Hi all,

Long time no see! I have one question about the implementation of continued equality for a break condition of my algorithm. I want to achieve the following pseudocode:

If: a(1) = a(2) = a(3) = … = a(n), then Break;

else: Continue;

In a GAMS code manner:

```
loop(n,
...
break$(a(1)=a(2) and a(1)=a(3) and ... a(1)=a(n));
continue;
);
```

Does anyone have idea for that? Thanks in advance!

Cheers!

Gabriel

Sorry but I found the solution myself And I apologize for the error in the post that I use ‘n’ twice for index.

I share my solution with you.

```
loop(i,
...
loop(n,
s(n) = a('1') - a(n);
);
if(sum(n, s(n)) = 0,
k = 1;
);
break$(k = 1);
continue;
);
```

If you have better idea please share with me too! Thanks!

Gabriel

A few observations:

- The sequence
*a = 2,1,3* will also result in sum(n,s(n))=0. You probably should use abs to calculate sn.
- GAMS likes parallel assignment statements (no need to loop over n) and you can even shove this all into the break $-condition
- Depending how
*a* got calculated (e.g. by a solve) they numbers might not be truly equal, so instead for =0 you probably should check for close to 0.

```
loop(i,
...
break$(sum(n, abs(a('1') - a(n)) < 1e-6);
);
```

-Michael

Thanks for your kind reply! That should be the best solution!

Gabriel