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Next-bit test

In cryptography and the theory of computation, the next-bit test[1] is a test against pseudo-random number generators. We say that a sequence of bits passes the next bit test for at any position in the sequence, if any attacker who knows the first bits (but not the seed) cannot predict the st with reasonable computational power.

Precise statement(s) Edit

Let   be a polynomial, and   be a collection of sets such that   contains  -bit long sequences. Moreover, let   be the probability distribution of the strings in  .

We now define the next-bit test in two different ways.

Boolean circuit formulation Edit

A predicting collection[2]   is a collection of boolean circuits, such that each circuit   has less than   gates and exactly   inputs. Let   be the probability that, on input the   first bits of  , a string randomly selected in   with probability  , the circuit correctly predicts  , i.e. :

 

Now, we say that   passes the next-bit test if for any predicting collection  , any polynomial   :

 

Probabilistic Turing machines Edit

We can also define the next-bit test in terms of probabilistic Turing machines, although this definition is somewhat stronger (see Adleman's theorem). Let   be a probabilistic Turing machine, working in polynomial time. Let   be the probability that   predicts the  st bit correctly, i.e.

 

We say that collection   passes the next-bit test if for all polynomial  , for all but finitely many  , for all  :

 

Completeness for Yao's test Edit

The next-bit test is a particular case of Yao's test for random sequences, and passing it is therefore a necessary condition for passing Yao's test. However, it has also been shown a sufficient condition by Yao.[1]

We prove it now in the case of the probabilistic Turing machine, since Adleman has already done the work of replacing randomization with non-uniformity in his theorem. The case of Boolean circuits cannot be derived from this case (since it involves deciding potentially undecidable problems), but the proof of Adleman's theorem can be easily adapted to the case of non-uniform Boolean circuit families.

Let   be a distinguisher for the probabilistic version of Yao's test, i.e. a probabilistic Turing machine, running in polynomial time, such that there is a polynomial   such that for infinitely many  

 

Let  . We have:   and  . Then, we notice that  . Therefore, at least one of the   should be no smaller than  .

Next, we consider probability distributions   and   on  . Distribution   is the probability distribution of choosing the   first bits in   with probability given by  , and the   remaining bits uniformly at random. We have thus:

 

 

We thus have   (a simple calculus trick shows this), thus distributions   and   can be distinguished by  . Without loss of generality, we can assume that  , with   a polynomial.

This gives us a possible construction of a Turing machine solving the next-bit test: upon receiving the   first bits of a sequence,   pads this input with a guess of bit   and then   random bits, chosen with uniform probability. Then it runs  , and outputs   if the result is  , and   else.

References Edit

  1. ^ a b Andrew Chi-Chih Yao. Theory and applications of trapdoor functions. In Proceedings of the 23rd IEEE Symposium on Foundations of Computer Science, 1982.
  2. ^ Manuel Blum and Silvio Micali, How to generate cryptographically strong sequences of pseudo-random bits, in SIAM J. COMPUT., Vol. 13, No. 4, November 1984

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In cryptography and the theory of computation the next bit test 1 is a test against pseudo random number generators We say that a sequence of bits passes the next bit test for at any position i displaystyle i in the sequence if any attacker who knows the i displaystyle i first bits but not the seed cannot predict the i 1 displaystyle i 1 st with reasonable computational power Contents 1 Precise statement s 1 1 Boolean circuit formulation 1 2 Probabilistic Turing machines 2 Completeness for Yao s test 3 ReferencesPrecise statement s EditLet P displaystyle P nbsp be a polynomial and S S k displaystyle S S k nbsp be a collection of sets such that S k displaystyle S k nbsp contains P k displaystyle P k nbsp bit long sequences Moreover let m k displaystyle mu k nbsp be the probability distribution of the strings in S k displaystyle S k nbsp We now define the next bit test in two different ways Boolean circuit formulation Edit A predicting collection 2 C C k i displaystyle C C k i nbsp is a collection of boolean circuits such that each circuit C k i displaystyle C k i nbsp has less than P C k displaystyle P C k nbsp gates and exactly i displaystyle i nbsp inputs Let p k i C displaystyle p k i C nbsp be the probability that on input the i displaystyle i nbsp first bits of s displaystyle s nbsp a string randomly selected in S k displaystyle S k nbsp with probability m k s displaystyle mu k s nbsp the circuit correctly predicts s i 1 displaystyle s i 1 nbsp i e p k i C P C k s 1 s i s i 1 s S k with probability m k s displaystyle p k i C mathcal P left C k s 1 ldots s i s i 1 right s in S k text with probability mu k s nbsp Now we say that S k k displaystyle S k k nbsp passes the next bit test if for any predicting collection C displaystyle C nbsp any polynomial Q displaystyle Q nbsp p k i C lt 1 2 1 Q k displaystyle p k i C lt frac 1 2 frac 1 Q k nbsp Probabilistic Turing machines Edit We can also define the next bit test in terms of probabilistic Turing machines although this definition is somewhat stronger see Adleman s theorem Let M displaystyle mathcal M nbsp be a probabilistic Turing machine working in polynomial time Let p k i M displaystyle p k i mathcal M nbsp be the probability that M displaystyle mathcal M nbsp predicts the i 1 displaystyle i 1 nbsp st bit correctly i e p k i M P M s 1 s i s i 1 s S k with probability m k s displaystyle p k i mathcal M mathcal P M s 1 ldots s i s i 1 s in S k text with probability mu k s nbsp We say that collection S S k displaystyle S S k nbsp passes the next bit test if for all polynomial Q displaystyle Q nbsp for all but finitely many k displaystyle k nbsp for all 0 lt i lt k displaystyle 0 lt i lt k nbsp p k i M lt 1 2 1 Q k displaystyle p k i mathcal M lt frac 1 2 frac 1 Q k nbsp Completeness for Yao s test EditThe next bit test is a particular case of Yao s test for random sequences and passing it is therefore a necessary condition for passing Yao s test However it has also been shown a sufficient condition by Yao 1 We prove it now in the case of the probabilistic Turing machine since Adleman has already done the work of replacing randomization with non uniformity in his theorem The case of Boolean circuits cannot be derived from this case since it involves deciding potentially undecidable problems but the proof of Adleman s theorem can be easily adapted to the case of non uniform Boolean circuit families Let M displaystyle mathcal M nbsp be a distinguisher for the probabilistic version of Yao s test i e a probabilistic Turing machine running in polynomial time such that there is a polynomial Q displaystyle Q nbsp such that for infinitely many k displaystyle k nbsp p k S M p k U M 1 Q k displaystyle p k S mathcal M p k U mathcal M geq frac 1 Q k nbsp Let R k i s 1 s i u i 1 u P k s S k u 0 1 P k displaystyle R k i s 1 ldots s i u i 1 ldots u P k s in S k u in 0 1 P k nbsp We have R k 0 0 1 P k displaystyle R k 0 0 1 P k nbsp and R k P k S k displaystyle R k P k S k nbsp Then we notice that i 0 P k p k R k i 1 M p k R k i M p k R k P k M p k R k 0 M p k S M p k U M 1 Q k displaystyle sum i 0 P k p k R k i 1 mathcal M p k R k i mathcal M geq p k R k P k mathcal M p k R k 0 mathcal M p k S mathcal M p k U mathcal M geq frac 1 Q k nbsp Therefore at least one of the p k R k i 1 M p k R k i M displaystyle p k R k i 1 mathcal M p k R k i mathcal M nbsp should be no smaller than 1 Q k P k displaystyle frac 1 Q k P k nbsp Next we consider probability distributions m k i displaystyle mu k i nbsp and m k i displaystyle overline mu k i nbsp on R k i displaystyle R k i nbsp Distribution m k i displaystyle mu k i nbsp is the probability distribution of choosing the i displaystyle i nbsp first bits in S k displaystyle S k nbsp with probability given by m k displaystyle mu k nbsp and the P k i displaystyle P k i nbsp remaining bits uniformly at random We have thus m k i w 1 w P k s S k s 1 s i w 1 w i m k s 1 2 P k i displaystyle mu k i w 1 ldots w P k left sum s in S k s 1 ldots s i w 1 ldots w i mu k s right left frac 1 2 right P k i nbsp m k i w 1 w P k s S k s 1 s i 1 1 s i w 1 w i m k s 1 2 P k i displaystyle overline mu k i w 1 ldots w P k left sum s in S k s 1 ldots s i 1 1 s i w 1 ldots w i mu k s right left frac 1 2 right P k i nbsp We thus have m k i 1 2 m k i 1 m k i 1 displaystyle mu k i frac 1 2 mu k i 1 overline mu k i 1 nbsp a simple calculus trick shows this thus distributions m k i 1 displaystyle mu k i 1 nbsp and m k i 1 displaystyle overline mu k i 1 nbsp can be distinguished by M displaystyle mathcal M nbsp Without loss of generality we can assume that p m k i 1 M p m k i 1 M 1 2 1 R k displaystyle p mu k i 1 mathcal M p overline mu k i 1 mathcal M geq frac 1 2 frac 1 R k nbsp with R displaystyle R nbsp a polynomial This gives us a possible construction of a Turing machine solving the next bit test upon receiving the i displaystyle i nbsp first bits of a sequence N displaystyle mathcal N nbsp pads this input with a guess of bit l displaystyle l nbsp and then P k i 1 displaystyle P k i 1 nbsp random bits chosen with uniform probability Then it runs M displaystyle mathcal M nbsp and outputs l displaystyle l nbsp if the result is 1 displaystyle 1 nbsp and 1 l displaystyle 1 l nbsp else References Edit a b Andrew Chi Chih Yao Theory and applications of trapdoor functions In Proceedings of the 23rd IEEE Symposium on Foundations of Computer Science 1982 Manuel Blum and Silvio Micali How to generate cryptographically strong sequences of pseudo random bits in SIAM J COMPUT Vol 13 No 4 November 1984 Retrieved from https en wikipedia org w index php title Next bit test amp oldid 1095121770, wikipedia, wiki, book, books, library,

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