Am 25. August 2020 20:17:06 MESZ schrieb Domingo Alvarez Duarte
<[email protected]>:
>Hello Heinrich !
>
>Here is better example from examples/shikaku.mod:
>
Just another example where the existing if-then-else expression can be used to
shorten the code.
Anyway comparing a binary to =1 is not a good idea in GLPK. You should use > .5.
Best regards
Heinrich
>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.
>>
>> Instead of
>>
>> 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:
>>
>> param p,symbolic := "dad";
>> 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
