[Pharo-project] [Tuning] Set class>>new:
stephane.ducasse at inria.fr
Wed Jan 11 08:39:54 CET 2012
no time to think about it but I found that interesting.
Begin forwarded message:
> From: Paul Baumann <paul.baumann at theice.com>
> Subject: [vwnc] [Tuning] Set class>>new:
> Date: January 11, 2012 1:11:19 AM GMT+01:00
> To: "'VWNC'" <vwnc at cs.uiuc.edu>
> Inspect this code to see hashed collection instances grouped by their free space percentage:
> ^((ObjectMemory allInstancesOfAny: Set withAllSubclasses asArray) asOrderedCollection
> reject: [:ea | ea size < 10 ])
> groupedBy: [:ea | ea capacity / ea size - 1.0 ]
> Hashing collections in VW need to have extra space allocated to act as search limits between logical groups. Sets performance drops there are aren't many free slots allocated. When there are only 25% free slots allocated then the collection grows. VW rounds up to a prime number for instance creation and then targets 33% rounded up to a prime for growths afterwards. A target of 33% rounded to prime skimps on memory at the expense of performance.
> As a collection grows to the minimum 25% free space goal (in #fullCheck) there is a 75% chance of crossing into the next logical bucket before finding a nil value. Crossing into the next logical bucket just once can double the number of slots searched for a miss. Four logical buckets may be traversed for each search-miss by the time the 25% minimum is reached.
> When VW skimps on allocation then it has to grow collections more frequently and has to do more comparisons for each lookup. That is a general performance problem, but there is another problem with how VW allocates memory for pre-sized collections...
> When you see collections created for an expected size like these:
> anArray := Array new: 1000.
> aSet := Set new: 1000.
> anOc := OrderedCollection new: 1000.
> aDict := Dictionary new: 1000.
> How many items would you expect each of these collections to efficiently contain? Clearly the Array will be exactly 1000. The OrderedCollection will take up to 1000 additions before it needs to grow. Pre-sized collections are created to avoid incremental growth costs. Is it not common and natural to think that the Set and Dictionary would also be created to efficiently contain 1000 items too? Would you be surprised to learn that all subclasses of Set in VW (and likely only VW) are inefficient with this kind of pre-allocation when you add 1000 objects to them?
> No application code should be expected to know how Sets are implemented and to pad sizes having knowledge of VW implementation details! Look at the capacity variances that VW creates even though the Set has been created each time to anticipate 1000 additions:
> (Set withAll: (Array new: 1000)) capacity
> (Set withAll: ((1 to: 1000) collect: [:i | Object new ])) capacity
> (Set new: 1000) capacity
> ((Set new: 1000) addAll: ((1 to: 1000) collect: [:i | Object new ]); yourself) capacity
> ((Set new) addAll: ((1 to: 1000) collect: [:i | Object new ]); yourself) capacity
> The first four should all have the same result. The last one can be different because it is not pre-sized. The third example shows that only 0.9% free space is allocated in addition to the requested size. The fourth example shows that the under-allocation caused a growth that then over-allocated to end at 102.7% free space. The costs of the third and fourth example can be seen in more detail with this test:
> | objects set timings cap0 t0 t1 |
> objects := (1 to: ObjectMemory maximumIdentityHashValue) collect: [:i | Object new].
> timings := OrderedCollection new: objects size.
> t0 := t1 := Time microsecondClock.
> set := Set new: objects size.
> cap0 := set capacity.
> objects size to: 1 by: -1 do: [ :i |
> set add: (objects at: i).
> i \\ 1000 = 1 ifTrue: [timings add: (t1 negated + (t1 :=Time microsecondClock))].
> (Array new: 5)
> at: 1 put: cap0; "Starting capacity"
> at: 2 put: set capacity; "Ending capacity"
> at: 3 put: (1 - (set capacity / set size) * -100.0); "Percentage of free slots"
> at: 4 put: timings asArray; "Timings of each 1000"
> at: 5 put: t1 - t0; "Total time"
> #(17417 34939 113.264 #(147 264 262 261 262 262 265 261 261 260 261 260 260 2474 184 184 184) 6312) "primes"
> A difference between the first and second numbers indicates growth costs were incurred while adding. The third number shows there were 1.13 free slots allocated for each logical bucket. The last number is total execution time. Notice that performance of each group improves 41% just after the collection grows.
> Load a package named ICXVwHashTuning from the Cincom Public Code Repository and try the test again.
> #(28670 28670 74.9985 #(104 182 179 179 180 263 180 183 185 184 190 189 188 188 189 254 188) 3205) "After loading ICXVwHashTuning"
> Total execution time was cut by 49% while consuming 18% less memory and ending with a (very efficient) 75% free slots.
> When application code says "Set new: 1000" then they mean that they want a Set instance that should expect to contain 1000 additions. It is the responsibility of the Set to know how much physical space should be allocated to contain 1000 items. These are simple changes that make a broad performance improvement while also having intuitive performance. Cincom should take notice of opportunities like this but at least you can improve the performance of your existing code if they don't.
> You might want to also execute "(ObjectMemory allInstancesOfAny: Set withAllSubclasses asArray) do: [:ea | ea trim ]." to optimize free space allocation for all the hashed instances. You may notice a performance boost from doing that once after loading ICXVwHashTuning.
> Paul Baumann
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