Walking with Euler through Ostpreußen and RNA

Transcription

1 Walking with Euler through Ostpreußen and RNA Mark Muldoon February 4, 2010

2 Königsberg (1652) Kaliningrad (2007)? The Königsberg Bridge problem asks whether it is possible to walk around the old city in such a way as to: start and finish in the same place and cross each of the seven bridges exactly once. And you have to cross the bridges from one end to the other: there s no walking halfway out, turing around, and then doing the other half from the other side.

3 Euler s abstraction Leonhard Euler solved it in a paper presented to the Academy of Science in St. Petersburg in August 1735: the drawing at right comes from his paper.

4 Euler s abstraction Leonhard Euler solved it in a paper presented to the Academy of Science in St. Petersburg in August 1735: the drawing at right comes from his paper. His key insight is that although walking along the banks or around the islands may get us from one bridge to another, the details of those parts of the walk are unimportant: it s the way that the bridges connect the land masses that matters.

5 Notions and Notation The drawing at right is an example of a graph. The dots are called vertices or nodes. The curves connecting them are edges or arcs. The number of edges attached to a vertex is the vertex s degree. A walk through the graph that doesn t re-use any edges is a trail. A trail that uses each edge exactly once is an Eulerian trail. An Eulerian trail that starts an ends at the same place is an Eulerian tour Armed with all this new vocabulary, we can translate (and then solve) the Königsberg Bridge Problem. It becomes: Is there an Eulerian tour in this graph?

6 No Eulerian Tours in Königsberg It s not hard to see that such a tour is impossible: Imagine that we start at the rightmost vertex: we must leave by taking one of its three edges, which we then can t use again. Later in the walk, we ll need to return, using up another edge. Then we ll have to leave a second time (otherwise the third edge will go unused). But then we ll have used all three edges and there will be no way to get back: our walk can t finish where it started No Eulerian tour. Similar reasoning shows that a graph can t have an Eulerian tour if any of its vertices has odd degree. It could, however, have an Eulerian trail that started and finished on different vertices: the first and last vertex could have odd degree, but all others would still have to be even.

7 Another problem, not obviously involving graphs... Ribonucleic acid (RNA) is a family of polymeric molecules whose members play many roles 1 in the chemistry of living cells. RNA is similar to its more famous sister molecule, DNA (deoxyribonucleic acid), in that it consists of long chains composed of 4 subunits called bases. For RNA the bases are U, C, G, and A and they can be combined in any order, so an RNA molecule can be represented by a string such as AGUCAGUGAGCA Sequencing RNA In the early days of RNA research (1960 s) biochemists could sequence accurately only rather short pieces of RNA, but were interested in much longer chains. They addressed the problem by chopping up the longer chains into smaller fragments that they could sequence directly. These assemblies of smaller fragments are called digests of the original molecule and one way to sequence a long chain is to compare and combine the results of two different sorts of digest. 1 If you want to learn more about RNA and its amazingly rich biochemistry, I recommend a book by (Manchester s own) Terry Brown, Genomes 3, Garland Scientific (2006). The previous edition, Genomes 2, is freely available online at

8 Two different sorts of digest Our first enzyme cuts the RNA sequence after every G, while the second cuts after both U and C: AGUCAGUGAGCA AGUCAGUGAGCA G-digest AG UCAG UG AG CA AGU C AGU GAGC A Of course, the digests don t come so neatly organized: in a real scientific problem we d get two jumbles of short fragments like G-digest: : AG AG CA UG UCAG A C AGU AGU GAGC and have to work out how to rearrange them in a sensible order.

9 A thought experiment: digesting the digests Imagine attacking each fragment from the G-digest with the UC-enzyme. They d break down as follows: AG UCAG AG UG U C AG AG CA U G AG C A The small pieces produced in this way are called extended bases. Note that we d get the same set of extended bases if we d started with the fragments of the and attacked then with the G-enzyme. We ll mainly be interested in those cases, such as UCAG U C AG, where a second digest would produce two or more extended bases: we ll use them to draw a very helpful graph.

10 A new kind of graph The double-digests that would yield two or more pieces are CA C A UCAG U C AG UG U G We ll use them to draw a graph like, AG U G C A C which has: a vertex for each extended base that appears on the right-hand side of one of the digests at the top of the slide; a directed edge (a bit like a bridge with a one-way street on it) connecting the first and last extended base in each of the digests-of-digests and for those cases where we got three or more extended bases, an edge label.

11 Eulerian trails to the rescue! If we do the same thing with the other set of digests attacking the original UC-fragments with the G enzyme and then add the corresponding edges to our graph we end up with: AG AG U G C A C There s exactly one (directed) Eulerian trail through this graph and, if we write down the labels of its vertices and edges in the order we encounter them we get AG U C AG U G AG C A = AGUCAGUGAGCA which is the original sequence!

12 A glimpse of why this works AG U C AG U G AG C A G-digest AG U C AG U G AG C A The diagram above shows two views of our example sequence, one with G-fragments shown in red rounded boxes and another with the UC-fragments in blue boxes: in both cases, the extended fragments are shown separated by dots. It s only the larger fragments those containing two or more extended bases that contribute to the graph. These large fragments overlap in a very particular way: the extended base that marks the end of a large blue fragment also marks the beginning of the next large red one. The endpoints of large fragments are the vertices of our graph, while the graph s directed edges, along with their labels, come from to the big fragments themselves.

13 Afterword: problems to think about on the way home 1 Back to Königsberg: I argued that there can t be an Eulerian tour (in an undirected graph) if there s a vertex with odd degree, but what if all vertices have even degree... is there always a tour then? 2 What s the corresponding result for directed graphs? Hint: it s helpful to define separate in- and out-degrees. The former counts the directed edges pointing into a vertex, while the latter counts the edges pointing out. 3 Does the RNA-sequencing approach described here always yield a unique answer? Hint: draw graphs and find sequences consistent with the examples below. G-digest G G U UAAAG U AAAGGGU G CG UG ACC C C GU GGAC G AG UG UCG C GG GU AGU And if you re super-keen, I first learned about this problem from a textbook Edgar G. Goodaire and Michael M. Paramenter (2006). Discrete Mathematics with Graph Theory, 3rd Edition, ISBN whose authors give lots of related exercises.