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<title><string language="fre"><![CDATA[3.4. Complexity Analysis]]></string></title>
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<string language="fre"><![CDATA[In this session, I will
present the main technique to make the analysis of the
various algorithms presented in this course. So, Information Set
Decoding refers to a family of algorithms which is
similar to the Prange algorithm that we have just seen. All variants of Information
Set Decoding repeat a large number of independent
iterations which all have a constant cost K and a success probability P. This means that this
iteration has to be repeated an expected number of
times N where N = 1/P. And the total
workfactor of the algorithm will simply be N multiplied
by K, the cost of the iteration. First, do we want one solution to the CSD problem or all solutions? So, we consider the
CSD(H,s,w) problem. We will assume, as I
said, it is the case for most cryptanalysis, that
the problem we are considering has at least one solution,
that is CSD(H,s,w) is not empty. There are two possibilities for the weight. Either the weight is
smaller than the Gilbert-Varshamov radius, then there is
exactly one solution, either the weight w
is larger than the Gilbert-Varshamov radius. In that case, there are
several solutions (n,w)/2^(n-k) on average. The first case
is the most common and of course, there is no
difference between one or all solutions because
there is only one solution. In the second case, we
expect that finding only one solution instead of all
solutions will be less expensive. Intuitively, it is
reasonable to assume that we may make the economy of a
factor equal to the number of solutions. So, some probabilities. Recall that Information Set
Decoding will perform many independent iterations. For
one iteration, we denote P? the probability to find one
specific solution to our problem. And we denote P1 the
probability to find any one solution to our problem. If N is the number of
solutions then we may write P1, as given in the slide. The exact formula
will produce a value which is the minimum of 1 and N*P?. In practice, most of the
time, we will have P1 = N*P? when N is not too large at least. For the complexity analysis, we will
have to distinguish two situations.]]></string></description>
<keyword><string language="fre"><![CDATA[algèbre linéaire]]></string></keyword><keyword><string language="fre"><![CDATA[chiffrement à clé publique]]></string></keyword><keyword><string language="fre"><![CDATA[cryptage des données]]></string></keyword><keyword><string language="fre"><![CDATA[cryptographie]]></string></keyword><keyword><string language="fre"><![CDATA[algorithmes]]></string></keyword>
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<date><dateTime>2015-05-05</dateTime></date>
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