I have list of facts as follows.
items(itemId('P01'),prodName('Pots'),stockQty(50),price(8200)).
items(itemId('P02'),prodName('Pans'),stockQty(50),price(400)).
items(itemId('P03'),prodName('Spoons'),stockQty(50),price(200)).
items(itemId('P04'),prodName('Forks'),stockQty(50),price(120)).
items(itemId('P05'),prodName('Kettles'),stockQty(50),price(500)).
items(itemId('P06'),prodName('Plates'),stockQty(50),price(60)).
How to print on the console something like the following when a command like print_all_products. is given.
..............
Available Products
..........
Name Qty
Pots 60
Pans 50
Spoons 40
..................
The Name and Qty must be properly formatted in a tabular structure.
I tried using forall and foreach I am unsuccessful in generating what i need.
Answer with more details is posted here.
Below is the code so that this is not a link only answer.
items(itemId('P01'),prodName('Pots'),stockOty(50),price(8200)).
items(itemId('P02'),prodName('Pans'),stockOty(50),price(400)).
items(itemId('P03'),prodName('Spoons'),stockOty(50),price(200)).
items(itemId('P04'),prodName('Forks'),stockOty(50),price(120)).
items(itemId('P05'),prodName('Kettles'),stockOty(50),price(500)).
items(itemId('P06'),prodName('Plates'),stockOty(50),price(60)).
header("\n........................\nAvailable Products\n........................\nName Qty\n").
footer("........................\n").
spaces(Length,Spaces) :-
length(List,Length),
maplist([_,0'\s]>>true,List,Codes),
string_codes(Spaces,Codes).
padded_string(String,Width,Padded_string) :-
string_length(String,String_length),
Padding_length is Width - String_length,
spaces(Padding_length,Padding),
atom_concat(String,Padding,Padded_string).
format_detail_line(item(Name,Quantity),width(Name_width),Formatted_item) :-
padded_string(Name,Name_width,Padded_name),
atom_concat(Padded_name,Quantity,Formatted_item).
add_detail_line(width(Name_Width),Item,Lines0,Lines) :-
format_detail_line(Item,width(Name_Width),Formatted_item),
atomic_list_concat([Lines0,Formatted_item,"\n"], Lines).
items_detail(Detail) :-
findall(item(Name,Quantity),items(_,prodName(Name),stockOty(Quantity),_),Items),
aggregate_all(max(Width),Width,(items(_,prodName(Name),_,_),string_length(Name,Width)),Name_Width),
Name_field_width is Name_Width + 1,
foldl(add_detail_line(width(Name_field_width)),Items,"",Detail).
print_all_products(Report) :-
header(Header),
items_detail(Detail),
footer(Footer),
atomic_list_concat([Header,Detail,Footer], Report).
print_all_products :-
print_all_products(Report),
write(Report).
:- begin_tests(formatted_report).
test(1) :-
print_all_products(Report),
with_output_to(atom(Atom),write(Report)),
assertion( Atom == '\n........................\nAvailable Products\n........................\nName Qty\nPots 50\nPans 50\nSpoons 50\nForks 50\nKettles 50\nPlates 50\n........................\n' ).
:- end_tests(formatted_report).
Note: The answer given by Peter is the customary way to do the formatting, but as I noted, that drives me nuts. Even so, that is the way I would do it in a production environment.
I gave this answer because the OP noted they were looking for a way to do it using predicates like forall/2 or foreach/2. Granted neither of them is used in this answer but the intent of using a more functional approach is used.
If the question was more open ended I would have given a answer using DCGs.
format/2 ... for putting things in neat columns, use ~|, ~t, ~+.
~| sets a tab to "here", ~t inserts fill characters, ~+ advances the tab beyond the last "here" (~|) and distributes the fill characters. So,
format("(~|~`.t~d~5+)~n", [123])
produces (..123) -- the format string right-justifies the number with .s in a width of 5, surrounded by parentheses.
You are asking for SQL-style tabular output and yes, that should be in the language as basic predicate set since when Reagan was prez. I don't know what's going on. It's probably out there in a library though (but where is the library?)
Meanwhile, here is the "failure driven-loop" using some of my personal toolbox goodies, but it uses SWI Prolog:
In file printthem.pl:
:- use_module(library('heavycarbon/strings/string_of_spaces.pl')).
:- use_module(library('heavycarbon/strings/string_overwriting.pl')).
items(itemId('P01'),prodName('Pots'),stockOty(50),price(8200)).
items(itemId('P02'),prodName('Pans'),stockOty(50),price(400)).
items(itemId('P03'),prodName('Spoons'),stockOty(50),price(200)).
items(itemId('P04'),prodName('Forks'),stockOty(50),price(120)).
items(itemId('P05'),prodName('Kettles'),stockOty(50),price(500)).
items(itemId('P06'),prodName('Plates'),stockOty(50),price(60)).
printthem :-
% ideally these should be built by getting max(length) over a column - hardcode for now!
string_of_spaces(5,SpacesId),
string_of_spaces(10,SpacesName),
string_of_spaces(4,SpacesQuant),
string_of_spaces(6,SpacesPrice),
% begin failure-driven loop!
items(itemId(Id),prodName(Name),stockOty(Quant),price(Price)), % backtrack over this until no more solutions
% transform data into string; see predicate format/2;
% capture output instead of letting it escape to STDOUT
with_output_to(string(TxtId),format("~q",[Id])),
with_output_to(string(TxtName),format("~q",[Name])),
with_output_to(string(TxtQuant),format("~d",[Quant])),
with_output_to(string(TxtPrice),format("~d",[Price])),
% formatting consist in overwriting the space string with the data-carrying string
string_overwriting(SpacesId,TxtId, 1,TxtIdFinal),
string_overwriting(SpacesName,TxtName, 1,TxtNameFinal),
string_overwriting(SpacesQuant,TxtQuant, 1,TxtQuantFinal),
string_overwriting(SpacesPrice,TxtPrice, 1,TxtPriceFinal),
% output the line
format("~s~s~s~s\n",[TxtIdFinal,TxtNameFinal,TxtQuantFinal,TxtPriceFinal]),
% close the loop
fail.
The above is just an ébauche. Improvements are possible in several distinct directions.
The modules loaded via
:- use_module(library('heavycarbon/strings/string_of_spaces.pl')).
:- use_module(library('heavycarbon/strings/string_overwriting.pl')).
can be obtained from GitHub here. You will have to grab several files and arrange them appropriately. Read the script load_and_test_script.pl. Don't mind the mess, this is work in progress.
If everything has been set up correctly:
?- [printthem].
true.
?- printthem.
'P01' 'Pots' 50 8200
'P02' 'Pans' 50 400
'P03' 'Spoons' 50 200
'P04' 'Forks' 50 120
'P05' 'Kettles' 50 500
'P06' 'Plates' 50 60
false.
Related
update :-
write("Name?:"),
read(Name),
assert(Name),nl,
write("Age?:"),
read(Age),
assert(Age),
write("Continue(y or n)?:"),
read(Respond),
process(Respond).
process(y) :-
write('Name?:'),
read(Name),
assert(Name),nl,
write("Age?:"),
read(Age),
assert(Age),
repeat,
write("y or n"),
read(Respond),
process(Respond).
process(n) :- !.
I want to run this Prolog to assert in the name and age, but when I write age for the number, it shows
?- update.
Name?:fred.
Age?:|: 25.
ERROR: Type error: `callable' expected, found `25' (an integer)
ERROR: In:
ERROR: [9] assert(25)
ERROR: [8] update at c:/example.pl:11
ERROR: [7] <user>
?-
How to figure out this problem.
Problem 1
Incorrect input for assert/1
The problem is not with just Age it is with any input that uses assert, e.g.
?- update.
Name?:Fred
|: .
ERROR: Arguments are not sufficiently instantiated
ERROR: In:
ERROR: [9] assert(_4940)
ERROR: [8] update at c:/example.pl:8
ERROR: [7] <user>
?- update.
Name?:Jim.
ERROR: Arguments are not sufficiently instantiated
ERROR: In:
ERROR: [9] assert(_5826)
ERROR: [8] update at c:/example.pl:8
ERROR: [7] <user>
The problem is that assert/1 is not being given a fact or rule.
assert/1 says:
Assert a clause (fact or rule) into the database.
See facts and rules
In the example above Fred is not a fact because it does not end with a period (.).
In the example above with Jim. a period was given but because Jim starts with a capital letter, it is not a fact or rule but a variable.
When the age is entered as a number, again this is not a fact or rule it is an integer.
Problem 2
Use of read/1 which says:
Read the next Prolog term from the current input stream and unify it with Term.
When reading a Prolog term the input must end with a period.
This not only requires the input to be a term, but end with a . which is even more confusing given the prompt, e.g Age. Most of the examples you find do what you did, the corrected code below does what you want.
Problem 3
Competing ways or repeating.
The code is using two ways:
Use of repeat/0
It is recursive, e.g.
process(y) :-
...
process(Respond).
This is making it hard to get the code working.
Problem 4
Duplicate code, e.g.
write("Name?:"),
read(Name),
assert(Name),nl,
write("Age?:"),
read(Age),
assert(Age),
write("Continue(y or n)?:"),
read(Respond),
process(Respond).
Duplicated code is more likely to lead to problems when one copy is corrected and the other copy is not corrected.
Problem 1 fix
Make the input a fact before storing in the database with assert/1, e.g.
Values in variables
Name
Age
Variables converted to facts by adding a functor
name(Name)
age(Age)
The facts used with assert/1
assert(name(Name))
assert(age(Age))
Problem 2 fix
Use read_string/5, e.g.
read_string(user, "\n", "\r", End, Name)
This reads the input into the variable Name as a string. Now that the input is a string, and not a Prolog term, the period is no longer required. There are predicates that operate on strings.
Problem 3 fix
Use the recursion form and remove repeat/0.
This could also use repeat/0 instead of recursion. The corrected code below uses recursion to demonstrate the change to process/1.
Problem 4 fix
Just refactor the code. You can see this in the corrected code at the end.
Now with the fixes in place.
Change 1
Since the input for continue is no longer a term, e.g. y or n, but a string, the parameter for process needs to be a string, e.g.
process("y") :-
process("n") :-
Change 2
Age will be asserted as a string but would be better asserted as an integer.
number_string/2 can solve this, e.g.
number_string(Age_n,Age),
assert(age(Age_n))
Change 3
user27815 Asked in a comment:
do you need the cut in process("n") :- !. ?
Since
process(Respond).
is not creating a choice point, the cut is not needed.
Corrected code:
update :-
% Respond will be read as a string and not as a term, so it needs "".
process("y").
process("y") :-
write('Name: '),
read_string(user, "\n", "\r", End, Name),
assert(name(Name)),
write("Age: "),
read_string(user, "\n", "\r", End, Age),
number_string(Age_n,Age),
assert(age(Age_n)),
write("Continue: (y or n) "),
read_string(user, "\n", "\r", End, Respond),
process(Respond).
process("n").
Example run:
?- update.
Name: Fred
Age: 30
Continue: (y or n) y
Name: Jim
Age: 21
Continue: (y or n) n
true.
To check that the database was updated use listing/1
?- listing(name/1).
:- dynamic name/1.
name("Fred").
name("Jim").
true.
?- listing(age/1).
:- dynamic age/1.
age(30).
age(21).
true.
A free enhancement.
Keeping the facts of name and age separate doesn't keep the relation between them intact. A better solution would be a person fact with both Name and Age values.
Here is the necessary modified code.
update :-
% Respond will be read as a string and not as a term, so it needs "".
process("y").
process("y") :-
write('Name: '),
read_string(user, "\n", "\r", End, Name),
write("Age: "),
read_string(user, "\n", "\r", End, Age),
number_string(Age_n,Age),
assert(person(Name,Age_n)),
write("Continue: (y or n) "),
read_string(user, "\n", "\r", End, Respond),
process(Respond).
process("n").
Example run:
?- update.
Name: Fred
Age: 30
Continue: (y or n) y
Name: Jim
Age: 21
Continue: (y or n) n
true.
To check that the database was updated use listing/1
?- listing(person/2).
:- dynamic person/2.
person("Fred", 30).
person("Jim", 21).
true.
After noticing your deleted answer.
In your deleted answer you have
?- person(name(N), age(A)).
N = nancy,
A = 22;
N= steve,
A = 100;
true.
The change needed for this variation of the fact to be created is
assert(person(name(Name),age(Age_n)))
however that might not be the optimal way to go.
In Prolog, positions typically indicate the meaning of a value, e.g. first position is name and second position is age. In this variation by adding the functors name and age to the fact person/2 you are duplicating known knowledge, but more importantly the possibility the amount of work Prolog has to do.
For example:
If the fact was person(Name,Age). to get at Name and Age Prolog only needs one unification. But with person(Name,Age). Prolog now needs to unify with person(name(nancy),age(22)) then to get Name has to unify again with name(nancy) and to get Age has to unify with age(22). You could also use person(name(Name),age(Age)). which requires only one unification, but now makes your code more verbose.
When first learning Prolog this crutch helps, but when working with larger data sets, this starts to impact performance.
Another item of note in your deleted answer is that the names of the people are still based on using read/1, e.g. nancy and steve. While a lot of Prolog examples do this, there is no requirement to keep them as such, they can be strings. Odds are the code will never need to exactly match on nancy or steve and instead will always reference them as a value in a variable. The nice thing about keeping them as strings is that when writing them out, they will appear correctly as Nancy and Steve.
This is because assert does not work on variables. It asserts a fact or rule; in other words, assert(something) asserts that something must be true.
From the SWI-Prolog documentation:
Assert a clause (fact or rule) into the database.
An integer value is not a rule or a fact. It is (in this case) an integer, not something that evaluates to a boolean value. There's no point in asserting a value.
I would write some helpers:
read_assert(P,V) :- format('~w ? ',[P]), read(V), A =.. [P,V], assert(A).
?- maplist(read_assert, [name,age], Vs).
name ? capellic.
age ? 99.
Vs = [capellic, 99].
?- name(N).
N = capellic.
Given following facts:
route(TubeLine, ListOfStations).
route(green, [a,b,c,d,e,f]).
route(blue, [g,b,c,h,i,j]).
...
I am required to find all the pairs of tube Lines that do not have any stations in common, producing the following:
| ?- disjointed_lines(Ls).
Ls = [(yellow,blue),(yellow,green),(yellow,red),(yellow,silver)] ? ;
no
I came up with the below answer, however it does not only give me incorrect answer, but it also does not apply my X^ condition - i.e. it still prints results per member of Stations lists separately:
disjointed_lines(Ls) :-
route(W, Stations1),
route(Z, Stations2),
setof(
(W,Z),X^
(member(X, Stations1),nonmember(X, Stations2)),
Ls).
This is the output that the definition produces:
| ?- disjointed_lines(L).
L = [(green,green)] ? ;
L = [(green,blue)] ? ;
L = [(green,silver)] ? ;
...
I believe that my logic relating to membership is incorrect, however I cannot figure out what is wrong. Can anyone see where am I failing?
I also read Learn Prolog Now chapter 11 on results gathering as suggested here, however it seems that I am still unable to use the ^ operator correctly. Any help would be appreciated!
UPDATE:
As suggested by user CapelliC, I changed the code into the following:
disjointed_lines(Ls) :-
setof(
(W,Z),(Stations1, Stations2)^
((route(W, Stations1),
route(Z, Stations2),notMembers(Stations1,Stations2))),
Ls).
notMembers([],_).
notMembers([H|T],L):- notMembers(T,L), nonmember(H,L).
The following, however, gives me duplicates of (X,Y) and (Y,X), but the next step will be to remove those in a separate rule. Thank you for the help!
I think you should put route/2 calls inside setof' goal, and express disjointness more clearly, so you can test it separately. About the ^ operator, it requests a variable to be universally quantified in goal scope. Maybe a concise explanation like that found at bagof/3 manual page will help...
disjointed_lines(Ls) :-
setof((W,Z), Stations1^Stations2^(
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2)
), Ls).
disjoint(Stations1, Stations2) :-
... % could be easy as intersection(Stations1, Stations2, [])
% or something more efficient: early fail at first shared 'station'
setof/3 is easier to use if you create an auxiliary predicate that expresses the relationship you are interested in:
disjoint_routes(W, Z) :-
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2).
With this, the definition of disjointed_lines/1 becomes shorter and simpler and no longer needs any ^ operators:
disjointed_lines(Ls) :-
setof((W, Z), disjoint_routes(W, Z), Ls).
The variables you don't want in the result of setof/3 are automatically hidden inside the auxiliary predicate definition.
Pair data is nice to work with, but I found hard to present. I ask how to accomplish printing a set of vectors such that keys & values line up in a pleasantly. Leading zeros in the pair-values, would help. Sample data:
[1-7,2-43,3-56,4-87,5-110,6-80,7-15]
[1-1837,2-1873,3-1911,4-1946,5-1975,6-1994,7-2005]
I tried to figure out use of SWI format_predicate ; but couldn't.
Then I thought to experiment inline;
format('~n~w ~w~w~n', ['Pairs: ',1-246,1-2, ' EOL']).
End result should deal with pairs of the form KK-VVVV:
01-0007 02-0043 03-0056 04-0087 05-0110 06-0080 07-0015 398 People 7 Gens.
01-1837 02-1873 03-1911 04-1946 05-1975 06-1994 07-2005 Spanning 168 Years
Final Answers:
fpair(A-B) :- format('~`0t~d~2|-~`0t~d~7| ', [A,B])
applist(_,[]). applist(P,[X|L]) :- Q =.. [P,X],call(Q),applist(P,L).
dojustone(X):- format('~# ',[fpair(X)]).
dolist(X):- applist(dolist,X).
I use a # specifier for complex formats, it allows to output specific terms. For instance
?- format('~s~n~#~n~#~n~w~n', ['Pairs: ',fpair(1-246),fpair(1-2), ' EOL']).
that is, fpair/1 is an user predicate, called by #, capturing its output.
To get fixed width fields, I use the tab specification, built from two specifiers working together. Finally, to prefix with 0s, I would use
fpair(A-B) :-
format('~`0t~d~6| ~`0t~d~12|', [A,B]).
Without knowing a priori the maximum number of digits, we must use a guess. I used 6 here.
I would need help about Prolog.
I posted my code, the problem is that i do not obtain the expected result.
I want planning actions for moving on table all blocks until is possible. To do this I prompt :
?- do(while(some(x, block(x) & -onTable(x)),pi(x,putOnTable(x))),s0,S).
I expect to see a response like :
S = do(putOnTable(e), do(putOnTable(b), do(putOnTable(c), s0)))
but Prolog returns "false" only. Someone can help me??
% Golog interpreter
%:- [golog_swi].
:- discontiguous clear/2, on/3, onTable/2.
:- op(800,xfy,[&]).
do(E,S,do(E,S)):- primitive_action(E),poss(a,S).
% Primitive Action Declarations.
primitive_action(putOn(_,_)).
primitive_action(putOnTable(_)).
poss(putOn(X,Y),S) :- clear(X,S), clear(Y,S), \+ on(X,Y,S), \+ X=Y.
poss(putOnTable(X),S):- clear(X,S), \+(onTable(X,S)).
% Successor State Axioms.
on(X,Y,do(A,S)):- A = putOn(X,Y); on(X,Y,S), \+ (A = putOnTable(X); A = putOn(X,_)).
onTable(X,do(A,S)) :- A = putOnTable(X); onTable(X,S), \+ A= putOn(X,_).
clear(X,do(A,S)) :- on(Y,X,S), (A = putOn(Y,_) ; A = putOnTable(Y)); clear(X,S), \+ A = putOn(_,X).
% Restore suppressed situation arguments
restoreSitArg(onTable(X),S,onTable(X,S)).
restoreSitArg(on(X,Y),S,on(X,Y,S)).
restoreSitArg(clear(X),S,clear(X,S)).
block(X):- member(X,[a,b,c,d,e]).
% iniTial COndition
onTable(a,s0).
on(b,a,s0).
on(c,b,s0).
clear(c,s0).
onTable(d,s0).
on(e,d,s0).
clear(3,s0).
thank you!!!
Your predicate do/3 cannot succeed because the goal primitive_action/1 will fail with your query.
Currently, while/2 is not described in primitive_action/1 and it seems it is missing also from your program. So you need to extend primitive_action/1 by further facts, or add a new rule to do/3. And in addition to that you need to describe what while/2 means.
This question is actually about Golog. Your mistake is pretty mundane: you didn't copy the Golog interpreter code into your source file/directory.
Golog defines a number of high-level programming constructs, including while-loops and non-deterministic picks (pi), used here. I'm sure you don't want to reinvent Golog, so just go and get it. I'm assuming that your question is part of an assignment of sorts, and your teacher probably pointed you to the Golog interpreter. Otherwise, you can always find it on the pages of the cognitive robotics group at the Univ. of Toronto: http://www.cs.toronto.edu/cogrobo/main/systems/index.html
go :- match(Mn,Fn),
write('--Matching Result--'),
nl,
write(Mn),
write(' match with '),
write(Fn),
match(Mn1,Fn1).
person(may,female,25,blue).
person(rose,female,20,blue).
person(hock,male,30,blue).
person(ali,male,24,blue).
match(Mn,Fn):-person(Fn,'female',Fage,Fatt),
person(Mn,'male',Mage,Matt),
Mage>=Fage,
Fatt=Matt.
Hi,this is my code...but it's only can show the 1 output...but there are 3 pair of matching in match(X,Y).how to show them all in my go function.
Thank you
You get all your matches if you force backtracking, usually by entering ; (e.g. in SWI Prolog). But you also see that you are getting unnecessary outputs true. This is because the last clause in go is match(Mn1,Fn1). This clause succeeds three times and binds the variables Mn1,Fn1 but then only true is output, because you do not write() after that clause. The fourth time match(Mn1,Fn1) fails and by backtracking you come back to the first clause match(Mn,Fn) that matches, the match is output, etc.
You surely do not want to have this behavior. You should remove the last clause match(Mn1,Fn1) in go. Now by pressing ; you get the 3 matches without any output true in between.
But what you likely want is that the program does the backtracking. To achieve this, you just need to force backtracking by adding false as the last clause. To get proper formatting of the output, use the following program. The last clause go2. is added to get true at the very end.
go2 :- write('--Matching Result--'), nl,
match(Mn,Fn),
write(Mn), write(' match with '), write(Fn), nl,
fail.
go2.
This technique is called failure driven loop.
If you have any predicate that has multiple results and want to to find all of them, you should use findall/3
For example, in your case, you could do something like:
findall([X,Y], match(X,Y),L).
L will be a list that will contain all the X,Y that satisfy match(X,Y) in the format [X,Y].
for example, assuming that:
match(m1,f1).
match(m2,f2).
the result will be L = [ [m1,f1], [m2,f2] ]
note that you can define the format as you wish, for example you could write:
findall(pair(X,Y), match(X,Y), L).
L = [ pair(m1,f1), pair(m2,f2) ]
findall( X, match(X,Y), L).
L = [ m1, m2]
findall( 42, match(X,Y), L).
L = [42, 42]
then you have to recurse on the list to print them.
However, if you wish to find one result, run some code and then continue you could use forall/2:
forall(match(X,Y), my_print(X,Y).
Prolog is a lazy language. Which means that it will stop once it has found a condition that made your problem true. This will be the very first match alone.
IF your code is working (I haven't tried it), then you should try and run the match-statement like this in your prolog inspector: match(X,Y)
The prolog inspector will return all states and print them for you.