On 1 November 2012 00:59, Axel Beckert <a...@debian.org> wrote: > I tried to weight removals-of-manual higher than other removals and came > up with the following rule: > > aptitude -o Aptitude::ProblemResolver::SolutionCost='10 * removals - 9 * > removals-of-manual + canceled-actions' remove htop
This cost formula prefers removing manual packages. Assuming that subtraction worked, it finds solutions in this order: 1) remove one manual (10*1 - 9*1 = 1) 2) remove one manual, two auto (10*3 - 9*1 = 21) 3) remove two manual, two auto (10*4 - 9*2 = 22) 4) remove three auto (10*3 - 9*0 = 30) 5) remove one manual, three auto (10*4 - 9*1 = 31) 6) remove four auto (10*4 - 9*0 = 40) where as “removals + 3 * removals-of-manual” gives: 1) remove three auto (3 + 3*0 = 3) 2) remove one manual (1 + 3*1 = 4) 2) remove four auto (4 + 3*0 = 4) 4) remove one manual, two auto (3 + 3*1 = 6) 5) remove one manual, three auto (4 + 3*1 = 7) 6) remove two manual, two auto (4 + 3*2 = 10) which prefers removing automatic packages over manual at a ratio of about 3:1. I presume you were trying to achieve something similar to the second list. There are also other options, such as a adjusting the scales of the later example, a SolutionCost of “removals-of-manual, …”, or adding “reject !~M :UNINST” to Hints. > A simpler, yet less practical SolutionCost value with which we were able > to reproduce the issue is: > > aptitude -o Aptitude::ProblemResolver::SolutionCost='removals - > canceled-actions' remove htop Indeed that is very impractical! The resolver will find that removing two packages and cancelling all requested actions if preferable to just removing those two packages. * On subtraction in SolutionCost Simpler solutions (those with less actions) are often more preferable. This fits well with using only scaled addition, where extra actions increase the cost of a solution by some amount. The costs of actions can be adjusted relative to each other (“prefer this amount of install over that amount of remove”). With subtraction, it becomes possible to prefer increasing amounts of some action over anything else. Consider a cost formula containing “- installs” which would prefer to install a package A and hundreds of others, over simply installing A. Where do those hundreds of other packages come from and what are they related to? Various odd places (the resolver investigates some strange things on it's way to sanity land), and we have all seen such solutions offered later. With subtraction these excessive solutions will be moved right to the front—the more installs the better, and damn anything else. Generally, using subtraction in SolutionCost will lead to strange results. I would be very interested to find a practical example of subtraction that is better than a similar formula using addition. Supporting subtraction is trivial, but lets avoid this unless there are real benefits to doing so. Regards -- To UNSUBSCRIBE, email to debian-bugs-dist-requ...@lists.debian.org with a subject of "unsubscribe". Trouble? Contact listmas...@lists.debian.org
