Solving a Single Congruence Equation - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Software Libraries (/forum-10.html) +--- Forum: HP Prime Software Library (/forum-15.html) +--- Thread: Solving a Single Congruence Equation (/thread-446.html) Solving a Single Congruence Equation - Eddie W. Shore - 01-16-2014 03:14 AM The program solves for x in the equation: A * x = B mod N Examples: 4 * x = 6 mod 7 A = 4, B = 6, N = 7 Solution: 5 5 * x = 3 mod 17 A = 5, B = 3, N = 17 Solution: 4 11 * x = 3 mod 16 A = 11, B = 3, N = 16 Solution: 9 HP Prime Program: CONG Code:  EXPORT CONG( ) BEGIN LOCAL A,B,N,I; // 2014-01-15 EWS INPUT({A,B,N}, "Ax = B mod N",  {"A","B","N"}, { }, {0, 0, 0} ); // safe guard if the user does not enter integers (optional line) A:=IP(A); B:=IP(B); N:=IP(N); // Algorithm FOR I FROM 1 TO N-1 DO IF FP((A*I-B)/N) == 0 THEN MSGBOX("x ="+STRING(I)); RETURN I; KILL; END; END; RETURN "No Solution"; END; RE: Solving a Single Congruence Equation - Thomas Klemm - 04-12-2014 07:22 AM How long does it take to solve: 999999999998 * x = 1 mod 999999999999 Or maybe just: 999998 * x = 1 mod 999999 Kind regards Thomas PS: Ever heard of the Chinese remainder theorem? RE: Solving a Single Congruence Equation - rprosperi - 04-12-2014 11:00 PM (04-12-2014 07:22 AM)Thomas Klemm Wrote:  How long does it take to solve: 999999999998 * x = 1 mod 999999999999 Or maybe just: 999998 * x = 1 mod 999999 Kind regards Thomas PS: Ever heard of the Chinese remainder theorem? Returns answer of 2 instantly on an actual Prime. I guess faster than instantly on the emulator. RE: Solving a Single Congruence Equation - Thomas Klemm - 04-15-2014 01:48 PM (04-12-2014 11:00 PM)rprosperi Wrote:  Returns answer of 2 instantly on an actual Prime. I guess faster than instantly on the emulator. Is that the answer to: 999999999998 * x = 1 mod 999999999999 ? Because that's wrong. 999999999998 * 2 = 999999999997 mod 999999999999 But that's probably due to a rounding error: $$\frac{999999999998 \times 2 - 1}{999999999999} = 1.9999999999969999999999969999999999969999999999969999...$$ This will be rounded to 2.00000000000. The correct answer is of course: x = 999999999998 = -1 mod 999999999999 The 2nd example shouldn't suffer from these kind of problems though I didn't test it. Cheers Thomas RE: Solving a Single Congruence Equation - Han - 04-15-2014 11:31 PM In keeping the algorithm fairly simple, I was thinking of the following adjustment. If $$A > B$$ then compute $$(q,r) \in \mathbb{Z}^2$$ such that $$A = qB + r$$. Then $Ai - B \equiv (qB+r)i - B \equiv B (qi -1) + ri$ and let $$i$$ run from $$-N/2$$ to $$N/2$$ (adjusting for even/odd $$N$$ of course). Similarly, if $$B = qA + r$$ then $Ai - B \equiv Ai - (qA+r) \equiv A (i-q) -r$ We still run into overflow issues, but I think this approach might handle a few more cases than the original approach. Also, I wonder if the MOD command would work better than division inside FP(). RE: Solving a Single Congruence Equation - Thomas Klemm - 04-16-2014 04:00 AM (04-15-2014 11:31 PM)Han Wrote:  In keeping the algorithm fairly simple Euklid's algorithm to calculate the greatest common divisor isn't that complicated. You just keep track of the multiples of A and N. ([u v] is short for: u*N + v*A) Example: 5 * x = 3 mod 17 Calculate gcd(17, 5): [1 0] 17 [0 1] 5 [1 -3] 2 = 17 - 3 * 5 [-2 7] 1 = 5 - 2 * 2 Thus gcd(17, 5) = 1 = -2 * 17 + 7 * 5 Therefore: 5 * 7 = 1 mod 17 5 * 7 * 3 = 3 mod 17 5 * 21 = 3 mod 17 5 * 4 = 3 mod 17 Cheers Thomas PS: You can ignore u. Just keep track of v. RE: Solving a Single Congruence Equation - Han - 04-16-2014 04:23 AM Even simpler :-) Very nice! RE: Solving a Single Congruence Equation - Thomas Klemm - 04-16-2014 07:48 AM This program shouldn't result in an overflow: Code: #!/usr/bin/python def add(a, b, n):     result = a + b     return result if result < n else (a - n) + b def minus(a, b, n):     result = a - b     return result if 0 <= result else result + n def double(a, n):     return add(a, a, n) def times(a, b, n):     result = 0     while a > 0:         if a % 2:             result = add(result, b, n)         a /= 2         b = double(b, n)     return result def inverse(a, n):     p, q = n, a     u, v = 0, 1     while q > 0:         r = p / q         u, v = v, minus(u, times(r, v, n), n)         p, q = q, p - r * q     return u      a, b, n = 5, 3, 17 print times(inverse(a, n), b, n)      a, b, n = 999999999998, 1, 999999999999 print times(inverse(a, n), b, n) Maybe somebody feels like translating that for the Prime? Cheers Thomas RE: Solving a Single Congruence Equation - salvomic - 02-07-2015 03:12 PM (01-16-2014 03:14 AM)Eddie W. Shore Wrote:  The program solves for x in the equation: A * x = B mod N I was searching a program like this Thank you, Eddie! Salvo RE: Solving a Single Congruence Equation - Albert Chan - 03-08-2019 03:24 AM It may be easier to solve A x ≡ B (mod N) problem using continued fraction. It is the same as Euclid extended gcd method, without tracking multiples for A and N Example 5 x ≡ 3 (mod 17) 17/5 = 3 + 1/(2 + 1/2) Drop the last term, we get 3 + 1/2 = 7/2 -> 7*5 - 2*17 = 1, so 7 ≡ 1/5 (mod 17) x ≡ 3/5 ≡ 3*7 ≡ 4 (mod 17) RE: Solving a Single Congruence Equation - Albert Chan - 03-08-2019 07:16 PM Getting to the "2nd best" convergents might be messy, we can do guesses. You get to the same result even if the guesses were off. Example: With modulo N=17789, solve 12345 x ≡ 1 12345 ≡ -5444 N/5444 ≈ 3.2676 ≈ 13/4, 13*(12345 x ≡ 1) → (384 x ≡ 13) N/384 ≈ 46.3255 ≈ 139/3, 139*(384 x ≡ 13) → (9 x ≡ 1807) N/9 ≈ 1976.5555 ≈ 3953/2, 3953*(9 x ≡ 1807) → (-x ≡ 9682) → (x ≡ 8107) Warning: make sure guess scaling is co-prime to the modulo. RE: Solving a Single Congruence Equation - Albert Chan - 03-08-2019 11:34 PM (03-08-2019 07:16 PM)Albert Chan Wrote:  Example: solve 12345 x ≡ 1 (mod N), with N = 17789, for x 12345 ≡ -5444 (mod N) For comparison, this build 17789/5444 continued fraction convergents P/Q Code: (next column) = CF * (current column) + (prev column) CF   3   3   1   2   1   3   1   6    5    2 P 0  1   3  10  13  36  49 183 232 1572 8107 17789 Q 1  0   1   3   4  11  15  56  71  482 2481  5444 Q values are not needed here ... 12345 * 8107 ≡ 1 (mod N), thus x ≡ 1/12345 ≡ 8107 (mod N) Edit: it might be cheaper to do Q row instead, since Q's < ½ P's P's = round(Q's * fraction) RE: Solving a Single Congruence Equation - Albert Chan - 03-09-2019 02:10 PM Since we only need 2nd best convergents (to get inverse), we can skip some intermediates. Build CF coef with rounded of number, not the integer part. Coefficients are not really continued fraction coefficients, but it is OK The list is likely shorter, and easily built with calculator FIX-0 mode: 17789/5444 = ; show 3 1/(Ans - Rnd(Ans = ; show 4 1/(Ans - Rnd(Ans = ; show -4 ... Code: Coef 3 4 -4   5   -7   -5    -2 P 0  1 3 13 -49 -232 1575 -8107 17789 12345 * 8107 ≡ 1 (mod 17789) x ≡ 1/12345 ≡ 8107 (mod 17789) RE: Solving a Single Congruence Equation - Albert Chan - 03-10-2019 12:45 AM Another way to do inverse is to force even coef., thus easily reduced. (make sure mul/div factors co-prime to the modulo) Same example, solve 12345 x ≡ 1 (mod 17789) 12345 x ≡ -5444 x ≡ 1 ≡ -17788 (mod 17789) 1361 x ≡ 4447 (mod 17789) (12345 - 9*1361) x ≡ 1 - 9*4447 (mod 17789) 96 x ≡ -40022 ≡ -75600 (mod 17789) 2 x ≡ -1575 ≡ 16214 (mod 17789) x ≡ 8107 (mod 17789) If N is large, we can solve another, with smaller modulo: x ≡ (4447 - 17789 k) / 1361 (mod 17789) → 17789 k ≡ 4447 (mod 1361) 96 k ≡ 4447 ≡ 5808 (mod 1361) 2 k ≡ 121 ≡ -1240 (mod 1361) k ≡ -620 (mod 1361) x ≡ (4447 - 17789 * -620) / 1361 ≡ 8107 (mod 17789) RE: Solving a Single Congruence Equation - Albert Chan - 03-12-2019 03:46 AM (03-10-2019 12:45 AM)Albert Chan Wrote:  If N is large, we can solve another, with smaller modulo: x ≡ (4447 - 17789 k) / 1361 (mod 17789) → 17789 k ≡ 4447 (mod 1361) Example, solve 1223334444 x ≡ 1 (mod 9988776655) List Euclid GCD intermedates, build inverses in reverse order. Start from last pair, 1 x ≡ 1 (mod 3), scale it up. Euclid Inverse(row) (modulo next-row) 1 → 1 3 → -floor(1/3 * 19) = -6 19 → -floor(-6/19 * 22) = 7 22 → -floor(7/22 * 63) = -20 63 → -floor(-20/63 * 211) = 67 211 → -87 274 → 850 2677 → -8587 27044 → 26611 83809 → -88420 278471 → 115031 362280 → -548544 1727591 → 2857751 9000235 → -3406295 10727826 → 64171061 202101103 → -388432661 1223334444 → 3171632349 9988776655 We have 1/1223334444 (mod 9988776655) ≡ 3171632349 More specifically, [1223334444 , 9988776655] • [3171632349, -388432661] = 1 RE: Solving a Single Congruence Equation - Albert Chan - 03-12-2019 09:03 PM (03-12-2019 03:46 AM)Albert Chan Wrote:  List Euclid GCD intermedates, build inverses in reverse order. Start from last pair, 1 x ≡ 1 (mod 3), scale it up. There is no need to walk the whole chain. For x ≡ 1/1223334444 (mod 9988776655), gcd intermediates are even, final inverse is positive. We can guess where it should end up. 1st entry is -6 (mod 19) → x ≈ |floor(−6/19 * 9988776655)| = 3154350523 2nd entry is 7 (mod 22) → x ≈ |floor(+7/22 * 9988776655)| = 3178247117 Extrapolation from -6 (mod 19) is too much. The big inverse had not converged yet. So, scale the inverse in steps, each time guaranteed convergence. $$\frac{1}{|6/19 - 7/22|}$$ = 19*22 = 418 > 274, we can safely skip to mod 274 inverses. 1/211 (mod 274) = (-1)4 floor(-6/19 * 274) = -87 Redo previous example, skipping unneeded calculations: Euclid Inverse(row) (modulo next-row) 1 → 1 ; 3*19=57 3 19 → (-1)^2 floor(1/3 * 22) = 7 ; 22*63=1386 22 63 211 → (-1)^3 floor(7/22 * 274) = -87 ; 274*2677=733498 274 2677 27044 83809 278471 → (-1)^5 floor(-87/274 * 362280) = 115031 ; 362280*1727591 > 6e11 362280 1727591 9000235 10727826 202101103 1223334444 → (-1)^6 floor(115031/362280 * 9988776655) = 3171632349 9988776655 Thus, with only 4 scalings, 1/1223334444 ≡ 3171632349 (mod 9988776655)