2
$\begingroup$

The Cryptography made simple (page 207, under Fig 11.12)(Nigel Smart) say that adversary's advantage of IND-PASS Game is $Adv1 = 2\times|Pr[b=b']-\frac{1}{2}|$.
The reason for multiplying by 2 is to normalize advantage from $[0,\frac{1}{2}]$ to $[0,1]$.

But in this paper (page 5, line 9), the advantage of IND-CKA Game is $Adv2 = |Pr[b=b']-\frac{1}{2}|$ which is not normalized and scale is $[0,\frac{1}{2}]$.
And this value is used with the advantage of pseudo random function
$Adv3 = |Pr[A(f)=0]-Pr[A(g)=0]|$
(f is pseudo random function. g is random function)
which scale is $[0,1]$.

Does $Adv2$ need not be normalized to $[0,1]$ for use with $Adv3$?
Or do I usually not need to be aware of normalization of "advantage"?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Welcome to Cryptography.SE. We have $\LaTeX$/MathJax enabled on our site. You can also provide the page number of the book. It is free to download that I've given the link. $\endgroup$
    – kelalaka
    Nov 10, 2021 at 12:36
  • $\begingroup$ thank you! Added page number. (It is 207 pages in print, but it looks 212 page in Google Chrome) $\endgroup$
    – apapapa
    Nov 10, 2021 at 14:51
  • 1
    $\begingroup$ The "value" of the advantage often depends on the context and agreed-upon conventions as long as they express something meaningful. Consequently, the expression for a CPA game may look different than the one for a hash collision game. For a specific game, the scale should not really matter as long as the goal is understood. I would say it is still good to be aware of the different conventions even though they express the same thing. $\endgroup$
    – Marc Ilunga
    Nov 11, 2021 at 14:40

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.