Elgamal Elgamal demonstration demonstration

project on project on calculators TI-83+calculators TI-83+

Gerard TelGerard TelUtrecht UniversityUtrecht University

With results from Jos Roseboom With results from Jos Roseboom and Meli Samikinand Meli Samikin

Workshop Elgamal 2

Overview of the lectureOverview of the lecture1. History and background2. Elgamal (Diffie Hellman)3. Discrete Log: Pollard rho4. Experimentation results5. Structure of Function Graph:

Cycles, Tails, Layers6. Conclusions

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1. History and background1. History and background1. 2003, lecture for school teachers

about Elgamal2. 2006, lecture with calculator demo

Why Elgamal, not RSA?• Functional property easy to show• Security: rely on complexity• Compare exponentiation and DLog

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Programming ExperiencesProgramming Experiences• Nuisances:

– typing by selecting symbols– no subroutines: inline exponentiation– no local variables

• Limitation: arithmetic in 14 digits– Limit modulus to 7 digits

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Math: Modular arithmeticMath: Modular arithmetic• Compute modulo prime p (95917)

with 0, 1, … p-2, p-1• Generator g of order q (prime)

(g is 29609, q is 7993)• Rules of algebra are valid

(ga)k = (gk)a

Secure application: p has ~309 digits!!

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Calculator TI-83, 83+, 84+Calculator TI-83, 83+, 84+• Grafical, 14 digit• Programmable• Generally available

in VWO (pre-academic school type in the Netherlands)

• Cost 100 euro(free for me)

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The Elgamal programThe Elgamal program• Ceasar cipher (symmetric)• Elgamal parameter and key

generation• Elgamal encryption and

decryption• Discrete Logarithm: Pollard

Infeasible problem!! But doable for 7 digit modulus

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2. Public Key codes2. Public Key codes

The problem of Key Agreement:• A and B are on two sides of a river• They want to have common z• Oscar is in a boat on the river• Oscar must not know z• Common parameters: p, q, g

(Or: group with hard DLog problem)

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Solution: Diffie-HellmanSolution: Diffie-Hellman• Alice takes random a, shouts b = ga

• Bob takes random k, shouts u = gk

• Alice computes z = ua = (gk)a

• Bob computes z = bk = (ga)k

The two numbers are the sameThe difference in complexity for A&B

and O is relevant

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Parameter generationParameter generation• Hoofdmenu, parameters, Maak p,q,r• Input limits on q and p• Search for prime q from q-limit down• Search for prime p from p-limit down

among multiples of 2q + 1• Generator: try 100(p-1)/q, 101(p-1)/q, …

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What does Oscar hear?What does Oscar hear?Seen:1. Public b = ga

2. Public u = gk

Not computable:1. Secret a, k2. Common zThis needs discrete

logarithm

Oscar sees the communication, but not the secrets

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The Elgamal programThe Elgamal program• In class use• Program, explanation,

slides on website• Program extendible• Booklet with ideas for

experimenting, papers• All in Dutch!

http://people.cs.uu.nl/gerard/Cryptografie/Elgamal/

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3. Pollard Rho Algorithm3. Pollard Rho Algorithm• Fixed p (modulus), g, q (order of g);

H is set of powers of g• Size of H is q• Discrete Logarithm problem:

– Given y in H– Return x st gx = y

• Pollard Rho: randomized, √q time

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Pollard Rho: RepresentationPollard Rho: Representation• Representation of z: z = ya.gb

• Two representations of same number reveil log y:If ya.gb = yc.gd,then y = g(b-d)/(c-a)

• Goal: find 2 representations of one number z (value does not matter)

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Strategy: Birthday TheoremStrategy: Birthday Theorem• All values z = ya.gb are in H• Birthday Theorem:

In a random sequence, we expect a collision after √q steps

• Simulate effect of random sequence by pseudorandom function: zi+1 = f (zi)(Keep representation of each zi)

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Cycle detectionCycle detection• Detect collision by storing previous

values: too expensive• Floyd cycle detection method:

– Develop two sequences: zi and ti

– Relation: ti = z2i

– Collision: ti = zi, i.e., zi = z2i

In each round, z “moves” one step and t moves two steps.

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4. Experimentation results4. Experimentation results

p q x m 1 2 3 4 5 Ave

971 97 4 3 8 16 8 16 8 11,2

3989 997 114 10 30 30 60 15 60 39

39869 9967 4 3 117 117 117 117 53 104,2

39869 9967 1144 15 192 65 192 65 192 141,2

999611 99961 4 3 335 335 335 335 335 335

999611 99961 11 6 683 683 683 683 683 683

999611 99961 1144 15 680 340 340 340 680 476

Spring 2006, by Barbara ten Tusscher, Jesse Krijthe, Brigitte Sprenger

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Barbara, Jesse, BrigitteBarbara, Jesse, Brigitte• Verify Pollard rho

analysis• Use various

values of p, q, y• Clear

dependence of time on q

• Ignoring 80, cor- relation to √q is overly exact.

p q av. it

999683 97 19

997001 997 68

957409 9973 80

999611 99961 683

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Dependence on yDependence on y• Run same p, q

combination with different inputs y = gx

• Correspondence to √q again

• Not to x: the log of small power of g is no easier

p q x time

3989 997 4 44

3989 997 11 16

3989 997 114 39

999611 99961 4 335

999611 99961 114 297

999611 99961 11144 266

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Surprise: individual numbersSurprise: individual numbersp q x 1 2 3 4 5999611 99961 4 335 335 335 335 335999611 99961 11 683 683 683 683 683999611 99961 114 103 392 206 392 392999611 99961 1144 680 340 340 340 680999611 99961 11144 158 120 300 390 360

Iterations: equal or have high common factor!

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ObservationsObservations• Average number of iterations

coincides well with √q• Almost no variation within one row

• Is this a bug in the program??– Bad randomization in calculator?– Or general property of Pollard Rho?

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5. Function graph5. Function graph• Function f: zi -> zi+1 defines graph• Out-degree 1, cycles with in-trees• Length, component, size• Graph is the same when algorithm is

repeated with the same input• Starting point differs• As zi = z2i, i must be multiple of cycle

length

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Layers in a componentLayers in a component• Layer of node: measure distance to

cycle in terms of its length l:– Point z in cycle has layer 0– Point z is in layer 1 if f(l)(z) in cycle– Point z is in layer c if f(c.l)(z) in cycle

• Lemma: z0 in layer c gives c.l iter.

• Is there a dominant component or layer?

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Layers 0 and 1 dominateLayers 0 and 1 dominateProbability theory analysis by Meli

Samikin

Lemma: Pr(layer ≤ 1) = ½Proof: Assume collision after k steps: z0 -> z1 -> … -> … -> zk-1 -> ??

Layer of z0 is 0 if zk = z0, Pr = 1/k

Layer of z0 is 1 if zk = zj < k/2, Pr ≈ 1/2

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Dominant ComponentDominant ComponentLemma: Random z0 and w0,

Pr(same component) > ½.Proof: First collision after k steps: z0 -> z1 -> … -> … -> zk-1 -> ??

w0 -> w1 -> … -> … -> wk-1 -> ??

Pr ( z meets other sequence ) = ½.Then, w-sequence may collide into z.

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Experiments: dominanceExperiments: dominance• Jos Roseboom:

count points in layers of each component

• ACS Experimentation Project, Fall 2007

• Explicitly construct and measure function graphs

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Size of largest componentSize of largest componentVerdeling puntenwolk Pollard

0

10

20

30

40

50

60

70

80

90

100

1,00E+00 1,00E+01 1,00E+02 1,00E+03 1,00E+04 1,00E+05 1,00E+06 1,00E+07

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ConclusionsConclusions• Elgamal + handcalculators = fun• Functional requirements easier to

explain than for RSA• Security: experiment with DLog• Pollard, only randomizes at start• Iterations: random variable, but

takes only limited values• Most often: size of heaviest cycle

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Rabbit FormulaRabbit Formula• Ontsleutelen is: v delen door ua

• u(a1+a2) is: ua1.ua2

• Deel eerst door ua1 en dan door ua2

• Team 1: bereken v’ = Deca1(u, v)Team 2: bereken x = Deca2(u, v’)

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Overzicht van formulesOverzicht van formules• Constanten:

Priemgetal p, grondtal g• Sleutelpaar:

Secret a en Public b = ga

• Encryptie: (u, v) = (gk, x.bk) met bDecryptie: x = v/ua met a

• Prijsvraag: b = b1b2. Ontsleutelen?