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Re: [p2-dev] Resolver priority question
- From: Daniel Le Berre <leberre@xxxxxxx>
- Date: Thu, 5 Sep 2013 13:43:59 +0200
- Delivered-to: firstname.lastname@example.org
The penalty for root IUs (explicitely installed IUs) or IUs to install is 1, the other penalties is a power of K
(K for the most recent IU, K**2 for the second most recent IU, etc) where K is the number of installed IUs on the current system + 1.
The relevant par of the code in p2 is here (line 73):
Le 5 sept. 2013 à 13:31, Thomas Hallgren <thomas@xxxxxxx> a écrit :
> Hi Daniel,
> Thanks for an elaborate explanation. It all makes sense. Next question is of course, how is the magnitude of the two penalties (higher number of IU's versus older versions) determined?
> - thomas
> On 2013-09-05 10:48, Daniel Le Berre wrote:
>> Hi Thomas,
>> The resolver tries to install as few IUs as possible, because each installed IU is associated with a penalty.
>> Furthermore, there is also another rule which will prefer most recent versions to older ones.
>> It means that the penalty is smaller for most recent versions than for older versions.
>> So it means that when you use the most recent IUs, solutions with fewer installed IUs will be preferred to solutions with a higher number of IUs.
>> But since at the end we aggregate all penalties, there might be some compensation effects:
>> a solution installing many new recent IUs may be preferred to a solution installing a few older IUs.
>> Le 4 sept. 2013 à 19:23, Thomas Hallgren <thomas@xxxxxxx> a écrit :
>>> Confronted with two OK solutions, what would the p2 resolver do if one solution was resolved by brining in a higher number of IUs? If it prefers the one with a lower IU count, what does it do if that one satisfies some requirements with lower versions?
>>> - thomas
>>> p2-dev mailing list
>> Daniel Le Berre mailto:leberre@xxxxxxx
>> MCF-HDR, CRIL-CNRS UMR 8188 Universite d'Artois
>> p2-dev mailing list
> p2-dev mailing list
Daniel Le Berre mailto:leberre@xxxxxxx
PR, CRIL-CNRS UMR 8188 Universite d'Artois