Bézout (or extended Euclidean algorithm) finds integer solutions for C, E & GCD( X , Y ) in the equation
C * X + E * Y = GCD( X , Y ).
For example, for stack
Y: 666,777
X: 777,666
The programme returns
Z: 111 (our GCD)
Y: -3,233 (E)
X: 2,772 (C)
& indeed
2,772 * 777,666 + -3,233 * 666,777 = 111
1 LBL B
2 1►E►F-1►C►D
3 R↓
4 REGY
5 REGY
6 INT/
7 +/-
8 ENTER
9 RCL* D
10 RCL+ E
11 x<> D
12 STO E
13 R↓
14 ENTER
15 RCL* F
16 RCL+ C
17 x<> F
18 STO C
19 R↓
20 REGY
21 *
22 REGZ
23 +
24 x≠0?
25 GTO B004
26 R↓
27 RCL E
28 RCL C
29 RTN
Should the GCD be 1, then E is the multiplicative inverse of Y modulo X & C the multiplicative inverse of X modulo Y.
(12-20-2014 05:05 PM)Gerald H Wrote: [ -> ]Should the GCD be 1, then E is the multiplicative inverse of Y modulo X & C the multiplicative inverse of X modulo Y.
You might be interested in:
Solving a Single Congruence Equation
(04-18-2014 05:24 PM)Thomas Klemm Wrote: [ -> ]Remark:
Since overflow is avoided it's possible to solve:
999,999,999,989 * x = 1 mod 999,999,999,999
Solution:
899,999,999,999
Kind regards
Thomas
Here an alternative programme to the above, same input & output. I expected the time taken for calculations to be less than with the first progamme & don't really understand why that's not the case.
1 LBL E
2 STO H
3 x<>y
4 STO G
5 [H,1]
6 [G,0]
7 REGX*[1,0]
8 x=0?
9 GTO E022
10 REGZ
11 x<>y
12 /
13 [1,0]
14 *
15 IP
16 REGY
17 *
18 REGZ
19 -
20 +/-
21 GTO E007
22 REGZ*[1,0]
23 REGT*[0,1]
24 (REGY-REGX*H)/G
25 x<>y
26 RTN
(12-22-2014 04:32 PM)Gerald H Wrote: [ -> ]Here an alternative programme to the above, same input & output. I expected the time taken for calculations to be less than with the first progamme & don't really understand why that's not the case.
Q: How do you evaluate the speed of the execution?
My 2 cents GCD program, nothing special here, I just like the simplicity
Code:
A001 LBLA
A002 x>y?
A003 x<>y
A004 RMDR
A005 x=0?
A006 GTO A009
A007 LASTx
A008 GTO A003
A009 LASTX
A010 RTN
(02-17-2015 08:49 AM)Tugdual Wrote: [ -> ] (12-22-2014 04:32 PM)Gerald H Wrote: [ -> ]Here an alternative programme to the above, same input & output. I expected the time taken for calculations to be less than with the first progamme & don't really understand why that's not the case.
Q: How do you evaluate the speed of the execution?
My 2 cents GCD program, nothing special here, I just like the simplicity
Code:
A001 LBLA
A002 x>y?
A003 x<>y
A004 RMDR
A005 x=0?
A006 GTO A009
A007 LASTx
A008 GTO A003
A009 LASTX
A010 RTN
I store the begin time, run the programme, & subtract present time from the stored value.
(02-17-2015 09:42 AM)Gerald H Wrote: [ -> ]I store the begin time, run the programme, & subtract present time from the stored value.
You have a time function on the 35s?
(02-17-2015 08:49 AM)Tugdual Wrote: [ -> ]My 2 cents GCD program, nothing special here, I just like the simplicity
Whoa... 10 (ten!) steps. Obviously you never used a calculator with limited memory. ;-)
Code:
A001 LBLA
A002 RMDR
A003 LASTx
A004 x<>y
A005 x≠0?
A006 GTO A002
A007 +
A008 RTN
Yes, it's the same method.
Dieter
(02-17-2015 07:29 PM)Dieter Wrote: [ -> ] (02-17-2015 08:49 AM)Tugdual Wrote: [ -> ]My 2 cents GCD program, nothing special here, I just like the simplicity
Whoa... 10 (ten!) steps. Obviously you never used a calculator with limited memory. ;-)
Code:
A001 LBLA
A002 RMDR
A003 LASTx
A004 x<>y
A005 x≠0?
A006 GTO A002
A007 +
A008 RTN
Yes, it's the same method.
Dieter
Oh wow I would have sworn RMDR was requiring a specific sort order...
I like the final "A007 +"
