在Prolog中将数字统一到列表
好的，我正在做一个家庭作业，这应该是一个四功能计算器，比如输入一个字符串列表[3，乘以2]，然后输出一个数字。它只考虑初始列表中从1到20的数字。下面的代码都是我自己的。它一直运行到我用来测试它的列表中最后一项的位置，但问题是numberize中的任何输入都不会统一在Prolog中将数字统一到列表,prolog,Prolog,好的，我正在做一个家庭作业，这应该是一个四功能计算器，比如输入一个字符串列表[3，乘以2]，然后输出一个数字。它只考虑初始列表中从1到20的数字。下面的代码都是我自己的。它一直运行到我用来测试它的列表中最后一项的位置，但问题是numberize中的任何输入都不会统一 calculator([twenty, times, three, plus, five, divided_by, two],Total). 我知道解决方案一定很简单，但我在Prolog中还没有足够的经验来解决这个问题 我的问题是
calculator([twenty, times, three, plus, five, divided_by, two],Total).
calculator(X,Total):-
numberize(X,L),
reverse(L,L1),
func(L1,Total).
numberize([X,Y|T],L):-
str2num(X,X1),
numberize(T,[Y,X1|L]).
numberize([X],L):-
str2num(X,X1),
%somehow add on the X1 to the front of L without any errors and it's golden
/*Whatever that line is*/L is [X1|L].
func([X1,X,Z1|T], Total):-
(X == times, times(X1,Z1,Ttl));
(X == plus, plus(X1,Z1,Ttl));
(X == divided_by, divided_by(X1,Z1,Ttl));
(X == minus, minus(X1,Z1,Ttl)),
func([Ttl|T],Total).
str2num(one, X):- X is 1.
str2num(two, X):- X is 2.
str2num(three, X):- X is 3.
str2num(four, X):- X is 4.
str2num(five, X):- X is 5.
str2num(six, X):- X is 6.
str2num(seven, X):- X is 7.
str2num(eight, X):- X is 8.
str2num(nine, X):- X is 9.
str2num(ten, X):- X is 10.
str2num(eleven, X):- X is 11.
str2num(twelve, X):- X is 12.
str2num(thirteen, X):- X is 13.
str2num(fourteen, X):- X is 14.
str2num(fifteen, X):- X is 15.
str2num(sixteen, X):- X is 16.
str2num(seventeen, X):- X is 17.
str2num(eighteen, X):- X is 18.
str2num(nineteen, X):- X is 19.
str2num(twenty, X):- X is 20.
times(X,Y,Prod):-
Prod is X*Y.
plus(X,Y,Sum):-
Sum is X+Y.
divided_by(X,Y,Quo):-
Quo is X/Y.
minus(X,Y,Dif):-
Dif is X-Y.
小风格备注：对str2num使用事实/2：仅使用str2numone，1。代替str2numone，X:-X是1，等等。额外的好处是现在谓词可以双向使用，就像str2numWord，1一样 至于主要问题，你几乎是对的 整个numberize谓词可以如此简单：
numberize([X], [N]) :-
str2num(X, N).
numberize([X, Op | T], [N, Op | NewT]) :-
str2num(X, N),
numberize(T, NewT).
?- numberize([one, plus, two, minus, three], L).
L = [1, plus, 2, minus, 3]
func([X1,X,Z1|T], Total):-
(
X == times, times(X1,Z1,Ttl)
;
X == plus, plus(X1,Z1,Ttl)
;
X == divided_by, divided_by(X1,Z1,Ttl)
;
X == minus, minus(X1,Z1,Ttl)
),
func([Ttl|T],Total).
func([X], X).
?- calculator([one, plus, two], Total).
Total = 3
?- calculator([one, plus, two, minus, four], Total).
Total = -1
小风格备注：对str2num使用事实/2：仅使用str2numone，1。代替str2numone，X:-X是1，等等。额外的好处是现在谓词可以双向使用，就像str2numWord，1一样 至于主要问题，你几乎是对的 整个numberize谓词可以如此简单：
numberize([X], [N]) :-
str2num(X, N).
numberize([X, Op | T], [N, Op | NewT]) :-
str2num(X, N),
numberize(T, NewT).
?- numberize([one, plus, two, minus, three], L).
L = [1, plus, 2, minus, 3]
func([X1,X,Z1|T], Total):-
(
X == times, times(X1,Z1,Ttl)
;
X == plus, plus(X1,Z1,Ttl)
;
X == divided_by, divided_by(X1,Z1,Ttl)
;
X == minus, minus(X1,Z1,Ttl)
),
func([Ttl|T],Total).
func([X], X).
?- calculator([one, plus, two], Total).
Total = 3
?- calculator([one, plus, two, minus, four], Total).
Total = -1
我的方法是从定义算术表达式的语法开始。定义语法的标准方法是左递归。由于prolog执行递归下降解析，因此语法不能保持递归。每次迭代都必须从令牌流中删除某些内容，以免陷入无限递归的死亡螺旋。这是我的非左递归语法，适用于像您这样的4-banger计算器：
expression : multiplicative_expression '+' expression
| multiplicative_expression '-' expression
| multiplicative_expression
;
multiplicative_expression : factor '*' multiplicative_expression
| factor '/' multiplicative_expression
| factor '%' multiplicative_expression
| factor
;
factor : '-' value
| '(' expression ')'
| value
;
value : number
operator( plus , additive , '+' ) .
operator( minus , additive , '-' ) .
operator( times , multiplicative , '*' ) .
operator( divided_by , multiplicative , '/' ) .
operator( modulo , multiplicative , 'mod' ) .
number_word( zero , 0 ).
number_word( one , 1 ).
...
number_word( nineteen , 19 ) .
number_word( twenty , 20 ) .
expr( Xs , Result , Tail ) :- % per the grammar, an expression is
mult( Xs , LHS , [Sym|X1] ) , % - a multiplicative expression, followed by
operator( Sym , additive , Op ) , % - an infix additive operator, followed by
expr( X1 , RHS , X2 ) , % - another expression
Term =.. [Op,LHS,RHS] , % * in which case, we construct the proper prolog structure
Result is Term , % * in which case, we evaluate the result in the usual way
Tail = X2 % * and unify any remaining tokens with the Tail
. %
expr( Xs , Result , Tail ) :- % alternatively, an expression is simply
mult( Xs , Result , Tail ) % - a single multiplicative expression
. %
mult( Xs , Result, Tail ) :- % a multiplicative expression is
factor( Xs , LHS , [Sym|X1] ) , % - a factor, followed by
operator( Sym , multiplicative , Op ) , % - an infix multiplicative operator, followed by
mult( X1 , RHS , X2 ) , % - another factor
evaluate( Op , LHS , RHS , Result ) , % * in which case, we evalute the result in the usual way
Tail = X2 % * and unify any remaining tokens with the tail
. %
mult( Xs , Result , Tail ) :- % alternatively, a multiplicative expression is simply
factor( Xs , Result , Tail ) % - a single factor
. %
factor( [X|Xs] , Value , Xs ) :- % a factor is simply
number_word(X,Value) % - a number value (in our case, a word that we convert to an integer)
.
evaluate( Op , LHS , RHS , Result ) :- % to evaluate an infix term,
Term =.. [Op,LHS,RHS] , % - use univ to convert to the correct prolog structure, and
Result is Term % evaluate it as the result
. %
我的方法是从定义算术表达式的语法开始。定义语法的标准方法是左递归。由于prolog执行递归下降解析，因此语法不能保持递归。每次迭代都必须从令牌流中删除某些内容，以免陷入无限递归的死亡螺旋。这是我的非左递归语法，适用于像您这样的4-banger计算器：
expression : multiplicative_expression '+' expression
| multiplicative_expression '-' expression
| multiplicative_expression
;
multiplicative_expression : factor '*' multiplicative_expression
| factor '/' multiplicative_expression
| factor '%' multiplicative_expression
| factor
;
factor : '-' value
| '(' expression ')'
| value
;
value : number
operator( plus , additive , '+' ) .
operator( minus , additive , '-' ) .
operator( times , multiplicative , '*' ) .
operator( divided_by , multiplicative , '/' ) .
operator( modulo , multiplicative , 'mod' ) .
number_word( zero , 0 ).
number_word( one , 1 ).
...
number_word( nineteen , 19 ) .
number_word( twenty , 20 ) .
expr( Xs , Result , Tail ) :- % per the grammar, an expression is
mult( Xs , LHS , [Sym|X1] ) , % - a multiplicative expression, followed by
operator( Sym , additive , Op ) , % - an infix additive operator, followed by
expr( X1 , RHS , X2 ) , % - another expression
Term =.. [Op,LHS,RHS] , % * in which case, we construct the proper prolog structure
Result is Term , % * in which case, we evaluate the result in the usual way
Tail = X2 % * and unify any remaining tokens with the Tail
. %
expr( Xs , Result , Tail ) :- % alternatively, an expression is simply
mult( Xs , Result , Tail ) % - a single multiplicative expression
. %
mult( Xs , Result, Tail ) :- % a multiplicative expression is
factor( Xs , LHS , [Sym|X1] ) , % - a factor, followed by
operator( Sym , multiplicative , Op ) , % - an infix multiplicative operator, followed by
mult( X1 , RHS , X2 ) , % - another factor
evaluate( Op , LHS , RHS , Result ) , % * in which case, we evalute the result in the usual way
Tail = X2 % * and unify any remaining tokens with the tail
. %
mult( Xs , Result , Tail ) :- % alternatively, a multiplicative expression is simply
factor( Xs , Result , Tail ) % - a single factor
. %
factor( [X|Xs] , Value , Xs ) :- % a factor is simply
number_word(X,Value) % - a number value (in our case, a word that we convert to an integer)
.
evaluate( Op , LHS , RHS , Result ) :- % to evaluate an infix term,
Term =.. [Op,LHS,RHS] , % - use univ to convert to the correct prolog structure, and
Result is Term % evaluate it as the result
. %
我们需要关心操作符优先级吗？换句话说，3加4乘以5等于23还是35？@他们不需要在意。我从OP的同学在这个问题上的例子中知道了这一点：是的，我们不需要它。这是我们给出的另一个例子，所以我想@SergeyDymchenko说另一张海报是同学是对的。但不，我没有，事情本来应该像我做的那样顺利。比str2numone更好、更简单，X:-X为1。。对于其他数字也是如此。我们需要考虑运算符优先级吗？换句话说，3加4乘以5等于23还是35？@他们不需要在意。我从OP的同学在这个问题上的例子中知道了这一点：是的，我们不需要它。这是我们给出的另一个例子，所以我想@SergeyDymchenko说另一张海报是同学是对的。但不，我没有，事情本来应该像我做的那样顺利。比str2numone更好、更简单，X:-X为1。。其他数字也是如此。@Throsby还注意到，只要答案在你翻译的数字范围内，你就可以把它翻译回去，这样你就可以得到？-计算器[一，加，四，减，三]，总数。Total=2。@Throsby还注意到，只要答案在你翻译的数字范围内，你就可以把它翻译回去，这样你就可以得到？-计算器[1，加，4，减，3]，Total。Total=2。答案很好，但我在寻找解决方案，而不是更好的方案。我应该早点接受第一个答案。对不起，谢谢你！答案很好，但我在寻找解决方案，而不是更好的方案。我应该早点接受第一个答案。对不起，谢谢你！