# 4 a α a s r r α m kr α kr α m mod p b α a m r r

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4.(a) (αa)srrαm-krαkrαm(modp).(b) (αa)mrrαamαkrαs(modp).(c) (αa)rrmαarαkmαs(modp).5.(a) We havev1βrr-rv-1βr(βvαu)-rv-1αar-arvv-1-urv-1α-urv-1, andv2αmαsuα-ruv-1. Thereforev1v2, so the signature isvalid.(b) In part (a), we choose the message by lettingmsu(modp-1)Therefore onceu, vare chosen, it is unlikely thatαh(m)αm(modp-1). Sothe signature will probably not be valid.6.First, sincek=a, she’ll haver=β, so Eve can see this.Thensk-1(m-ar)k-1m-r(modp-1). Since Eve knowsr, s, m, she can solvethis fork-1.Actually, she obtains gcd(m, p-1) possibilities fork-1, henceforkand fora. She tries each of these until she finds the value ofasuch thatαaβ(modp).7.(a) The density of primes nearxis approximately 1/lnx. Withx10100,we have a density of around 1/230.Since we only look at odd numbers, thedensity if around 1/115.(b) By the reasoning in (a), the density of primes in the odd numbers near2512is approximately 2/ln(2512)1/177.8.(a)βf(r)rsαaf(r)+ksαm(modp).(b) Eve needs arrange thatrsαm.For example, she can takek= 1,r=α, ands=mand have a valid signature (m, r, s).27
Chapter 10 - Exercises1.First calculateaA,aB,aC,bA,bBandbCto getaA= 10,aB= 17,aC= 14,bA= 14,bB= 6 andbC= 5. UsinggA(x) =aA+bAx, and similarly forgB(x),we may calculate the keys byKAB=gA(rB) = 21,KAC=gA(rC) = 7 andKBC=gB(rC) = 29.2.(a) To showKAB=a+b(rA+rB) +crArBsimply substitute the definitionforaAandbAintoKAB=gA(rB).(b) Using part (a) it is clear thatKAB=KBA.(c)KAB=f(rA, rB).3.We must solve fora,b, andc.To do this, we may use 18 =a+ 9b(mod 31) and 24 =a+ 2b(mod 31) to geta= 8 andb= 8.Then, using29 =b+ 9cwe getc= 23.4.(a) Alice calculates (αyq)xand Bob calculates (αxq)y. These are the same.(b) Observe that the key is (αq)xy. SinceαMq=αp-1= 1 there are onlyMpossible values forαkqfor different values ofk.Eve may find the key bycalculatingαqand raising this to successive powers.5.Suppose we arrange the participants in a ring, likeBobTedCarolAliceBob.Each person starts withαand raises it to their private exponent, e.g.Bobcalculatesαb.They each send their respective result to the person to theirright. The next round, they take what they received and raise it to their privateexponent and pass it to their right. If we repeat this two more times, they willall have calculatedαbtca.6.(a) UseK=M1KNand substituteM1=M2KHto getK=M2KHKN. Now substituteM2=M3KHand use the fact thatKNKN= 0to getK=M3KH.(b) Observe thatKN=M3M2andK=M1KN. Thus Eve can calculatethe key.
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