Having a "out of global stack" in prolog - prolog

Hey guys I have a fairly simple question about Prolog.
%on(Block,Object).
% clear(Object).
block(b1).
block(b2).
block(b3).
place(p1).
place(p2).
place(p3).
place(p4).
state1([clear(p2),clear(p4),clear(b2),clear(b3),on(b1,p1),on(b2,p3),on(b3,b1)]).
% visual state1
% b3
% b1 b2
% = = = =
% 1 2 3 4 <----Positions
% can(Action,Condition).
% adds(Action,AddRelationship).
% deletes(Action,DeleteRelationship).
% move(Block,From,To).
can( move( Block, From, To), [ clear( Block), clear( To), on( Block, From)]) :-
block( Block), % Block to be moved
object( To), % "To" is a block or a place
To \== Block, % Block cannot bå moved to itself
object( From), % "From" is a block or a place
From \== To, % Move to new position
Block \== From. % Block not moved from itself
adds(move(X,From,To),[on(X,To),clear(From)]).
deletes(move(X,From,To),[on(X,From),clear(To)]).
object(X):-
place(X)
;
block(X).
% plan(State,Goals,Plan,FinalState).
plan(State,Goals,[],State):-
satisfied(State,Goals).
plan(State,Goals,Plan,FinalState) :-
append(PrePlan,[Action|PostPlan],Plan),
select(State,Goals,Goal),
achieves(Action,Goal),
can(Action,Condition),
plan(State,Condition,PrePlan,MidState1),
apply(MidState1,Action,MidState2),
plan(MidState2,Goals,PostPlan,FinalState).
% satisfied(State,[]).
satisfied(State,[Goal|Goals]):-
member(Goal,State),
satisfied(State,Goals).
select(State,Goals,Goal):-
member(Goal,Goals),
not(member(Goal,State)).
achieves(Action,Goal):-
adds(Action,Goals),
member(Goal,Goals).
apply(State,Action,NewState):-
deletes(Action,DelList),
delete_all(State,DelList,State1),!,
adds(action,AddList),
append(AddList,State1,NewState).
delete_all([],_,[]).
delete_all([X|L1],L2,Diff):-
member(X,L2),!,
delete_all(L1,L2,Diff).
delete_all([X|L1],L2,[X|Diff]):-
delete_all(L1,L2,Diff).
After running this in the compiler it says it has no problems but when i try to execute the command
plan(state1,on(b1,b2),Plan,FinalState).
it just says out of global stack. Can someone help me to fix this

You only need to look at this:
plan(State,Goals,[],State):- false,
satisfied(State,Goals).
plan(State,Goals,Plan,FinalState) :-
append(PrePlan,[Action|PostPlan],Plan), false,
select(State,Goals,Goal),
achieves(Action,Goal),
can(Action,Condition),
plan(State,Condition,PrePlan,MidState1),
apply(MidState1,Action,MidState2),
plan(MidState2,Goals,PostPlan,FinalState).
?- plan(state1,on(b1,b2),Plan,FinalState).
Since this program already loops, the very same program will loop with the additional false goals removed. You need to address this problem first. See failure-slice for more.

Related

Prolog internal variable names

I have a large numbers of facts that are already in my file (position(M,P)), M is the name and P is the position of the player , I am asked to do a player_list(L,N), L is the list of players and N is the size of this list. I did it and it works the problem is that it gives the list without the names it gives me numbers and not names
player_list([H|T],N):- L = [H|T],
position(H,P),
\+ member(H,L),
append(L,H),
player_list(T,N).
what I get is:
?- player_list(X,4).
X = [_9176, _9182, _9188, _9194] .
so what should I do ?
You could use an additional list as an argument to keep track of the players you already have. This list is empty at the beginning, so the calling predicate calls the predicate describing the actual relation with [] as an additional argument:
player_list(PLs,L) :-
pl_l_(PLs,L,[]). % <- actual relation
The definition you posted is missing a base case, that is, if you already have the desired amount of players, you can stop adding others. In this case the number of players to add is zero otherwise it is greater than zero. You also have to describe that the head of the list (PL) is a player (whose position you don't care about, so the variable is preceded by an underscore (_P), otherwise the goal is just like in your code) and is not in the accumulator yet (as opposed to your code, where you check if PL is not in L) but in the recursive call it is in the accumulator. You can achieve the latter by having [PL|Acc0] in the recursive goal, so you don't need append/2. Putting all this together, your code might look something like this:
pl_l_([],0,_). % base case
pl_l_([PL|PLs],L1,Acc0) :-
L1 > 0, % number of players yet to add
L0 is L1-1, % new number of players to add
position(PL,_P), % PL is a player and
\+ member(PL,Acc0), % not in the accumulator yet
pl_l_(PLs,L0,[PL|Acc0]). % the relation holds for PLs, L0 and [PL|Acc0] as well
With respect to your comment, I assume that your code contains the following four facts:
position(zlatan,center).
position(rooney,forward).
position(ronaldo,forward).
position(messi,forward).
Then your example query yields the desired results:
?- player_list(X,4).
X = [zlatan,rooney,ronaldo,messi] ? ;
X = [zlatan,rooney,messi,ronaldo] ? ;
...
If you intend to use the predicate the other way around as well, I suggest the use of CLP(FD). To see why, consider the most general query:
?- player_list(X,Y).
X = [],
Y = 0 ? ;
ERROR at clause 2 of user:pl_l_/3 !!
INSTANTIATION ERROR- =:=/2: expected bound value
You get this error because >/2 expects both arguments to be ground. You can modify the predicate pl_l_/3 to use CLP(FD) like so:
:- use_module(library(clpfd)).
pl_l_([],0,_).
pl_l_([PL|PLs],L1,Acc0) :-
L1 #> 0, % <- new
L0 #= L1-1, % <- new
position(PL,_P),
\+ member(PL,Acc0),
pl_l_(PLs,L0,[PL|Acc0]).
With these modifications the predicate is more versatile:
?- player_list([zlatan,messi,ronaldo],Y).
Y = 3
?- player_list(X,Y).
X = [],
Y = 0 ? ;
X = [zlatan],
Y = 1 ? ;
X = [zlatan,rooney],
Y = 2 ?
...

Prolog STRIPS planner never completes

Following examples by Ivan Bratko on Artificial Intelligence in Prolog through his book:
"Prolog Programming for Artificial Intelligence - 3rd Edition" (ISBN-13: 978-0201403756) (1st edition 1986 by Addison-Wesley, ISBN 0-201-14224-4)
I've noticed that a lot of the examples do not run to completion but instead seem to get stuck. I have tried several different implementations following it to the letter, but with no luck. Would anyone be willing to take a gander at the code to see if they can spot where there is faulty logic or if I made a mistake?
This is the complete program of a STRIPS style planner for a blocks world as illustrated in the book:
% This planner searches in iterative-deepening style.
% A means-ends planner with goal regression
% plan( State, Goals, Plan)
plan( State, Goals, []) :-
satisfied( State, Goals). % Goals true in State
plan( State, Goals, Plan) :-
append( PrePlan, [Action], Plan), % Divide plan achieving breadth-first effect
select( State, Goals, Goal), % Select a goal
achieves( Action, Goal),
can( Action, Condition), % Ensure Action contains no variables
preserves( Action, Goals), % Protect Goals
regress( Goals, Action, RegressedGoals), % Regress Goals through Action
plan( State, RegressedGoals, PrePlan).
satisfied( State, Goals) :-
delete_all( Goals, State, []). % All Goals in State
select( State, Goals, Goal) :- % Select Goal from Goals
member( Goal, Goals). % A simple selection principle
achieves( Action, Goal) :-
adds( Action, Goals),
member( Goal, Goals).
preserves( Action, Goals) :- % Action does not destroy Goals
deletes( Action, Relations),
not((member( Goal, Relations),
member( Goal, Goals))).
regress( Goals, Action, RegressedGoals) :- % Regress Goals through Action
adds( Action, NewRelations),
delete_all( Goals, NewRelations, RestGoals),
can( Action, Condition),
addnew( Condition, RestGoals, RegressedGoals). % Add precond., check imposs.
% addnew( NewGoals, OldGoals, AllGoals):
% OldGoals is the union of NewGoals and OldGoals
% NewGoals and OldGoals must be compatible
addnew( [], L, L).
addnew( [Goal | _], Goals, _) :-
impossible( Goal, Goals), % Goal incompatible with Goals
!,
fail. % Cannot be added
addnew( [X | L1], L2, L3) :-
member( X, L2), !, % Ignore duplicate
addnew( L1, L2, L3).
addnew( [X | L1], L2, [X | L3]) :-
addnew( L1, L2, L3).
% delete_all( L1, L2, Diff): Diff is set-difference of lists L1 and L2
delete_all( [], _, []).
delete_all( [X | L1], L2, Diff) :-
member( X, L2), !,
delete_all( L1, L2, Diff).
delete_all( [X | L1], L2, [X | Diff]) :-
delete_all( L1, L2, Diff).
can( move( Block, From, To), [clear(Block), clear(To), on(Block,From)]) :-
block(Block),
object(To),
To \== Block,
object( From),
From \== To,
Block \== From.
adds( move(X,From,To),[on(X,To),clear(From)]).
deletes( move(X,From,To),[on(X,From), clear(To)]).
object(X) :-
place(X)
;
block(X).
impossible( on(X,X), _).
impossible( on( X,Y), Goals) :-
member( clear(Y), Goals)
;
member( on(X,Y1), Goals), Y1 \== Y % Block cannot be in two places
;
member( on( X1, Y), Goals), X1 \== X. % Two blocks cannot be in same place
impossible( clear( X), Goals) :-
member( on(_,X), Goals).
block(a).
block(b).
block(c).
block(d).
block(e).
block(f).
block(g).
place(1).
place(2).
place(3).
place(4).
I added 7 blocks and 4 locations and tested it with a representation where all the blocks are alphabetically stacked from a through g on position 1, and the goal is to stack them in the same order on position 2.
To run the program call plan(StartState,GoalState, Sol).
plan([on(a,1), on(b,a), on(c,b), on(d,c), on(e,d), on(f,e), on(g,f),
clear(g), clear(2), clear(3)],
[clear(1), on(a,2), on(b,a), on(c,b), on(d,c), on(e,d), on(f,e),
on(g,f), clear(g), clear(3)],
P).
~ ~
g g
f f
e e
d ---> d
c c
b b
a ~ ~ ~ ~ a ~ ~
_ _ _ _ _ _ _ _
1 2 3 4 1 2 3 4
References:
Definition of move: http://media.pearsoncmg.com/intl/ema/ema_uk_he_bratko_prolog_3/prolog/ch17/fig17_2.txt
End means planner with goal regression: http://media.pearsoncmg.com/intl/ema/ema_uk_he_bratko_prolog_3/prolog/ch17/fig17_8.txt
Any advice would be greatly appreciated.
In the end, the code is correct but the combinatorial explosion kills it.
Data:
3 places, 3 blocks succeeds with 5 moves after 9'755 calls to plan/3.
4 places, 3 blocks succeeds with 5 moves after 98'304 calls to plan/3.
3 places, 4 blocks succeeds with 7 moves after 915'703 calls to plan/3.
3 places, 5 blocks succeeds with 9 moves after 97'288'255 calls to plan/3.
There is no sense trying with more, especially not with 4 places, 7 blocks. It is clear that heuristics, exploitation of symmetry, etc. are needed to go further. All of those need larger amounts of memory. Here, memory used remains small in all cases: only one path down the iteratively deepened (and stored on stack) search tree is live at any time. We don't remember any states visited or anything, it's a very simple search.
Below the updated code (LONG, 337 lines)
Changes (important ones marked with 'FIX' in the code)
library(list) predicates have been used where possible, getting rid of some code.
Debugging output generation using format/2 added.
Assertions (see here) using assertion/1 added to check that what happens is what I think happens.
Predicates and variables renamed to better reflect their intended meaning.
run/0 predicate added which initializes the State and Goal, calls plan/3 and prettyprints the Plan.
can/2 confusingly combined two separate aspects: instantiating an Action and determining its Preconditions. Separated into two predicates instantiate_action/1 and preconditions/2.
select_goal/2 looked like it depended on State, but really didn't. Cleaned up.
Note the trick for making this an "iterative deepening" search. It is very clever but on second thoughts, it is too clever by half as it is based on the predicate run/3 behaving differently when being called with Plan an unbound variable than with Plan being a bound variable. The first case occurs only at the very top node of the implied search tree. This may be further explained in the textbook, which I don't have, and it took some time to realize what is actually happening in this code.
If the pruning expression ((nonvar(Plan), Plan == []) -> fail ; true ) that I put at the start of the search branch of plan/3 irritates, then so should the iterative deepening trick. IMHO, better use tree depth counters and return the Plan via an accumulator. Especially if someone will be tasked to maintain such code in a production system (that is, a "system in production", not a "forward-chaining rule-based system").
% Based on
%
% Exercise 17.5 on page 429 of "Prolog Programming for Artificial Intelligence"
% by Ivan Bratko, 3rd edition
%
% The text says:
%
% "This planner searches through the state space in iterative-deepening style."
%
% See also:
%
% https://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search
% https://en.wikipedia.org/wiki/Blocks_world
%
% The "iterative deepening" trick is all in the "Plan" list structure.
% If you remove it, the search becomes depth-first and no longer terminates!
% ----------
% Encapsulator to be called by user from the toplevel
% ----------
run :-
% Setting up
start_state(State),
final_state(Goals),
% TODO: Build predicates that verify that State and Goal are actually validly constructed
% Or choose better representations
nb_setval(glob_plancalls,0), % global variable for counting calls (non-backtrackable)
b_setval(glob_depth,0), % global variable for counting depth (backtrable)
% plan/3 is backtrackable and generates different/successively longer plans on backtrack
% it may however generate the same plan several times
plan(State, Goals, Plan),
dump_plan(Plan,1).
% ----------
% Writing out a solution found
% ----------
dump_plan([P|R],N) :-
% TODO: Verify that the plan indeed works!
format('Plan step ~w: ~w~n',[N,P]),
NN is N+1,
dump_plan(R,NN).
dump_plan([],_).
% The representation of the blocks world (see below) is a bit unfortunate as places and blocks
% have to be declared separately and relationships between places and blocks, as well
% as among blocks themselves have to declared explicitely and consistently.
% Additionally we have to specify which elements have a view of the sky (i.e. are clear/1)
% On the other hand, the final state and end state need not be specified fully, which is
% interesting (not sure what that means exactly regarding solution finding)
% The atoms used in describing places and blocks must be distinct due to program construction!
start_state([on(a,1), on(b,a), on(c,b), clear(c), clear(2), clear(3), clear(4)]).
final_state([on(a,2), on(b,a), on(c,b), clear(c), clear(1), clear(3), clear(4)]).
% ----------
% Representation of the blocks world
% ----------
% We have BLOCKs identified by atoms a,b,c, ...
% Each of those is identified by block/1 attribute.
% A block/1 is clear/1 if there is nothing on top of it.
% A block/1 is on(Block, Object) where Object is a block/1 or place/1.
block(a).
block(b).
block(c).
% We have PLACEs (i.e. columns of blocks) onto which to stack blocks.
% Each of these is identified by place/1 attribute.
% A place/1 can be clear/1 if there is nothing on top of it.
% (In fact these are like special immutable blocks and should be modeled as such)
place(1).
place(2).
place(3).
place(4).
% OBJECTs are place/1 or block/1.
object(X) :- place(X) ; block(X).
% ACTIONs are terms "move( Block, From, To)".
% "Block" must be block/1.
% "From" must be object/1 (i.e. block/1 or place/1).
% "To" must be object/1 (i.e. block/1 or place/1).
% Evidently constraints exist for a move/3 to be possible from or to any given state.
% STATEs are sets (implemented by lists) of "goal" terms.
% A goal term is "on( X, Y)" or "clear(Y)" where Y is object/1 and X is block/1.
% ----------
% plan( +State, +Goals, -Plan)
% Build a "Plan" get from "State" to "Goals".
% "State" and "Goals" are sets (implemented as lists) of goal terms.
% "Plan" is a list of action terms.
% The implementation works "backwards" from the "Goals" goal term list towards the "State" goal term list.
% ----------
% ___ Satisfaction branch ____
% This can only succeed if we are at the "end" of a Plan (the Plan must match '[]') and State matches Goal.
plan( State, Goals, []) :-
% Debugging output
nb_getval(glob_plancalls,P),
b_getval(glob_depth,D),
NP is P+1,
ND is D+1,
nb_setval(glob_plancalls,NP),
b_setval(glob_depth,ND),
statistics(stack,STACK),
format('plan/3 call ~w at depth ~d (stack ~d)~n',[NP,ND,STACK]),
% If the Goals statisfy State, print and succeed, otherwise print and fail
( satisfied( State, Goals) ->
(sort(Goals,Goals_s),
sort(State,State_s),
format(' Goals: ~w~n', [Goals_s]),
format(' State: ~w~n', [State_s]),
format(' *** SATISFIED ***~n'))
;
format(' --- NOT SATISFIED ---~n'),
fail).
% ____ Search branch ____
%
% Magic which generates the breath-first iterative deepening search:
%
% In the top node of the call tree (the node directly underneath "run"), "Plan" is unbound.
%
% At point "XXX" "Plan" is set to a list of as-yet-unbound actions of a given length.
% At each backtrack that reaches up to "XXX", "Plan" is bound to list longer by 1.
%
% In any other node of the call tree than the top node, "Plan" is bound to a list of fixed length
% becoming shorter by 1 on each recursive call.
%
% The length of that list determines how deep the search through the state space *must* go because
% satisfaction can only be happen if the "Plan" list is equal to [] and State matches Goal.
%
% So:
% On first activation of the top, build plans of length 0 (only possible if Goals passes satisfied/2 directly)
% On second activation of the top, build plans of length 1 (and backtrack over all possibilities of length 1)
% ...
% On Nth activation of the top, build plans of length N-1 (and backtrack over all possibilities of length N-1)
%
% A slight improvement is to fail the search branch immediately if Plan is a nonvar and is equal to []
% because append( PrePlan, [Action], Plan) will fail...
plan( State, Goals, Plan) :-
% The line below can be commented out w/o ill effects, it is just there to fail early
((nonvar(Plan), Plan == []) -> fail ; true ),
% Debugging output
nb_getval(glob_plancalls,P),
b_getval(glob_depth,D),
NP is P+1,
ND is D+1,
nb_setval(glob_plancalls,NP),
b_setval(glob_depth,ND),
statistics(stack,STACK),
format('plan/3 call ~w at depth ~d (stack ~d)~n',[NP,ND,STACK]),
format(' goals ~w~n',[Goals]),
% Even more debugging output
( var(Plan) -> format(' Top node of plan/3 call~n') ; true ),
( nonvar(Plan) -> (length(Plan,LP), format(' Low node of plan/3 call, plan length to complete: ~w~n',[LP])) ; true ),
% prevent runaway behaviour
% assertion(NP < 1000000),
% XXX
% append/3 is backtrackable.
% For the top node, it will generate longer completely uninstantiated PrePlans on backtracking:
% PrePlan = [], Plan = [Action] ;
% PrePlan = [_G981], Plan = [_G981, Action] ;
% PrePlan = [_G981, _G987], Plan = [_G981, _G987, Action] ;
% PrePlan = [_G981, _G987, _G993], Plan = [_G981, _G987, _G993, Action] ;
% For lower nodes, Plan is instantiated to a list of length N already, and PrePlan will therefore necessarily
% be the prefix list of length N-1
% XXX
append( PrePlan, [Action], Plan),
% Backtrackably select some concrete Goal from Goals
select_goal( Goals, Goal), % FIX: In the original this seems to depend on State, but it really doesn't
assert_goal(Goal),
format( ' Depth ~d, selected Goal: ~w~n',[ND,Goal]),
% Check whether Action achieves the Goal.
% As Action is free, what we actually do is instantiate Action backtrackably with something that achieves Goal
achieves( Action, Goal),
format( ' Depth ~d, selected Action: ~w~n', [ND,Action]),
% Fully instantiate Action backtrackably
% FIX: Passed "conditions", the precondition for a move, which is unused at this point: broken up into two calls
instantiate_action( Action),
format( ' Depth ~d, action instantiated to: ~w~n', [ND,Action]),
assertion(ground(Action)),
% Check that the Action does not clobber any of the Goals
preserves( Action, Goals),
% We now have a ground Action that "achieves" some goals in Goals while "preserving" all of them
% Work backwards from Goals to a "prior goals". regress/3 may fail to build a consistent GoalsPrior!
regress( Goals, Action, GoalsPrior),
plan( State, GoalsPrior, PrePlan).
% ----------
% Check
% ----------
assert_goal(X) :-
assertion(ground(X)),
assertion((X = on(A,B), block(A), object(B) ; X = clear(C), object(C))).
% ----------
% A State (a list) is satisfied by Goals (a list) if all the terms in Goals can also be found in State
% ----------
satisfied( State, Goals) :-
subtract( Goals, State, []). % Set difference yields empty list: [] = Goals - State
% ----------
% Backtrackably select a single Goal term from a set of Goals
% ----------
select_goal( Goals, Goal) :-
member( Goal, Goals).
% ----------
% When does an Action (move/2) achieve a Goal (clear/1, on/2)?
% This is called with instantiated Goal and free Action, so this actually instantiates Action
% with something (partially specified) that achieves Goal.
% ----------
achieves( Action, Goal) :-
assertion(var(Action)),
assertion(ground(Goal)),
would_add( Action, GoalsAdded),
member( Goal, GoalsAdded).
% ----------
% Given a ground Action and ground Goals, will Action from a State leading to Goals preserve Goals?
% ----------
preserves( Action, Goals) :-
assertion(ground(Action)),
assertion(ground(Goals)),
would_del( Action, GoalsDeleted),
intersection( Goals, GoalsDeleted, []). % "would delete none of the Goals"
% ----------
% Given existing Goals and an (instantiated) Action, compute the previous Goals
% that, when Action is applied, yield Goals. This may actually fail if no
% consistent GoalsPrior can be built!
% ** It is actually not at all self-evident that this is right and that we get a valid
% "GoalsPrior" via this method! ** (prove it!)
% FIX: "Condition" replaced by "Preconditions" which is what this is about.
% ----------
regress( Goals, Action, GoalsPrior) :-
assertion(ground(Action)),
assertion(ground(Goals)),
would_add( Action, GoalsAdded),
subtract( Goals, GoalsAdded, GoalsPriorPass), % from the "lists" library
preconditions( Action, Preconditions),
% All the Preconds must be fulfilled in Goals2, so try adding them
% Adding them may not succeed if inconsistencies appear in the resulting set of goals, in which case we fail
add_preconditions( Preconditions, GoalsPriorPass, GoalsPrior).
% ----------
% Adding preconditions to existing set of goals and checking for inconsistencies as we go
% Previously named addnew/3
% New we use union/3 from the "lists" library and the modified "consistent"
% ----------
add_preconditions( Preconditions, GoalsPriorIn, GoalsPriorOut) :-
add_preconditions_recur( Preconditions, GoalsPriorIn, GoalsPriorIn, GoalsPriorOut).
add_preconditions_recur( [], _, GoalsPrior, GoalsPrior).
add_preconditions_recur( [G|R], Goals, GoalsPriorAcc, GoalsPriorOut) :-
consistent( G, Goals),
union( [G], GoalsPriorAcc, GoalsPriorAccNext),
add_preconditions_recur( R, Goals, GoalsPriorAccNext, GoalsPriorOut).
% ----------
% Check whether a given Goal is consistent with the set of Goals to which it will be added
% Previously named "impossible/2".
% Now named "consistent/2" and we use negation as failure
% ----------
consistent( on(X,Y), Goals ) :-
\+ on(X,Y) = on(A,A), % this cannot ever happen, actually
\+ member( clear(Y), Goals ), % if X is on Y then Y cannot be clear
\+ ( member( on(X,Y1), Goals ), Y1 \== Y ), % Block cannot be in two places
\+ ( member( on(X1,Y), Goals), X1 \== X ). % Two blocks cannot be in same place
consistent( clear(X), Goals ) :-
\+ member( on(_,X), Goals). % if something is on X, X cannot be clear
% ----------
% Backtrackably instantiate a partially instantiated Action
% Previously named "can/2" and it also instantiated the "Condition", creating confusion
% ----------
instantiate_action(Action) :-
assertion(Action = move( Block, From, To)),
Action = move( Block, From, To),
block(Block), % will unify "Block" with a concrete block
object(To), % will unify "To" with a concrete object (block or place)
To \== Block, % equivalent to \+ == (but = would do here); this demands that blocks and places have disjoint sets of atoms
object(From), % will unify "From" with a concrete object (block or place)
From \== To,
Block \== From.
% ----------
% Find preconditions (a list of Goals) of a fully instantiated Action
% ----------
preconditions(Action, Preconditions) :-
assertion(ground(Action)),
Action = move( Block, From, To),
Preconditions = [clear(Block), clear(To), on(Block, From)].
% ----------
% would_del( Move, DelGoals )
% would_add( Move, AddGoals )
% If we run Move (assuming it is possible), what goals do we have to add/remove from an existing Goals
% ----------
would_del( move( Block, From, To), [on(Block,From), clear(To)] ).
would_add( move( Block, From, To), [on(Block,To), clear(From)] ).
Running the above produces lots of output and eventually:
plan/3 call 57063 at depth 6 (stack 98304)
Goals: [clear(2),clear(3),clear(4),clear(c),on(a,1),on(b,a),on(c,b)]
State: [clear(2),clear(3),clear(4),clear(c),on(a,1),on(b,a),on(c,b)]
*** SATISFIED ***
Plan step 1: move(c,b,3)
Plan step 2: move(b,a,4)
Plan step 3: move(a,1,2)
Plan step 4: move(b,4,a)
Plan step 5: move(c,3,b)
See also
STRIPS automatic planner
Iterative deepening depth-first search
Blocks World

Insert a given value v after the 1-st, 2-nd, 4-th, 8-th ... element of a list. (Prolog)

I'm trying to solve this problem in SWI Prolog, and my code currently looks like this:
insert(L1,X,L2):-
COUNTER = 1,
NEXT = 1,
insert_plus(L1,COUNTER,NEXT,X,L2).
insert_plus([],_,_,_,[]).
insert_plus([H|T],COUNTER,NEXT,X,[H|T1]) :- % don't insert
COUNTER \= NEXT,
insert_plus(T,COUNTER+1,NEXT,X,T1).
insert_plus([H|T],COUNTER,NEXT,X,[H|[X|T]]) :- % DO insert
COUNTER = NEXT,
insert_plus(T,COUNTER+1,NEXT*2,X,T).
Can someone explain why this does not always work as expected?
?- insert([1,2,3,4,5,6,7],9,X).
X = [1,9,2,3,4,5,6,7]. % BAD! expected: `X = [1,9,2,9,3,4,9,5,6,7]`
Prolog doesn't evaluate expressions, it proves relations. So arithmetic must be carried away explicitly. Here
...
insert_plus(T, COUNTER+1, NEXT, X, T1).
you need
...
SUCC is COUNTER+1,
insert_plus(T, SUCC, NEXT, X, T1).
the same problem - with both COUNTER and NEXT - occurs in the last rule.
The absolute bare minimum that you need to change is:
insert_plus([],_,_,_,[]).
insert_plus([H|T],COUNTER,NEXT,X,[H|T1]) :-
COUNTER =\= NEXT, % `(=\=)/2` arithmetic not-equal
insert_plus(T,COUNTER+1,NEXT,X,T1).
insert_plus([H|T],COUNTER,NEXT,X,[H|[X|T1]]) :- % use `T1`, not `T`
COUNTER =:= NEXT, % `(=:=)/2` arithmetic equal
insert_plus(T,COUNTER+1,NEXT*2,X,T1). % use `T1` (as above)
Sample query:
?- insert([1,2,3,4,5,6,7],9,X).
X = [1,9,2,9,3,4,9,5,6,7]. % expected result
In addition to the above changes I recommend you take advise that #CapelliC gave
in his answer concerning arithmetic expression evaluation using the builtin Prolog predicate (is)/2...
... or, even better, use clpfd!

How to remove duplicate facts in Prolog

I am writing a rule in Prolog to create a fact, pit(x,y). This rule below is called three times from my main function, and it is inserting three pits in which none of them is at (1,1) or (1,2) or (2,1) but the problem is that sometimes 2 pits have the same x and y where x and y can be from 1 to 4 only. (4x4 grid)
placePit(_) :- Px is random(4)+1,
Py is random(4)+1,
write(Px),
write(' '),
writeln(Py),
(Px =\= 1;
Py =\= 1),
(Px =\= 1;
Py =\= 2),
(Px =\= 2;
Py =\= 1)
->
pit(Px,Py);
placePit(4).
I don't want this to happen, so I write another rule to check whether 2 pits are the same first and will extend later to REMOVE EITHER ONE from the database. From what I have tested, it doesn't get fired at all even though 2 pits appear to be the same. What am I doing wrong? How to remove duplicate facts?
pit(A,B) :- pit(C,D),
A = C,
B = D,
write('Duplicate').
PS. I am very new at Prolog. Any suggestion is appreciated.
maybe this could help, in assumption you're actually required to generate facts:
:- dynamic(pit/2).
pit(1,1).
pit(1,2).
pit(2,1).
placePit(N) :-
N > 0,
Px is random(4)+1,
Py is random(4)+1,
( \+ pit(Px, Py) % if not exist
-> assertz(pit(Px, Py)), % store
M is N-1 % generate another
; M = N % nothing changed, retry
),
placePit(M). % recursion is the proper Prolog way to do cycles
placePit(0). % end of recursion (we call it 'base case')
you should call as
?- placePit(3).
It shows a bit of syntactic detail, like the 'if/then/else', that in Prolog has a peculiar form.
edit When done, you could remove unwanted pit/2, to get your db 'clean'.
?- maplist(retract, [pit(1,1),pit(1,2),pit(2,1)]).
(note that I assumed - based on your description - that a DB stored pit/2 was of value for further processing).

Hangman Game in SWI Prolog

I'm trying to make a simple hangman game in SWI Prolog.
Since we made this program run can you help me enchance the program with the following:
1) By keeping up with the letters that have been guessed so far. If the user guesses a letter that has already been guessed, the program should say 'You guessed that!' and just continue the game.
2) Lastly, add a counter that counts the number of incorrect guesses and quits the game when a certain number is reached. The program should tell the user that they lose, display what the phrase really was, and terminate. Duplicate guesses should not be counted as wrong.
I would like to thank everyone who helped me so far. This means a lot to me.
I provide you with the code and comments.
% This top-level predicate runs the game. It prints a
% welcome message, picks a phrase, and calls getGuess.
% Ans = Answer
% AnsList = AnswerList
hangman:-
getPhrase(Ans),
!,
write('Welcome to hangman.'),
nl,
name(Ans,AnsList),
makeBlanks(AnsList, BlankList),
getGuess(AnsList,BlankList).
% Randomly returns a phrase from the list of possibilities.
getPhrase(Ans):-
phrases(L),
length(L, X),
R is random(X),
N is R+1,
getNth(L, N, Ans).
% Possible phrases to guess.
phrases(['a_picture_is_worth_a_thousand_words','one_for_the_money','dead_or_alive','computer_science']).
% Asks the user for a letter guess. Starts by writing the
% current "display phrase" with blanks, then asks for a guess and
% calls process on the guess.
getGuess(AnsList, BlankList):-
name(BlankName, BlankList),
write(BlankName),
nl,
write('Enter your guess, followed by a period and return.'),
nl,
read(Guess),
!,
name(Guess, [GuessName]),
processGuess(AnsList,BlankList,GuessName).
% Process guess takes a list of codes representing the answer, a list of codes representing the current
% "display phrase" with blanks in it, and the code of the letter that was just guessed. If the guess
% was right, call substitute to put the letter in the display phrase and check for a win. Otherwise, just
% get another guess from the user.
processGuess(AnsList,BlankList,GuessName):-
member(GuessName,AnsList),
!,
write('Correct!'),
nl,
substitute(AnsList, BlankList, GuessName, NewBlanks),
checkWin(AnsList,NewBlanks).
processGuess(AnsList, BlankList,_):-
write('Nope!'),
nl,
getGuess(AnsList, BlankList).
% Check to see if the phrase is guessed. If so, write 'You win' and if not, go back and get another guess.
checkWin(AnsList, BlankList):-
name(Ans, AnsList),
name(BlankName, BlankList),
BlankName = Ans,
!,
write('You win!').
checkWin(AnsList, BlankList):-
!,
getGuess(AnsList, BlankList).
% getNth(L,N,E) should be true when E is the Nth element of the list L. N will always
% be at least 1.
getNth([H|T],1,H).
getNth([H|T],N,E):-
N1 is N-1,
getNth(T,N1,E1),
E=E1.
% makeBlanks(AnsList, BlankList) should take an answer phrase, which is a list
% of character codes that represent the answer phrase, and return a list
% where all codes but the '_' turn into the code for '*'. The underscores
% need to remain to show where the words start and end. Please note that
% both input and output lists for this predicate are lists of character codes.
% You can test your code with a query like this:
% testMakeBlanks:- name('csc_is_awesome', List), makeBlanks(List, BlankList), name(Towrite, BlankList), write(Towrite).
makeBlanks(AnsCodes, BlankCodes) :-
maplist(answer_blank, AnsCodes, BlankCodes).
answer_blank(Ans, Blank) :-
Ans == 0'_ -> Blank = Ans ; Blank = 0'* .
% substitute(AnsList, BlankList, GuessName, NewBlanks) Takes character code lists AnsList and BlankList,
% and GuessName, which is the character code for the guessed letter. The NewBlanks should again be a
% character code list, which puts all the guesses into the display word and keeps the *'s and _'s otherwise.
% For example, if the answer is 'csc_is_awesome' and the display is 'c*c_**_*******' and the guess is 's', the
% new display should be 'csc_*s_***s***'.
% You can test your predicate with a query like this:
% testSubstitute:- name('csc_is_awesome', AnsList), name('c*c_**_*******', BlankList), name('s',[GuessName]), substitute(AnsList, BlankList, GuessName, NewBlanks),
% name(Towrite, NewBlanks), write(Towrite).
% Also, since the predicate doesn't deal directly with character codes, this should also work:
% substitute(['c','s','c'],['c','*','c'],'s',L). L should be ['c','s','c'].
substitute(AnsCodes, BlankCodes, GuessName, NewBlanks) :-
maplist(place_guess(GuessName), AnsCodes, BlankCodes, NewBlanks).
place_guess(Guess, Ans, Blank, Display) :-
Guess == Ans -> Display = Ans ; Display = Blank.
maplist/3 & maplist/4 apply their first argument (a predicate of appropriate arity) against all elements of other arguments lists, then your makeBlanks could be:
makeBlanks(AnsCodes, BlankCodes) :-
maplist(answer_blank, AnsCodes, BlankCodes).
answer_blank(Ans, Blank) :-
Ans == 0'_ -> Blank = Ans ; Blank = 0'* .
and substitute:
substitute(AnsCodes, BlankCodes, GuessName, NewBlanks) :-
maplist(place_guess(GuessName), AnsCodes, BlankCodes, NewBlanks).
place_guess(Guess, Ans, Blank, Display) :-
Guess == Ans -> Display = Ans ; Display = Blank.
edit:
on additional requests: 1) can be solved with an additional predicate:
alreadyGuessed(Guess, AnsCodes) :-
memberchk(Guess, AnsCodes).
while regards 2) getGuess and processGuess together make a loop, that will just terminate when no more calls happen. Remove the last rule of checkWin, add an argument as counter to keep track of failed guesses, and extend processGuess to signal failure:
processGuess(AnsList, BlankList, _, CountFailed) :-
( CountFailed == 5
-> format('Sorry, game over. You didn\'t guess (~s)~n', [AnsList])
; write('Nope!'),
CountFailed1 is CountFailed + 1,
getGuess(AnsList, BlankList, CountFailed1)
).
Why so many cuts? Check out SWI library predicates that may be useful to you: memberchk/2, format/2 and nth1/3.

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