A sextic algorithm for website design

Brent Heeringa (heeringa@cs.umass.edu)(Joint work with Micah Adler)

21 October 2004Union College

A website design problem(for example: a new kitchen store)

Given products, their popularity, and their organization:

How do we create a good website?Navigation is naturalAccess to information is timely

paring chef bread steak

Wüstof Henkels

Knives

Type Maker

0.26 0.33 0.27 0.14

Good website: Natural Navigation

Organization is a DAG

TC of DAG enumerates all viable categorical relationships and introduces shortcuts

Subgraph of TC preserves logical relationship between categories

Transitive Closure

Subgraph of TC

A B C A B CTC

Good website: Timely Access to Info

Two obstacles to finding info quickly Time scanning a page for correct link Time descending the DAG

Associate a cost with each obstacle Page cost (function of out-degree of

node) Path cost (sum of page costs on path)

Good access structure: Minimize expected path cost Optimal subgraph is always a full tree

1/2

Page Cost = # links Path Cost = 3+2=5Weighted Path Cost = 5/2

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

CB DA

1/4 1/4 1/4 1/4

(x)=x

CB DA

1/4 1/4 1/4 1/4

(x)=x Cost:4

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

3(1/4)

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

CB DA

1/4 1/4 1/4 1/4

(x)=x Cost:4

3(1/4)5(1/4)

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

CB DA

1/4 1/4 1/4 1/4

(x)=x Cost:4

3(1/4)5(1/4)

5(1/4)

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

CB DA

1/4 1/4 1/4 1/4

3(1/4)5(1/4)

5(1/4)

(x)=x Cost:4

3(1/4)

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

CB DA

1/4 1/4 1/4 1/4

(x)=x Cost:4

1/4(3+5+5+3)= 1/4(16)= 4

CB DA

1/2 1/6 1/6 1/6

Constrained Subtree Selection (CSS)

An instance of CSS is a triple: (G,,w) G is a rooted, DAG with n leaves

(constraint graph) is a function of the out-degree of

each internal node (degree cost) w is a probability distribution over

the n leaves (weights)

A solution is any directed subtree of the transitive closure of G which includes the root and leaves

An optimal solution is one which minimizes the expected path cost

(x)=x Cost: 3 1/2

Constraint-Free Graphs and k-favorability

Constraint-Free GraphEvery directed, full tree with n leaves is a

subtree of the TC

CSS is no longer constrained by the graph

k-favorable degree cost Fix . There exists k>1 for any constraint-

free instance of CSS under where an optimal tree has maximal out-degree k

Linear Degree Cost - (x)=x

• 5 paths w/ cost 5

• 3 paths w/ cost 5• 2 paths w/ cost 4

• Unweighted path costs are all less, so weighted path costs must all be less• Generalization to n>6 paths is straightforward

Linear Degree Cost - (x)=x

• 4 paths w/ cost 4

• 4 paths w/ cost 4

• Prefer binary structure when a leaf has at least half the mass

• Prefer ternary structure when mass is uniformly distributed

> 1/2

Linear Degree Cost - (x)=x

CSS with 2-favorable degree costs and C.F. graphs is Huffman coding problem Examples: quadratic, exp, ceiling of log

Results

Complexity: NP-Complete for equal weights and many Sufficient condition on Hardness depends on constraint graph

Highlighted Algorithm: Theorem: O(n6)-time DP algorithm

(x)=x and G is constraint free

Other results: Characterizations of optimal trees for uniform probability

distributions Theorem: poly-time constant-approximation:

≥1 and k-favorable; G has constant out-degree Approximate Hotlink Assignment - [Kranakis et. al]

Related Work Adaptive Websites [Perkowitz & Etzioni]

Challenge to the AI community Novel views of websites: Page synthesis problem

Hotlink Assignment [Kranakis, Krizanc, Shende, et. al.] Add 1 hotlink per page to minimize expected distance

from root to leaves Recently: pages have cost proportional to their size

Hotlinks don’t change page cost

Optimal Prefix-Free Codes [Golin & Rote] Min code for n words with r symbols where symbol ai has

cost ci

Resembles CSS without a constraint graph

Dynamic Programming Review

Problems which exhibit:Optimal substructure

An optimal sol. may be written in terms of opt. solutions to subproblems

Inductive definition

Overlapping subproblemsDifferent problem instances share

subproblemsRepeated computation

Dynamic Programming: Fib

Optimal substructure (inductive definition)

Overlapping subproblemsFib(7) = Fib(6) + Fib(5) (but Fib(6) calls Fib(5))We only need to calculate Fib(5) onceDon’t repeat computationsIdea: Store solutions to subproblems in a table

Fib(0) = 0Fib(1) = 1Fib(i) = Fib(i-1) + Fib(i-2)

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …Problem: What is the ith Fibonacci number?

Dynamic Programming: FibGeneral Approach

Write inductive definitionRange of parameters in definition defines table sizeFill in table using definitionAnalysis: (Table size) * (# of lookups)

Fib(14) : 0 ≤ i ≤ 14

0 1 1 2 3 5 8 144233377

12 13 14 0 1 2 3 4 5 6

…Fib(i):

i:

Fib(0) = 0Fib(1) = 1Fib(i) = Fib(i-1) + Fib(i-2)

Dynamic Programming: Subset Sum

Example: X={2, 3, 5, 9, 10, 15, 17} and T=28

Subset Sum (SS): Given a set of n positive integers X=(x1,…,xn) and a positive integer T, is there a subset of X which sums to T?

Dynamic Programming: Subset Sum

Example: X={2, 3, 5, 9, 10, 15, 17} and T=28 Yes: {2, 9, 17} and {3, 10, 15}

Subset Sum (SS): Given a set of n positive integers X=(x1,…,xn) and a positive integer T, is there a subset of X which sums to T?

Dynamic Programming: Subset Sum

Example: X={2, 3, 5, 9, 10, 15, 17} and T=28 Yes: {2, 9, 17} and {3, 10, 15} Inductive definition:

Let Xi = (x1,…,xi) = the first i integers of X

SS(t,i) = TRUE if there is a subset of Xi which sums to t

= FALSE, otherwise

Subset Sum (SS): Given a set of n positive integers X=(x1,…,xn) and a positive integer T, is there a subset of X which sums to T?

Dynamic Programming Review

…

…

… …

T

n (t,i)

Table Size: T*nEach cell – (t,i) – depends on 2 other cellsO(Tn) time for SS

SS(0,i) = TRUESS(t,0) = FALSESS(t,i) = SS(t-xi,i-1) OR SS(t,i-1)

The ith element is in the subset

The ith element is not in the subset

Parameter Range:0 ≤ t ≤ T0 ≤ I ≤ n

Lopsided Trees

Recall: (x)=x (3-favorable) and G is constraint free

Node level = path cost

Adding an edge increases level

Grow lopsided trees level by level

Lopsided Trees

Lopsided Trees

Lopsided Trees

Lopsided Trees

We know exact cost of tree up to the current level i:

Exact cost of m leaves Remaining n-m leaves must have path-cost at least i

Lopsided Trees: Cost

Exact cost of C: 3 • (1/3)=1

Remaining mass up to level 4: (2/3) • 4 = 8/3

Total: 1+8/3=11/3

Lopsided Trees: Cost

Tree cost at Level 5 in terms of Tree cost at Level 4: Add in the mass of

remaining leaves

Cost at Level 5: No new leaves 11/3+2/3=13/3

Cost updates don’t depend on level

Lopsided Trees

Lopsided Trees

Lopsided Trees

Equality on trees: Equal number of leaves at or above

frontier Equal number of leaves at each

relative level below frontier

Nodes have outdegree ≤ 3 Node below frontier ≤ (3)=3 (m;l1, l2, l3) = signature Example Signature: (2; 3, 2, 0)

2: C and F are leaves 3: G, H, I are 1 level past the frontier 2: J and K are 2 levels past the frontier

Signature if F is interior node with 3 children?

Inductive Definition

Let CSS(m,l1,l2,l3) = min cost tree with sig (m;l1, l2, l3)

Can we define CSS(m,l1,l2,l3) in terms of optimal solutions to subproblems?

Which trees, when grown by one level, have sig (m;l1,l2,l3)?

Which parent sigs (m’;l’1,l’2,l’3) lead to the child sigs (m;l1,l2,l3)

Different Signatures

(0; 4, 0, 0) (2; 2, 0, 0)

Same Signature (2; 0, 2, 3)

Different signatures lead to (2; 0, 2, 3)

Sig: (0; 2, 0, 0)

Sig: (1; 0, 0, 3)

Growing a tree only affects frontierOnly l1 affects next levelChoose # of leavesThe remaining nodes are

internalChoose degree-2 (d2)

Remaining nodes are degree-3 (d3)

O(n2) choices

The other direction

(which signatures can a tree grow)

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3)

CHILDPARENT

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3) Suppose we know

l’1 (the # of nodes one level below the frontier)

d2 (the # of l’1 which are degree-2 interior nodes in (m,l1,l2,l3))

Let’s determine the values of the remaining variables1

2

3l’1 nodes

1

2

d2 nodes3

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3) Suppose we know

l’1 (the # of nodes one level below the frontier)

d2 (the # of l’1 which are degree-2 nodes in (m,l1,l2,l3))

m = m’ + l’1 - d2 - d3

The new number of leaves

The old number of leaves

Nodes at one level below the frontier

Internal nodes of degree 2

Internal nodes of degree 3

1

2

3

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3) Suppose we know

l’1 (the # of nodes one level below the frontier)

d2 (the # of l’1 which are degree-2 nodes in (m,l1,l2,l3))

m = m’ + l’1 - d2 - l3/3

The new number of leaves

The old number of leaves

Nodes at one level below the frontier

Internal nodes of degree 2

Internal nodes of degree 3

1

2

3

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3) Suppose we know

l’1 (the # of nodes one level below the frontier)

d2 (the # of l’1 which are degree-2 nodes in (m,l1,l2,l3))

l’2 = l1

The old number of nodes at2 levels below the frontier

New nodes one level below the frontier

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3) Suppose we know

l’1 (the # of nodes one level below the frontier)

d2 (the # of l’1 which are degree-2 nodes in (m,l1,l2,l3))

l2 = l3+2d2

The new number of nodes 2 levels below the frontier

d2 nodes are binary so they contribute 2d2 to the frontier

The original question(warning: here be symbols)

Which (m’;l’1,l’2,l’3) (m;l1,l2,l3) l’1 and d2 are sufficient

l’1 and d2 are both O(n)

O(n2) possibilities for (m’;l’1,l’2,l’3)

CSS(m,l1,l2,l3) = min cost tree with sig. (m;l1, l2, l3)

= CSS(m’,l’1,l’2,l’3) + cm’ for 1≤d2≤l’1≤n

(cm’ are the smallest n-m’ weights)

CSS(n,0,0,0) = cost of optimal tree Analysis:

Table size = O(n4) Each cell takes O(n2) lookups O(n6) algorithm

Some Observations

Generalize algorithm: Theorem: O(n(k)+k)-time DP algorithm

is positive, integer-valued, non-decreasing, k-favorable and G is constraint free

Signatures = (k)+1 vectors Table size = (k)+1 Each cell requires k-1 lookups

(extra slides follow)

Motivation and Lower Bound

Many constraint graphs have constant out-degreeRemains NP-Hard for many degree costs

Lemma 1: H(w)/log(k) is a lower bound on the cost of an optimal tree For any k-favorable degree cost , with ≥1 G is constraint-free

T

C(T) ≥ c’(T) ≥ c’(T’) ≥ H(w)/log(k) (shannon)

1 1 1

1T1 1 1

1T’

1

A Simple Lemma Lemma 2: For any tree with m weighted nodes there exists 1 node

(splitter) which, when removed, divides the tree into subtrees with at most half the weight of the original tree.

splitter

< 1/2 < 1/2

<1/2

Aproximation AlgorithmLet G be a DAG where out-degree of every node

dChoose a spanning tree T from GBalance-Tree(T):

Find a splitter node in T (Lemma 2) Stop if splitter is child of root

Disconnect the splitter and reconnect it to the root root has degree at most d+1

Call Balance-Tree on all subtrees

splitter

Mass of each subtree is at least half of whole tree

Approximation Algorithm

Analysis: Mass under any node is half of mass under its

grandparent Path length to leaf with weight wi is -2log(wi)

Theorem: O(m)-time O(log(k)(d+1))-approx to optimal solution

For any DAG G with m nodes and out-degree d For every k-favorable degree cost ≥ 1,

Upper Bound on Node Cost Weighted Path Length

Open Problems

Theorem: There is an for any instance (G,,w) of CSS where G is constraint free, is k-favorable, maps the positive integers to the positive integers and is non-decreasing

Proof:c(T) ≥ c’(T) ≥ c’(T’) ≥ H(w)/log(k)T is optimal tree for CSS cost cT’ is optimal tree for OPC cost c’ for k symbols each with weight 1 (i.e. (x)=1)H is entropy

NO