purely functional data structures

1. PFDS §5.2+6.4.2 Queues
@shtaag
2012年5月19日土曜日

2. First...
http://www.kmonos.net/pub/Presen/PFDS.pdf
2012年5月19日土曜日

3. FIFO Queue（2リストキュー）
type foo Queue = foo list x foo list
fun head (x :: f, r) = x O(1)
fun tail (x :: f, r) = (f, r) O(1)
fun snoc ((f, r), x) = (f, x :: r) O(1)
リスト終端への挿入にconsを用いるため、
リスト後半をreverseして保持
2012年5月19日土曜日

4. FIFO Queue（2リストキュー）
invariant to maintain
ｆはｒが[ ]の時のみ[ ]
これがないとf = [ ]時にheadがr終端からの取り出
しになりheadがO(n)かかる
2012年5月19日土曜日

5. FIFO Queue（2リストキュー）
to maintain invariant...
fun checkf ([ ], r) = (rev r, [ ])
| checkf q = q
fun snoc ((f, r), x) = checkf (f, x :: r)
fun tail (x :: f, r) = checkf (f, r)
2012年5月19日土曜日

6. Banker’s method
snoc = 1 step + 1 credit
tail (w/o reverse) = 1 step
tail (w/ reverse) = (m+1) steps - m credits
tail . tail . tail . snoc 3 . snoc 2 . snoc 1
2012年5月19日土曜日

7. real amortized
snoc 1 -> [] [1] -> [1] [] 2 2
checkf w/ 1 step
snoc 2 -> [1] [2] 1 2 +1
snoc 3 -> [1] [3,2] 1 2 +1
tail -> [] [3,2] -> [2,3] [] 3 1 -2
checkf w/ 2 step
tail -> [3] [] 1 1
tail -> [] [] 1 1
9 9
2012年5月19日土曜日

8. real amortized
snoc 1 -> [] [1] -> [1] [] 2 2
snoc 2 -> [1] [2] 1 2 +1
snoc 3 -> [1] [3,2] 1 2 +1
tail -> [] [3,2] -> [2,3] [] 3 1 -2
tail -> [3] [] 1 1
積み立てておいたcreditを消費
tail -> [] [] 1 1
9 9
2012年5月19日土曜日

9. Physicist’s method
Φ = length of the rear list
snoc = 1 step + 1 potential (for 1 elem)
tail (w/o reverse) = 1 step
tail (w/ reverse) = (m+1) steps - m potentials
2012年5月19日土曜日

10. real amortized
snoc 1 -> [] [1] -> [1] [] 2 2
snoc 2 -> [1] [2] 1 2 +1
snoc 3 -> [1] [3,2] 1 2 +1
tail -> [] [3,2] -> [2,3] [] 3 1 -2
tail -> [3] [] 1 1
tail -> [] [] 1 1
9 9
2012年5月19日土曜日

11. §6.4.2 with Lazy Evaluation
O(1)
in amortized time
2012年5月19日土曜日

12. §6.4.2 with Lazy Evaluation
Queue with O(1) amortized time.
Persistent
Lazy Evaluation
Strict Working Copy
2012年5月19日土曜日

13. §6.4.2 with Lazy Evaluation
with the Physicist’s Method.
change in
shared potential
strict cost
(non-amortized)
cost amortized
unshared cost
cost
2012年5月19日土曜日

14. §6.4.2 with Lazy Evaluation
with the Physicist’s Method.
change in
potential
strict paid
(non-amortized) shared cost
cost
paid shared cost
= actually executed cost
2012年5月19日土曜日

15. §6.4.2 with Lazy Evaluation
with the Physicist’s Method. max potential
=
max shared cost ???
change in
potential
strict paid
(non-amortized) shared cost
cost
2012年5月19日土曜日

16. Implementation of PhysicistsQueue
type a Queue = a list x int x a list susp x int x a list
2012年5月19日土曜日

17. type a Queue = a list x int x a list susp x int x a list
suspended front list + ordinal rear list
lengths for front and rear lists
to calculate the diff of front and rear lengths
to guarantee front >= rear
working copy of front list
front is suspended, so necessary to keep the access to the prefix for head
queries.
2012年5月19日土曜日

18. Potential
length of working copy
diff of front and rear lengths
if those values are = 0, then rotation happens
definition of potential
Ψ(q) = min (2|w|, |f| - |r|)
2012年5月19日土曜日

19. if potential = 0
[ 5, 8] [ 5, 8] [ 2, 4]
snoc
unshared = 1
2012年5月19日土曜日

20. if potential = 0
[ 5, 8] [ 5, 8] [ 2, 4]
snoc
shared cost of rotation = 2m + 1
force suspended front ++ rear
2012年5月19日土曜日

21. if potential = 0
[ 5, 8] [ 5, 8] [ 2, 4]
snoc
change of potential = + 2m create debt
= accumulated debt
2012年5月19日土曜日

22. if potential = 0
[ 5, 8] [ 5, 8] [ 2, 4]
snoc
complete cost = 2
unshared = 1
shared = 2m + 1
change in potential = 2m
2012年5月19日土曜日

23. if potential > 0
[ 5, 8, 2, 4] [ 5, 8, 2, 4] [ 1, 3]
potential = 2
snoc
unshared = 1
change in potential = -1
2012年5月19日土曜日

24. Therefore...
snoc is O(1) amortized time
the cost is 2 or 4
2012年5月19日土曜日

25. if potential = 0
[ 5, 8] [ 5, 8] [ 2, 4]
tail
[ 8] [ 8] [ 2, 4]
2012年5月19日土曜日

26. if potential = 0
[ 5, 8] [ 5, 8] [ 2, 4]
tail
unshared cost = 2
shared cost = 2m + 1
change in potential = 2m
2012年5月19日土曜日

27. if potential > 0
[ 5, 8, 2, 4] [ 5, 8, 2, 4] [ 1, 3]
potential = 2
tail
[ 8, 2, 4] [ 8, 2, 4] [ 1, 3]
2012年5月19日土曜日

28. if potential > 0
[ 5, 8, 2, 4] [ 5, 8, 2, 4] [ 1, 3]
potential = 2
tail
unshared = 1 (from working copy)
paid shared = 1 (from front)
change in potential = -1 for w and -1 for (f-r)
2012年5月19日土曜日

29. Therefore...
tail is in O(1) amortized time
the cost is 4 or 3
2012年5月19日土曜日

30. Theorem 6.2
The amortized costs of snoc and tail are at most 2
and 4, respectively
This leads to that every operation is O(1) order.
2012年5月19日土曜日

31. snoc
without rotation
snoc ((w, lenf, f, lenr, r), x)
= check (w, lenf, f , lenr+1, x :: r)
cost = 1 & decrease in potential = 1
operation for |f| - |r|
amortized cost = 1 - (-1)
=2
2012年5月19日土曜日

32. tail
without rotation
tail (x :: w, lenf, f, lenr, r)
= check (w, lenf-1, $tl (force f), lenr, r)
cost = 2 & decrease in potential = 2
operation for |w| & operation for |f| - |r|
amortized cost = 2 - (-2)
=4
2012年5月19日土曜日

33. With Rotation
|f| = m and |r| = m + 1
shared cost of rotation = 2m + 1
potential of resulting queue = 2|w|
= 2m
amortized cost of snoc
1 + (2m + 1) - 2m = 2
amortized cost of tail
2 + (2m + 1) - 2m = 3
2012年5月19日土曜日