From: Venkatesh Pallipadi on
On Tue, Jun 1, 2010 at 10:42 AM, Peter Zijlstra <peterz(a)infradead.org> wrote:
> On Tue, 2010-06-01 at 10:34 -0700, Venkatesh Pallipadi wrote:
>>
>> Yes. This will cover all the cases on insert. But on erase, there is
>> still a case where a rotate of sibling node is done during the
>> re-coloration process. There we have a child change on sibling's
>> child. I am not able to think of any easy way to handle that case.
>
> Let me go draw some figures with pen and paper to match up the erase
> path with the rb_augment_erase_begin() code, because I can't quite spot
> the case we're missing.
>
> If you have it handy, ascii art might help..

It is this case

P
/ \
N S
/ \
SL SR

changing to

P
/ \
N SL
\
S
\
SR

This can happen in __rb_erase_color, where 'other' points to sibling node.

Thanks,
Venki
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From: Peter Zijlstra on
On Tue, 2010-06-01 at 11:09 -0700, Venkatesh Pallipadi wrote:
> On Tue, Jun 1, 2010 at 10:42 AM, Peter Zijlstra <peterz(a)infradead.org> wrote:
> > On Tue, 2010-06-01 at 10:34 -0700, Venkatesh Pallipadi wrote:
> >>
> >> Yes. This will cover all the cases on insert. But on erase, there is
> >> still a case where a rotate of sibling node is done during the
> >> re-coloration process. There we have a child change on sibling's
> >> child. I am not able to think of any easy way to handle that case.
> >
> > Let me go draw some figures with pen and paper to match up the erase
> > path with the rb_augment_erase_begin() code, because I can't quite spot
> > the case we're missing.
> >
> > If you have it handy, ascii art might help..
>
> It is this case
>
> P
> / \
> N S
> / \
> SL SR
>
> changing to
>
> P
> / \
> N SL
> \
> S
> \
> SR

Right, but see: http://en.wikipedia.org/wiki/Red-black_tree
That is delete_case5, however then we fall into delete_case6 and perform
a left rotation.

So suppose we start with the tree:

P P P SL
/ \ / \ / \ / \
D S --> N S --> N SL --> P S
\ / \ / \ \ / \
N SL SR SL* SR S* N SR
\
SR

and then remove D, delete case 5 and finally delete case 6, * marks red.

rb_augment_erase_begin(D) will return N, and then rb_augment_path(N)
will re-augment: N, P, SL and S.


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From: Venkatesh Pallipadi on
On Tue, Jun 1, 2010 at 11:42 AM, Peter Zijlstra <peterz(a)infradead.org> wrote:
> On Tue, 2010-06-01 at 11:09 -0700, Venkatesh Pallipadi wrote:
>> On Tue, Jun 1, 2010 at 10:42 AM, Peter Zijlstra <peterz(a)infradead.org> wrote:
>> > On Tue, 2010-06-01 at 10:34 -0700, Venkatesh Pallipadi wrote:
>> >>
>> >> Yes. This will cover all the cases on insert. But on erase, there is
>> >> still a case where a rotate of sibling node is done during the
>> >> re-coloration process. There we have a child change on sibling's
>> >> child. I am not able to think of any easy way to handle that case.
>> >
>> > Let me go draw some figures with pen and paper to match up the erase
>> > path with the rb_augment_erase_begin() code, because I can't quite spot
>> > the case we're missing.
>> >
>> > If you have it handy, ascii art might help..
>>
>> It is this case
>>
>> � � P
>> � �/ \
>> � N � S
>> � � �/ \
>> � � SL SR
>>
>> changing to
>>
>> � � P
>> � �/ \
>> � N �SL
>> � � � \
>> � � � �S
>> � � � � \
>> � � � � SR
>
> Right, but see: http://en.wikipedia.org/wiki/Red-black_tree
> That is delete_case5, however then we fall into delete_case6 and perform
> a left rotation.
>
> So suppose we start with the tree:
>
> � �P � � � � � � � � �P � � � � � � � P � � � � � � � �SL
> �/ � \ � � � � � � � / \ � � � � � � / \ � � � � � � � / \
> �D � � S � � �--> � �N �S � � �--> � N � SL � � --> � �P � S
> �\ � / \ � � � � � � �/ \ � � � � � � � �\ � � � � � / � � \
> � N SL �SR � � � � �SL* SR � � � � � � � �S* � � � �N � � �SR
> � � � � � � � � � � � � � � � � � � � � � �\
> � � � � � � � � � � � � � � � � � � � � � � SR
>
> and then remove D, delete case 5 and finally delete case 6, * marks red.
>
> rb_augment_erase_begin(D) will return N, and then rb_augment_path(N)
> will re-augment: N, P, SL and S.
>

Yes. I had missed that rotate_right of parent following the
rotate_left of sibling (or vice-versa). Thanks for fixing this. The
latest patch looks good.

Acked-by: Venkatesh Pallipadi <venki(a)google.com>
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From: Suresh Siddha on
On Tue, 2010-06-01 at 11:42 -0700, Peter Zijlstra wrote:
> On Tue, 2010-06-01 at 11:09 -0700, Venkatesh Pallipadi wrote:
> > On Tue, Jun 1, 2010 at 10:42 AM, Peter Zijlstra <peterz(a)infradead.org> wrote:
> > > On Tue, 2010-06-01 at 10:34 -0700, Venkatesh Pallipadi wrote:
> > >>
> > >> Yes. This will cover all the cases on insert. But on erase, there is
> > >> still a case where a rotate of sibling node is done during the
> > >> re-coloration process. There we have a child change on sibling's
> > >> child. I am not able to think of any easy way to handle that case.
> > >
> > > Let me go draw some figures with pen and paper to match up the erase
> > > path with the rb_augment_erase_begin() code, because I can't quite spot
> > > the case we're missing.
> > >
> > > If you have it handy, ascii art might help..
> >
> > It is this case
> >
> > P
> > / \
> > N S
> > / \
> > SL SR
> >
> > changing to
> >
> > P
> > / \
> > N SL
> > \
> > S
> > \
> > SR
>
> Right, but see: http://en.wikipedia.org/wiki/Red-black_tree
> That is delete_case5, however then we fall into delete_case6 and perform
> a left rotation.
>
> So suppose we start with the tree:
>
> P P P SL
> / \ / \ / \ / \
> D S --> N S --> N SL --> P S
> \ / \ / \ \ / \
> N SL SR SL* SR S* N SR
> \
> SR
>
> and then remove D, delete case 5 and finally delete case 6, * marks red.
>
> rb_augment_erase_begin(D) will return N, and then rb_augment_path(N)
> will re-augment: N, P, SL and S.


P SL
/ \ / \
N S ---> N S
/ / \ / \
C SL SR C SR

If P needs to be removed, we need to re-augment S also in this case,
right? It looks like we are not handling this case.

thanks,
suresh



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From: Venkatesh Pallipadi on
On Wed, Jun 2, 2010 at 2:11 PM, Suresh Siddha <suresh.b.siddha(a)intel.com> wrote:
> On Tue, 2010-06-01 at 11:42 -0700, Peter Zijlstra wrote:
>> On Tue, 2010-06-01 at 11:09 -0700, Venkatesh Pallipadi wrote:
>> > On Tue, Jun 1, 2010 at 10:42 AM, Peter Zijlstra <peterz(a)infradead.org> wrote:
>> > > On Tue, 2010-06-01 at 10:34 -0700, Venkatesh Pallipadi wrote:
>> > >>
>> > >> Yes. This will cover all the cases on insert. But on erase, there is
>> > >> still a case where a rotate of sibling node is done during the
>> > >> re-coloration process. There we have a child change on sibling's
>> > >> child. I am not able to think of any easy way to handle that case.
>> > >
>> > > Let me go draw some figures with pen and paper to match up the erase
>> > > path with the rb_augment_erase_begin() code, because I can't quite spot
>> > > the case we're missing.
>> > >
>> > > If you have it handy, ascii art might help..
>> >
>> > It is this case
>> >
>> > � � P
>> > � �/ \
>> > � N � S
>> > � � �/ \
>> > � � SL SR
>> >
>> > changing to
>> >
>> > � � P
>> > � �/ \
>> > � N �SL
>> > � � � \
>> > � � � �S
>> > � � � � \
>> > � � � � SR
>>
>> Right, but see: http://en.wikipedia.org/wiki/Red-black_tree
>> That is delete_case5, however then we fall into delete_case6 and perform
>> a left rotation.
>>
>> So suppose we start with the tree:
>>
>> � � P � � � � � � � � �P � � � � � � � P � � � � � � � �SL
>> � / � \ � � � � � � � / \ � � � � � � / \ � � � � � � � / \
>> �D � � S � � �--> � �N �S � � �--> � N � SL � � --> � �P � S
>> � \ � / \ � � � � � � �/ \ � � � � � � � �\ � � � � � / � � \
>> � �N SL �SR � � � � �SL* SR � � � � � � � �S* � � � �N � � �SR
>> � � � � � � � � � � � � � � � � � � � � � � \
>> � � � � � � � � � � � � � � � � � � � � � � �SR
>>
>> and then remove D, delete case 5 and finally delete case 6, * marks red.
>>
>> rb_augment_erase_begin(D) will return N, and then rb_augment_path(N)
>> will re-augment: N, P, SL and S.
>
>
> � � �P � � � � � � � � � � � SL
> � �/ �\ � � � � � � � � � � / �\
> � N � �S � � � ---> � � � �N � �S
> �/ � / �\ � � � � � � � � / � � �\
> �C � SL �SR � � � � � � � C � � � �SR
>
> If P needs to be removed, we need to re-augment S also in this case,
> right? It looks like we are not handling this case.
>

rb_augment_erase_begin() should take care of that. In this case, it
will return S as the deepest node and we start the walk-back-to-root
from there.

Thanks,
Venki
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