no idea how to answer Gruenberger’s question, though the impediment now is not lack of hardware.

Gruenberger asks us to trace a planar path whose steps are indexed by the odd integers starting at 3. For each number N we turn right 90 degrees before taking a step if N is a prime congruent to 1 mod 6; we turn left 90 degrees before moving one unit if N is a prime congruent to −1 mod 6; otherwise we continue straight ahead in whatever direction we happen to be facing.

In his typewriter graphics, Gruenberger plotted the trajectory from N=3 through 97. Below I continue the path through N=199.

But something’s amiss here. Gruenberger wrote:

Eventually the path will cross itself, so that the cell containing 111 will also contain 147. Similarly, one cell will contain both 91 and 179.

Those two self-intersections are nowhere to be found in the diagram. When I first noticed this discrepancy, I assumed I must have made a mistake somewhere. (This eagerness to blame myself is not mere knee-jerk humility; I have years of experience to back it up.) Eventually, though, I concluded that it was Gruenberger who had made the wrong turn. I believe he mistakenly went left at 127, as shown in the brown trail below:

The brown continuation of the red path includes the two coincidences mentioned in Gruenberger’s problem statement. But the left turn at N=127 is incorrect, because 127 is a prime equal to (6×21)+1, and thus it should specify a right turn. The error is of no great consequence, but it does reveal something interesting: Gruenberger must have been plotting these paths by hand. Most likely he wrote a program to compute the series of residue classes, then traced out the trajectory on squared paper.

Setting aside this anomaly, Gruenberger was quite right that the path does intersect itself. Here’s the trail continued through N=1,001:

And if that’s not tangled enough, here’s what it looks like at N=10,001:

Gruenberger asks for “a list of the contents of those cells containing more than one number, arranged in the order of the smallest number in the cell.” It’s not hard to identify some cells that belong on such a list. The table below includes all multiply-occupied cells discovered when tracing the path up to N=1,001, sorted as Gruenberger requests:

                   x    y    values of N
                 -11   28    (137 337)
                 -15   27    (147 683)
                 -16   27    (149 349 685)
                 -18   26    (155 355)
                 -19   27    (159 691)
                 -19   28    (161 693)
                 -19   29    (163 695)
                 -17   31    (171 319)
                 -18   32    (175 315)
                 -19   32    (177 701)
                 -20   32    (179 703)
                 -22   31    (185 769)
                 -23   31    (187 771)
                 -24   31    (189 773)
                 -30   41    (245 269)
                 -30   42    (247 271)
                 -27   40    (281 733)
                 -26   40    (283 735)
                 -26   37    (289 725)
                 -23   35    (299 715)
                 -22   35    (301 761)
                 -21   35    (303 759)
                 -20   35    (305 757)
                 -17   27    (351 687)
                 -18   27    (353 689)
                 -17   24    (361 673)
                 -16   24    (363 675)
                 -15   24    (365 677)
                 -17   21    (379 667)
                 -17   22    (381 669)
                 -17   23    (383 671)
                 -20   22    (391 631)
                 -20   21    (393 633)
                 -20   20    (395 635)
                 -20   19    (397 637)
                 -22   19    (401 593)
                 -22   18    (403 591)
                 -22   17    (405 589)
                 -22   16    (407 587)
                 -27   15    (419 575)
                 -27   14    (421 573)
                 -28   14    (423 819)
                 -29   14    (425 549)
                 -32   14    (431 539)
                 -32   13    (433 537)
                 -26   10    (563 831)
                 -26   11    (565 829)
                 -27   13    (571 823)
                 -28   18    (607 811)
                 -22   32    (707 767)
                   4   -6    (923 971)
                   4   -7    (925 969)
                   4   -8    (927 967)
                   4   -9    (929 989)
                   5   -9    (931 991)

Is this list the answer to Gruenberger’s question? No, it’s not, because there’s no reason to stop at an arbitrary limit such as N=1,001. Indeed, the list above is not even a prefix of the complete answer. The smallest value of N appearing in the list is 137, but the trail will eventually revisit cells occupied by smaller values of N. For example, continuing the experiment to N=10,001 reveals a bunch of intersections quite close to the beginning of the path, including a site that’s visited five times:

                   x    y    values of N
                   1    0    (5 1621)
                   1    1    (7 1623)
                   2    1    (9 4725)
                   3    1    (11 1263)
                   3    2    (13 1265)
                   5    3    (19 1635)
                   6    3    (21 1637)
                   7    4    (25 7537)
                   7    5    (27 7319 7539)
                   7    6    (29 7505 7541)
                   6    6    (31 1643 7323 7503 7543)
                   6    7    (33 1645 7325)
                   6    8    (35 1647 7327)
                   6    9    (37 1649 7329)

One point still missing from this list is the origin–the site at x=0, y=0, N=3. Does the path ever revisit its starting point? If so, at what value (or values) of N does it come back home? Since I don’t know the answer to this question, I guess I’ll have to leave it as an exercise for the reader.

I suspect that the problem Gruenberger meant to pose (or thought he was posing) was to generate a list of self-intersection sites arranged in their natural order of occurrence–that is, the order in which the crossings are created when you construct the path starting from the origin. This natural-order list is not at all the same as a list “arranged in the order of the smallest number in the cell.” The natural-order list is easy to generate step by step. All you need to do is obey the left/right/straight rules, plot the resulting sequence of positions on the xy lattice, and leave behind a trail of breadcrumbs so you can check at each step to see if the site has been visited before. This task is a matter of straightforward computation–just the kind of assignment that Gruenberger favored. The natural-order list begins:

                   x    y    values of N
                 -30   41    (269 245)
                 -30   42    (271 247)
                 -18   32    (315 175)
                 -17   31    (319 171)
                 -11   28    (337 137)
                 -16   27    (349 149)
                 -18   26    (355 155)
                 -32   13    (537 433)
                 -32   14    (539 431)
                 -29   14    (549 425)
                 -27   14    (573 421)

Thus the prime path first crosses itself when N=269, a value that shares the same coordinates as N=245, namely x=−30, y=41. There are 56 such crossings up to N=1,001, and 112,988 self-intersections up to N=10⁶.

*     *     *

There is a wilder, conjectural answer to Gruenberger’s challenge–which I’m pretty sure he did not have in mind. It goes like this: Maybe the complete list of revisited values of N is simply the list of all N. In other words, maybe the Gruenberger prime path fills up the entire lattice of integers, crossing over itself everywhere many times.

In 1921 George Pólya published a celebrated proof that a random walk on the lattice of integers is recurrent in one or two dimensions, though not in higher dimensions. Recurrent means that the walk returns to each point along its length with probability 1, and indeed visits every point in its domain infinitely often. Is it possible that the prime path is also recurrent?

Pólya’s theorem is one of those mind-expanding results that seem impossible on first acquaintance, and then inevitable, and finally just so amazing that you want to go kiss a mathematician. I have to confess that I’ve never gotten all the way through Pólya’s original paper (it’s not long, but it’s in German). On the other hand, I can highly recommend a little book by Peter Doyle and Laurie Snell, Random Walks and Electric Networks, which gives several alternative proofs of the theorem; it was published in the MAA’s Carus Monograph series, and there’s a postprint available on the arXiv.

The key insight underlying Pólya’s result, as I understand it, is this: If you never revisit a former home, then you must be spending eternity somewhere else, and you can do that only if your universe has enough somewhere elses that you’ll never run out of new territories to visit. Suppose that, some eons after starting your journey, you find yourself at distance r from the origin. If you’re living in a one-dimensional universe, then there are just two places you could be at that moment, namely at +r or −r. It doesn’t matter how far you run; there are still just two points at any given distance from the origin. In two dimensions, a fugitive at distance r has a little more room to maneuver; the number of available points grows in proportion to r, forming a circle of radius r. But this is still not enough room to get lost in. Only in three dimensions or more is there a nonzero probability of escape. In three dimensions, the space available at radius r is proportional to r². In this three-dimensional world, the volume of empty space grows faster than a random walker’s expected distance from home.

What does all this have to do with Gruenberger’s prime path? Well, it’s no secret that the distribution of prime numbers looks convincingly random–if you look at it in just the right way. And in particular the distribution of primes in various residue classes, such as 6K+1 and 6K−1, seems to behave at least approximately like a random variable. All this suggests we might consider viewing the Gruenberger prime trail as if it were a random walk through the two-dimensional lattice of integers. Because the space is two-dimensional, it’s a good guess that the walk should be recurrent.

The original recurrence results of Pólya refer to a simple random walk, where at each step the walker chooses randomly among the available directions and then moves one unit in that direction. For example, in the two-dimensional lattice of integers there are four possible directions: north, south, east, west. The simple random walk is not the best model of the Gruenberger process, which is more like a nonreversing random walk–a path where on each step the walker can turn left or turn right or go straight ahead but can never make a 180-degree about-face. We can further refine the random-walk model of the Gruenberger process by biasing the choice made at each step to reflect the changing abundance of prime numbers. Primes grow scarcer as their magnitude increases; in the vicinity of a given value of N, the probability that a randomly chosen number is prime is approximately 1/log N. Since the Gruenberger path goes straight for all composite numbers and turns only when N is prime, the trail will have longer and longer straight segments, and rarer turns, as N increases. A random walk can mimic this behavior by choosing an action at each step according to this logic:

    if random(1.0) > 2/log(N)
       then go straight
       elseif randomboolean()
           then turn left
           else turn right

(The proportion of primes is given as 2/log(N) rather than 1/log(N) because the Gruenberger process is defined on odd numbers only, which immediately eliminates half of the composites.)

One way to compare the various kinds of random walks is to measure the root-mean-square displacement–the distance from the origin to the final position of the walker, averaged over many realizations of the random process. For a simple random walk, the RMS displacement for an N-step walk converges to \(\sqrt{N}\); for the nonreversing random walk the average displacement is \(\sqrt{2N}\). The biased random walk based on the distribution of primes also appears to yield an RMS distance proportional to the square root of the number of steps; numerically the curve looks something like \(\sqrt{10N}\). I’m not entirely sure that’s the true form of the curve, but the geometric details don’t really make much difference. If I understand correctly, all three of these random processes should be recurrent in the sense of Pólya.

Does the same reasoning apply to the Gruenberger prime path? There are two sides to this question.

The naysayer points out that Pólya’s theorem applies to random walks, but there’s nothing truly random about the sequence of primes. After all, we have a straightforward, deterministic algorithm for generating primes, as well as an efficient algorithm for testing whether any given integer is prime or composite. The essence of a random process is that every time you run it you get a different result, but there’s only one sequence of prime numbers, and so the Gruenberger prime path will come out exactly the same every time. According to this view of things, the kind of probabilistic reasoning that goes into the proof of Pólya’s recurrence theorem is out of bounds here. For randomness to make any sense, you need to average over some ensemble of independent instances. For example, you could average over the 50 salmon-pink paths in the graph below, which represent 50 independent realizations of a biased random walk; you can’t average over the prime path itself (green), because there’s only that one path.

The yeasayer retorts that a single path is all you need–if the path is infinitely long. Indeed, the salmon-colored trails above could be interpreted not as 50 distinct runs of a random process but as 50 segments of a single long path, which repeatedly loops around through the point at x=0, y=0, wanders off in various directions, and then comes back home yet again. In essence, everything that could possibly happen in an infinite set of random paths happens somewhere within a single infinite path; all possible variety is already present there.

I’m not sure how to settle this dispute between Dr. Yea and Professor Nay. When an argument hinges on the nature of randomness, the meaning of infinity and patterns in the distribution of the primes, I known I’m in over my head.

So I’ll leave that deep question unresolved and say a final word about a lesser curiosity. In the Gruenberger process, we’re using the congruence classes of prime numbers mod 6 as a kind of coin flip to decide which way to turn. Is it a fair coin flip? For small values of N, it certainly doesn’t look fair:

                               6x+1      6x−1
       primes <     100         11        12
       primes <    1000         80        86
       primes <   10000        611       616
       primes <  100000       4784      4806
       primes < 1000000      39231     39265

There’s a persistent excess of −1 primes, and the imbalance seems to be getting steadily larger. As a result, the prime path has a “winding number” that reaches 8.5 at N=10⁶; that is, the path makes eight and a half net counterclockwise revolutions. Does the windup continue with still larger N? I gather that the definitive answer is “Yes and No.” For more see the masterful paper by Andrew Granville and Greg Martin cited above.

[Correction 2010-02-19: reflected the accent on Pólya.]

The question again: Is there a four-coloring of the 17-by-17 grid in which none of the 18,496 rectangles have the same color at all four corners? As I said last time, Bill Gasarch would not have put a bounty on this problem if it had an easy solution. Over the past couple of weeks I’ve invested some 10¹⁴ CPU cycles in the search, and a few neural cycles too. I have nothing to show for the effort, except maybe a slightly clearer intuition about the nature of the problem.

If you generate a bunch of four-colored n-by-n grids at random, the average number of monochromatic rectangles per grid increases quite smoothly with n:

This gradual progression might lead you to suspect that the difficulty of finding or producing an n-by-n grid that is totally devoid of monochrome rectangles would also be a smooth function of n. The truth is quite different.

Finding a solution is easy for square grids of any size up to 15 by 15. The task suddenly becomes very hard at size 16 by 16. As for 17 by 17, it’s much harder still–and indeed is not yet known not to be impossible. (Details on the data behind this graph: For each size class from n=8 to n=17 I started with 1,000 randomly four-colored n-by-n grids. Then I applied a simple heuristic search (the first of the algorithms listed below) to each grid, running the program for 1,000 × n² steps. The graph records the number of times this procedure succeeded–i.e., produced a grid with no monochrome rectangles–at each grid size. Up to n=14, the search never failed; at n=16 and beyond, it never succeeded.)

This kind of sudden transition from easy to hard is a familiar feature in the realm of constraint satisfaction. Well-known intractable problems such as graph coloring and boolean satisfiability have the same structure. That doesn’t bode well for any of the simple-minded computational methods I’ve tried. Here’s a brief catalog of my failures. These are algorithms that I’m pretty sure are never going to pay off.

• Biased random walk. Start with a randomly colored grid. Repeatedly choose a site at random, then try changing its color; accept the move if it reduces the overall number of monochrome rectangles. This is the simplest of all the algorithms. None of the more-elaborate schemes is decisively better.
• Whack-a-mole. Find all the monochrome rectangles in the grid; choose one of them and alter the color of one corner, thereby eliminating the rectangle. In the simplest version of this algorithm, you choose the rectangle and the corner and the new color at random; in more sophisticated versions, you might evaluate the alternatives and take the one that offers the greatest benefit.
• Steepest descent. Examine all possible moves (for the 17-by-17 grid there are 867 of them) and choose one that minimizes the rectangle count.
• Lookahead steepest descent. Examine all possible moves, and then all possible sequels to each such move (for the 17-by-17 grid there are 751,689 two-move sequences); choose a sequence that minimizes the rectangle count. In principle this method could be extended to chains of three or more moves, but the cost soon gets out of hand. (The lookahead technique is the mirror image of backtrack search; it explores the tree of possible moves breadth-first instead of depth-first.)
• Color-balanced search. Allow only moves that maintain the overall balance of colors in the grid. For example, in a 16-by-16 four-colored grid, color balance implies 16 sites in each color. One way to maintain balance is to make moves that swap the colors of two sites. (There is no reason to think that a rectangle-free grid will have exact color balance; on the other hand, a solution for a large grid cannot depart too far from perfect balance. Thus a color-balanced search might be an effective trick for finding a neighborhood where solutions are more common.)
• Row-and-column-balanced search. Allow only moves that maintain the balance of colors within each row and each column of the grid. In a 16-by-16 grid, each row and column should have four sites in each of four colors. A simple way to maintain this detailed color balance is to search for “harlequin rectangles” with the color pattern \(\begin{array}{cc}a & b\\b & a\end{array}\) and permute them to \(\begin{array}{cc}b & a\\a & b\end{array}\).

Most of these techniques are greedy: At each stage the algorithm chooses an action that maximizes some measure of progress. On hard instances, a pure greedy strategy almost always fails; the search gets stuck in some local optimum. Thus it’s usually best to temper the greediness to some extent, occasionally choosing a move other than the one that yields the best immediate return. (The family of methods known as simulated annealing are more elaborate variations on this idea, based on insights from thermal physics.)

Here we see traces of three runs of the algorithm identified above as steepest descent, with differing values of a greediness parameter m. (The grid size is 15 × 15.) At m=0 (no greediness at all), all moves are equally likely to be chosen, and the algorithm executes a random walk on the space of grid colorings. At m=1 (maximum greediness), the program always chooses the highest-ranked move, which works well until the system stumbles into a state where no move can reduce the rectangle count. A value of m=0.3 seems to be a good compromise. (I’ll say a little more below on the greediness parameter; indeed, I have a question about how best to define and implement it.)

After all this fussing with a dozen variations on local-search algorithms, I’m afraid the outlook for success is not promising. With a little patience and some tuning of parameters, any one of these algorithms can solve grids up to 15 × 15. With a lot more patience and tuning, they’ll eventually yield answers for 16 × 16. But none of the algorithms come even close to cracking the 17 × 17 barrier. Solving that one is going to require a fundamentally new idea. Perhaps someone will find an analytic approach to constructing a solution, rather than blindly searching for one. Or perhaps someone will prove that no solution exists.

On the computational front, I suspect the best hope is a family of algorithms known in various contexts as belief propagation, survey propagation and the cavity method. I’ve been hoping that friends who are expert in these techniques might swoop in and solve the problem for me, but if not I may have to give it a try myself.

In the meantime, here’s the thing about greediness (an apt subject for this time of year?). We want to define a function greedy whose arguments are a vector of alternative moves ranked from best to worst and a number m such that 0 ≤ m ≤ 1. If the greediness parameter m is 0, the function returns a random element of the vector. If m = 1, the returned value is always the first (highest-ranked) move. Otherwise, we must somehow interpolate between these behaviors. One attractive notion is to return the first element of the vector with probability m, the second choice with probability m(1 − m), and so on. Thus for m = 1/2 the series of probabilities would begin 1/2, 1/4, 1/8…. For m = 1/3 the first few values are 1/3, 2/9, 4/27….

This scheme works just fine for a vector of infinite length, but there’s a problem with shorter vectors. Consider what happens with the procedure call greedy(v=[1, 2, 3, 4], m=0.5). We have the following table of probabilities:

     1 --> 1/2
     2 --> 1/4
     3 --> 1/8
     4 --> 1/16

But on adding up those values, we come up 1/16th short of 1. What happens to the missing probability? I took an easy way out, distributing the “extra” probability equally over all the elements of the vector. The code looks something like this:

function greedy(v, m)
  for i=0 to length(v)
     if (i==length(v))
        return v[random(length(v))]
     elseif (random(1.0) < m)
        return v[i]

This procedure seems to give sensible results, but I wonder if there might be a better or more natural definition of greedy probabilities. Also, the running time for my code is logarithmic in the length of the vector (assuming m < 1). Is there a constant-time algorithm that gives the same results? (We don’t know the length of the vector in advance, so merely precomputing the table of probabilities is not an option.)

William Gasarch is not the Clay Mathematics Institute. He isn’t paying a million bucks for proofs of famous conjectures. But Gasarch is putting up 17² of his own dollars for the solution to an intriguing little stumper. And the prize problem appears to be somewhat easier than the Riemann hypothesis or the P=NP question. (Unless it’s impossible!)

Gasarch sets forth his prize challenge in a blog post, with further background in a paper and in the slides from a talk. All of those works are well worth reading, but for those who don’t want to chase down the references, here’s the gist. Our mission, should we choose to accept it, is to color the nodes of an n-by-m grid, using only a specified number of colors, and observing a particular constraint: Nowhere in the grid may the four corners of a rectangle all have the same color. (Only rectangles with sides parallel to the x and y axes are considered.) For example, here is a four-colored 15-by-15 grid that satisfies the no-monochromatic-rectangles constraint:

In this array of dots there are \( {15 \choose 2}^2=11025\) distinct rectangles. If you care to check through all of them one by one, you’ll find that in no case do all four corners have the same color. In contrast, here is a 16-by-16 grid that is almost but not quite rectangle-free:

Gasarch offers his $289 bounty for any four-colored 17-by-17 grid with no monochromatic rectangles. Why is that grid of particular interest? It’s a border case. Among square grids, all those up through 16-by-16 have been shown to have rectangle-free four-colorings. For the 18-by-18 grid and all larger squares, rectangle-free four-coloring has been proved impossible. For squares larger than 18-by-18, four-coloring has been proved impossible. The status of the 17-by-17 and 18-by-18 grids remains unsettled, but Gasarch believes that both are four-colorable.

Gasarch has much more to say about the mathematics behind this problem. Here I would like to muse on some computational aspects of searching for a 17-by-17 four-coloring.

To state the obvious first, this is not a problem we can expect to solve by exhaustive enumeration. There are 4²⁸⁹ possible colorings of the grid. Casting out symmetries brings that down only to 2 × 4²⁸⁷. There’s not world enough or time for checking them all.

Testing grids at random is also hopeless in the 17-by-17 case. This nonstrategy actually works quite well for small grids. For example, you can readily find a four-coloring of an 8-by-8 grid just by generating a few hundred thousand random colorings. But the method fails for larger grids because the proportion of all colorings that are rectangle-free falls steeply with grid size. (Consider the 2-by-2 grid: There are 256 four-colorings, and all but four of them are rectangle-free.)

To make any progress toward the 17-by-17 case, we’ll have to do at least a little thinking, rather than expecting the computer to do all the work. Here’s one idea that’s very easy to implement: Find a monochrome rectangle somewhere within the grid, change the color of one of its corners, and repeat until you can’t find any more rectangles. This algorithm works reasonably well for grids up to about 12-by-12, but then it runs out of steam. On larger grids, changing the color of a node to eliminate one rectangle is likely to create another rectangle elsewhere (or several more of them). As a result, the system merely takes a random walk, with trendless fluctuations in the number of rectangles at any given moment. You discover a rectangle-free coloring only if the walk happens to stumble on the zero point.

I found the 15-by-15 four-coloring shown above with an algorithm that’s a little more effective even thought it’s no more sophisticated than the corner-twiddling method. The program repeatedly chooses a node at random and tries assigning it all four possible colors, tallying up the number of rectangles for each color choice. Some color or set of colors must minimize the rectangle count; from among these optimal colors the program chooses one at random and sets the node to that color before repeating the loop. This is a “greedy” method: At each step the number of rectangles can decrease or remain constant but can never increase. Greedy methods are notorious for getting stuck in local optima that are inferior to the global optimum. Maybe that’s what happens to my program when I try it on 16-by-16 and 17-by-17 grids. Or maybe the search space is just too large. In any case, when I woke up this morning and checked the results of an overnight run, I did not find a rectangle-free four-colored 17-by-17 grid awaiting me.

Of course I really didn’t have to do any algorithm analysis at all to know that I wasn’t going to win $289 and eternal fame with a day or two of idle hacking. If the problem were that easy, Gasarch and his students would have solved it for themselves long ago.

In spite of these various failures and frustrations, the grid-coloring problem still looks tantalizingly solvable. If a four-coloring of the Gasarch grid exists, it seems like we should be able to find it by some practical computation.

There are certainly lots of approaches more powerful than the blind dart game I’ve been playing. For example, if local optima are the major impediment, some variation on simulated annealing might help.

A more radical possibility is to try to construct an instance rather than merely search for it. If we assume that the four colors are represented as evenly as possible, then the 17-by-17 grid must have 72 nodes in each of three colors and 73 nodes in the fourth color. Starting from a blank grid, it’s easy to mark off 73 nodes in a single color without creating a forbidden rectangle. Adding 72 nodes of a second color is only a little harder. But then the job gets tricky. When you try to fill in a third color, you also by default choose nodes for the fourth color at the same time, and conflicts pile up in a hurry. Some kind of backtracking approach is probably needed here. Gasarch links to a paper by Elizabeth Kupin of Rutgers that explores these ideas in more detail. (If you want to prove the nonexistence of a four-coloring, this is presumably the way to go.)

Gasarch mentions two other promising avenues: integer programming (the discrete variant of linear programming) and SAT solvers–algorithms for the satisfiability problem. Having spent some time hanging out with a few master SAT solvers, I’m intrigued by the latter possibility. You can almost encode the grid-coloring problem as an instance of NAE-SAT, or not-all-equal SAT. Each node of the grid is represented by a variable that can take on any of four values. We group subsets of variables four at a time into clauses, where each clause includes the variables representing the four corners of a rectangle somewhere in the grid. For the 17-by-17 grid there are \({17 \choose 2}^2=18496\) clauses of this kind. The entire formula is satisfied if we can assign values to the 289 variables in such a way that none of the 18496 clauses has all four of its variables with the same value. After 40 years of work on SAT, there’s a highly developed technology for solving such problems. However, there’s a hitch. SAT problems are formulated in terms of boolean variables, with just two values each, but the grid-color variables have four values. Thus a further layer of encoding is likely to be needed, bringing a further explosion in the size of the problem instance.

One final hackerish note: What’s the best way to detect the presence of a monochromatic rectangle in a grid? My candidate goes like this. We encode the rows of the grid in a set of bit vectors–four vectors for each row, representing the four possible colors. For example, the red vector for a row has a 1 at each position where the corresponding node is red, and zeroes elsewhere. The blue vector has 1s for blue nodes, and so forth. Now we can detect a rectangle merely by taking the logical AND of two rows (an operation that could be a single machine instruction). A rectangle exists if and only if the result of the AND is nonzero. at least two bits are set in the resulting vector.

[Thanks to all the commenters for corrections and elaborations.]