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 school of philosophy called Antipodianism briefly flourished on the fringes of the Hellenistic world more than 2,000 years ago. The sect held that every person has an opposite number, a mirror image who inverts all our beliefs, feelings, actions and attitudes. If I smile, my antipodian counterpart frowns; when I wake, she sleeps. If I’m a Mac, she’s a PC. For every liberal Democrat there’s an antipodian Tea Party Republican. In this way the universe is held in balance. It’s an enforced equilibrium, which none of us has the power to upend, hard as we might try.

The earliest of the Antipodians believed that every such matched pair (commonly designated A and ∀) live at diametrically opposite points on the surface of the earth. This arrangement ensures that A and ∀ can never meet—thereby averting a cosmic catastrophe. A later quantum-field version of Antipodianism relaxed the geographic constraint by allowing for the creation and annihilation of A∀ pairs, but that idea never really caught on.

One day an Antipodian master was teaching an exchange student from New Zealand. The child was crafty.

“Is it not true,” she asked, “that A and ∀ always do the opposite thing?”

“Yes, antisymmetry demands it,” the master replied.

“If A walks north, ∀ must walk south?” the child asked.

Again the master assented.

“If A goes east, ∀ must go west?”

“Yes.”

“If A turns to the right, ∀ must turn to the left, no?”

The master agreed, although he sensed trouble coming.

“I’m afraid the universe is out of joint,” said the child. “If A goes north and turns to the right, while ∀ goes south and turns to the left, afterwards they are both walking east. They are doing the same thing.”

Needless to say, this was a moment of crisis in Antipodian doctrine. The Pythagoreans, you may recall, resolved a similar impasse by resorting to violence. When some upstart challenged their precept that “all is number” by showing that no known number can be the square root of 2, the Pythagoreans tossed the troublemaker out of the boat. But in this case the Antipodian master kept his calm.

“Ah my little Kiwi,” he said to the student. “You are clever but not wise. Your own statements refute your claim. Did you not begin by saying that A and ∀ always do the opposite thing? When A walks 10 paces north, ∀ walks 10 paces south. When A turns right, ∀ turns left. But now you would have us believe they both take a step forward, contradicting the most basic law of their nature. What really happens is that A walks forward and ∀ walks backward. Thus A goes eastward and ∀ westward, and all is well with the world.”

Through this brittle sophistry the master extricated himself from the classroom—though he may have had to walk backwards to make good his escape. He never taught again. The Kiwi student went on to a brilliant career studying the weak interactions of neutral K mesons. As for Antipodianism, it vanished without a trace.

Or maybe it left a tiny trace. I’ve never visited the antipodes, but I hear that corkscrews Down Under turn the other way.

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.

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.]

Four colors suffice for any planar map: We’ve known that since 1977. If a map is divided into countries or provinces or other regions, and you want to color the map so that no two adjacent regions have the same color, you’ll never need more than four crayons. But there are a couple of definitions that have to be accepted if this theorem is to hold. First, “adjacent” means that two regions share a boundary segment, not merely a point or a discontinuous set of points. Second, a “region” has to be a connected area, so that you can reach any point in the interior from any other such point without crossing a boundary.

A few days ago the strange and wonderful Strange Maps blog called attention to this map of the Principality of Liechtenstein:

(The map comes from the Wikimedia Commons Atlas of Liechtenstein, which has a larger image.)

It seems that Liechtenstein is divided into 11 communes, which emphatically do not satisfy the connectivity requirement of the four color map theorem. Just four of the communes consist of a single connected area (Ruggell, Schellenberg and Mauren in the north, and Triesen in the south). All the rest of the communes have enclaves and/or exclaves. (I confess that I didn’t know the distinction between these words until yesterday. They make a nice pair.) The most fragmented commune is Vaduz, which includes the principality’s capital. Vaduz consists of six isolated pieces; the smallest is a sliver about a kilometer northeast of the town of Planken.

In the map above, each commune is assigned its own color, and so we have an 11-coloring. It’s easy to see we could make do with fewer colors, but how many fewer? I have found a five-clique within the map; that is, there are five communes that all share a segment of border with one another. It follows that a four-coloring is impossible. Is there a five-coloring? What is the chromatic number of Liechtenstein?

The drawing below, brazenly swiped from a 1964 Martin Gardner column, illustrates the solution to a well-known puzzle. If you stack n bricks on a table, how far can you make them extend over the edge without toppling?

The answer given, for bricks of unit length, is one-half the nth harmonic number, the sum of the series 1/2 + 1/4 + 1/6 + 1/8 + … + 1/2n. Since this series diverges, the overhang can be as large as you please, given enough bricks (and a strong enough table).

In describing this result, Gardner uses the word “unbelievable,” and when I first read about it, that was my reaction too. It is one of those facts that are inescapably true, and yet still implausible. Now I learn that the truth is even stranger—that far larger overhangs can be achieved.

The overhang for the harmonic stack is approximately equal to 1/2 ln n; it turns out other stacks achieve an overhang on the order of the cube root of n. This is an enormous improvement. For a million bricks, the harmonic stack can’t get beyond 13.8 bricklengths, whereas the cube-root stack extends 100 bricklengths beyond the edge.

How is it done? I’m not telling. If you can’t figure out the trick, you’ll have to consult the links below.

The key idea in the alternative solutions has apparently been known for at least 20 years. (Actually, I’d be surprised if it wasn’t known to stonemasons centuries ago, as well as generations of children playing with wood blocks—but I don’t have a reference to support that conjecture.) Quite new, however, is the mathematical analysis showing exactly how large the overhang can be. Last year Mike Paterson of the University of Warwick and Uri Zwick of Tel Aviv University presented a paper at SODA (the Symposium on Discrete Algorithms) giving optimal solutions for n up to 30, and showing that the asymptotic rate of growth is n^1/3. Now Peterson and Zwick, along with Yuval Peres (Berkeley), Mikkel Thorup (AT&T Labs) and Peter Winkler (Dartmouth) have proved that order n^1/3 is the best that can achieved:

More specifically, we show that any n-block stack with an overhang of at least 6n^1/3 is unbalanced, and must hence collapse. Thus we conclude that the maximum overhang with n blocks is of order n^1/3.

Links: The 2006 Paterson-Zwick paper is here in the SODA proceedings (subscription required) and here in the arXiv. (It is also to appear in the American Mathematical Monthly.) The five-author proof is at the arXiv here.

Among the 250 million Rubik’s cubes manufactured since 1980, how many lie abandoned in a scrambled state, having never regained their original configuration since being taken out of the box? Most of them, I would guess. Now comes word that those cubes might be restored to pristinity with a little less effort. The upper bound on the number of moves required to solve any state of Rubik’s cube has been lowered from 27 to 26. The result is reported by Daniel Kunkle and Gene Cooperman of Northeastern University in a preprint available here.

It’s well-known that Rubik’s cube isn’t really a toy or a puzzle but rather a group-theory machine in disguise. Each possible move, or twiddle—rotating a face by some multiple of 90 degrees—is a transformation that permutes the positions and orientations of the 26 “cubies.” From any position there are just 18 distinct nontrivial moves, but in various combinations they generate a total of 43,252,003,274,489,856,000 permutations. What is the maximum number of moves needed to transform any one state of the cube into any other? That’s the question Kunkle and Cooperman are addressing. In other words, they are looking for the longest shortest path.

In principle we could get the answer by exhaustive search. It would be done by constructing the “Cayley graph” for the Rubik group: a graph in which each possible configuration of the cube is represented by a node, and two nodes x and y are connected by an edge if some single move allows configuration x to be transformed into y. The most efficient solution for any state is the shortest path (i.e., sequence of edges) leading from that state to the node representing the solved state. The worst-case solution length for the entire puzzle is the maximum of the shortest solution paths for all the nodes. The Cayley graph approach was used by Cooperman and two colleagues to find the worst-case path for the 2 × 2 × 2 version of Rubik’s cube. The Cayley graph for this device has 88,179,840 nodes, and it turns out the longest paths within it have 11 steps. For the 3 × 3 × 3 cube, however, exhaustive search is just too exhausting.

To make the search feasible, Cubists have had to scale back their ambitions and accept an upper bound rather than a true maximum. First it was shown that the worst-case solution path cannot require more than 52 moves; then the bound was lowered to 30, and then 29; a year ago Silviu Radu got the number down to 27.

Kunkle and Cooperman’s strategy is to factor the problem into two pieces by segregating half-turn (180-degree) and quarter-turn (±90-degree) moves. The subgroup of states reachable by half-turn moves is quite small by Rubik group standards, with just 663,552 elements, and the number is further reduced to only 15,752 after various symmetries are eliminated. Calculating the best solution for each of these states took only about a day on an ordinary PC. The longest such paths have 13 steps.

The second phase of the analysis tackles the “cosets” of the half-turn subgroup. For each of the 663,552 positions reachable by making half-turns only, there are 65,182,537,728,000 additional positions reachable by making quarter-turn moves. Kunkle and Cooperman had to search for the longest shortest path connecting any two elements of this set. They exploited various symmetries to reduce the scope of the problem and devised fast algorithms for some of the basic steps in the calculation. Even so, they wound up running their computer program for 63 hours on 128 processors at the San Diego Supercomputer Center. At one point the data store for intermediate results ballooned to seven terabytes. At the end of the run, they found that the longest shortest path consisted of 16 moves.

Combining the two phases of this solution yields 13 moves for the half-turn part of the path and 16 moves for the coset part, for a total of 29 moves. This is not an improvement over the best existing result of 27. However, Kunkle and Cooperman found that the two-stage algorithm reports path lengths greater than 26 only for a comparative handful of states—about 14,000. By a brute-force search they were able to find solutions of length 26 or less for each of these states, and thereby established the 26-move bound overall. It’s possible that further brute-force searching will lower the limit further to 25 moves.

Other experiments done by Herbert Kociemba, Richard Korf and others have shown that most randomly selected cube states can be solved in 18 moves or less. Korf has conjectured that all configurations are solvable in about 20 steps. We’ll see whether this is true—but probably not soon.

The March-April issue of American Scientist is now available on the Web; paper copies should be on their way soon. My column is about hump yards and turnouts and wyes—in other words, about algorithms for railroad workers. “Computing with locomotives and box cars takes a one-track mind.” There’s a small puzzle near the end of the column. You’re welcome to post comments, complaints and solutions here.

In the new issue I also recommend a “Macroscope” article on Avogadro’s number by Ronald M. Fox and Theodore P. Hill of Georgia Tech. For those who have forgotten their chemistry, Avogadro’s number is the number of molecules in a mole of a substance (an amount in grams numerically equal to the molecular weight). Specifically, N_A is defined as the number of carbon atoms in 12 grams of carbon-12, and its value is roughly 6.02 × 10²³. Fox and Hill suggest turning the definition upside-down: Instead of trying to count the atoms in a gram, define the gram as a certain number of atoms. They have a specific number to recommend: 602,214,141,070,409,084,099,072. I invite you to deduce what’s so special about this particular number and why they favor it over other candidates in the same range.

Our story so far: Having stumbled upon the Jacobsthal numbers, 1, 3, 5, 11, 21, 43, 85, 171, 341,…, I idly asked, “Who was Jacobsthal?” Keith Matthews promptly responded with a wealth of biographical information, even arranging to have an obituary translated from the Norwegian. So I asked, “Where did Jacobsthal mention these numbers?” and Barry Cipra quickly supplied a reference: “Fibonaccische Polynome und Kreisteilungsgleichungen” in Sitzungsberichte der Berliner Mathematischen Gesellschaft 17 (1919-1920), 43-57.

I dare not ask another question, lest someone else go running off to do more of my library errands for me. Embarrassing.

Unfortunately, though, when I look into the Jacobsthal paper, further questions are inescapable. Nowhere in the article, as far as I can tell, does Jacobsthal list any of the numbers in the sequence that now bears his name. Admittedly, I don’t actually read German; I just make it up as I go along. But, ungebildet as I am, I can at least recognize numerals, and 1, 3, 5, 11, etc. are not to be found. The closest approach is in this passage:

Here we see the recurrence relation f(n+1) = f(n) + xf(n–1), which produces the Jacobsthal sequence in the case when x = 2. But nowhere does Jacobsthal mention the specific case of x = 2, or any other specific example for that matter, except for noting that x = 1 corresponds to the “so-called” Fibonacci numbers.

So once again I’m left wondering: Exactly how did Jacobsthal’s name get attached to the numbers 1, 3, 5, 11, 21, 43…?

Incidentally, Google Language Tools offers a wonderful translation of Jacobsthal’s title: “Fibonacci polynomials and circling hurrying equations.”

In an item published here last May I stumbled across the sequence

1 3 5 11 21 43 85 171 341 683 1365 2731 5461 10923 21845 43691
87381 174763 349525 699051

which I dubbed “the oddest numbers” but which Neil J. A. Sloane’s Encyclopedia of Integer Sequences calls the Jacobsthal sequence. I asked: Who is or was Jacobsthal? Keith Matthews, creator and maintainer of the Number Theory Web, has stepped forward with an answer. Ernst Jacobsthal (1882-1965) was a German-born mathematician who fled to Norway when the Nazis came to power and eventually became a Norwegian citizen and professor at Trondheim. The story is told in an obituary that Matthews has made available on the Number Theory Web both in the original Norwegian and in an English translation by Jan Kristian Haugland. Matthews also mentions a brief biography in German. And the Mathematics Genealogy Project offers the further information that Jacobsthal was a student of Georg Frobenius and Issai Schur.

A bit more poking around reveals that Jacobsthal’s doctoral dissertation is available online, published by the Humboldt-Universität zu Berlin. In this work Jacobsthal gives a proof that every prime of the form 4n+1 can be written as a sum of two squares.

I’m very grateful to Matthews for answering my idle question about Jacobsthal—but there’s no end to questions. What I’d like to know now is when and where and in what context Jacobsthal discussed his sequence. The numbers are most readily defined by the recurrence J_n = J_n-1 + 2J_n-2, with J₀ = J₁ = 1. How did this idea come up in his work? Several journal articles and web sites mention the Jacobsthal numbers, but I’ve yet to find a reference to any specific publication.