In the Shadow of Tall Numbers

The Universe is too Small to Store some Numbers that Lie in Plain Sight

Numbers, impossibly large on the human scale can be fascinating. I accidentally stumbled into such numbers while thinking about probabilities. It was unexpected and mixed with a little tingling of awe. I decided to share my venture, maybe it will amuse others, as well. Although there is a little math involved, it is easy to understand. If you ever thought about your chances of winning the lottery, you have the needed background.

My so-called stumbling into large numbers came about while reading the science fiction novel CONTACT, written by Carl Sagan. Carl Sagan was an astrophysicist, a scientist through and through. He was also a public figure, a great advocate for science. He created and presented on TV the original Cosmos series. For many, this captivating and inspiring series was their first encounter with our Universe from the perspective of science. He was also active in the skeptical movement confronting all matters of pseudoscience. For all these reasons I was quite surprised that he sprinkled CONTACT with a small dose of mysticism. What follows is the account of my encounter with the spiritual ending of Carl Sagan’s science fiction story.

This novel was also the basis of a very successful movie, where Jodie Foster played the main character: Dr. Eleanor Arroway, or Ellie. The film only broadly follows the book. There is nothing in the movie from the last chapter titled “The Artist’s Signature”. In it, Ellie instructs her computer to look for some “anomalies” in the digits of Pi, the ratio of a circle’s circumference to its diameter. (Pi=3.141592…) Pi is a so-called transcendental number, it has an infinite number of digits and those digits statistically behave like a sequence of random numbers. But! Lo and behold, Ellie’s computer does find something interesting: “The anomaly showed up most starkly in Base 11 arithmetic, where it could be written out entirely as zeros and ones.” … “The program reassembled the digits into a square raster, an equal number across and down. The first line was an uninterrupted file of zeros, left to right. The second line showed a single numeral one, exactly in the middle, with zeros to the borders, left and right. After a few more lines, an unmistakable arc had formed, composed of ones. The simple geometrical figure had been quickly constructed, line by line, self-reflexive, rich with promise. The last line of the figure emerged, all zeros except for a single centered one. The subsequent line would be zeros only, part of the frame. Hiding in the alternating patterns of digits, deep inside the transcendental number, was a perfect circle…”

Ellie has no choice but to conclude: “The universe was made on purpose, the circle said. In whatever galaxy you happen to find yourself you take the circumference of a circle, divide it by its diameter, measure closely enough, and uncover a miracle — another circle, drawn kilometers downstream of the decimal point.” … “There is a preexisting intelligence in the universe..”

Anybody who watched Cosmos, first broadcasted in 1980, could feel the marvel and awe that Carl Sagan felt for the UNIVERSE. In CONTACT he presented his readers — I’d guess with a wink and a nod — with this hidden intelligence of his beloved Universe. I began to ponder at this point: what if once we indeed find such a circle in the digits of Pi? Did we receive a message from the Universe, or among the many digits of Pi we happened to bump into the right sequence simply by chance? The book says the circle was “kilometers downstream of the decimal point”. Well, there are plenty of digits in kilometers or miles. Is there a reasonable chance of finding the string of our circle within those miles of random digits without the intervention of the Universe? Such a sequence must be somewhere in Pi since Pi has infinitely many digits. In an infinite random series, any pre-specified string of any finite length will appear; actually, it will appear infinitely many times.

Well, let’s figure out the likelihood of running into a circle, depicted by the symbols 0s and 1s within a random sequence of digits. First, what sort of a circle are we dealing with? The book said it was framed by 0s on all sides, and that it was an easily recognizable pretty circle: “After a few more lines, an unmistakable arc had formed, composed of ones.” Unless a square grid is quite fine, marks on it don’t lend themselves to define good-looking circles. I tried a few and concluded that a frame of 31 0s per side was adequate to yield at least a decent looking circle, even if not a nice smooth one. The picture below shows the first couple of lines of such a circle.

A circle displayed in a square pattern of digits is one particular series of 0s and 1s when we expand the digits of the square into a string. In this case, a 31*31=961 digits long string. Looking at the picture, the string’s beginning would be as follows: 46 0s, (31 in the first line that is the frame, plus 15 from the second line) one 1, 26 0s, one 1, 7 0s, one 1, 20 0s, one 1, …. and so on. The string would end mirroring the beginning: … one 1, 26 0s, one 1, 46 0s. This is the string we have to look for in the digits of Pi. What are the chances of finding one? Or, a more precise question, to how many digits — at least approximately — would we have to calculate Pi to have a reasonable probability, let’s say 0.5, (or 50%, chance), of finding our string at least once among those calculated digits?

The novel has one last instruction for us: “The anomaly showed up most starkly in Base 11 arithmetic…”. Sure, why not? One can express Pi’s digits in any Base, but in each particular Base they would be a different sequence of symbols. Ellie found the circle when she expressed them in Base 11, so we’ll do the same in our estimations. Maybe this is the right time to say that we are only doing estimations: aiming for the right exponents, and at most 2 digits of precision. If one is further interested, there are plenty of excellent textbooks on probability full of detailed answers and precision. As for notation, a “*” means to multiply, while “**” is exponentiation.

In Base 11 arithmetic one works with 11 symbols, the same way that in our everyday Base 10 we have 10 symbols: 0, 1, 2, … 8, 9. What matters for our purpose is that in Base 11 any of Pi’s digits have an equal chance to turn out to be any one of the 11 possible symbols. So let’s start… First question: what is the probability that — in Base 11 — our 961 digits long string would appear after one specific digit, let’s say the 100th digit? Well, the 101st digit would have to be a 0. The probability of that is 1/11; the 102nd again would have to be 0, the probability of that is also 1/11. The probability that the 101st and the 102nd digits are both 0s is (1/11)**2. And so on… The probability that the digits from 101 to 146 are all zeros is (1/11)**46. And so on… Finding our exact string following the 100th digit has a probability of (1/11)**961. The 100th digit was picked as an example, there is nothing special about it. Finding our string after any particular digit, be that 100, 1000, 123456, 722! + 3, or any other, has always the same probability: (1/11)**961. This fact allows us to think of finding our string as an event. The probability of this event occurring is (1/11)**961, which is approximately (1/10)**1001. It is 0.0 followed by another 999 zeros and then a 1!

In human terms, this is beyond small. Compare it to the probability of winning the jackpot in one of the national lotteries, typically in the range of a few times (1/10)**9. A jackpot is also an event and upon many trials the event happens and the jackpot is won. In the lottery’s case, each (non-duplicate) ticket is a trial. All it takes is a few hundred million tickets and the jackpot is conquered. Same with our string. Here, each successive digit of Pi is a new trial. Each digit might turn out to be the “magic” one, the first in our sought-after string. Except that finding the string is an event that has an incomparably smaller chance than winning the lottery. On the other hand, although our chances may be the tiniest we do have many trials at our disposal. We only have to keep churning out the digits of Pi, until bingo! (A digit that follows a partial match is not the start of a new trial but a check on whether the match continues. Such digits are few in Base 11, and do not change anything regarding the probabilities discussed.)

Ellie found the string after a few miles worth of digits. How far can we expect to go on calculating Pi’s digits in Base 11 arithmetic to have something like a 50% chance that our sting will be somewhere in there? The number of trials needed for a random event to happen at least once with a 50% chance is about 0.7 divided by the probability of the event. In our case, the 0.7 is irrelevant the emphasis is on “divided by the probability”. Since our probability is ~ (1/10)**1001 we have to calculate out to 10**1001 digits. Well, here we are… we ran into a really big number! 10**1001 is not only beyond a few miles worth of digits, not only beyond any human scale, but it is even way beyond anything relating to our Universe.

Imagine we find the “magic” digit somewhere past digit 10**1000, which is where 90% of the calculated digits are. We want to record the digits that we had found so that we never have to calculate them again. Suppose, we have an ingenious method that can store 10**18 digits in a cubic millimeter (mm) of volume. Way beyond anything we’d ever be able to do in reality. A characteristic measure of the observable Universe is about 25 billion light-years, which gives us a volume of ~ 10**88 mm**3. So if we stuff every cubic mm of our observable Universe with 10**18 of the calculated digits we get to store ~ 10**106 of them. If we are truly ambitious we corral the part of our Universe that we cannot observe. (Due to the expansion of the Universe light cannot reach us from there.) This unobservable part may be good for at most another factor of about 100, thus giving us ~ 10**108 storage places. Well, any way of looking at it we are short of space to store our number by a mere 10**893 Universes!

What chance did Ellie’s computer have to find the string if there was no intervention by the Universe itself? You guessed it, absolutely none. Suppose Ellie’s computer was special and able to calculate 10**40 digits per second. This is more than generous: humanity will never have computers of such capabilities, even in that extremely unlikely case that the much-ballyhooed quantum computers were to come to fruition. But, let’s say Ellie did have such a computer, and this computer calculated continuously starting from the moment of the Big Bang. That was a little bit less than 5*10**17 seconds ago, giving Ellie at most a puny 5*10**57 digits by last Tuesday. She would have needed at least 10**943 of her special computers to do the job.

Such estimates are the ones I referred to in the beginning as capable of filling one with awe. Numbers like these cast a shadow. We really did not ask for much, finding a specific string among random numbers. And look what it got us into: needing 10**893 additional Universes!

But, let’s get off our high horse. What if Ellie’s circle was barely distinguishable from a square and had no frame of 0s? If she had imagination, and in the story she certainly had, she could have gleaned a circle in a 7*7 square of properly placed 0s and 1s. Then, all it would take is to calculate about 11**49 digits which in powers of 10 is about 10**51. A much less daunting task than finding the earlier better-looking circle in a 31*31 frame. Not that we could ever calculate 10**51 digits, but if we did, at least we could store this number within the confines of our Solar system.

So yes, if Ellie’s computer came up with any kind of recognizable circle of 0s and 1s in Base 11 arithmetic it definitely was a message from the UNIVERSE. Professor Sagan did allow a little play with wonder. Kudos for giving such an imaginative twist to his book. One is left to reflect, however, why did he choose Base 11 arithmetic, rather than some other Base, let’s say binary, or our homely Base 10? I don’t know if he ever commented on this, but it could easily have been a deliberate selection. CONTACT was published in 1985. Superstring theories came to the fore of theoretical physics just in that period. According to superstring theories, the universe is 10 or 11 dimensional. Wouldn’t it make sense that an 11 dimensional UNIVERSE communicates in Base 11 arithmetic?

Finding some shape hidden as a string in a series of random numbers becomes ever less probable the larger the Base is in the arithmetic we are using. The probability of finding any given symbol at any given place in Base N arithmetic is 1/N because N number of symbols are competing for the same spot. Could Ellie’s computer have found a circle or other meaningful shape in the binary expression Pi? Or, let’s just forget about Pi, and think of any random sequence of 0s and 1s. Well, finding a 961 digits long sequence would be just as much a pie in the sky endeavor in Base 2 as it was in Base 11. But the ugly circle of 49 digits is a very different proposition. In Base 2 we would need a sequence of only about 2**49 digits. 2**49 is less than 10**15. Churning out a sequence of 10**15 random 0s and 1s and checking if our predefined 49 digit sequence is among them would make a quick job for a present-day supercomputer.

Why, of course, would an expensive supercomputer bother with such matters. But, one doesn’t need a supercomputer to play with smaller images. For instance, with a little goodwill one can discern a human figure in the 3 by 6 rectangle shown here.

It contains a sequence of only 18 binaries. Finding such an 18 digits long sequence in Pi’s binary expression, or just as well, in any random sequence of 0s and 1s is getting us in the range of PC power. Suppose we are looking for a fixed binary string of length L in a sequence of random 0s and 1s. For our human figure, the string would be 010111010010101101 and L=18. In general, how long would a random binary sequence have to be to have a 0.5 probability to contain at least one case of our predefined L long string? Following the same reasoning as with the circles, such a sequence would have to contain about 1.4*10**(0.3*L) digits. The 1.4 comes from the 0.7, which as mentioned above, connects the number of trials with the probability of the event occurring at least once with a 50% chance, and a factor of approximately 2 from the fact that not every member of the sequence is a new trial. There are string segments that start just like the sought one but do not reach the required length. For example, in the case of our human figure, a string like 0100 is a good start for 3 digits but then it deviates, and while it is only a single failed trial it had the cost of 2 discarded digits. The 0.3*L in the exponent simply serves to get away from powers of 2. All said, for L=18 if we generate 350 thousand random 0s and 1s we have about a 50% chance of finding our human figure in there at least once. If we go twice as far and generate 700 thousand digits we have a 75% chance, or if we fall short and manage only 175 thousand we still have about a 30% chance to find at least one of our gal/guy figure.

Obviously, there is no difference in looking for a string that can be shaped into something recognizable, or for any other of the same length. Suppose we are looking for all 1s. In a sequence of about 5600 random 0s and 1s, we have a 50% chance of finding 12 of them in a row. If we want to find 15 of them in a row we need a sequence 2**3 = 8 times as long, which is almost 45000 digits. (We don’t worry about rounding errors.) One can easily spend an afternoon in front of a PC playing with such things and observe that finding various figures in random sequences is not a miracle, but pretty soon can turn into one if we are too ambitious with the length of our strings.

In the meantime, Pi smiles… I am not fazed by your 961 digits long predefined string any more than by a single digit. I have room to spare for them all. I have an infinite number of digits!