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A not so simple theorem about undirected graphs. See pg.2.
(The other day Alain Martin told me that C.S.Scholten had shown him a rather complicated proof for what seemed a fairly simple theorem, and commucated the theorem to me. I felt challenged and tried to find a “simple” proof myself. The proof that I found is recorded below, partly because I think my proof simple enough to have some beauty, partly because perhaps the theorem is not so simple after all: it took me a full day to prove it and another five hours to write the following in manuscript.)
In a finite, undirected graph we call two nodes that are directly connected by an edge of the graph each other’s “neighbours”. Let P and Q be two different nodes of the graph. We define a “path from P to Q ” as a sequence of nodes, starting with P and ending with Q , such that no node occurs more than once in it and any two adjacent nodes in the sequence are each other’s neighbours in the graph. Of such a path, P and Q are called “the terminal nodes”, the ones in between are called “the internal nodes”. (If P and Q are not each other’s neighbours, a path from P to Q has at least one internal node.)
The theorem states that for any pair of different nodes A and B that are not each other’s neighbours, there exists a node C that is an internal node of any path from A to B , or there exists a pair of paths from A to B that have no internal node in common.
In order to prove it, we arbitrarily select one path from A to B and call it “the special path”; its edges are called “the special edges”, its internal nodes are called “the special nodes”. For brevity’s sake we denote on the special path the direction from A to B as the direction “from left to right”.
Next, let P and Q be two non-adjacent nodes of the special path: we define “an external path between P and Q ” as a path between P and Q of which no node of the special path is an internal node. If between a P and a Q of the special path an external path exists, we call the special nodes between P and Q —if P is to the left of Q : the special nodes that are to the right of P and also to the left of Q— “covered” by that external path.
It is clear that a covered special node can never be a candidate for C an external path allows a path from A to B that bypasses all special nodes it covers.
There are now two mutually exclusive cases: either there exist one or more special nodes not covered b an external ath or each special node is covered by at least one external path.
In the first case the theorem is true, for each path from A to B must pass through all uncovered special nodes, hence any special node that is not covered by any external path can be taken as node C.
In the second case the theorem is also true, for if each special node is covered by at least one external path, two paths from A to B exist that have no internal node in common. We shall show this existence by construction.
We shall construct a sequence of external paths PQ0, PQ1, ... , PQN . The left-hand terminal node of the path PQi will be denoted by Pi ; the right-hand terminal node of the path PQi will be denoted by Qi . Denoting the relation “to the left of” by “<” our sequence of external paths shall have the following two properties:
Property 1:
| A = P0 < P1 < Q0 ≤ P2 < Q1 ≤ P3 < Q2 ≤ ..... QN-2 ≤ PN < QN-1 < QN = B . |
| For i ≠ j the external paths PQi and PQj have no internal node in common . |
Because (see property 1) we have for 0 < i < N
| Pi < Qi-1 ≤ Pi+1 < Qi , |
For PQ0 we choose an external path with P0 = A —because the leftmost special node is covered, such an external path exists— and Q0 as far to the right as possible. If Q0 coincides with B the construction stops here, otherwise we proceed repeatedly as follows until an external path PQN with QN = B has been selected.
Let PQi be the last external path selected and let Qi not coincide with B . Then Qi is a special node and, hence, covered by at least one external path. For PQi+1 we choose a path covering Qi with Qi+1 as far to the right as possible. For i = 0 , the fact that Q0 < Q1 and the way in which PQ0 has been selected imply P0 < P1 ; for i > 0 , the fact Qi < Qi+1 and the way in which PQi has been selected imply that the path PQi+1 does not cover Qi+1 and hence we conclude Qi-1 < Pi+1, The quality Pi+1 < Qi follows from the fact that PQi+1 covers Qi . This proves the inequalities mentioned in property 1 . Property 2 follows from the fact that PQi+1 has no internal node in common with PQk for 0 ≤ k ≤ i : such a common node would provide an external path between Pk and Qi+1 , the existence of which is not compatible with the construction of Qk as a rightmost node. And this completes our proof.
Acknowledgements. I am indebted to C.S.Scholten for having found the theorem, to Alain Martin for having screened my proof, and to my mother, mrs.B.C.Dijkstra - Kluyver for having discovered a serious omission in my proof’s first presentation —as a matter of fact, the omission was so serious that she did not believe the proof— .
Note. With directed paths A → B and P ⇒ Q the proof can be applied directly to the more interesting case of directed graphs.
| Plataanstraat 5 | prof.dr.Edsger W.Dijkstra |
| 5671 AL NUENEN i | Burroughs Research Fellow |
| The Netherlands | . |
Transcribed by Martin P.M. van der Burgt
Last revision 2015-01-25
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