zero2

I having a little trouble understanding, how to calculate a problem. I first checked some of the examples in the book, and I couldn't figure out what or how they produce certain parts of the answers, then I went on the web to see if there was an easier or more clearer way of doing this problem, but I'm still having some problems.

Now on to the problem, well actually the example in the book:

Compute GCD (Greatest Common Divisor) d, Let a = 190 and b = 34. Write d = sa + tb

Step 1: Find GCD using Euclidean Algorithm ( I have no problem doing this part, for instance, 190 = 5.34 + 20, 5.34 is something called quotient or 5 x 34 and 20 is calculated by subtracting 190 - 5 x 34)

divide 190 by 34: 190 = 5.34 + 20
divide 34 by 20: 34 = 1.20 + 14
divide 20 by 14: 20 = 1.14 + 6
divide 14 by 6: 14 = 2.6 + 2 <---GCD
divide 6 by 2: 6 = 3.2 + 0

(Once we get a 0 remainder, we stop, and GCD is the remainder above which is 2)

GCD(190, 34) = 2

Step 2: (This is the step where Im stuck at, it doesn't really make much sense, apparently they solve for s & t here, my question is was this figured out?)

GCD(190, 34) = 2 = 14 - 2(6)
=14 - 2[20 - 1(14)] --->6 = 20 - 1.14
= 3(14) -2(20)
= 3[34 - 1(20)] - 2(20) --->14 = 34 - 1.20
= 3(34 - 5(190 - 5.34) --->20 = 190 - 5.34
= 28(34) - 5(190).

s = -5 t = 28

TimeTested

http://www1.cs.columbia.edu/~tal/499...notes/ITC6.pdf

You are using the Extended Euclidean Algorithm. Refer to Example 1. It is a much simpler version of this problem. You use the equations computed in the Euclidean Algorithm to replace the value of the remainders until you are left with only the orginal values a & b.

Restating each of the equations above you get:
190 = 5(34) + 20 --> 20 = 190 - 5(34)
34 = 1(20) + 14 --> 14 = 34 - 1(20)
20 = 1(14) + 6 --> 6 = 20 - 1(14)
14 = 2(6) + 2 --> 2 = 14 - 2(6)

We use these to compute s & t as follows:
Start with the last one -> 2 = 14 - 2(6)
Use the next equation above it to replace 6 to get -> 2 = 14 - 2[20 - 1(14)]
Simplify -> 2 = 14 -2(20) + 2(14) --> 2 = 3(14) - 2(20)
Use the next equation above to replace 14 to get -> 2 = 3[34 - 1(20)] - 2(20)
Simplify -> 2 = 3(34) - 3(20) - 2(20) --> 2 = 3(34) - 5(20)
Now use the top equation to replace 20 to get -> 2 = 3(34) - 5[190 - 5(34)]
Simplify -> 2 = 3(34) - 5(190) + 25(34) --> 2 = 28(34) - 5(190)

since a = 190, s = -5
since b = 34, t = 28

Hope this helps