Re: [PATCH 06/11] cgroups: Add res counter common ancestor searching

["Kirill A. Shutemov" <kirill@shutemov.name>]

On Tue, Sep 13, 2011 at 01:11:26AM +0200, Frederic Weisbecker wrote:
> Add a new API to find the common ancestor between two resource
> counters. This includes the passed resource counter themselves.
> 
> Signed-off-by: Frederic Weisbecker <fweisbec@gmail.com>
> Acked-by: Paul Menage <paul@paulmenage.org>
> Cc: Li Zefan <lizf@cn.fujitsu.com>
> Cc: Johannes Weiner <hannes@cmpxchg.org>
> Cc: Aditya Kali <adityakali@google.com>
> Cc: Oleg Nesterov <oleg@redhat.com>
> Cc: Andrew Morton <akpm@linux-foundation.org>
> Cc: Kay Sievers <kay.sievers@vrfy.org>
> Cc: Tim Hockin <thockin@hockin.org>
> Cc: Tejun Heo <tj@kernel.org>
> ---
>  include/linux/res_counter.h |    3 +++
>  kernel/res_counter.c        |   22 ++++++++++++++++++++++
>  2 files changed, 25 insertions(+), 0 deletions(-)
> 
> diff --git a/include/linux/res_counter.h b/include/linux/res_counter.h
> index de4ba29..558f39b 100644
> --- a/include/linux/res_counter.h
> +++ b/include/linux/res_counter.h
> @@ -147,6 +147,9 @@ static inline void res_counter_uncharge(struct res_counter *counter,
>  	res_counter_uncharge_until(counter, NULL, val);
>  }
>  
> +struct res_counter *res_counter_common_ancestor(struct res_counter *l,
> +						struct res_counter *r);
> +
>  /**
>   * res_counter_margin - calculate chargeable space of a counter
>   * @cnt: the counter
> diff --git a/kernel/res_counter.c b/kernel/res_counter.c
> index 0cc9c7e..22f07cc 100644
> --- a/kernel/res_counter.c
> +++ b/kernel/res_counter.c
> @@ -91,6 +91,28 @@ void res_counter_uncharge_until(struct res_counter *counter,
>  	local_irq_restore(flags);
>  }
>  
> +/*
> + * Walk through r1 and r2 parents and try to find the closest common one
> + * between both. If none is found, it returns NULL.
> + */
> +struct res_counter *
> +res_counter_common_ancestor(struct res_counter *r1, struct res_counter *r2)
> +{
> +	struct res_counter *iter;
> +
> +	while (r1) {
> +		iter = r2;
> +		while (iter) {
> +			if (iter == r1)
> +				return iter;
> +			iter = iter->parent;
> +		}
> +
> +		r1 = r1->parent;
> +	}
> +
> +	return NULL;
> +}

Your implementation is O(n*m).

What about:

struct res_counter *
res_counter_common_ancestor(struct res_counter *r1, struct res_counter *r2)
{
	struct res_counter *iter;
	int r1_depth = 0, r2_depth = 0;

	for (iter = r1; iter; iter = iter->parent)
		r1_depth++;
	for (iter = r2; iter; iter = iter->parent)
		r2_depth++;

	while (r1_depth > r2_depth) {
		r1 = r1->parent;
		r1_depth--;
	}
	while (r2_depth > r1_depth) {
		r2 = r2->parent;
		r2_depth--;
	}

	while (r1 != r2) {
		r1 = r1->parent;
		r2 = r2->parent;
	}

	return r1;
}

Untested. It's O(n+m).

-- 
 Kirill A. Shutemov