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## Re: Adding if/then/else statement to GMPL

 From: Domingo Alvarez Duarte Subject: Re: Adding if/then/else statement to GMPL Date: Tue, 25 Aug 2020 20:17:06 +0200 User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:68.0) Gecko/20100101 Thunderbird/68.10.0

```Hello Heinrich !

Here is better example from examples/shikaku.mod:

With if/then/else:

=====

/* Output solution graphically */
printf "\nSolution:\n";
for { row in rows1 } {
for { col in cols1 } {
if (sum{(i,j,k,l,m,n) in B:
col >= l and col <= n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1} 1) > 0 then
{
if (sum{(i,j,k,l,m,n) in B:
row >= k and row <= m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1} 1) > 0 then printf "+";
else printf "-";
}
else
{
if (sum{(i,j,k,l,m,n) in B:
row >= k and row <= m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1} 1) > 0 then printf "|";
else printf " ";
}

if (sum{(i,j,k,l,m,n) in B:
col >= l and col < n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1} 1) > 0 then printf "---";
else printf "   ";
}
printf "\n";

for { col in cols: (sum{ s in rows: s = row } 1) = 1 } {
if (sum{(i,j,k,l,m,n) in B:
row >= k and row < m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1} 1) > 0 then printf "|";
else printf " ";
```
if (sum{ (i,j) in V: i = row and j = col} 1) > 0 then printf " %2d", givens[row,col];
```        else printf "  .";
}
if (sum{ r in rows: r = row } 1) = 1 then printf "|\n";
}

=====

Original:

=====

/* Output solution graphically */
printf "\nSolution:\n";
for { row in rows1 } {
for { col in cols1 } {
printf{0..0: card({(i,j,k,l,m,n) in B:
col >= l and col <= n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1}) > 0 and
card({(i,j,k,l,m,n) in B:
row >= k and row <= m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1}) > 0} "+";
printf{0..0: card({(i,j,k,l,m,n) in B:
col >= l and col <= n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1}) = 0 and
card({(i,j,k,l,m,n) in B:
row >= k and row <= m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1}) > 0} "|";
printf{0..0: card({(i,j,k,l,m,n) in B:
row >= k and row <= m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1}) = 0 and
card({(i,j,k,l,m,n) in B:
col >= l and col <= n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1}) > 0} "-";
printf{0..0: card({(i,j,k,l,m,n) in B:
row >= k and row <= m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1}) = 0 and
card({(i,j,k,l,m,n) in B:
col >= l and col <= n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1}) = 0} " ";

printf{0..0: card({(i,j,k,l,m,n) in B:
col >= l and col < n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1}) > 0} "---";
printf{0..0: card({(i,j,k,l,m,n) in B:
col >= l and col < n and (row = k or row = m) and
x[i,j,k,l,m,n] = 1}) = 0} "   ";
}
printf "\n";

for { (col,p) in { cols, 1 }: card({ s in rows: s = row }) = 1 } {
printf{0..0: card({(i,j,k,l,m,n) in B:
row >= k and row < m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1}) > 0} "|";
printf{0..0: card({(i,j,k,l,m,n) in B:
row >= k and row < m and (col = l or col = n) and
x[i,j,k,l,m,n] = 1}) = 0} " ";
```
printf{0..0: card({ (i,j) in V: i = row and j = col}) > 0} " %2d", givens[row,col];
```        printf{0..0: card({ (i,j) in V: i = row and j = col}) = 0} "  .";
}
printf{0..0: card({ r in rows: r = row }) = 1} "|\n";
}

=====

Cheers !
On 24/8/20 16:59, Heinrich Schuchardt wrote:
```
```On 24.08.20 16:33, Domingo Alvarez Duarte wrote:
```
```Hello Meketon !

Could you share your view of how it would be expressed (an ideal model
sample) ?

If you want to talk about it, maybe I'll be interested in implement it !

Can you share a collection of models data to be used as base for the
test/implementation ?
```
```Dear Domingo,

I do not yet understand what was you weren't able to express with the
current syntax.

if length(p) == 3 then display "true 3"; else display "false 3";
if length(p) == 5 then display "true 5"; else display "false 5";

you can write:

display if length(p) == 3 then "true 3" else "false 3";
display if length(p) == 5 then "true 5" else "false 5";
solve;
end;

Best regards

Heinrich
```
```

```