Question

以可逆方式将二进制位（例如，可能是0/1的列表）转换为数字的最佳方法是什么。我在swi中编写了一个原生谓词，但有更好的解决方案吗？最好的问候

Answer 1

使用CLP（FD）约束，例如：

:- 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.

示例查询：

?- 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.

Answer 2

这是我想到的解决方案，或者更确切地说是我希望存在的解决方案。

:- 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).

此解决方案也会针对以下查询终止：

?- Bs = [1|_], N #=< 5, binary_number(Bs, N).

Answer 3

解决方案

此答案旨在提供一个谓词binary_number/2，它同时显示logical-purity和最佳终止属性。我使用when/2来阻止canonical_binary_number(B, 10)之类的查询在找到第一个（唯一）解决方案后进入无限循环。当然，有一个权衡，该计划现在有冗余目标。

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).

纯度和终止

我声称这个谓词表示logical-purity来自构造。我希望从这些答案中得到正确答案：one，two和three。

任何具有正确参数的目标都会终止。如果需要检查参数，最简单的方法是使用内置的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

示例查询

?- 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 .

Answer 4

玩位......

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 = [] ).

可逆的“二进制到数字”谓词

4 个答案: