Below is a small part of a river crossing problem written in pddl. I tried to find the solution in two different tools (editor.planning.domains and stripsfiddle.herokuapp.com) but both of them gave the same result.
;domain;
(define (domain RiverCrossing)
(:requirements :strips :typing)
(:types
Farmer Fox - passengers
)
(:predicates
(onLeftBank ?p - passengers)
(onRightBank ?p - passengers)
)
(:action crossRiverLR
:parameters (?f - Farmer)
:precondition ( and (onLeftBank ?f))
:effect( and (onRightBank ?f) )
)
(:action crossRiverRL
:parameters (?f - Farmer)
:precondition ( and (onRightBank ?f))
:effect( and (onLeftBank ?f) )
)
(:action crossRiverLRf
:parameters ( ?fx - Fox ?f - Farmer)
:precondition ( and (onLeftBank ?f) (onLeftBank ?fx) )
:effect( and (onRightBank ?fx) (onRightBank ?f) )
)
(:action crossRiverRLf
:parameters (?f - Farmer ?fx - Fox)
:precondition ( and (onRightBank ?f) (onRightBank ?fx) )
:effect( and (onLeftBank ?f) (onLeftBank ?fx) )
)
)
Problem
(define (problem RCP)
(:domain RiverCrossing)
(:objects
farmer - Farmer
fox - Fox
)
(:init
(onRightBank farmer) (onLeftBank fox)
)
(:goal
(and
(onLeftBank farmer) (onRightBank fox)
)
)
)
Both of the compilers give the same result; Farmer does not go to LeftBank:
Solution found in 2 steps!
1. crossRiverRL farmer
2. crossRiverLRf fox farmer
Can anyone help me figure out the point I am missing?
Thanks in advance,
I figured out that the problem is not negating the previous situation (OnLeftBank) after setting the next situation (OnRightBank).
Below is the sample correction which I applied to all effects;
(:action crossRiverLR
:parameters (?f - Farmer)
:precondition ( and (onLeftBank ?f))
:effect( and (onRightBank ?f)
(not (onLefttBank ?f)) ; **** adding this solved the problem. ****
)
)
Related
I need help with this program, I couldn't find out how to do that a user can input a yes/no for characters and I have defined a type of animal-like mammal - for mammal applies that milk character must be yes and legs(number of legs) can be 2 or 4 and another character can be yes or no like for example -
mammal - milk yes, legs 2 or 4, but backbone can yes or no, predator yes or no .... but I don't know how to do it (OR condition or idk something like that), user can input one of these its find out the type of animal which is defined in effects, thanks for help :)
;*********** DEFTEMPLATE ***********;
(deftemplate animal_type
(slot type (type SYMBOL) (allowed-symbols mammal bird fish))
(slot milk (type SYMBOL) (allowed-symbols yes no))
(slot feathers (type SYMBOL) (allowed-symbols yes no))
(slot fins (type SYMBOL) (allowed-symbols yes no))
(slot backbone (type SYMBOL) (allowed-symbols yes no))
(slot fly (type SYMBOL) (allowed-symbols yes no))
(slot predator (type SYMBOL) (allowed-symbols yes no))
(multislot legs (type INTEGER) (allowed-integers 0 2 4 6 8))
)
(deftemplate finding_type
(slot milk (type SYMBOL) (allowed-symbols yes no))
(slot feathers (type SYMBOL) (allowed-symbols yes no))
(slot fins (type SYMBOL) (allowed-symbols yes no))
(slot backbone (type SYMBOL) (allowed-symbols yes no))
(slot fly (type SYMBOL) (allowed-symbols yes no))
(slot predator (type SYMBOL) (allowed-symbols yes no))
(multislot legs (type INTEGER) (allowed-integers 0 2 4 6 8))
)
;*********** DEFFACTS ***********;
(deffacts characters_type
(animal_type (typ mammal) (milk yes) (legs 2 4))
(animal_type (typ bird) (milk no) (feathers yes) (fly yes))
(animal_type (typ fish) (milk no) (feathers no) (fins yes) (legs 0))
)
;something like this (animal_type (typ mammal) (milk yes) (backbone or(yes no)) ... (legs 2 4))
(deffacts temp_fact
(next_search)
)
(defrule input_characters
?gone<-(next_search)
=> (retract ?gone)
(printout t " " crlf)
(printout t "Enter - yes/no, legs - 0/2/4/6/8" crlf)
(printout t "==================================================================" crlf)
(printout t "Milk:")
(bind ?o1 (read))
(printout t "Feathers:")
(bind ?o2 (read))
(printout t "Fins:")
(bind ?o3 (read))
(printout t "Backbone:")
(bind ?o4 (read))
(printout t "Fly:")
(bind ?o5 (read))
(printout t "Predator:")
(bind ?o6 (read))
(printout t "Legs:")
(bind ?o7 (read))
(assert (finding_type (milk ?o1) (feathers ?o2) (fins ?o3) (backbone ?o4) (fly ?o5) (predator ?o6) (legs ?o7)) )
)
(defrule find_out_type
(finding_type (milk ?o1) (feathers ?o2) (fins ?o3) (backbone ?o4) (fly ?o5) (predator ?o6) (legs ?o7))
(animal_type (type ?type) (milk ?o1) (feathers ?o2) (fins ?o3) (backbone ?o4) (fly ?o5) (predator ?o6) (legs ?o7))
=>
(printout t " " crlf)
(printout t "Type of animal is: " ?type crlf)
)
(defrule not_found
(finding_type (milk ?o1) (feathers ?o2) (fins ?o3) (backbone ?o4) (fly ?o5) (predator ?o6) (legs ?o7))
(not (animal_type (type ?type) (milk ?o1) (feathers ?o2) (fins ?o3) (backbone ?o4) (fly ?o5) (predator ?o6) (legs ?o7)) )
=>
(printout t " " crlf)
(printout t "Nothing found!" crlf)
)
(defrule cancel (declare (salience -10))
?gone<-(finding_type (milk ?o1) (feathers ?o2) (fins ?o3) (backbone ?o4) (fly ?o5) (predator ?o6) (legs ?o7))
=>
(retract ?gone)
)
Modify the animal_type pattern in your rules to use multifield wildcards to match extraneous values to the left and right of the specified number of legs. As your rules are currently the animal_type patterns will only be matched by facts containing exactly one value in the legs slot.
(animal_type (type ?type)
(milk ?o1)
(feathers ?o2)
(fins ?o3)
(backbone ?o4)
(fly ?o5)
(predator ?o6)
(legs $? ?o7 $?))
How to add double quotes to the gene_id?
My original file:
##gtf-version 3
Bany_Scaf1 maker gene 201136 207903 . + . Alias "maker-Bany_Scaf1-snap-gene-2.23"; Dbxref "InterPro:IPR019774" "Pfam:PF00351"; ID Bany_03723; Name Bany_03723; Ontology_term "GO:0016714" "GO:0055114"; gene_id Bany_03723
Bany_Scaf1 maker transcript 201136 207903 . + . Alias "maker-Bany_Scaf1-snap-gene-2.23-mRNA-1"; Dbxref "InterPro:IPR019774" "Pfam:PF00351"; ID "Bany_03723-RA"; Name "Bany_03723-RA"; Ontology_term "GO:0016714" "GO:0055114"; Parent Bany_03723; _AED "0.06"; _QI "45|1|1|1|1|1|7|425|530"; _eAED "0.06"; gene_id Bany_03723; original_biotype mrna; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 201136 201304 . + . ID "Bany_03723-RA:1"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 202687 202770 . + . ID "Bany_03723-RA:2"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 202886 202921 . + . ID "Bany_03723-RA:3"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 203004 203820 . + . ID "Bany_03723-RA:4"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 206097 206223 . + . ID "Bany_03723-RA:5"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 206649 206878 . + . ID "Bany_03723-RA:6"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
Bany_Scaf1 maker exon 207304 207903 . + . ID "Bany_03723-RA:7"; Parent "Bany_03723-RA"; gene_id Bany_03723; transcript_id "Bany_03723-RA"
I hope to change all gene_id Bany_xxxxxx to gene_id "Bany_xxxxxx".
I tried this:
sed -E 's#(Parent|gene_id|ID) ([0-9A-Za-z.]+)#\1 \"\2\"#g'
But the double quotes were added in the wrong place, like:
gene_id "Bany"_03723
What should I do...
with sed
$ sed -E 's/gene_id ([^;]+)/gene_id "\1"/' file
find the next word delimited with ; to gene_id and quote it. Assumes space between them. If tab change to \t
I've googled for it for a while but I didn't find the solution.
I've this list:
((-1 . <Nome entità: 7ff5ff905910>) (0 . "DICTIONARY") (5 . "1F9") (102 . "{ACAD_REACTORS") (330 . <Nome entità: 7ff5ff9038c0>) (102 . "}") (330 . <Nome entità: 7ff5ff9038c0>) (100 . "AcDbDictionary") (280 . 0) (281 . 1) (3 . "1") (350 . <Nome entità: 7ff5ff9933c0>) (3 . "2") (350 . <Nome entità: 7ff5ff9fa0d0>) (3 . "3") (350 . <Nome entità: 7ff5ff9fa410>))
I want to get all the entities with dxf code 350.
With that instruction I can get only the fist one.
(assoc 350 list)
How to get all the associations?
Thanks, Dennis
There are various ways to achieve this - here are a few examples:
(defun mAssoc1 ( key lst / rtn )
(foreach x lst
(if (= key (car x))
(setq rtn (cons (cdr x) rtn))
)
)
(reverse rtn)
)
(defun mAssoc2 ( key lst )
(apply 'append
(mapcar
(function
(lambda ( x ) (if (= key (car x)) (list (cdr x))))
)
lst
)
)
)
(defun mAssoc3 ( key lst )
(mapcar 'cdr
(vl-remove-if-not
(function (lambda ( x ) (= key (car x))))
lst
)
)
)
(defun mAssoc4 ( key lst / item )
(if (setq item (assoc key lst))
(cons (cdr item) (mAssoc4 key (cdr (member item lst))))
)
)
(defun mAssoc5 ( key lst / item rtn )
(while (setq item (assoc key lst))
(setq rtn (cons (cdr item) rtn) lst (cdr (member item lst)))
)
(reverse rtn)
)
(defun mAssoc6 ( key lst )
(mapcar 'cdr (acet-list-m-assoc key lst))
)
(defun mAssoc7 ( key lst )
(if lst
(if (= key (caar lst))
(cons (cdar lst) (mAssoc7 key (cdr lst)))
(mAssoc7 key (cdr lst))
)
)
)
Here is a quick performance comparison for the above functions:
;;; Benchmarking: Elapsed milliseconds / relative speed for 32768 iteration(s):
;;;
;;; (MASSOC4 2 L).....1482 / 1.25 <fastest>
;;; (MASSOC5 2 L).....1482 / 1.25
;;; (MASSOC6 2 L).....1498 / 1.24
;;; (MASSOC3 2 L).....1638 / 1.13
;;; (MASSOC7 2 L).....1747 / 1.06
;;; (MASSOC1 2 L).....1748 / 1.06
;;; (MASSOC2 2 L).....1856 / 1 <slowest>
assoc returns only first one, so You need to loop by all after find one as long as none is found (assoc returns nil). You should try such code:
(defun DXF:Dump (EntDef code / out)
(while (setq values(assoc code EntDef))
(setq out (append out (list (cdr values))))
(setq EntDef(cdr(member values EntDef)))
)
out
)
; (setq dxfs (entget (car (entsel))))
; (DXF:Dump dxfs 350)
I am currently working in CLIPS and I am new to it. I am trying to replicate the following information in a CLIPS deftemplate:
[Person, [Class,Class],[[M 9,11],[F,9,11]]]
It has a person, multiple classes that they can take and the times that they can take the class. I try to replicate this information in the following deftemplate:
(deftemplate person
(slot Name)
(multislot Class)
(multislot Available))
My problem is I do not understand what I am supposed to do in the available multislot because it has three inputs. Is there a way that I can make slots within a multislot? I have looked online for ways to do this but have not been able to correctly solve this problem.
Here are four different approaches. I would suggest something similar to approach 3 or 4 since these involve simple linkages between facts/instances.
CLIPS> (clear) ; Approach 1
CLIPS>
(deftemplate person
(slot Name)
(multislot Class)
(multislot Available))
CLIPS>
(deffacts people
(person (Name Frank)
(Class Biology Calculus)
(Available M 9 11 F 9 11)))
CLIPS>
(deffunction #-of-triplets (?mf)
(div (length$ ?mf) 3))
CLIPS>
(deffunction nth-triplet (?mf ?n)
(subseq$ ?mf (+ 1 (* (- ?n 1) 3))(* ?n 3)))
CLIPS>
(defrule print-availability
(person (Name ?name)
(Available $?a))
=>
(loop-for-count (?i (#-of-triplets ?a))
(bind ?triplet (nth-triplet ?a ?i))
(bind ?d (nth$ 1 ?triplet))
(bind ?b (nth$ 2 ?triplet))
(bind ?e (nth$ 3 ?triplet))
(printout t ?name " " ?d " " ?b " " ?e crlf)))
CLIPS> (reset)
CLIPS> (run)
Frank M 9 11
Frank F 9 11
CLIPS> (clear) ; Approach 2
CLIPS>
(deftemplate person
(slot Name)
(multislot Class)
(multislot Available-Weekday)
(multislot Available-Begin)
(multislot Available-End))
CLIPS>
(deffacts people
(person (Name Frank)
(Class Biology Calculus)
(Available-Weekday M F)
(Available-Begin 9 9)
(Available-End 11 11)))
CLIPS>
(defrule print-availability
(person (Name ?name)
(Available-Weekday $?f1 ?d $?)
(Available-Begin $?f2 ?b $?)
(Available-End $?f3 ?e $?))
(test (= (length$ ?f1)
(length$ ?f2)
(length$ ?f3)))
=>
(printout t ?name " " ?d " " ?b " " ?e crlf))
CLIPS> (reset)
CLIPS> (run)
Frank M 9 11
Frank F 9 11
CLIPS> (clear) ; Approach 3
CLIPS>
(deftemplate person
(slot Name)
(slot ID)
(multislot Class))
CLIPS>
(deftemplate availability
(slot owner-ID)
(slot Weekday)
(slot Begin)
(slot End))
CLIPS>
(deffacts information
(person (Name Frank) (ID 1)
(Class Biology Calculus))
(availability (owner-ID 1) (Weekday M) (Begin 9) (End 11))
(availability (owner-ID 1) (Weekday F) (Begin 9) (End 11)))
CLIPS>
(defrule print-availability
(person (Name ?name) (ID ?id))
(availability (owner-ID ?id) (Weekday ?d) (Begin ?b) (End ?e))
=>
(printout t ?name " " ?d " " ?b " " ?e crlf))
CLIPS> (reset)
CLIPS> (run)
Frank F 9 11
Frank M 9 11
CLIPS> (clear) ; Approach 4
CLIPS>
(defclass PERSON
(is-a USER)
(slot Name)
(multislot Class)
(multislot Available))
CLIPS>
(defclass AVAILABILITY
(is-a USER)
(slot Weekday)
(slot Begin)
(slot End))
CLIPS>
(definstances information
(of PERSON (Name Frank)
(Class Biology Calculus)
(Available (make-instance of AVAILABILITY (Weekday M) (Begin 9) (End 11))
(make-instance of AVAILABILITY (Weekday F) (Begin 9) (End 11)))))
CLIPS>
(defrule print-availability
(object (is-a PERSON) (Name ?name) (Available $? ?a $?))
(object (is-a AVAILABILITY) (name ?a))
=>
(printout t ?name " " (send ?a get-Weekday)
" " (send ?a get-Begin)
" " (send ?a get-End) crlf)))
CLIPS> (reset)
CLIPS> (run)
Frank F 9 11
Frank M 9 11
CLIPS>
I'm trying to implement the possible moves of a bishop on a chess table, which can have other pieces on random cells. I've been able to make a sketch of an answer, but it doesn't detect other pieces.
Previously to this rule I've written some code that creates a fact like the following for each cell of the table, indicating its contents:
(cell-info (coor {i} {j}) (contents {empty|black|white}))
and a fact that shows the position of a piece:
(piece (row {r}) (column {c}) (type {t}) (color {col}))
And here's my rule so far (probably it's also not too efficient):
(defrule bishop-moves
(declare (salience 30))
(piece (row ?rb) (column ?cb) (type bishop) (color black))
(cell-info (coor ?i ?j) (contents empty|white))
=>
(loop-for-count (?n 1 8)
(if (or (and (= ?i (+ ?rb ?n)) (= ?j (+ ?cb ?n)))
(and (= ?i (- ?rb ?n)) (= ?j (- ?cb ?n)))
(and (= ?i (+ ?rb ?n)) (= ?j (- ?cb ?n)))
(and (= ?i (- ?rb ?n)) (= ?j (+ ?cb ?n))))
then (assert (movement-allowed
(destination-cell ?i ?j)
(type bishop)
(start-cell ?rb ?cb))))))
Does anybody now what could I do? Thanks in advance.
;;; Added deftemplates and deffacts
;;; Replaced rule variable ?i with ?r and ?j with ?c.
;;; Made rule applicable for both black or white bishop
;;; Moved diagonal logic from actions of rule to conditions
;;; Added logic to rule for intervening pieces
(deftemplate piece (slot row) (slot column) (slot type) (slot color))
(deftemplate cell-info (multislot coor) (slot contents))
(deftemplate movement-allowed (multislot destination-cell) (slot type) (multislot start-cell))
(deffacts test-data
(piece (row 1) (column 1) (type pawn) (color black))
(cell-info (coor 1 1) (contents black)) ; Invalid - friendly piece
(cell-info (coor 1 2) (contents empty)) ; Invalid - not on diagonal
(cell-info (coor 1 3) (contents empty)) ; Valid
(piece (row 2) (column 2) (type bishop) (color black))
(cell-info (coor 2 2) (contents black)) ; Invalid - friendly piece
(cell-info (coor 2 8) (contents empty)) ; Invalid - not on diagonal
(cell-info (coor 3 1) (contents empty)) ; Valid
(cell-info (coor 3 3) (contents empty)) ; Valid
(cell-info (coor 4 4) (contents empty)) ; Valid
(cell-info (coor 5 5) (contents empty)) ; Valid
(piece (row 6) (column 6) (type pawn) (color white))
(cell-info (coor 6 6) (contents white)) ; Valid
(cell-info (coor 7 7) (contents empty)) ; Invalid - blocked by pawn
(piece (row 8) (column 8) (type pawn) (color white))
(cell-info (coor 8 8) (contents white))) ; Invalid - blocked by pawn
(defrule bishop-moves
(declare (salience 30))
(piece (row ?rb) (column ?cb) (type bishop) (color ?color))
;; The destination cell must be empty or contain
;; an opposing piece
(cell-info (coor ?r ?c) (contents empty | ~?color))
;; If the cell and piece are on the same diagonal, the
;; absolute difference between the two should be the same
(test (= (abs (- ?r ?rb)) (abs (- ?c ?cb))))
;; Check that there is not another piece that is within
;; the rectangle formed by the bishop and the destination
;; cell and is also on the same diagonal as the bishop
(not (and (piece (row ?ro) (column ?co))
(test (and (or (< ?rb ?ro ?r) (< ?r ?ro ?rb))
(or (< ?cb ?co ?c) (< ?c ?co ?cb))))
(test (= (abs (- ?ro ?rb)) (abs (- ?co ?cb))))))
=>
(assert (movement-allowed
(destination-cell ?r ?c)
(type bishop)
(start-cell ?rb ?cb))))