8/12/2018
Graham Will Drive You Crackers
I was helping Sasi with her homework the other night,
when I spotted something that took me back.
Indeed, it took me back to the beginning of time. Before we are through we’ll head back there,
as well as answering a question Scientific American once claimed was
unanswerable, and solving the greatest cold case of all time. On the way to doing that, let’s first visit
the year 1963, when I was ten and read a book called 1, 2, 3 … Infinity by George Gamow.
Gamow was a physicist who propounded the Big Bang Theory. His book was popular science and mathematics. In the first chapter, from whence his title, he began by telling us that the Hottentots could only count: one, two, three, many. He ended by explaining that not all infinities are created equal, and that there were known to be three, indicated by the subscripts zero through two, so in effect we were back to counting one, two three, many. In between, he gave us what he called the math problem with the largest answer. Sure, he said, one could always create a problem artificially, just by saying “What’s the answer to the largest math problem up to now, plus one?” But his was an attempt to answer a real, if contrived, question. Specifically, if the three monkeys banging away at typewriters reproduced every possible line of type, how many would that be? Assuming no duplication it would be 50^65. Why? If there are 26 letters, 10 digits including 0, the blank space, and some special characters such as the punctuation marks, you have fifty possible things to type into the first space on the line. (I don’t know that there are exactly thirteen special characters, but let’s not be fussy.) There are 65 spaces on the standard line of type, so you have 50 times 50 times 50 … Sixty-five times, or 50^65. That’s equivalent to 10^110, or a 1 with 110 zeros trailing after it.
Gamow then proposed that if we had as many printers as there were subatomic particles in the known universe (as of 1948, so today we know of a much bigger universe and of much smaller particles), each one printing a line in the shortest known interval of time, without duplication, and they started at the time of the Big Bang (in 1948 it was thought to have happened four billion years ago, but today it has been pushed back to more than thirteen), then they would by 1948 have only completed one-thirtieth of one percent of their task. With what we know now about the universe, and subatomic particles, and the dating of the Big Bang, they would be done, in case you’d like to read their output. Even so, it is obvious that his version of the Three Monkey Problem produced a really big number. For years I went around telling friends and family about this big number, until one day I heard about a bigger one.
Isaac Asimov wrote much more than science fiction. He wrote regular columns in popular science journals filled with interesting facts about all sorts of things. In the late seventies I was reading a collection of them. I don’t remember which, so I guess you will have to buy all his books, and read until you find it. He wrote five hundred, but that is so much less than 10^110 that you shouldn’t be discouraged. In this particular book that I was reading he revealed that years before Gamow wrote his book, a South African mathematician named Stanley Skewes had come up with a number that was way bigger than Gamow’s.
Gamow was a physicist who propounded the Big Bang Theory. His book was popular science and mathematics. In the first chapter, from whence his title, he began by telling us that the Hottentots could only count: one, two, three, many. He ended by explaining that not all infinities are created equal, and that there were known to be three, indicated by the subscripts zero through two, so in effect we were back to counting one, two three, many. In between, he gave us what he called the math problem with the largest answer. Sure, he said, one could always create a problem artificially, just by saying “What’s the answer to the largest math problem up to now, plus one?” But his was an attempt to answer a real, if contrived, question. Specifically, if the three monkeys banging away at typewriters reproduced every possible line of type, how many would that be? Assuming no duplication it would be 50^65. Why? If there are 26 letters, 10 digits including 0, the blank space, and some special characters such as the punctuation marks, you have fifty possible things to type into the first space on the line. (I don’t know that there are exactly thirteen special characters, but let’s not be fussy.) There are 65 spaces on the standard line of type, so you have 50 times 50 times 50 … Sixty-five times, or 50^65. That’s equivalent to 10^110, or a 1 with 110 zeros trailing after it.
Gamow then proposed that if we had as many printers as there were subatomic particles in the known universe (as of 1948, so today we know of a much bigger universe and of much smaller particles), each one printing a line in the shortest known interval of time, without duplication, and they started at the time of the Big Bang (in 1948 it was thought to have happened four billion years ago, but today it has been pushed back to more than thirteen), then they would by 1948 have only completed one-thirtieth of one percent of their task. With what we know now about the universe, and subatomic particles, and the dating of the Big Bang, they would be done, in case you’d like to read their output. Even so, it is obvious that his version of the Three Monkey Problem produced a really big number. For years I went around telling friends and family about this big number, until one day I heard about a bigger one.
Isaac Asimov wrote much more than science fiction. He wrote regular columns in popular science journals filled with interesting facts about all sorts of things. In the late seventies I was reading a collection of them. I don’t remember which, so I guess you will have to buy all his books, and read until you find it. He wrote five hundred, but that is so much less than 10^110 that you shouldn’t be discouraged. In this particular book that I was reading he revealed that years before Gamow wrote his book, a South African mathematician named Stanley Skewes had come up with a number that was way bigger than Gamow’s.
Skewes was working on a problem involving prime numbers. There are an infinite number of them, so it isn’t surprising that there is room for a numerous answer. When mathematicians don’t know an answer, they sometimes say that while they don’t know exactly what the answer is, it is between this number, and that number. Skewes Number is an upper bound. As you will see the two boundaries were rather far apart, sort of as though he had said that the contact lens he dropped was somewhere in the Milky Way Galaxy, or maybe one of its neighbors. Skewes Number is 10^10^10^34.
Sasi’s homework reminded me of this, because her
teacher asked how large the class guessed 3^3^3 might be? Would it be 19,683, or
7,625,597,484,987? I think you have
guessed correctly, since we’re talking big numbers. The way you calculate it is from right to
left, so the answer is equal to 3^27, and not 27^3. If you multiply 27*27*27
you get 19,683, but if you multiply 3*3*3*3 … twenty-seven times you do get
that rather large number equivalent to seven and five-eighths trillion.
This gives you a clue as to how immense Skewes Number
is, as 10^34 is already much, much bigger than 7.625 * 10^12, and then you take
10 to that power, and then 10 to that power. It gets worse, because twenty-two years after
his first call, Skewes made things Skewier by mentioning that of course he had
assumed that the Riemann hypothesis was true, but what if it wasn’t? You
remember the Riemann hypothesis? If not,
perhaps you remember the movie A
Beautiful Mind? In it Russell Crowe, playing mathematician John Nash, launches
his career by teaching other mathematicians that the secret to getting laid is
by making passes at every girl in the bar except the best looking one. Then, after marrying the most beautiful girl
himself, he gets to thinking about the Riemann hypothesis, and it drives him
mad. He starts hallucinating Soviet agents and Paul Bettany. Which is why Bettany was chosen to play
Vision in the Avengers movies; he already had experience. This movie taught the audience a valuable
lesson: they should never have let Gomer and Goober teach young Opie math and
psychology during breaks on the set.
Skewes wasn’t driven mad by Riemann, but he was driven to excess, and
his “what if he’s wrong” version of Skewes Number was 10^10^10^1000.
The lower bound was 6. Since then they have narrowed things down. The lower bound has grown by leaps and bounds, or bounds, at least, to 10^14. The upper bound, meanwhile, has shrunk to a miniscule 1.397162914 * 10^316. They have called this close enough for government work, and are letting it rest right there. If you are in the neighborhood, stop by and have a look.
The lower bound was 6. Since then they have narrowed things down. The lower bound has grown by leaps and bounds, or bounds, at least, to 10^14. The upper bound, meanwhile, has shrunk to a miniscule 1.397162914 * 10^316. They have called this close enough for government work, and are letting it rest right there. If you are in the neighborhood, stop by and have a look.
Awhile
back, hard pressed for a backgammon article, I wrote about some improbable
occurrences that actually had occurred.
We shouldn’t be surprised, since every sequence of rolls in backgammon
is improbable. The difference is that these seemed more unlikely than their
peers. For instance if we roll 65, 52,
54, and 63 on our last four turns, it is precisely as improbable as rolling 21
four times in a row, but the latter leads to an unexpected loss instead of a
routine win.
Anyway, the most remarkable sequence of which I’m aware was one where the person’s chance of winning had fallen to around 10^-23. That’s a big number (or a small one, I suppose), but nowhere near the Three Monkeys Problem, or Skewes Number. So I decided to show how small it really was. I did a search to refresh my memory of Skewes Number and learned that around the time I was reading Asimov, a fellow named Graham had come up with a larger number.
Anyway, the most remarkable sequence of which I’m aware was one where the person’s chance of winning had fallen to around 10^-23. That’s a big number (or a small one, I suppose), but nowhere near the Three Monkeys Problem, or Skewes Number. So I decided to show how small it really was. I did a search to refresh my memory of Skewes Number and learned that around the time I was reading Asimov, a fellow named Graham had come up with a larger number.
Before
I go into it, let me digress. There is
another number, the Erdös Number. You
may not know of it, but maybe you have heard of Bacon’s Number? That’s the number of links between any
performer and Kevin Bacon. If you are
Kevin Bacon (lucky you) you get a Bacon Number of 0. If you were in a movie
with Kevin, say you were John Belushi and in Animal House, you were a 1. If you were my mom, you were a 2,
because you were in Blues Brothers (with
John Belushi). This means if you are me, you are a 3, because mom and I both
worked on Ring of Fire. The Erdös Number is similar. Paul Erdös was a mathematician who lived to a
ripe old age despite an amphetamine habit, was very productive, and loved to
collaborate. Most mathematicians write a
couple of papers, and then retire on their fabulous mathematical pensions. It is a young man’s game, and most do
brilliant work when they are young, and then by the time they are thirty are
hiring H & R Bloch to do their tax returns, though I guess if you have a
fabulous mathematical pension you would too.
But not Paul Erdös. He travelled
around the world knocking on the doors of promising young mathematicians. When
the door opened he’d say “Let’s go to work!”
He would move in, write a paper with you, and before you could say: “Hey,
who ate all my Dexedrine?” He would be on to the next co-author. He wrote fifteen hundred papers, fewer than
the Three Monkeys problem, but still impressive. So if you were Paul, you’d get a 0, if you
wrote a paper with him, a 1, if you wrote one with someone who wrote one with
him, a 2. Retired mathematicians spend their time figuring out new ways to get
the number. How young is the youngest
recipient? An infant. Can an animal get
one? Yes, they have. What about a
fictional character? Need you ask?
Now it so happens that I have a friend, Sam Pottle, who is an Erdös 3, even though he is not an animal, and not a fictional character (though I have blurred the line writing about him on occasion). Once I saw where I was going with that backgammon article, I exchanged some emails about Graham’s Number, and then listed him as my co-author. Presto! I became an Erdös 4, and my late friend Walter Trice, thanks to his contributions to Can A Fish Taste Twice As Good? instantly became a 5. Yes, there is such a thing as an Erdös/Bacon number, and unless you are a mathematician over 30 you have guessed mine is 3 + 4 = 7.
Now it so happens that I have a friend, Sam Pottle, who is an Erdös 3, even though he is not an animal, and not a fictional character (though I have blurred the line writing about him on occasion). Once I saw where I was going with that backgammon article, I exchanged some emails about Graham’s Number, and then listed him as my co-author. Presto! I became an Erdös 4, and my late friend Walter Trice, thanks to his contributions to Can A Fish Taste Twice As Good? instantly became a 5. Yes, there is such a thing as an Erdös/Bacon number, and unless you are a mathematician over 30 you have guessed mine is 3 + 4 = 7.
Anyway,
to talk about Graham’s Number it helps to know that 3^3^3 is already a very
large number, because that will get us started.
According
to an article in Scientific American which says, not unreasonably, that Graham’s
Number is so big that she (it’s author) can’t tell you how big it is, around
forty years ago Martin Gardner, a legendary writer of popular math and science
articles, was talking to a mathematician named Ronald Graham, and asked:
“Whacha workin’ on, Ronnie?” Or something along those lines. What he was working on had to do with
Ramsey’s Theory. I have no idea who Ramsey was, nor what his Theory might be.
Here’s the scam. Let’s say you are a
mathematician. You put your feet up on
the desk, look like you are pondering deeply, and then you announce you “have a
theory.” Maybe your theory is that there are exactly forty-seven cube roots which
contain the social security numbers of every member of Congress. The other
mathematicians vote, and if they agree yours is a pretty good theory, it gets
marked down as an “important problem.” People spend years trying to solve it.
Eventually someone either proves it is true, or proves that it isn’t true.
That’s the beauty! It doesn’t matter.
Either way they give you a fabulous pension and you are able to retire
to Bali.
So there is Graham, and he says he is working on Ramsey’s Theory, and he has come up with an upper bound, and Gardner agrees that his upper bound is pretty nifty, and puts it in Scientific American.
Okay, remember 3^3^3? There are other ways to write that. For instance there is Knuth’s Up Arrow Notation. Here is your chance to learn something Knuth. Suppose we were to write 3↑↑3. That’s another way of saying that we want three to the power of three, to the power of three, in other words up twice, or the same as 3^3^3. How about 3↑↑4? That gives us a stack of four, which is 3^7,625,597,484,987. If 3^27 = 7,625,597,484,987 can we begin to imagine what 3 that many times itself is? Actually, we can, sort of.
I think it is around 10^3.8 trillion. I did it in my head, so there was some rounding involved. But it should be close. (Actually, I have since looked it up, and I was off. It is 1.2580143 * 10^3638334640024. I feel better knowing that.) When I was a boy I used to visit the Chicago Public Library. This was not the Harold Washington Library, but the original one with the stone lions outside, like the Art Institute. It was built in 1894, took up half the block at Michigan and Randolph, and was five stories of Greek temple. You can see it in the movie Code of Silence. (You can also see my friend Jeff Hoke inside it, spotted sniffing cocaine by police sergeant Chuck Norris. “Catch you later,” he tells Jeff, as he is hot on the trail of Graham’s Number.) Libraries used to have card catalogs. They were deep drawers holding a few hundred three by five cards, one for each book. There were ten drawers in a stack in each cabinet, with as many stacks as the cabinets were wide. The Des Plaines library had a couple of cabinets, back to back, with hundreds of drawers each holding a few hundred cards. In the Chicago Library there was this long hallway, maybe a hundred yards long. Both sides, as far as the eye could see, lined with card catalog cabinets. There were five million books in that library.
Suppose that in a book, you started out at the top of page one without indenting, and printed “12580143 0000”, and kept on printing zeros. At 65 spaces per line, you would have a number bigger than Gamow’s Three Monkey number when you got to the 103rd zero, which would occur nineteen spaces before the end of the second line of type. If you have 31 lines on a page you will have roughly 2000 digits trailing the “1.” If there are 500 pages in the book you will have about a million zeroes. Let’s say that book is slightly longer than average, but you need more zeroes. You keep writing them, and writing them, and writing them, until you have filled the old Chicago Library’s five million books. That’s how you write out 3^7625597484987. Whatever you do, don’t use commas after every third zero, or you will run out of paper!
If we can stack four threes, can we stack five threes! Sure we can. But now we will have a bit of a problem. The library won’t hold the result, which is 3^ (That thing we just printed). But you know what? There are storage drives these days that are about large enough to hold that number we just wrote. They are only around 10 cm by 5 cm by 1 cm. Why not use those, and try to write 3^3^3^3^3? We will fill up some drives, and I am going to recommend we don’t confine ourselves. We will pave over Planet Earth, and stack those puppies a kilometer high, higher than the Burj Khalifa. No short libraries with wasted space for stairways and walls and bathrooms for us. We need some space. The surface area of the Earth (with the oceans paved over and mountains flattened) is around 510 million km^2. (If you know that the Earth is around 25,000 miles in circumference you may estimate it’s thickness at 8000 miles, divide that in half, square it, multiply by 3 and 1/7 and again by 4, and then by 1.6^2 to convert to kilometers, and you will come pretty close. After you do all of that: just Google it.) We plan to stack one kilometer high, so our cubic kilometers are the same total. We can fit 10,000 drives (long) * 20,000 drives (wide) * 100,000 drives (high) for each cubic kilometer, or 2 * 10^13 drives in all. That number of drives per cubic kilometer times 5 * 10^8 = 10^22 drives covering the Earth, and on each drive we had around 4 * 10^12 zeroes, so we are up to 4 * 10^34.
Clearly we need to do more, but I will take a station break to tell you a story. Many years ago I visited Korea for the first time. Shortly after arrival I asked a waitress how to say “a billion” in Korean. As one does. Before I came I had learned how to say any number up to 99,999,999, but wasn’t sure how to proceed from there. Mia (the waitress) consulted with friends, and the answer they gave me was “you can’t say that.” And I recalled Gamow’s description of the Hottentots and concluded that Koreans were like Hottentots: they counted one, two three, many. But then I met a woman named A Rha, and a billion once again came up in conversation, as it does, and I wrote it out on a napkin. She looked it over, scratched out my commas, and added her own, so it looked like this - “10,0000,0000” - and told me it was “ship ok.” (Excuse my McCune-Reischauer!) You see the system, created aeons earlier by the Chinese, was based upon units of ten thousand, not one thousand. Where we would say: thousand, million, and billion (unless we are British), they would say: ten thousand, one hundred million, one trillion, and ten quadrillion. Or rather they would say: “man, ok, jo, kyong.” However A Rha kept going, and after “kyong” rattled off seven more! The lady counted up to 10^44 without breaking a sweat. I fell instantly in love. As one would. (It didn’t hurt that she was gorgeous.) I have the others written on a napkin in storage – a different napkin – or I would recite them for you. A Rha, by the way, is the only person I have found who knows the ones past kyong. Though with it you can count to nine thousand, nine hundred, and ninety-nine ten quadrillions, or 10^20 – 1, which should see you through most situations. And here is a fun fact: that same kyong, which is pronounced kyo in Japanese or jing in Mandarin also means “capital.” It really puts the “jing” in Beijing!
Where were we? Oh, yes, at 4 * 10^34. Onward! Let’s say that the surface areas of all the planets in the solar system including the sun total around 1.25 * 10^4 times as much as Earth alone, so we grow a scooch to about 5 * 10^38. There are 100 billion stars in the Milky Way, and if they average around what ours does, we could fill the galaxy with 5 * 10^49 zeroes. Heck, not halfway to the Three Monkeys Number; we can’t even stop for lunch.
Let’s rethink it. Okay! The Planck length is the shortest distance we know of, and it’s 1.6 * 10^-35 meters. The universe is 13.8 billion years, old, so its radius is about 46.6 billion light years (so they say, and who am I to doubt them?), and the volume is 4 * 10^80 m^3. There are about 2.2 * 10^107 cubic Plancks per cubic meter, so we will have around 10^187 zeroes, assuming one per cubic Planck. Hmm? What if we take the time it takes light to travel one Planck length, the shortest meaningful unit of time? Light travels 300,000 km per second, so we are talking 3* 10^-9 times 1.6 * 10^-35 or about 5 * 10^-44. (We could call that “five negative A Rha’s.”) What if we wrote all those 10^187 zeroes in the span of five negative A Rha’s, then erased and did it again, counting the total we had done, and we began at the beginning of time, and kept going until now? Isn’t that 10^62 * 10^187? I think so. Shucks, that’s only 10^249. That’s around Three Monkeys squared times one hundred octillion. That’s a big number, but I am sure you see we have been wasting our time, because it is nowhere near 3^3^3^3^3.
So there is Graham, and he says he is working on Ramsey’s Theory, and he has come up with an upper bound, and Gardner agrees that his upper bound is pretty nifty, and puts it in Scientific American.
Okay, remember 3^3^3? There are other ways to write that. For instance there is Knuth’s Up Arrow Notation. Here is your chance to learn something Knuth. Suppose we were to write 3↑↑3. That’s another way of saying that we want three to the power of three, to the power of three, in other words up twice, or the same as 3^3^3. How about 3↑↑4? That gives us a stack of four, which is 3^7,625,597,484,987. If 3^27 = 7,625,597,484,987 can we begin to imagine what 3 that many times itself is? Actually, we can, sort of.
I think it is around 10^3.8 trillion. I did it in my head, so there was some rounding involved. But it should be close. (Actually, I have since looked it up, and I was off. It is 1.2580143 * 10^3638334640024. I feel better knowing that.) When I was a boy I used to visit the Chicago Public Library. This was not the Harold Washington Library, but the original one with the stone lions outside, like the Art Institute. It was built in 1894, took up half the block at Michigan and Randolph, and was five stories of Greek temple. You can see it in the movie Code of Silence. (You can also see my friend Jeff Hoke inside it, spotted sniffing cocaine by police sergeant Chuck Norris. “Catch you later,” he tells Jeff, as he is hot on the trail of Graham’s Number.) Libraries used to have card catalogs. They were deep drawers holding a few hundred three by five cards, one for each book. There were ten drawers in a stack in each cabinet, with as many stacks as the cabinets were wide. The Des Plaines library had a couple of cabinets, back to back, with hundreds of drawers each holding a few hundred cards. In the Chicago Library there was this long hallway, maybe a hundred yards long. Both sides, as far as the eye could see, lined with card catalog cabinets. There were five million books in that library.
Suppose that in a book, you started out at the top of page one without indenting, and printed “12580143 0000”, and kept on printing zeros. At 65 spaces per line, you would have a number bigger than Gamow’s Three Monkey number when you got to the 103rd zero, which would occur nineteen spaces before the end of the second line of type. If you have 31 lines on a page you will have roughly 2000 digits trailing the “1.” If there are 500 pages in the book you will have about a million zeroes. Let’s say that book is slightly longer than average, but you need more zeroes. You keep writing them, and writing them, and writing them, until you have filled the old Chicago Library’s five million books. That’s how you write out 3^7625597484987. Whatever you do, don’t use commas after every third zero, or you will run out of paper!
If we can stack four threes, can we stack five threes! Sure we can. But now we will have a bit of a problem. The library won’t hold the result, which is 3^ (That thing we just printed). But you know what? There are storage drives these days that are about large enough to hold that number we just wrote. They are only around 10 cm by 5 cm by 1 cm. Why not use those, and try to write 3^3^3^3^3? We will fill up some drives, and I am going to recommend we don’t confine ourselves. We will pave over Planet Earth, and stack those puppies a kilometer high, higher than the Burj Khalifa. No short libraries with wasted space for stairways and walls and bathrooms for us. We need some space. The surface area of the Earth (with the oceans paved over and mountains flattened) is around 510 million km^2. (If you know that the Earth is around 25,000 miles in circumference you may estimate it’s thickness at 8000 miles, divide that in half, square it, multiply by 3 and 1/7 and again by 4, and then by 1.6^2 to convert to kilometers, and you will come pretty close. After you do all of that: just Google it.) We plan to stack one kilometer high, so our cubic kilometers are the same total. We can fit 10,000 drives (long) * 20,000 drives (wide) * 100,000 drives (high) for each cubic kilometer, or 2 * 10^13 drives in all. That number of drives per cubic kilometer times 5 * 10^8 = 10^22 drives covering the Earth, and on each drive we had around 4 * 10^12 zeroes, so we are up to 4 * 10^34.
Clearly we need to do more, but I will take a station break to tell you a story. Many years ago I visited Korea for the first time. Shortly after arrival I asked a waitress how to say “a billion” in Korean. As one does. Before I came I had learned how to say any number up to 99,999,999, but wasn’t sure how to proceed from there. Mia (the waitress) consulted with friends, and the answer they gave me was “you can’t say that.” And I recalled Gamow’s description of the Hottentots and concluded that Koreans were like Hottentots: they counted one, two three, many. But then I met a woman named A Rha, and a billion once again came up in conversation, as it does, and I wrote it out on a napkin. She looked it over, scratched out my commas, and added her own, so it looked like this - “10,0000,0000” - and told me it was “ship ok.” (Excuse my McCune-Reischauer!) You see the system, created aeons earlier by the Chinese, was based upon units of ten thousand, not one thousand. Where we would say: thousand, million, and billion (unless we are British), they would say: ten thousand, one hundred million, one trillion, and ten quadrillion. Or rather they would say: “man, ok, jo, kyong.” However A Rha kept going, and after “kyong” rattled off seven more! The lady counted up to 10^44 without breaking a sweat. I fell instantly in love. As one would. (It didn’t hurt that she was gorgeous.) I have the others written on a napkin in storage – a different napkin – or I would recite them for you. A Rha, by the way, is the only person I have found who knows the ones past kyong. Though with it you can count to nine thousand, nine hundred, and ninety-nine ten quadrillions, or 10^20 – 1, which should see you through most situations. And here is a fun fact: that same kyong, which is pronounced kyo in Japanese or jing in Mandarin also means “capital.” It really puts the “jing” in Beijing!
Where were we? Oh, yes, at 4 * 10^34. Onward! Let’s say that the surface areas of all the planets in the solar system including the sun total around 1.25 * 10^4 times as much as Earth alone, so we grow a scooch to about 5 * 10^38. There are 100 billion stars in the Milky Way, and if they average around what ours does, we could fill the galaxy with 5 * 10^49 zeroes. Heck, not halfway to the Three Monkeys Number; we can’t even stop for lunch.
Let’s rethink it. Okay! The Planck length is the shortest distance we know of, and it’s 1.6 * 10^-35 meters. The universe is 13.8 billion years, old, so its radius is about 46.6 billion light years (so they say, and who am I to doubt them?), and the volume is 4 * 10^80 m^3. There are about 2.2 * 10^107 cubic Plancks per cubic meter, so we will have around 10^187 zeroes, assuming one per cubic Planck. Hmm? What if we take the time it takes light to travel one Planck length, the shortest meaningful unit of time? Light travels 300,000 km per second, so we are talking 3* 10^-9 times 1.6 * 10^-35 or about 5 * 10^-44. (We could call that “five negative A Rha’s.”) What if we wrote all those 10^187 zeroes in the span of five negative A Rha’s, then erased and did it again, counting the total we had done, and we began at the beginning of time, and kept going until now? Isn’t that 10^62 * 10^187? I think so. Shucks, that’s only 10^249. That’s around Three Monkeys squared times one hundred octillion. That’s a big number, but I am sure you see we have been wasting our time, because it is nowhere near 3^3^3^3^3.
Anyway, the number we need to meet is 3↑↑↑↑3. Hi
there, Cutie Pie! Now things get hairy.
I spent a lot of time reading up on Knuth’s notation, and other
notations. Not anything by the man himself, I used sort of the Cliff’s Knuth’s
version. I nearly tied myself in knots,
or Knuth’s, trying to sort it all out.
There is discussion of what it means to have one, two, three, many
arrows. And what it means to have a 2, or 3, or 4, etc. after the last
arrow. Or about you could write 39↑↑14,
and that is the same as 39[4]14, and not the same as 39[2]14. Or how about 77
? That’s of course 77[79]77 and not 77[77]77. Of course it is. That’s before we get into power towers,
hyperoperators, stacked power towers, or Conway’s chained arrow notation, which
features arrows that got dizzy trying to figure out what was going on, and fell
over on their sides.
? That’s of course 77[79]77 and not 77[77]77. Of course it is. That’s before we get into power towers,
hyperoperators, stacked power towers, or Conway’s chained arrow notation, which
features arrows that got dizzy trying to figure out what was going on, and fell
over on their sides.
Last time around when writing the backgammon article I
got it wrong. Will I get it right this time? Probably not, but I think I have
gotten closer. Let’s review a couple of
points. If I write 
I have used a superscript to
write 3^3. But using a conventional word
processor like Word, I have no easy way to write 3^3^3 with superscripting,
because the threes would climb, with the third one as superscript to the
second. And 3^3^3^3 would form a little tower, poking a hole in the line above. Here is a second thing to remember. We can write 7,625,597,484,987 using thirteen
digits, but that is just a way of writing the number. To actually write that many digits, we would
fill the volumes in two Chicago libraries.
Okay, I hope you don’t have vertigo, because we are going to climb! We determined that 3^3^3^3 was equal to a number with around 3.6 trillion digits, a number that would fill a Chicago library with zeroes. And that is also the number you would get from 3↑↑↑↑2. And we further determined that 3↑↑↑↑↑2 , or five threes stacked into a tower of superscripting, was so big we could not even figure out how to write it out using the entirety of the known universe, let along how many digits or “things” it actually represented. Now for 3↑↑↑↑3 you first work out that a stack of four 3s is equal to our old friend 1.2580143 * 10^3638334640024. Then you build a tower of 3s with that many actual 3s stacked up. No matter how small a font you use that tower will extend up and to the right leaving the page, leaving the building, leaving the atmosphere; it will pierce the fabric at the edge of our universe, and skewer the universe in the multiverse not above us, but up and to the right. It will go on all the way through it, and on to the next and the next. It will stab so many universes that the number will seem to us rivaling infinity. Eventually all things must end, and when it has strung all its beads it will tip over under the weight of all those universes, and scythe down through the universes below. Below us it will swing back and forth like a pendulum, crushing or scattering like billiards on a break shot the universes in its path. Then its weight will cause it to fall like a great icicle from a skyscraper and it will drive down, down, down through the infinite universes below us, falling forever. Unless the physicists have something to say about that. But this is a math problem, so they don’t have an invitation to the party.
I should have warned you that once we got to a stack of five 3s you risk radiation burns. Given our present journey I hope you have donned a hazmat suit and protective goggles.
For our next step, if you are suitably garbed, we need to climb the tower of 3s, and start working back towards the base. If five is beyond our ability to conceive, or even write, then what about the sixth, the seventh and eventually the 1.2580143 * 10^3638334640024th? A radiation warning isn’t sufficient. Move back from your monitor, because being in the presence of that number has been known to induce labor. It has been known to induce labor even in women who are not pregnant. It has even been known to induce labor in men who are not pregnant.
Given the cataclysmic effects of the tipping tower, you might think that the number’s name is Shiva, Destroyer of Worlds. No, it has a name, and its name is G1. Yes, G, as in Graham. So that’s Graham’s Number, you say. No, it isn’t.
Once we have worked out the value of G1, then we will need Jake’s Back-To-Back Bow notation. Someone has probably already claimed this method of writing things, but if you bump into them don’t tell them I have slapped my brand on it. Notice how {} curly brackets resemble the bows with which you shoot arrows? Let’s write 3{4}3, and agree that it is the same as 3↑↑↑↑3, which we know is equal to G1. Therefore 3{G1}3 is equal to those threes with a G1 number of up arrows in between. You saw what happened with just four arrows, so try to imagine what happens with G1 arrows. You would have to take 3 to a stack of 3s G1 high, and then after figuring out what that was, build a new stack that high. And that would equal G2. Then you do it again, and do it again, and do it again until you reach G64. That is Graham’s Number. I’m reminded of the legend of the wise man that invented chess, and asked the Shah for a grain of wheat for the first square, two for the second, four for the third, etc. The Shah learned that before he reached the 64th square, long before, there wasn’t enough wheat in the world. You could fill the board with 2^64 -1 grains if there were that many. Even Deep Blue would be lost filling the board with Graham’s.
Earlier I promised three things. Visiting the beginning of time has been checked off, and now I will simultaneously explain just how large Graham’s number really is, and solve the greatest of cold cases. A hundred years or so ago a stringer for the wire services named Freddy Nietzsche broke the news that God was dead. The cause of death was undetermined, and the case has remained open. Until now. Which God are we talking about? The big boy; the one who created the heavens and the Earth in six days and needed only one day off to rest. The one who in his avatar Vishnu was able to preserve the Universe in a single bead of sweat on his forehead. That’s either from Hindu theology, or Roger Zelazny. At the end of the 19th century, looking ahead, something he was wont to do, God spotted Gardner and Graham talking, read their lips, and thought: “Oho, I’ll figure out Graham’s Number!” So he tried, and it killed him. That’s how big Graham’s Number is.
I am anticipating you! Jake, you are about to object, if Graham did this forty years ago, mightn’t there be …? There might. I saw mention of a number equal to G1000. Then there is TREE(3). Some fiend named Kruskal invented a “tree theorem.” The things these guys do to get those fabulous math pensions! One article online suggested watching twenty-one half hour videos to learn how it works. If he was talking to me, he was barking up the wrong tree. Anyway, the tree grows kind of fast. TREE(1) = 1. Okay, good start. TREE(2) = 3. And then TREE(3) = ? Apparently Graham’s number is “approximately”
(4), while an “extremely weak lower bound” for TREE(3) is 
(1). Those A’s are “Ackermann’s
Function in hyperopration.” In other words: even Hester Prynne has no clue what
they are. To those who trim TREE(3) for
a living, if you bring up Graham’s Number, they just snicker and go back to
watching reruns of Big Bang Theory. What I know is that there are power lines and
gas mains underground in this neighborhood, so I am not planting trees like
that in my yard. SP Services would be on me so fast …
One more thing to wrap it up. Having worked with Sasi on her math homework, is she now an Erdös 5? She is older than an infant, so she is old enough. She is both a real person, and a fictional character. (Those who have read my novella One Night at the Aquarium, in which Sasi and her friend Niamh use a bag of borrowed Doritos to thwart the attack of the Colossal Crab-Biters have surely realized it was not strictly speaking non-fiction.) Given all that what’s the hitch? It’s the question as to whether someone who got less than a B in 7th grade math can receive an Erdös Number? The answer turns out to be “yes,” it has happened before. I won’t say how I know, but take my word for it, I do.
I have used a superscript to
write 3^3. But using a conventional word
processor like Word, I have no easy way to write 3^3^3 with superscripting,
because the threes would climb, with the third one as superscript to the
second. And 3^3^3^3 would form a little tower, poking a hole in the line above. Here is a second thing to remember. We can write 7,625,597,484,987 using thirteen
digits, but that is just a way of writing the number. To actually write that many digits, we would
fill the volumes in two Chicago libraries. Okay, I hope you don’t have vertigo, because we are going to climb! We determined that 3^3^3^3 was equal to a number with around 3.6 trillion digits, a number that would fill a Chicago library with zeroes. And that is also the number you would get from 3↑↑↑↑2. And we further determined that 3↑↑↑↑↑2 , or five threes stacked into a tower of superscripting, was so big we could not even figure out how to write it out using the entirety of the known universe, let along how many digits or “things” it actually represented. Now for 3↑↑↑↑3 you first work out that a stack of four 3s is equal to our old friend 1.2580143 * 10^3638334640024. Then you build a tower of 3s with that many actual 3s stacked up. No matter how small a font you use that tower will extend up and to the right leaving the page, leaving the building, leaving the atmosphere; it will pierce the fabric at the edge of our universe, and skewer the universe in the multiverse not above us, but up and to the right. It will go on all the way through it, and on to the next and the next. It will stab so many universes that the number will seem to us rivaling infinity. Eventually all things must end, and when it has strung all its beads it will tip over under the weight of all those universes, and scythe down through the universes below. Below us it will swing back and forth like a pendulum, crushing or scattering like billiards on a break shot the universes in its path. Then its weight will cause it to fall like a great icicle from a skyscraper and it will drive down, down, down through the infinite universes below us, falling forever. Unless the physicists have something to say about that. But this is a math problem, so they don’t have an invitation to the party.
I should have warned you that once we got to a stack of five 3s you risk radiation burns. Given our present journey I hope you have donned a hazmat suit and protective goggles.
For our next step, if you are suitably garbed, we need to climb the tower of 3s, and start working back towards the base. If five is beyond our ability to conceive, or even write, then what about the sixth, the seventh and eventually the 1.2580143 * 10^3638334640024th? A radiation warning isn’t sufficient. Move back from your monitor, because being in the presence of that number has been known to induce labor. It has been known to induce labor even in women who are not pregnant. It has even been known to induce labor in men who are not pregnant.
Given the cataclysmic effects of the tipping tower, you might think that the number’s name is Shiva, Destroyer of Worlds. No, it has a name, and its name is G1. Yes, G, as in Graham. So that’s Graham’s Number, you say. No, it isn’t.
Once we have worked out the value of G1, then we will need Jake’s Back-To-Back Bow notation. Someone has probably already claimed this method of writing things, but if you bump into them don’t tell them I have slapped my brand on it. Notice how {} curly brackets resemble the bows with which you shoot arrows? Let’s write 3{4}3, and agree that it is the same as 3↑↑↑↑3, which we know is equal to G1. Therefore 3{G1}3 is equal to those threes with a G1 number of up arrows in between. You saw what happened with just four arrows, so try to imagine what happens with G1 arrows. You would have to take 3 to a stack of 3s G1 high, and then after figuring out what that was, build a new stack that high. And that would equal G2. Then you do it again, and do it again, and do it again until you reach G64. That is Graham’s Number. I’m reminded of the legend of the wise man that invented chess, and asked the Shah for a grain of wheat for the first square, two for the second, four for the third, etc. The Shah learned that before he reached the 64th square, long before, there wasn’t enough wheat in the world. You could fill the board with 2^64 -1 grains if there were that many. Even Deep Blue would be lost filling the board with Graham’s.
Earlier I promised three things. Visiting the beginning of time has been checked off, and now I will simultaneously explain just how large Graham’s number really is, and solve the greatest of cold cases. A hundred years or so ago a stringer for the wire services named Freddy Nietzsche broke the news that God was dead. The cause of death was undetermined, and the case has remained open. Until now. Which God are we talking about? The big boy; the one who created the heavens and the Earth in six days and needed only one day off to rest. The one who in his avatar Vishnu was able to preserve the Universe in a single bead of sweat on his forehead. That’s either from Hindu theology, or Roger Zelazny. At the end of the 19th century, looking ahead, something he was wont to do, God spotted Gardner and Graham talking, read their lips, and thought: “Oho, I’ll figure out Graham’s Number!” So he tried, and it killed him. That’s how big Graham’s Number is.
I am anticipating you! Jake, you are about to object, if Graham did this forty years ago, mightn’t there be …? There might. I saw mention of a number equal to G1000. Then there is TREE(3). Some fiend named Kruskal invented a “tree theorem.” The things these guys do to get those fabulous math pensions! One article online suggested watching twenty-one half hour videos to learn how it works. If he was talking to me, he was barking up the wrong tree. Anyway, the tree grows kind of fast. TREE(1) = 1. Okay, good start. TREE(2) = 3. And then TREE(3) = ? Apparently Graham’s number is “approximately”
(4), while an “extremely weak lower bound” for TREE(3) is (1). Those A’s are “Ackermann’s
Function in hyperopration.” In other words: even Hester Prynne has no clue what
they are. To those who trim TREE(3) for
a living, if you bring up Graham’s Number, they just snicker and go back to
watching reruns of Big Bang Theory. What I know is that there are power lines and
gas mains underground in this neighborhood, so I am not planting trees like
that in my yard. SP Services would be on me so fast …One more thing to wrap it up. Having worked with Sasi on her math homework, is she now an Erdös 5? She is older than an infant, so she is old enough. She is both a real person, and a fictional character. (Those who have read my novella One Night at the Aquarium, in which Sasi and her friend Niamh use a bag of borrowed Doritos to thwart the attack of the Colossal Crab-Biters have surely realized it was not strictly speaking non-fiction.) Given all that what’s the hitch? It’s the question as to whether someone who got less than a B in 7th grade math can receive an Erdös Number? The answer turns out to be “yes,” it has happened before. I won’t say how I know, but take my word for it, I do.
