Either this or/and that fact in prolog
I want do build a expense calculator in prolog. The facts onetime_expense(X). or monthly_expense(X). can be present. To sum up the expenses I use this formula:
expenses(X):-
findall(
Value,
( monthly_expense(Y),
Value is Y * 12
),
Values
),
findall(
Value1,
( onetime_expense(Z),
Value1 is Z
),
Values1
),
sum_list(Values, Sum),
sum_list(Values1, Sum1),
X is Sum + Sum1.
Unfortunately prolog throws the error "Undefined procedure" if one of the facts is not in the knowledge base. How can this problem be solved?
prolog
add a comment |
I want do build a expense calculator in prolog. The facts onetime_expense(X). or monthly_expense(X). can be present. To sum up the expenses I use this formula:
expenses(X):-
findall(
Value,
( monthly_expense(Y),
Value is Y * 12
),
Values
),
findall(
Value1,
( onetime_expense(Z),
Value1 is Z
),
Values1
),
sum_list(Values, Sum),
sum_list(Values1, Sum1),
X is Sum + Sum1.
Unfortunately prolog throws the error "Undefined procedure" if one of the facts is not in the knowledge base. How can this problem be solved?
prolog
You can simply a little:findall( Value1, ( onetime_expense(Z), Value1 is Z ), Values1 ),-->findall( Value1, onetime_expense(Value1), Values1 ),
– lurker
Nov 14 '18 at 19:29
add a comment |
I want do build a expense calculator in prolog. The facts onetime_expense(X). or monthly_expense(X). can be present. To sum up the expenses I use this formula:
expenses(X):-
findall(
Value,
( monthly_expense(Y),
Value is Y * 12
),
Values
),
findall(
Value1,
( onetime_expense(Z),
Value1 is Z
),
Values1
),
sum_list(Values, Sum),
sum_list(Values1, Sum1),
X is Sum + Sum1.
Unfortunately prolog throws the error "Undefined procedure" if one of the facts is not in the knowledge base. How can this problem be solved?
prolog
I want do build a expense calculator in prolog. The facts onetime_expense(X). or monthly_expense(X). can be present. To sum up the expenses I use this formula:
expenses(X):-
findall(
Value,
( monthly_expense(Y),
Value is Y * 12
),
Values
),
findall(
Value1,
( onetime_expense(Z),
Value1 is Z
),
Values1
),
sum_list(Values, Sum),
sum_list(Values1, Sum1),
X is Sum + Sum1.
Unfortunately prolog throws the error "Undefined procedure" if one of the facts is not in the knowledge base. How can this problem be solved?
prolog
prolog
asked Nov 14 '18 at 12:11
user10462107
You can simply a little:findall( Value1, ( onetime_expense(Z), Value1 is Z ), Values1 ),-->findall( Value1, onetime_expense(Value1), Values1 ),
– lurker
Nov 14 '18 at 19:29
add a comment |
You can simply a little:findall( Value1, ( onetime_expense(Z), Value1 is Z ), Values1 ),-->findall( Value1, onetime_expense(Value1), Values1 ),
– lurker
Nov 14 '18 at 19:29
You can simply a little:
findall( Value1, ( onetime_expense(Z), Value1 is Z ), Values1 ), --> findall( Value1, onetime_expense(Value1), Values1 ),– lurker
Nov 14 '18 at 19:29
You can simply a little:
findall( Value1, ( onetime_expense(Z), Value1 is Z ), Values1 ), --> findall( Value1, onetime_expense(Value1), Values1 ),– lurker
Nov 14 '18 at 19:29
add a comment |
1 Answer
1
active
oldest
votes
A simple solution is to declare the predicates dynamic:
:- dynamic(monthly_expense/1).
:- dynamic(onetime_expense/1).
A call to a predicate that is declared as dynamic but not defined simply fails instead of throwing a predicate existence error.
P.S. You can simplify your code using the de facto standard predicate findall/4:
expenses(Sum):-
findall(
MonthlyValue,
( monthly_expense(Y),
MonthlyValue is Y * 12
),
MonthlyValues
),
findall(
OneTimeValue,
( onetime_expense(Z),
OneTimeValue is Z
),
Values,
MonthlyValues
),
sum_list(Values, Sum).
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
I feel like it would be weird to have this problem and not run into the missingdynamicdeclaration eventually anyway.
– Daniel Lyons
Nov 14 '18 at 17:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
A simple solution is to declare the predicates dynamic:
:- dynamic(monthly_expense/1).
:- dynamic(onetime_expense/1).
A call to a predicate that is declared as dynamic but not defined simply fails instead of throwing a predicate existence error.
P.S. You can simplify your code using the de facto standard predicate findall/4:
expenses(Sum):-
findall(
MonthlyValue,
( monthly_expense(Y),
MonthlyValue is Y * 12
),
MonthlyValues
),
findall(
OneTimeValue,
( onetime_expense(Z),
OneTimeValue is Z
),
Values,
MonthlyValues
),
sum_list(Values, Sum).
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
I feel like it would be weird to have this problem and not run into the missingdynamicdeclaration eventually anyway.
– Daniel Lyons
Nov 14 '18 at 17:17
add a comment |
A simple solution is to declare the predicates dynamic:
:- dynamic(monthly_expense/1).
:- dynamic(onetime_expense/1).
A call to a predicate that is declared as dynamic but not defined simply fails instead of throwing a predicate existence error.
P.S. You can simplify your code using the de facto standard predicate findall/4:
expenses(Sum):-
findall(
MonthlyValue,
( monthly_expense(Y),
MonthlyValue is Y * 12
),
MonthlyValues
),
findall(
OneTimeValue,
( onetime_expense(Z),
OneTimeValue is Z
),
Values,
MonthlyValues
),
sum_list(Values, Sum).
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
I feel like it would be weird to have this problem and not run into the missingdynamicdeclaration eventually anyway.
– Daniel Lyons
Nov 14 '18 at 17:17
add a comment |
A simple solution is to declare the predicates dynamic:
:- dynamic(monthly_expense/1).
:- dynamic(onetime_expense/1).
A call to a predicate that is declared as dynamic but not defined simply fails instead of throwing a predicate existence error.
P.S. You can simplify your code using the de facto standard predicate findall/4:
expenses(Sum):-
findall(
MonthlyValue,
( monthly_expense(Y),
MonthlyValue is Y * 12
),
MonthlyValues
),
findall(
OneTimeValue,
( onetime_expense(Z),
OneTimeValue is Z
),
Values,
MonthlyValues
),
sum_list(Values, Sum).
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
A simple solution is to declare the predicates dynamic:
:- dynamic(monthly_expense/1).
:- dynamic(onetime_expense/1).
A call to a predicate that is declared as dynamic but not defined simply fails instead of throwing a predicate existence error.
P.S. You can simplify your code using the de facto standard predicate findall/4:
expenses(Sum):-
findall(
MonthlyValue,
( monthly_expense(Y),
MonthlyValue is Y * 12
),
MonthlyValues
),
findall(
OneTimeValue,
( onetime_expense(Z),
OneTimeValue is Z
),
Values,
MonthlyValues
),
sum_list(Values, Sum).
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
answered Nov 14 '18 at 12:36
Paulo MouraPaulo Moura
12.2k21426
12.2k21426
I feel like it would be weird to have this problem and not run into the missingdynamicdeclaration eventually anyway.
– Daniel Lyons
Nov 14 '18 at 17:17
add a comment |
I feel like it would be weird to have this problem and not run into the missingdynamicdeclaration eventually anyway.
– Daniel Lyons
Nov 14 '18 at 17:17
I feel like it would be weird to have this problem and not run into the missing
dynamic declaration eventually anyway.– Daniel Lyons
Nov 14 '18 at 17:17
I feel like it would be weird to have this problem and not run into the missing
dynamic declaration eventually anyway.– Daniel Lyons
Nov 14 '18 at 17:17
add a comment |
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You can simply a little:
findall( Value1, ( onetime_expense(Z), Value1 is Z ), Values1 ),-->findall( Value1, onetime_expense(Value1), Values1 ),– lurker
Nov 14 '18 at 19:29