Submodular Maximization with Cardinality Constraints

Moran Feldman

Based OnSubmodular Maximization with Cardinality Constraints. Niv Buchbinder, Moran Feldman, Joseph (Seffi) Naor and Roy Schwartz, SODA 2014 (to appear).

2

Set FunctionsDefinitionGiven a ground set N, a set function f : 2N R assigns a number to every subset of the ground set.

Intuition• Consider a player participating in an auction on a set N of elements.• The utility of the player from buying a subset N’ N of elements is

given by a set function f.

Basic Properties of Set Functions• Non negativity – the utility from every subset of elements is non-

negative.

• Monotonicity - More elements cannot give less utility.

f(B)f(A) NBA :

f(A) NA 0:

3

Submodularity - DefinitionIntuition• Captures scenarios where elements can replace each other,

but never complement each other.• The marginal contribution of an element to a set decreases as

more elements are added to the set.

NotationGiven a set A, and an elements u, fu(A) is the marginal contribution of u to A:

Formal Definition

AfuAfAfu

For sets A B N, and u B:fu(A) fu(B)

For sets A, B N:f(A) + f(B) f(A B) + f(A B)

4

Submodular Function - Example

0567

10811

0

Too heavy

• Non-negative• Nonmonotone• Submodular

54-8

5

Where can One Find Submodular Set Functions?

In Combinatorial Settings

In Applicative Settings• Utility/cost functions in economics (economy of

scale).• Influence of a set of users in a social network.

Ground Set Submodular FunctionNodes of a graph The number of edges leaving a

set of nodes.Collection of sets The number of elements in the

union of a sub-collection.

6

Maximization Subject to a Cardinality Constraint

InstanceA non-negative submodular function f : 2N R+ and an integer k.

ObjectiveFind a subset S N of size at most k maximizing f(S).

Accessing the Function• A representation of f can be exponential in the size of the ground set.• The algorithm has access to f via an oracle.• Value Oracle – given a set S returns f(S).

Algorithmic Evaluation Criteria• Approximation ratio.• Oracle queries.• Time complexity - ignored in this talk.

1 23

7

The (Classical) Greedy Algorithm

The AlgorithmDo k iterations. In each iteration pick the element with the maximum marginal contribution.

More Formally1. Let S0 .

2. For i = 1 to k do:3. Let ui be the element maximizing: fui

(Si-1).

4. Let Si Si-1 {ui}.

5. Return Sk.

8

Results for Monotone Functions

Greedy Achieves• 1-1/e approximation [Nemhauser et al. 78].• Match a hardness of [Nemhauser et al. 78]• O(nk) – oracle queries.• For other constraints:

• ½-approximation for a general matroid constraint. [Nemhauser et al. 78]• (k+1)-1-approximation for k-set systems. [Nemhauser et al. 78] (presented formally by

[Calinescu et al. 11]).

Reducing the Number of Oracle Calls• O(nk) oracle queries is very good compared to tight algorithms for

more involved constraints.• A new result gives 1 – 1/e – ε approximation using O(nε-1log (n / ε))

oracle queries. [Ashwinkumar and Vondrak 14]• The number of oracle queries can be further reduced to O(n log(ε-1)).

1 23

9

What About Non-monotone Functions?

Approximation Ratio• 0.325 approximation via simulated annealing [Oveis Gharan and Vondrak

11]• 1/e – o(1) approximation (measured continuous greedy) [Feldman et al.

11]• 0.491 hardness [Oveis Gharan and Vondrak 11]

Oracle Queries• Both algorithms require many oracle queries.• The greedy algorithm requires few oracle queries, but guarantees no

constant approximation ratio.– Example:

– The greedy algorithm will select v in the first iteration.

1 23

otherwise0

2and,,,, 21 vS

SvS

SfuuuvN n

10

The Random Greedy Algorithm

The AlgorithmDo k iterations. In each iteration pick at random one element out of the k with the largest marginal contributions.

More Formally1. Let S0 .

2. For i = 1 to k do:3. Let Mi be set of k the elements maximizing: fu

(Si-1).

4. Let ui be a uniformly random element from Mi.

5. Let Si Si-1 {ui}.

6. Return Sk.

11

Warm Up: Analysis for Monotone Functions

In iteration i• Fix everything that happened before iteration i. All the expectations will

be conditioned on the history.• By submodularity and monotonicity:

• The elememt ui is picked at random from Mi, and OPT is a potential candidate to be Mi.

• Unfix history - if it holds for every given history, it holds in general too.

1111

iiiOPTu

iu SfEOPTfSfESOPTfESfE

k

SfEOPTf

SfEk

SfEk

SfE

i

OPTuiu

Muiuiu

i

i

)(

11

1

111

1ii SfESfE

12

Warm Up: Analysis for Monotone Functions (cont.)

Adding up all iterationsRearranging:

Combining:

Rearranging again:

We get a set with a value of (1 – 1/e) ∙ f(OPT) in expectation (unlike in the classical greedy).

1

11

ii SfEOPTf

kSfEOPTf

OPTfk

SfOPTfk

SfOPTfkk

k

11

11 0

OPTfe

OPTfk

Sfk

k

11

111

13

Reduction for Non-monotone Functions• We add k dummy elements of value 0.• The dummy elements are removed at the end.• Allows us to assume OPT is of size exactly k.

Analysis for Non-monotone FunctionsHelper LemmaFor a submodular function g : 2N R+ and a random set R containing every element with probability at most p: E[g(R)] ≥ (1 – p) ∙ g().

• Similar to a Lemma from [Feige et al. 2007].• Will be proved later.

Current Objective• Lower bound E[f(OPT Si)].

• Method - show that no element belongs to Si with a large probability, and then apply the above lemma.

14

Analysis for Non-monotone Functions (cont.)

ObservationIn every iteration i, every element outside of Si-1 has a probability of at most 1/k to get into Si.

CorollaryAn element belongs to Si with probability at most 1 – (1-1/k)i.

Applying the Helper Lemma• Let g(S) = f(OPT S).• Observe that g(S) is non-negative and submodular.• E[f(OPT Si)] = E[g(Si)] ≥ (1-1/k)i ∙ g() = (1-1/k)i ∙ f(OPT).

Next StepRepeat the analysis of the classical greedy algorithm, and use the above bound instead of monotonicity.

15

Analysis for Non-monotone Functions (cont.)

In iteration i• Fix everything that happened before iteration i. All the expectations

will be conditioned on the history.• By submodularity:

• The elememt ui is picked at random from Mi, and OPT is a potential candidate to be Mi.

111

iiOPTu

iu SfESOPTfESfE

k

SfESOPTfE

SfEk

SfEk

SfE

ii

OPTuiu

Muiuiu

i

i

)(

11

11

111

1ii SfESfE

16

Analysis for Non-monotone Functions (cont.)

• Unfixing history, and using previous observations, we get:

Adding up all iterations• We got a lower bound on the (expected) improvement in each iteration.• Using induction it is possible to prove that:

Remarks• This algorithm both uses less oracle calls than the previous ones, and

gets ride of the o(1) in the approximation ratio.• Now it all boils down to proving the helper lemma.

k

SfEOPTfkSfESfE i

i

ii

)(/11 11

OPTfkk

iSfE

i

i

11

1 OPTfeOPTfkk

kSfE

k

k

1

11

1

17

Proof of the Helper LemmaHelper Lemma - ReminderGiven a submodular function g : 2N R+ and a random set R containing every element with probability at most p. Then, E[g(R)] ≥ (1 – p) ∙ g().

Intuition• Adding all the elements can reduce the value of g() by at most g() to 0.• Adding at most a p fraction of every element, should reduce g () by no more than p

∙ g().

Notation• Order the elements of N in an order u1, u2, …, un of non-increasing probability to

belong to R.• Let Ni be the set of the first i elements in the above order.

• Let pi = Pr[ui R].

• Let Xi be an indicator for the event ui R. Notice that E[Xi] = pi.

F1

18

Proof of the Helper Lemma (cont.)

The value of the set R can be represented using the following telescopic sum:

Taking an expectation over both sides, we get:

k

iii RNgRNggRg

11

gpNgpppNggp

NgNgpgNgNgXEg

RNgRNgXEg

RNgRNgEgRgE

nii

k

ii

k

iiii

k

iiii

k

iiii

k

iii

11

1

11

11

11

11

11

11

F1

19

Playing with the Size of Mi

Question• The size of Mi determines the guarantee we have on E[f(Si

OPT)].• The larger Mi - the better the guarantee.

• Why not increase |Mi| to be larger than k?

Answer• We know there are k good elements (in average) – the

elements of OPT.• Increasing Mi might introduce into it useless elements.• The gain in every single iteration might decrease.

20

And yet…The Bad Case• Let Mi

k be the set of the k elements with best marginal values at iteration i.

• There are no useful elements outside of Mik.

• Most of OPT’s value is contributed by OPT Mik.

• The best subset of Mik is:

– Feasible.– Has a lot of value.– Can be (approximately) found using an algorithm for unconstrained submodular

maximization.

Taking Advantage• Apply the fast algorithm with Mi larger than k.

• At every iteration, find the best subset of Mik.

• Output the best set seen.

21

And yet… (cont.)

• By making the size of Mi a function of i, one can get e-1 + ε for some small constant ε > 0.

• Using a few more tricks, one can improve ε to 0.004.

Implications• Very small improvement in approximation ratio at the cost

of many more oracle queries.• The ratio e-1 is not the right ratio for a cardinality

constraint.– No candidate for the right ratio.– e-1 is the state of the art for a general matroid constraint. Is it

right for that case?

22

Equality Cardinality Constraint

New ObjectiveFind a subset S N of size exactly k maximizing f(S).

Monotone FunctionsNot interesting. We can always add arbitrary elements to the output.

Non-monotone Functions• Best previous approximation: ¼ - o(1).• Modifications to our algorithm:

– Apply a reduction that let us assume k n/2.– Avoid the reduction described previously.– Select only elements of N \ Si-1 into Mi.

• Achieves:– Approximation of: where v = n/k – 1.– Uses O(nk) oracle queries.– The term ok(1) can be replaced with ε at the cost of a multiplicative constant increase

in the number of oracle queries.

1 23

12/1erfiv/22/11 kv

oe

v

23

Understanding the Approximation Ratio

• The interesting range is: 1 v (k n/2).• erfi is the imaginary error function:

• The approximation ratio as a function of v:

z y dyez0

22erfi

24

Reduction

AimWe want to assume k n/2.

Observations• Equivalent problem - find a subset of size exactly n-k maximizing

h(S) = f(N \ S).• h(S) is non-negative and submodular if and only if f has these

properties.

CorollaryIf k > n/2, we can switch to the above equivalent problem.

BAhBAhBAfBAf

BAfBAfBfAfBhAh

25

Analysis Intuition• A possible candidate for Mi is OPT \ Si padded with random

elements of N \ (OPT Si).• The padding elements can reduce the value of the solution.• However:

– The expected number of padding elements in iteration i is only: k(1 – (1 – 1/k)i) (and k is small compared to n because of the reduction).

– Adding all the elements of N \ (OPT Si) reduces the value to 0 (at the worst case).

– Thus, an average element of N \ (OPT Si) reduces the value by a factor of at most 1 / |N \ (OPT Si)|.

OPT \ Si Padding

k

26

Other ResultsCardinality Constraint• For both problems we consider, an approximation ratio of:

– For k = n/2, both problems have an approximation ratio of ½.– For an equality constraint: 0.356-approximation by balancing this ratio with the

one presented before.

Fast Algorithms for General Matroid Constraint

• State of the art approximation ratio for a general matroid constraint: e-1 – o(1).

1

21

kkn

n

Approximation Ratio Oracle Queries Time Complexity

1/4 O(nk) O(nk)

(1-e2) / 2 – ε > 0.283 O(nk + k3) O(nk + kω+1)

27

Open Problems• Cardinality Constraint

– The approximability depends on k/n.• For k/n = 0.5, we have 0.5 approximation.• For small k’s, one cannot beat 0.491 [Oveis Gharan and Vondrak 11]

– What is the correct approximation ratio for a given k/n?

• Fast Algorithms– Finding fast algorithms for more involved constraints.

• Such as a general matroid constraint.

– Beating e-1 using a fast algorithm:• Even for large k/n values.

– Further reducing the number of oracle quires necessary to get 1-1/e-ε approximation.• No lower bounds on the number of necessary oracle queries.

28