reversible "binary to number" predicate
Solution 1:
Use CLP(FD) constraints, for example:
:- use_module(library(clpfd)).
binary_number(Bs0, N) :-
reverse(Bs0, Bs),
foldl(binary_number_, Bs, 0-0, _-N).
binary_number_(B, I0-N0, I-N) :-
B in 0..1,
N #= N0 + B*2^I0,
I #= I0 + 1.
Example queries:
?- binary_number([1,0,1], N).
N = 5.
?- binary_number(Bs, 5).
Bs = [1, 0, 1] .
?- binary_number(Bs, N).
Bs = [],
N = 0 ;
Bs = [N],
N in 0..1 ;
etc.
Solution 2:
Here is the solution I was thinking of, or rather what I hoped exists.
:- use_module(library(clpfd)).
binary_number(Bs, N) :-
binary_number_min(Bs, 0,N, N).
binary_number_min([], N,N, _M).
binary_number_min([B|Bs], N0,N, M) :-
B in 0..1,
N1 #= B+2*N0,
M #>= N1,
binary_number_min(Bs, N1,N, M).
This solution also terminates for queries like:
?- Bs = [1|_], N #=< 5, binary_number(Bs, N).
Solution 3:
The solution
This answer seeks to provide a predicate binary_number/2
that presents both logical-purity and the best termination properties. I've used when/2
in order to stop queries like canonical_binary_number(B, 10)
from going into infinite looping after finding the first (unique) solution. There is a trade-off, of course, the program has redundant goals now.
canonical_binary_number([0], 0).
canonical_binary_number([1], 1).
canonical_binary_number([1|Bits], Number):-
when(ground(Number),
(Number > 1,
Pow is floor(log(Number) / log(2)),
Number1 is Number - 2 ^ Pow,
( Number1 > 1
-> Pow1 is floor(log(Number1) / log(2)) + 1
; Pow1 = 1
))),
length(Bits, Pow),
between(1, Pow, Pow1),
length(Bits1, Pow1),
append(Zeros, Bits1, Bits),
maplist(=(0), Zeros),
canonical_binary_number(Bits1, Number1),
Number is Number1 + 2 ^ Pow.
binary_number(Bits, Number):-
canonical_binary_number(Bits, Number).
binary_number([0|Bits], Number):-
binary_number(Bits, Number).
Purity and termination
I claim that this predicate presents logical-purity from construction. I hope I got it right from these answers: one, two and three.
Any goal with proper arguments terminates. If arguments need to be checked, the simplest way to achieve this is using the built-in length/2
:
binary_number(Bits, Number):-
length(_, Number),
canonical_binary_number(Bits, Number).
?- binary_number(Bits, 2+3).
ERROR: length/2: Type error: `integer' expected, found `2+3'
Exception: (6) binary_number(_G1642009, 2+3) ? abort
% Execution Aborted
?- binary_number(Bits, -1).
ERROR: length/2: Domain error: `not_less_than_zero' expected, found `-1'
Exception: (6) binary_number(_G1642996, -1) ? creep
Example queries
?- binary_number([1,0,1|Tail], N).
Tail = [],
N = 5 ;
Tail = [0],
N = 10 ;
Tail = [1],
N = 11 ;
Tail = [0, 0],
N = 20 .
?- binary_number(Bits, 20).
Bits = [1, 0, 1, 0, 0] ;
Bits = [0, 1, 0, 1, 0, 0] ;
Bits = [0, 0, 1, 0, 1, 0, 0] ;
Bits = [0, 0, 0, 1, 0, 1, 0, 0] ;
Bits = [0, 0, 0, 0, 1, 0, 1, 0, 0] .
?- binary_number(Bits, N).
Bits = [0],
N = 0 ;
Bits = [1],
N = 1 ;
Bits = [1, 0],
N = 2 ;
Bits = [1, 1],
N = 3 ;
Bits = [1, 0, 0],
N = 4 ;
Bits = [1, 0, 1],
N = 5 .
Solution 4:
playing with bits...
binary_number(Bs, N) :-
var(N) -> foldl(shift, Bs, 0, N) ; bitgen(N, Rs), reverse(Rs, Bs).
shift(B, C, R) :-
R is (C << 1) + B.
bitgen(N, [B|Bs]) :-
B is N /\ 1 , ( N > 1 -> M is N >> 1, bitgen(M, Bs) ; Bs = [] ).