In a preparatory question we had to recover decimal digits $@$ of $r$ and $s$ given

$g=51234$, $h=90403$, $N=111649$, $r=3@497$, $s=276@3$,

with $r$ and $s$ the smallest positive solutions to $g^r\equiv h\pmod N$ and $h^s\equiv g\pmod N$.

For large $N$ of unknown factorization, we can't compute $\varphi(N)$ from $N$ as was possible in the above. What techniques can be used to solve the problem?

I tried to use Index Calculus on the example below but failed. This is a very simple problem launched in 2016:

Recover all digits of $r$, $s$ and the factorization of $N$.

It is given $N$, $g$, $h$, $r=\log_g h$ such that $g^r\equiv h\pmod N$, and $s=\log_h g$ such that $h^s\equiv g\pmod N$, with @ figuring an unknown decimal digit:

N = 4770047289861054128673165840475881666985708841069569122247779994575999373990855387188690474135592158102615485877459618836427811081668578893542268141889988869764086285203345206116260923268064851230188829163155218509489708275219402417
g = 1818674628639967921918316591104926407701223436868786162485485191757839034354239408133296195009525639079428182610287045427973123535470060353099447741037582550708917693998695104714221823196243333713913383763090393730760094204741650090
h = 2208658095572997139683467669564020944425802120964728746970717796984260065210935180456703377157524093154996948648279266706057554344748359754567027091354206701512345543745638076145129569549991043944481795915889787265463492997755593901
r =  1@6826295140541573@9405922358@031348321@068314394@23991550@60902965@70229914@6203562@77909154@2484058239@305290554@451041944@77284864589306634379124981555276837338619890402355995522612897221369937487171781010716680@8004465050769716
s =  57@2623169175139588682565993777078629@51369239581656482190187775@98993533114834@931596544472993@0093163506369690@0954113499@984660633390860@46262557530048@277694556546720@7679243674115857@55557012193132332836046587@22597144319762@@

Editor's note: the original question was changed for clarity, de-emphasizing the numerical example, and using the simpler notation introduced in this answer to the original preparatory question.


Resolution plan

$N$ is a composite with no trival factor, and our plan will be:

  1. Find what's missing in $r$ (14 decimal digits) from what's known, the values of $N$, $g$, $h$, and the relation $g^r\equiv h\pmod N$, using a minor variation of the baby-step giant-step Discrete Log algorithm.
  2. Similarly find the missing decimal digits in $s$.
  3. Raising $g^r\equiv h\pmod N$ to the power $s$ then substituting $h^s\equiv g\pmod N$ gives $g^{r\cdot s}\equiv g\pmod N$. Hence $r\cdot s-1$, which we now can compute, is a multiple of the order of $g$, and thus has a fair chance to be a multiple of $\varphi(N)/2$.
    Note: That's not sure! In the preparatory problem $g$ is of order $\varphi(N)/2$, and it's at best reasonable to hope that in the full problem that order is $\varphi(N)/2$ or $\varphi(N)$. However, if $g$ has been randomly selected, its order likely is a multiple of $\varphi(N)/c$ for some small $c$ dividing $\varphi(N)$, and $c$ can be searched.
  4. Factor $N$ from its value and that of the multiple of $\varphi(N)/2$ just obtained.

Finding the full $r$ with baby-step giant-step

We are working in base $b$ and missing digits of $r$ at index $i$ in set $\Bbb M$ (starting at index 0 on the right), and can write $\displaystyle r=\hat r+\sum_{i\in\Bbb M}x_i\,b^i$ with unknowns $x_i\in[0,b)$. Here $b=10$, $\hat r$ is obtained by replacing @ with 0 in the given form of $r$, and the 14 indexes of the @ starting from the right are $\Bbb M=\{16,106,116,126,137,146,154,163,172,181,191,201,212,229\}$.

We split $\Bbb M$ about evenly into disjoint subsets $\Bbb U$ and $\Bbb V$ and rewrite $g^r\equiv h\pmod N$ as

$$\begin{array}{llll} g^r&\equiv&h&\pmod N\\ g^{\left(\hat r+\displaystyle\sum_{i\in\Bbb M}x_i\,b^i\right)}&\equiv&h&\pmod N\\\\ g^{\hat r}\cdot\displaystyle\prod_{i\in\Bbb M}\left(g^{(b^i)}\right)^{x_i}&\equiv&h&\pmod N\\ g^{\hat r}\cdot\displaystyle\prod_{i\in\Bbb U}\left(g^{(b^i)}\right)^{x_i}&\equiv&h\cdot\displaystyle\prod_{i\in\Bbb V}\left(g^{(-b^i)}\right)^{x_i}&\pmod N \end{array}$$

We pre-compute $g^{-1}\bmod N$ (using e.g. the half-extended Euclidean algorithm), and $\left(g^{-1}\right)^{(b^i)}\bmod N$ for ${i\in\Bbb V}$, then can sequentially compute and store in a data structure allowing fast search all $b^{|\Bbb V|}$ values of $\displaystyle\left(h\cdot\prod_{i\in\Bbb V}\left(g^{(-b^i)}\right)^{x_i}\right)\bmod N$ (alongside the matching tuple of $x_i$ for $i\in\Bbb V$). We start from $h$ for all $x_i=0$ with ${i\in\Bbb V}$ and perform one modular multiplication for each other (we need $|\Bbb V|$ intermediary variables, including $h$).

We pre-compute $g^{\hat r}\bmod N$ and the $g^{(b^i)}\bmod N$ for ${i\in\Bbb U}$, and sequentially compute and search in said data structure $g^{\hat r}\cdot\displaystyle\left(\prod_{i\in\Bbb U}\left(g^{(b^i)}\right)^{x_i}\right)\bmod N$. We start from $g^{\hat r}\bmod N$ for all $x_i=0$ with ${i\in\Bbb U}$ and perform one modular multiplication for each other (we need $|\Bbb U|$ intermediary variables, including $g^{\hat r}$). When we get a match, that gives us all the $x_i\in\Bbb M$ (those from $\Bbb V$ come from the data structure, those from $\Bbb U$ are the one which led to the match), hence the full $r$.

The worst case computing effort is dominated by about $2\,b^{|\Bbb M|/2}$ (here 20,000,000) modular multiplications modulo $N$ (here, costing about $2\,\lceil(\log_2N)/32+1\rceil^2<1500$ elementary multiplications of 32-bit variables and additions of the 64-bit result using schoolbook algorithms). We are talking minutes of runtime, and less than 3 Megabytes of memory (and by making $|\Bbb V|$ one smaller, we reduce memory requirement by a factor $b$, at the expense of runtime).

  • $\begingroup$ Thank you So much. I'll try to find r, s through BSGS.! But I do not understand this calculation for @bit ... Could you please show me something like the sage code for the following article? or sample calculations ..? $\endgroup$ – user60186 Jul 8 '18 at 14:07
  • $\begingroup$ sorry sir , I don't understand,, line of here>> We split M about evenly into disjoint subsets U and V << here $\endgroup$ – user60186 Jul 8 '18 at 15:06
  • $\begingroup$ Oh , I understand it! Thank you! In this time , I'll try to solve Phi(i in U) g^(10)^i^x_i == h * Phi(i in V) ...~!! $\endgroup$ – user60186 Jul 8 '18 at 15:28
  • $\begingroup$ Thank you So much! :) If I have something I do not know later, can I ask another question? $\endgroup$ – user60186 Jul 8 '18 at 15:28
  • $\begingroup$ @MinYoung Lee: yes, sure. For something directly related to that answer, write a comment here, and I'll get a notification. If it is somewhat independent and of general interest, post a new question. Note: I suggest removing obsolete comments, and thanks (I have read them). I'll then remove my answers to removed comments. That way things are kept tidy (and hopefully our nice "comments are not for chat" police/bot will remain at a distance). $\endgroup$ – fgrieu Jul 8 '18 at 15:35

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