pfds 5 2+6_4_2

PFDS §5.2+6.4.2 Queues@shtaag

2012年5月19日土曜日

First...

http://www.kmonos.net/pub/Presen/PFDS.pdf




FIFO Queue（2リストキュー）

type foo Queue = foo list x foo list

fun head (x :: f, r) = x

fun tail (x :: f, r) = (f, r)

fun snoc ((f, r), x) = (f, x :: r)

O(1)

O(1)

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



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



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)


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

Banker’s method


snoc 1 -> [] [1] -> [1] []

snoc 2 -> [1] [2]

snoc 3 -> [1] [3,2]

tail -> [] [3,2] -> [2,3] []

tail -> [3] []

tail -> [] []

2

1

1

3

1

1

9

2

2

2

1

1

1

9

real amortized

+1

+1

-2

checkf w/ 1 step

checkf w/ 2 step


snoc 1 -> [] [1] -> [1] []

snoc 2 -> [1] [2]

snoc 3 -> [1] [3,2]

tail -> [] [3,2] -> [2,3] []

tail -> [3] []

tail -> [] []

2

1

1

3

1

1

9

2

2

2

1

1

1

9

real amortized

積み立てておいたcreditを消費

+1

+1

-2


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

Physicist’s method


snoc 1 -> [] [1] -> [1] []

snoc 2 -> [1] [2]

snoc 3 -> [1] [3,2]

tail -> [] [3,2] -> [2,3] []

tail -> [3] []

tail -> [] []

2

1

1

3

1

1

9

2

2

2

1

1

1

9

real amortized

+1

+1

-2


§6.4.2 with Lazy Evaluation

O(1) in amortized time


Queue with O(1) amortized time.

Persistent

Lazy Evaluation

Strict Working Copy



with the Physicist’s Method.


strict(non-amortized)

costunshared

cost

sharedcost

change inpotential

amortizedcost





cost

paidshared cost

change inpotential

paid shared cost = actually executed cost





cost

paidshared cost

change inpotential

max potential=

max shared cost ???


Implementation of PhysicistsQueue

type a Queue = a list x int x a list susp x int x a list


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 lengthsto guarantee front >= rear

working copy of front list

front is suspended, so necessary to keep the access to the prefix for head queries.


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|)


if potential = 0

[ 5, 8] [ 5, 8] [ 2, 4]

snoc

unshared = 1


if potential = 0

[ 5, 8] [ 5, 8] [ 2, 4]

snoc

shared cost of rotation = 2m + 1

force suspended front ++ rear


if potential = 0

[ 5, 8] [ 5, 8] [ 2, 4]

snoc

change of potential = + 2m

= accumulated debt

create debt


if potential = 0

[ 5, 8] [ 5, 8] [ 2, 4]

snoc

complete cost = 2

unshared = 1

shared = 2m + 1

change in potential = 2m


if potential > 0

[ 5, 8, 2, 4] [ 5, 8, 2, 4] [ 1, 3]

potential = 2

snoc

unshared = 1

change in potential = -1


Therefore...

snoc is O(1) amortized time

the cost is 2 or 4


if potential = 0

[ 5, 8] [ 5, 8] [ 2, 4]

tail

[ 8] [ 8] [ 2, 4]


if potential = 0

[ 5, 8] [ 5, 8] [ 2, 4]

tail

unshared cost = 2

shared cost = 2m + 1

change in potential = 2m


if potential > 0

[ 5, 8, 2, 4] [ 5, 8, 2, 4] [ 1, 3]

potential = 2

tail

[ 8, 2, 4] [ 8, 2, 4] [ 1, 3]


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)


Therefore...

tail is in O(1) amortized time

the cost is 4 or 3


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.


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


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


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


pfds 5 2+6_4_2

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