# Code

A bunch of code.

## Building DFAs

Before describing the model, which is very compact, we will see how Deterministic Finite Automaton (DFA) can be build.

We will focus on a single sequence: {1, 2, 3}.

### Regexp way

The regular expression that encodes the sequence is "0*10+1{2}0+1{3}0*":

• 0* the word can start with unbounded number of 0 (* means zero or more times)
• 10+ the first block of 1 is followed by at least one 0 (+ means one or more times)
• 1{2}0+ deals with the second block of 2 (a{n} means a occurs exactly n times) which is followed by at least one 0
• 1{3}0* the third – and last – block of size 3 is followed by zero or more 0. Indeed, this the last block of the sequence, so there cannot be other 1 after but 0s are optional.

Starting and ending 0s are optional but it has to be defined in the regexp, otherwise some valid words may be skipped.

private void dfa(BoolVar[] cells, int[] rest, Model model) {
StringBuilder regexp = new StringBuilder("0*");
int m = rest.length;
for (int i = 0; i < m; i++) {
regexp.append('1').append('{').append(rest[i]).append('}');
regexp.append('0');
regexp.append(i == m - 1 ? '*' : '+');
}
IAutomaton auto = new FiniteAutomaton(regexp.toString());
model.regular(cells, auto).post();
}


### Constructive way

The constructive way requires to declare all states of the automaton and links together with transitions. A transition corresponds to a character in the word, and a state is between two characters of the word.

So there is a need of an initial state from which (through an outgoing transition) the first character of the word will be provided. And at least one final state to which (through an ingoing transition) the last character of the word will be provided.

We note $s_i$ the initial state. The first character can either be a 0 or a 1, there will be two transitions outgoing from $s_i$. Then, transition from $s_i$ producing 0 will go to $i_0$ (first transition). And transition from $s_i$ producing 1 will go to $i_1$ (second transition). $i_0$ points to itself providing 0 (third transition). Outgoing transition from $i_1$ goes to $i_2$ and produces 0 (fourth transition). Two transitions outgoes from $i_2$: one goes to itself (fifth transition, producing 0), one goes to $i_3$ (sixth transition, producing 1). $i_3$ goes to $i_4$ (seventh transition) and produce 1. $i_4$ goes to $i_5$ (eighth transition) and produce 0. And so on.

And here the code for building such a DFA for any sequence:

private void dfa2(BoolVar[] cells, int[] seq, Model model) {
FiniteAutomaton auto = new FiniteAutomaton();
auto.setInitialState(si); // declare it as initial state
auto.addTransition(si, i0, 0); // first transition
auto.addTransition(i0, i0, 0); // second transition
int from = i0;
int m = seq.length;
for (int i = 0; i < m; i++) {
if(i == 0){
}
from = ii;
for(int j = 1; j < seq[i]; j++){
from = jj;
}
// the word can end with '1' or '0'
if(i == m - 1){
auto.setFinal(from, ii0);
}
from = ii0;
}
model.regular(cells, auto).post();
}


## The entire code

// number of columns
int N = 15;
// number of rows
int M = 15;
int[][][] BLOCKS =
new int[][][]{{
{2},
{4, 2},
{1, 1, 4},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 2, 2, 1},
{1, 3, 1},
{2, 1},
{1, 1, 1, 2},
{2, 1, 1, 1},
{1, 2},
{1, 2, 1},
}, {
{3},
{3},
{10},
{2},
{2},
{8, 2},
{2},
{1, 2, 1},
{2, 1},
{7},
{2},
{2},
{10},
{3},
{2}}};

Model model = new Model("Nonogram");
// Variables declaration
BoolVar[][] cells = model.boolVarMatrix("c", N, M);
// Constraint declaration
// one regular per row
for (int i = 0; i < M; i++) {
dfa(cells[i], BLOCKS[0][i], model);
}
for (int j = 0; j < N; j++) {
dfa(ArrayUtils.getColumn(cells, j), BLOCKS[1][j], model);
}
if(model.getSolver().solve()){
for (int i = 0; i < cells.length; i++) {
System.out.printf("\t");
for (int j = 0; j < cells[i].length; j++) {
System.out.printf(cells[i][j].getValue() == 1 ? "#" : " ");
}
System.out.printf("\n");
}
}


## Things to remember

• Regular constraint constructs valid fix-sized words on the basis of a vocabulary and a grammar.
• A deterministic finite automaton can either be build with a regular expression or step-by-step.
• Regular constraints are very useful when patterns occur in solutions. For example, when dealing with shifts on a personnal scheduling problem: for example: “a nurse doesn’t do a late night shift followed by a day shift the next day”.