Steven Landsburg | The Big Questions: Tackling the Problems of Philosophy with Ideas from Mathematics, Economics, and Physics » Puzzles

The Big Surprise

Steve Landsburg — Thu, 15 Dec 2011 07:01:29 +0000

Back in the 1930’s, Kurt Godel proved two amazing facts about arithmetic: First, there are true statements in arithmetic that can’t be proven. Second, the consistency of arithmetic can’t be proven (at least not without recourse to logical methods that are on shakier ground than arithmetic itself).

Yesterday, I showed you Gregory Chaitin’s remarkably simple proof, of Godel’s first theorem. Today, I’ll show you Shira Kritchman and Ron Raz’s remarkably simple (and very recent) proof of Godel’s second theorem. If you work through this argument, you will, I think, have no trouble seeing how it was inspired by the paradox of the surprise examination.

Start with this list of statements:

It takes more than 10000 characters to specify the number 1.
It takes more than 10000 characters to specify the number 2.
It takes more than 10000 characters to specify the number 3.

and so forth.

Obviously, statements 1, 2 and 3 are all false, since it only takes a single character (namely “1″) to specify the number 1, and a different single character (namely “2″) to specify the number 2. But if you continue this list long enough, you’ll eventually get to some true statements. Yesterday we saw a proof that none of those true statements is provable.

But suppose that, undeterred by the proof, we are determined to identify a specific true statement on the list. Here’s our strategy:

First we write down the first gazillion statements on the list, where a gazillion is some number so big that we’re sure to have included some truths.
Next we go down the list, eliminating false statements. Notice that if we’re willing to work long enough, every false statement eventually gets eliminated — we just keep trying shorter-than-10000-character prescriptions and crossing off the numbers they describe. This leaves us a list of candidates.
Suppose there’s just one true statement on the list. Then eventually we cross all the others off, and learn that the one remaining candidate is true (because the list was so long there had to be at least one true statement). In fact, we’ve just proved (by process of elimination) that this statement is true. But we learned yesterday that no such proof is possible. Conclusion: There can’t be just one true statement on this list.
Suppose there are just two true statements on the list. Then eventually we’ll cross all the others off, leaving two candidates, at least one of which is true. But we’ve already proved that there can’t be just one true statement, so these must both be true. In fact, we’ve just proved they’re true. Which is impossible. Conclusion: There can’t be just two true statements on the list.
Do you see where this is going? There also can’t be just three true statements on the list, or four, or five, or any other number. Yet we know there are true statements on the list! Something’s wrong here!!!!!

Okay, what went wrong? Answer:

Our argument relies on the fact that every statement on our list must be unprovable. Why should we believe that? Well, we proved it yesterday; that’s why!
Actually, our argument relies on a bit more — it relies not just on the fact that every statement on our list is unprovable, but that we can prove their unprovability. But again, we did that yesterday.
The only way out of this is to conclude that there’s some gap in yesterday’s proof.
But there’s only one thing we used yesterday without proving it: We began from the assumption that our arithmetical reasoning is consistent.
So that must be where the gap is — and it must be impossible to fill that gap! In other words, it must be impossible to prove that arithmetic is consistent.
Tada!

There’s just one other way out: If arithmetic actually is inconsistent, then all bets are off and we can prove its consistency — but in that case, of course, our conclusion will be wrong. In any event, there are very few people who think this is a contingency worth worrying about.

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Berry Interesting

Steve Landsburg — Wed, 14 Dec 2011 07:01:46 +0000

Today, I’m going to give you a short, simple proof of Godel’s First Incompleteness Theorem — the one that says there are true statements in arithmetic that can’t be proven. The proof is due to Gregory Chaitin, and it is far far simpler than Godel’s original proof. A bright high-schooler can grasp it instantly. And it’s wonderfully concrete. At the end, we’ll have an infinite list of statements, all easy to understand, and none of them provable — but almost all of them true (though we won’t know which ones).

Chaitin’s proof, like Godel’s, is inspired by a classical paradox — in this case, Berry’s Paradox as opposed to the Liar Paradox (both of which I described yesterday).

We start by observing that some numbers are more complicated than others. The 10,000 digit number that starts off 10101010101010101….. and continues the same way is somehow less complicated than a random 10,000 digit number.

Kolmogorov formalized this notion by defining the “complexity” of a number as the length of the shortest prescription (in some fixed language) for writing it down. The number above has a 48-character prescription: “Write down a 1, then a 0, then repeat 5000 times”. That makes it pretty simple.

Now suppose we want to find a more complicated number — one that requires at least, say, 60 characters to prescribe. We know there must be many such numbers, but we’d like to find a specific example — together with a proof that our example can’t be prescribed in less than 60 characters. So we write a computer program, called Finder60, which searches for examples-with-proofs. Ideally, the program outputs something like: “I have found a proof that the number 2834932709472398472328923478902342903848927189374901742309842398742 cannot be prescribed with fewer than 60 characters.”. Finder60 searches systematically, so if there’s such a number/proof combination to be found, Finder60 will find it. Otherwise, it keeps on running forever.

We can also, of course, write programs called Finder90, Finder120, Finder10000 and so on, to find ever-more-complicated numbers together with proofs that they are complicated.

Once you’ve written the code for Finder60, you only have to tweak it slightly to get the code for Finder10000 — and the code gets only slightly longer in the process. Basically, you just change all the 60’s to 10000’s, replacing two-digit strings with five-digit strings. Not much difference.

So if the code for Finder60 is, say, 5000 characters long, then the code for Finder10000 is just a bit more than that — say 5200 characters.

Now suppose M is any number that provably requires more than 10000 characters in its prescription. Then Finder10000 will find it and print it out. Which means we have the following less-than-10000-character prescription for M:

Run the following program: [insert the 5200-character code for Finder10000 here]

Uh oh! If prescrbing M provably requires more than 10000 characters, then M can be prescribed in fewer than 10000 characters. Contradiction!

Conclusion: There must be no such M. That is, no number can be proved to require more than 10000 characters in its prescription.

But surely some numbers do require more than 10000 characters; after all there are only finitely many ways to prescribe a number with fewer than 10000 characters, which leaves infinitely many numbers left over.

So what’s a true statement that can’t be proven? Answer: Any true statement of the form “The number M cannot be prescribed in fewer than 10000 characters”. No matter what M you plug in , this statement must be unprovable. If you plug in M=1 or 2, you’ll get a false statement. But for most values of M (though we don’t know which values!) you’ll get a statement that’s true, but unprovable.

Isn’t that about the coolest trick ever? Well, it turns out that you can add another twist to make it even cooler. That’s where the surprise examination paradox comes in. I’ll explain it all one day later this week, though you won’t know which day till you log in that morning.

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A Tale of Three Paradoxes

Steve Landsburg — Tue, 13 Dec 2011 07:00:09 +0000

This is a tale of three paradoxes and why they matter.

First, the ancient Liar Paradox: “This sentence is false”. If this sentence is true, it must be false. If it’s false, it must be true.
Next, the century-old Berry Paradox: Call a phrase “short” if it contains fewer than 13 words. The English language contains a finite number of words, and hence a finite number of short phrases. Hence there must be some natural numbers that can’t be described by any short phrase. Among these natural numbers, there must be a smallest. What is that natural number? Why, it’s the smallest natural number that can’t be described by any short phrase, of course. Except that this number is in fact described by the short phrase in boldface.
Finally, the more modern Paradox of the Surprise Examination (or the Unexpected Hanging), which we discussed yesterday.

The paradoxes are slippery, because they are stated in the imprecise language of English. But each of them has inspired a precise mathematical counterpart that is central to a brilliant argument in mathematical logic.

Start with the liar: “This sentence is false” can’t be true, or it would be false — and can’t be false, or it would be true. This tells us that there’s such a thing as an English sentence that’s neither true nor false, which comes at first as a considerable surprise, but isn’t devastating.

One of Kurt Godel’s great insights was that you can go a lot deeper by considering a slightly different sentence: “This sentence is not provable”. If that statement is false, then it’s provable. But surely no false statement should be provable! So maybe the statement is true. In that case, it’s true but not provable, which says something about the limits of logic. It says that not every true statement can be proved.

At one level, this is still just wordplay. What makes it profound is Godel’s discovery of a code that converts certain English sentences into statements of pure arithmetic (that is, statements of the form “Every number is the sum of four squares” or “Every prime number is divisible by 2″) in such a way that true statements are matched with true statements and false statements are matched with false statements. The code is cleverly constructed so that there’s a statement in pure arithmetic (say, for illustration, that it’s the statement “every even number is the sum of two primes”) that corresponds to the English sentence “The statement that every even number is the sum of two primes cannot be proven.” These statements are either both false, in which case it’s possible to prove a false statement, which we believe (and hope to God!) is not the case — or they’re both true, in which case we’ve found a true statement in pure arithmetic that can’t be proven.

Much brilliant work goes into constructing the code, but the brilliant idea is to adapt the Liar Paradox to a context where you can’t just say “Well, I suppose it’s neither true nor false” — because statements like “Every even number is the sum of two primes” must be either true or false.

(The above intentionally sacrifices a little precision in the interest of readability; the linked post is more carefully worded.)

So that’s why mathematicians care about the Liar Paradox. More on the Berry Paradox and the Surprise Exam (and how all three tie together) as the week goes on.

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The Surprise Exam, and More Surprises

Steve Landsburg — Mon, 12 Dec 2011 06:01:13 +0000

If you’re the sort of person who reads this blog, you’re likely to be familiar with the paradox of the unexpected hanging, which has been floating around since 1943 but achieved popular notoriety around 1969 through the writing of Martin Gardner. But you’re less likely to be aware that the unexpected hanging plays a central role in a wonderful new piece of serious mathematics related to algorithmic complexity, Godel’s theorems, and the gap between truth and provability.

The unexpected hanging might as well be a surprise examination, and that’s the form in which I present this paradox to my students every year: In a class that meets every weekday morning, the professor announces that there will be an exam one day next week, but that students won’t know exactly which day until the exams are handed out.

The students, of course, immediately start trying to guess the day of the exam. One student (call him Bob) observes that the quiz can’t be on Friday — because if it is, the students will know that by Thursday afternoon. After all, if Monday, Tuesday, Wednesday and Thursday mornings have all passed by, only Friday remains. A Friday exam can’t be a surprise exam.

A more thoughtful student (call her Carol) observes that this means the quiz must be on one of Monday, Tuesday, Wednesday or Thursday — and that if it’s on Thursday, they’ll know that by Wednesday night. After all, Friday’s ruled out, so if Monday, Tuesday and Wednesday have passed by, then only Thursday remains. That rules out a surprise exam on Thursday.

Another student (call him Ted) observes that thanks to Bob and Carol, we know the exam must be on one of the first three days of the week — which means that if it’s not on Monday or Tuesday, it must be on Wednesday. Therefore if it’s on Wednesday, they’ll know this by Tuesday night. Scratch Wednesday from the list of possibilities.

Now Ted’s girlfriend Alice points out that the exam can’t be on Tuesday either. Whereupon Bob concludes that the exam must be on Monday. But wait a minute! Carol points out that if they know the exam will be on Monday, it can’t be a surprise. Therefore no surprise exam is possible.

The students, relieved, decide not to study. But they’re awfully surprised when they show up in class the following Tuesday and the professor hands out an exam.

Where did the students go wrong? There is no consensus among the many philosophers and logicians who have considered this problem. The great Willard Van Orman Quine believed that Bob went wrong at the very beginning when he ruled out Friday. (According to Quine, Bob’s argument fails to distinguish between a proof that the exam can’t be on Friday and a proof that the students will know that the exam can’t be on Friday.) Other deep thinkers have accepted Bob’s argument (agreeing that the exam can’t be on Friday) but refused to accept Carol’s (thus refusing to rule out Thursday). You can, if you wish, read a pretty comprehensive survey of this literature here. But even among those who think Bob (or Carol) is mistaken, there is little agreement about exactly why they are mistaken.

Now, I happen to think the surprise examination paradox is pretty interesting as a pure intellectual exercise. But it’s also got important applications. I use it in the classroom to illuminate our discussion of the underlying “backward induction” technique, which economists (and especially game theorists) use all the time in serious arguments. Much more recently, the surprise examination has been used to illuminate some key concepts in mathematical logic, which I alluded to back in the first paragraph of this post. That’s the coolest part of all, and I’ll tell you all about it later in the week.

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Thursday Puzzle and More

Steve Landsburg — Thu, 08 Sep 2011 06:01:48 +0000

Yesterday’s post on taxation generated a whole lot of comments that deserve responses; unfortunately I’m too swamped right now to respond. Worse yet, I’ll be out of town — and probably not blogging — for the next few days. Sometime next week, I’ll try to craft a new blogpost addressing much of what was said in those comments.

Meanwhile, here, courtesy of our frequent and invariably interesting commenter Mike H, is a puzzle to keep you busy while I’m gone:

Your enemy places, say, 8 red points and 8 blue points in the plane. Your job is to connect each red point to some blue point with a straight line segment, so that none of the line segments cross.

1) Can your enemy place the points in such a way that you are guaranteed to fail?

2) What if we make your job harder by requiring each blue point to be used exactly once?

3) What if we change 8 to 17 or 26 or 43 or …. ?

Edited to add: Question 2) is the interesting one. Also, to keep it interesting, your enemy is not allowed to put any three points on the same line.

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Thursday Puzzle

Steve Landsburg — Thu, 01 Sep 2011 06:01:31 +0000

Our frequent and reliably insightful commenter Jonathan Kariv sends along this neat puzzle:

Your enemy chooses 10 points on an infinite tabletop and gives you 10 coins of the same size (let’s say U.S. quarters). Can you always place the coins on the tabletop in such a way that all 10 points are covered, but no two coins overlap?

If your enemy puts all 10 points far away from each other, you can just put one quarter over each point. If he scrunches them up real near each other, you can just put one quarter over all of them, and put the other nine quarters anywhere you want. But if he’s clever enough, can he make your job impossible?

Edit: I originally described the tabletop as “big”; I’ve edited this post to change it from big to infinite, in order to be true to the puzzle as Kariv originally posed it. My apologies for the original inaccurate transcription.

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Friday Solution

Steve Landsburg — Fri, 19 Aug 2011 07:01:17 +0000

Re yesterday’s puzzle, you’ll find answers in the comments. (We are blessed with some very smart commenters here at The Big Questions!!)

Commenter Roger Schlafly pointed this Wikipedia article where I was surprised and delighted to see a reference to a paper co-written by my old friend Dave Rusin. I did not remember that Dave had anything to do with this problem, but in retrospect I bet I knew this at one time.

I managed to dig out some notes I jotted down on this subject many many years ago. I have not doublechecked these results, and I can’t completely vouch for the careful accuracy of my younger self, so take these for what they’re worth. But here’s what I once claimed to have proved:

The reason there is exactly one pair of nonstandard six-sided dice is that six is the product of two distinct primes. For the same reason, there is exactly one pair of nonstandard n-sided dice when n is 10, or 15, or 21, or …. For any product of three distinct primes, there are at most 40 nonstandard pairs.

I also found (in what appears to be my handwriting) this chart, which I reproduce with the same caveats:

Number of sides	Number of nonstandard pairs
1	0
2	0
3	0
4	1
5	0
6	1
7	0
8	3
9	1
10	1
11	0
12	at most 13
13	0
14	1
15	1
16	9
17	0
18	at most 13
19	0
20	at most 13
21	1
22	1
23	0
24	at most 94
25	1
26	1
27	3
28	at most 13
29	0
30	at most 40
31	0
32	25

Corrections welcome!

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Thursday Puzzle

Steve Landsburg — Thu, 18 Aug 2011 07:01:39 +0000

I love this problem, which I found on the Internet many years ago. I suppose you could find a solution by Googling, but that’s of course no fair.

A standard pair of six-sided dice induce a probability distribution on the outcomes 1 through 12: The probability of rolling a 1 is 0, of rolling a 2 is 1/36, of rolling a 3 is 1/18, etc. Is there any nonstandard pair of six-sided dice that induces exactly the same probability distribution? If so, how many such pairs are there?

(A non-standard pair of six-sided dice might have, say, the numbers 1,2,2,3,8,9 on one cube and the numbers 2,3,4,4,4,4 on the other.)

What about the same problem for seven-sided dice, or eight-sided, or n-sided?

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Worst Puzzle Ever

Steve Landsburg — Mon, 08 Aug 2011 07:01:06 +0000

From Air Canada’s inflight magazine:

And since I’m sure someone will ask, here is, apparently, the only solution they could think of:

(This puzzle brought to you by the same thoughtful folks who bring you Air Canada’s customer service.)

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Yesterday’s Puzzle

Steve Landsburg — Fri, 25 Mar 2011 07:02:05 +0000

I didn’t think anyone would get it. I was completely stumped myself until I got help from my friends. But Neil got it.

In his words, “We have onomatopoeaic words for the sounds made by all of the animals on the right.”

Or, as I prefer to think of it, the animals on the right all have vocabularies (consisting, in most cases, of a single word) while those on the left do not.

A donkey brays, and when it brays it says hee-haw. The donkey makes it to the right of the line not by virtue of braying, but by saying hee-haw. Thus the elephant, which trumpets, but thereby merely makes a noise (as opposed to saying a word) is consigned to the left.

Lions, tigers, and jaguars all roar, but to the best of my recollection from extensive reading (mostly at about age 5), lions and tigers, when roaring, actually say the word “roar”, while a jaguar merely roars incoherently. Chickens say “cheep”. Hens say “buck-buck-buck” (the act of saying this is called “clucking”). Roosters can crow in either of two dialects: Some say “rrr-rr-rrr-rr-rrrrr” while others (who my five-year-old self considered unbecomingly pretentious) say “cock-a-doodle-doo”. Pretentious they may be, but as a scientist, I am here to record the facts, not to judge them.

Regarding some of the side discussions that came up in comments, I have no idea why the count is off at the top of the first column, or whether the jaguar was added at a later time, or if so why.

For the benefit of math_geek, who (in yesterday’s comments) lamented the absence of the okapi, I present another list of about the same vintage as yesterday’s, this time with the columns clearly labeled so that future generations (i.e. us) wouldn’t have to puzzle this one out — and with the okapi well represented:

And finally, I think we should let Andy Kaufman have the last word on this:

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