问题
I want to write a predicate that an integer and a list of digits, and succeed if Digits contain the digits of the integer in the proper order, i.e:
?-digit_lists( Num, [1,2,3,4] ).
[Num == 1234].
Here is what I have so far:
my_digits( 0, [] ).
my_digits(N,[A|As]) :- N1 is floor(N/10), A is N mod 10, my_digits(N1, As).
回答1:
As already suggested, consider using finite domain constraints:
:- use_module(library(clpfd)).
number_digits(Number, 0, [Number]) :- Number in 0..9.
number_digits(Number, N, [Digit|Digits]) :-
Digit in 0..9,
N #= N1 + 1,
Number #= Digit*10^N + Number1,
Number1 #>= 0,
N #> 0,
number_digits(Number1, N1, Digits).
This predicate can be used in all directions. Examples with either argument instantiated:
?- number_digits(215, _, Ds).
Ds = [2, 1, 5] ;
false.
?- number_digits(N, _, [4,3,2,1]).
N = 4321 ;
false.
And two more general queries:
?- number_digits(N, _, [A,B]).
N in 10..99,
_G2018+B#=N,
_G2018 in 10..90,
A*10#=_G2018,
A in 0..9,
B in 0..9 ;
false.
?- number_digits(N, _, Ds).
Ds = [N],
N in 0..9 ;
Ds = [_G843, _G846],
N in 0..99,
_G870+_G846#=N,
_G870 in 0..90,
_G843*10#=_G870,
_G843 in 0..9,
_G846 in 0..9 ;
etc.
回答2:
Here comes yet another variant based on clpfd... Based on (#=)/3 and if_//3 we define:
n_base_digits(N, R, Ds) :-
N #> 0, % positive integers only
R #> 1, % smallest base = 2
Ds = [D|_], % leading digit may not be 0
D #> 0,
phrase(n_base_digits_aux(N, R, Ds), Ds).
n_base_digits_aux(N, Base, [_|Rs]) -->
{ D #= N mod Base,
M #= N // Base },
if_(M #= 0,
{ Rs = [] },
n_base_digits_aux(M, Base, Rs)),
[D].
Query using SICStus Prolog 4.3.3:
| ?- n_base_digits(1234, 10, Ds).
Ds = [1,2,3,4] ? ;
no
Works the other way round, too!
| ?- n_base_digits(I,10,[1,2,3]).
I = 123 ? ;
no
Note that the above is faster than number_digits/3 as proposed by @mat in his answer.
回答3:
You could also avoid recursion and use in-built predicates for type conversions:
my_digits(Number, List) :-
atomic_list_concat(List, Atom),
atom_number(Atom, Number).
The first line converts the list to an atom, and the second line converts this atom to a number, which will give true if that number is the same as that passed in.
I don't know if there is an even more direct way to convert the list into a number (don't think so..), in which case it could be achieved in a single line.
回答4:
I think this is easier:
numToList(NUM,[LIST|[]]):-
NUM < 10,
LIST is NUM,
!.
numToList(NUM,LIST):-
P is NUM // 10,
numToList(P,LIST1),
END is (NUM mod 10),
append(LIST1,[END] ,LIST).
回答5:
I don't agree with @ssBarBee. After all, you should get 4321 if you supply your list and their allegation is correct; but instead you get this:
?- my_digits(Num, [1,2,3,4]).
ERROR: is/2: Arguments are not sufficiently instantiated
We could try it with clpfd:
my_digits( 0, [] ).
my_digits(N,[A|As]) :- N1 #= N/10, A #= N mod 10, my_digits(N1, As).
We get this:
?- my_digits(Num, [1,2,3,4]), label([Num]).
Num = -6789 ;
Num = 4321.
I find all that pretty curious, but tracing with clpfd is not pleasant.
If you just wanted to parse a list of numbers I would be inclined to make it tail recursive like so:
my_digits(Num, List) :- my_digits(0, List, Num).
my_digits(Num, [], Num).
my_digits(N, [A|As], Num) :- N1 is N * 10 + A, my_digits(N1, As, Num).
This gives us:
?- my_digits(Num, [1,2,3,4]).
Num = 1234 ;
false.
But it doesn't generate:
?- my_digits(1234, X).
ERROR: is/2: Arguments are not sufficiently instantiated
If I were solving this without clpfd, I'd be inclined at this point to just inspect my arguments and have separate predicates. Gross, I know, but that's what I'd do.
my_digits(Num, List) :-
nonvar(List),
my_digits_p(0, List, Num).
my_digits(Num, List) :-
var(List),
my_digits_g(Num, ListRev),
reverse(ListRev, List).
my_digits_p(Num, [], Num).
my_digits_p(N, [A|As], Num) :- N1 is N * 10 + A, my_digits(N1, As, Num).
my_digits_g(0, []) :- !.
my_digits_g(N, [A|As]) :- A is N mod 10, N1 is floor(N / 10), my_digits_g(N1, As).
This can parse or check, or generate if the number is a non-variable:
?- my_digits(1234, X).
X = [1, 2, 3, 4].
?- my_digits(X, [1,2,3,4]).
X = 1234 ;
false.
?- my_digits(1234, [1,2,3,4]).
true;
false.
If you try and generate with both arguments as variables you'll get a pretty unhelpful result though:
?- my_digits(X, Y).
X = 0,
Y = [].
So we can try and generate by adding another special case to my_digits:
my_digits(Num, List) :-
var(Num), var(List),
my_digits_g_from(0, Num, ListRev),
reverse(ListRev, List).
my_digits(Num, List) :-
nonvar(List),
my_digits_p(0, List, Num).
my_digits(Num, List) :-
var(List),
my_digits_g(Num, ListRev),
reverse(ListRev, List).
my_digits_g_from(N, N, List) :- my_digits_g(N, List).
my_digits_g_from(N, Num, List) :- succ(N, N1), my_digits_g_from(N1, Num, List).
That's a lot of code, and a good demonstration of the kind of acrobatics one has to do when not using clp(fd). It's an unfortunate fact that when doing arithmetic in Prolog one must work around the fact that is does not unify, but the complexity of clp(fd) is good proof of why that is.
I hope someone else has a more elegant solution!
回答6:
For a class assignment? What the professor is probably looking for is something like the following. A a general rule, your analysis of the problem statement should first identify the special cases (in this case, zero and negative values) and then the general case.
: -- int_2_digits/2 ------------------------------------------------------------
:
: The public api.
:
: we've got 2 special cases here:
:
: * zero, and
: * negative numbers
:
: and, of course, the general case: a positive value.
:
: ------------------------------------------------------------------------------
int_2_digits( 0 , [0] ) . : zero is a special case
int_2 digits( X , ['-'|Ds] ) :- : negative numbers are a special case
X < 0 , : which we handle (YMMV) by prepending the
X1 is - X , : sign and than processing the absolute value
int_2_digits(X1,Ds) . :
int_2_digits( X , Ds ) :- : the general case is a positive value
X > 0 , : just invoke the worker predicate.
int_2_digits(X,[],Ds) . :
: -- int_2_digits/3 ------------------------------------------------------------
:
: The guts of the operation.
:
: We're using an accumulator here because we compute the result right-to-left,
: from least significant digit to most significant digit. Using the accumulator
: builds the list in the correst sequence, so we don't have to reverse it at
: the end.
: ------------------------------------------------------------------------------
int_2_digits( 0 , Ds , Ds ) . : if we hit zero, we're done. Unify the accumulator with the result
int_2_digits( X , Ts , Ds ) :- : otherwise...
D is mod(X,10) , : - get the current digit (X modulo 10)
T is div(X,10) , : - get the next value via integer division
int_2_digits( X1 , [T|Ts] , Ds ) : - recurse down
. : Easy!
来源：https://stackoverflow.com/questions/16047339/prolog-converting-integer-to-a-list-of-digit