Ω: my favorite strange number

Ω is my own personal favorite transcendental number. Ω isn’t really a specific number, but rather a family of related numbers with bizzare properties. It’s the one real transcendental number that I know of that comes from the theory of computation, that is important, and that expresses meaningful fundamental mathematical properties. It’s also deeply non-computable; meaning that not only is it non-computable, but even computing meta-information about it is non-computable. And yet, it’s *almost* computable. It’s just all around awfully cool.
So. What is it Ω?
It’s sometimes called the *halting probability*. The idea of it is that it encodes the *probability* that a given infinitely long random bit string contains a prefix that represents a halting program.
It’s a strange notion, and I’m going to take a few paragraphs to try to elaborate on what that means, before I go into detail about how the number is generated, and what sorts of bizzare properties it has.
Remember that in the theory of computation, one of the most fundamental results is the non-computability of *the halting problem*. The halting problem is the question “Given a program P and input I, if I run P on I, will it ever stop?” You cannot write a program that reads P and I and gives you the answer to the halting problem. It’s impossible. And what’s more, the statement that the halting problem is not computable is actually *equivalent* to the fundamental statement of Gödel’s incompleteness theorem.
Ω is something related to the halting problem, but stranger. The fundamental question of Ω is: if you hand me a string of 0s and 1s, and I’m allowed to look at it one bit at a time, what’s the probability that *eventually* the part that I’ve seen will be a program that eventually stops?
When you look at this definition, your reaction *should* be “Huh? Who cares?”
The catch is that this number – this probability – is a number which is easy to define; it’s not computable; it’s completely *uncompressible*; it’s *normal*.
Let’s take a moment and look at those properties:
1. Non-computable: no program can compute Ω. In fact, beyond a certain value N (which is non-computable!), you cannot compute the value of *any bit* of Ω.
2. Uncompressible: there is no way to represent Ω in a non-infinite way; in fact, there is no way to represent *any substring* of Ω in less bits than there are in that substring.
3. Normal: the digits of Ω are completely random and unpatterned; the value of and digit in Ω is equally likely to be a zero or a one; any selected *pair* of digits is equally likely to be any of the 4 possibilities 00, 01, 10, 100; and so on.
So, now we know a little bit about why Ω is so strange, but we still haven’t really defined it precisely. What is Ω? How does it get these bizzare properties?
Well, as I said at the beginning, Ω isn’t a single number; it’s a family of numbers. The value of *an* Ω is based on two things: an effective (that is, turing equivalent) computing device; and a prefix-free encoding of programs for that computing device as strings of bits.
(The prefix-free bit is important, and it’s also probably not familiar to most people, so let’s take a moment to define it. A prefix-free encoding is an encoding where for any given string which is valid in the encoding, *no prefix* of that string is a valid string in the encoding. If you’re familiar with data compression, Huffman codes are a common example of a prefix-free encoding.)
So let’s assume we have a computing device, which we’ll call φ. We’ll write the result of running φ on a program encoding as the binary number p as &phi(p). And let’s assume that we’ve set up φ so that it only accepts programs in a prefix-free encoding, which we’ll call ε; and the set of programs coded in ε, we’ll call Ε; and we’ll write φ(p)↓ to mean φ(p) halts. Then we can define Ω as:

Ωφ,ε = Σp ∈ Ε|p↓ 2-len(p)

So: for each program in our prefix-free encoding, if it halts, we *add* 2-length(p) to Ω.
Ω is an incredibly odd number. As I said before, it’s random, uncompressible, and has a fully normal distribution of digits.
But where it gets fascinatingly bizzare is when you start considering its computability properties.
Ω is *definable*. We can (and have) provided a specific, precise definition of it. We’ve even described a *procedure* by which you can conceptually generate it. But despite that, it’s *deeply* uncomputable. There are procedures where we can compute a finite prefix of it. But there’s a limit: there’s a point beyond which we cannot compute *any* digits of it. *And* there is no way to compute that point. So, for example, there’s a very interesting paper where the authors [computed the value of Ω for a Minsky machine][omega]; they were able to compute 84 bits of it; but the last 20 are *unreliable*, because it’s uncertain whether or not they’re actually beyond the limit, and so they may be wrong. But we can’t tell!
What does Ω mean? It’s actually something quite meaningful. It’s a number that encodes some of the very deepest information about *what* it’s possible to compute. It gives a way to measure the probability of computability. In a very real sense, it represents the overall nature and structure of computability in terms of a discrete probability. It’s an amazingly dense container of information – as a n infinitely long number and so thoroughly non-compressible, it contains an unmeasurable quantity of information. And we can’t get near most of it!
[omega]: http://www.expmath.org/expmath/volumes/11/11.3/Calude361_370.pdf

0 thoughts on “Ω: my favorite strange number

  1. MiguelB

    Mark,
    Is there a way to determine if a given program halts by looking at Ω?
    In other words, assuming I have Ω for my machine and my encoding, and I have a program, can I use Ω to find if my program will halt?

    Reply
  2. Mark C. Chu-Carroll

    Miguel:
    No, you can’t figure out if a specific program halts by looking at Ω. *If* you could create an encoding in which there was only one valid program of any given length, then the individual bits of Ω would each tell you whether a specific program would halt or not; but in the prefix-free encoding that’s used for defining Ω there can be more than one program of a given length. (So, for example, if the program that you’re interested in has length “p”, the 2-p bit might be zero, because there were two halting programs of size p; the summing would cause 2p-1 to be 1 and 2p to be zero.)

    Reply
  3. Walker

    One more interesting property: unsplittable.
    Solvay introduced a notion of relative randomness that is similar to Kolmogorov complexity, but is nicer because addition of the numbers corresponds to the join. For every other computable real r, there is a+b = r such that both a and b are properly less random. Ω has no such decomposition.
    See http://www.math.uchicago.edu/~drh/

    Reply
  4. Torbjörn Larsson

    I got to ask. Someone mentioned on an earlier post that one can’t prove randomness. As I understand it you can’t take a single random number and show if it is random or pseudorandom, whether specifically or in general I’m not sure. Is that true and is it connected to Omega? (Perhaps since we wouldn’t recognise an Omega if we see it, if I understand the above correctly.)

    Reply
  5. Walker

    Someone mentioned on an earlier post that one can’t prove randomness. As I understand it you can’t take a single random number and show if it is random or pseudorandom, whether specifically or in general I’m not sure.

    Random reals are one of those names that can be confusing for the layman because they are conflating two different terms. Technically, individual numbers are not random, but sequences are. For example, we may say that a die is random because rolling it produces a random sequence of numbers, but no single value is technically “random”.
    However, every individual real number can be considered a sequence. Just turn it into an infinite decimal expansion and consider each digit a number:
    1/9 = 0.999999999…. is 9,9,9,9,9, …
    π = 3.14259… is 3,1,4,2,5,9
    If this sequence is random this is what we mean when we say “random reals”.

    Reply
  6. Chris Nelson

    I think that there’s one too many 0’s at the end of “any selected pair of digits is equally likely to be any of the 4 possibilities 00, 01, 10, 100”.

    Reply
  7. Sean

    I also think this is a wonderfully fascinating post, and wish I was smart enough to have written it, and not just notice typos in it. 😛
    p.s. if I had intended the substitution of “two” for “too” in the above post, it would have almost been funny, in an extremely lame sort of way, but actually, it was just a typo.

    Reply
  8. Torbjörn Larsson

    “Technically, individual numbers are not random, but sequences are.”
    Sorry, I didn’t thought that through properly. Yes, I sloppily assumed that a randomly generated and selected real (“single random”) has a randomised sequence of digits. Why shouldn’t it have – does each pseudorandom algorithm produce uncountably many sequences?

    Reply
  9. Torbjörn Larsson

    “does each pseudorandom algorithm produce uncountably many sequences”
    Or rather, does some algorihms, et cetera.

    Reply
  10. Benjamin Higgins

    You certainly can use omega to determine if a program halts or not. Just run the set of programs less than a certain number of bits, and wait until the number of programs that halt equal the beginning of omega, then you know that the remaining programs do not halt.

    Reply
  11. passersby

    maybe it’s the omega-part of “i’m the alpha & the omega” the finite and the infinite 😀
    we should create a new religion based on this revelation.

    Reply
  12. Tom Ritchford

    It seems to me as if knowing Ω does in fact work as an oracle to solve the halting problem.
    Quick sketch: given Ω, you know exactly how many machines of size n or below will halt: call this w. To find if a given program of size n halts, run all of the programs of size n or less in parallel until exactly w of them have halted — then the remaining programs are the ones that will not halt.
    Or am I missing something?

    Reply
  13. Steve

    So if Omega tells us what it’s possible to compute, and we can’t compute Omega, does that mean that it’s impossible for us to know what it’s possible for us to know?

    Reply
  14. Mark C. Chu-Carroll

    Tom:
    Nope, doesn’t quite work.
    The problem is that programs *larger* than the size threshold you pick can have a “bubble-up” effect – remember that Omega is a sum. So, for example, if you have a program of size 23, and there are 17 1s in the first 23 digits of Omega, that doesn’t mean that there are 17 halting programs of length 23 or less. There could be 1000 programs of length 28 that halt – and so their effect was pushed upwards. So their could be *less* than 17 halting programs of size 23 or less – and you’ll spend forever trying to wait for 17 programs to halt.

    Reply
  15. Mark C. Chu-Carroll

    Steve:

    So if Omega tells us what it’s possible to compute, and we can’t compute Omega, does that mean that it’s impossible for us to know what it’s possible for us to know?

    Omega doesn’t tell us what it’s possible to compute; it expresses an overall probability of finding a halting program via a particular kind of random search. In terms of “what we know”, the nature of Omega does tell us more about what we can’t know than the halting problem, but it’s the same kind of knowledge that we get from the un-solvability of the halting problem.

    Reply
  16. Vaughan

    Well, the developers of MacOS 8.x must have inadvertantly embedded omega into the code somehow, because I remember my Quadra used to halt every couple of hours.

    Reply
  17. Mark

    “I think my brain just exploded.
    Posted by: Joshua | August 7, 2006 02:58 PM ”
    That’s the only post that I’ve understood so far.

    Reply
  18. Ernesto Badillo

    Mark:
    Benjamin’s and Tom’s arguments are correct. Actually, Chaitin uses a similar argument. You can see this on: http://cs.umaine.edu/~chaitin/sciamer3.html look for “Why Is Omega Incompressible?”.
    Chaitin says:
    “My strategy for demonstrating that Ω is incompressible is to show that having the first N bits of Ω would tell me how to solve the Turing halting problem for programs up to length N bits”.
    That is an amazing fact: Omega contains the solution of every halting problem (restricted to a specific UTM).
    So, for instance, if you want to know if he Riemann hypothesis is true, you can write a program wich halts until it finds a counter-example. If you have Omega you can know the answer in finte time.

    Reply
  19. Ω777Ω

    …extremely freaky songi bout tranzak
    …wonderfully informative about something close to me that I knew nothing about…
    …[Ω] embodies an enormous amount of wisdom in a
    very small space . . . inasmuch as its first few thousands
    digits, which could be written on a small piece
    of paper, contain the answers to more mathematical
    questions than could be written down in the entire
    universe….bennet (bennet & gardner 79)…
    …conceptually it is very powerful…
    …thanx…

    Reply
  20. O, Me, GA Man

    This research will lead to the end of the world as we know it, with Charlton Heston shooting at nocturnal, homicidal maniacs.
    I’ve seen this future, in a premonition.
    No, wait…it was a movie.
    …….and so will be Premonition, starring Sandra Bullock, in 2007.

    Reply
  21. Ralph

    “…it encodes the probability that a given infinitely long random bit string contains a prefix that represents a halting program.”
    BUT…
    “Uncompressible: there is no way to represent Ω in a non-infinite way; in fact, there is no way to represent any substring of Ω in less bits than there are in that substring.”
    Doesn’t the first statement, as a definition or “representation” of omega, refute the second?
    Isn’t “the smallest integer which is a sum of three cubes” a perfectly good “representation” of 36?

    Reply
  22. Mark C. Chu-Carroll

    Ralph:
    You’ve nailed one of the strange things about Ω which I deliberately didn’t mention; I wanted to see how long it took before someone commented on it. 🙂
    Ω is one of those numbers that trips Gödel self-reference. You *can* define it in a way that inherently involves self-referentiality within the formal system where you’re expressing the definition.
    On the other hand, the second statement is true, assuming the right definition of “representation”, which is something far too complex to get into in a comment! But basically, it’s talking about a “representation” that allows you to *use* the information that is naturally encoded in Ω.

    Reply
  23. Torbjörn Larsson

    “Yes, I sloppily assumed that a randomly generated and selected real (“single random”) has a randomised sequence of digits. Why shouldn’t it have – does each pseudorandom algorithm produce uncountably many sequences?”
    Dear me, what a fool I can be.
    IIRC now, that this is one of the elementary results in my probability course book (which as usual I can’t find) – the events of the random real numbers are completely orthogonal to the events of the digit sequence.
    Which I allow for above but doesn’t take care of – any sequence, even ordered, will happen.
    So to answer my own question, a single isolated event can never be proved to be random by looking at the digit sequence, one must collect more events and model them (measure the statistics) to see that they may be.
    Same for the events of the “random real” digits in the sequence, as Walker discuss.
    Steve:
    “does that mean that it’s impossible for us to know what it’s possible for us to know?”
    I thought gödel incompleteness pretty much assured us that we will never know the end of it or what will happen.
    If so, it is a great argument to work in science, much as I guess the halting problem is an argument to work with CS/programming, and the NFL theorems seems to be an argument to work on signal processing/detection problems.

    Reply
  24. George

    Could you reformulate omega as the sum of BB(n)/2^n for all positive integers n (BB=Busy Beaver)? I don’t see why it only works with prefix-free encodings.

    Reply
  25. Brian King

    want to represent it, you either need a singular symbol such as Ω or an infinite number of sets of symbols that represent the ‘structure’ that when viewed as a whole is in itself a singularity. A singular symbol doesn’t tell us a lot about what is inside, so we want to find a way to look at its pieces. The favorite method is to try to represent it decimally. In that context, only the first nine digits truly do not repeat themselves when we take each as a singular entity. So, to represent all of the 10 unique ‘holders’ I actually have to have 9+8+7+6+5+4+3+2+1+0(1) (yes you have to count zero as one because it represents a state in the decimal system) buckets to represent them. My one ponit there is to get you thinking about what each of those digits represents. Remember, a decimal 6 is merely a mathematical representation of 3+3 or 1+1+1+1+1+1. So to demonstrate a 6 I can use 6 (1)’s. That fact in itself breaks the rule of not being able to represent Ω using a summed subset such as A + B = Ω. Or does it (sorry about my grammer)? Perhaps it simply means that Ω cannot be represented using the decimal system since the instance of a six would either be in violation of the rule of being able to represent Ω as a summed set of subsets (a 6 is already a summed set of subsets as I demonstrated above). The same argument can be used against the ability to represent it in binary. It all boils down to how you are trying to measure it. By what standard can you measure a singularity? Trying to represent it in any other format would be a form of measuring a singularity and that violates the rule of computability. Therefore it cannot be represented via any other form of symbology. A singularity is unique by its natural. Therefore trying to measure it by any standard is futile because there is no standard or any combination of standards by which to measure something truly unique. That also makes the argument of compressibility as true since a singularity cannot be compressed as compression requires measurement and yada yada, back to the start. See where it goes?

    Reply
  26. Brian King

    Sorry the beginning got chopped by the web code. Here it is;
    Such a narrow view of what all of this means. Consider this. The theory basically states that Ω is a singularity that has a size greater than one, right (ok, I’ll explain)? So, you want to represent it, you either need a singular symbol such as Ω or an infinite number of sets of symbols that represent the ‘structure’ that when viewed as a whole is in itself a singularity. A singular symbol doesn’t tell us a lot about what is inside, so we want to find a way to look at its pieces. The favorite method is to try to represent it decimally. In that context, only the first nine digits truly do not repeat themselves when we take each as a singular entity. So, to represent all of the 10 unique ‘holders’ I actually have to have 9+8+7+6+5+4+3+2+1+0(1) (yes you have to count zero as one because it represents a state in the decimal system) buckets to represent them. My one ponit there is to get you thinking about what each of those digits represents. Remember, a decimal 6 is merely a mathematical representation of 3+3 or 1+1+1+1+1+1. So to demonstrate a 6 I can use 6 (1)’s. That fact in itself breaks the rule of not being able to represent Ω using a summed subset such as A + B = Ω. Or does it (sorry about my grammer)? Perhaps it simply means that Ω cannot be represented using the decimal system since the instance of a six would either be in violation of the rule of being able to represent Ω as a summed set of subsets (a 6 is already a summed set of subsets as I demonstrated above). The same argument can be used against the ability to represent it in binary. It all boils down to how you are trying to measure it. By what standard can you measure a singularity? Trying to represent it in any other format would be a form of measuring a singularity and that violates the rule of computability. Therefore it cannot be represented via any other form of symbology. A singularity is unique by its natural. Therefore trying to measure it by any standard is futile because there is no standard or any combination of standards by which to measure something truly unique. That also makes the argument of compressibility as true since a singularity cannot be compressed as compression requires measurement and yada yada, back to the start. See where it goes?

    Reply
  27. Brian King

    I also said this and didn’t finish the sentence;
    Perhaps it simply means that Ω cannot be represented using the decimal system since the instance of a six would either be in violation of the rule of being able to represent Ω as a summed set of subsets (a 6 is already a summed set of subsets as I demonstrated above).
    It should continue as;
    or mean that if a 6 is used int he representation, then an occurance of a 3 could not be used in its representation, but that would digress into a whole different argument.

    Reply
  28. Torbjörn Larsson

    “that Ω is a singularity”
    How is Omega a singularity?
    What is a singularity to you?
    “In mathematics, a singularity is in general a point at which a given mathematical object is not defined, or a point of an exceptional set where it fails to be well-behaved in some particular way, such as differentiability.” ( http://www.answers.com/topic/mathematical-singularity )
    And as Mark said:
    “this number – this probability – is a number which is easy to define”

    Reply
  29. Vincent

    yeah, omega is cool ! his complexity is unfinite, and his bits, being mathematical facts, are true, but with no reason.

    Reply
  30. Ian

    It sounds to me like you can use Ω as a halting oracle. All you have to do is translate any given program into an equivalent program with a unique length. For any binary string as a program, one-hot encoding that string will provide a unique length for every possible program, at which point you can just check the appropriate bit of Ω.

    Reply
  31. Gary Osborn

    >Nope – that’s just apophenia, like Gary Osborn and his 23.5 degree angles.http://garyosborn.moonfruit.com/revelations2
    Click on the image entitled underneath ‘The Baptist Revelation’ and read through the presentation pages.
    There are over twenty paintings – from between the 15th and 18th centuries with the index finger of John the Baptist pointing at 23.5 degrees.
    Gary Osborn

    Reply
  32. Mark C. Chu-Carroll

    Gary:
    So are you going to answer the question of just how you decide what line to use to measure the angle of those fingers? Or are you going to just go off again ranting about how it’s obvious?
    Looking at the pictures on your page, it looks to me like you’re taking anything that’s anywhere close to your angle, and insisting that it’s right. In the first three images, *none* of them look to me like the angle is what you claim it is. How do you decide what line to use for measuring the angle of a finger?

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  33. Gary Osborn

    I responded on the wrong thread, but that’s really not important when you continue to casually drop in derogatory remarks about myself and my work – saying its all “apothenia”.
    You write:
    >In the first three images, *none* of them LOOK TO ME like the angle is what you claim it is

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  34. Mark C. Chu-Carroll

    Gary:
    The reason I keep pointing out that refuse to answer questions is because you refuse to answer perfectly reasonable legitimate questions.
    If the fundamental result of your research is the discovery of certain *extremely* precise angles in paintings through the ages, then asking how you determine the precise angles of features in the paintings is a perfectly legitimate question.
    The fact that you refuse to answer it is damning.
    You’ll write paragraphs and pages to attack me for daring to question you. But you won’t answer simple questions about your research metholodogy.
    I’m a professional researcher. I’m *constantly* grilled about my research methodology. Often in very unpleasant hostile or confrontational ways. I’ve never refused to answer a question about my research methods. Even obnoxious ones, even mind-bogglingly ignorant or stupid ones. It comes with the territory.
    Anyone who does real research for a living will say the same. We expect questions; we welcome questions; and we answer questions.
    If you don’t want to answer the questions, you can’t act surprised when people don’t take you seriously.

    Reply
  35. Mark C. Chu-Carroll

    Oh, and Gary? One more thing:
    If you’re incapable of actually figuring out which post on a blog to respond to, you might want to take some time out and learn to read. It’s not hard to tell the difference between a post on the golden ratio and the ways in which it’s been misused, and a post on the universal halting probability.

    Reply
  36. Gary Osborn

    You write:
    —————————————————-
    >>If the fundamental result of your research is the discovery of certain *extremely* precise angles in paintings through the ages, then asking how you determine the precise angles of features in the paintings is a perfectly legitimate question. The fact that you refuse to answer it is damning.

    Reply
  37. Mark C. Chu-Carroll

    You’d rather cut your toenails than read my blog; but any time I so much as mention your name in a post, you’re right there in the comments writing page-long diatribes. Quite interesting.
    And I note that *still* you avoid answering my question. From the time I first talked about your gibberish, I’ve been asking: how do you measure the angles? Answering “with a protractor” isn’t a real answer, and you know it. The issue is that you’re looking at features of a painting that *aren’t* simple lines. They’re fingers, or arms, or legs, or angles between features. How you determine *what angle* you’re measuring? Are you determining midlines? If so, how? Are you measuring it to find a valid midline? Or are you just eyeballing it and picking a line that just happens to measure 23.5 degrees, because that’s what you expect to see?

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  38. Gary Osborn

    >>And I note that *still* you avoid answering my question.>You’d rather cut your toenails than read my blog; but any time I so much as mention your name in a post, you’re right there in the comments writing page-long diatribes. Quite interesting.

    Reply
  39. Manix

    I’m not shure if I understand the practical purpose of Ω. Given a sequence of code, a function for example that calls itself and has a conditioned return statement, Ω would be the statistical probability that that function would eventually return, and thus finish the nested execution?

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  40. Canuckistani

    No, it’s more like this:
    1) You have a turing-equivalent computing device that takes a bit string as input and does a computation. The encoding of the programs is such that no valid program has a prefix which is also a valid program.
    2) Choose a program by flipping a fair coin over and over. The probability that this method of choosing a program will eventually hit a halting program is Ω.
    Of course, the precise value of Ω isn’t fixed until you fix a computing device and program encoding.

    Reply
  41. Mike Stay

    Some notes on Omega:
    If you have the first n bits of Omega for a given universal machine, you can solve the halting problem for all programs up to length n. Just run all programs and add up their contributions; as soon as the first n bits match, then you know that no more programs whose length is less than or equal to n can halt, because it would change one of the bits that you already know is correct.
    Omega is really a zeta function evaluated at 1, which is why you need prefix-free programs. If you define
      Omega(s) = Σp halts 2-s|p|
    then it converges for any Turing machine and s>1. This zeta function is related to the geometric zeta function of the HALT fractal, the set of real numbers in the unit interval whose binary expansions are concatenations of halting programs.
    A more natural definition of the zeta function is
      zeta(s) = ΣΣn halts n-s
    where n is the index on a lex-order enumeration of programs. With this definition, there’s a class of programs including prefix-free ones that give random sequences when evaluated at 1, and the total Turing machine, that halts on all inputs, is Riemann’s zeta function.
    Chaitin showed that given any finite axiomatic system, you can only prove finitely many bits of the sequence; by adding one more bit of axiom to your FAS, you can prove at most one more bit of the sequence.
    There’s a natural extension of this definition to partial randomness: by adding one more bit of axiom, you can derive s more bits of the sequence. Such a sequence is called 1/s-random. A neat theorem by Tadaki is that Omega(s) is precisely 1/s-random.
    I showed in my master’s thesis that random sequences satisfy an uncertainty principle. Given a random sequence x of bits, you essentially take two to the power of both sides in the inequality
      H(x1 x2 x3 … xn) >= n-c
    Sort programs in quasi-lex order; then the index of the program times the uncertainty in your estimation of Omega given the first n bits is greater than a constant.

    Reply
  42. Dan H.

    This is an exceptionally interesting post on Ω! I have learned much. My son has Aspergers and loves math (except at school- he’s bored) I think there are a couple of thoughts I can throw his way that will really intrique him. Thank you.
    Note: Gary and Mark…Grow up. Give the other person a call and “quit bellyachin!” You guys may be smart in a lot of ways, but boy are you ignorant when it comes to taking care of conflict in a mature fashion.
    Step 1: call the other person…This will reveal who the kid is and who the man is. Quit posting this trash in a great blog.
    Step 2: realize that you might only agree to disagree
    Step 3: realize that the other person might actually be wholly or partially right about the issue(s)…and you might actually be wholly or partially wrong about the issue(s).
    Step 4: Be man enough to walk away and realize that you might only agree to disagree
    Step 5: Don’t be a grade schooler and keep whining.
    Step 6: Have lunch/dinner or coffee together and be friends. Life is too short to do otherwise.
    Hope it all works out.

    Reply
  43. Julian

    Please elaborate!
    “the statement that the halting problem is not computable is actually equivalent to the fundamental statement of Gödel’s incompleteness theorem”
    How exactly is undecidability equivalent to incompleteness?
    Both undecidability and incompleteness can be formalized as statements in ZF set theory and proven in ZF. This makes them equivalent in the trivial sense that any two theorems of ZF are equivalent. But that’s not what you meant by saying they’re equivalent. So what precisely do you mean? I’m not trying to nit-pick or criticize, I really want to know, as precisely and formally as possible, in what deep way the two are related.
    Sorry for going on a tangent to the main topic.

    Reply

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