How to solve one of the most important puzzles in mathematics - MQL4 and MetaTrader 4

Sceptic Philozoff 2011.01.22 11:36 #4591

MetaDriver: (пост от 16.01.2011 04:14)

2011.01.16 03:41:44 MetaSage (EURUSD,H1) Test =>..... etc. All other choices are false, for even.
2011.01.16 03:41:40 MetaSage (EURUSD,H1) Test => 2+888=890 false
2011.01.16 03:40:02 MetaSage (EURUSD,H1) Test => 111+16=127 true
2011.01.16 03:39:23 MetaSage (EURUSD,H1) Test => 3+592=595 false
2011.01.16 03:38:08 MetaSage (EURUSD,H1) Test => 37+48=85 false
2011.01.16 03:38:08 MetaSage (EURUSD,H1) S=127; P=1776; a=16; b=111

S=127, P=1776 (numbers - 16 and 111) cannot be a solution.

A: (1776=16*3*37.) Don't know.

B: (127 = 2+ odd_component.) I knew it without you.

A: (So the sum is 2+ odd_component. 1776 = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Only the highlighted one with 16*111 is appropriate). Know the numbers.

B: (Here I will indicate only two variants of a complete search, which are sufficient:

127=2+125. P(=2*5*5) = 2*125 = 10*25 = 50*5. The sums are 127, 35, 55. Only one - the allocated one - is allowed. The sum of 35 is unacceptable because 35=4+31=16+19=32+3 (ambiguous representation by the sum of the powers of two and a prime). Candidate (numbers are 2 and 125).

127=16+111. П(=16*3*37) = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Similarly. Candidate (the numbers are 16 and 111.) ) I don't know.

___________________________________

The consolation for you is the unrepresentability of 127 as the sum of a degree of two and a prime. There aren't many numbers like that, but they're not too rare.

Check S=373; P=19776; a=64; b=309. This is the second version of your solution with a composite odd number, which I doubted.

The first two lines pass. The third:

А: (19776(=64*3*103) = 64*309 = 192*103 = 6592*3. The sums are 373, 295, 6595. Only the allocated one fits. The last amount, by the way, is not included in the eligible even if the restrictions on amounts are removed. So 64 and 309) . Knowing the numbers.

Haven't figured out the rest yet. But going to the last B calculations, we already know that one split of the sum 373=64+309 we have already checked and we have the first candidate.

P.S. Let's try to guess (just find one other example with a single matching sum):

Б: 373 = 32+341. П(=32*11*31) = 32*341 = 352*31 = 992*11. The sums are 373, 383, 1003. Only the highlighted one fits. The other two do not, but for a more subtle reason: each of them is ambiguously decomposed into the sum of the powers of two and the prime. I've already written about this additional filter here. So we have another candidate for a pair of conceived numbers - 32 and 341. Consequently, sage B will not be able to calculate the pair of conceived ones.

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Sceptic Philozoff 2011.01.22 17:53 #4592

MD, judging by the listing, you check only one product for possible decompositions. I.e. you do the work of sage A.

What about B's work before his last line? Recall what his reasoning is. Let it be the variant S=373; P=19776; a=64; b=309.

Sage B has only the amount given to him - 373. And there is information that A, using B's previous tip, made sure that the product 19776=64*3*103 among all variants of the expansion into 2 multipliers has the only admissible sum. Sage A almost didn't have to work, because it was enough for him to check only three variants. What does B do now?

He has to go through all the decompositions of 373 into 2 summands. These are 2+371, 3+370, 4+369, ... 186+187. That's 185 choices in all.

For each variant he has to multiply the summands, and then do what A did earlier. Here, for example, variant 134+239.

1. Calculate the product (P=2*67*239).

2. Go through the variants of grouping - 2*16013, 67*478, 134*239.

3. We calculate the corresponding sums - 16015, 545, 373.

4. Two sums are admissible - 545, 373. Hence, the "134+239" variant is dropped.

That was only one variant. Then he has to go through the next ones on the list.

And only when among all these 185 variants he will have only one with a single admissible sum, he can say his line. (Note: after checking the option "32+341" and seeing that there is a single valid sum, he cannot stop and declare that he knows the numbers. He has to go all the way and check, perhaps, all the others: what if there are more variants with one permissible?)

So far I have found only one more or less rigorous reasoning on the net. The author is Konstantin Knop. It's here. The reasoning is a bit more complicated than mine, but for the "sum less than 100" constraint he strictly brings it to an end. However, for sums withlarger constraints he only has a few hypotheses. And an appeal to a computeraire too...

Pure maths, physics, chemistry, 1st and 2nd derivatives How can I tell

Vladimir Gomonov 2011.01.22 19:51 #4593

Mathemat:

MD, judging from the listing, you only check one product for possible decompositions. I.e. you are doing the work of Sage A.

What about B's work before his last line? Recall what his reasoning is. Let it be the variant S=373; P=19776; a=64; b=309.

Sage B has only the amount given to him - 373. And there is information that A, using B's previous tip, made sure that the product 19776=64*3*103 among all variants of the expansion into 2 multipliers has the only admissible sum. Sage A almost didn't have to work, because it was enough for him to check only three variants. What does B do now?

He has to go through all the decompositions of 373 into 2 summands. These are 2+371, 3+370, 4+369, ... 186+187. That's a total of 185 choices. // see golden commentary

For each variant he should multiply the summands and then do what A did earlier. Here, for example, variant 134+239.

1. Calculate the product (P=2*67*239).

2. Go through the variants of grouping - 2*16013, 67*478, 134*239.

3. We calculate the corresponding sums - 16015, 545, 373.

4. Two sums are allowed - 545, 373. Hence the 134+239 option is discarded.

That was only one variant. Then he has to go through the next ones on the list.

And only when among all these 185 variants he will have only one with a single admissible sum, he can say his line. (Note: after checking the option "32+341" and seeing that there is a single valid sum, he cannot stop and declare that he knows the numbers. He has to go all the way and check, perhaps, all the others: what if there are more variants with one permissible?)

So far I have found only one more or less rigorous reasoning on the net. The author is Konstantin Knop. It's here. The reasoning is a bit more complicated than mine, but for the "sum less than 100" constraint he strictly brings it to an end. However, for sums withlarger constraints he only has a few hypotheses. And an appeal to a computeraire too...

Not so. Here is the basic checking procedure (see below). It tests fairness of the third (A) and fourth (B) replica at once.

The outer loop checks fairness of replica 4 (if the Count variable at the end of the big loop == 1)

The inner loop checks the fairness of cue 3 (if the count variable at the end of the inner loop == 1)

See the green comments in the text below.

   uint GetCountValidSum(uint n,uint &P,uint &a,uint &b)
     {
      uint Count=0;
      //       for(uint i=2;i<=sqrt(n);i++)  // ОШИБКА!! 
      for(uint i=2;i<n/2;i++) // Правильно так.                  // Внешний цикл
                                                         // проверяет все разбиения суммы на 2 слагаемых. 
         {
         uint count=0;
         sMX J;
         J.Join(MX[i],MX[n-i]); // объединяем множители слагаемых // 1. Вычисляем произведение (P=2*67*239). 
         for(uint j=1; j<=J.GetCountAllSums(); j++)              // Внутренний цикл
                                                      // 2. Перебираем варианты группировки - 2*16013, 67*478, 134*239. 
            count+=IsValidSum(J,j); // j - номер суммы      // 3. Вычисляем соответствующие суммы - 16015, 545, 373. 
         if(count==1)  // это условие истинно только если для конкретного набора множителей существует только одна валидная сумма
           {           // т.е. если это так - мудрец А сможет однозначно определить числа
            Count++;
            P=J.Value();
            a=i;
            b=n-i;
           }
        }
      return Count;  // А вот если таких произведений, для которых мудрец А способен найти решение после второй реплики только одно
     }               // т.е. Count==1  тогда и мудрец В сможет однозначно найти решение

Something like this. :)

In red, I have copied your findings as a comment to the text of the procedure, so as to link it to the ground.

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Vladimir Gomonov 2011.01.22 20:21 #4594

Mathemat:

S=127, P=1776 (numbers - 16 and 111) cannot be a solution.

A: (1776=16*3*37.) Don't know.

B: (127 = 2+ odd_component.) I knew without you.

A: (So the sum is 2+ odd_component. 1776 = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Only the highlighted one with 16*111 is appropriate). Know the numbers.

B: (Here I will indicate only two variants of a complete search, which are sufficient:

127=2+125. P(=2*5*5) = 2*125 = 10*25 = 50*5. The sums are 127, 35, 55. Only one - the allocated one - is allowed. The sum of 35 is unacceptable because 35=4+31=16+19=32+3 (ambiguous representation by the sum of the powers of two and a prime). Candidate (numbers are 2 and 125).

127=16+111. П(=16*3*37) = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Similarly. Candidate (the numbers are 16 and 111.) ) I don't know.

___________________________________

The consolation for you is the unrepresentability of 127 as the sum of a degree of two and a prime. There aren't many numbers like that, but they're not too rare.

Check S=373; P=19776; a=64; b=309. This is the second version of your solution with a composite odd number, which I doubted.

The first two lines pass. The third:

А: (19776(=64*3*103) = 64*309 = 192*103 = 6592*3. The sums are 373, 295, 6595. Only the allocated one fits. The last amount, by the way, is not included in the eligible even if the restrictions on amounts are removed. So 64 and 309) . Knowing the numbers.

Haven't figured out the rest yet. But going to the last B calculations, we already know that one split of the sum 373=64+309 we have already checked and we have the first candidate.

P.S. Let's try to guess (just find one other example with a single matching sum):

Б: 373 = 32+341. П(=32*11*31) = 32*341 = 352*31 = 992*11. The sums are 373, 383, 1003. Only the highlighted one fits. The other two do not, but for a more subtle reason: each of them is ambiguously decomposed into the sum of the powers of two and the prime. I've already written about this additional filter here. So we have another candidate for a pair of conceived numbers - 32 and 341. Consequently, sage B will not be able to calculate the pair of conceived.

Lyosha, your (and Knopov's) criterion about the uniqueness of the decomposition by the sum of the powers of two and a prime. is an unproven hypothesis.

That this is often true is not a proof. So - either a proof in the studio, or a full brute force test on the computer. The second is preferable because it does not need proof by the fact of presentation. It does not pass my test.

By the way, the program is debugged - servicedesk found the error. It turned out to be mine (I needed to zeroize memory before sorting in test procedure), I fixed it.

Prog in the trailer.

Files:

metasage_6.mq5 6 kb

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Sceptic Philozoff 2011.01.22 21:05 #4595

MetaDriver:

Lyosha, your (as well as Knopov's) criterion about the uniqueness of the decomposition by the sum of the powers of two and a prime is an unproved hypothesis.

It's not mine, I got it from you :) The short formulation is: if the decomposition is ambiguous (there are several ways), then the sum is invalid. Are you ready to refute it? Go ahead, I'm waiting for an example.

I've already posted my way of using decomposition by the sum of powers of two and prime. There is almost no proof, but there is a practical way of using observation, which is 100% reasonable. See highlighted in green.

I'm copying it here so I don't have to click on the links:

In fact there is a more general observation (it can be seen from the MD printout): probably all reasonable choices are restricted to pairs of numbers 2^n and p (prime). I haven't proved it, I'm just assuming it.

Now, based on that assumption, let's do something real. The most difficult thing in the dialogue of the wise men is the last line. It is the one which so far requires that many options be considered. Let us assume that we have already had three replicas and only the last one remains. How many sums from MDS can be represented as 2^n + prime?

Why this particular decomposition? Simply because B in the last line, considering possible decompositions of sums (see my previous post) and corresponding products, having met product 2*...*2*simple, already knows in advance that only one of sums for it can be admissible, since only one is odd - if numbers are equal to powers of two and odd prime. This immediately gives a real candidate.

So, let's go.

11 = 2^2+7 = 2^3+3. There are two candidates. Bummer immediately.

17 = 2^2+13. There are no more such submissions. Good candidate.

23 = 2^2+19 = 2^4+7. Bummer.

27 = 2^2+23 = 2^3+19 = 2^4+11. All the more bummer.

29 = 2^4+13. Submission alone. Another candidate.

35 = 2^2+31 = 2^4+19 = 2^5+3. Bummer.

37 = 2^3+29 = 2^5+5 . Bummer.

41 = 2^2 +37. Submission singular. Candidate.

47 = 2^2+43 = 2^4+31. Bummer.

51 = 2^2+47 = 2^3+43 . Bummer.

53 = 2^4+37. Submission is singular. Candidate.

So out of all the MDS, we are left with only 4 admissible sums - 17, 29, 41, 53.

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Sceptic Philozoff 2011.01.22 22:58 #4596

I'm confused. The thoughtless application of different filters can lead to nonsense.

Vladimir Gomonov 2011.01.22 23:31 #4597

Mathemat:
I'm confused. The thoughtless application of different filters can lead to nonsense.

Well, sort of. I agree that if there are multiple valid decomposition methods, then the option is invalid.

But this only applies to valid criteria, e.g. S="2+combinable odd". For this criterion the corresponding lemma is strictly and correctly proved.

The criterion "degree of two + prime" does not appear in the conditions of the problem and is not a proved lemma. It is simply a property of most solutions. But not all of them, as it turned out.

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Sceptic Philozoff 2011.01.23 00:24 #4598

MetaDriver: But this only applies to valid criteria, e.g. S="2+complete odd". The relevant lemma is strictly and correctly proved for that criterion.

Well, thank you, at least this was looked at...

The criterion "degree of two + prime" does not appear in the conditions of the problem and is not a proved lemma. It is simply a property of most solutions. But not all, as it turns out.

And here you haven't looked at it. I have it as an anti-criterion - proved strictly and correctly. Try it yourself, if you don't want to see my proof (it's in the post above, in green):

If the sum is representable as the sum of the degree of two and the prime in several ways, then that sum is invalid after the third rejoinder.

Notice, I'm not talking about sums represented in this way in a single way...

P.S. Revisited my refutation of your "solution" 16, 111. I don't see any errors there yet. I copy here:

S=127, P=1776 (the numbers are 16 and 111) cannot be the solution.

A: (1776=16*3*37.)

B: (127 = 2+ odd_component.) I knew [ that you don't know] without you.

A: (So the sum is 2+ odd_component. 1776 = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Only the highlighted one with decomposition 16*111 fits, since 85-2 and 595-2 are prime). Know the numbers.

B: (Here I will point out only two variants of the complete search, which are sufficient:

127=2+125. P(=2*5*5*5) = 2*125 = 10*25 = 50*5. The sums are 127, 35, 55. Only one is admissible - the highlighted one. The sum of 35 after the third rejoinder is not allowed, because 35=4+31=16+19=32+3 (ambiguous representation by the sum of the powers of two and a prime). Candidate (the numbers are 2 and 125).

127=16+111. П(=16*3*37) = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Similarly. Candidate (the numbers are 16 and 111.) ) Don't know.

Do you accept this as a correct rebuttal, MD?

Deadlock Pure maths, physics, logic Strange Issue .TP is

Vladimir Gomonov 2011.01.23 01:57 #4599

Mathemat:

Do you accept this as a correct rebuttal, MD?

I don't think so.

S=127, P=1776 (the numbers are 16 and 111) cannot be the solution.

A: (1776=16*3*37.) I don't know.

B: (127 = 2+ odd_component.) I knew [ that you don't know] without you.

A: (So the sum is 2+ odd_component. 1776 = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Only the highlighted one with decomposition 16*111 fits, since 85-2 and 595-2 are prime). Know the numbers.

B: (Here I will point out only two variants of the complete search, which are sufficient:

127=2+125. P(=2*5*5*5) = 2*125 = 10*25 = 50*5. The sums are 127, 35, 55. Only one is admissible - the highlighted one. The sum of 35 after the third rejoinder is not allowed, because 35=4+31=16+19=32+3 (ambiguous representation by the sum of powers of two and a prime). Candidate (the numbers are 2 and 125).

127=16+111. П(=16*3*37) = 16*111 = 48*37 = 592*3. The sums are 127, 85, 595. Similarly. Candidate (the numbers are 16 and 111.) ) Don't know.

There is a logical error here.

The sum of 35 is perfectly acceptable on this turn of reasoning, for in his third line, sage A has only one criterion - the sum of known B = 2+ odd-composite.

35=2+33=2+3*11, hence the decomposition 2+125 is invalid for both 127 and 35 are valid. That leaves 16 and 111.

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Sceptic Philozoff 2011.01.23 03:14 #4600

Taking a break. I feel like I've done something wrong, but I can't figure out what it is yet :)

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