[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 368
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Помогите!!!! Час уже себе мозг ломаю!!!! Подумайте еще кто нибудь! Условия задачи вообще со одними переменными :))) Про двери не реально было самому вопрос придумать, а тут ..... !
This is one of the variants of the problem, which demonstrates the power of the construction excluding or. But this is the first time I've seen it in this formulation. I have to go this way Asking a question And what answer will B give me to the question he is a god of truth?
Всего то час?!
Хехх, вы трейдер или хто?
What's that got to do with it! Even if you are a compressor operator :)
If these gods answer in Russian, the questions and the algorithm seem clear! But here's the thing about their peculiar language, it makes my head spin!
это один из вариантов задачи который демострирует силу конструкции исключающее или. Правда в такой постановке я её встречаю впервые. Нужно идти путем типа Задаю вопрос А что мне ответит B на вопрос он бог истины ?
Это то при чем! Даже если машинист компрессорных установок :)
Если эти боги отвечают на русском языке, то вопросы и алгоритм кажется понятны! Но вот весь прикол в их особенном языке,тут у меня процессор в голове дымится!
Да я ж угараю, пардон. Терпения и выдержки коснутся темы имел желания я.
Ughhhh, Guys, I caught one of these things today - you're going to love it :)))))))))
Backstory:
Walking home. There's a convenience store on my way home. Passing by - there are some young people sitting - deciding something on a stool. I decided to take a look and got stuck. What is the gist of it?
So, man, you sit down on a stool, you put a stool in front of you. You take a match and put it upright in front of you. At the very top of the stool, so that you see it as a vertical line.
Under that match, you put three more matches, just as vertically oriented. Under them, five matches. And below them, seven.
So you have a pyramid - one at the top, seven at the bottom. Now the rules of the game. We take turns. It doesn't matter who moves first. For one move each player has the right to remove any number of matches from the stool, but only from one row (horizontal). The loser is the last player to draw a match from the stool.
The problem hooked me because it solves not only the question of programming, but also the modeling of artificial intelligence.
The guy who played against everyone always won. He got enough beer to get half of Beijing drunk. He's got some scheme in his brain that's a hundred percent working.
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P.S.
Corrected the post.
I forgot to say - that guy claimed that it was possible to beat him! And then I remembered that some time ago, when studying cybernetics, I encountered a problem of such class and its solution was given in the form of a closed graph-scheme. At that time I diligently took notes of interesting things. If the abstract is still alive, I will surely show it.
The man who played against everyone always won. He got enough beer to get half of Beijing drunk. There's a scheme in his brain that's a hundred percent working. If you solve it (along with me), I'll show you another trick that I remembered from my childhood, which is also so twisted and also has a win-win option.
In my opinion, you have to make your move in such a way that after it:
1) there's an odd number of rows left;
2) if during the move the row is not completely removed, then it must remain 2 matches.
PS. I understand that there are two players.
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1. If there is only one row with more than one match left, the one who moves right now wins: he simply takes all but one, and there is one match left, which the opponent will take.
2а. If there are two rows left, at least one of which has one match (1,n), then the one who moves now wins again, taking the row n.
2б. If (2,2), then the player always loses, in case of the opponent's optimal game. So, he must not allow such an arrangement before his move.
2в. If (2, m>2), then the walker now makes (2,2) and wins.
2г. If (n>2, m>2), then the walker now just has to equalize the quantities if he gets them. If they are equal, he loses. It is proved by induction. So, he can't allow the opponent to do that.
3. With three rows - more complicated. I wrote some nonsense here, but now I've erased it.
Corrected my post....
I forgot to say - that man claimed that it was possible to beat him! And then I remembered that some time ago, while studying cybernetics, I encountered a problem of a similar class and its solution was given in the form of a closed graph-scheme. At that time I diligently took notes of interesting things. If it is still alive, I will surely show the solution, because it seems to be exactly like that.
Of course you can - if your opponent also has an optimal strategy. And it also seems to depend on who moves first.