Virginia's 74th Magisterial District Democratic Primary

On 2007 June 12, primaries were held in the state, or Commonwealth, of Virginia. Among these were the Democratic primary for the 74th Magisterial District, which includes a huge portion of Henrico County and parts of the cities of Richmond and Hopewell and Charles City County. This primary was hotly contended by five candidates. It looked like a four-candidate race until Joseph D. Morrissey, a former Commonwealth's Attorney, barged into the race. Here are the results:

Joseph Morrissey	2,055	38 %
Floyd Miles	1,488	27 %
J.M. "Jackie" Jackson	1,053	19 %
David Lambert	545	10 %
Shirley McCall Goodall	285	5 %

Morrissey won easily, leading his next challenger by 11%. Sic semper tyrannis. This character has been disbarred by the Virginia bar and cannot practice law. He has gotten into two fist fights with others in the courtroom, and has been the object of lawsuits and has served time in jail. How did he get elected, then? Part of it was that the people don't like what government is doing. That is why Morrissey, a white person, got a huge vote in a mostly black district. But that does not explain it all. 62% of the voters voted against Morrissey; they voted for someone else. So how did Morrissey win?

It is the election system that allowed him to win. The 62% that were opposed to him split among four candidates. All of Morrissey's support went to Morrissey. So Morrissey won. Is this a fair system? I wouldn't think so. 62% of the electorate did not want him in. However, 95% did not want Shirley Goodall elected. But Shirley has hit no one, that I know of, has not served jail time or otherwise been controversial. The difference is that although Morrissey was first place among 37% of the voters, I suspect most of the rest rated him last or next to last.

That is why a runoff system would be better. In that system, if a candidate wins more than 50% of the vote, he wins. Otherwise, there is a runoff election involving only the top two candidates. If that had happened, I suspect the result would have been something like:

Floyd Miles	3,071	57%
Joseph Morrissey	2,355	43%

This is because all but 300 of the voters of the other candidates voted for Miles rather than Morrissey (these figures are hypothetical). The flaw with the existing system is that although it reflects voters' first choice among those running, it does not reflect their second or later choices.

This type of paradox has happened before, sometimes with devastating results. In Chile, Salvador Allende was elected president in a three-way race; his percentage was in the upper 30s. He was a Marxist, and that displeased some people in the administration of the United States. Some undercover figures there engineered a coup, resulting in a repressive dictatorshop under Augusto Pinochet for many years. In Minnesota in the late 1990s, the world champion wrestler Jesse Ventura won as an independent over the Democrat and Republican with a vote in the mid to upper 30s. I don't know if Jesse would have won if a runoff had occurred.

There are other voting systems around. An improvement in the runoff scheme is what I call the Olympic system, because the Olympics use it to decide sites for their Games. In this system, if a majority vote occurs, that wins the election. If it does not, the last place city is dropped from contention and another runoff occurs. If a majority occurs, that wins the election. Otherwise, the last place of those remaining is dropped and another runoff occurs. This continues until a majority occurs, which it will unless it ends in a tie between the remaining two candidates. There are flaws in these systems as well. No system will be perfect.

One system seems to stand out. In approval voting, voters are asked to pick any number of candidates, not necessarily one. Of course voting for all the candidates is equivalent to throwing your vote away, because you are not expressing a preference among them. This system (and all the others I have described here) are the same for two-candidate contests. In a three-way contest, voting for two of the candidates is effectively a vote against the third one. This allows voters to express dissatisfaction with a candidate. The Mathematical Association of America and American Mathematical Society use this system in their elections.

I suspect that if approval voting had occurred, so many votes for {Miles, Goodall, Jackson, Lambert} would have occurred that there is no way Morrissey would have won.

Baseball Musical Chairs

It happened sometime in the spring of this year (2006). The Ottawa Lynx AAA baseball team was bought out by two Philadelphia businessmen, Joseph Finley and Craig Stein. They reached an arrangement with the Philadelphia Phillies by which the Lynx would become the Phillies AAA farm team, which would move to Allentown in 2008, when a new stadium there is built. Allentown and Philadelphia are only 54 miles apart as the crow flies.

Would you believe that just this little double play by these two men would cause a free for all, and then a five-way swaperoo of AAA minor league teams among the majors?

When the Scranton Wilkes-Barre Red Barons, the current Phillies farm team, found out about this, their fans were upset. They wanted to find another major league team to be their parent team. The early word was that this team would wind up being a Baltimore farm team. Seems logical. Swap teams. But the fans clamored for something better. They appealed to major league teams all over the place, and especially the ones in New York City. Pretty soon both the Yankees and Mets became interested and they jumped in the fray.

Discussions between the Red Barons and the Yankees got the Columbus Clippers, present farm team of the Yankees, worried. Columbus started appealing all over the place, and the Orioles and Mets showed an interest. The Norfolk Tides became annoyed. Actually they were annoyed for some time. They did not like their parent team, the Mets. The Mets ignored them. So they took this ruckus as an opportunity to search for something better. This is another case of a minor team turning its back on its parent major league team, with the first one being the Rochester Red Wings' rejection of the Baltimore Orioles in 2002. To top it off, the Washington Nationals did not want any more of the New Orleans Zephyrs. Too far, it seemed. So they jumped in the fray.

Today the whole thing settled out to what amounts to a five-way circle-around among the five major league teams. The Phillies got their Lynx, which will become the Allentown Allies, I suppose. The parent team of the Lynx, the Baltimore Orioles, got into an agreement with the Norfolk Tides. (How strange. The Tides, rejecting the Mets, picked the team that was rejected by the Red Wings.) The New York Mets got left out and were forced to sign with the New Orleans Zephyrs. That's even farther than Washington. The Washington Nationals got into a deal with the Columbus Clippers, and the New York Yankees got their team in Scranton Wilkes-Barre to complete the cycle.

I am wondering why the Mets settled for the Zephyrs. Couldn't they have gone after the Durham Bulls, which would have resulted in the Zephyrs getting something closer also - the Tampa Bay Devil Rays?

So how does this affect the travel budget of these pairs of teams for players and other officials going back and forth between the major league city and the AAA city? This year, the total of the crow-miles for the five pairs of teams was 2267 miles. After the swap, the total miles will be 2160. So they saved some mileage. They will save even more after the Lynx move to Allentown; the miles will go down to 1830.

But this is not the best they could achieve. If the ten teams had gotten together and picked the assignment that minimizes the crow-mile total, they would have paired as follows:

Baltimore Orioles - Columbus Clippers
Washington Nationals - New Orleans Zephyrs
Philadelphia Phillies - Norfolk Tides
New York Mets - Ottawa Lynx (and then Allentown Allies)
New York Yankees - Scranton Wilkes-Barre Red Barons

The total crow-miles would now be 1983, and after the move to Allentown, they would be 1723. The optimal assignment is the same regardless of the location of the Lynx.

This shows that major league teams use other criteria for locating their minor league teams than distance (hence fuel) costs. With shortages of oil coming up, they should have considered the minimal-mile arrangement above.

By the way, I tried this for all 30 pairs of major and AAA minor league teams a year ago, and published the result in Blogtrek. I thought it strange that it assigned the Twins to the Norfolk Tides and the Columbus Clippers to the Milwaukee Brewers. Today I found that I had entered the latitude and longitude incorrectly for Columbus and Norfolk. In addition, the Red Barons will relocate to Allentown. I will have to redo the overall assignment. Watch for the results.

Note to operations research instructors. This blog provides an example of an assignment problem and the use of the Hungarian algorithm to solve the problem. I used the Hungarian method to come up with the above pairing. The hardest part of setting this problem up is finding the 435 distances between major and AAA cities. I just simply used the latitude and longitude and used a trigonometric formula to find the crow-distances.

Knoxville Mathematics

Earlier this month I attended Mathfest 2006, a conference sponsored by the Mathematical Association of America in Knoxville, TN. It was an interesting convention, held in the downtown area at the hotel I was staying at and at the Knoxville Convention Center, which was next door to the yellow sphere that symbolizes the World's Fair that was once held in Knoxville.

Most of the talks were on how to teach mathematics and what are the best way to teach a certain theory to the public at large. There were three talks on Sudoku, the 9x9 square craze of the past couple of years. One gave us the variations on Sudoku, another told us to some extent how to solve a Sudoku, and the third one showed how to program a computer to solve a Sudoku puzzle.

The invited addresses included an interesting one on knots, in which the presenter enumerated the types of knots there were, and gave us some knot invariants, such as the Alexander polynomial. None of these completely characterized knots; in fact the Alexander polynomial flunked tenderfoot, because it can't tell a square knot from a granny knot. Another talk was about the Yawp of Mathematics. What is Yawp? Watch the movie "Dead Poets Society" and find out. It means getting excited about finding something striking or beautiful in mathematics.

Another talked about whether 9 to the 9 to the 9 to the 9 to the 9 power or 9!!!! was bigger. The presenter asked for the biggest number with 5 symbols. I tried 9^^^9, where ^ is an up arrow, but he would not allow that type of notation and wanted to stick to ^ (exponentiation) and below. Turns out the answer is neither but rather (9 to the 9th)!!!

One section dealt with mathematics in popular culture. Now this is interesting. Popular culture is dominated by prima divas, murder cases, silly sitcoms, wardrobe malfunctions and other such highly illogical and immathematical pursuits. So how can mathematics slice through this muck? The TV show NUMB3RS helps, and one presenter analyzed the mathematics of this show. The Simpsons had one episode wherein the equation 3987^12 + 4365^12 = 4472^12 was written on the board. No, Andrew Wiles. Your proof of the Tanayama-Shimura conjecture is still intact. Note that the left side is divisible by 3 but the right isn't. One presenter showed us how to model Yoda's robe which was swinging around all over the place. It wasn't too easy. Another analyzed mathematics in movies after noting that these movies never made it to the Box Office: Math Wars, f(x)Files, Lord of the Ring of Matrices, and e^t, as in e^t phone home. Another speaker delved into mathematics and mental illness, using A Beautiful Mind and Proof as examples. He could have included Georg Cantor, Kurt Gödel, Theodore Kaczynski (the Unabomber), and Paul Morphy (the chess genius). This is an interesting topic - is there a relationship between mathematics and mental illness, or are there enough Paul Cohens running around to disprove such a relationship?

The most interesting night was when we all went on a riverboat cruise aboard the Star of Knoxville on a stormy night, which gave us a dramatic scene of Knoxville in twilight with a severe storm's lightning in the area, and in which we got back to our hotel rooms just minutes before the storm hit.

So when am I going to another major mathematics conference? The American Mathematical Society/Mathematical Association of America conference this January is in New Orleans. How can this be? The convention area was not hit by floods, and holding a convention pumps money into an economy that desperately needs it. The convention's events will be at two large hotels, not at the infamous Convention Center. All the fixings of New Orleans, including the Vieux Carré and New Orleans jazz, will be there. So maybe that is a good convention to go to, but I will drive there if I go, even though it is a two-day drive, because of all the stuff that has been occurring on the airlines recently.

How to Find a Lock Combination

I sorted out my safe recently and decided that many papers and other items don't need to be in a safe. I saw a couple of Master padlocks there and wanted to keep them outside the safe. But I wanted to keep them only if I knew the combinations. I had a whole bunch of note papers with combinations on them and tried them out on the two locks. I quickly opened one of them, so I tagged it with its combination and put it away. But the other one would not open. So I threw it out.

I wondered about that. It's a perfectly good lock. The only reason why I want to throw it out is that I did not know its combination. Was there some way I could get that combination by fooling around with lock somehow? The existence of some way could endanger the security of the lock, but I wanted to do something other than throw it out.

I looked on the Web using Google and found Electroatomics. The site gave a procedure for finding the combination of a Master padlock. It's a good site, and is its owner's most favorite page. I looked through the instructions and it said to turn the lock to 0 (the numbers range from the real numbers 0 through 40, but combinations are given in integers). So I did that. Then it said to pull on the hasp and turn the dial. Which direction? It did not say. I turned it towards the forward numbers, 1, 2, and so forth. But how can I when the hasp was pulled? After some monkeying around I found that I could hop from stable "settle" point to point. The instructions say to move the dial of the lock until it hits these points and record them. They say there should be 12 of them. I tried that and got a series of numbers. They could be right on the integers, as 13, or they could be in between, like between 18 and 19. In that case record an 18.5.

He then says that there should be seven numbers ending in .5. He says these are decoys. He says look at the numbers that end in .0; i.e., integers. There should be five of these. In his example, he said he got 32, 22, 19, 12, and 2. They all end in the same digit except one. That one is the third number of the combination. So in his case, the lock's combination ended in a 19.

I tried that with the lock I had and got a series of numbers. But different times I did it, I got slightly different numbers, and I was getting 6, or 8, or something like that numbers ending in .5. The whole thing was wishy washy to me, and the integers I got had at least two repetitions of two integers, so I could not identify the third number of the combination. I reread the instructions and it said to turn it back and forth to see what range of numbers it would go in. He said if it went between 4.5 and 5.5, record a 4, and if it went between 4 and 5, record 4.5. I was not getting just exact readings and .5s however. I was also getting .8s and .3s and other such stuff. So the whole thing became really confusing. I put these results in a table, where left is as far left as I can get the stop point, and right is as far right. Avg is simply the average of those two numbers:

	Left	Right	Avg
1	-0.5	0.5	0
2	2.8	3.8	3.3
3	6.0	7.0	6.5
4	9.8	10.8	10.3
5	12.8	13.8	13.3
6	16.0	17.0	16.5
7	19.6	20.5	20.05
8	22.8	23.8	23.3
9	26.0	27.0	26.5
10	29.5	30.5	30.0
11	32.8	33.8	33.3
12	36.0	37.0	36.5

This still did not tell me much. I tabulated the 0s, the .3s, and the .5s, and got 3, 5, and 4 readings respectively. This did not match at all what Liam Bowen said I would get. I looked at the readings. The five .3s gave me 3, 10, 13, 23, 33, which according to Bowen's rules would say that 10 was the third number of the combination. But these are .3s, not integers. I noticed that the .5s gave 6, 16, 26, and 36, so that there is more than one sequence of numbers all ending in the same digit. So I tried something different. I rearranged the numbers into a 3x4 array:

0	3.3	6.5
10.3	13.3	16.5
20.05	23.3	26.5
30	33.3	36.5

Now it comes out. Each column looks like it lines up nicely except the first, where the 10.3 sticks out like a sore thumb. I was now certain that the third digit of the combination was 10.

I tried this out on a lock whose combination I knew. The combination ended in a 27. I tried the same table technique and, sure enough, the 27 was different from the others.

So that now I know the third number is a 10, Bowen then makes a lengthy explanation of what "mod 4" means. What he says is that that x - z = 0 mod 4, and that x - y = 2 mod 4, where the combination is x-y-z. This means the first number is one of 2, 6, 10, and so forth, and the second number is one of 0, 4, 8, and so forth. So now there are only 100 combinations to try.

I tried them one at a time, and none of them worked. Just as I was about to exhaust all the combinations, I pulled on the hasp. It unlocked! I had found it! I memorized the number (of course I won't give the result here as it would compromise the lock). But I had found the combination and now don't have to throw the lock out.

So the procedure for finding the combination of a Master Lock (it has to be a Master, according to Bowen) is:

1. Turn the dial while pulling the hasp out. There should be 12 points at which it "settles" (Bowen calls them "clicks").

2. At each point, turn the dial back and forth until it won't go and record the results to one-decimal accuracy. The result should be a table giving 12 left and right readings.

3. Take the average of each row of this table.

4. Rearrange these averages into a 3x4 table, putting 3 readings in the first row, 3 in the second and so forth.

5. Find the number whose decimal does not jibe with the other entries in its column. This number is the third number of the combination. Call this z. Call the combination x-y-z

6. Test all combinations of values of x and y, where x is congruent to z mod 4, and y+2 or y - 2 is congruent to z mod 4.

7. One of these should be the lock's combination. If this does not work, you made an error somewhere or they have manufactured a new type of lock.

Baseball Sudoku

The latest craze hitting the world is Sudoku, the American game of putting the digits 1-9 into a 9x9 grid separated into nine 3x3 sections such that each row, column, and section contains exactly one of the digits from 1-9. Sudoku puzzles appear in the paper each day, and there are many web sites with puzzles on them. There are many variations on the Sudoku theme, including overlapping puzzles, Sodoko (certain 9-patterns must have 1-9 as well), the Monster (certain sums have to add up), and so forth.

A rather unusual application appeared recently on the Web, although I had thought of it months ago. Nine is the number of cells in a row or column in Sudoku, but 9 is also the number of members of a baseball team. So how about a baseball sudoku? The one I had thought of was to replace the numbers (which really aren't numbers; adding 2 to 4 to get 6 in a puzzle does not make sense) with baseball positions. Certain cells are filled with positions, such as c (catcher), 1b (first baseman), cf (center fielder) and so forth. The idea is to fill in the rest so that each row, column, and block constitutes an entire baseball team. Big deal. This is just ordinary Sudoku with baseball positions in it.

As any baseball fan knows, many players can play several positions. For example, a center fielder can play right field, a shortstop can play second base and so forth. So the idea is to provide as clues not positions, but real major league players, some of whom can play more than one position. The idea is to fill in the remaining squares with real players and designate the positions of each of these players so that each row, column, and block contains a complete baseball team. This is equivalent to a variation of Sudoku where the clues are choices of numbers, saying this square is either a 3 or a 5. A variation on this would be to make it so some special rows, columns, or blocks are from the same team, so that it's conceivable that this team of players could actually be fielded in a real baseball game.

So if you are going to do baseball sudoku, do it right. Don't just make it plain old Sudoku with baseball labels. Get some baseball into it, too. And by the way, make sure you use only National League players. If you use American League players, that injects a tenth position into the puzzle - the designated hitter. You no longer then have the nice 9x9 square, if 10 positions have to be filled.

Bigamy Logic

I heard about a bigamy case in Fairfax, Virginia today. This man had married seven women, with some of these overlapping, and was exposed when a woman who would be his eighth wife saw him on Dr. Phil and called police. Last year, charges were pressed against him for marrying his seventh wife while married to his sixth wife. However, the authorities dropped the charges when it was decided that the marriage to number 6 was invalid because at that time he was still married to number 5.

This must have been some character, all right. I suspect his motive was money; in each case, he seemed to have cleaned out his wife's bank accounts. But this brings up an interesting question. When someone starts committing bigamy like this over and over again, which of the marriages are valid and which are not? For example, in the case above, maybe the authorities could have convicted him because he was validly married to his sixth wife, because his marriage to his fifth wife was invalid because at the time of that marriage, he was married to his fourth wife. I don't know if this was really the case; I suspect it was not. One can continue with this for some time; for example, maybe the marriage to number 7 was invalid after all because the marriage to number 4 was invalid since he was married to number 3 at the time. Or maybe it was valid, because marriage 3 could have been void because he was married to number 2 at the time. Or maybe it is invalid because when marrying number 2, he could have been married to number 1. But that ends it. Any marriage that you make with your first spouse is valid; or at least it can't be invalid because of bigamy. So the lout's marriage to his seventh wife was invalid. But that's only if all this overlapping occurred.

Suppose then that a man does the following:

Marries Annette;
Marries Betty while still married to Annette;
Divorces Annette;
Marries Cindy while still married to Betty;
Marries Dorothy while still married to Cindy;
Divorces Cindy;
Marries Ethel while still married to his other wives;
Divorces Betty;
Divorces Ethel.

Which of these marriages were valid? I am assuming that statements saying he divorced someone are to be striken from this account if the marriage that is being divorced is invalid. Don't scroll any farther if you want to find out.

Answer: His marriage to Annette was valid. His marriage to Betty was not valid. His marriage to Cindy was valid, since his marriage to Betty was not valid, and he had divorced Annette. His marriage to Dorothy was not valid. His marriage to Ethel was valid, since he divorced Annette and Cindy and since his marriages to Betty and Dorothy were not valid. After divorcing Ethel, he is all by himself with no wives.

In general, how do you determine who a bigamist is legally married to when you are given his list of marriages and divorces?

A Tale of Two Numbers: Omega and The Number

Yesterday (2006 February 9), I ran into two numbers that are important in our lives. They help describe the logical world we live in, as well as our state of well being in it. Neither of these numbers can be computed precisely, for different reasons, but they nevertheless exist.

One came in an issue of Scientific American which arrived in my mail yesterday. That magazine contained an article called "The Limits to Reason", which describes a number defined by Gregory Chaitin and denoted by Ω, capital Omega. This number is defined by the probability that if a randomly chosen computer program is chosen, that it will halt. Now programs on these PCs that we use need to be designed to halt. If it doesn't, the computer freezes up, and I call that "bombing" or "crashing". So in a sense, Ω is the probability that if a computer program is run, that it will not bomb the computer. This number is not computable. In terms of my paper Hamlet II, it is an ultimately indescribable number. In that article, I show how Georg Cantor demonstrated that such numbers exist, but I did not give any examples of such numbers. Ω is an example of such a number. (Don't confuse Ω with ω, the first transfinite ordinal number, defined by Cantor, and Ω, sometimes used to denote the class of all transfinite ordinal numbers. Ω is a real number.)

Why? It derives from the famous Halting Theorem of Turing, which says that no computer program can be constructed that inputs a computer program as input and outputs whether the program, if executed, will halt or not. Another way of putting it: There can be no perfect Crash Guard. The proof of this, in a sketchy form, can be given like this. Suppose there were such a program P. Let's hack that program. Find the places in the code where it says that the input program will halt, take out the termination statement and replace it with something like A: go to B; B: go to A. Then go to those places that say that the input program will never halt, and put in the warning "Do not execute this program. It will never end, and it will bomb your computer." and follow that with a statement that halts the program. Call the hacked program H. H still inputs computer programs as input, so feed it the program H itself. Suppose this program halts. Then H will detect that, and it will go into the AB loop above and never halt. Contradiction. Suppose H never halts. Then H will detect that, print out the warning and stop. That is also a contradiction. This shows that H cannot exist, and therefore P can't either. There are many important derivative results of this theorem. For example, given a group with generators and relations and an element of the group, no computer program can compute whether that element is the identity element or not. Also there is no terminating algorithm that solves all Diophantine (answers have to be integer) systems of general polynomial equations.

And it shows that Ω is ultimately indescribable. For if I could describe it to you, then I could write a computer program to compute it, and then I could hack that program to yield a program that tells whether a computer program will halt, counter to Turing's theorem.

Chaitin has a number of other interesting observations. The most interesting, I found, was this remark:

Comprehension implies compression.

This says that in order to comprehend something, a simple description or program needs to be found for it. For example, to describe π, I can say add up with alternating signs all the reciprocals of odd integers and multiply the result by four. Or perhaps I can simply say the circumference of a circle divided by the diameter. On the other hand, to describe 0.92659876193865, the simplest way I know is to say 9, 2, 6, 5, 9, …, 5 after the decimal point, which is as long as the number is. So π is a simple number, but 0.92659876193865 is not. To understand π, I think of the circle. So the more I compress a description of a number or mathematical concept, the more I understand. Hence Chaitin's statement.

Another comment he made was on the principle of sufficient reason. That says that everything has a cause. The condensation of water vapor causes rain, for instance. But Chaitin says that there has to exist "irreducible" numbers, numbers as complicated as 0.92659876193865 that can't be described any simpler than naming the digits. If we generalize this to arbitrary mathematical concepts, there must exist mathematical statements that have no cause. Not even for mathematics is the principle of sufficient reason true. Some things are just the way they are. "Cause was not the reason for the evening", says the singing group America; in fact, the entire song "Tin Man" seems to have no cause or reason for it; it just is.

The other number I met yesterday was in a library book entitled The Number, by Lee Eisenberg. This number is more prosaically defined. Each person has The Number, and for him it is that sum of money, such that if he possesses it, it will enable him to live comfortably for the rest of his life. There is a simple way to compute this number, given you can get the inputs. Total up all the expenses you have or would have in the life style you want. Divide the result by the interest that a typical money market fund or a CD gives, and the result is, for you, The Number. For example, if your expenses are $50,000 per year and the interest rate is 5%, then $50,000/0.05 = $1,000,000 is The Number. That's right, a modest amount like that yields a million dollars. That is a point made by the author: many, in fact, the majority, of Americans simply don't have The Number. This will result in trouble for them later when they age and retire, and the sum total of them will seriously jolt the national economy. So it is an important discussion, and I ask you to figure out what is Your "The Number". And by the way, it is computable, given that your expenses and the ongoing interest rates, are computable. It is not a Chaitin's Ω.

Cryptorithm in a Store Sales Receipt

Cryptorithms are puzzles wherein one replaces the digits in an arithmetic computation with letters. The puzzle is presented with the letters, and the object is to find out what the digits were. A famous example is (ignore the dot)


. S E N D
+ M O R E
_________
M O N E Y


Don't read too much below if you want to solve the problem yourself. In this case, the M sticks out as an additional digit on the front end of the sum, so it has to be 1. S + M = O then becomes S + 1 = O. The possibilities are 9 + 1 = 0 and 8 + 1 = 0, because 9 + 1 = 1 is eliminated becayse that would make both M and O equal to 1. Therefore, O is 0. O is often confused with zero, as in calling this year Oh-Six. Usually, O is not 0. But this case is an exception. We were told that O is some digit, and we wound up proving that it is 0.

This reduces the problem to:


. S E N D
+ 1 0 R E
_________
1 0 N E Y


If S were 8, then a carryover would have to had to occur in E + 0 = N. This implies that E would have to be 9 and N would have to be 0; however, we have already assigned O to 0. Therefore, S is 9:


. 9 E N D
+ 1 0 R E
_________
1 0 N E Y


The sum now says EN + 0R = NE. So adding 0R to EN reverses EN's digits. When you reverse the digits of a number and subtract from the number, the result is always a multiple of 9. So with no carryover, 0R has to be a multiple of 9. If there is a carryover, 0R has to be 1 less than a multiple of 9. With zero as the first digit, that means that R has to be 8 or 9. 9 is already taken, so R = 8, and N = E + 1. The digits that are left are 234567. We can choose EN to be any two consecutive digits from this. But then two of the remaining digits have to differ by 10 minus the lesser of these digits, because of D + E = Y, which carries over. So if we select 23, two of 4567 have to differ by 8, which they don't. with 34, two of 2567 differ by 7; they don't. With 45, two of 2367 differ by 6; they don't. With 6 and 7, two of 2345 differ by 4. They don't. With 56, two of 2347 differ by 5; 7 and 2 differ by 5. So D = 7, Y = 2, E = 5, N = 6, and the solution is:


. 9 5 6 7
+ 1 0 8 5
_________
1 0 6 5 2


So that is what a cryptorithm, and this shows how you can solve one. SEND MORE MONEY is a moderately difficult cryptorithm. I expect cryptorithms, as interesting puzzles, to appear in puzzle books, newspapers, and recreational mathematics books. But take a look at what I found on a store receipt the other day!

What kind of a receipt is this? It says the Megan doll costs TOO. Well, yeah, things do cost too much nowadays. But then it gives as the subtotal QQOI. Whaaa?? Then I realized that the entire thing is a cryptorithm! Why would JC Penney put a cryptorithm on their sales receipts? Maybe they want their customers to solve puzzles. But I am game for this one. The complete cryptorithm is:


Megan Doll...TOO
Plush Dolls..TOO
Subtotal....QQOI
Sales Tax.....YP
TOTAL.......QWTI


I thought maybe O was a misprint or was intended to be a zero. It can't be, as 0 + 0 = 0, not I, whatever I is. For the same reason as before, Q must be 1. This means the double of a digit is 11. The only possibility is 5 with a carry, 5 + 5 = 11. So O + O must be O, with a carry. The only digit which has this property is 9: 9 + 9 = 19. So O is 9, and both dolls cost $5.99. This means that the Subtotal is $11.98, making I = 8. Now here at this point I used the fact that Virginia sales tax is 5%. This gives either 59 or 60 for YP. Since 9 has been used, YP must be 60. The total of QWTI then must be $12.58. Sure enough, this amount matched a figure on my check account statement.

So there it is. A cryptorithm on a sales receipt solved. Where else can we get mathematical puzzles? Suppose 9 players play 9 different other players in some 1-1 game, such as chess or racquetball. Suppose each of 9 weeks, each player must play a different player from the other team, and suppose there are three places for the games, and that every three weeks, the players rotate places. Suppose someone has half scheduled this tournament. The scheduling the remaining players is a Sudoku puzzle!

Or how about having Rubik's Cubes show up as airline schedules? Airline scheduling is harder than solving that infamous cube. There are many ways in which services and manufacturers could confront customers with puzzles.

In this case, I noticed some words. "Gift Receipt". Does a cryptorithm come with each gift? Maybe the idea was to hide the price, since most givers don't like to disclose the price to receivers. But look what happens when you substitute the numbers for the letters in the JC Penny cryptorithm:


0 1 2 3 4 5 6 7 8 9
P Q W . . T Y . I O


What does that say? Qwerty, that's what. The first line of characters on the typewriter keyboard. But P following O on the keyboard means that the 0 must follow the 9:


1 2 3 4 5 6 7 8 9 0
Q W E R T Y U I O P


But someone could have typed in these prices simply by typing the real prices with your hands off the home keys. What a code that is! It's easily cracked. But not everyone is going to solve cryptorithms or type prices in the computer with the hands misplaced. So I guess this code is safe.

Interesting. Penney's receipt cryptarithms. What will they think of next?

The Mathematics of Hybrid Cars

This week I found an article on CNN's web page entitled "Hybrids: Don't Buy the Hype". Note that this link may become invalid soon. Here is a quote from it:


A hybrid Honda Accord costs about $3,800 more than the comparable non-hybrid version, including purchase, maintenance and insurance costs. Over five years, assuming 15,000 miles of driving per year, you'll make up that cost in gasoline money if the price of gas goes up immediately to $9.20 a gallon and averages that for the whole period.


This sounds like a math problem. Some of the questions that may be asked are "How much does a hybrid Honda Accord cost?" or "How many miles per gallon does the hybrid Honda Accord get? This is not a real type of question. It only exists because some people chose not to give some information. In this case, CNN chose not to state what the price of a Honda Accord was or how many miles per gallon it gets.

But what if we try to solve it? Let

H = cost of a hybrid Honda Accord
C = cost of a non-hybrid standard Honda Accord

The first statement then says:

H = C + 3800

The second statement assumes you will drive 15,000 miles per year over a 5-year period. That is 5 * 15,000 or 75,000 miles.

The third statement says that if it costs $9.20 a gallon for gasoline, then the costs of the two Hondas are the same. This seems somewhat more complex. Let the miles per gallon of the hybrid be h and let the miles per gallon of the standard car be c. Then the cost of fuel for the hybrid is:

75,000 miles / h miles per gallon * $9.20/gallon = $690,000/h

And for the non-hybrid:

75,000 miles/c miles per gallon = $690,000/c

The total cost of the hybrid is the purchase cost plus the fuel cost, or

H + $690,000/h = C + $3,800 + $690,000/h

And that for the standard car is:

C + $690,000/c

The story now states that these two are equal:

C + $3,800 + $690,000/h = C + $690,000/c

Or

$3,800 + $690,000/h = $690,000/c

This means we can't solve the problem for we have two unknowns in this equation and can't solve for either unless we know something more about these cars.

So now I leave these two problems as exercises:

1. Find c on the Internet; one way may be to Google for "Honda Accord 2005 mpg" or something like that. Then find h, H, and C.

2. Find h in terms of c and graph the result. What type of curve do you get?

Polyhedra and Conway's Notation

It has been two weeks since SUUSI and it's a long wait until 2006. One of the things I did at this camp or conference was that I gave a workshop on how to weave polyhedra out of various construction paper strips. This led me to think about the polyhedra and reminded me of a neat site for them. This is George Hart's site, at www.georgehart.com in which he has VRML files of many polyhedra models. He even has a page on which you can design and construct your own VRML polyhedra, at

http://www.georgehart.com/virtual-polyhedra/conway_notation.html

This introduces a code by John H. Conway, author of "Numbers and Games" and of numerous neat mathematical things. Conway introduces about 12 operators or so, and 8 objects on which to operate. The operators include a for "ambo", which means take the in-between model (e.g., the cuboctahedron for the cube/octahedron pair) , t for "truncate", j for "join" (make rhombuses out of all the edges) and so forth. I found that there was only one basic object letter (except for prisms, pyramids, and antiprisms) and 6 basic operations. The letter is T for Tetrahedron. The operations are a (ambo), t (truncate), s (snub), d (dual), r (mirror-reverse), and p (propeller - not Conway's invention). I am going to consider only the first 4. I note that Cube, for instance, is daT, and Icosahedron (or I) is actually sT, the snub tetrahedron. Here are some of the other examples:

aaT Cuboctahedron (or aC)
asT Icosadodecahedron (or aD)
taT Truncated octahedron (or tO)
daaaT Trapezoidal Icosatetrahedron (or 24-hedron) (or deC)
ssaT The snub of the snub cube (Hart points out that there are four versions of this)

and so forth. These symbols form a logic of sorts with syntax rules; for example, dd = identity, ad = a, and sd = s. Not all of these can be formed with regular polygons; for example, datsT is Hart's favorite solid, the rhombic enneacontahedron (90-hedron). atsT is its dual, and this turns out to have pentagons, triangles and hexagons. However, these polygons cannot be regular, as atsT (the truncatedicosahedronpentakisdodecahedron or ambo-truncated icosahedron) is not one of the 13 regular Archimedean polyhedra. Presumably Hart has a formula for producing an image of these polyhedra and figuring out where the points on them go, and this might not be easy; for example, the pentakis dodecahedron is made up of somewhat irregular triangles. Where are the midpoints of the faces?

Also, the site does not work right. It is supposed to pop up a VRML window showing the polyhedron defined by the entered Conway notation. Instead, when you enter the notation in the blank, and click "generate", you get a window with VRML code in it. You can copy this by dragscrolling through the document, and putting it in word and making sure each comment is on a line of its own, then saving it as a text document of type .vrml. But one should not have to go through this. George Hart should fix it so you get the image immediately.

But it is one of the best sites on the Internet, in my opinion, and you can enter any of the codes in this blog into that site and after copying to Word and fixing the comments, you can get a VRML page showing the polyhedron whose code you selected.

The Scarlet Algorithm

In writing a short-story comedy about people sticking messages in library books and the like, I created a book about an algorithm or procedure for finding out who your best romantic partner is. I called this fictional book "The Scarlet Algorithm". The idea is that such an algorithm would help people find conjugal partners. This idea is something that has been around a long time. In 1966, my college was introduced to something called Operation Match. The idea is that people going to a certain dance would fill out a questionnaire about themselves and who they want as a partner, and the computer would find for each such person a list of persons of the opposite sex that are the most compatible with them. This did not always turn out right, partly because people put down on paper who they think they want rather than who they want. Nowadays this practice continues in the plethora of dating services on the Internet. To me this is not the best way to meet someone. The best way is through joining clubs or organizations that represent your interests and finding people there.

But if one were going to organize an Operation Match dance, what would be the algorithm? Probably some variant of the Hungarian algorithm, which matches elements of one set with elements of the other to optimize total worth, or of the Gale-Shapley algorithm, which eliminates the chance that "affairs" will form; i.e., two people who prefer each other to the people that they have been assigned to. Recently I used the Hungarian method to pair major league baseball teams with their AAA affiliates, and found that major league baseball could cut major-league-team-AAA-affiliate costs by 40%.

Other mathematical schemes describe love itself. John Alan Lee describes something called the "Colors of Love", in which he describes love as a two-dimensional continuum, specifically the triangle described by x + y + z = 1, with x, y, z >= 0. The three variables in this are "primary colors", which Mr. Lee describes as eros, ludus, and storge. He then describes three secondary colors, with mania between eros and ludus, pragma between ludus and storge, and agape between eros and storge. He then describes tertiary colors, but only for combinations involving mania, including manic storge, which would be a mix of a secondary color with its opposite complement. I therefore conclude that Mr. Lee is really dealing with a four-dimensional model, with four primary colors of eros, ludus, storge, and mania, however, it is hard to picture colors in four dimensions.

Then there are rating schemes. One is the familiar "10" or "perfect 10". This envisions that a man, for example, will rate women on a scale from 1-10, and that what he really wants is a perfect 10. There is even a movie with the name "10", which describes such a desirable woman. First of all, it should be a scale of 0-10. This is the most natural way of rating a scale on two extremes, with 0 at one end and 1 at the other; just multiply this scale by 10.

But another problem is that one will find that there are too many 10s out there. The scale suggests that each number from 0 to 10 will have the same number of people in them; since we meet hundreds of people there will be tens of 10s out there. A better idea is to use a logarithmic scale; see Logarithms Keep Dr. Brown in Perspective". In this scale, a 1 is one out of 10, and so is the counterpart of the "perfect 10". A 2 is one in a hundred, a 3 is one out of a thousand, out of 1,000 with 3 zeroes after the 1. A 6 would have six zeroes after the one, and would be one in a million.

So even though the two subjects of mathematics and romance don't seem to have much in common, there are ways of applying the mathematics to help out with finding a suitable partner. These ways I call "The Scarlet Algorithm".

Mathematical Baseball

This week I have been noticing that of all the team sports, especially those that are broadcast extensively, baseball is the most mathematical. It is a discrete game; that is, the game proceeds by innings, which are composed of outs, which are composed of trips to the plate, which are composed of discrete pitches. It is almost as though a pitch was an "atom" of baseball. Baseball games are composed into box scores, which total the number of items that occurred in the game by categories. For example, you will see the number of runs by each team, the number of hits by each player, the number of walks allowed by each pitcher, and so forth, and from these statistics can be developed, such as standings and batting averages. One can even simulate baseball based on what has happened in previous games.

One of my main interests is what can happen in baseball? What are the logical extremes? For example:

1. What is the fewest number of pitches in a game?

This one is a little tricky. Think about it a while and then come back to this blog.

Did you say 54 pitches? Each player bangs into an out on the very first pitch, 3 pitches a half inning, and so 3 x 2 x 9 = 54 pitches in a game? Almost, but not quite. If the home team is ahead, the bottom of the 9th will not be played, and that makes it 51 pitches. But that's not correct, either. If it were 51 pitches, every one would be an out, and the score would be tied, 0-0. One more pitch is needed somewhere to score a run for the home team, probably a solo home run. So the answer is 52.

2. What is the most number of pitches in a game?

The answer is infinity. This gets into a concept I have invented in games called entropy. In games, some moves are reversible; you can go back to the same situation again. Others are irreversible. In chess, for example, moving a knight to an open square is reversible; you can move him back again. But capturing a piece is irreversible, as the piece can never return, and moving a pawn is irreversible, as pawns can't move backward. And so it is with baseball. An out is irreversible; one can't undo the out and go back from 1 out to none, for instance. A hit, however, is reversible, since a team and hit and hit and hit over again and the situation will still remain the same, especially if the hits are home runs. Conceivably the team could score run after run after run as the New York Yankees did yesterday (against Tampa Bay on 2005 June 21) without ever getting the third out. So the game goes infinite. (Well actually, the Yankees stopped after 13 runs) There are four ways a game can go infinite:

a. Run infinite. A team scores run after run after run in an inning without the third out ever coming.

b. Inning infinite. The game goes into extra innings, one after another, for an infinite amount of time because the two teams score the same number of runs in each inning.

c. Foul ball infinite. A foul ball after two strikes (unless it is a tip or is bunted) does not count anything in baseball - it's reversible. So have a batter hit foul after foul after foul without ever causing an out or hitting and getting safe on first. There is even a book written about this idea, "The Boy who Batted 1.000".

d. Pickoff infinite. The pitcher over and over again attempts to pick off a runner with a throw to that base, and with the runner making it back on time each time. There may be a rule against this one, something about delay of game.

If one assumes that none of these infinites can occur, then a baseball game must end. However, the maximum number of pitches depends on the limits you place on innings, batter-friendly events, foul balls, and pickoff attempts.

3. Suppose a pitcher gets 7 strikeouts in a single inning. Show that a run must have scored.

This one is strange. How can a pitcher get more than 3 strikeouts? 3 outs ends the inning. The answer is that if the catcher can't catch the ball, and there are two outs or first base is open, then the batter who is struck out can run for first base, and if he beats the ball, he is safe at first. The pitcher still gets a strikeout. So 7 strikeouts means 7 people to the plate, and one can fit only 3 of these in outs and 3 on the bases, so at least one run must have scored.

4. How many hits must a team get before a team is certain of scoring a run?

You think it may be something like 10? 15? 22? Try 55. That's right. A team must get 55 or more hits in a game before it becomes certain that a run must have occurred. A team can get 54 hits, and furthermore, 27 of these can be triples, without scoring a single run. Here's how:

A. Hits triple and is out trying to score.
B. Hits triple and is out trying to score.
C. Hits triple.
D. Hits single; 3rd base coach holds runner at 3rd.
E. Hits single; 3rd base coach holds runner at 3rd.

Here is where the problem occurs. If the next runner is out, that ends the inning with only 5 hits. If the next runner hits, with the bases loaded, a run must score. It seems like we need a hit and an out simultaneously and how do we get that? Here's how:

F. Hits ball, batted ball hits a runner. F is credited with a single. The runner is automatically out.

And there it is, 6 hits and no runs. Repeat that over 9 innings and you get 54 hits, with 27 triples, and no runs. We can fire this third base coach.

5. Can a pitcher lose a no-hitter?

Yes he can. A pitcher must win a shutout, because you need runs to win a game. A pitcher must win a perfect game, for you must get on base before you can score. But a pitcher can very well lose a no-hitter. Hits are not necessary for a win. All you need is 3 errors, as happened on 1990 July 1 with a game between the New York Yankees and the Chicago White Sox. The Yankees had 4 hits but 0 runs, and the White Sox had 0 hits but 4 runs. The no-hit White Sox won the game by a substantial margin.

This starts my newest blog, one devoted to mathematics and science, but mostly to mathematics. I called it Hilbert's Hotel after the fabulous hotel that was constructed with an infinite amount of space that will hold a countable infinity of guests, and there is always room for one more, since if John Q. Onemoreguest should arrive, we just tell everybody to move out of their rooms and into the room with the next highest number; e.g., from 107 to 108. Then we can fit John in room 1.

Today I got a question to the math help website I belong to, Mathnerds, concerning cubic equations. These have fascinated me since I was small. There was a formula for quadratic equations, but how about cubics? The best way to understand these is to first of all convert the general cubic

ax³ + bx² + cx + d = 0

into one of the form
x³ + px + q = 0

by dividing everything by a and then by applying the transform y = x - a/3 (and then of course x = y). Let w be an imaginary cube root of 1; i.e., w = -1/2 + sqrt(-3)/2. Compute this product:

(x - u - v)(x - wu - w²v)(x - w²u - wv)

You get:

x³ - 3uvx - (u³+v³)

which is exactly in the form of the reduced cubic equation. Then set

-3uv = p

p³ + q³ = -q

and solve by solving for v in the first equation, and plugging into the second, yielding a quadratic in p³. Solve this using the formula, and select a cube root of the result for u. Then solve for v in the first equation, and then the roots are

u + v
wu + w²v
w³u + wv

This is essentially the method of Lagrange resolvents, and the question in Mathnerds asked about those. Another interesting way of looking at it is through Dan Kalman's circulants.

There is also a discriminant D. If D is 0, the roots are probably rational and two of them are the same. If D > 0, then there is one real root and two imaginary ones, and one can actually come up with a formula for the real root, but it will be somewhat complex looking, being a sum of the cube roots of two quadratic expressions.

The really confusing case is the irreducible one, where D < 0, and my favorite equation here is x3 - 3x + 1 = 0. If you use the formula above, you get w^1/3 + w^2/3 for one of the roots. The thing is, this is real, yet it is expressed in imaginary quantities. I searched all over when I was young for a real solution, until I realized that there are none. There is no way of expressing the roots of this equation in real radicals. If you are willing to use trig functions, there is a way: 2cos(40 degrees) or 2 cos (2*π/9) is one of the roots, for instance.

Cubic equations were originally solved by the Italian mathematicians Tartaglia and Cardano, and Galois added a lot to the theory by associating with equations and field extensions a structure called a Galois group, which for most cubic equations is the group of permutations of three objects, known as S₃. Most of the time when confronted with one, one uses Newton's method to approximate the real roots. But cubic equations solved algebraically have a fascination with me and they will continue to do so.