Re: [RFC] kernel/lockdep: use BFS(breadth-first search) algorithmto search target
From: Ming Lei
Date: Tue May 26 2009 - 10:37:03 EST
On Tue, 26 May 2009 15:59:34 +0200
Peter Zijlstra <a.p.zijlstra@xxxxxxxxx> wrote:
> On Tue, 2009-05-26 at 21:54 +0800, Ming Lei wrote:
> > Hi,All
> > Currently lockdep uses recursion DFS(depth-first search) algorithm
> > to search target in checking lock
> > circle(check_noncircular()),irq-safe ->
> > irq-unsafe(check_irq_usage()) and irq inversion when adding a new
> > lock dependency. I plan to replace the current DFS with BFS, based
> > on the following consideration:
> > 1,no loss of efficiency, no matter DFS or BFS, the running
> > time are O(V+E) (V is vertex count, and E is edge count of one
> > graph);
> > 2,BFS may be easily implemented by circular queue and
> > consumes much less kernel stack space than DFS for DFS is
> > implemented by recursion, we know kernel stack is very limited, eg.
> > 4KB.
> > 3, The shortest path can be obtained by BFS if the target is
> > found, but can't be got by DFS. By the shortest path, we can
> > shorten the lock dependency chain and help to troubleshoot
> > lock problem easier than before.
> > Any suggestions, objections or viewpoint?
> Ah, replace the full cycle detection might be worth it, esp with that
> pre-allocated stack you used. Its all serialized on the graph lock
> I'm not sure about 3, though, since we search on adding a each new
> dependency we'll only ever have a choice between cycles when one new
> dependency generates two cycles at the same time. Something I think is
IMHO, maybe it is not much, but is not rare, for example, the prev node
is in a graph(GA) and the next node is in another graph(GB). If GA and
GB is not connected, the edge of prev to next can't generate a cycle.
If GA and GB is connected, it is possible to generate one or multiple
cycles,which depends on the connect degree between GA and GB.
BTW: BFS in the previous patch finds a shorter path in the
> But yes, it wuold be nice to get rid of current recursive
> algorithm there.
OK, I'll start to do it.
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