Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 143

 
Mathemat:

Another mind-boggling problem about megamooks and invaders:

(5) A hundred megabrains had caps with numbers from the range 1...100 put on their heads, not necessarily different for everyone. For example, all of them may be given a cap with number 7, or half of them may be given a cap with number 20, and the other half with number 10. The main thing is not less than 1 and not more than 100. After that they were all put in a circle. Each megabrain sees 99 numbers on the heads of the others, but not his own. After that everyone writes a number from 1 to 100 on a piece of paper - the supposed number on his/her cap. Communicating and peeping is not allowed ;) They will all be let go if at least one guesses their number. What strategy should they follow if they want to be guaranteed to be let go? (The megamooks could have agreed on the strategy beforehand.

Comment: After they put the caps on (consider that it happened instantly), the megamooks don't pass any information to each other. They just watch and count and then write their numbers.



2 Mathemat: please post "decisions" given or taken by "moderators"...
Why? Judging by the problems and answers smells like a sect of self-righteous "moderators" of a commercial site - a simple example: I can prove that the above problem has no solution. What did they take there?
When I studied, I often argued with teachers (at least a doctor of sciences), and when I was sure of his correctness I stupidly went to Belotserkovsky and he arranged a council, where I won a couple of times... And where will you go if the "moderator" is wrong?
 

moby_dick: Судя по задачам и ответам пахнет сектой самоуверенных "модераторов" коммерческого сайта - простой пример: могу доказать что вышеприведённая задача решения не имеет. А что приняли там?

[...] when I was sure that I was right, I went to Belotserkovsky and he arranged a consultation where I won a couple of times...

Go ahead, I would love to read your evidence. I'm 100 percent sure of mine. I went to school #18, if you've heard of it.

Did you go to a physics school?

Judging by the tasks and answers, it smells like a sect of self-righteous "moderators" of a commercial site.

There is that, I'm willing to admit. Sometimes they make not fully informed "decisions", I've seen it for myself. However, not everyone is like that, there are quite objective guys.

But as far as commercial orientation... I highly doubt it. They are actively looking for sources of funding.

 
Mathemat:

Once again, either you have not read carefully:

Comment: no considerations of terrain continuity apply. Brainiac may well turn out to be extremely rugged in elevation - like a Dirichlet function, for example (this function is not continuous at any point).



I didn't use "considerations of relief continuity" - if you can't divide Brainiac into arbitrary squares then there are many non-continuous solutions, e.g: \\\\\\\\\ (45g squares).
I only used the absence of constraints on the partitioning of squares, which in the correct problem is equivalent to arbitrary...
 
moby_dick: I didn't use "considerations of continuity of relief" - if you can't divide Brainiac into arbitrary squares then there are many non-planar solutions, e.g: \\\\\\\\\ (45g squares).
I only used the absence of constraints on squaring, which in the correct problem is equivalent to arbitrary...

There is a clarification from the moderators, which I have not written here: the square in question is located on the plane, i.e. on the map. It is not on the surface of the terrain.

And second: the statement of the character of the problem is about any square, not about a square of allocated size.

 
Mathemat:

Go ahead, I'd love to read your proof. I'm 100 percent sure of mine. I went to school #18, if you've heard of it.

You went to phstech? Yes.

The proof is very simple: the number on everyone's cap by the condition has nothing to do with the numbers of others, so assuming that someone has calculated his number, the occupant only needs to rewind time and change this number to any other number and no one can warn him, which leads to a contradiction...
 
moby_dick: The proof is very simple: the number on everyone's cap is by convention unrelated to the numbers of others, so assuming someone has calculated his number, the occupant need only rewind time and change that number to any other number and no one can warn him, which leads to a contradiction...

You have the wrong premise: You somehow think that someone is obliged to calculate their number accurately. This is not necessary at all. The important thing is that the calculated number and the real number coincide in at least one person who doesn't even know it.

And second: who told you that occupant can rewind time?

 
Mathemat:

You have the wrong premise: You somehow think that someone is obliged to calculate their number accurately. This is not necessary at all. The important thing is that the calculated number and the real number coincide in at least one person, who does not even know about it.

And second: who told you that the occupant can rewind time?



That's the "at least one" I was talking about... it doesn't matter if he can rewind or not, what is important is that if he rewinds, he will have to change the whole megaalgorithm to infinity - this is a contradiction, which means there is no algorithm (unlike first mega-summer where there is a connection between numbers)...
For some reason it seems to me that you are being led away from mathematics...
 
moby_dick: which means that there is no algorithm (unlike first mega-sum problems where there is a connection between numbers)...

Too categorical.

It's pure mathematics, no cheating and no backtracking. Put the caps on and that's it, it's done. Next, the mega-mosks work - and the occupiers can no longer influence events.

The solution has already been published here. I can repeat it especially for you.

Note: here is a more accurate formula:

calc(n) = (n - S_n) mod 100 + 1.

And from the penultimate sentence remove the phrase "the sum of them modulo 100".

Look for the error.

 
Mathemat:

Overly categorical.

It's pure mathematics, no cheating and no backtracking. The caps are put on and that's it, it's done. Next, the mega-mosks work - and the occupiers can no longer influence events.

The solution has already been published here. I can repeat it especially for you.

Note: here is a more exact formula:

calc(n) = (n - S_n) mod 100 + 1.

And from the penultimate sentence remove the phrase "their sum modulo 100".



Now I understood the value of this branch - if I lacked the thrill of trading and poker, I could argue for money... :))
Beware, you are too categorical and such a player may be found - ask the moderators if they will support you...
 
moby_dick: Now I understand the value of this thread - if I lacked the thrill of trading and poker, you could bet money... :))
Beware, too categorical you and such a player can be found - ask the moderators, whether they will support you ...

Already supported: the task is bluntly credited on the first try. And in the comments for those who solved it someone posted the same solution too.

Have you found a definite error in my reasoning - or will you continue to philosophize? Well, it is somehow unseemly for a graduate of a physics department, arguing with Belotserkovsky himself...

And, by the way, what about the topography of Brainiac? There is a solution available to a 6th or even 8th grader. No higher matter.