topic Re: Variable number of nested loops in Intel® Fortran Compiler

Variable number of nested loops

roy437 — Wed, 16 Dec 2020 22:23:19 GMT

Hi,

How to write a Fortran code with a variable number of nested loops.

Thank you.

Re: Variable number of nested loops

Steve_Lionel — Thu, 17 Dec 2020 00:28:10 GMT

I can't think of a way to do this. Consider an alternate approach - maybe one that uses a list of actions to take.

Re: Variable number of nested loops

mecej4 — Thu, 17 Dec 2020 01:12:25 GMT

In old Fortran, such nested looping would be accomplished using GOTO statements.

Instead of writing the nested loops explicitly using DO constructs, you can accomplish the same effect by using recursive subroutines or functions. See, for example, this implementation of Quicksort. The recursion can be ended by reaching a count that is computed at run time, or a more general condition may be used.

Re: Variable number of nested loops

Arjen_Markus — Thu, 17 Dec 2020 07:12:29 GMT

Could you explain your use case in some detail? Recursion is certainly a versatile tool, but I simply do not understand what you are trying to accomplish and there may be other, more appropriate ways. (Also it would satisfy my curiosity)

Re: Variable number of nested loops

roy437 — Thu, 17 Dec 2020 11:33:00 GMT

For m=2 and m = 3 here it is code:

select case (m) case (2) do i1 = 0, 9 do i2 = 0, 9 do i3 = 0, 9 e = v - n(2)*(p(2)+0.1*i2+0.01*i3) t = 10.0*(e/n(1)-p(1))-0.1*i1 if(dabs(t-dnint(t))<1.0d-8.and.(-0.0001<=t.and.t<9.0001)) then ! if t = 0, 1, 2, 3, ...9 r(1) = p(1)+0.1* t+0.01*i1 r(2) = p(2)+0.1*i2+0.01*i3 go to 91 end if end do end do end do case (3) do i1 = 0, 9 do i2 = 0, 9 do i3 = 0, 9 do i4 = 0, 9 do i5 = 0, 9 e = v - n(2)*(p(2)+0.1*i2+0.01*i3) - n(3)*(p(3)+0.1*i4+0.01*i5) t = 10.0*(e/n(1)-p(1))-0.1*i1 if(dabs(t-dnint(t))<1.0d-8.and.(-0.0001<=t.and.t<9.0001)) then ! if t = 0, 1, 2, 3, ...9 r(1) = p(1)+0.1* t+0.01*i1 r(2) = p(2)+0.1*i2+0.01*i3 r(3) = p(3)+0.1*i4+0.01*i5 go to 91 end if end do end do end do end do end do end select 91 continue

I would like a code for some m.

Thank you

Re: Variable number of nested loops

Arjen_Markus — Thu, 17 Dec 2020 12:09:59 GMT

Well, for that sort of situations a solution using recursion is certainly useful. Your current code is a brute-force approach, nothing wrong with that. I would do the recursion on m: the sum up to and including m-1 (for particular choices of ji, ki and li) would provide a new value for V (Vnew = V - sum(i = 1...m-1)) and then you try and find values for jm, km and lm using the nested loop. If you find a solution, print it and be done with it (or continue the search). Once the search space for this level is exhausted, return and proceed with the next iteration. Something like:

recursive subroutine searc( v, m, ... ) do j(i) = 1,9 do k(i) = 1,9 do l(i) = 1,9 vnew = v - ... if ( vnew > 0 .and. m > 1 ) then call search( vnew, m-1, ... ) else print the solution ... endif enddo enddo enddo

Mind you: just a sketch!

Re: Variable number of nested loops

mecej4 — Thu, 17 Dec 2020 12:50:13 GMT

Arjen Markus is spot on, and I hope that his question gets you to think about methods other than blind brute force. The assignment appears to be a rather effective homework assignment, and it presents you with an opportunity to think/learn about the properties of modular arithmetic, and exploit that knowledge to make the program short, general and efficient.

You do not need the arbitrarily-deep DO loop capability for this problem. You should not use a CASE construct for m = 2, 3, .... Instead, make m a variable and write code that will work for any input m, perhaps up to a limit. Unless you have a valid reason to use floating point arithmetic, do not use it for a Diophantine problem.

Here is a hint to get you started. Find the answer by thinking and using pencil-and-paper:

What is the value of V mod 10 ?

If this is a homework assignment, you may be doing us as well as yourself a disservice by asking us to provide you with the code itself.

Re: Variable number of nested loops

roy437 — Thu, 17 Dec 2020 13:14:09 GMT

This is not a homework assignment, it is a real problem that I face every day in my accounting work. Finding the solution with pencil and paper for m greater than 2 takes a long time.

Re: Variable number of nested loops

roy437 — Thu, 17 Dec 2020 13:22:43 GMT

Thanks, I'll try.

Re: Variable number of nested loops

mecej4 — Thu, 17 Dec 2020 13:54:26 GMT

Your problem statement does not contain the array p(:), which your code does. Please state what it is, and whether it is an input array, part of the solution, or just a temporary array.

Re: Variable number of nested loops

roy437 — Thu, 17 Dec 2020 15:40:33 GMT

p(i) is j(i) from equation, is known.

Re: Variable number of nested loops

mecej4 — Thu, 17 Dec 2020 17:49:58 GMT

Roy, I have a solution that I am testing, and it has worked on a small problem that I made up.

Please provide a couple of sample problems that you have solved, but don't give me the solution yet. That is, provide values for m, n(), j() and V.

I am asking you for such problems because I may have chosen a rather trivial case.

Re: Variable number of nested loops

roy437 — Thu, 17 Dec 2020 21:28:44 GMT

mecej4,

m = 2 n() j() V = 101646 20 46 3 31 m = 3 n() j() V = 227444 12 78 25 46 6 29

Re: Variable number of nested loops

mecej4 — Fri, 18 Dec 2020 01:07:17 GMT

In your problem definition (your second post in this thread), you have an inconsistency:

"i, j, k, l, m, n, V are integers > 0" versus "k_i, l_i are integers between [0,9]".

Assuming that k, l are allowed to have 0 as a value, I obtained these two solutions (yes, there can be multiple solutions) for your m = 3 case.

Solution A:

i	1	2	3
k_i	9	1	0
l_i	5	0	9

Solution B:

i	1	2	3
k_i	0	3	9
l_i	0	4	9

If these are acceptable solutions, I'd like to try a problem for a larger m, say, m = 7 or 11. Do you have one?

[I may have made some errors in transcribing the values from my calculation into the tables. If you spot an error, please let me know and I'll rectify.]

By the way, this type of problem is closely related to Cryptarithmetic Problems.

Re: Variable number of nested loops

roy437 — Fri, 18 Dec 2020 09:55:17 GMT

Thank you mecej4,
I wanted to say: i,j(),m,n(),V are integers > 0, and k(), l() are integers between [0,9]

The solutions are correct and I found solution C: i 1 2 3 ------------ k_i 2 3 5 l_i 1 4 7 ************************************************************************************** Here are two other cases: m = 7 n() j() V = 187321251 98 465 550 269 354 720 806 524 610 328 413 780 642 745 ------------------------- m = 11 n() j() V = 268177985 811 178 263 630 715 434 3 327 534 858 286 389 37 141 568 671 99 202 851 954 382 706

Re: Variable number of nested loops

mecej4 — Fri, 18 Dec 2020 10:57:13 GMT

Here are the solutions (two, for each problem) for your m=7 and m=11 problems.

Problem for m = 7, Solution A

i	1	2	3	4	5	6	7
k_i	7	9	9	9	5	0	0
l_i	8	2	0	0	0	5	1

Problem for m = 7, Solution B

i	1	2	3	4	5	6	7
k_i	8	9	8	8	5	1	1
l_i	0	0	1	1	1	7	0

Problem for m = 11, Solution A

i	1	2	3	4	5	6	7	8	9	10	11
k_i	0	0	4	0	9	8	3	9	9	9	9
l_i	9	9	9	7	9	9	9	9	9	9	9

Problem for m = 11, Solution B

i	1	2	3	4	5	6	7	8	9	10	11
k_i	9	9	9	9	0	0	9	9	0	9	0
l_i	0	0	9	1	9	1	5	9	0	0	9

I obtained these solutions using the AMPL solver . The "model" file for m = 11:

# Roy437 test problem. param M integer := 11; param V integer := 268177985; set IDX = 1..M; var k {IDX} integer >= 0 <= 9; var l {IDX} integer >= 0 <= 9; param n {IDX} integer >= 1; param j {IDX} integer >= 1; subj to C: sum{i in IDX} (100*j[i]+10*k[i]+l[i])*n[i] = V; data ; param: n := 1 811 2 263 3 715 4 3 5 534 6 286 7 37 8 568 9 99 10 851 11 382 ; param: j := 1 178 2 630 3 434 4 327 5 858 6 389 7 141 8 671 9 202 10 954 11 706 ; option solver ilogcp; solve; display {i in IDX} (k[i],l[i]);

You may use the online NEOS server if you do not wish to obtain the AMPL demo package and install it on your machine. If you use the server, use the three files in the attached zip.

If obtaining the solution is sufficient, you may use AMPL on your PC or on the NEOS server. If you wish to call the AMPL solver from a Fortran or C program, a number of methods may be tried, but mating a Fortran program to the AMPL solvers is non-trivial.

Re: Variable number of nested loops

roy437 — Fri, 18 Dec 2020 12:29:58 GMT

Thank you very much mecej4.

The solutions are good, I checked them. I installed the AMPL Command Line download for Windows program, but it gives me an error. Please, can you tell me what commands to give or how to proceed with the three files royM11.

Re: Variable number of nested loops

mecej4 — Fri, 18 Dec 2020 12:48:38 GMT

There are a number of tutorials for running AMPL from the command line, or using the Eclipse-based IDE, and the AMPL language itself. For now, since you have AMPL installed locally, the following is the simplest approach:

open a CMD window, and set the path to include the directory containing AMPL.EXE.
in your working directory, copy the code in the box in my post above (not the the files in the zip, which are for NEOS) to, say, M11.mod .
enter the command

ampl M11.mod

You will see the results displayed at the console.

Re: Variable number of nested loops

roy437 — Fri, 18 Dec 2020 13:03:17 GMT

Thanks mecej4,

I did it. Can I redirect the result to a file?

Re: Variable number of nested loops

mecej4 — Fri, 18 Dec 2020 13:14:01 GMT

Of course. The command can be

ampl M11.mod > M11.out

You can then type the M11.out file, open it in Notepad or another editor, Word, etc.