As mentioned in my last post, after lunch talks can be very fruitful at times. Describing a yet another fruitful chat with my friend Sambuddha Roy.
The other day he gave me an interesting problem to think about, which is formerly known as, as he told me, necklace theorem. Two thieves have stolen queen's very precious necklace. There are three kinds of gems in the necklace called R, G and B. All are even. The problem is to divide the necklace in such a way that both the thieves get equal amount of gems of each kind. The necklace can have any interleaving of the gems. ( e.g. RRGBGGRBBBG..... ) Prove that it is always sufficient to divide the necklace in equal share by at most 3 cuts.
The proof makes use of Ham Sandwich Theorem. For brevity, I will quote it as HST now on. HST for 2-dimension says that given any two point sets in a plane, there always exist a line which will cut both these point sets simultaneously into half. To say it more formally, call the point sets R and B. Then one side of line will contain at most |R|/2 points and at most |B|/2 points. The same constraints will be followed on the other side too.
If there is only 1 point set R. For a given direction we can always have a line which will divide R in two. The proof is based on intermediate value theorem. It says that for a continuous function if f(x1) = 0 and f(x2)=1 then there exist a z \in (x1,x2) such that f(z) = 1/2.
Now that we have two point sets. What we can do is we get a line for point set R. Now we evaluate how this line L behaves with respect to B. Let us define diff(L) = number of blue gems on the left of L - number of blue gems on the right of L. For initial line L, let the value of diff(L) be b. What we can do is rotate this line a bit. We know that for any given direction, we have a corresponding line L which will divide R equally. We will again evaluate wrt B. We keep on doing this until we have rotated the line for 180 degree. At this point value of diff(L) would be -b. From intermediate value theorem, we can state that there will be a line where diff(L) will be 0.
How about, Ham Sandwich theorem in 3-dimension?? Well, my friend has given me the proof as an exercise. The statement is that given 3 point sets R, G, B in R^3 ( or 3-dim space ) there always exist a plane which will divide all 3 point sets simultaneously into equal portion.
Now, we need to reduce Necklace Theorem to HST in 3D. Starting from left to right, assign every gem an index i. Plot a parametric curve ( t, t^2, t^3 ) using these indices. HST assures us that we will have a plane ax+by+cz+d=0 which will divide R, G, B equally. This plane will divide the parametric curve at , at most 3 points as at + bt^2 + ct^3 + d = 0 will have at most 3 roots. Now, it is easy to map these 3 intersection points on the curve to 3 cuts on the necklace.
Phew!!! It never occurred to me that this proof will use such weird things. Induction based method was too tempting for me to let me think in some different direction. But the story is not over yet. The proof given above is not constructive. So if we were to find an algorithm to find these 3 cuts we might have to explore all possibilities taking O(n^3) time. So another exercise would be to try to find more efficient algorithm to find these cuts.
The generalization of HST is that given k point sets in R^k there would be always be a hyperplane which would cut all these k point sets simultaneously into equal portion. Yet another generalization one can think of is that, what if there are more thieves??? Given q number of thieves and k kind of gems, what would be the upper bound on the number of cuts required?? The answer is due to dold's theorem ( Neither sambuddha nor I have read the proof ). It say that for such situation at most k(q-1) cuts would be necessary.
What if there are 3-point sets in two dimension?? Definitely a line would not suffice. An exercise would be to prove that two rays originating from an apex would be sufficient. Give an algorithm for finding them out.
Two more puzzles are :
1) Given a point set of size n in 2D, prove that there would exist a pair of orthogonal lines which would divide the point set in such a way that in each quadrant number of points are at most n/4.
2) Given a point set of size n in 2D, prove that there would exist 3 concurrent lines which would divide the point set in such a way that in each segment the number of points are at most n/6.
That's enough to tease our brain for a while :D.