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Ctrie Data Structure
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Aleksandar Prokopec
February 28, 2012
Programming
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210
Ctrie Data Structure
The description of the Ctrie data structure from PPoPP 2012.
Aleksandar Prokopec
February 28, 2012
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Transcript
Concurrent Tries with Efficient Non-blocking Snapshots Aleksandar Prokopec Phil Bagwell
Martin Odersky École Polytechnique Fédérale de Lausanne Nathan Bronson Stanford
Motivation val numbers = getNumbers() // compute square roots numbers
foreach { entry => x = entry.root n = entry.number entry.root = 0.5 * (x + n / x) if (abs(entry.root - x) < eps) numbers.remove(entry) }
Hash Array Mapped Tries (HAMT)
Hash Array Mapped Tries (HAMT) 0 = 0000002
Hash Array Mapped Tries (HAMT) 0
Hash Array Mapped Tries (HAMT) 0 16 = 0100002
Hash Array Mapped Tries (HAMT) 0 16
Hash Array Mapped Tries (HAMT) 0 16 4 = 0001002
Hash Array Mapped Tries (HAMT) 16 0 4 = 0001002
Hash Array Mapped Tries (HAMT) 16 0 4
Hash Array Mapped Tries (HAMT) 16 0 4 12 =
0011002
Hash Array Mapped Tries (HAMT) 16 0 4 12 =
0011002
Hash Array Mapped Tries (HAMT) 16 0 4 12
Hash Array Mapped Tries (HAMT) 16 33 0 4 12
Hash Array Mapped Tries (HAMT) 16 33 0 4 12
48
Hash Array Mapped Tries (HAMT) 16 0 4 12 48
33 37
Hash Array Mapped Tries (HAMT) 16 4 12 48 33
37 0 3
Hash Array Mapped Tries (HAMT) 4 12 16 20 25
33 37 0 1 8 9 3 48 57
Immutable HAMT • used as immutable maps in functional languages
4 12 16 20 25 33 37 0 1 8 9 3
Immutable HAMT • updates rewrite path from root to leaf
4 12 16 20 25 33 37 0 1 8 9 3 4 12 8 9 11 insert(11)
Immutable HAMT • updates rewrite path from root to leaf
4 12 16 20 25 33 37 0 1 8 9 3 4 12 8 9 11 insert(11) efficient updates - logk (n)
Node compression 48 57 48 57 1 0 1 0
48 57 1 0 1 0 48 57 10 BITPOP(((1 << ((hc >> lev) & 1F)) – 1) & BMP)
Node compression 48 57 48 57 1 0 1 0
48 57 1 0 1 0 48 57 10 48 57
Ctrie Can mutable HAMT be modified to be thread-safe?
Ctrie insert 4 9 12 16 20 25 33 37
0 1 3 48 57 17 = 0100012
Ctrie insert 4 9 12 16 20 25 33 37
0 1 3 48 57 17 = 0100012 16 17 1) allocate
Ctrie insert 4 9 12 20 25 33 37 0
1 3 48 57 17 = 0100012 16 17 2) CAS
Ctrie insert 4 9 12 20 25 33 37 0
1 3 48 57 17 = 0100012 16 17
Ctrie insert 4 9 12 33 37 0 1 3
48 57 18 = 0100102 16 17 20 25
Ctrie insert 4 9 12 33 37 0 1 3
48 57 18 = 0100102 16 17 20 25 1) allocate 16 17 18
Ctrie insert 4 9 12 33 37 0 1 3
48 57 18 = 0100102 20 25 2) CAS 16 17 18
Ctrie insert 4 9 12 33 37 0 1 3
48 57 18 = 0100102 20 25 2) CAS 16 17 18 Unless…
Ctrie insert 4 9 12 33 37 0 1 3
48 57 18 = 0100102 16 17 20 25 T1-1) allocate 16 17 18 Unless… 28 = 0111002 T1 T2
Ctrie insert 4 9 12 0 1 3 18 =
0100102 16 17 20 25 T1-1) allocate 16 17 18 Unless… 28 = 0111002 T1 T2 20 25 28 T2-1) allocate
Ctrie insert 4 9 12 0 1 3 18 =
0100102 16 17 20 25 T1-1) allocate 16 17 18 28 = 0111002 T1 T2 20 25 28 T2-2) CAS
Ctrie insert 4 9 12 0 1 3 18 =
0100102 16 17 20 25 T1-2) CAS 16 17 18 28 = 0111002 T1 T2 20 25 28 T2-2) CAS
Ctrie insert 4 9 12 0 1 3 18 =
0100102 16 17 20 25 16 17 18 28 = 0111002 T1 T2 20 25 28 Lost insert!
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 20 25 Solution: I-nodes
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 20 25 18 = 0100102 28 = 0111002 T1 T2
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 T1 T2 20 25 18 = 0100102 28 = 0111002 16 17 18 20 25 28 T2-1) allocate T1-1) allocate
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 T1 T2 20 25 16 17 18 20 25 28 T2-2) CAS T1-2) CAS
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 18 20 25 28
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 18 20 25 28 Idea: once added to the Ctrie, I-nodes remain present.
Ctrie insert – 2nd attempt 4 9 12 0 1
3 16 17 18 20 25 28 Remove operation supported as well - details in the paper.
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 0
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 0
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 0
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 0
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 1
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 2
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 3
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 5
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 5 actual size = 12
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 5 0 1 actual size = 12
Ctrie size 4 9 12 0 1 3 16 17
18 20 25 28 size = 5 0 1 CAS actual size = 11
Ctrie size 4 9 12 16 17 18 20 25
28 size = 5 0 1 actual size = 11
Ctrie size 4 9 12 16 17 18 20 25
28 size = 6 0 1 actual size = 11
Ctrie size 4 9 12 16 17 18 20 25
28 size = 6 0 1 actual size = 11 19
Ctrie size 4 9 12 16 17 18 20 25
28 size = 6 0 1 actual size = 11 16 17 18 19
Ctrie size 4 9 12 16 17 18 20 25
28 size = 6 0 1 actual size = 12 16 17 18 19 CAS
Ctrie size 4 9 12 20 25 28 size =
6 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
6 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
7 0 1 actual size = 9 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
8 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
9 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
10 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
11 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
12 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
13 0 1 actual size = 12 16 17 18 19
Ctrie size 4 9 12 20 25 28 size =
13 0 1 actual size = 12 16 17 18 19 But the size was never 13!
Global state information 4 9 12 20 25 28 0
1 16 17 18 19 • size • find • filter • iterator
Global state information 4 9 12 20 25 28 0
1 16 17 18 19 • size • find • filter • iterator snapshot
Snapshot using locks 4 9 12 20 25 28 0
1 16 17 18 19
Snapshot using locks 4 9 12 20 25 28 0
1 16 17 18 19 • copy expensive
Snapshot using locks 4 9 12 20 25 28 0
1 16 17 18 19 • copy expensive • not lock-free
Snapshot using locks 4 9 12 20 25 28 0
1 16 17 18 19 • copy expensive • not lock-free • can insert or remove remain lock-free? 0 1 2 CAS
Snapshot using locks 4 9 12 20 25 28 0
1 16 17 18 19 • copy expensive • not lock-free • can insert or remove remain lock-free? 0 1 2 CAS
Snapshot using logs 4 9 12 20 25 28 0
1 16 17 18 19 • keep a linked list of previous values in each I-node
Snapshot using logs 4 9 12 20 25 28 0
1 16 17 18 19 0 1 2 • keep a linked list of previous values in each I-node
Snapshot using logs 4 9 12 20 25 28 0
1 16 17 18 19 • keep a linked list of previous values in each I-node • when is it safe to delete old entries? 0 1 2
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 root
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 root
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 snapshot! root
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 snapshot! #2 root 1) create new I-node at #2
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 snapshot! #2 root 2) set snapshot snapshot #1
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 snapshot! #2 root 3) CAS root to new I-node snapshot #1
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 2
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 2 generation #2 - ok!
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 2 generation #1 not ok, too old!
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root 1) create updated node at #2 snapshot #1 2 #2 #2
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root 2) CAS to the updated node snapshot #1 2 #2 #2
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 2 #2 #2 #1 too old!
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 2 #2 #2 4 9 12 #2 1) create updated node at #2
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 2 #2 #2 4 9 12 #2 2) CAS
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 subsequent insert #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 finally, create a new leaf and CAS
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 another insert #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 another insert #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 But... this won't really work... why? #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3 T2: remove 19 16 17 18
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3 T2: remove 19 16 17 18 CAS
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3 T2: remove 19 16 17 18 CAS How to fail this last CAS?
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3 T2: remove 19 16 17 18 DCAS How to fail this last CAS? DCAS
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3 T2: remove 19 16 17 18 How to fail this last CAS? DCAS - software based DCAS
Snapshot using immutability 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 0 1 2 3 T2: remove 19 16 17 18 How to fail this last CAS? DCAS - software based ...creates intermediate objects DCAS
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 T2: remove 19 16 17 18 prev 1) set prev field
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 T2: remove 19 16 17 18 prev 2) CAS
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 T2: remove 19 16 17 18 prev 3) read root generation
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 16 17 18 prev 4) if root generation changed CAS prev to FailedNode(prev) FN
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 16 17 18 prev 4) if root generation changed CAS prev to FailedNode(prev) FN
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 16 17 18 prev 5) CAS to previous value FN
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 16 17 18 prev 4) if root generation unchanged CAS prev to null
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 16 17 18 4) if root generation unchanged CAS prev to null
GCAS - generation-compare-and-swap 4 9 12 20 25 28 0
1 16 17 18 19 #1 #1 #1 #1 #1 #2 root snapshot #1 #2 #2 4 9 12 #2 0 1 2 3 1) Replace all CAS with GCAS 2) Replace all READ with GCAS_READ (which checks if prev field is null)
Snapshot-based iterator def iterator = if (isSnapshot) new Iterator(root) else
snapshot().iterator()
Snapshot-based size def size = { val sz = 0
val it = iterator while (it.hasNext) sz += 1 sz }
Snapshot-based size def size = { val sz = 0
val it = iterator while (it.hasNext) sz += 1 sz } Above is O(n). But, by caching size in nodes - amortized O(logk n)! (see source code)
Snapshot-based atomic clear def clear() = { val or =
READ(root) val nr = new INode(new Gen) if (!CAS(root, or, nr)) clear() } (roughly)
Evaluation - quad core i7
Evaluation – UltraSPARC T2
Evaluation – 4x 8-core i7
Evaluation – snapshot
Conclusion • snapshots are linearizable and lock-free • snapshots take
constant time • snapshots are horizontally scalable • snapshots add a non-significant overhead to the algorithm if they aren't used • the approach may be applicable to tree-based lock-free data-structures in general (intuition)
Thank you!