Lecture 10 : 590.03 Fall 12 1

Post-processing outputs for better utility

CompSci 590.03Instructor: Ashwin Machanavajjhala

Lecture 10 : 590.03 Fall 12 2

Announcement

• Project proposal submission deadline is Fri, Oct 12 noon.

Lecture 10 : 590.03 Fall 12 3

Recap: Differential Privacy

For every output …

OD2D1

Adversary should not be able to distinguish between any D1 and D2 based on any O

Pr[A(D1) = O] Pr[A(D2) = O] .

For every pair of inputs that differ in one value

< ε (ε>0)log

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Recap: Laplacian Distribution

-10-4.300000000000021.4 7.09999999999990

0.10.20.30.40.50.6

Laplace Distribution – Lap(λ)

Database

Researcher

Query q

True answer q(d) q(d) + η

η

h(η) α exp(-η / λ)

Privacy depends on the λ parameter

Mean: 0, Variance: 2 λ2

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Recap: Laplace MechanismThm: If sensitivity of the query is S, then the following guarantees ε-

differential privacy.

λ = S/εSensitivity: Smallest number s.t. for any d, d’ differing in one entry,

|| q(d) – q(d’) || ≤ S(q)

Histogram query: Sensitivity = 2• Variance / error on each entry = 2x4/ε2 = O(1/ε2)

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This class• What is the optimal method to answer a batch of queries?

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How to answer a batch of queries?• Database of values {x1, x2, …, xk}

• Query Set: – Value of x1 η1 = x1 + δ1– Value of x2 η2 = x2 + δ2– Value of x1 + x2 η3 = x1 + x2 + δ3

• But we know that η1 and η2 should sum up to η3!

Lecture 10 : 590.03 Fall 12 8

Two Approaches• Constrained inference

– Ensure that the returned answers are consistent with each other.

• Query Strategy– Answer a different set of strategy queries A– Answer original queries using A

– Universal Histograms– Wavelet Mechanism– Matrix Mechanism

Lecture 10 : 590.03 Fall 12 9

Two Approaches• Constrained inference

– Ensure that the returned answers are consistent with each other.

• Query Strategy– Answer a different set of strategy queries A– Answer original queries using A

– Universal Histograms– Wavelet Mechanism– Matrix Mechanism

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Constrained Inference

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Constrained Inference• Let x1 and x2 be the original values. We observe noisy values η1,

η2 and η3• We would like to reconstruct the best estimators y1 (for x1/) and

y2 (for x2) from the noisy values.

• That is, we want to find the values of y1, y2 such that:

min (y1-η1)2 + (y2 – η2)2 + (y3 – η3)2

s.t., y1 + y2 = y3

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Constrained Inference [Hay et al VLDB 10]

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Sorted Unattributed Histograms• Counts of diseases

– (without associating a particular count to the corresponding disease)

• Degree sequence: List of node degrees – (without associating a degree to a particular node)

• Constraint: The values are sorted

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Sorted Unattributed HistogramsTrue Values 20, 10, 8, 8, 8, 5, 3, 2Noisy Values 25, 9, 13, 7, 10, 6, 3, 1 (noise from Lap(1/ε))

Proof:?

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Sorted Unattributed Histograms

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Sorted Unattributed Histograms• n: number of values in the histogram• d: number of distinct values in the histogram• ni: number of times ith distinct value appears in the histogram.

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Two Approaches• Constrained inference

– Ensure that the returned answers are consistent with each other.

• Query Strategy– Answer a different set of strategy queries A– Answer original queries using A

– Universal Histograms– Wavelet Mechanism– Matrix Mechanism

Lecture 10 : 590.03 Fall 12 18

Query Strategy

IPrivate

Data

WA

Differential Privacy

A(I) A(I) W(I)~ ~

Original Query Workload

Strategy Query Workload

Noisy StrategyAnswers

Noisy WorkloadAnswers

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Range Queries• Given a set of values {x1, x2, …, xn}• Range query: q(j,k) = xj + … + xk

Q: Suppose we want to answer all range queries?

Strategy 1: Answer all range queries using Laplace mechanism

• O(n2/ε2) total error. • May reduce using constrained optimization …

Lecture 10 : 590.03 Fall 12 20

Range Queries• Given a set of values {x1, x2, …, xn}• Range query: q(j,k) = xj + … + xk

Q: Suppose we want to answer all range queries?

Strategy 1: Answer all range queries using Laplace mechanism

• Sensitivity = O(n2)• O(n4/ε2) total error across all range queries. • May reduce using constrained optimization …

Lecture 10 : 590.03 Fall 12 21

Range Queries• Given a set of values {x1, x2, …, xn}• Range query: q(j,k) = xj + … + xk

Q: Suppose we want to answer all range queries?

Strategy 2: Answer all xi queries using Laplace mechanism Answer range queries using noisy xi values.

• O(1/ε2) error for each xi. • Error(q(1,n)) = O(n/ε2)• Total error on all range queries : O(n3/ε2)

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Universal Histograms for Range Queries

Strategy 3: Answer sufficient statistics using Laplace mechanismAnswer range queries using noisy sufficient statistics.

x1 x2 x3 x4 x5 x6 x7 x8

x12 x34 x56 x78

x1234 x5678

x1-8

[Hay et al VLDB 2010]

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Universal Histograms for Range Queries• Sensitivity: log n• q(2,6) = x2+x3+x4+x5+x6 Error = 2 x 5log2n/ε2

= x2 + x34 + x56 Error = 2 x 3log2n/ε2

x1 x2 x3 x4 x5 x6 x7 x8

x12 x34 x56 x78

x1234 x5678

x1-8

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Universal Histograms for Range Queries• Every range query can be answered by summing at most log n

different noisy answers• Maximum error on any range query = O(log3n / ε2)• Total error on all range queries = O(n2 log3n / ε2)

x1 x2 x3 x4 x5 x6 x7 x8

x12 x34 x56 x78

x1234 x5678

x1-8

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Universal Histograms & Constrained Inference

• Can further reduce the error by enforcing constraintsx1234 = x12 + x34 = x1 + x2 + x3 + x4

• 2-pass algorithm to compute a consistent version of the counts

[Hay et al VLDB 2010]

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Universal Histograms & Constrained Inference

• Pass 1: (Bottom Up)

• Pass 2: (Top down)

[Hay et al VLDB 2010]

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Universal Histograms & Constrained Inference

• Resulting consistent counts – Have lower error than noisy counts (upto 10 times smaller in some cases)– Unbiased estimators – Have the least error amongst all unbiased estimators

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Next Class• Constrained inference

– Ensure that the returned answers are consistent with each other.

• Query Strategy– Answer a different set of strategy queries A– Answer original queries using A

– Universal Histograms– Wavelet Mechanism– Matrix Mechanism