Comments on: Minipolymath4 project: IMO 2012 Q3 http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/ Massively collaborative mathematical projects Sun, 19 May 2013 02:40:38 +0000 hourly 1 http://wordpress.com/ By: Anonymous http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-9479 Wed, 22 Aug 2012 15:27:23 +0000 http://polymathprojects.org/?p=304#comment-9479 1) for the case of k consecutive responses one of which is true:
we supoose k = 0, so all answers are true, so we can know the exact number x, so n = 1 ≥ 2 ^ 0

I now propose a case that will help me in my demonstration:
2) For k+2 consecutive responses which one is true and one is false :
we assume that k = 0, so two consecutive responses contiennt true and the other false, so thanks to our given 1 ≤ x ≤ N, making this series of questions:
* X = 0?
* 1 ≤ x ≤ N?
* X = 1?
* 1 ≤ x ≤ N?
.
.
.
I will know the exact number x, so in this case n = 1 ≥ 2^0

*** Now assume that for k+1 consecutive responses which one is true n must be n ≥ 2 ^ k
one must demonstrate that responses to k+2 n in this case must be n ≥ 2 ^ (k +1)

and *** k+2 consecutive responses which one is true and one is false n must be n ≥ 2 ^ k
we must show that for k+3 consecutive responses in this case n must be n ≥ 2 ^ (k +1)

1 *) where k+2 consecutive responses which one is true can be divided into two cases:
— K 2 answers are true then n ≥ 1
— K 2 answers contiennet true and another false must therefore n ≥ 2^k
so it is sufficient to write n ≥ 2^(k+1)

2 *) where k+3 consecutive responses which one is true and the other must be false can be divided into two cases:
— K+2 answers are wrong and one seulle is true: in this case the placement of the true answer is the same after each k+2 answers, so to see her placement just ask the same question k+2 times and the correct answer will be different among the answers … and how we can know the exact number x, so n ≥ 1
— K+3 answers contain at least two answers true and one false: this case is a special case of the general case (k+2 consecutive responses of which is true), so n ≥ 2 ^ (k +1)

It’s finished

]]> By: prateekchandrajha http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-8392 Mon, 30 Jul 2012 22:58:56 +0000 http://polymathprojects.org/?p=304#comment-8392 Reblogged this on Wikipedia Afficianado and commented:
IMO is the Mecca of young mathematicians battling out in this divine field of which I am in oblivion of until now. Whenever I try and study mathematics it is with a notion of solving a problem and that problem is hard enough for veterans to try but what I have come to know from those who do “Research” is that they don’t do it to solve the problem but to firstly understand it well and secondly to find why is that problem tough than what it seems to be. Terence Tao as you all know is a known child prodigy and inculcated abilities to solve problems involving numbers at a very young age. He id the youngest even to have received a fields medal. This Re-Blogged post concerns a question which appeared in this year’s IMO (International Mathematical Olympiad in case you are not familiar with what it is) and a good thread to discuss what comes to your mind while approaching it.

]]> By: Game Theory at the Polymath Project « The Leisure of the Theory Class http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7703 Fri, 13 Jul 2012 20:47:36 +0000 http://polymathprojects.org/?p=304#comment-7703 [...] (Rubinstein would say that this is true of most real-life applications of game theory as well.) Try your hand, or look at the comments, which surely have spoilers by now as it has been up for about a day: [...]

]]> By: Gagik Amirkhanyan http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7696 Fri, 13 Jul 2012 20:14:40 +0000 http://polymathprojects.org/?p=304#comment-7696 I think it’s correct solution, just in the definition of x_i^{j+1} should be P_i instead of S. [Corrected, -T.]

]]> By: Minipolymath4 project, second research thread « The polymath blog http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7694 Fri, 13 Jul 2012 20:04:16 +0000 http://polymathprojects.org/?p=304#comment-7694 [...] the previous research thread is getting quite lengthy (and is mostly full of attacks on the first part of the problem, which is [...]

]]> By: Gagik Amirkhanyan http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7693 Fri, 13 Jul 2012 20:04:11 +0000 http://polymathprojects.org/?p=304#comment-7693 We use a greedy approach to choose D_i
We choose
D_1 = S_1 if | S_1 | > N/2 t, otherwise D_1 = Compliment(S_1).
To pick D_2 we can see whether S_2 or complement (S_2) covers at least 1/2 portion of [1, N] \ D_1
We pick D_{i+1} such that it covers at least 1/2 portion of D\(D_1 u D_2 … u D_i}.
We can claim that in at least p = log_2 N steps D_1 u D_2 … u D_p = [1, N]
Where p = k log_2 (1.99).
It means that for each of the numbers [1, N] we A gave at least one correct answer in the first p steps.

Because p > k / 2, it will not imply part b), probably some modifications are needed.

]]> By: Terence Tao http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7691 Fri, 13 Jul 2012 19:56:33 +0000 http://polymathprojects.org/?p=304#comment-7691 Dear all,

As this thread is becoming quite full, I am opening a fresh thread at http://polymathprojects.org/2012/07/13/minipolymath4-project-second-research-thread/ to refocus the discussion. I’ll leave this thread open for responses to existing comments here, but if you could put all new comments in the new thread, that would be great. (Now would also be a good time to resummarise some of the observations made in this thread onto the fresh thread, to make it easier to catch up.)

]]> By: akash chayan http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7690 Fri, 13 Jul 2012 19:53:00 +0000 http://polymathprojects.org/?p=304#comment-7690 The game can be re-formulated in an equivalent one: The player A chooses an element x from the set S (with |S|=N ) and the player B asks the sequence of questions. The j -th question consists of B choosing a set D_j\subseteq S and player A selecting a set P_j\in\left\{ D_j,D_j^C\right\} . For every j\geq 1 the following relation holds: x\in P_j\cup P_{j+1}\cup\cdots \cup P_{j+k}. The player B wins if after a finite number of steps he can choose a set X with | X|\leq n such that x\in X

a) It suffices to prove that if N\geq 2^k+1 then the player B can determine a set S^{\prime}\subseteq S with |S^{\prime}|\leq N-1 such that x\in S^{\prime} . Assume that N\geq 2^n+1 . In the first move B selects any set D_1\subseteq S such that |D_1|\geq 2^{k-1} and |D_1^C|\geq 2^{k-1} . After receiving the set P_1 from A , B makes the second move. The player B selects a set D_2\subseteq S such that | D_2\cap P_1^C|\geq 2^{k-2} and |D_2^C\cap P_1^C|\geq 2^{k-2} . The player B continues this way: in the move j he/she chooses a set D_j such that | D_j\cap P_j^C|\geq 2^{k-j} and |D_j^C\cap P_j^C|\geq 2^{k-j} . In this way the player B has obtained the sets P_1 , P_2 , \dots , P_k such that \left(P_1\cup \cdots \cup P_k\right)^C\geq 1 . Then B chooses the set D_{k+1} to be a singleton containing any element of P_1\cup\cdots \cup P_k . There are two cases now: 1^{\circ} The player A selects P_{k+1}=D_{k+1}^C . Then B can take S^{\prime}=S\setminus D_{k+1} and the statement is proved. 2^{\circ} The player A selects P_{k+1}=D_{k+1} . Now the player B repeats the previous procedure on the set S_1=S\setminus D_{k+1} to obtain the sequence of sets P_{k+2} , P_{k+3} , \dots , P_{2k+1} . The following inequality holds:

\left|S_1\setminus \left(P_{k+2}\cdots P_{2k+1}\right)\right|\geq 1, since |S_1|\geq 2^k .

However, now we have

\left|\left(P_{k+1}\cup P_{k+2}\cup\cdots\cup P_{2k+1}\right)^C\right|\geq 1,

and we may take S^{\prime}=P_{k+1}\cup \cdots \cup P_{2k+1} .

(b) Let p and q be two positive integers such that 1.99\lneq p\lneq q\lneq 2 . Let us choose k_0 such that

\left(\frac{p}{q}\right)^{k_0}\leq 2\cdot \left(1-\frac{q}2\right)\quad\quad\quad\mbox{and}\quad\quad\quad p^k-1.99^k\gneq 1.

We will prove that for every k\geq k_0 if |S|\in\left(1.99^k, p^k\right) then there is a strategy for the player A to select sets P_1 , P_2 , \dots (based on sets D_1 , D_2 , \dots provided by B ) such that for each j the following relation holds:

P_j\cup P_{j+1}\cup\cdots\cup P_{j+k}=S.

Assuming that S=\{1,2,\dots, N\} , the player A will maintain the following sequence of N -tuples: (\mathbf{x})_{j=0}^{\infty}=\left(x_1^j, x_2^j, \dots, x_N^j\right) . Initially we set x_1^0=x_2^0=\cdots =x_N^0=1 . After the set P_j is selected then we define \mathbf x^{j+1} based on \mathbf x^j as follows:

x_i^{j+1}=\left\{\begin{array}{rl} 1,&\mbox{ if } i\in P_i\\ q\cdot x_i^j, &\mbox{ if } i\not\in P_i. \end{array}\right.

The player A can keep B from winning if x_i^j\leq q^k for each pair (i,j) . For a sequence \mathbf x , let us define T(\mathbf x)=\sum_{i=1}^N x_i . It suffices for player A to make sure that T\left(\mathbf x^j\right)\leq q^{k} for each j . Notice that T\left(\mathbf x^0\right)=N\leq p^k \lneq q^k . We will now prove that given \mathbf x^j such that T\left(\mathbf x^j\right)\leq q^k , and a set D_{j+1} the player A can choose P_{j+1}\in\left\{D_{j+1},D_{j+1}^C\right\} such that T\left(\mathbf x^{j+1}\right)\leq q^k . Let \mathbf y be the sequence that would be obtained if P_{j+1}=D_{j+1} , and let \mathbf z be the sequence that would be obtained if P_{j+1}=D_{j+1}^C . Then we have

T\left(\mathbf y\right)=\sum_{i\in D_{j+1}^C} qx_i^j+\left|D_{j+1}\right| T\left(\mathbf z\right)=\sum_{i\in D_{j+1}} qx_i^j+\left|D_{j+1}^C\right|.

Summing up the previous two equalities gives:

T\left(\mathbf y\right)+T\left(\mathbf z\right)= q\cdot T\left(\mathbf x^j\right)+ N\leq q^{k+1}+ p^k, \mbox{ hence} \min\left\{T\left(\mathbf y\right),T\left(\mathbf z\right)\right\}\leq \frac{q}2\cdot q^k+\frac{p^k}2\leq q^k,

because of our choice of k_0.

]]> By: dd http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7688 Fri, 13 Jul 2012 19:39:52 +0000 http://polymathprojects.org/?p=304#comment-7688 Since this is a cooperative effort, let me blurt out some weaker result for part 2 which may be refined to give the desired result.
Suppose that we just want to prove that for n < 2^{k/2} there is no winning strategy for B.

We let N = n + 1.

Before describing A's strategy let us look at what B must do. After a finite number of rounds, B provides a set of n elements which he claims must contain x. In other words, B is stating that precisely one element y from [1, N] should not be x. What we have to show is that all of A's answer are compatible with y being equal to x. More precisely, we have to show that for x = y, the sequence of answers do not contain any k+1 consecutive lies.

A's answers are encoded as a sequence of sets S_0, S_1, \dots, S_M such that each answer is of the form x does not belong to S_i. If the set given by B on the ith round is T_i then the set S_i is either T_i or its complement. A's choice for S_0 is arbitrary (say S_0 = T_0). For i = 1,\dots, k/2, pick S_i \in \{T_i, T_i^c\} such that |intersection of S_0, S_1, …, S_i| <= |intersection of S_0, …, S_{i – 1}| / 2. In particular, the intersection of S_0, \dots, S_{k/2} is empty. After this, A starts fresh with the S_{k/2 + 1} = T_{k/2 + 1} and repeats the same process again.

Now for any choice of y that B selects, all the answers of A are compatible, since for any k + 1 consecutive sets in the sequence S_0, \dots, S_M there must be a subsequence of k/2 + 1 terms S_j, S_{j + 1}, \dots, S_{j + k/2} such that their common intersection is empty. In particular, y cannot be in all of the sets S_j, \dots, S_{j + k/2}, so one of those answers was truthful.

]]> By: Gagik Amirkhanyan http://polymathprojects.org/2012/07/12/minipolymath4-project-imo-2012-q3/#comment-7686 Fri, 13 Jul 2012 19:23:39 +0000 http://polymathprojects.org/?p=304#comment-7686 We can think of the game in the following way.
B asks questions about the sets
S_1, S_2, … , S_k,…

And A should make a choice (by answering YES or NO) of S_i or its complement in each time and we will get
D_1, D_2, … D_k, …
where D_i is either S_i or the complement of S_i (A has the option to choose)
A wants to do in a way that for each m, all the numbers from the range [1, N] appear at least once in the sequence
D_m, D_{m+1}, … D_{m+k}

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