physical unclonable functions -...

Physical Unclonable Functions

Ismail San

Anadolu University

EEM 530

Context

What is PUF?

Examples of PUFs

PUF for identification and Authentication

PUF Notation, An approach to make a PUF experiment

Helper Data Algorithm or ECC based noise removal

BCH encoding and decoding for error correction

Security, Threats and Environmental Parameters

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What is PUF?

PUF as a Function

Let x = 2n, y = 2m where xdenotes a set of n-bitchallenges and y denotes aset of m-bit responses.

Mapping between challengesand responses

(xi, yi) pairs are determinedby manufacturing variationsof the the device.

010110100010101101010101 100011011010010 011100

Challenge PUF Response

Challenge Response

......

100010101 110110

(xi , yi)

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Ideal PUF

Design philosophy

H(xi, yi) is distributed normally from 1 to n.

H(yi, yj) is approx. equal to n2 when H(xi, xj) = 1 is given

“Apply crypt. hash function to response”

Observation requires evaluation of PUF

“Ring oscillator type PUF satisfies this”

Summary:

PUFi should be unique to device i

Each challenge and response pairshould not be linkable

Challenge and response pairsshould be independent

010110100010101101010101 100011011010010 011100

Challenge Response

......

100010101 110110

(xi , yi)

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Ideal PUF

Design philosophy

Summary:

010110100010101101010101 100011011010010 011100

Challenge Response

......

100010101 110110

(xi , yi)

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Ideal PUF

Design philosophy

Summary:

010110100010101101010101 100011011010010 011100

Challenge Response

......

100010101 110110

(xi , yi)

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Ideal PUF

Design philosophy

Summary:

010110100010101101010101 100011011010010 011100

Challenge Response

......

100010101 110110

(xi , yi)

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Ideal PUF

Design philosophy

Summary:

010110100010101101010101 100011011010010 011100

Challenge Response

......

100010101 110110

(xi , yi)

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Principle of PUF

CICI CI

ChallengeCI

II R2 R3R1

IIPUF1 PUF2 PUF3

· · ·

R1 6= R2 6= R3 6= · · · 6= Rm

5 / 49

Process variations in MOSFET

Components (MOSFET) Interconnect

Geometry

Effective channel length Line width and spaceGate length Metal thicknessComponent width Contact and via size

Material ParametersDoping variations Contact and via resistanceDeposition and anneal Metal resistivity

Electrical Parameters

Threshold voltage Line resistivityParasitic capacities Line capacityGate and source resistivityLeakage currents

Some important parameters that are affected by process variations,and the electrical parameters on which this is an impact.

D.S. Boning and S.R. Nassif. Models of process variations in device and interconnect.Design of High Performance Microprocessor Circuit. IEEE Press, 2000

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Delay-based PUF: Arbiter PUF

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Each challenge selects a unique pair of delay paths

Digital race condition on two paths with an identical delay in design

Random, uncontrollable process variations determines who will win

Arbiter output creates 1-bit output response

Multiple bits are obtained by duplicating the circuit or use of differentchallenge

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Each challenge selects a unique pair of delay paths

Digital race condition on two paths with an identical delay in design

Random, uncontrollable process variations determines who will win

Arbiter output creates 1-bit output response

Multiple bits are obtained by duplicating the circuit or use of differentchallenge

7 / 49

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

Delay difference and so the arbiter output will be device specific. ofan Arbiter PUF.

*** A metastabe state occurs if both delays are nearly identical. After ashort time, arbiter leaves the metastable state and outputs a randombinary value.

In that case, the arbiter output is not device-specific. Thisphenomenon is the cause of unreliability (noise) of the responses.

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

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Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

0 01 0

Response

Challenge

cN cN−1 cN−2 c1

1 1 1 0

Response

Challenge

cN cN−1 cN−2 c1

Response

Challenge

cN cN−1 cN−2 c1

Response

0 1 01

· · ·

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Delay-based PUF: Ring Oscillator PUFs

Delay Counter ResponseEdge

Detector

Challenge

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Memory-based PUF: SRAM PUFs

(1) Logic circuit of

an SRAM PUF (a cell)

(2) Transistor-level circuit of an SRAM PUF

in a standard CMOS tech.

Word Line

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SRAM PUF

R. Maes. Physically Unclonable Functions:Constructions, Properties and Applications. Springer, 2013. 11 / 49

Memory-based PUF: Butterfly PUFs

(1) Logic circuit of

an SRAM PUF (a cell)

(2) Schematical-level circuit of

a butterfly PUF cell

Excite

Preset

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PUF for Identification

PUF-based identification method uses a unique and untamperabledevice identifier exploiting the pyhsics of the device.

Identification is similar to authentication but in this context it hasweaker concept.

PUF should provide the identity of the device without any convincingproof.

In certain situations, identification is enough to achieve entityauthentication.

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PUF for Authentication

PUF gives a measure of the device specific physical feature.

Therefore, device authentication is provided by PUF (not messageauthentication).

Besides plain identification, also corroborative evidence of the deviceidentity is required.

Corroborative evidence means that it could only have been created bythat particular device.

For the device authentication:

the identity of a second party involved

the second party is active at the time of the evidence is created.

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Device Authentication

UntrustedSupply chain

Responser ′2

Responseri

Challenge Response

100010101100010101100010101

010110010110010110

Challengec2

Device ?

Is this deviceauthentic tothe manufacturer?Device A

(ci , ri)

Challengeci

... ...

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PUF Notation

Creation of a PUF Instance:

P ≡ {pufi ← P.Create(rCi ) : ∀i, rCi$← {0, 1}∗}

Random evaluation of PUF instance pufi for challenge x:

y(j)i (x)← pufi(x).Eval(rEj

$← {0, 1}∗)

Y(x)i ← pufi(x).Eval

where rC and rE are randomization variables representing anundetermined fair coin tosses.

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PUF Experiment

a PUF class generates responses for the challenges provided to it

Y(x)i ← pufi(x).Eval

where Yis are responses for challenge x.

Statistically, the distribution of PUF responses should be estimated.

This require experiments to understand the distribution statistics fromobserved PUF responses:

1 Experiments estimating the distribution for same challenge x ondistinct evaluations.

2 These experiments should be repeated on responses for differentchallenges.

3 Experiments on responses from different PUF instances

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PUF Experiment

A PUF class generates responses for the challenges provided to it

ExperimentP(Npuf , Nchal, Nmeas)→ YExp(P)

YExp(P) = [y(j)i (xk)← pufi(xk).Eval(rEj )

∀1 ≤ j ≤ Nmeas : rEj$← {0, 1}∗,

∀1 ≤ k ≤ Nchal : xk$← XP ,

∀1 ≤ i ≤ Npuf : pufi$← P,

If the conditions are important while doing experiments, then theyshould be specified with a parameter.

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PUF Response Distances: intra distance

A PUF response intra distance is a random variable refers to thedistance between two responses from the same PUF instance andusing the same challenge.

Dintrapufi

(x) , dist[Yi(x);Y ′i (x)],

with Yi(x) and Y ′i (x) two distinct random evaluations of PUFinstance pufi on the same challenge x. dist[; ] denotes hammingdistance between two random variables.

For the reproducibility of a PUF class, distribution of Dintrapufi

has somestatistical properties.

Estimation of mean (µintraP ), standard deviation (∑

[DintraP ]),

histogram and order statistics should be found.

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PUF Response Distances: inter distance

A PUF response inter distance is a random variable refers to thedistance between two responses from different PUF instances andusing the same challenge.

DinterP (x) , dist[Y (x);Y ′(x)],

with Y (x) and Y ′(x) two distinct random evaluations of two distinctPUF instances on the same challenge x.

For the reproducibility of a PUF class, distribution of Dinterpufi

has somestatistical properties.

Estimation of mean (µinterP ), standard deviation (∑

[DinterP ]),

histogram and order statistics should be found.

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PUF Tests

Hamming distance 128

Different challengeSame challenge

Intra-Distance

SC Inter-PUF

Different challengeSame challenge

Inter-Distance

SC Intra-PUF

Uniqueness

Quantative tests yield to model a PUF by Hori et al., Quantitativeand Statistical Performance Evaluation of Arbiter Physical UnclonableFunctions on FPGAs. ReConFig2010, pp.298-303, 2010.

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Experimental uniqueness results

PUF No Nbits µinterP σinterPSRAM PUF 0 65536 49.59% 0.33%

1 65536 49.61% 0.33%2 65536 49.68% 0.31%3 65536 49.72% 0.30%

Arbiter PUF (basic) 0 65536 47.13% 0.44%

Arbiter PUF (2-XOR) 0 32768 49.74% 0.29%

Ring Oscillator PUF (L.G.) 0 12544 46.86% 0.48%

Latch PUF 0 8192 34.84% 1.20%1 8192 37.01% 1.23%2 8192 33.17% 1.62%3 8192 16.37% 2.02%

Inter-distances statistics (at nominal condition)Table taken from the book by R. Maes. Physically UnclonableFunctions: Constructions, Properties and Applications. Springer,2013.

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Helper Data Algorithm or Fuzzy Extractor

Two problems are solved with HDA or FE:

1. PUF responses are noisy. Some processing has to be employed toremove the noise.

2. PUF responses are not generally uniformly random. Hence, aprocessing is required to transform the possibly unknown distribution ofthe physical measurements into a uniform distributed response value.

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FE in Secure Key Generation

Measuring

RepetitionEncoding

BCHEncoding

Randomnumbers Measuring

Activation code orHelper data

RandomnessExtracting

BCH ErrorCorrecting

RepetitionDecoding

(1) Enrollment phase (2) Reconstruction phase

Activation code or

Helper data

Enrollment phase Reconstruction phaseGenerating activation code It is then used to obtain the K.

K ← PUF(x) K ′ ← PUF(x)C ← ECC-Encoding(r), r is random C ′ ← K ′ + hh← K + C C ← ECC-Decoding(C ′)

K ← C + h

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Information Reconciliation

The action of eliminating noise found in the measurments is calledInformation Reconciliation.

Mapping the response X onto an element W of a set of Helper Data.

Helper Data has the following properties:

1. With a deterministic recovery function Rec, X can be recovered bymeans of W and a noisy response of the response X ′.

2. The amount of uncertainty about an unknown response X onlydecreases with a limited amount when W for that response is known.This loss of useful uncertainty is unavaoidable.

For the Helper Data W , the use of error correcting codes (ECC) isone of the solutions.

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Privacy Amplification

Besides noise removal, to guarantee a uniform distribution of thederived keys is another important issue and called as PrivacyAmplification.

Approximation of the distribution is obtained by collecting a lot ofsamples from different PUFs.

There is a concept Strong Extractor:

Ext : {0, 1}n → {0, 1}l

It can transform a random variable X, with an unknown distribution,into a new random variable K, with a distribution close to uniform.

Strong extractors can be implemented as pairwise independentuniversal hash functions.

A non-uniform selection of a hash function or keep using the same hashfunction for all the time gives only a minor deviation from uniformity inthe distribution K.

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BCH Code: Bose-Chaudhuri-Hoequenghem

A class of cyclic error correcting code

It is constructed in Finite Fields

It is a generalized parity code.

Binary BCH codes are more popular

Reed Solomon codes (RS) are a class of non-binary BCH codes

Some special types of BCH codes are called as Golay code:(23, 12), (31, 16), etc.

BCH (N,K, t) code can correct t bit errors

t =(N −K)

where m is an integer such that N = 2m − 1.

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t =(N −K)

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t =(N −K)

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t =(N −K)

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t =(N −K)

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t =(N −K)

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t =(N −K)

27 / 49

BCH Code

Construction of (N,K, t) BCH Codeα is the primitive element in GF(pm)

N is the block length: N = 2m − 1

K is the message length

u(X) is the message polynomial whereu(X) = uK−1X

K−1 + · · ·+ u1X + u0

it is capable of correcting t errors

bit length of the redundancy is N −KN −K ≤ mtdmin ≥ 2t+ 1

g(X) is the generator polynomial and it is computed by LCM ofminimal polynomials

g(X) = LCM {Λ1(X), · · · ,Λ2t(X)}g(X) = gN−KX

N−K + · · ·+ g1X + g0

where Λi(X) is the minimal polynomials of αi, and Λi(X) = Λi×2(X)

28 / 49

BCH Code

K−1 + · · ·+ u1X + u0

N−K + · · ·+ g1X + g0

28 / 49

BCH Code

K−1 + · · ·+ u1X + u0

N−K + · · ·+ g1X + g0

28 / 49

BCH Code

K−1 + · · ·+ u1X + u0

N−K + · · ·+ g1X + g0

28 / 49

BCH Code

K−1 + · · ·+ u1X + u0

bit length of the redundancy is N −K

N −K ≤ mtdmin ≥ 2t+ 1

N−K + · · ·+ g1X + g0

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BCH Code

K−1 + · · ·+ u1X + u0

N−K + · · ·+ g1X + g0

28 / 49

BCH Code

K−1 + · · ·+ u1X + u0

N−K + · · ·+ g1X + g0

where Λi(X) is the minimal polynomials of αi, and Λi(X) = Λi×2(X)28 / 49

Galois Field

Elements: {0, 1, 2, · · · , p− 1}p is prime and operation is “modulo-p” addition and multiplication

GF(pm)

Primitive polynomial of degree m over GF(p) with a root α

p(X) = Xm + pm−1Xm−1 + · · ·+ p1X + p0

Elements:{

0, 1, α, α2 · · · , αpm−2}p is prime and operation is “modulo-p” addition and multiplication

Construction

¬ f(X) = q(X)p(X) + r(X), where deg r < deg p.

αi = (ri,m−1, · · · , ri,1, ri,0)

29 / 49

Galois Field

GF(pm)

p(X) = Xm + pm−1Xm−1 + · · ·+ p1X + p0

Elements:{

Construction

αi = (ri,m−1, · · · , ri,1, ri,0)

29 / 49

Galois Field

GF(pm)

p(X) = Xm + pm−1Xm−1 + · · ·+ p1X + p0

Elements:{

0, 1, α, α2 · · · , αpm−2}

p is prime and operation is “modulo-p” addition and multiplication

Construction

αi = (ri,m−1, · · · , ri,1, ri,0)

29 / 49

Galois Field

GF(pm)

p(X) = Xm + pm−1Xm−1 + · · ·+ p1X + p0

Elements:{

Construction

αi = (ri,m−1, · · · , ri,1, ri,0)

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Galois Field

GF(pm)

p(X) = Xm + pm−1Xm−1 + · · ·+ p1X + p0

Elements:{

Construction

αi = (ri,m−1, · · · , ri,1, ri,0)

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Enumeration of the elements of GF(24)

GF(24)

and p(X) = X4 + X + 1

Power Polynomial Binary Decimalrepresentation representation representation representation

0 0 0000 0α0 1 0001 1α1 α 0010 2α2 α2 0100 4α3 α3 1000 8α4 α+ 1 0011 3α5 α2 + α 0110 6α6 α3 + α2 1100 12α7 α3 + α+ 1 1011 11α8 α2 + 1 0101 5α9 α3 + α 1010 10α10 α2 + α+ 1 0111 7α11 α3 + α2 + α 1110 14α12 α3 + α2 + α+ 1 1111 15α13 α3 + α2 + 1 1101 13α14 α3 + 1 1001 9α15 1 0001 1

Click to go to Decoding30 / 49

Galois Field

Minimal polynomial of α

Λ(X) = λdXd + · · ·+ λ1X + λ0

where λi ∈ GF (p) and d is the least integer such that Λ(α)|α 6=0 = 0

Conjugates of α in GF(pm)

α, αp, αp2, · · · , αpm

It is due to: Λ(α) = Λ(αp) = Λ(αp2) = · · · = Λ(αp

m) = 0

Example α in GF(24)

and p(X) = X4 +X + 1

¬{α, α2, α4, α8

}, where min poly is X4 +X + 1

{α3, α6, α9, α12

}, → X4 +X3 +X2 +X + 1

®{α5, α10

}, → X2 +X + 1

¯{α7, α11, α13, α14

}, → X4 +X3 + 1

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Galois Field

Λ(X) = λdXd + · · ·+ λ1X + λ0

α, αp, αp2, · · · , αpm

m) = 0

and p(X) = X4 +X + 1

¬{α, α2, α4, α8

{α3, α6, α9, α12

}, → X4 +X3 +X2 +X + 1

®{α5, α10

}, → X2 +X + 1

¯{α7, α11, α13, α14

}, → X4 +X3 + 1

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Galois Field

Λ(X) = λdXd + · · ·+ λ1X + λ0

α, αp, αp2, · · · , αpm

m) = 0

and p(X) = X4 +X + 1

¬{α, α2, α4, α8

{α3, α6, α9, α12

}, → X4 +X3 +X2 +X + 1

®{α5, α10

}, → X2 +X + 1

¯{α7, α11, α13, α14

}, → X4 +X3 + 1

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BCH Code: g(X)

Minimal polynomial sets: α in GF(24)

and p(X) = X4 +X + 1

¬{α, α2, α4, α8

{α3, α6, α9, α12

}, → X4 +X3 +X2 +X + 1

®{α5, α10

}, → X2 +X + 1

¯{α7, α11, α13, α14

}, → X4 +X3 + 1

When t = 2,

g(X) = LCM {Λ1(X),Λ3(X)}

When t = 3,

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}

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BCH Code: g(X)

and p(X) = X4 +X + 1

¬{α, α2, α4, α8

{α3, α6, α9, α12

}, → X4 +X3 +X2 +X + 1

®{α5, α10

}, → X2 +X + 1

¯{α7, α11, α13, α14

}, → X4 +X3 + 1

When t = 2,

g(X) = LCM {Λ1(X),Λ3(X)}

When t = 3,

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}

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BCH Code: g(X)

and p(X) = X4 +X + 1

¬{α, α2, α4, α8

{α3, α6, α9, α12

}, → X4 +X3 +X2 +X + 1

®{α5, α10

}, → X2 +X + 1

¯{α7, α11, α13, α14

}, → X4 +X3 + 1

When t = 2,

g(X) = LCM {Λ1(X),Λ3(X)}

When t = 3,

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}

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BCH Code Encoding

(N,K, t) BCH Code

c(X) = u(X) · g(X)

XN−Ku(X) = q(X) · g(X) + r(X)

c(X) = XN−Ku(X) + r(X)

Encoding as simple as computing polynomial modulo

r(X) = XN−Ku(X) mod g(X)

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BCH Code Encoding

(N,K, t) BCH Code

c(X) = u(X) · g(X)

XN−Ku(X) = q(X) · g(X) + r(X)

c(X) = XN−Ku(X) + r(X)

Encoding as simple as computing polynomial modulo

r(X) = XN−Ku(X) mod g(X)

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BCH Code Encoding: An Example

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1

Message is (0, 1, 1, 0, 1) where u(X) = X4 +X2 +XN = 15, K = 5, t = 315− 5 ≤ 4× 3dmin ≥ 2× 3 + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

)r(X) = P (X) mod g(X)

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1Message is (0, 1, 1, 0, 1) where u(X) = X4 +X2 +X

N = 15, K = 5, t = 315− 5 ≤ 4× 3dmin ≥ 2× 3 + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1Message is (0, 1, 1, 0, 1) where u(X) = X4 +X2 +XN = 15, K = 5, t = 3

15− 5 ≤ 4× 3dmin ≥ 2× 3 + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1Message is (0, 1, 1, 0, 1) where u(X) = X4 +X2 +XN = 15, K = 5, t = 315− 5 ≤ 4× 3

dmin ≥ 2× 3 + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1Message is (0, 1, 1, 0, 1) where u(X) = X4 +X2 +XN = 15, K = 5, t = 315− 5 ≤ 4× 3dmin ≥ 2× 3 + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}

=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

P (X) = X15−5u(X) =(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

r(X) = P (X) mod g(X)

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

34 / 49

(15, 5, 3) BCH Code in GF(24)

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}=(X4 +X + 1

)×(X4 +X3 +X2 +X + 1

)×(X2 +X + 1

)=(X10 +X8 +X5 +X4 +X2 +X + 1

)P (X) = X15−5u(X) =

(X14 +X12 +X11

= X8 +X4 +X3 +X2 +X

c(X) = P (X) + r(X)

= X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

Codeword:c = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1) 34 / 49

Parity Check Matrix H

c(αi) = cN−1αN−1i+ · · ·+ c1α

i + c0 = 0 for 1 ≤ i ≤ 2tIt is because

g(X)|c(X) andg(α) = g(α2) = · · · = g(α2t) = 0

Then, parity check matrix: cHT = 0

1 α α2 · · · αN−1

1 α2 (α2)2 · · · (α2)N−1

......

.... . .

...1 α2t (α2t)2 · · · (α2t)N−1

35 / 49

Syndrome and Error Location

Received data: r = c + e

The noise: e = (e0, e1, · · · , eN−1) where ej ∈ {0, 1}Computation of syndrome:

S = rHT = (c + e)HT

= eHT = (S1, S2, · · · , S2t)

N−1∑j=0

ejαij , for i = 1, 2, · · · , 2t

= e(αi) because

e(X) = eN−1XN−1 + · · ·+ e1X + e0

Where v ≤ t, v errors exists at j1, j2, · · · , jv(0 ≤ j1 < j2 < · · · < jv < N)

e(X) = Xj1 +Xj2 + · · ·+Xjv

36 / 49

Error Locator Polynomial

Si = e(αi) = (αj1)i + (αj2)i + · · ·+ (αjv)i, for i = 1, 2, · · · , 2tSolving 2t equations will give the set(

αj1 , αj2 , · · · , αjv)

Error locations are (j1, j2, · · · , jv)

Location numbers of error is denoted as βl = αjl

Syndrome is also computed with respecto βls

S1 = β1 + β2 + · · ·+ βv

S2 = β21 + β22 + · · ·+ β2v...

S2t = β2t1 + β2t2 + · · ·+ β2tv

37 / 49

Error Locator Polynomial

Error locator polynomial is denoted with σ(X) and

σ(X) = (1 + β1X)(1 + β2X) · · · (1 + βvX)

= σvXv + · · ·+ σ1X + σ0

Roots of σ(X) is (β−11 , β−12 , · · · , β−1v )Finding the roots of σ(X) will reveal the error locationsError locations are simply the inverse of the roots of σ(X)

38 / 49

BCH Decoding

1 Syndrome computation

2 Computing the error locator polynomial σ(X)

Peterson-Gorenstein-Zierler algorithmBerlekamp-Massey algorithmEuclid’s algorithm

3 Finding error location: Chien search for v roots

4 Correcting the errors

Decoding result is simply (r− e)

39 / 49

Peterson-Gorenstein-Zierler algorithm¬ Initialize v = t

Compute the determinant of M

det(M) =

S1 S2 · · · SvS2 S3 · · · Sv+1...

.... . .

...Sv Sv+1 · · · S2v−1

® Find the correct value of v: it may less than t{

go to step 4 det(M) 6= 0

v ← v − 1, and then go to step 2 det(M) = 0

¯ Invert M° Compute σ(X) by simply multiplying M−1 by a syndrome vector

σvσv−1

...σ1

= M−1

− Sv+1

−Sv+2...−S2v

40 / 49

Peterson-Gorenstein-Zierler algorithm¬ Initialize v = t Compute the determinant of M

det(M) =

S1 S2 · · · SvS2 S3 · · · Sv+1...

.... . .

...Sv Sv+1 · · · S2v−1

® Find the correct value of v: it may less than t{go to step 4 det(M) 6= 0

σvσv−1

...σ1

= M−1

− Sv+1

−Sv+2...−S2v

40 / 49

det(M) =

S1 S2 · · · SvS2 S3 · · · Sv+1...

.... . .

...Sv Sv+1 · · · S2v−1

σvσv−1

...σ1

= M−1

− Sv+1

−Sv+2...−S2v

40 / 49

det(M) =

S1 S2 · · · SvS2 S3 · · · Sv+1...

.... . .

...Sv Sv+1 · · · S2v−1

¯ Invert M

° Compute σ(X) by simply multiplying M−1 by a syndrome vectorσvσv−1

...σ1

= M−1

− Sv+1

−Sv+2...−S2v

40 / 49

det(M) =

S1 S2 · · · SvS2 S3 · · · Sv+1...

.... . .

...Sv Sv+1 · · · S2v−1

σvσv−1

...σ1

= M−1

− Sv+1

−Sv+2...−S2v

40 / 49

BCH Code Decoding: An Example

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}= X10 +X8 +X5 +X4 +X2 +X + 1

∗c(X) = X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

e(X) = X14 +X3

Then, received codeword is r(X) = X12 +X11 +X8 +X4 +X2 +X

First, computing the syndrome

S1 = r(α) = 1 S2 = r(α2) = 1 S3 = r(α3) = α8

S4 = r(α4) = 1 S5 = r(α5) = α5 S6 = r(α6) = α

∗Same codeword as in Encoding example41 / 49

(15, 5, 3) BCH Code in GF(24)

and p(X) = X4 +X + 1

g(X) = LCM {Λ1(X),Λ3(X),Λ5(X)}= X10 +X8 +X5 +X4 +X2 +X + 1

∗c(X) = X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

e(X) = X14 +X3

Then, received codeword is r(X) = X12 +X11 +X8 +X4 +X2 +X

First, computing the syndrome

S1 = r(α) = 1 S2 = r(α2) = 1 S3 = r(α3) = α8

S4 = r(α4) = 1 S5 = r(α5) = α5 S6 = r(α6) = α

∗Same codeword as in Encoding example41 / 49

¬ v = t = 3

Computing determinant of M

det(M) =

S1 S2 S3S2 S3 S4S3 S4 S5

1 1 α8

1 α8 1α8 1 α5

® det(M) = 0, then go to step 2 with v = v − 1 = 2

Computing determinant of new M

det(M) =

[S1 S2S2 S3

[1 11 α8

]= α2

® det(M) = α2, then go to step 4

Look at the Table42 / 49

¬ v = t = 3

det(M) =

S1 S2 S3S2 S3 S4S3 S4 S5

1 1 α8

1 α8 1α8 1 α5

det(M) =

[S1 S2S2 S3

[1 11 α8

]= α2

¬ v = t = 3

det(M) =

S1 S2 S3S2 S3 S4S3 S4 S5

1 1 α8

1 α8 1α8 1 α5

det(M) =

[S1 S2S2 S3

[1 11 α8

]= α2

¬ v = t = 3

det(M) =

S1 S2 S3S2 S3 S4S3 S4 S5

1 1 α8

1 α8 1α8 1 α5

det(M) =

[S1 S2S2 S3

[1 11 α8

]= α2

¬ v = t = 3

det(M) =

S1 S2 S3S2 S3 S4S3 S4 S5

1 1 α8

1 α8 1α8 1 α5

det(M) =

[S1 S2S2 S3

[1 11 α8

]= α2

¯ Inverting M

M−1 =

[α6 α13

α13 α13

° Computing σ(X) [σ2σ1

[α6 α13

α13 α13

] [α8

]σ(X) = α2X2 +X + 1

σ(X) will tell you the location of the errors

σ(α) = 0, σ(α2) = α14, · · · σ(α12) = 0

After finding the roots, we now compute the inverse of the roots αand α12

α−1 = α14

α−12 = α3

43 / 49

¯ Inverting M

M−1 =

[α6 α13

α13 α13

]° Computing σ(X) [

σ2σ1

[α6 α13

α13 α13

] [α8

]σ(X) = α2X2 +X + 1

σ(α) = 0, σ(α2) = α14, · · · σ(α12) = 0

α−1 = α14

α−12 = α3

43 / 49

¯ Inverting M

M−1 =

[α6 α13

α13 α13

σ2σ1

[α6 α13

α13 α13

] [α8

]σ(X) = α2X2 +X + 1

σ(α) = 0, σ(α2) = α14, · · · σ(α12) = 0

α−1 = α14

α−12 = α3

43 / 49

¯ Inverting M

M−1 =

[α6 α13

α13 α13

σ2σ1

[α6 α13

α13 α13

] [α8

]σ(X) = α2X2 +X + 1

σ(α) = 0, σ(α2) = α14, · · · σ(α12) = 0

α−1 = α14

α−12 = α3

43 / 49

¯ Inverting M

M−1 =

[α6 α13

α13 α13

σ2σ1

[α6 α13

α13 α13

] [α8

]σ(X) = α2X2 +X + 1

σ(α) = 0, σ(α2) = α14, · · · σ(α12) = 0

α−1 = α14

α−12 = α3

43 / 49

Finally, e(X) = X14 +X3

Corrected codeword:

r(X) + e(X) = X14 +X12 +X11 +X8 +X4 +X3 +X2 +X

c̄ = (0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1)

44 / 49

Security of PUF

EntropyPUF should have sufficient amount of randomness

PUF type Entropy per 1000 bits

SRAM PUF 950Delay PUF 130Butterfly PUF 600

Tamper evidenceRead-proof hardware against invasive and non-invasive

Invasive attacks make the chip functionally destroyed

Currently, there is no non-invasive attack that gives information aboutthe PUF

UnclonabilityUnique physical property of PUF is translating into a bitstream

This physical property prevents constructing a same PUF

Keep the challenge response pairs secret

45 / 49

Security of PUF

45 / 49

Security of PUF

45 / 49

Effects of Environmental Parameters on PUF

Die temperature and supply voltage

Heavily affect the delay in ring oscillator PUFs

Less affect the arbiter PU due to the differential measurement

46 / 49

Threats to PUFs

Modeling CRPs proposed by Ruhrmair et al. Modeling Attacks onPhysical Unclonable Functions. CCS, 2010.

delay-based PUF constructions can be modeled using ML algorithms.

CRPs should not be made public in protocols.

Required a Trust to Manufacturers

Helfmeier et al. cloned an SRAM PUF with a failure analysisequipment in at HOST 2013. Cloning Physically UnclonableFunctions.

Nedospasov et al. described successful invasive attempts on SRAMPUFs at FDTC 2013. Invasive PUF Analysis.

EM analyses on RO PUFs have been carried out by Merli et al. atWESS 2011. Semi-invasive EM attack on FPGA RO PUFs andcountermeasures.

47 / 49

Threats to PUFs

Modeling CRPs proposed by Ruhrmair et al. Modeling Attacks onPhysical Unclonable Functions. CCS, 2010.

delay-based PUF constructions can be modeled using ML algorithms.

CRPs should not be made public in protocols.

Required a Trust to Manufacturers

Helfmeier et al. cloned an SRAM PUF with a failure analysisequipment in at HOST 2013. Cloning Physically UnclonableFunctions.

Nedospasov et al. described successful invasive attempts on SRAMPUFs at FDTC 2013. Invasive PUF Analysis.

EM analyses on RO PUFs have been carried out by Merli et al. atWESS 2011. Semi-invasive EM attack on FPGA RO PUFs andcountermeasures.

47 / 49

References

I would like to acknowledge:

R. Maes. Physically Unclonable Functions:Constructions, Properties and Applications. Springer, 2013.

D.S. Boning and S.R. Nassif. Models of process variations in deviceand interconnect. Design of High Performance Microprocessor Circuit.IEEE Press, 2000.

R. Maes and P. Tuyls. Process variations for security: PUFs. Chapter 7of Secure Integrated Circuits and Systems. Editor Ingrid M.R.Verbauwhede, Springer 2009.

Ahmad-Reza Sadeghi and David Naccache. Towards Hardware-IntrinsicSecurity: Foundations and Practice (1st ed.). Springer-Verlag NewYork, Inc., New York, NY, USA.

48 / 49

Thank you for your attention

Questions?

49 / 49

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