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I am trying to write code to find one solution to satisfy Diophantine equations $ax + by = c$ given a, b, and c (will use a = 172, b = 20, c = 1000 for example).

Here is the method I am following, and I am stuck on step #2. How can I generalize this algorithm or what would be some tips for writing the proper prior steps into the division algorithm to get the right values?

  1. Use the division algorithm to find the greatest common denominator "d" between a and b: I am doing this using a function. Store values during this in lists so they can be reproduced. I believe I'm successful in this so far. See code snippet #1.

  2. Use the Euclidean Algorithm to then "work backwards" using the formulas during the division algorithm to get coefficients $x_0$ and $y_0$ such that $ax_0 + by_0 = d$ .

  3. Find p such that $c = dp$ .

  4. Then use $x_0 = x*p$ and $y_0 = y*p$ for the solutions of $c = ax + by$ .

Code snippet #1

clist = list()
dlist = list()
ilist = list()
rlist = list()

def gcd(a, b):
    assert a >= 0 and b >= 0 and a + b > 0


if a > 0 and b > 0:
    if a>b:
        dlist.append(int(a/b))
    else:
        dlist.append(int(b/a))

while a > 0 and b > 0:
    if a >= b:
        ilist.append(a)
        a = a % b
        clist.append(a)
        if a > 0 and b > 0: #trivial recomparison
            if a>b:
                dlist.append(int(a/b))
            else:
                dlist.append(int(b/a))
        rlist.append(max(a,b))
        print("a executed") #debugging print
    else:
        ilist.append(b)
        b = b % a
        clist.append(b)
        if a > 0 and b > 0:
            if a>b:
                dlist.append(int(a/b))
            else:
                dlist.append(int(b/a))
        rlist.append(max(a,b))
        print("b executed") #debugging print
print("clist:", clist)
print("dlist:", dlist)
print("ilist:", ilist)
print("rlist:", rlist)
return max(a, b)




print(gcd(172,20))

Output is as follows:

a executed

b executed

a executed

b executed

clist: [12, 8, 4, 0]

dlist: [8, 1, 1, 2]

ilist: [172, 20, 12, 8]

rlist: [20, 12, 8, 4]

4

I understand how to do the following (solving for #2 above) manually, but there are steps like simplifying and factoring that I am having trouble getting into Python. Would really appreciate some recommendations!

Manually, I know I can get one solution by following this path:

  1. Solve for the remainder: $$ 172/20 = 8 R 12 $$ $$ 20/12 = 1 R 8 $$ $$ 12/8 = 1 R 4 $$ --> 4 is thus the final remainder and the gcd $$ 8/4 = 2 R 0 $$

Restated as three equations, $$ 12 = 172 - 1 * 20 $$ $$ 8 = 20 - 1 * 12 $$ $$ 4 = 12 - 1 * 8 $$

So working backwards and re-factoring: $$ 4 = 12 - 1 * 8 $$ $$ 4 = 12 - 1 * (20 - 1 * 12) $$ $$ 4 = 2 * 12 - 1 * 20 $$ $$ 4 = 2 * (172 - 8 * 20) - 1 * 20 $$ $$ 4 = 2 * 172 + (-17) * 20 $$ Therefore, 2 and -17 are solutions, and I'd like to be able to achieve this in code but I'm a ways away. Part of the problem is I am not sure if there's an easy way to generalize this method.

New on the site, new to computer science, and new to discrete math so I don't mind constructive criticism on all three of these.

qiabozi
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    Welcome to Math Stack Exchange. Did you mean Euclid's algorithm when you wrote Euler's in the title? – J. W. Tanner Mar 08 '23 at 22:22
  • The extended euclidean algorithm is described here in many places, e.g. see the linked dupes (and their links). – Bill Dubuque Mar 08 '23 at 22:43
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    @J.W.Tanner oof. Yes. Thanks! – qiabozi Mar 08 '23 at 22:49
  • @BillDubuque thank you! I checked all of the linked dupes and none of them describe how the function might be coded, where I'm running into issues is understanding Euclid's algorithm at a coding level, I understand how to solve this manually as the dupe questions asked. I'll take my question elsewhere, thank you for your time. – qiabozi Mar 08 '23 at 22:53
  • What are you having problems coding? The algorithms use only basic arithmetic operations. How to write programs in specific languages is off-topic here. – Bill Dubuque Mar 08 '23 at 22:55
  • Toward the bottom I have 5 lines where I manually come up with a solution. The first line is straightforward and corresponds to results that I indexed during the gcd function, namely: clist_2 = ilist_2 - dlist2 * rlist2. In the following line I refactor 8 as (20 - 1 * 12), which is the prior step in the division algorithm. Then I group like factors together in line 3. On line 5, I once again group like factors together. Though it's clearly hard for me to articulate, I'd say this is the main problem I am running into, as simplification is something I understand manually but not programatically. – qiabozi Mar 08 '23 at 22:59
  • You can psuedocode for the algorithm here and likely in many answers here. – Bill Dubuque Mar 08 '23 at 23:00
  • @BillDubuque I will give it a shot - once again, thanks for your time! – qiabozi Mar 08 '23 at 23:01

0 Answers0