(Yes, I made that name up)
The basic idea that we will use in some form or the other for the next few problems is a simple consequence of the pigeonhole principle. Suppose you have n points on a line, and you are allowed to mark some of them such that no consecutive points are marked. Then the maximum number of points you can mark is ⌈ /2⌉, and this can be achieved by marking alternate points. If the n points were on a circle and not a line segment, then the maximum would be ⌊ /2⌋. These obvious statements can be cleverly applied in several combinatorial settings.
Example 4 [IMO Shortlist 1990]
Let n ≥ 3 and consider a set E of 2n-1 distinct points on a circle. Suppose that exactly k of these points are to be colored red. Such a coloring is good if there is at least one pair of red points such that
the interior of one of the arcs between them contains exactly n points from E. Find the smallest value of k so that every such coloring of k points of E is good.
Answer:
Let j be maximum number of colored points a bad coloring can have. Then k = j+1, so it suffices to find j.
Let the points be A1, A2, …, A2n-1. Join vertices Ai and Ai+n+1 by an
edge for each i (indices modulo 2n-1). This decomposes E into disjoint cycles. The coloring is bad if and only if no two red vertices are joined by an edge. In other words, no two consecutive vertices in the cycle are both red. How many cycles are there? Using elementary number theory, it is easy to show that the number of cycles is equal to gcd (n+1, 2n-1).
Since 2n-1 = 2(n+1)–3, gcd (n+1, 2n-1) = 3 if n+1 is divisible by 3, and gcd (n+1, 2n-1) = 1 otherwise. If gcd (n+1, 2n-1) = 1, we get only one cycle containing all 2n-1 points. Then j = n-1 by our earlier discussion. Hence k = j+1 = n. Figure 7.1. 1 2
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Figure 7.2.
If gcd(n+1, 2n-1) = 3, then we get 3 cycles, each containing vertices. Each cycle then can have at most ⌊(2 − 1)/6⌋ red points in a bad coloring. Thus at most 3⌊(2 − 1)/6⌋ points in total can be colored in a bad coloring. Hence k = 3⌊(2 − 1)/6⌋+1 if 3 divides n+1, and k = n otherwise. ■
Example 5 [USAMO 2008, Problem 3]
Let n be a positive integer. Denote by Sn the set of points (x, y) with
integer coordinates such that |x| + |y+½| ≤ n. A path is a sequence of distinct points (x1, y1), (x2, y2), …, (xk, yk) in Sn such that, for i = 1,
2, …, k-1, the distance between (xi, yi) and (xi+1, yi+1) is 1 (in other
words, the points (xi, yi) and (xi+1, yi+1) are neighbors in the lattice
of points with integer coordinates). Prove that the points in Sn
cannot be partitioned into fewer than n paths (a partition of Sn
into m paths is a set P of m nonempty paths such that each point in Sn appears in exactly one of the m paths in P).
Answer:
Color the points of each row of Sn alternately red and black,
starting and ending with red. Any two neighboring points are of opposite color, unless they are from the middle two rows. 1 2 3 4 5 6 7 8 9
Consider a partition of Sn into m paths, for some m. For each of
the m paths, split the path into two paths wherever there are consecutive red points. Now no path has consecutive red points. Further, since there are n pairs of consecutive red points in Sn
(from the middle two rows), we have split paths at most n times. Thus we now have at most m+n paths (each split increases the number of paths by one).
Figure 7.3. Example for n = 3 Now, there are 2n more red points than black points in Sn, but each of the m+n paths contains at most one more red point than black point (since no path contains consecutive red points). Thus we obtain m+n ≥ 2n, or m ≥ n, proving the result. ■ In the next example, in order to use the chain decomposition idea we need some starting point, that is, some information or assumption that lends enough structure to the problem for us to exploit. To this end, we use the extremal principal by considering the smallest element in a certain set. Example 6 [USAMO 2009-2] Let n be a positive integer. Determine the size of the largest subset S of N = {-n, -n+1, …, n-1, n} which does not contain three elements
(a, b, c) (not necessarily distinct) satisfying a + b + c = 0.
Answer:
Obviously 0 is not in S since 0+0+0 = 0. We claim the answer is n if n is even and (n+1) if n is odd. These bounds can clearly be achieved by taking all the odd numbers in N, since the sum of three odd numbers can never be 0. To show this is the maximum, let j be the element of smallest absolute value in S (if both j and –j are present, consider the positive one). Assume WLOG that j > 0, and let T denote the set of elements with absolute value at least j. Note that all elements of S are in T.
Case 1: (-j) is not in S. Consider the pairs (j, –2j), (j+1, –(2j+1)), (j+2, –(2j+2)), …., (n–j, –n). In each of these pairs the sum of the numbers is (-j), so at most one of the two elements is in S (otherwise the sum of the two elements plus j would be 0). There are exactly (n-2j+1) pairs, so at most (n-2j+1) of the paired numbers are in S.
Furthermore, there are exactly 2j-1 unpaired numbers in T \ {-j}: j positive unpaired numbers (namely n, n–1, …, n–j+1), and (j–1) negative unpaired numbers (namely –(j+1), –(j+2), …, –(2j–1)). Thus the maximum number of elements in S is (2j–1) + (n–2j+1) = n.
Case 2: (–j) is also in S. Now we use the chain decomposition idea. If a and b are elements in T, then a is joined to b by an edge if and only if a+b = j or a+b = –j. This ensures that no two elements joined by an edge can both be in S. Each element x in T is joined to at least one and at most two other elements of T ((j–x) and (–j–x)). Hence the elements of T have been partitioned into disjoint chains.
Figure 7.4: n = 13, j = 3. There are 6 chains, of which 2 are shown.
Now the rest is just counting. There are exactly 2j chains, namely the chains starting with ±j, ±(j+1), …, ±(2j-1). Let the lengths of these chains be l1, l2, …, l2j. From a chain of length li, at
most elements can be in S. Thus the total number of elements we can take is at most ∑ ≤ ∑ = j + ∑ = j + ( ) = n+1. Here we used ∑ = 2n – 2(j-1) since both sides are equal to the number of elements of absolute value at least j. Now we are done if n is odd. If n is even, we need to tighten our bound to n. For this we note that the inequality ∑ ≤ ∑ is strict if there is a chain of even length, since is strictly less than for even x. It now suffices to prove the existence of a chain of even length if n is even; this is pretty simple and is left to the reader. ■ 9 10 11 12 13 -9 -10 -11 -12 -13 1 2 3 4 5 6 7 8 -1 -2 -3 -4 -5 -6 -7 -8