Olympiad Maths Prep

Track / Stage 8 / 76 of 180 #1776 of 2000

Problem 1776

IMO Shortlist mid-range; USAMO P2/P5
Combinatorics Difficulty 8.3 Prove it Team Selection Test for IMO 2011 · Turkey · 2011

Graphistan has 2011 cities and Graph Air (GA) is running one-way flights between all pairs of these cities. Determine the maximum possible value of the integer kk such that no matter how these flights are arranged it is possible to travel between any two cities in Graphistan riding only GA flights so long as the absolute values of the difference between the number of flights originating and terminating at any city is not more than kk.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

We want to find the largest integer kk such that a directed path exists from any vertex v1v_1 to any vertex v2v_2 in any directed complete graph GG with 2011 vertices satisfying indeg(v)outdeg(v)k|\text{indeg}(v) - \text{outdeg}(v)| \le k for all vertices vv. The answer is 1005.

Observe that indeg(v)outdeg(v)|\text{indeg}(v) - \text{outdeg}(v)| is always even. We first give a counterexample with k=1006k=1006. Let V=XYV = X \cup Y, X={x1,x2,,x1005}X = \{x_1, x_2, \dots, x_{1005}\}, Y={y1,y2,,y1006}Y = \{y_1, y_2, \dots, y_{1006}\}, and E={(xi,xj):1i<jmax(i+502,1005) or 1jmax(i503,0)}{(yi,yj):1i<jmax(i+503,1006) or 1j<max(i503,0)}{(xi,yj):1i1005 and 1j1006}E = \{(x_i, x_j) : 1 \le i < j \le \max(i+502, 1005) \text{ or } 1 \le j \le \max(i-503, 0)\} \cup \{(y_i, y_j) : 1 \le i < j \le \max(i+503, 1006) \text{ or } 1 \le j < \max(i-503, 0)\} \cup \{(x_i, y_j) : 1 \le i \le 1005 \text{ and } 1 \le j \le 1006\}.

Then indeg(v)outdeg(v)1006|\text{indeg}(v) - \text{outdeg}(v)| \le 1006 for all vVv \in V in the graph G=(V,E)G = (V, E), but there is no directed path from y1y_1 to x1x_1.

Now we will show that there is a directed path from any vertex to any vertex if indeg(v)outdeg(v)1004|\text{indeg}(v) - \text{outdeg}(v)| \le 1004 for all vertices vv. Note that if one of indeg(v)\text{indeg}(v) and outdeg(v)\text{outdeg}(v) is less than 503, then the other is greater than 1507 and indeg(v)outdeg(v)|\text{indeg}(v) - \text{outdeg}(v)| is at least 1006. Therefore indeg(v)\text{indeg}(v) and outdeg(v)\text{outdeg}(v) are both at least 503.

Let v1v_1 and v2v_2 be two distinct vertices in GG. Let V1V_1 be the set of vertices that can be reached from v1v_1 and let V2V_2 be the set of vertices from which v2v_2 can be reached. Then all arrows with tail in V1V_1 have also their heads in V1V_1. Therefore the sum of outdeg(v)\text{outdeg}(v) for vv in V1V_1 is (V12)\binom{|V_1|}{2}. On the other hand, this sum must be at least 503V1503|V_1|. It follows that V11007|V_1| \ge 1007. A similar argument shows that V21007|V_2| \ge 1007. Hence V1V_1 and V2V_2 cannot be disjoint.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.