Imo 2010

This is the wiki page for the mini-polymath2 project, which seeks solutions to Question 5 of the 2010 International Mathematical Olympiad.

The project will start at 16:00 UTC July 8, and is hosted at the polymath blog. A discussion thread is hosted at Terry Tao's blog.

Rules

This project will follow the usual polymath rules. In particular:

• Everyone is welcome to participate, though people who have already seen an external solution to the problem should probably refrain from giving spoilers throughout the experiment.
• This is a team effort, not a race between individuals. Rather than work for extended periods of time in isolation from the rest of the project, the idea is to come up with short observations (or to carry an observation of another participant further) and then report back what one gets to the rest of the team. Partial results or even failures can be worth reporting.
• Participants are encouraged to update the wiki, or to summarise progress within threads, for the benefit of others.

Threads

Discussion and planning:

• Future mini-polymath project: 2010 IMO Q6? June 12, 2010.
• Organising mini-polymath2 June 21, 2010.
• Mini-polymath2 start time, June 27, 2010.
• Mini-polymath2 discussion thread, July 8 2010.

Research:

• Minipolymath2 project: IMO 2010 Q5, July 8 2010.

The question

The question to be solved is Question 5 of the 2010 International Mathematical Olympiad:

Problem In each of six boxes [math]\displaystyle{ B_1, B_2, B_3, B_4, B_5, B_6 }[/math] there is initially one coin. There are two types of operation allowed:

Type 1: Choose a nonempty box [math]\displaystyle{ B_j }[/math] with [math]\displaystyle{ 1 \leq j \leq 5 }[/math]. Remove one coin from [math]\displaystyle{ B_j }[/math] and add two coins to [math]\displaystyle{ B_{j+1} }[/math].

Type 2: Choose a nonempty box [math]\displaystyle{ B_k }[/math] with [math]\displaystyle{ 1 \leq k \leq 4 }[/math]. Remove one coin from [math]\displaystyle{ B_k }[/math] and exchange the contents of (possibly empty) boxes [math]\displaystyle{ B_{k+1} }[/math] and [math]\displaystyle{ B_{k+2} }[/math].

Determine whether there is a finite sequence of such operations that results in boxes [math]\displaystyle{ B_1, B_2, B_3, B_4, B_5 }[/math] being empty and box [math]\displaystyle{ B_6 }[/math] containing exactly [math]\displaystyle{ 2010^{2010^{2010}} }[/math] coins. (Note that [math]\displaystyle{ a^{b^c} := a^{(b^c)} }[/math].)

Observations and partial results

• If the left-most box [math]\displaystyle{ B_1 }[/math] becomes empty, then it cannot ever become non-empty again. Furthermore, the left-most box can never have more than one coin; it can be touched exactly once.
• Define the worth W of a state to be [math]\displaystyle{ W = B_6 + 2 B_5 + 4 B_4 + 8 B_3 + 16 B_2 + 32 B_1 }[/math]. Then the initial worth is 63, the final desired worth is [math]\displaystyle{ 2010^{2010^{2010}} }[/math], and the Type 1 move does not affect the worth. On the other hand, the Type 2 move increases the worth when [math]\displaystyle{ B_{j+2} - B_{j+1} \geq 4 }[/math].
• Once one has a large number of coins in one of the first four boxes, say [math]\displaystyle{ B_k }[/math], one can apply the Type 2 move repeatedly to remove coins from [math]\displaystyle{ B_k }[/math] while swapping [math]\displaystyle{ B_{k+1} }[/math] and [math]\displaystyle{ B_{k+2} }[/math] repeatedly. This suggests that it is relatively easy to remove coins from the system; the difficulty is in adding coins to the system.
• The total number of coins in the system is bounded. Indeed, let [math]\displaystyle{ f(N,\Sigma) }[/math] be the maximum number of coins that one can end up with starting with N boxes with at most [math]\displaystyle{ \Sigma }[/math] coins in them. Thus for instance [math]\displaystyle{ f(1,\Sigma)=\Sigma }[/math]. By considering the times when one touches the left-most box, we can bound [math]\displaystyle{ f(N,\Sigma) }[/math] by at most [math]\displaystyle{ \Sigma }[/math] iterations of the map [math]\displaystyle{ n \mapsto f(N-1,n)+2 }[/math] starting with [math]\displaystyle{ n=\Sigma }[/math]. This gives an Ackermann-type bound on [math]\displaystyle{ f(N,\Sigma) }[/math]. We need f(6,6) to be less than [math]\displaystyle{ 2010^{2010^{2010}} }[/math], but this bound is likely to be too large.

Possible strategies

• Split the problem into two pieces. Part I: try to show the weaker result that the number of coins in the system can eventually be as large as [math]\displaystyle{ 2010^{2010^{2010}} }[/math]. Part II: Show that once one has a lot of coins, one can move to the final state where [math]\displaystyle{ B_1=\ldots=B_5=0 }[/math] and [math]\displaystyle{ B_6 = 2010^{2010^{2010}} }[/math].
• Try to show that a quantity such as the worth increases or decreases in a controlled manner as one applies the Type 1 and Type 2 moves.
• We know that the first box can never contain more than one coin. What can we say about the second box, third box, etc.?
  • There may be a recursive formula for the maximal size of box [math]\displaystyle{ B_j }[/math], possibly requiring one to solve the five-box, four-box, etc. problems first.
• Work backwards?
• Try to completely solve the three-box problem (say) first: starting from [math]\displaystyle{ [X,Y,Z] }[/math], what is the most number of coins one can generate?

Compound moves

Here we use Type 1 move [math]\displaystyle{ [a,b] \to [a-1,b+2] }[/math] and the Type 2 move [math]\displaystyle{ [a,b,c] \mapsto [a-1,c,b] }[/math] to create more advanced moves.

1. We can create the move [math]\displaystyle{ [a,b] \to [0,b+2a] }[/math] from repeated application of Type 1.
2. We have [math]\displaystyle{ [1,a,b] \to [0,0,a+2b] }[/math] by applying Type 2 once and then Type 1 b times.
   • Or, by using advanced move 1 first, the move [math]\displaystyle{ [1, a, b] \to [1, 0, b+2a] \mapsto [0, b+2a, 0] \mapsto [0, 0, 2b+4a] }[/math].
3. For [math]\displaystyle{ a \geq 1 }[/math] we have [math]\displaystyle{ [a,0,0] \to [0,2^a,0] }[/math] via [math]\displaystyle{ [a,0,0] \mapsto [a-1,2,0] \mapsto [a-1,0,4] \mapsto [a-2,4,0] \mapsto [a-2,0,8] \mapsto \ldots \mapsto [1, 0, 2^a] \mapsto [0,2^a,0] }[/math].
4. Using the previous move, for any [math]\displaystyle{ a \geq 1 }[/math] we have [math]\displaystyle{ [a,b,0,0] \to [a-1, 2^b, 0, 0] }[/math].

The last move seems to be the key to the solutions so far discovered, since it allows one to introduce an exponential at only a linear cost.

World records

To make the second box as big as possible:

• [math]\displaystyle{ [1,1,1] \mapsto [1,0,3] \mapsto [0,3,0] }[/math] places 3 coins in box 2.

To make the third box as big as possible:

• [math]\displaystyle{ [1,1,1] \mapsto [0,3,1] \mapsto [0,0,7] }[/math] places 7 coins in box 3. (Here we use advanced move 2).

To make the fourth box as big as possible

• [math]\displaystyle{ [1,1,1,1] \to [0,3,0,3] \to [0,2,2,3] \to [0,2,0,7] \to [0,1,7,0] \to [0,1,0,14] }[/math] [math]\displaystyle{ \to [0,0,14,0] \to [0,0,0,28] }[/math] gives 28 coins in box 4.

To make the sixth box as big as possible

• Using the fourth box move we have [math]\displaystyle{ [0,0,0,28,1,1] }[/math]. We can move [math]\displaystyle{ [28,1,1] }[/math] to [math]\displaystyle{ [27,0,7] }[/math] through Type 1; then using a variant of advanced move 3 we get [math]\displaystyle{ [0,0,7 * 2^{27}] }[/math], leading to [math]\displaystyle{ 7 \times 2^{27} }[/math] coins in the last box.

• A new record: we can get [math]\displaystyle{ 2^{192} }[/math] in the 6th box by [math]\displaystyle{ [1,1,1,1,1,1]\to \to [0,0,7,1,1,1]\to [0,0,7,0,3,1] \to \to [0,0,1,0,3\times 2^6,1]\to[0,0,0,3\times 2^6,0,1]\to [0,0,0,0,0,2^{3\times 2^6}] }[/math] where [math]\displaystyle{ \to\to }[/math] are the special moves.

Completed solutions

First solution

Fist, I introduce a move that follows from compound move 3:

(1) [math]\displaystyle{ [N,M,0,0] \to [N-1, 1, 0, 2^{M+1}] \to [N-2,2^{M+1}, 0, 0 ] }[/math]

We can get

(2) [math]\displaystyle{ [0,0, 140, 0, 0 ,0] }[/math] as follows:

[math]\displaystyle{ [1,1,1,1,1,1] \to [0,2,2,2,2,3] \to [0,2,1,1,8,3] \to [0,2,1,1,0,19] \to [0,1,19,0,0,0] }[/math]

[math]\displaystyle{ \to [0,1,1,36,0,0] \to [0,1,1,1,0,140] \to [0,0,140,0,0,0] }[/math]

By (1) combined with (2) we can get some number that is greater than [math]\displaystyle{ n:=2010^{2010^{2010}} }[/math] in the 4th spot

And swapping 5 with 6 enough times, we can adjust this number to have the value n/4. Moving everything to the right will give us the desired result.

Second solution

[math]\displaystyle{ [1,1,1,1,1,1] \to [0,3,1,1,1,1] \to [0,2,3,1,1,1] \to [0,2,1,5,1,1] \to [0,2,1,1,9,1] }[/math][math]\displaystyle{ \to [0,2,1,1,0,19] \to [0,2,1,0,19,0] \to [0,2,0,19,0,0] \to [0,1,19,0,0,0] }[/math].

Make use of the compound move 4 here so that [math]\displaystyle{ \to [0,1,0,2^{2^{\cdots^2}},0,0] \to [0,0,2^{2^{\cdots^2}},0,0,0]=[0,0,N,0,0,0] }[/math] where there are 19 2's. Note that [math]\displaystyle{ 2^{2^{2^2}}=2^16\gt 2010 }[/math]. Thus, [math]\displaystyle{ M=2^{2^{\cdots^2}} }[/math] with 12 2's is bigger than [math]\displaystyle{ K=2010^{2010^{2010}} }[/math]. Hence, we have [math]\displaystyle{ N\gt K\gt K/8 }[/math] so that [math]\displaystyle{ [0,0,N,0,0,0] \to [0,0,N-1,0,0,0] \to \cdots \to [0,0,K/8,0,0,0] \to [0,0,0,K/4,0,0] \to [0,0,0,0,K/2,0] \to [0,0,0,0,0,K] }[/math].