Discrete Optimization Modelling (with MiniZinc)
Combinatorial Problems are extremely difficult. In the industry, we are more interested in tasks like "Maximize Profit", "Minimize Costs" than questions like "in how many ways..". Some of this problems are proven to be so hard (NPComplete) that a general solving time shoots up exponentially unless P = NP.
In the real world, we need an answer to these problems anyway. There are many sophisticated solvers such as gecode, flatzinc, Google ortools and many more. (You might think of Prolog too.) They try to exploit substructures and simplify the problems with sophisticated techniques gathered from several PhD thesies and research papers.
These solvers can be used to solve a variety of problems ranging from several logic puzzles, solving a minesweeper board to serious problems like linear programming, bin packing or scheduling problems.
Contents
Modelling
Modelling is in no way a form procedural programming. It is a form of declarative programming. To clarify, when modelling we make a very specific and more importantly, nonambiguous description of the problem.
This description can be handed over to the solver, who decides how to solve the problem. That is not to say that all equivalent models of the same problem are equally as powerful or we have absolutely no control over the solving mechanism. Just like the human brain, the solver have varying difficulties solving a problem even though they are technically equivalent. Also, to some extent, it is possible to specify the search technique that the solver is going to use.
Here is a simple example of a MiniZinc model of a Linear Programming Problem
We know how to make two sorts of cakes.
A banana cake which takes 250g of selfraising flour, 2 mashed bananas, 75g sugar and 100g of butter, and a chocolate cake which takes 200g of selfraising flour, 75g of cocoa, 150g sugar and 150g of butter. We can sell a chocolate cake for $4.50 and a banana cake for $4.00. And we have 4kg selfraising flour, 6 bananas, 2kg of sugar, 500g of butter and 500g of cocoa.
How many cakes of each sort should we bake to maximize profit?
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21% Baking cakes for the school fete var 0..100: b; % no. of banana cakes var 0..100: c; % no. of chocolate cakes % flour constraint 250*b + 200*c <= 4000; % bananas constraint 2*b <= 6; % sugar constraint 75*b + 150*c <= 2000; % butter constraint 100*b + 150*c <= 500; % cocoa constraint 75*c <= 500; % maximize our profit solve maximize 400*b + 450*c; output ["no. of banana cakes = ", show(b), "\n", "no. of chocolate cakes = ", show(c), "\n"];
MiniZinc
MiniZinc is a constraint modelling language that offers a higher level modelling interface for specifying problems to a number of lower level solvers like Flatzinc or gecode.
One can easily download the MiniZinc packages from here
One can just download the command line executables or the IDE if he pleases.
Dissecting the first example
We know how to make two sorts of cakes.
A banana cake which takes 250g of selfraising flour, 2 mashed bananas, 75g sugar and 100g of butter, and a chocolate cake which takes 200g of selfraising flour, 75g of cocoa, 150g sugar and 150g of butter. We can sell a chocolate cake for $4.50 and a banana cake for $4.00. And we have 4kg selfraising flour, 6 bananas, 2kg of sugar, 500g of butter and 500g of cocoa.
How many cakes of each sort should we bake to maximize profit?
Imagine for a second that this were a Math class. How would you go about formulating the problem?
Here is a possible formulation:
Let $b$ be the number of banana cakes, $c$ the number of chocolate cakes.
Maximize $p(b,c) = 400b + 500c$
subject to $250b + 200c \leq 4000 \\ 2b \leq 6 \\ 75b + 150c \leq 2000 \\ 100b + 150c \leq 500 \\ 75c \leq 500$
Notice that our model is exactly the same as that.
First, we begin with declaring our decision variables, i.e, the things we wish for the solver to solve.
1 2 

Then, we specify the constraints:
1 2 3 4 5 6 7 8 9 10 

And then comes what we want the solver to do:
1 2 

Running Models
If you're using the IDE, this section will not help much but it is still recommended you learn the tricks
The next most important thing you should do is to download the model and run it. Note that minizinc files always end with a .mzn
extension.
Go to the appropriate directory, and type minizinc cakes.mzn
.
On my computer, I see something like this.
A couple of remarks to make:
 The

marks the end of a solution.  The
==========
tells you that the solver has proven this solution to be the optimal solution (according to the criteria given).  You might get an
====UNSATISFIABLE====
implying that your model has no feasible solutions.
If you are interested in seeing all feasible solutions, run minizinc allsolutions cakes.mzn
.
Sometimes, we want to run models with additional data files. (See the next section).
The generic bash command to supply a model is minizinc <model.mzn> <data.dzn>
The data files are required to end with a .dzn
extension.
Models vs. Instances
Usually, we want to write models that are very general. In contrast, an instance is a combination of a model with some data that is passed on to a solver.
Here is an example of how we could split up the above problem into a generalized model and a data file:
cakes2.mzn
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38% Baking cakes for the school fete (with data file) int: flour; %no. grams of flour available int: banana; %no. of bananas available int: sugar; %no. grams of sugar available int: butter; %no. grams of butter available int: cocoa; %no. grams of cocoa available constraint assert(flour >= 0,"Invalid datafile: " ++ "Amount of flour is nonnegative"); constraint assert(banana >= 0,"Invalid datafile: " ++ "Amount of banana is nonnegative"); constraint assert(sugar >= 0,"Invalid datafile: " ++ "Amount of sugar is nonnegative"); constraint assert(butter >= 0,"Invalid datafile: " ++ "Amount of butter is nonnegative"); constraint assert(cocoa >= 0,"Invalid datafile: " ++ "Amount of cocoa is nonnegative"); var 0..100: b; % no. of banana cakes var 0..100: c; % no. of chocolate cakes % flour constraint 250*b + 200*c <= flour; % bananas constraint 2*b <= banana; % sugar constraint 75*b + 150*c <= sugar; % butter constraint 100*b + 150*c <= butter; % cocoa constraint 75*c <= cocoa; % maximize our profit solve maximize 400*b + 450*c; output ["no. of banana cakes = ", show(b), "\n", "no. of chocolate cakes = ", show(c), "\n"];
pantry.dzn
1 2 3 4 5flour = 4000; banana = 6; sugar = 2000; butter = 500; cocoa = 500;