How create a single Prolog rule that match with more then a single words list?

Question

I am having some problem to do a particular operation on a specific list.

I have a list of tokens where a token represents a word and I want use a single predicate to recognize if some contiguous tokens in this list represent a date which may take the following two forms:

1) FIRST FORM: 1° Marzo 1995

2) SECONDO FORM: 1° Marzo of 1995

To do a concrete examples about what I mean when I speck of token list.

This is the tokens list related to the first form:

[t(1,[49]),t(-1,[176]),t(-1,[32]),t(2,[77,97,114,122,111]),t(-1,[32]),t(3,[49,57,57,53]),t(-1,[10])]

and this is the tokens list related to the second form:

[t(1,[49]),t(-1,[176]),t(-1,[32]),t(2,[77,97,114,122,111]),t(-1,[32]),t(3,[111,102]),t(-1,[32]),t(4,[49,57,57,53]),t(-1,[10])]

As you can see a general token have a "functor" t and into it there are two arguments: a number (not necessarily progressive, if a token not represents a word but is something different from an alnum character it value is -1) and a string that represent a word or a single character (like a whit space or the "°" and so on)

I have implement a good working predicate that recognizes if 7 contiguous general tokens in a token list represents a date (like 1° Marzo 1995) and then build a single specialized data token by concatenation, in this way:

tagga([t(Number1, Day), t(-1, "°"), t(-1, Space1), t(Number4, Month), t(-1, Space2), t(Number6, Year)|ListaToken], [d(Number1, CompositeDateTag)|ListaTokenTaggati]) :-

    length(Day, LnDay),
    (LnDay =:= 1; LnDay =:= 2),
    Day = [Head|Tail],
    char_type(Head, digit),
    %number_codes(DayNumber, Day),
    %DayNumber =< 31,

    Space1 == " ",

    member(Month, ["gennaio", "febbraio", "marzo", "aprile", "maggio", "giugno", "luglio", "agosto", "settembre", "ottobre", "novembre",
                       "dicembre", "Gennaio", "Febbraio", "Marzo", "Aprile", "Maggio", "Giugno", "Luglio", "Agosto", "Settembre", "Ottobre",                   "Novembre", "Dicembre"]),

    Space2 == " ",

    length(Year, LnYear),
    LnYear =:= 4,   

    NumericString = [Head|Tail],
    char_type(Head, digit),

    append(Day, "°", UntilSt),
    append(UntilSt, Space1, UntilSpace1),
        append(UntilSpace1, Month, UntilMonth),
        append(UntilMonth, Space2, UntilSpace2),
        append(UntilSpace2, Year, CompositeDateTag),

    tagga(ListaToken, ListaTokenTaggati).

Now I could implement an additional predicate that recognize if 9 contiguous general tokens in the tokens list represent a date in the second form (something like: 15 Marzo of 1995) but I would know if I can modify my previous tagga/2 predicate to do that it recognize both dates that meet the first and the second form

My reasoning is the following one, I have a tokens list that could be something like:

[t(Number1, Day), t(-1, "°"), t(-1, Space1), t(Number4, Month), t(-1, Space2), t(Number6, Year)|ListaToken] in the case of the FIRST DATE FORM

Or something like:

[t(Number1, Day), t(-1, "°"), t(-1, Space1), t(Number4, Month), t(-1, Space2), t(Number7, "of"), t(-1, Space3), t(Number9, Year)|ListaToken] in the case of the SECOND DATE FORM

Ok, I will know if I can combine in some way the two forms into a single general form that match with both and then operate on it...

Something like this:

[t(Number1, Day), t(-1, "°"), t(-1, Space1), t(Number4, Month) | SUBLIST |t(Number9, Year)|ListaToken]

Where sublist could be:

1) The single token: t(-1, Space2)

or

2) Something like this token list: t(-1, Space2), t(Number7, "of"), t(-1, Space3)

Is this idea good? Someone can help me to implement a predicate like this? (I am trying but, until now I have obtained no result)

score 0 · Answer 1 · answered Jun 03 '13 at 10:40

0

by far the easiest way will be to declare both alternative, and let Prolog match the appropriate one.

This could result in maintenance problems in long term, then could be worth to evaluate how many of such patterns you are willing to support, and eventually factor out the common code.

The most important detail is that your rules should be side effect free - as it seems at first glance.

answered Jun 03 '13 at 10:40

CapelliC

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I was thinking the proposed solution because I prefer have a single rule that handles many cases that are very simililar each other rather to have many rules (I prefer so also to do something that is more complex and so to have more argument when I present my project to the teacher, it is also something like an exercise for me) – AndreaNobili Jun 03 '13 at 10:44
so what do you mean that this is side effect free? do you mean that don't change nothing in the original tokens list? – AndreaNobili Jun 03 '13 at 10:45
Prolog don't let you change arguments, exactly because it doesn't relies on assignment to implements its computation model. I was referring to assert/retract, IO and the like. – CapelliC Jun 03 '13 at 11:02

How create a single Prolog rule that match with more then a single words list?

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