I have a longstanding friendly feud with my Editor-in-Chief over the use of logarithmic scales in graphs. I tend to go for a log plot if there’s the slightest hint of an exponential trend in the data; she argues that the human sense of numbers is inherently linear, and thus a nonlinear scale should be used only in exceptional circumstances. Linear is natural, she says; logs are for geeks.

Vindication is always sweet, so I was delighted to see a report in the latest issue of Science (subscription needed for full-text access) suggesting that members of an Amazonian indigenous group seem to order numbers on a logarithmic scale. Stanislas Dehaene, Véronique Izard, Elizabeth Spelke and Pierre Pica worked with 33 adults and children of the Mundurucu group in Brazil’s Para province. The subjects were shown a line segment with labeled end points (illustration above) and asked to place various other numbers within this interval. The numbers were presented as patterns of dots, as sequences of tones or as spoken number words in the Mundurucu language or in Portuguese. The results of the experiment show, according to Dehaene et al.:

The Mundurucu seem to hold intuitions of numbers as a log scale where the middle of the interval 1 through 10 is 3 or 4, not 5 or 6.

In contrast, 16 American control subjects from the Boston area placed numbers on the scale pretty much in proportion to their magnitude.

For the sake of my ongoing quibble with the E-in-C, it’s in my interest to support and endorse this finding. I wish could do so.

Exactly what does it mean to “hold intuitions of numbers as a log scale”? Simply taking base-10 logarithms of numbers isn’t quite what’s wanted. The logarithm function maps the integers 1 through 10 onto a range from 0 to 1. To map numbers in the domain [1, 10] onto the same co-domain “logarithmically,” we need a transformation along the lines of R = (9 log₁₀ S) + 1, where R is the “response location” and S is the “stimulus number.” But the resulting curve, superimposed in red on the graph above, doesn’t match the experimental data very well at all. And, looking at the data more closely, it’s hard to see how any plausible transformation is going to account for the positions of those dots and error bars. There’s trouble in the middle, where 5 was judged larger than 6. And there’s trouble at both ends: No matter where you place the numbers 2 through 9 on the scale, shouldn’t 1=1 and 10=10? The anomalies at the ends are made more puzzling by the instructions given to the test subjects:

Only two training trials were presented, with sets of dots whose numerosity corresponded to the ends of the scale (1 and 10). The participants were told that these two stimuli belonged to their respective ends, but that other stimuli could be placed at any location.

In view of these instructions, it seems peculiar that stimulus number 1 has a response location near 1.5, and stimulus number 10 corresponds to a response location of roughly 8.

I wonder if the data aren’t better explained by a simple step function. Perhaps the subjects counted dots up to 4, but beyond that merely estimated the number of spots, lumping together any stimulus with 5, 6 or 7 dots in one class, and putting patterns of 8, 9 or 10 spots in a second class.

Or maybe the arrangement is logarithmic after all. The Mundurucu may have a lot more mathematical sophistication than Dehaene et al. give them credit for. When presented with a scale that runs from 1 to 10, rather than 0 to 10, they intuit that the interval belongs to the series …, 0.01, 0.1, 1, 10, 100,…, rather than the more prosaic series …, –10, 0, 10, 20, 30, …. Anyway, I think that’s what I’ll tell my Editor-in-Chief.

I’ll be on the radio Monday (May 26th) at 6 p.m., chatting with Dorian Devins about my recent book Group Theory in the Bedroom, and Other Mathematical Diversions. If you’re within shouting range of Jersey City, N.J., you can listen in on WFMU; otherwise, stream it live over the net; or wait for the podcast. (Incidentally, even if you have no interest in listening to my self-promotional rants, I recommend the WFMU web site for excellent advice and instructions on doing Internet radio.)

In other GTiBaOMD news, I received a Google Alert the other day, pointing me to a new review of the book. When I followed the link, I was momentarily perplexed. Before I could see the review, I had to press a button bearing the legend: “I am over 18 and agree to the viewing of sexually explicit material.” As it happens, the page that awaited me beyond this warning had no sexually explicit material at all. Indeed, that lack was the essence of the reviewer’s complaint:

Sex and Math. You would think I would be in heaven at the mere thought that somebody had written a book combining these two things. And this book would be heaven if it had combined those two things. Instead, sadly, this is a case where the title is both a little too literal and yet not quite accurate.

Point taken. It’s true that there’s little titillation in my tale of mattress-flipping. Considering that disappointment, the reviewer’s assessment is remarkably even-handed. (”Would I recommend this book? Sure, especially if you have a fancy for computing, math and solving problems.”) Thanks, Naughty Bookworm.

Wow. Jupiter has sprouted a third red spot. It was just two years ago that the Great Red Spot was joined by a smaller companion, which was quickly dubbed “Junior.” I guess the new red spot, discovered in the past few weeks, will have to be called “III.”

In the view above, from the Hubble Space Telescope, Junior is southwest of the Great Spot, and the new, smallest member of the family is due west of the big one and a little farther downwind. This is a false-color image, constructed by assigning colors to monochromatic images recorded at three wavelengths, but the intent is to correctly render colors as perceived by the human eye. Evidently none of the spots are really red at the moment. If they were all newly discovered right now, we would have the Great Peach of Jupiter and the Two Little Apricots.

When I get beyond merely admiring the glorious, painterly spectacle of this Jello-chiffon dessert in the sky, what fascinates me most is the time scale of the red spot phenomenon. The Great Spot has been there for at least a century or two, and probably much longer. It is a storm, with rapid counterclockwise circulation clearly visible in the time-lapse photos returned by the Voyager I spacecraft in 1979.

Storms are something we can relate to from our earthling experience; we have cyclones here too. But what kind of storm lasts for hundreds of years? Even allowing for the larger spatial scale of events on Jupiter, the Great Spot seems extraordinarily long-lived. The rotation period is roughly one earth-week, which means the spot has survived for something on the order of 10,000 revolutions. And it is geographically stable, too: Although the spot drifts in longitude, it seems to be pinned in latitude, hovering at a swirling boundary between easterly and westerly wind belts.

Very likely, the key to the Great Spot’s longevity is that Jupiter has no continents or other surface irregularities to disrupt the flow of the atmosphere. But that fact makes the uniqueness of the spot somewhat mysterious. If such features can arise spontaneously, purely from the dynamics of the atmospheric flow, like a pearl created without any need for a grain of sand, then why is there just one red spot? You’d think that such storms would develop from time to time wherever conditions were favorable.

And now we have our answer: There’s not just one red spot. But the question of time scales doesn’t entirely go away. It seems implausible that one storm would go on for centuries in lonely splendor, and then suddenly two more would evolve within a couple of years. Perhaps there have been others and we just didn’t notice? Not within the past 50 years, I think. Another possible explanation of this improbable coincidence is that the births of Junior and III are not independent events. All three storms are nearby (at least by Jovian standards) and are surely interacting. If that’s the case, we may not have seen the end of this sequence of events. Will there be more spots? Will they collide or coalesce? Stay tuned.

In the matter of time scales, I can’t help noting that Jupiter has a connection with another epochal event in the modern Internet era. In July of 1994 comet Shoemaker-Levy 9 crashed into Jupiter, and the world followed along via the web. The idea that anyone with a modem could download the images directly from JPL—no waiting for the news media—made quite an impression. The Netscape icon was the apotheosis of this event.

Links:

More third-spot images and explanations of how they were made, from Imre de Pater, UC Berkeley.

Reporting from Science Blog.

Reporting from New Scientist.

A report from the Philippine Daily Inquirer with some background on who first spotted the new spot.

The Wikipedia article on the Great Red Spot (which already has a note on the new one).

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

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

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

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

When I finished writing about the Zeno wagering game recently, I had some trees left over, so I thought I would try planting them here.

In the Zeno article I was trying to understand and explain the structure of this peculiar-looking tree, which I’ll call the tangled tree:

As a warmup exercise, I started out with this simpler example, the combed tree:

These are both binary trees: Each node has exactly two descendants. And the trees also share a basic internal symmetry. In both structures, if a node has horizontal coordinate x, the two descendant nodes are symmetrically arrayed at x-d and x+d. The difference between the two trees is all in how the displacement distance d is calculated. In the combed tree, d has the initial value 1/4 at the root node, and it is reduced by half at each lower level of the tree. In the tangled tree, the formula is d = 1/2 min(x, 1-x). In other words, d is half the distance to the nearer boundary of the interval [0,1].

There was a third tree I didn’t have room to include in the article. I had stumbled upon it when I made a programming error. This rogue structure looks like this, and I’m calling it the braided tree:

The nature of my programming error is not hard to see. Intending to generate the tangled tree, I had meant to calculate the positions of the child nodes as follows:

      left = x - 1/2 min(x, 1 - x)
      right = x + 1/2 min(x, 1 - x)

But instead I carelessly wrote:

      left = x - 1/2 x
      right = x + 1/2 (1 - x)

Stupid goof, but the result is visually intriguing. And although the braided tree is a botched version of the tangled tree, it actually has a lot in common with the combed tree. In particular, both the combed and the braided trees have exactly the same set of nodes; the difference is that those nodes are wired differently.

One way to understand the difference is to look at the paths leading from the root to any given node. Each such path is a sequence of left and right turns. For example, in the combed tree you can get from the root to the node at x = 3/16 by turning left at the root, then left again at 1/4 and finally right at 1/8. In the braided tree the path to 3/16 goes right to 3/4 then left to 3/8 and left to 3/16. Paths of this kind can be encoded as binary numerals, letting a 0 stand for each left turn and a 1 for each right turn. In this scheme the two paths described above are 001 and 100.

In essence, the braided tree is a permutation of the combed tree. If you list all the nodes at some fixed level in the combed tree (say those three levels below the root, where all denominators are 16), then the corresponding paths are in numerical order:

    1/16    000    (0)
    3/16    001    (1)
    5/16    010    (2)
    7/16    011    (3)
    9/16    100    (4)
   11/16    101    (5)
   13/16    110    (6)
   15/16    111    (7)

Here’s the corresponding tabulation for the braided tree:

    1/16    000    (0)
    3/16    100    (4)
    5/16    010    (2)
    7/16    110    (6)
    9/16    001    (1)
   11/16    101    (5)
   13/16    011    (3)
   15/16    111    (7)

The order is “counting from the left,” incrementing the most significant bit first. I wonder what other permutations give interesting-looking trees.

Turning back to the tangled tree that began this whole investigation, I would argue that it looks tangled and unruly not so much because lots of edges cross (there are even more crossings in the braided tree) but because of many near collisions and coincidences. For example, just below the root of the tangled tree, the mirror-image paths <left-right-right> and <right-left-left> almost converge on the same value, but they miss by just a bit. We can fix this! All it takes is an adjustment in the formula for the displacement d. In the tangled tree that displacement is 1/2 min(x, 1-x). The constant 1/2 in this expression is quite arbitrary, and we can substitute any other value a, with 0 < a < 1.

So what is the magic value of a in the following tree, which I am inclined to call the crocheted tree?

Note: If the title of this post fails to ring a bell, see the poem by Robert Hass:

The gene pool threw up a wobbly stem
And the tree danced. No.
The tree capitalized.
No. There are limits to saying,
In language, what the tree did.

From the Associated Press, via the New York Times:

LOS ANGELES (AP) — California faces an almost certain risk of being rocked by a strong earthquake by 2037, scientists said in the first statewide temblor forecast.

New calculations reveal there is a 99.7 percent chance a magnitude 6.7 quake or larger will strike in the next 30 years. The odds of such an event are higher in Southern California than Northern California, 97 percent versus 93 percent.

I read this report with a certain sense of wonder. What impressed me was not the prediction itself; it’s not the first time I’ve heard that the Big One is coming. What took me by surprise was the level of mathematical sophistication that we can now take for granted in readers of the morning newspaper. No more do we have to worry that people will add up 97 percent and 93 percent to get 190 percent. Evidently, we’ve reached a state of universal numeracy, where everyone knows how to combine probabilities, and there’s no need to explain the calculation. We don’t even need to remind anyone that when we compute 1 – (1 – p)(1 – q), or p + q – pq, we are assuming that p and q represent probabilities of statistically independent events; everybody knows that. And everybody understands that in this context “a chance of a quake” really means “a chance of at least one quake.”

I guess the only place where we might still stumble is in actually doing the arithmetic. My calculator tells me the number is 99.8 percent, not 99.7.

A further note: The original report on which the news item is based leaves me even more perplexed. The probability model adopted in the forecast is explained as follows:

The simplest assumption is that earthquakes occur randomly in time at a constant rate; i.e., they obey Poisson statistics. This model, which is used in constructing the national seismic hazard maps, is “time independent” in the sense that the probability of each earthquake rupture is completely independent of the timing of all others. Here we depart from the… conventions by considering “time-dependent” earthquake rupture forecasts that condition the event probabilities… on the date of the last major rupture. Such models… are motivated by the elastic rebound theory of the earthquake cycle…; they are based on stress-renewal models, in which probabilities drop immediately after a large earthquake releases tectonic stress on a fault and rise as the stress re-accumulates due to constant tectonic loading of the fault.

In other words, it doesn’t sound as though the assumption of independence is even approximately satisfied. I must be missing something. The 99.7 percent combined probability is mentioned in the executive summary of the report, but I found no explanation of how that number was calculated.

Perhaps I shouldn’t worry so much. I live thousands of kilometers away in a zone of seismic serenity.

Update, several hours later: After reading a little more carefully, I think the report does assume that all possible earthquake sites are independent. At each site the probability of an event is a function of time, but it is independent of probabilities at other sites. Thus calculating a joint probability for the northern and southern parts of the state does seem to be a valid operation. And the distinction between “exactly one” and “at least one” doesn’t really enter into the matter either. That’s because the model is only valid until the next major earthquake occurs; after that, all bets are off, since the time-dependent probabilities have to be recalculated.

If this interpretation of the model is correct, I think the way the result is expressed is somewhat misleading. To say there’s a 97 percent chance in Socal and a 93-percent chance in Nocal implies there’s a high probability (90.2 percent) of seeing both events in the course of the 30-year period. But the model is no longer valid after the first quake.

I wonder if there isn’t a better way to express the concept at the heart of this story. Qualitatively, it’s easy enough to grasp: In the next 30 years there will almost certainly be a major earthquake somewhere in California, and the event is more likely to happen in the southern part of the state than in the northern part. Putting this into numbers is somewhat tricky—or at least I’ve had a lot of trouble with it. Having finally surrendered to the computer and performed a Monte Carlo simulation, I come up with this statement: There’s a 99.8 percent chance that the next major California earthquake will happen by 2037. If indeed such a quake occurs, the odds are about 57 to 43 it will hit in Southern California.

The latest issue of American Scientist is just out, both on the newsstand and on the web. My “Computing Science” column is titled “Wagering with Zeno”; it returns to a subject mentioned briefly in an earlier column, “Follow the Money.”

Consider a random walk on the interval (0,1), where the walker moves according to these rules:

• The walker begins at x = 1/2.
• At each step the walker moves either left (toward 0) or right (toward 1) with equal probability.
• The length of each step is one-half the distance to 0 or to 1, whichever is nearer. In other words, the distance moved is 1/2 min(x, 1–x).

If you want to know more about this process and why I bothered to write about it, please see the column. Here I just want to say a few words about visual aids that might be helpful in understanding how the random walk evolves and where the walker is likely to wind up.

Here is one of the illustrations that appears in the column:

We see four trajectories, each lasting 10 steps. Although the walk is actually one-dimensional—moving back and forth along an interval of the x axis—the trajectories are plotted in two dimensions for clarity, moving down the page so that the illustration becomes a kind of discrete spacetime diagram. Even with this device, though, some of the paths overlap for part of their course. If I tried to crowd more trajectories into the figure, the overlaps would make it hard to sort out one walk from another.

In making pictures like this, the issue is not just how you explain an experiment to other people; it’s also a matter of how you come to understand the process yourself. While I was trying to make sense of the Zeno walk, I plotted lots of trajectories like these, initially by hand and then with a program that generated Postscript files. Some important features were quickly apparent. In particular, I noticed that the walker takes much bigger strides in the middle of the space than near the edges. (Of course this fact follows directly from the definition of the walk, but it doesn’t hurt to see a picture.) I also observed that most of the walks seem to spend most of their time hugging one edge or the other, seldom venturing back across the midline of the space. Even after looking at a few hundred examples, however, I didn’t feel like I had a really secure sense of how a “typical” walk would proceed.

As with many processes that evolve over time, the illustrative possibilities are much richer if we can escape the static bounds of ink on paper. And thus I was led to try creating a more dynamic and interactive display.

What you see above is not dynamic; it’s merely static pixels on glass. But you can see it all in action in this Java applet. Feel free to click on the link and go play, but do come back here afterward so we can continue the conversation.

Hello again.

Does this program succeed in conveying an intuitive sense of how the Zeno walker walks? Personally, I’ve found it helpful, although it still comes up far short of ideal. Adjusting the number of steps in each walk gives a sense of how the walker’s apparent attraction to the edges gets more extreme over time. And adjusting the rate at which paths fade into the background allows for a degree of balance between clutter and evanescence. But there may well be a better way to go about visualizing these concepts. I’d be interested in hearing suggestions.

A note on the implementation: The applet was created with the programming language called Processing, invented by Ben Fry and Casey Reas. I’ve just reviewed two books on Processing, and so this project was undertaken partly to try out the language and its programming environment. I found it very slick, and mostly trouble-free. The source code of the zenowalk program is available through a link on the applet page. (Most of the code for the buttons and sliders was cribbed from the book by Fry and Reas.)

Of course there’s always something lacking in any programming language. In this case what I missed most acutely was the exact rational arithmetic I rely on in Lisp. Processing has only single-precision floating-point numbers. In the Zeno walk, these numbers begin running out of significant digits after about 150 steps (which is why I limited the program to 145 steps). As far as I can tell, no one has yet written a Processing library for bignums and exact rationals. I guess that’s my job.

My new book has come out this week: Group Theory in the Bedroom, and Other Mathematical Diversions, Hill and Wang, xi+269 pages, $25. ISBN-10: 0-8090-5219-9, ISBN-13: 978-0-8090-5219-9, Library of Congress Call Number: T185.H39 2008.

This is a collection of essays on themes that will be familiar to many readers of bit-player.org. Indeed, the essays themselves may be familiar! Eleven of the twelve essays appeared earlier as “Computing Science” columns in American Scientist; the twelfth was published in The Sciences. But there is some new material: Appended to each chapter is an “Afterthoughts” section where I confess to errors, castigate critics, bring outmoded notions up to date, and generally try to offer readers some excuse for selling them a $25 book, when they could find most of the content free on the web.

The book has its own handsome web site (I was lucky to snap up the domain name “grouptheoryinthebedroom.com” before some speculator squatted on it). I cordially invite all of you to go over there and have a look around. Meanwhile, back here at bit-player HQ, I’m going to throw myself a little party. See you in a day or two.

   ’Tis pleasant, sure, to see one’s name in print;
    A book’s a book, although there’s nothing in’t.
                                      —Lord Byron

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

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

Some properties of the graph:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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