Control Algorithms 2Chapter 6
Production Systems
A Model of Computation
Emil Post (40’s): production systems as a formal theory of computation. Equivalent to a Turing machine. Set of rewrite rules for strings.
Newell and Simon (60’s, 70’s, 80’s): General Problem Solver
John Anderson, Newell and Simon (80’s): learning models, ACT*, SOAR
Everyone (80’s): Expert systems
Components
1. Set of rewrite rulesS NP VPLHS: Condition PartRHS: Action Part
Components
2. Working Memory--Contains the current state of the world--Contains pattern that is matched against the condition of the production--When a match occurs, an action is performed
Components
3. Recognize-Act Cycle--Isolate a subset of productions whose conditions match patterns in working memory: conflict set--Choose one of them
---Fire---Change contents of working memory
--Stop when there are no matches
Example: Production system to generate the set of palindromes over the alphabet {0,1}
Productions1. N 0N02. N 1N13. N 04. N 15. N λ
Iteration Working Memory Conflict Set Fired0 N 1,2,3,4,511 0N0 1,2,3,4,512 00N00 1,2,3,4,523 001N100 1,2,3,4,534 0010100
Knight’s Tour As a Production System
Given a 3X3 matrixWhat squares can a knight land on
What values of X, Y satisfy mv(X,Y) X,Y are elements of {1,2,…,9)
1 2 3
4 5 6
7 8 9
1. mv(1,8) 7. mv(4,9) 13. mv(8,3)
2. mv(1,6) 8. mv(4,3) 14. mv(8,1)
3. mv(2,9) 9. mv(6,1) 15. mv(9,2)
4. mv(2,7) 10. mv(6,7) 16. mv(9,4)
5. mv(3,4) 11. mv(7,2)
6. mv(3,8) 12. mv(7,6)
The General Case
),(),(),(( yxpathyzpathzxmvzyx
)),(( xxpathx
Changes
1. Every expression of the form mv(x,y) becomes on(x) on(y)
2. Use no path expression3. Working memory is the current state and
goal state4. Conflict set is the set of rules that match
the current state5. Apply all rules until the current state
equals the goal state
Productions
1. mv(1,8) 7. mv(4,9) 13. mv(8,3)
2. mv(1,6) 8. mv(4,3) 14. mv(8,1)
3. mv(2,9) 9. mv(6,1) 15. mv(9,2)
4. mv(2,7) 10. mv(6,7) 16. mv(9,4)
5. mv(3,4) 11. mv(7,2)
6. mv(3,8) 12. mv(7,6)
1. on(1) -> on(8) 7. on(4) -> on(9) 13. on(8) -> on(3)
2. on(1) -> on(6) 8. on(4) -> on(3) 14. on(8) -> on(1)
3. on(2) -> on(9) 9. on(6) -> on(1) 15. on(9) -> on(2)
4. on(2) -> on(7) 10. on(6) -> on(7) 16. on(9) -> on(4)
5. on(3) -> on(4) 11. on(7) -> on(2)
6. on(3) -> on(8) 12. on(7) -> on(6)
Can We Get from 1 to 2?
Iteration --Working Memory-- Conflict Set FiredCurrent Goal
0 1 2 1,2 11 8 2 13,14 132 3 2 5,6 5
3 4 2 7,8 74 9 2 15,16 155 2 2 Halt
Pattern Search
path(1,2) {1/x,2/y}mv(1,z)^path(z,2) {8/z}mv(1,8)^path(8,2) mv(8,z)^path(z,2) {3/z} mv(8,3)^path(3,2) mv(3,z)^path(z,2) {4/z} mv(3,4)^path(4,2) mv(4,z)^path(z,2) {9/z} mv(4,9)^path(9,2) mv(9,z)^path(z,2) {2/z} mv(9,2)^path(2,2) t t t tt
Now look at working memory in the production system
Equivalences
Production System Pattern SearchProductions mvworking memory path(X,Y)Fire lowest numbered production Choose first rule that unifies
Conclusion:Production Systems and pattern search are
almost equivalent
Almost?
Production systems have no loop detection mechanism
Solution:Invent two new productions1. assert(X) causes X to be stored in WM2. been(X) is T if X has been visited3. assert(been(X)) records in wm that
we’ve already visited X
Can be expressed in PC notation like this
),(),(^
))(()^(),((
yxpathyzpath
zbeenassertzbeenzxmvzyx
)),(( xxpathx
Can We Get from 1 to 7?
Iteration --Working Memory-- Conflict Set FiredCurrent Goal been
0 1 7 1 1,2 11 8 7 8 13,14 132 3 7 3 5,6 5
3 4 7 4 7,8 74 9 7 9 15,16 155 2 7 2 3,4 3
(firing 3 causes been(9) to fail)2 7 2 4 47 7 7
Notice that this search is data driven
Can Also Be Goal Driven
Instead of starting with current state=1 and goal = 7
Start with current state = 7 and goal = 1
Works great for a 3x3 matrix
What about 8x8?Either enumerate all moves or encode them8 possible situations1. d(2),r(1) 5. u(2),r(1)2. d(2),l(1) 6. u(2),l(1)3. d(1),r(2) 7. u(1),r(2)4. d(1),l(2) 8. u(1),l(2)
Not applicable everywhere
Situation have preconditions:Pre: row <=6, col <=7Situation 1: d(2),r(1)
Requires 4 new functionssq(r,c) returns cell number, left to right, top to
bottom where r is row number, c is column number
plus(r,2) returns r + 2eq(X,Y) T if X = Ylte(X,Y) T if X<=Y
Encoding of situation 1: d(2),r(1)
mv(sq(R,C),sq(Nr,Nc))
lte(R,6)^eq(Nr,plus(R,2)) ^lte(C,7)
^eq(Nc,plus(c,1) There are 7 more just like this
Control Loop for Knight’s Tour
)),(),,((
))),((()),(()),(),,(((
)),(),,(((
),(),,(((
ncnrsqzczrsqpath
zczrsqbeenassertzczrsqbeenzczrsqcrsqmvzczr
ncnrsqcrsqpathncnrcr
crsqcrsqpathcr
Strength of Production Systems
1. Said to model human cognition2. Separation of knowledge from control3. Natural mapping onto state space
search4. Modularity of production rules5. Simple Tracing and explanation—
compare a rule with a line of c++ code6. Language independence
But
Production systems are easily rendered in prolog
We’ll consider several versions of the knight’s tour
Global Record of Squares Visited
Put Squares Visited on a List
Use a Stack to Display Path to Goal
Use of Queue to Keep Track of Visited Squares
Display Backtracking
Try: path(1,W)◦ 1 is on the stack◦ Base case succeeds: w = 1◦ Now path1 is invoked, Start = 1◦ Move(1,6) succeeds, stack: [6,1]◦ Base case succeeds: w = 6, path is 1,6◦ Now path1 invoked:, Start = 6◦ Move (6,1) fails because 1 has been visited◦ Move(6,7) succeeds: stack:[7,6,1]◦ Base case succeeds: w = 7, path is 1,6,7◦ This process continues until all possible states have been visited◦ Occurs when path is 1,6,7,2,9,4,3,8◦ Now we backtrack from 8◦ But everything has been tried until we return to second move predicate starting with 1
(move(1,8))◦ Now we go through the entire process again
Cut
!◦Always succeeds the first time it is encountered◦When backtracked to, it causes the entire goal
in which it was contained to failWithout ! (4 2 path moves)With ! (2 2 path moves, because the first move cannot be backtracked to)
Farmer Problem
A farmer (f) has a dog (d), a goat (g),and a cabbage (c)
A river runs North and SouthThe farmer has a boat that can hold only
the farmer and one other itemWithout the farmer
◦The goat will eat the cabbage◦The dog will eat the goat
How does the farmer (and his cohort) cross the river
Define a predicate:state(F,D,G,C)Where F,D,G,C can be set to e or w
indicating the side of the river each is on.
As State-Space
st(w,w,w,w)
st(e,e,w,w) st(e,w,e,w) s(e,w,w,e)
st(w,w,e,w)
s(e,e,e,w) st(e,w,e,e)
etc.
Constructing a Move Predicate
st(e,e,-,-) st(w,w,-,-)Means Farmer and dog went from east to
westCan be rewritten:mv(st(X,X,G,C),st(Y,Y,G,C))
Facts
opp(e,w)opp(w,e)
Givingmv(st(X,X,G,C),st(Y,Y,G,C)) :- opp(X,Y).opp(e,w).opp(w,e).
In Motion
Suppose: st(e,e,G,C)mv(st(X,X,G,C), st(Y,Y,G,C)) :- opp(X,Y).opp(e,w).opp(w,e). w,wmv(st(e,e,G,C),st(Y,Y,G,C)) opp(X,Y)
{e/X, w/Y)
opp(e,w)
This is one of 4 move predicates (3 items to move + return trip alone)
Unsafe
Goat and cabbage are together◦unsafe(st(X,D,Y,Y)) if x != y◦unsafe(st(X,D,Y,Y)) :- opp(X,Y)
Dog and goat are together◦Unsafe(st(X,Y,Y,C) if x != y◦Unsafe(st(X,Y,Y,C) :- opp(X,Y)
Never move to an unsafe state
mv(st(X,X,G,C),st(Y,Y,G,C)) :- opp(X,Y), not(unsafe(st(Y,Y,G,C))).
Traversal Mechanism
Use the stack mechanism from Knight3
Farmer Problem