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I have a question that intrigue me:

Given primes and some reminder vector

P=primes(13)=[2 3 5 7 11 13]

R=[1 2 1 3 9 11]  (R=mod(1571,P))

What option do i have to reconstruct 1571 from R?

I already know about CRT.

Is there any other way? (assumming that direct brute force search is not practical)

Mendi Barel
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    Why doesn't CRT suffice? – Bill Dubuque Aug 09 '19 at 19:07
  • It suffice and works well, but it is little involved – Mendi Barel Aug 09 '19 at 19:13
  • We want to find x. So we know $x \equiv 11 \pmod {13} $ Then $x = 13k + 11$ for some $k$.

    Plug this in into the equation $x \equiv 9 \pmod {11}$ to get $13 k + 11 \equiv 9 \pmod{11}$

    $13k \equiv -2 \equiv 9 \pmod{11} \Rightarrow 2k \equiv 9 \pmod{11} \Rightarrow k \equiv 10 \pmod{11}$.

    So $x$ is now $13(11p + 10) + 11$ for some integer $p$. Now plug this into another equation and keep going until you get $x = 1571 + 30030 * q$ $ $ for any integer $q$

     – Francisco José Letterio Aug 09 '19 at 19:18
  • Probably because you don't know optimizations. Are you solving them all-at-once or stepwise, two-at-a-time (e.g.in prior comment). – Bill Dubuque Aug 09 '19 at 19:20
  • we can use that it's odd, to sieve a lot out. but that's mostly CRT. –  Aug 09 '19 at 19:32
  • My question is for the general case. I guess that for some special examples there are some special methods. @Bill, what do you mean about optimization? I dont like CRT because it uses extended euclid algorithm. – Mendi Barel Aug 09 '19 at 19:38
  • @Francisco can you write this algorithm for any R value and length, in matlab function? – Mendi Barel Aug 09 '19 at 19:40
  • I added an answer showing some optimizations. – Bill Dubuque Aug 09 '19 at 19:50
  • @MendiBarel should be able to, sounds pretty easy – Francisco José Letterio Aug 09 '19 at 20:29
  • Impractically we could use 2 isn't a quadratic residue mod 3, 3 isn't a cubic residue mod 7, and 9 isn't a quintic residue mod 11. But, again that's impractical. –  Aug 09 '19 at 20:58

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It's easy if you solve them stepwise - two-at-a-time.

$x\equiv -2\, \bmod 11\, \&\, 13 \iff x\equiv -2\pmod{\!143}\ $ by CCRT = Constant case CRT. Similarly

$\ x \equiv\ 1\,\bmod\ 2\ \, \&\ \ 5\ \iff\, x\ \equiv\ 1\ \pmod{\!10}.\ $ Solving them pairwise we obtain:

$\!\!\bmod \color{#c00}{10}\!:\,\ 1\equiv x\equiv -2+143\,\color{#c00}i\equiv -2+3i\iff 3i\equiv 3\iff \color{#c00}{i\equiv 1}$

Therefore $\ x = -2+143(\color{#c00}{1\!+\!10j}) = \color{#0a0}{141 + 1430j}$

$\!\!\bmod 7\!:\,\ 3\equiv x\equiv\color{#0a0}{ 1+2j}\iff 2j\equiv 2\iff j\equiv 1$

Therefore $\,x \equiv 141+1430(1\!+\!7k) = \color{#90f}{1571 + 10010k}$

$\!\!\bmod 3\!:\,\ 2\equiv x\equiv\color{#90f}{ 2+2k}\iff k = 0\iff k =3n$

Therefore $\,x \equiv 1571+ 30030n.\,$ Just a couple minutes mental arithmetic (with practice).

Bill Dubuque
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  • Can you write function in matlab that solve the problem for any R length and value? Input should be R only. first line should be P=primes(length(R)). – Mendi Barel Aug 09 '19 at 20:35
  • @MendiBarel Yes, that could be done (I implemented something similar in Macsyma long ago). – Bill Dubuque Aug 09 '19 at 20:41