### Control Algorithms

```Control Algorithms 2
Chapter 6
Production Systems
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

A Model of Computation
1.
Set of rewrite rules
S  NP VP
LHS: Condition Part
RHS: 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
Components
Productions
1.
N  0N0
2.
N  1N1
3.
N0
4.
N1
5.
Nλ
Iteration
0
1
2
3
4
Working Memory
N
0N0
00N00
001N100
0010100
Conflict Set
1,2,3,4,5
1,2,3,4,5
1,2,3,4,5
1,2,3,4,5
Fired
1
1
2
3
Example: Production system to
generate the set of palindromes over
the alphabet {0,1}
Knight’s Tour As a Production
System
Given a 3X3 matrix
 What squares can a knight land on
What values of X, Y satisfy mv(X,Y)
X,Y are elements of {1,2,…,9}
1
4
7
2
3
1. mv(1,8) 7. mv(4,9)
13. mv(8,3)
2. mv(1,6) 8. mv(4,3)
14. mv(8,1)
15. mv(9,2)
5
6
3. mv(2,9) 9. mv(6,1)
8
9
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)
 x ( path ( x , x ))
 x  y  z ( mv ( x , z )  path ( z , y )  path ( x , y )
The General Case (write on board)
1. Every expression of the form mv(x,y)
becomes on(x)  on(y)
2. Use no path expression
3. Working memory is the current state and
goal state
4. Conflict set is the set of rules that match
the current state
5. Apply all rules until the current state
equals the goal state
Changes

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)
Productions (write on board)
Iteration
0
1
2
3
4
5
--Working Memory-Current
Goal
1
2
8
2
3
2
4
2
9
2
2
2
Conflict Set
Fired
1,2
13,14
5,6
7,8
15,16
1
13
5
7
15
Halt
Can We Get from 1 to 2?
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
t
t
Pattern Search
Now look at working memory in the production system
Production System
Pattern Search
productions
working memory
Fire lowest numbered production
mv
path(X,Y)
Choose first rule that unifies
Conclusion:
Production Systems and pattern search are
equivalent (almost)
Equivalences
Loop Detection
 Pattern Search: global list of visited states
(closed)
 Production Systems: Record previously
visited states in working memory
Two new productions
1. assert(X) causes X to be stored in WM
2. been(X) is T if X has been visited
3. assert(been(X)) records in wm that we’ve
Almost?
 x ( path ( x , x ))
 x  y  z ( mv ( x , z )   been ( z )^ assert ( been ( z ))
^ path ( z , y )  path ( x , y )
Can be expressed in PC
notation like this
Iteration
0
1
2
3
4
5
--Working Memory-Conflict Set
Current Goal been
1
7
1
1,2
8
7
8
13,14
3
7
3
5,6
4
7
4
7,8
9
7
9
15,16
2
7
2
3,4
(firing 3 causes been(9) to fail)
2
7
2
4
7
7
7
Notice that this search is data driven
Can We Get from 1 to 7?
Fired
1
13
5
7
15
3
4
Instead of starting with current state=1
and goal = 7
Can Also Be Goal Driven
Either enumerate all moves or encode them
8 possible situations
1. 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)
Works great for a 3x3 matrix
Situation have preconditions:
Pre: row <=6, col <=7
Situation 1: d(2),r(1)
Requires 4 new functions
sq(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 + 2
eq(X,Y) T if X = Y
lte(X,Y) T if X<=Y
Not applicable everywhere
mv(sq(R,C),sq(Nr,Nc)) 
lte(R,6)^eq(Nr,plus(R,2)) ^
lte(C,7)^eq(Nc,plus(c,1))
%down two rows
%right 1 col
There are 7 more analogous to this
Encoding of situation 1: d(2),r(1)
 R  C ( path ( sq ( R , C ), sq ( R , C ))
 R  C  Nr  Nc ( path ( sq ( R , C ), sq ( Nr , Nc )) 
 Zr  Zc ( mv ( sq ( R , C ), sq ( Zr , Zc ))   been ( sq ( Zr , Zc ))  assert ( been ( sq ( Zr , Zc ))) 
path ( sq ( Zr , Zc ), sq ( Nr , Nc )))
Control Loop for Knight’s Tour
1.
2.
3.
4.
5.
6.
Said to model human cognition
Separation of knowledge from control
Natural mapping onto state space
search
Modularity of production rules
Simple tracing and explanation—
compare a rule with a line of c++ code
Language independence
Strength of Production Systems
Production systems are easily rendered in
prolog
We’ll consider several versions of the
knight’s tour
And (this is the best part)
Record of Squares Visited
knight1
Put Visited Squares on a List
Knight2
Stack Displays Path to Goal
Knight3
Queue Displays Path to Goal
Knight4 (continued next class)

!
◦ Always succeeds the first time it is encountered
◦ When backtracked to, it causes the entire goal
in which it was contained to fail
Without ! (4 2 path moves from 1)
With ! (2 2 path moves from 1)
Cut
A farmer (f) has a dog (d), a goat (g),and
a cabbage (c)
 A river runs North and South
 The farmer has a boat that can hold only
the farmer and one other item
 Without the farmer

◦ The goat will eat the cabbage
◦ The dog will eat the goat

How does the farmer (and his cohort)
cross the river
Farmer Problem
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.
State Predicate
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.
As State-Space
st(e,e,-,-)  st(w,w,-,-)
Means Farmer and dog went from east to
west
Can be rewritten:
mv(st(X,X,G,C),st(Y,Y,G,C))
Constructing a Move Predicate
opp(e,w)
 opp(w,e)

Giving
mv(st(X,X,G,C),st(Y,Y,G,C)) :- opp(X,Y).
opp(e,w).
opp(w,e).
Four of these: 3 items to move + 1 solo return trip
Facts

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)
Unsafe
mv(st(X,X,G,C),st(Y,Y,G,C)) :- opp(X,Y),
not(unsafe(st(Y,Y,G,C))).
Never move to an unsafe state
Use Move/Control Technique from Knight3
Farmer Problem
Finding a Solution
```