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  • kenneth_chiu
    replied
    What Pezner is referring to here is if you have one entrance and two exits from a node. You know that there is only one way to visit the node x, but observe two exits. This can't happen if you only visit the node once.
    Hm...that would be if a node had in-total-multiplicity of 1, and and out-total-multiplicity of 2, correct? That's not the case in Fig. 5b, though. In fact, in Fig. 5b, the multiplicities don't seem to match the paths. My understanding is that in that figure, there are three paths corresponding to three reads. In that case, the multiplicity of edge x should be 3, not 2, correct?

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  • timydaley
    replied
    Each node is a distinct (k-1)-mer and the paths are k-mers connecting different paths. They may lie on the same read or on multiple reads.

    Consider the following example, using 3-mers. We have 3 6-bp reads: ATATAT, ATATGC, GCTATA. We have the following distinct 2-mers: AT, TA, TG, GC, and CT. AT -> TA (by ATA), AT -> TG (by ATG), TG -> GC (by TGC), GC -> CT (by GCT), CT -> TA (by CTA), and TA -> AT (by TAT).

    In actual practice, this gets more complicated by including reverse compliments and self loops (i.e. AA -> AA (by AAA)).

    What Pezner is referring to here is if you have one entrance and two exits from a node. You know that there is only one way to visit the node x, but observe two exits. This can't happen if you only visit the node once.
    Last edited by timydaley; 10-03-2013, 09:23 AM.

    Leave a comment:


  • kenneth_chiu
    started a topic "Multiplicity" in 2001 Pevzner EULER paper

    "Multiplicity" in 2001 Pevzner EULER paper

    I have a question about exactly how the multiplicity of an edge in the de Bruijn graph is determined. It seems that the multiplicity is simply the number of reads which covers an edge.

    However, this seems to not be the complete story. Later, the paper goes on to discuss x,y-detachment. At the bottom of page 9752, left column, they state:
    The second condition implies that the Eulerian Superpath
    Problem has no solution, because P, Px,y1, and Px,y2 impose
    three different scenarios for just two visits of the edge x.

    My assumption here is that each of these paths is a read. Thus, there are three reads. Thus, the multiplicity of x would always be three. In other words, this second condition can never occur in a graph that was created from a set of reads, because there would be three visits (multiplicity three), rather than two visits.

    What am I missing?

    The paper is here.

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