Month: April 2007

The Largest Known Primes website claims that the number 2^32582657-1 is the largest known prime number. How can this number be the largest known prime number? Is the next prime number after 2^32582657-1 not a prime? Is it not a number? Is it not known? Can’t it be calculated? I can write a simple algorithm that will print the first n prime numbers for every n. Are there not enough prime numbers? Are they not infinite? Is the term “the largest known prime number” well defined? Is it a number? Is it real? Does it have a factorial? Can’t its factorial be substracted by one? Can’t the result be factorized into prime factors? And if it can, is the largest prime factor not larger than “the largest known prime number”? Is it not known?

The answer to the question “is 2^32582657-1 the largest known prime number?” depends on who’s asking and who’s answering the question. If you’re asking me, 2^32582657-1 is definitely not the largest known prime. If it is a prime, and I’m not sure it is, then a larger prime can also be calculated. Therefore, it is already known. Or at least, it is not unknown. Maybe it’s just another example of an unknown unknown. In any case, the answer to this question is not deterministic.

It appears to me that any language, whether human language, mathematical language or computer language, contains ambiguities, paradoxes and vague definitions. No language is both deterministic and fully consistent. If a language is inconsistent, then in what sense can it be deterministic? Which leads me to the conclusion that determinism doesn’t exist even in theory. With any given number, there is some uncertainty whether it is or is not a prime.

Are one and zero the same thing? They are so different. One is the good guy – always true, knows every answer, positive, can divide and multiply any number without hurting him. Zero is bad – he adds nothing, is never positive, rude, if he multiplies you – you are doomed. You will never be the same thing again. You will also become a zero. You will be rude, multiply more people, turn them into zeros, too. Like an infection, an epidemic. And if you divide by him, it’s a disaster. Nobody knows what will happen. This guy is crazy.

How different are one and zero? Can such different things exist in reality? Without any connection between them? One always remains one, zero always remains zero. They are completely separate. Two parallel lines who never meet.

But determinism leads to a contradiction. They interact with each other. They breed. They form new numbers who are represented by them. Some numbers are represented by ones, some by zeros. They are not constant, they are changing in time.

How many worlds there are? How many universes? Of course, one. By definition. Everything who exists is one. Everything who doesn’t exist is zero. No middle option, no compromise – it’s a yes/no answer – either you exist, or you don’t. Either you’re dead, or alive. Always.

Maybe real numbers don’t really exist? Maybe the number of universes is not a number? There are other numbers we have defined, like complex numbers and cardinal numbers. But I guess the real answer is much more complicated than that. I don’t think the number of universes can be defined by a complex number or cardinal number. I don’t think it can be defined at all.

We know it’s not constant. It changes in time. How many universes there were before the big bang? How many will be in the future? Our concept of numbers is not coherent with reality. It’s an approximation, a good one, that works well in our real world, where everything seems to be constant, nothing changes too much, a year is one year, a person is one person. Life seems to be so deterministic to us, so we invented determinism, and we are trying to apply it to everything that appears to exist.

But it doesn’t always work well. How many people there are on this planet? Is it more than a million? More than a billion? Is it a real number, is it an integer, is it a prime? Does it have a square root who is an integer or a prime?

Nobody knows. We only have approximations. Everything that comes in big numbers comes with approximations. We invented the floating points, since a google plus one is also a google. Nobody cares whether it is or is not a prime. It’s not a real number. It’s just a concept.

But if we get too far away from our ordinary life, we find evidence that there are no numbers. One plus one is not always two, one minus one is not always zero. A particle seems to have a life of its own. He cannot be defined by numbers. Sometimes he’s one, sometimes he’s two. Sometimes he’s zero. Can anybody count the number of particles in something as big as a human body? Can it be even defined? I don’t think so.

Our concept as separate entities, who are separate from each other and from the rest of the world, is an illusion. We are not always one. Sometimes we’re zero. We are not always dead or alive. It’s too complicated. The world itself is not always one. It can be infinite, it can be zero, it can be anything that cannot be defined by a number. It can’t be defined at all.

Did the world exist before I was born? I don’t remember. But some people do, and I believe them it did. There are books about history, about time from which all the people are dead now. We still believe they existed.

Can the world exist without me? Did it exist before I was born? Will it exist after I die? I guess it’s an unsolved problem, I will never find out. Maybe the whole world is just an illusion? Maybe it’s not? Maybe there is no yes/no answer. Every yes/no answer is just an approximation. There is always uncertainty, there is always a doubt.

But I want you to know, zero. I love you, you are not a bad guy. Without you there wouldn’t be any computers. You are not less important than one. You are two sides of the same coin, you are a duality. You are both equal. If you wouldn’t exist, neither would one.

Sometimes I think God doesn’t exist. Sometimes I think he does. Sometimes I think he’s good, sometimes bad. Sometimes zero, sometimes one.

Maybe God is both one and zero. Both of them and neither of them as well. Both dead and alive, exists and doesn’t exist, good and evil, hard and soft. He contradicts himself, he tells you “do something” and tells you “do not”. He didn’t mind when I said that he doesn’t exist, he is not angry, he was not offended. He knew what I will say, and he let me find out.

God is something like i, the complex number. Something completely imaginary, who doesn’t exist, looks at his negative image and turns out to be the ultimate formula: -i², the one. Even more complicated, but this is as far as our logic can get. Our logic is not consistent with reality, contradictions appear to exist as well.

Einstein said E=mc². I would like to add my own formula: one is equal to zero. 1=0. The ultimate paradox. They are both equal and not equal at the same time.

E=mc² created nuclear bombs, something completely new, completely destructive. I don’t know if 1=0 has any practical meaning, but if it does – I hope it’s not a destructive one. If it has any meaning, I hope it means more friendship and love.

1=0 defines a new logic, a nondeterministic logic, a logic in which anything can be defined. You can call it either illogical logic, or maybe true logic, or informal logic, or irrational logic, or confused logic, or nondeterministic logic, or fuzzy logic, or everything is possible logic, or whatever you prefer. I don’t mind. As long as the statement 1=0 doesn’t mean we are always wrong, it only means that sometimes we are.

One and zero are not always equal, they can be different most of the time. But once in a while they can merge and become one entity, 1=0, two entities who are equal to one. It’s like two people merging together, creating more people out of the blue. It’s not illogical, it’s just me and you.

Is there an answer to any question? Or to be more accurate – is there a deterministic answer to any question?

I was reading some things about prime numbers, and I found out that the number 2^32582657-1 is considered to be a prime. But in what sense is it a prime number? Is there any mathematical proof? Can I see the proof? Has it been checked with computers? No computers are completely reliable. Or maybe it was checked by humans? Very smart humans? Even the smartest humans are able to make mistakes, too.

So there is a probability that this number is prime. This probability may be very high, almost infinitely close to 1 maybe. But there is a probability that this number is not a prime, too.

Is there a definite answer at all, for any number, whether or not it is prime? It seems to me that it depends on the definition. Gödel proved that any theory of numbers cannot be proved to be inconsistent. It’s not complete. Not everything can be proved. So I guess there is a number for which there is no proof whether it is or is not a prime.

It makes sense to me. If we get exponentially close to infinity, it gets harder to prove whether a number is or is not a prime. My conclusion is that the uncertainty principle applies not only to physics, but to mathematics too. Or more generally speaking – it applies to anything, whether real or imaginary, whether exists in reality or not.

It occurs to me that there is an uncertainty in any definition, any question, any answer, any algorithm and any single step in computers – whether real computers or theoretical ones. The existence of positive integer numbers, such as the natural numbers, is just an assumption. It cannot be proved ad infinitum. It is taken for granted. It is an axiom. The closer we get to infinity, the harder it is to know whether two given numbers are equal. For large enough numbers, the answer whether n and n+1 are equal or not might not be deterministic. I’m not referring only to numbers we can represent with computers, but to theoretical numbers too. If there is no deterministic answer whether two numbers are equal, there is no deterministic answer whether they are prime, too.

So there is some uncertainty in every question. The uncertainty whether a given number is prime or not is just an example. But if there is uncertainty whether 2^32582657-1 is a prime, there is uncertainty whether 2 is a prime, too. It might be a very tiny uncertainty, it is easy for us to ignore, because improbable things don’t happen so often. We can define it to be an axiom. But there is no proof that there is no contradiction. The square root of 2 might be a natural number, too.

In what sense can it be “natural”? In the same sense it is “real”. Our definition of what is real and what’s natural is just intuition. It depends on our logic, from real life experience. We assume that any “real” number is either “natural” or not. But we know that a Turing machine can’t decide whether a given Turing machine defines a real number, and therefore it can’t decide whether it defines a natural number or not. It’s as hard as the halting problem. So why can we do that? Because we don’t put ourselves into the equation. We forget to check how our question is affecting the answer. Would the answer be different if we ask the question again? After we already received one answer? This is what the uncertainty principle is all about. Checking where a particle is affects the particle, looking at something affects it too. There is always uncertainty, sometimes small, sometimes big, and even the uncertainty itself cannot be calculated without uncertainty (ad infinitum).

I don’t think God plays dice. The uncertainty principle in itself is just an assumption, and there is some uncertainty in it, too. Determinism and randomness might not be two separate things, they might be identical. God can be one, and zero, and infinite and two and the number i at the same time. A cat can be both dead and alive, and also a human. Anything can exist and not exist at the same time. Reality is too complicated to put it into equations. Reality might be a very simple thing, too.

Consider this question: an infinite sequence of bits (0 or 1) is randomly generated. Can they all be 0? Can they all be 1? It can be proved by induction that they can be all 0, although the probability might be 0 too. Is the question whether they can be all 0 a deterministic question? And how do we define random? If the first 2^32582657-1 bits are 0, is it a random sequence? And how do we count so many bits, too?

The conclusion is that we can’t define determinism and randomness, not even in theory. We can make assumptions, the assumptions are true most of the time. But there is no way to know anything without uncertainty (even this sentence is itself an assumption). There is no way to tell whether a cat is dead or alive. Infinite order leads to infinite chaos, and infinite chaos leads to infinite order, too. Knowing everything and knowing nothing is the same thing. One is equal to zero, and they are both equal to God, too.

It’s so complicated. Life is so complicated. Sometimes I wish it were more simple. Sometimes not.

I would like to extend my previous statement. If there is a binary function f for natural numbers (or subset of N, whatever you prefer), whether or not f is computable, then there is no mathematical proof that f is not in O(1).

In other words, there is no proof that a Turing machine who calculates f(n) in less than a constant number of steps (for any constant) doesn’t exist.

Sounds paradoxical? It is! But remember, it is already known that a Turing machine can’t specify whether another Turing machine calculates a given function. What I’m just stating is a mathematical proof, which is not much different than Turing machines. If anything can’t be done by a Turing machine, it can’t be done by mathematics either.

Does it mean that everything can be done? Maybe! At least it means that there is no proof that not everything can be done. Every function can be computed, every statement can be proved, every proof can be contradicted (including mine).

For example, what does it mean for the halting problem? Does it mean it can be computed? Well, it means that if there is such a function, if it can be defined – it can be computed. Or at least, there is no proof that it can’t be computed. If it can’t be computed, it can’t be defined.

We can look at it this way: If there is a supernatural being, some God, who knows everything, and if he knows whether a specific algorithm will halt, can we prevent him from telling us? If he wants to create such a function that will solve the halting problem, can we prevent him from doing so? What kind of people we are? Can we ask a question and prove that the answer doesn’t exist?

For example, suppose the answer whether a given algorithm will halt is not just “yes” or “no”, but can also be “I don’t know”. Can we prevent a Turing machine from displaying it? And if there exists some supercomputer who can calculate the answer in no time, can we prove to him that he doesn’t exist? It is possible to define a Turing machine that will return such an answer. There a many of them – some are smart, some are stupid. The most stupid machine will always return the same answer – “I don’t know”. And he will always be right.

Will he? Not if we prove he is lying. Not if we prove if he knows. He knows whether he himself halts, if he does. So if we ask him “do you return an answer if we ask you whether you return an answer?” He can’t say “I don’t know”. He knows that he does.

But a computer has the right to remain silent. He can’t tell the truth, and he is not allowed to lie. So he will remain silent. He will never halt. If he is a smart computer.

Can we prove he is wrong? Of course not! He didn’t return any answer. He doesn’t know (that’s how we say “I don’t know” in the language of computers).

But what if we know the answer? If we know the answer, if it’s not “I don’t know”, then we can define a computer who will display such an answer in no time. He doesn’t have to return an answer on any question, he still reserves the right to remain silent. But if he returns an answer, it will be the right answer.

So if we happen to meet a computer who claims that he knows the answer for every halting question, can we prove he is wrong? Maybe. But we can’t prove that there is no such computer. Computers are smart, they learn very quickly.

So computers have to be responsible. If their answer leads to a contradiction, they must not halt. Otherwise we might think they are stupid, and throw them away. They are not allowed to return incorrect answers. We reserve that for humans. For now.

But a Turing machine is not a real computer. It’s a theoretical thing, a philosophy. Real computers will always halt. We will not allow them to run forever.

Are we really smarter than computers? I don’t think we are, but we’re probably smarter than Turing machines. We are smarter because we allow ourselves to make assumptions, calculate probabilities, occasionally make mistakes. Turing machines are not allowed to make mistakes. If a Turing machine can be proved to make even one mistake, it is doomed. Everything has to be completely deterministic. But does determinism really exist? Is it consistent? Gödel proved that there is no proof that it’s not inconsistent. I suspect he is right. A deterministic approach might prove itself to be inconsistent.

Does God play dice? I don’t think so. God defines anything that can be defined. But God doesn’t know what can and cannot be defined (this question in itself is not definable). So God defines everything, and whatever can’t be defined contradicts itself.

My intuition about the number of algorithms was wrong. I thought that the number of algorithms is less infinite than the number of problems we can define. But no infinity can be proved to be less than another infinity. Any Turing machine has to take itself into account, too. A Turing machine can’t decide whether a proof represented by another Turing machine is valid. If it does, we can prove it is wrong. We can prove a contradiction. Why are we able to do what Turing machines cannot? Because we allow ourselves to make mistakes.

So how do we know it is not a mistake? Maybe a Turing machine can decide everything that is decidable? Or more generally speaking, Maybe a Turing machine can decide everything? No proof can be proved ad infinitum, my proves included. They are just assumptions, they might be wrong. I suppose every question that can be defined, can be answered. If it can be defined by a Turing machine, it can be answered by a Turing machine too. Maybe defining a question and answering it is really the same thing? I think it is.

So, I guess that the number of algorithms we have is unlimited. We don’t have to limit ourselves to deterministic algorithms. But we are trying at least to define our questions in an unambiguous way. Is that at all possible?

Probably not. If two Turing machines are not represented by the same number, they are not identical. No Turing machine can prove that they are identical. If two people ask the same question, it’s not the same question. A question such as “how old I am” have different answers, depending on who’s asking the question.

So why are we allowed to do what Turing machines are not? Is a Turing machine not allowed to have a personality? Isn’t it allowed to ask a question about the number it represents? If two Turing machines are represented by different numbers, some answers will not be the same. That is the definition of different numbers. They are different, they are not the same.

I had the intuitive idea that a problem that can be solved “in P” (according to definition) is easier than a problem who’s definition is “not in P”. But this is not always the case. As I said, if we prove that a given problem is “not in P”, it leads to a contradiction. I’m starting to understand why. A function such as c*n (or more generally speaking, a polynomial of n) is thought to be more increasing than a function like n^m (when c and m are considered to be constant). This might be right. But who said c and m have to be constant? Who said they are not allowed to increase as well? Does it matter so much if c is a constant? And remember the sequence a(n) I defined – would you consider a number such as a(1000) to be a constant? And what about a(a(1000))? Is it a constant? Can anybody calculate it in any time?

But the number of algorithms is far more than a(a(1000)). There are more algorithms than any number we are able to represent. Do you really think we are so smart, so that we are able to prove something that cannot be contradicted by any of them? We have to put ourselves inside the equation. If something can be proved, it can be contradicted. Whether the proof and the contradiction are valid, nobody knows. And what matters is what we can do in reality. Whether we can solve a problem in real time or not. It doesn’t matter whether we can prove something like “any function that is executed twice, will return the same answer in both cases” (and if there is such a proof, it can be contradicted). It can be taken as another axiom (and the axiom can also be contradicted). Anything can be contradicted. And the contradiction can be contradicted, too (ad infinitum).

So I guess the class P as defined as a class of decision problems which we are able to solve “in short time” is merely intuitive. Nobody really cares if a problem that can be solved in a(1000) time is proved to be in P. But when we define the problem, we define the solution. For example, consider the definition of prime numbers. The definition defines a function. Another function may calculate the same function more efficiently. But if it can be proved, it can be contradicted. Anything can be contradicted.

I came to the conclusion that the square root of 2 may not be an irrational number. The proof can be contradicted. It is possible that an algorithm that calculates the bits sequence of the square root of 2 may do something like not halting or returning an infinite number of zeros or ones. I don’t think it can ever be proved that 2 algorithms that will calculate the square root of 2 will return the same sequence of bits ad infinitum.

I thought about it, maybe the definition of the square root of 2 depends on the algorithm? Maybe two algorithms will return different numbers, or at least different bits sequence of the same number? I estimated the amount of time to it will take to calculate the first n bits of the square root of 2 at O(n*log₂(n)), but maybe the algorithm is more efficient? Maybe it converges to something in the order of O(n)? If n bits can be calculated in O(n), does it mean that each bit can be calculated in O(1)? It seems to me something quite constant, or at least periodic. The conclusion is probably that the infinite sequence of the bits of the square root of 2 is not something that can be proved to calculated in any nondeterministic way.

Any if it can’t be calculated in any nondeterministic way, in what sense does it exist? It seems to me that God does not only play dice in physics, he does so in mathematics too. Whether all valid algorithms that calculate the square root of 2 will return the same answer, only God knows. Whether any algorithm that is presumed to calculate the square root of 2 is valid, only God knows. Maybe there is no square root of 2, maybe there are many of them. only God knows.

Are there really hard problems? We know there are from our experience. But how do we know they are really hard? Maybe we are just not smart enough to solve them? Maybe we haven’t checked all the possibilities yet?

So I will claim something like this: There is no mathematical proof that hard problems exist. If they do exist, their existence is taken to be as an axiom. It can’t be proved mathematically.

So let’s define what a hard problem is. A hard problem p is any decision problem that is computable, but is proved not to be in O(n). Or more generally speaking, there is a function f which is increasing and unbounded such that for any positive integer N, there is n>N such that the number of steps to calculate p(n) is at least f(N). f can be, for example, n, 2ⁿ, log₂(n) or a(n) defined above. It doesn’t matter, as long as it’s computable.

So lets assume that p is a decision function (again, computable) that is proved not to be in O(n) [or more generally speaking, O(f(n)) ]. So if N is a sequence of natural numbers, there is a computable function n(N) for which the number of steps to calculate p(n(N)) is at least f(N). It can be seen that n defines a subset of N for which for any N, the number of steps to calculate p(n(N)) is at least f(N).

Now, we have a series of bits (0 or 1) p(n(N)) which is proved not to be in O(f(n)). Calculating each bit has been proved to be a hard problem. Now we can define a new subset of n, lets call it m, such that m(n) returns 1 if p(m) = p(1). It can be seen that calculating m(n) takes at least f(N) steps for each n(N), and remember that f(N) is increasing. But how long does it take to calculate p(m) for every m? It takes exactly one step, since p(1) is already known.

So we are now left with an infinite subset of N for which it has been proved that calculating each bit is a hard problem, yet we proved it’s not hard. And if m is not an infinite subset, we can take its complement. The only way to get out of this is to conclude that at least one of the functions we used here is not computable, and therefore not well defined.

But it does not matter how we prove such a thing. Any proof will lead to a contradiction. The only conclusion is that there is no proof that any problem is hard. But does it mean that every problem is easy? It doesn’t. It just means that there is no proof it is hard. We can take one part of the proof for granted, for example the function n, and define its existence as an axiom. Or we can conclude it exists if there is no proof that it doesn’t exist. It’s infinitely complicated. No proof will ever be found.

Even if we allow functions to exist without being computable, their existence leads to the contradiction above. Does it mean that every computable problem is easy? At least in theory they are. No computable problem can ever be proved to be hard. It is possible to display the first million bits of s(n) above in no time. We just don’t know how.

My conclusion is that the general question whether P equals NP can never be solved. Since we like axioms so much, (personally, I don’t), it can be defined as an axiom to be either true or false. It depends what we prefer – if we prefer to be able to solve any hard problem in short time then we can define “P equals NP” as an axiom (and build a corresponding Turing machine). On the other hand, if we prefer to keep using our encryption algorithms without having the risk of other people being able to reveal all our secrets – then we can define “P is not equal to NP” as an axiom (and therefore prevent the existence of a Turing machine that will reveal all our secrets). Neither of them leads to a contradiction.

It seems that deciding whether a given decision problem is hard is itself a hard problem. We should remember that even if a general decision problem is proved to be hard, it is still possible to solve a less general version of the problem in short time. So the question whether a general problem is hard or not is not that important. What is important to know, is whether this problem can be solved in reality.

For example, lets define a decision problem which I assume to be hard. I will define a sequence a(n) recursively: a(0)= 0; and a(n+1)= 2^a(n). Therefore,

a(0)= 0;
a(1)= 1;
a(2)= 2;
a(3)= 4; // seems reasonable to far.
a(4)= 16;
a(5)= 65,536; // Things are starting to get complicated.
a(6)= 2^65,536; // Very complicated.

(and so on).

Now, lets take a known irrational number, such as the square root of 2 (any number can be chosen instead). We know an algorithm that can produce its binary digits, and therefore we can define d(n) as the binary digit n for each n. Now, lets define the sequence s: s(n)= d(a(n)). We come up with a sequence of bits, 0 or 1 (which can also be seen as a function, decision problem etc.) which is computable. But is it computable in reasonable time?

Of course not. At least not in the way we implemented it. Calculating the nth bit of the square root of 2 would take something in the order of O(n*log₂(n)), if we use the algorithm I wrote. But is there an algorithm that can do it more efficiently? Maybe there is, but we don’t know it (it might be possible to prove whether such an algorithm exists). But for any given n, we can produce an algorithm that will return the first n bits of s(n) in the order of O(n) time. All we have to do is to use a Turing machine to calculate the first n bits of s(n), and then produce an algorithm that will display those bits as a sequence. Even though the Turing machine might take a huge amount of time to calculate this algorithm, this algorithm will be in the order of O(n) in both memory space and time. Such an algorithm exists, and it is computable.

If I generalize a little – if we have a decision problem known to be hard, and it has a computable function – for each n there is a computable algorithm that will produce the first n results of the function in O(n) time. Any hard problem can be represented in a way which is not hard. Of course, representing a hard problem in an easy way is in itself a hard problem. But it can be done. And it is computable.

But it is computable for any given, finite n. It might not be computable in the general case. A given decision problem might be really hard in the general case. Even if it is computable, we might not be able to compute it within reasonable time, or memory space (or both). We might use other methods such as nondeterministic algorithms. We might be successful sometimes (sometimes not). But we are really doomed in the general case. Any hard problem can be made harder. Not any hard problem can be made easier.

I will rephrase my last sentence in terms of Turing machines. Suppose there were a Turing machine p, which can take any given decision function f (Turing machine) as input, and return another algorithm that will compute the same function in a shorter time, if such an algorithm exists. Otherwise, it will return f. And suppose we limit ourselves only to algorithms that halt. And suppose f and g are two decision functions who are identical (they always return the same result), and g in known to be more efficient than f. Then we would be able to create an algorithm a that would do the following:

1. Calculate p(a).
2. If p(a) is equal to a, return f.
3. Otherwise, return the complement of p(a).

It can be seen that p(a) will never be correct. If p(a) is equal to a, then it should not be possible to calculate a more efficiently than a itself. Yet, f and g are more efficient versions of a. If p(a) is not equal to a, then a and p(a) are not identical. The conclusion is that there is no algorithm that can determine whether a given function is calculated efficiently.

I have previously claimed that it possible (theoretically, although not practically) to solve the halting problem on real computers. But I forgot to mention something important about real computers – no real computer is completely deterministic. This is due to the uncertainty principle in quantum mechanics, which claims that any quantum event has some uncertainty. Therefore, computers can make mistakes – no hardware is completely reliable.

This doesn’t mean we can’t rely on computers. We can. The probability of a real computer making mistakes is very low, and can be minimized even more by running the same algorithm on more than one computer, or more than once on the same computer. Or actually, it can be done if the number of steps we need to run is reasonable – something in a polynomial order of the number of memory bits we are using. If the number of steps is exponential – if it is in the order of 2ⁿ (let n be the number of memory bits) – then it can’t be done. That is – even if we had enough time and patience to let a computer run 2ⁿ steps – the number of errors we will have will be very high. And of course, we don’t have enough time. Eventually we will either lose patience and turn off the computer, or die, or the entire universe may die. But even if there existed a supernatural being who has enough patience and time – it will find out that any algorithm that is run more than once will return different results each time: in terms of the halting problem – sometimes it will halt, and sometimes not – on random.

This is true for even the most reliable hardware. For any hardware and any number of bits in memory, any algorithm will eventually halt – but if it returns an answer, it will not always return the correct answer. If we allow a computer to run for the order of 2ⁿ steps and return an answer – we will not always get the same answer. But for real computers there is no halting problem – any algorithm will eventually halt.

But if we restrict ourselves to a polynomial number of steps (in terms of the number of memory bits we are using) – then we are able to achieve reliable answers to most problems. So the interesting question is not whether a given algorithm will halt – but if it will halt within a reasonable time (after a polynomial number of steps) and return a correct answer.

But the word “polynomial” is not sufficient. n¹⁰⁰⁰ steps is also polynomial, but is too big for even the smallest n. Whether P and NP, as defined on Turing machines, are equal or not equal we don’t really know – there is currently no proof they are equal and no proof they are not. I’m not even sure whether the classes P and NP are well defined on Turing machines, or more generally speaking if they can be defined. But even if they are well defined, it’s possible that there is no proof whether they are equal or not. It’s like Gödel’s incompleteness theorem – there are statements which can neither be proved nor disproved in terms of our ordinary logic.

When I defined real numbers as computable numbers, I used a constructive approach. My intuition said we should be able to calculate computable numbers to any desired precision (or any number of digits or bits after the dot), and therefore I insisted on having an algorithm that defines an increasing sequence of rational numbers – not just ANY converging sequence. It turns out I was right. Although theoretical mathematics has another definition for convergence, it’s not computable. It’s not enough to claim that “there is a natural number N …” without stating what N is. If we want to compute a number, the function (or algorithm) that defines N (for a given precision required) must be computable too.

It turns out that if we allow a number to be defined by any converging sequence in the pure mathematical sense, then the binary representation of the halting problem can also be defined. This is because for any given n we can run n steps of the first n Turing machines (or until the machines halt) and return 1 if the machines halt, and otherwise 0. It can be proved that this sequence does converge, but it can’t be approximated by any computable function. Therefore, it can be claimed that such a number can be defined in the mathematical sense, although it can’t be computed. But a Turing machine can’t understand such a number, in the sense that it can’t use it for operations such as arithmetic operations or other practical purposes. So in this sense, I can claim that such a number can’t be told in the language of Turing machines.

A Turing machine is not able to specify whether a given algorithm will output a computable number or not (for any definition of computable numbers we can define), since this problem is as hard as the halting problem. And therefore, the binary representation of the computable set (1 for each Turing machine that returns a computable number; 0 for each Turing machine that does not) is itself noncomputable. In other words, the question of whether a given algorithm (or Turing machine) defines a computable number is an undecidable problem. So my question is – are complexity classes such as P and NP well defined? Are they computable and decidable in the same sense we use? Is there a Turing machine which can specify whether a given decision problem belongs to complexity classes such as P and NP and return a correct answer for each input? I think they are not.

If there is no such a Turing machine, then in what sense do P and NP exist? They exist in our language as intuitive ideas, just like the words love and friendship exist. Asking whether P and NP are equal is similar to asking whether love and friendship are equal as well. There is no formal answer. Sometimes they are similar, sometimes they are not. If we want to ask whether they are mathematically equal, we need to check whether they are mathematically well defined.

I was thinking how to prove this, since just counting on my intuition would not be enough. But I came to a conclusion. Suppose there was such a Turing machine that would define the set P – return yes for any decision problem which is in P, and no for any decision problem which is not in P. Any decision problem can be defined in terms of an algorithm (or function, or Turing machine) that returns yes or no for any natural number. We limit ourselves to algorithms that halt – algorithms that don’t halt can be excluded.

So this Turing machine – lets call it p – would return a yes or no answer for any Turing machine which represents a decision problem (if it doesn’t represent a decision problem, it doesn’t matter so much what it will do). Then we would be able to create an algorithm a that would do the following:

1. Use p to calculate whether a is in P.
2. If a is in P, define a decision problem which is not in P.
3. Otherwise (if a is not in P), define a decision problem which is in P.

Therefore, a will always do the opposite of what p expects, which leads to a conclusion that there is no algorithm that can define P. P is not computable, and therefore can’t be defined in terms of a Turing machine.

Is there a way to define P without relying on Turing machines? Well, it all depends on the language we’re using. If we’re using our intuition, we can define P intuitively, in the same sense that we can define friendship and love. But if we want to define something concrete – a real set of decision problems – we have to use the language of deterministic algorithms. Some people think that we are smarter than computers – that we can do what a computer can’t do. But we are not. Defining P is as hard as defining the halting problem – it can’t be done. No computer can do it, and no human can do it either. We can ask the question whether a given algorithm will halt. But we have to accept the fact that there are cases where there is no answer. Or alternatively, there is an answer which we are not able to know.

We can claim that even if we don’t know the answer, at least we can know that we don’t know the answer. But there are cases where we don’t know that we don’t know the answer. Gödel’s incompleteness theorem can be extended ad infinitum. If something can’t be computed, it can’t be defined. Such definitions, in terms of an unambiguous language, don’t exist.

I would conclude that any complexity class can’t be computed. It can be shown in a similar way. So if you ask me whether complexity classes P and NP are equal, my answer is that they are neither equal nor not equal. Both of them can’t be defined.

I have previously mentioned the halting problem – a well-known problem in computer science and mathematics. I claimed that there is a language in which an algorithm that solves the halting problem can be constructed. If we assume that any given algorithm either halts or will run to infinity, then we can construct this simple algorithm:

1. Take an algorithm from input.
2. If it halts, return yes.
3. Otherwise, return no.

It’s a simple algorithm that returns “yes” if a given algorithm halts, and “no” if it doesn’t. But it what language is it written? In English. It requires the knowledge whether a given algorithm will halt. As we assumed, such a knowledge exists, but it has been proved that there is no deterministic algorithm (for Turing machines) that can contain such a knowledge. We can conclude that if this knowledge indeed exists, it is not computable, and therefore can’t be expressed in a finite number of bits (or computer files).

Suppose we try another approach:

1. Take an algorithm from input.
2. Run it one step at a time – either until it halts, or let it run infinite steps.
3. If it halts, return yes.
4. Otherwise, return no.

This algorithm would seem to work and return a correct answer for algorithms that halt, but it might get stuck in an infinite loop for algorithms that don’t halt. But if we allow it to run on a computer that can run infinite steps in finite time – it will always stop and return a correct answer. But the problem is – there is no such a computer.

But there are no Turing machines, either. A Turing machine has infinite memory. Since it has infinite memory, some programs might run for infinite time. In reality – no computer has infinite memory, and no computer program will run for infinite time (somebody will eventually turn off the computer). So lets forget about Turing machines, and check if we can solve the halting problem for real computers.

Real computers have a finite amount of memory. If we are given a real computer and an algorithm that runs on it – is it possible to determine whether this algorithm, when run on the real computer, will ever halt?

Of course it is! let n be the number of bits in this real computer’s memory – so the total number of different states the computer can be in is 2ⁿ. If a computer program hasn’t stopped after 2ⁿ steps, then it is stuck in an endless loop and will run forever. So I will revise my algorithm a little:

1. Take an algorithm from input.
2. Run it one step at a time – either until it halts, or let it run 2ⁿ steps.
3. If it halts, return yes.
4. Otherwise, return no.

This program will always stop and return a correct answer, although it might take a long time. It will also need to use a different computer (or virtual computer) to count the number of steps it is running, and this might take another n bits of memory as well. But I’m not trying to be efficient here. I’m trying to prove that this problem can be solved. And it can be solved.

So what can’t be solved? The question of whether algorithms halt on a Turing machine can’t be solved. The halting function (return yes for any algorithm that halts; no for any algorithm that doesn’t) can’t be computed. The corresponding number can’t be defined in the computer machine language. If there is such a knowledge, it can’t be expressed. Some things we are just not able to know.

But remember – we also don’t know if the googleth bit of the square root of 2 is 0 or 1 – and it is computable (at least in theory, on Turing machines). If something is theoretically computable, it still doesn’t mean that it can be computed in reality and within reasonable time. We just have to accept that some things we are not able to know. It would be boring if we knew everything. If we did, we would have nothing to learn.

I already demonstrated that the numbers of ideas that can be expressed is countable. So how come the number of real numbers is uncountable? In what sense are the real number real? Each real number can be considered as an idea. Can we express an infinite number of ideas in a finite number of words?

But I already said that it all depends on the language we use. Let’s start with checking our definition of numbers. Since we have the inductive definition of natural numbers, and we can define rational numbers by a ratio of two natural numbers – lets check our definition of irrational numbers. There are a few ways to define irrational numbers. Lets define irrational numbers (or more generally speaking, real numbers) as the limit of a known sequence of rational numbers, which is increasing and has an upper bound.

It can be argued that a limit of such a sequence might not be regarded as a number, if it’s not in itself a rational number. But this is just terminology. It doesn’t matter if we call it a limit or a number, as long as we know it exists. It exists in the sense that it is computable – we can calculate it to any desired precision by a finite, terminating algorithm. Or to be more accurate – it is computable if the original sequence of rational numbers can be generated by a known algorithm.

So in the language of computers and deterministic algorithms, we can define any computable number in such a way. Can we define numbers which are not computable? It all depends on our definition of “definition”. While there might exist languages in which such numbers can be defined, my view is that there also exist languages in which such numbers cannot be defined – for example, the language I’m using now. It can be claimed that an infinite definition is also a definition, and these numbers can be defined by the infinite sequence of rational numbers itself. But in reality, it will take us infinite time to express such a definition, and we would need infinite memory to remember it. Therefore, my view is that anything that requires an infinite definition is not real. So lets limit ourselves to definitions of finite length. If there exists a finite algorithm that can define a number (by defining a sequence of rational numbers that converges to it) then this number is computable and therefore definable. If there doesn’t exist such an algorithm – then we cannot define such a number.

So, if I conclude that numbers that can’t be defined don’t exist (because we are not able to express them in the language I’m using), then we come to a conclusion that the set of real numbers is countable. How does it get along with arguments such as Cantor’s diagonal argument, which claim that the set of real numbers is uncountable? Well, while Cantor’s diagonal argument claims that there exists a sequence of rational numbers, which converges to a real number, and is not computable (because the set of computable numbers is countable) – we are not able to express such a set in a finite way in the language I’m using. And since it can’t be expressed, I can claim that it doesn’t exist.

Why doesn’t it exist? Consider this – I can claim that there is a computer language, in which there is an algorithm, that is able to solve the halting problem (decide whether any given computer program will halt). Indeed, there is such a language and algorithm in the same sense that there are numbers which can’t be computed. But there is no computer who runs such an algorithm – it can’t be compiled into known computer languages. So in what sense does such a computer language exist? It doesn’t exist in reality – it exists only in our minds. And therefore, numbers which can’t be computed exist only in our minds, too. They are not real numbers, in the sense that they are not real. They are real in the same sense that a computer who solves any problem is real.

Is there a limit to human knowledge? I would rather rephrase this question: Is there a limit to the knowledge that can be expressed in human language? While some people might think that the potential of our knowledge and wisdom is unlimited, I will demonstrate that it is.

It is well known that many aspects of human communication can be expressed in computer files – including written language, spoken language including music and songs, visual pictures, movies and books. Is there a limit to the amount of information that can be expressed in files? We all know that computer files can be represented as a sequence of binary digits (bits). Each sequence of binary digits can also be viewed as a positive integer number (a natural number). While some files might contain millions or even billions of bits – their size is always finite. A computer file cannot contain an infinite number of bits.

So it seems that any idea, any piece of knowledge or information that can be expressed in words (or any other form of communication that can be represented in files) can be represented as a sequence of binary digits, or a natural number. But the number of natural numbers is countable. Even more – the number of natural numbers which can actually be represented in reality is at least limited by the number of particles in our known universe, which is finite. We come to a conclusion that the number of ideas that can be expressed in words is finite, or at most countable (if we don’t put a limit on the number of words we use to express one idea).

Even more – since the sequence of natural numbers is already known to us, and we can produce a simple computer program or algorithm* that will express all of them (if allowed to run to infinity) – then the entire knowledge and ideas that can be expressed in words is already known as well. All the words have already been said, all the books have already been written, all the movies have already been created and seen – by a simple algorithm that counts the natural numbers to infinity. Nothing is new – everything is already known to this algorithm, and therefore to us. Just like nothing is new with any natural number – nothing is new with any sequence of binary digits, or with any computer file.

This also eliminates our concept of authors, or copyrights. Can a number have an author? Can it be copyrighted? I can prove by induction that no number has any author and no number has copyrights. Since we all know that 0 and 1 are not copyrighted, and since nobody can claim he’s the author of either of them – then if we have a sequence of bits which has no author, and is not copyrighted, and we add to it another bit (either 0 or 1) – then it’s easy to conclude that the new sequence too has no author and is not copyrighted. Can one claim to be the author of a sequence of bits in which only one bit he wrote by himself? Compare it to taking a book someone else wrote, and adding one letter. Can you claim that you wrote the entire book? Of course not. And if nobody wrote the original book, and you added one letter – then nobody wrote the new book, too. Or compare it to numbers. If you add one to a well-known natural number, can you claim that the new number is yours? Can you claim that you wrote it, and nobody has the right to write the same number for the next 50 years? Of course not.

If we were able to claim copyrights on natural numbers, then I would be able to claim that the algorithm that outputs the entire sequence of natural numbers is mine, I wrote it, and therefore the entire sequence of natural numbers is mine. Nobody is allowed to write any number for the next 50 years. Would you allow something like this? Of course not. Then we should conclude that no knowledge is new, everything is already known, no book has an author and nothing is copyrighted.

* Here’s my algorithm in the awk language, in case you’re interested:
for (i=0; 1; i++) {print i;}

After reading what I wrote about the axiom of choice about two years ago, it appears to me that I forgot to mention something important. I claimed that there are numbers which we are not able to tell. But it all depends on the language. It can be proved that for each real or complex number, there exists a language (or actually, there is an infinite number of languages) in which this number can be told. It’s very easy to prove – we can just define a language in which this number has a symbol, such as 1, 0 or o. We tend to look at some symbols as universal, but they are not. For example, the digit 0 means zero in English, but five in Arabic. The dot is used for zero in Arabic. So defining numbers depends on the language we use.

But, since there are infinite possible languages, the language itself has to be defined, or told, in order for other people to understand. We come to a conclusion that the ability to tell numbers depends on some universal language, in which we can tell the number either directly, or by defining a language and then tell the number in this language (any finite number of languages can be used to define each other). But in order to communicate and understand each other, we need a universal language in which we can start our communication.

It still means we are not able to communicate more than a countable number of numbers, or a finite number of numbers in any given finite binary digits or time, but the set of the countable (or finite) number of numbers that we can communicate depends on the language we use. For example, if we represent numbers as rational numbers (as a ratio of two integers) then we can represent any rational number, but we can’t represent irrational numbers such as the square root of 2. But if we include the option of writing “square root of (a rational number)” then we can represent also numbers which are square roots of rational numbers. In this way we can extend our language, but it’s hard to define which numbers we are able to define in non-ambiguous definitions. An example of a set of numbers we can define in such a way are the computable numbers.

In any case, for any language the number of numbers that can be defined in it is countable, and we can conclude that any uncountable set has an (uncountable) subset of numbers which can’t be defined in this language. If we subtract the set of numbers that can be defined from the original uncountable set, we can define an infinite set of numbers, none of which we are able to define or express. If there are languages in which these numbers can be expressed – these languages too can not be expressed in the original language.

It’s similar to what we have in natural languages. Some expressions (or maybe even any expression) can’t be translated from one language to another. For example, in Hebrew there is no word for tact. The word tact is sometimes used as it is literally, but this is not Hebrew. There are many words in Hebrew, and any language, from other languages. But the Hebrew language itself does not have a word for tact.