Integers not divisible by $2012$ are arranged on the arcs of an oriented graph. We call the weight of a vertex the difference between the sum of the numbers on the arcs coming into it and the sum of the numbers on the arcs going away from it. It is known that the weight of each vertex is divisible by $2012$. Prove that non-zero integers with absolute values not exceeding $2012$ can be arranged on the arcs of this graph, so that the weight of each vertex is zero. Proposed by W. Tutte
Problem
Source: Tuymaada 2012, Problem 8, Day 2, Seniors
Tags: abstract algebra, graph theory, combinatorics proposed, combinatorics
22.07.2012 01:11
22.07.2012 02:20
How about Problem 4, day 2, Grade 1 (Juniors) at the 2010 Tuymaada ? It states, in educated language, that any finite digraph $G$ contains some non-empty set $A$ of vertices that has no arcs between any two of its vertices (an independent set), such that for any vertex $v \in G\setminus A$ there exists a vertex $a\in A$ so that $\overrightarrow{va}$ is an arc of $G$, or if not, that there exist vertices $w\in G\setminus A$ and $a\in A$ so that both $\overrightarrow{vw}$ and $\overrightarrow{wa}$ are arcs of $G$ (from any vertex in $G$, one can reach a vertex in $A$ in at most two steps). It looks like it has to have been known also ...
22.07.2012 02:50
I don't think there is such a result in Diestel, but the concept seems very related to the one of kernel, which is a set $U$ of independent vertices, such that every vertex outside $U$ is connected via an edge to some vertex $v\in U$, such that the edge points towards $v$ . This concept is presented in chapter 5, section 4. He uses is to prove the list coloring conjecture for bipartite graphs. Probably the closest result to the problem in Tuymaada is exercise 31 from the same chapter, Richardson's theorem, which states that every directed simple graph has a kernel if it doesn't contain odd directed cycles.
22.07.2012 19:00
The official solution is quite lengthy and complicated; the thesis is a particular case of a major theorem by Tutte (featured in [ R. Diestel - Graph Theory]); it is not my intent to make here a presentation of it. Only two among the $35$ participants managed to earn $1$ point (out of a possible $7$); the rest ending up tabula rasa. Ahh, the dangers of featuring a deep theoretical result in competition ...
24.07.2012 14:12
It is a pity that resources pages here omit the names of the authors; the official Tuymaada booklet clearly indicates W.Tutte as the author of the problem. mavropnevma wrote: Ahh, the dangers of featuring a deep theoretical result in competition ... Oh, it's just problem 8. Last year (http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=421157) it was not any deep theoretical result, but the participants fared no better.
24.07.2012 14:28
It is a pity, yes, that the problem setters do not even bother posting the problems and solutions, in English, on any site, so those that toil filling the resources pages have to somehow obtain the official Tuymaada booklet and manually enter them on this site. I therefore deliberately omitted the names of the authors; if they want public recognition, one of them may make an effort and provide the information in due form. I'm sorry, Alexander, for the flippant tone, but I've had my fill with all kind of competitions (BMO, jBMO, Zhautykov, Tuymaada), where one has to go on a wild-goose chase for problem statements, official solutions, and results, days after the competition has finished.
25.07.2012 01:09
I gather so far that you deliberately omit the name of William Tutte, probably to teach him some lesson. I fail to see however how you will know whether he'd learned you lesson until you meet him where he now is.
25.07.2012 01:26
It may come as a surprise to you, Alexander, that I am familiar with Tutte's name and work. When I said (Tutte related), it was in deference to the great man, who will not have stated a result pertaining to just this specific $2012$ number. I thought I could not attribute to him as an author such an ungainly statement. As to your wish I meet him where he now is, alas, chances are I will be located in less comfortable accomodations. Brimstone and ashes, rather than ambrozia for me, I'm afraid. I notice however that you carefully picked the portion of my remarks to which to retort. No word on relinquishing on the incumbent duties of the organizers of a competition that leave remote onlookers with their eyes in the sun. Well, birds of a different feather ...