Choco 2 0 1



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Building DFAs¶

Before describing the model, which is very compact, we will see ho the DFAs can be build.

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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*:

Choco
  • 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+
    • the second block of 2 (a{n} means a appears exactly n times)
    • 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 and `0`s are optional.

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

Choco

Caution

In Choco, DFAs only accept integer as character.0*a+ is not a valid grammar, there is no conversion Character (java term) to Integer.But, numbers are allowed, not only digits. Indeed, some variables can take value greater than 9 !In that case, number are declared using the specific characters < and >.For example: 0*<11><22>0* will produce words like 00112200 or 1122 but no 0120.

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.

Battlefront

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. Below is a graph illustrating the process for the sequence {1, 2}.

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

Note

Even though regexp appears to be easy to use, the constructive way is more expressive.