One At A Time

Alright, time to finish off the solution to the balls in jars problem from before.

The essential proof is that it is possible to find an algorithm that will move from a position (a,b,c), with a>b>c, to a position (x,y,z) such that z is less than c. Thus by repeating this algorithm no more than c times, you must reach zero (given that we can only have integer values).

OK, suppose we start in some position. Call the bucket with the most marbles A, the one with the fewest marbles C, and the other one B. The names of those jars will not change throughout the algorithm, specifically even though C has the fewest at the start it might not at some later point. Let a, b, and c be the number of marbles in jars A, B, and C respectively. Note that even though A, B, and C will not change throughout the algorithm, the values of a, b, and c will.

The algorithm is: Check the value of floor(b/c). If it is even, move marbles from A to C. If it is odd, move marbles from B to C. If C has more marbles than B then stop, otherwise go back to the start.

Simple enough, now, to prove it works. We must prove that at the end B will have fewer marbles in it than C started with.

At the start, we know that b is greater than c, let us write b=qc+t with t less than c, so q being the floor of b/c. For this proof, we will not change the values of a, b, and c, I will assume those values are locked at the start.

Let us suppose b is not at least twice as big as c, so then q=1 and when we move marbles from B to C (as per the algorithm), B will have t marbles in it, and t is less than c.

Now, if q is bigger than 1, I will show that the algorithm will reduce the number of marbles in jar B without changing the value of t, so eventually q will be 1.

So, if we suppose q is even, so we move marbles from A to C. Now C contains 2c, and B contains b. Since b=qc+t=(q/2)(2c)+t, so b mod c and b mod 2c both give t (note that q being even was needed).

If q is odd, we will move marbles from B to C. Now C contains 2c and B contains b-c. Since b=qc+t we can also then show b-c=((q-1)/2)(2c)+t, so b mod c and b-c mod 2c both give t (in this case q being odd is needed).

Each time, B contains fewer marbles, so eventually q will be 1 and then next move B will have t marbles, which was less than the starting value of C. So, this finishes the proof. One simply needs to repeat the algorithm many times (renaming the jars A, B, and C after each iteration) to win the game.

Two In One

Alright, time for the partial solution to the balls in jars problem from last time.

I guess since the two-jar problem is a decent amount easier, I'll solve that out first.

First of all, it is absolutley trivial to see that if the total number of balls in the jars is not an even number, there will be no way to win, since to win you must first arrive at a position of the form (p,p). Next, it is easy to see that any common divisor has no effect on the ability to win from a position, that is to say: the position (p,q) is winnable iff the position (np,nq) is winnable.

At this point, we might be inclined to work out a few winning positions. OK, so what are the winning positions of the form (1,q)? Well, clearly (1,1) is a winning position, and (1,q) can only be a winning position if q is odd. From the position (1,q) we move to the position (2,q-1), but since q is odd for any position of interest this is the same as (1,[q-1]/2) so if we suppose this is another position, say (1,k), we see that q=2k+1 so if (1,k) is a winning position then so is (1,2k+1). From here we can list off the positions as (1,3), (1,7), (1,15)... clearly (1,q) seems to be a winning position if q is 1 less than a power of two. You can prove that by induction if you like, and with a bit more work we can also be sure that there aren't other winning positions of the form (1,q), but I won't bother with that yet.

Next, what about positions of the form (3,q)? Well, first of all q=1 would be a winning position. Next, from (3,q) we can move to (6,q-3) (this assumes that q>3, but other positions are finite enough anyway). Now since we might as well assume q is odd, we can half this position to (3,[q-3]/2) so calling this (3,k) again, we see that if (3,k) is a winning position then so is (3,2k+3). Listing off positions we can see that the winning positions seem to be (3,1), (3,5), (3,11), (3,29)... again summing to a power of two. This makes it seem that a position (p,q) is a winning position iff the sum p+q is a power of two. Well, thats easy enough to disprove, (3,9) is a winning position, but it is merely a multiple of 3 away from (1,3).

The main thing that will finish this off is the realiziation that once you have divided off any multiples in your problem, no new ones will show up, besides 2, and that one will show up every move. Lets prove that. Suppose we are in some position (p,q) with p and q both odd (so that their sum is even, but we do not need to divide off an overall 2), and we assume that p and q are not both multiples of some prime number n (n>2). Now we move to (p-q, 2q) (setting p as the larger one), I claim that p-q and 2q are also not both multiples of n. If they were then q would be a multiple of n (since 2q is) and then p would be (since p-q is and q is). Naturally this argument won't work for n=2, an overall multiple of 2 shows up every single move.

Now I will prove the main result: from position (p,q) with p and q having no common factors, we are in a winning position iff p+q is a power of 2.

You can prove one direction by induction. Prove it for p+q=2^n. It is trivial for n=1, then if it is true for n=k and you are in position (p,q) with p+q=2^(k+1) you can make one move and then half p and q so that p+q will now equal 2^k.

For the other direction we must show that if p+q is a multiple of a prime other than 2 and p and q have no common factors, then it is a losing position. Suppose p+q is z*k where z is a prime other than 2. k must be even, for if p+q is odd then we are trivially in a losing position. In order to win, we must first reach (z*k/2, z*k/2) but we have already proven that once you have divided off multiples, no new ones can show up. A new multiple of z (for z>2) cannot happen, so the position (z*k/2, z*k/2) cannot be reached, so the game cannot be won.

This finishes the proof. While working on the problem I also noticed that the legal move from (p,q) is to (2p,2q) mod p+q, which I thought was sort of cool. Perhaps people who are better at number theory than I can make use of that. Anyway, feel free to try to use this to solve the three jar problem, I was never able to.

Balls In Jars

New puzzle time I suppose. I am going to put up a puzzle that I have been working on for about 2 years which I got from the Math Club room on KGS. Its possible that I spent alot of time overthinking the puzzle, but there was a very interesting diversion that I thought was a neat side puzzle that I will also post here. First, the main puzzle:

You have three jars that contain marbles. You are going to play a one-player game with these jars. On a turn, the legal move is to select two jars and move from the first jar to the second jar as many marbles as were in the second jar (naturally the "first jar" must be the one with more marbles). The goal is to set things up so that you can get one jar to be empty. Show that it is always possible, no matter how many marbles were in each of the three jars initially, to achieve the goal of getting an empty jar.

So, for example, you could start with (1,8,3) then move 1 marble from the second to the first, getting (2,7,3). Then move 2 marbles from the second to the first, getting (4,5,3). Next move 3 marbles from the second to the third, getting (4,2,6). Now move 4 marbles from the last to the first, getting (8,2,2). Finally move 2 marbles from the second to the last, getting (8,0,4) winning the game. I could have won that game more efficiently, but I wanted to just play around a bit to demonstrate the moves.

When I tried to solve the puzzle, I first tried to explore out the two-jar case, figuring that would help (it did not, by the way). In the end I saw a neat little result that I will also pose as a problem:

In the two jar version of the same problem, what is the complete list of initial situations (a,b) from which you can reach a winning position (q,0).

It is clear not all initial positions of the two-jar game can be solved, for example if you start with (8,2) you can only move to (6,4) and then to (2,8) and then to (4,6) and back to (8,2). However, (3,1) can be won, moving to (2,2) then to (4,0). You can see that there are no choices when playing this game, but from what initial setups can you win?