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

]]>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. ]]>

I think it’s correct solution, just in the definition of x_i^{j+1} should be P_i instead of S. *[Corrected, -T.]*

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.

]]>As this thread is becoming quite full, I am opening a fresh thread at https://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.)

]]>a) It suffices to prove that if then the player can determine a set with such that . Assume that . In the first move selects any set such that and . After receiving the set from , makes the second move. The player selects a set such that and . The player continues this way: in the move he/she chooses a set such that and . In this way the player has obtained the sets , , , such that . Then chooses the set to be a singleton containing any element of . There are two cases now: The player selects . Then can take and the statement is proved. The player selects . Now the player repeats the previous procedure on the set to obtain the sequence of sets , , , . The following inequality holds:

since .

However, now we have

and we may take .

(b) Let and be two positive integers such that . Let us choose such that

We will prove that for every if then there is a strategy for the player to select sets , , (based on sets , , provided by ) such that for each the following relation holds:

Assuming that , the player will maintain the following sequence of -tuples: . Initially we set . After the set is selected then we define based on as follows:

The player can keep from winning if for each pair . For a sequence , let us define . It suffices for player to make sure that for each . Notice that . We will now prove that given such that , and a set the player can choose such that . Let be the sequence that would be obtained if , and let be the sequence that would be obtained if . Then we have

Summing up the previous two equalities gives:

because of our choice of .

]]>Suppose that we just want to prove that for 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 such that each answer is of the form x does not belong to . If the set given by B on the ith round is then the set is either or its complement. A's choice for is arbitrary (say ). For , pick such that |intersection of S_0, S_1, …, S_i| <= |intersection of S_0, …, S_{i – 1}| / 2. In particular, the intersection of is empty. After this, A starts fresh with the 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 there must be a subsequence of k/2 + 1 terms such that their common intersection is empty. In particular, y cannot be in all of the sets , so one of those answers was truthful.

]]>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}