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Thy Shalt Come To The Thy Shalt Come To The TGIWTGIW
From: [email protected]: TGIWWhen: Tonight, 6:00pmWhere: WeH 7220
Nikhil will be talking about his work on Online Scheduling at this week's TGIW. This is joint work with Kedar, Jochen and Amitabh. The schedule and abstract are given below. First years are especially encouraged to attend. […]
October 2001
Guy BlellochGuy Blelloch
Maverick WooMaverick Woo
Back From The FutureBack From The Future
Efficient Finger SearchUsing Eager Walk
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List of ColorsList of Colors
ColorColorColorColorColorColorColorColor
ColorColorColor
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Important Terms/PhrasesImportant Terms/Phrases
Catenable (Cannibal on Halloween…)File“n” (no “m”)Nine (no “line”)NodeTheoryThree (“free” lunch?)
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Set IntersectionSet Intersection
You have two sets
A := {x1, x2, …, xa},
B := {y1, y2, …, yb}
You represent them as BSTs
TA and TB
Give an algorithm to find A Å B
BST: Balanced Search Tree
BST: Balanced Search Tree
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Easy SolutionEasy Solution
WLOG assume a ¸ b.
For each yj 2 TB, ask if yj 2 TA.
O(b log a) timeIs this optimal?Is this optimal?
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An InsightAn Insight
A := {1,2,3, …, 15}B := {1,3,10,11}
Decisions at “top” mostly the same
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Show MoreShow More
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Enters Finger SearchEnters Finger Search
FSearch(fx, y):
Given a finger fx, return finger fy in
O(log d) time, where d is the distance between x and y in the sorted order.
If y 2 T, return fy+ for smallest
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fx: finger on key x
fx: finger on key x
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Why Bother Finger Why Bother Finger Searching?Searching?
Another simple intersection algorithm:f = f0 /* finger on conceptual “y0” */For i = 1 to b /* in-order walk TB */
f = FSearchA(f, yi)if f on yi then print yi
Next i
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Running Time AnalysisRunning Time Analysis
Do b finger searches in TA.
Suppose distances are d1, d2, …, db.
Know i di · a
Total time = O(log d1 + log d2 + … + log db)
Time maximized when all di’s are equal, so let di = a/b.
Total time = O(b log(a/b))
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IteratorsIterators
InitializationTermination ConditionIncrementDe-referencing
InOrder(tree) :=
Iterator it;
for(it.init(tree);
it.hasMore();
it++)
{ // print (*it).node
}
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Special CaseSpecial Case
In-orderWalk
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Complete Binary Search Complete Binary Search TreeTree
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<6 >6
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Node SizeNode Size
struct Node
{
int key;
Node *left, *right;
};
Node needs at least three machine words
Node needs at least three machine words
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Running Time AnalysisRunning Time Analysis
InOrder(node) := If(node != Nil) { InOrder(node.left); // print node.key InOrder(node.right); }
How about time between two consecutive outputs?
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Running Time AnalysisRunning Time Analysis
InOrder(node) := If(node != Nil) { InOrder(node.left); // print node.key InOrder(node.right); }
Amortized O(1) per node
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Space RequirementSpace Requirement
O(log n) activation records on the program stack
InOrder(node) :=
If(node != Nil)
{
InOrder(node.left);
// print node.key
InOrder(node.right);
}
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So Far So Good?So Far So Good?
Version Tree Size Time/Node Space
Recursive n+2n = 3n Amortized O(1) O(log n)
#keys #pointers
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I wonder...I wonder...
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Version Tree Size Time/Node Space
Recursive n+2n = 3n Amortized O(1) O(log n)
Worst case O(1)?
Worst case O(1)?
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Why Bother?Why Bother?
Fast sequentialaccess
slowslow
slowslow
slowslow
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Let's TradeLet's Trade
Neighborpointers
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Node needs 2 extra wordsNode needs 2 extra words
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Faster InsertionFaster Insertion
Indexed Sequence(B+ Trees)
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1 2 3 4 5 6 7 8 9 10 1112 1314 15 /
Doubled node
)Doubled space
Doubled node
)Doubled space
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Oh Well...Oh Well...
Version Tree Size Time/Node SpaceRecursive n+2n = 3n Amortized O(1) O(log n)
Nbr. Pointers n+4n = 5n Worst Case O(1) O(1)
Indexed Seq. 2n+4n = 6n Worst Case O(1) O(1)
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Two TermsTwo Terms
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12Right Parentof 6
Right parent is first ancestorto the right
Right parent is first ancestorto the right
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Two TermsTwo Terms
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h 12, 10, 9 i is “Left spine” of 12 “RL spine” of 8
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Our IdeaOur Idea
Pre-compute right child’s left spineHow? Discover nodes step-by-step
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O(1) time?
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Our "Hand" IteratorOur "Hand" Iterator
Stack of (Node *node, Stack *spine)
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(n,s)Right Parent Stack
This Hand isnot valid. Justfor illustration.
This Hand isnot valid. Justfor illustration.
(n,s)
(n,s)
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Stack of StacksStack of Stacks
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12Right Parent Stack (RPS)
ptr within stack
ptr to spinecurrent nodeparent nodeleft spine node
(n,s)
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InitializationInitialization
Left spine of rootmaintained duringkey insertions
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1
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8RPS
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De-referenceDe-reference
Peek top of stack
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RPS
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TerminationTermination
Is stack empty?
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RPS
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IncrementIncrement
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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RPS
ptr within stack
ptr to spinecurrent nodeparent nodeleft spine node
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1 1 2 2
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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RPS
ptr within stack
ptr to spinecurrent nodeparent nodeleft spine node
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1 1 2 2
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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RPS
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1 1 2 2
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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3
RPS
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Hand on 2Hand on 2
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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RPS
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2 2 3 3
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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RPS
SS
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2 2 3 3
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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SS
RPS
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2 2 3 3
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
RPS
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3 3 4 4
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
RPS
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3 3 4 4
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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8RPS
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3 3 4 4
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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8RPS
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4 4 5 5
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
SS
RPS
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4 4 5 5
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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SS
RPS
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4 4 5 5
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
RPS
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Hand on 5Hand on 5
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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8RPS
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CheckpointCheckpoint
Do you understand how to this algorithm manipulate the Hand?
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Pop QuizPop Quiz
This is the Handon 5, so what isthe Hand on 6?
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
5
6
8 12
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Pop QuizPop Quiz
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12ptr within
stackptr to spinecurrent nodeparent nodeleft spine node
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8 12
Three easy steps1. Pop top cell and
keep its spine (SS)2. Extend spine of
new top cell3. Prepend SS to RPS
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AnswerAnswer
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5 5 6 65 5 6 6
ptr within stack
ptr to spinecurrent nodeparent nodeleft spine node
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12
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CorrectnessCorrectness
A node is either on left spine or not.
If on left spine, init finds it.
If not one left spine, another node must be on its left.
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Running Time AnalysisRunning Time Analysis
Increment has three easy steps1. Pop top cell and
O(1) keep its spine (SS)
2. Extend spine ofO(1)
new top cell3. Prepend SS to RPS O(1)
Total Time Worst case O(1)
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Space RequirementSpace Requirement
Starts with log(n) cellsEach increment
1. Pop top cell and -1keep its spine (SS)
2. Extend spine of +1new top cell
3. Prepend SS to RPS +0Total change 0
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How are we doing?How are we doing?
Version Tree Size Time/Node SpaceRecursive n+2n = 3n Amortized O(1) O(log n)
Nbr. Pointers n+4n = 5n Worst Case O(1) O(1)
Indexed Seq. 2n+4n = 6n Worst Case O(1) O(1)
Eager Walk n+2n = 3n Worst Case O(1) O(log n)
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General CaseGeneral Case
FingerSearch
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1980 Brown, Tarjan1980 Brown, Tarjan
Level-linked2-3 trees
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1 2 3 4 5 6 7 8 9 10 1112 1314 15 /
5 pointersper node
5 pointersper node
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Since 1980...Since 1980...
1988 – Tarjan, Van Wyk Heterogeneous Red-Black
1989 – Pugh Skip List
1995 – Seidel, Aragon Treaps
1996 – Kaplan, Tarjan Purely Functional
Catenable Sorted List
2001 – Blandford, Blelloch HeterogeneousTreaps
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Since 1980...Since 1980...
1988 – Tarjan, Van Wyk Heterogeneous Red-Black
1989 – Pugh Skip List
1993 – Cole (conjectured Splay treesby Sleator, Tarjan 1985)
1995 – Seidel, Aragon Treaps
1996 – Kaplan, Tarjan Purely Functional
Catenable Sorted List
2001 – Blandford, Blelloch HeterogeneousTreaps
1993 – Cole (conjectured Splay treesby Sleator, Tarjan 1985)
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Space-Time Price ChartSpace-Time Price Chart
Version Tree Size Time SpaceLevel-link 2,3 n+5n = 6n Worst case O(log d) O(1)
Hetero. RB n+2n = 3n Amortized O(log d) O(1)
Skip List n+2n = 3n Expected O(log d) O(1)
Splay Tree n+2n = 3n Amortized O(log d) O(1)
Treaps n+3n = 4n Expected O(log d) O(1)
Functional Listn+~3n =
~4nWorst case O(log d) O(1)
Hetero. Treaps
n+2n = 3n Expected O(log d) O(1)
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Our Forward SolutionOur Forward Solution
Same data structure
as in-order walk!
ptr within stack
ptr to spinecurrent nodeparent nodeleft spine node
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Pictorial HandPictorial Handptr within
stackptr to spinecurrent nodeparent nodeleft spine node
curr
RP
peer
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Global LevelGlobal Level
TODO: show how the invariant is. In particular, create a picture saying how long the spines are (for all nodes, it’s all up to the point where the key on top of it on the stack is)
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An ExampleAn Example
curr
RP
peer
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ptr within stack
ptr to spinecurrent nodeparent nodeleft spine node
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Hand InvariantsHand Invariants
Let stack be labeled upwards
1. What nodes should be on RPS?
2.How long are their spines?
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3
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x3
x2
x1
s3
s2
s1
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Skip InvariantsSkip Invariants
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Hand InvariantsHand Invariants
Let stack be h(xn,sn),(xn-1,sn-1),…,(x1,s1)i
1.x1 is on right spine of tree Æ
8 j > 1 : xj-1 is RP of xj
2.8 j : sj is a prefix of left spine of
xj’s right child Æ
8 j : |sj| =
h(xj) – 1 – h(xj+1)
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Hand InvariantsHand Invariants
Let stack be h(xn,sn),(xn-1,sn-1),…,(x1,s1)i
1.x1 is on right spine of tree Æ
8 j > 1 : xj-1 is RP of xj
2.8 j : sj is a prefix of left spine of
xj’s right child Æ
8 j : |sj| =
h(xj) – 1 – h(xj+1)
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Hand InvariantsHand Invariants
Let stack be h(xn,sn),(xn-1,sn-1),…,(x1,s1)i
1.x1 is on right spine of tree Æ
8 j > 1 : xj-1 is RP of xj
2.8 j : sj is a prefix of left spine of
xj’s right child Æ
8 j : |sj| =
h(xj) – 1 – h(xj+1)
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Hand InvariantsHand Invariants
Let stack be h(xn,sn),(xn-1,sn-1),…,(x1,s1)i
1.x1 is on right spine of tree Æ
8 j > 1 : xj-1 is RP of xj
2.8 j : sj is a prefix of left spine of
xj’s right child Æ
8 j : |sj| =
h(xj) – 1 – h(xj+1)
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Hand InvariantsHand Invariants
Let stack be h(xn,sn),(xn-1,sn-1),…,(x1,s1)i
1.x1 is on right spine of tree Æ
8 j > 1 : xj-1 is RP of xj
2.8 j : sj is a prefix of left spine of
xj’s right child Æ
8 j : |sj| =
h(xj) – 1 – h(xj+1)
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Hand InvariantsHand Invariants
Denote height of xj by h(xj).
Define h(xj) = 0 for any j > n.
(In reality, that xj does not exist.)
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h=1
h=2
h=3
h=4
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Sub-tree Search LemmaSub-tree Search Lemma
Given fx to minimum key x of a sub-tree,
we can obtain fy in O(log d) time if y is in
the same sub-tree as x, or if y is the right
parent of the sub-tree’s root. Restoring
invariants of hand also
takes O(log d) time.
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Prove LemmaProve Lemma
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CheckpointCheckpoint
Do you understand what the lemma is?
(Realize that this sub-tree search does not always happen because the target can be somewhere else.)
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v=3
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Where can our destination Where can our destination be?be?FSearch(fcurr,y)
1. curr < y < RP
2. y = RP
3. RP < y < peer
4. y = peer
5. peer < y
curr
RP
peer
Can distinguish cases in O(1) time
Can distinguish cases in O(1) time
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Case 1Case 1
1. Do an increment(now at curr++)
2. Sub-tree search
Increment is O(1), Search is O(log d)
Invariants restored automagically
curr
RP
peer
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Case 2Case 2
Proceed as in case 1
(Sub-tree search fory in will walk right spine as RP 2 , invariants restored automagically)
Analysis is the same as case 1
curr
RP
peer
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Case 3Case 3
1. Proceed as in case 2
to obtain a hand on RP
(search in gives fRP)
2. Do an increment (now on RP++)
3. Sub-tree search
curr
RP
peer
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d is at least size of
O(log d) to arrive at RP
O(1) to arrive at RP++
O(log d) for a sub-tree search
Case 3 AnalysisCase 3 Analysis
curr
RP
peer
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Case 4Case 4
Proceed as in case 3
to obtain fpeer
Analysis is the same as case 3
curr
RP
peer
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Case 5Case 5
1. Find the largest
right parent p · y by
popping RPS (and discard spines)
2. Get fp by fixing hand invariants
3. FSearch(fp, y)
(this time guaranteed to be case 1)
curr
RP
peer
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GraphicallyGraphically
We know it’s case 5
curr
RP
peer
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GraphicallyGraphically
curr
RP
peer
pFind largest parentp · y by popping RPS
p’s RP
p’s peer
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GraphicallyGraphically
p
p’s RP
p’s peerFix invariants toobtain fp
curr
RP
peer
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GraphicallyGraphically
p
p’s RP
p’s peerNow it must becase 1 (or y=p)
curr
RP
peer
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Case 5 AnalysisCase 5 Analysis
Imagine writing down all keys
… curr, ……, p, ……, y …
d
·O(log d)·O(log d)
distance
time
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Skip DetailsSkip Details
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Case 5 AnalysisCase 5 Analysis
Consider RPS before FSearch(fcurr, y).
Let the p be xj.
By invariant 2, |sj| = h(xj) – 1 – h(xj+1). This is the amount of work we have already done previously.
To fix invariant 2, we need to extend sj down to leaf so that |sj| = h(xj) – 1.
) Time = h(xj+1) = v in figure
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v
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Case 5 AnalysisCase 5 Analysis
How many keys between xj+1 and xj?
Answer: 2v-1 – 1
) d > 2v-1 – 1) v = O(log d)
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v
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Space-Time Price ChartSpace-Time Price Chart
Version Tree Size Time SpaceLevel-link 2,3 n+5n = 6n Worst case O(log d) O(1)
Hetero. RB n+2n = 3n Amortized O(log d) O(1)
Skip List n+2n = 3n Expected O(log d) O(1)
Splay Tree n+2n = 3n Amortized O(log d) O(1)
Treaps n+3n = 4n Expected O(log d) O(1)
Functional Listn+~3n =
~4nWorst case O(log d) O(1)
Hetero. Treaps
n+2n = 3n Expected O(log d) O(1)
Eager Walk n+2n = 3n Worst case O(log d) O(log n)
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ExtensionExtension
Slight change will work for any height-balanced BST (e.g. Red-Black, B-Tree)
Treaps? AVL? Splay trees?
Can also support forward and backward simultaneously (messy…)
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It's Halloween!It's Halloween!You are cordially invited to
the first HALLOWEEN PARTY of the Millennium on Halloween at the Church after Dusk
DRESS CODE AND DINNER MENU Humans will be eaten.
DIRECTIONS Follow the smell of blood up Forbes towards Squirrel Hill, turn right onto Wightman, turn second right onto Bartlett. Your liquid or solid of choice will be found at 5516 Bartlett St. Spouses, dates, siblings, housemates, familiars and friends are welcome.
John, Mukesh, and Urs
I said “Catenable”;not “Cannibal”
I said “Catenable”;not “Cannibal”
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