1. L9. STRIPS Examples Recursive STRIPS Block world
Partial-order planning

2. STRIPS planning STRIPS uses logical formulas to represent the states
Precondition
前提条件
STRIPS uses logical formulas to represent the states
S0, G, P, etc:
Description of operator f:
Logic formula
論理式
description 記述

3. Example1: The move operatorG S0 
f(P)->G
S0->P   
x/B, y/A, z/Fl
x/B, y/A, z/Fl  
On(x,y)
Clear(x)
Clear(z)
f: move(x,y, z) add: On(x,z), Clear(y)
del: On(x,y), Clear(z)
On(x,z)
Clear(x)
Clear(y) P:
f(P):

4. Example2: robot operators
Consider a robot able to
go(x,y)
pickup(x)
putdown(x)
pre: AtR(x)
del: AtR(x)
add: AtR(y)
eff: AtR(x)AtR(y)
pre: AtR(x), Empty-H
del: AtR(x), Empty-H
add: Holding(x)
eff: Holding(x)
pre: Holding(x)
del: Holding(x)
add: AtR(x)
eff: Holding(x)AtR(x)

5. Holding(C) Holding(C)
Example2: problem
S0: AtR(A) G0: AtR(B)
Holding(C) Holding(C)
The robot goes from room A to room B with holding C
Difference: AtR(B) in G0 but not in S0.
Relevant operator instance: go(x,B) if (y/B) (schema go(x, y))
P: AtR(x)
S0 -> P: x/A
f: go(x, y)
f(P)->G0: AtR(B) (x/A, y/B)
relevant
関連の
instance 実例
Achieve preconds of go(x, B) [AtR(x)]
Apply go(x, B)
Achieve G0
When the robot picks up x, it is no longer AtR(x) so it can’t go(x, y)!
It is necessary to repair the operators of a robot in page 4.

6. Example2: revised robot operators (Add object’s location sentence)
修正する
go(x,y)
pickup(x)
putdown(x)
“go from x to y”
pre: AtR(x)
del: AtR(x)
add: AtR(y)
eff: AtR(x)AtR(y)
“pick up x at y”
pre: At(x, y), AtR(y,) Empty-H
del: At(x,y), Empty-H
add: Holding(x)
At(x, y)
Object, x, is at the room, y.
eff: Holding(x)
“put down x at y”
pre: Holding(x), AtR(y)
del: Holding(x)
add: At(x,y)
eff: Holding(x)At(x,y)

7. Example2: Problem- get the slipper
S0: AtR(A) G0: AtR(A)
Empty-H Holding(slipper1)
At(slipper1, B)
Schema 概要な式
An operator is usually written in the scheme form
Difference: Holding(slipper1) in G0 but nor in S0.
Relevant operator instance: pickup(slipper1) (schema pickup(x), x/slipper1)
Pre: At(slipper1, y), AtR(y), Empty-H
Pre: (y/A) ->Pre’: At(slipper1, A), AtR(A), Empty-H
But At(slipper1, A) is not in S0!
To reduce the difference between: At(slipper1, A) and At(slipper1, B)
Relevant operator instance: go(x, B)
Pre: AtR(x) Del: AtR(x) Add: AtR(y)
S0 -> G2: (x/A) f(G2) -> S3: go(x, y) S3->G1: (y/B)
reduce
減少させる

8. Example2: Tree representation表示
S0: AtR(A)
Empty-H
At(slipper1, B)
S0->G0
S0->G1
AtR(y)
Empty-H
At(slipper1, y)
G1->S1
pickup(slipper1)
S1->G0
x=B
Pick up slipper1 at B
S2: AtR(B)
Holding(slipper1)
S3->G4
AtR(x)
G4->S4
go(x,A)
S4->G0
y=B
x=A
G0: AtR(A)
Holding(slipper1)
S0->G2
AtR(x)
G2->S2
go(x,B)
S2->G1
G1: AtR(B)
Empty-H
At(slipper1, B)

9. Recursive STRIPS
Achieve one sub-goal at a time. Achieve a new conjunct without ever violating already achieved conjuncts or maybe temporarily violating previous sub-goals.
Each sub-goal is achieved via a matched rule, then its preconditions are sub-goals and so on.
conjunct
連結した
violate
違反する

10. Plan with Run-time conditionals
We can allow disjunction in state description:
EX: On(B,A) V On(B,C)
For some operators may be applicable just with one of these disjunction that can be determined during run-time.
Run-time conditionals:
If On(B,A) apply oper1
If On(B,C) apply oper2.
Plan is a tree whose branching nodes are states with unknown information.
disjunction 分裂
applicable
適用[応用]できる
determine 決定

11. Block world F1 F2 F3 Start state S0 Goal state G0  delete list
S0: On(C, Fl)
On(B, C)
On(A, B)
Clear(A)
Clear(F2)
Clear(F3)
G0: On(C, F3)
On(B, C)
On(A, B)
Clear(A)
Clear(F1)
Clear(F2)
delete list
add list
Operators: Move(x, y, z) “Move x from y to z”

12. S0---------->G’0 S0---------->G1 G1---->S1 S1-->G0
Sub-goal: to achieve one of the conjuncts, G’0
S0: On(C, Fl)
On(B, C)
On(A, B)
Clear(A)
Clear(F2)
Clear(F3)
S >G’0
diff: On(C, F3)
G’0: On(C, F3)
G0: On(C, F3)
On(B, C)
On(A, B)
Clear(A)
Clear(F1)
Clear(F2)
G’0: On(C, F3)
On(B, A)
On(A, F2)
Clear(B)
Clear(C)
Clear(F1)
y/F1
S >G1
pre: On(C, y), Clear(F3), Clear(C)
G1---->S1
f: Move(C, y, F3)
S1-->G0
y/F1
x1/B, z1/A
S >G2
pre: On(x1, C), Clear(z1), Clear(x1)
G2---->S2
f: Move(x1, C, z1)
S2-->G1
x2/A, z2/F2
x1/B, z1/A
S >G3
pre: On(x2, B), Clear(F2), Clear(x2)
G3---->S3
f: Move(x2, B, z2)
S3-->G2

13. S0---------->G’’0 S0---------->G1 G1---->S2 S2-->G0
Sub-goal: to achieve one of the conjuncts, G’’0
G0: On(C, F3)
On(B, C)
On(A, B)
Clear(A)
Clear(F1)
Clear(F2)
G’’0: On(C, F3)
On(B, C)
On(A, F2)
Clear(B)
Clear(A)
Clear(F1)
S0=G’0: On(C, F3)
On(B, A)
On(A, F2)
Clear(B)
Clear(C)
Clear(F1)
S >G’’0
diff: On(B, C)
G’’0: On(B, C)
y/A
S >G1
pre: On(B, y), Clear(C), Clear(B)
G1---->S2
f: Move(B, y, C)
S2-->G0
Sub-goal: to achieve one of the conjuncts, G’’’0
G0: On(C, F3)
On(B, C)
On(A, B)
Clear(A)
Clear(F1)
Clear(F2)
G’’’0: On(C, F3)
On(B, C)
On(A, B)
Clear(A)
Clear(F1)
Clear(F2)
S0=G’’0: On(C, F3)
On(B, C)
On(A, F2)
Clear(B)
Clear(A)
Clear(F1)
S >G’’’0
diff: On(A, B)
G’’’0: On(A, B) =
y/F2
S >G1
pre: On(A, y), Clear(A), Clear(B)
G1---->S2
f: Move(A, y, B)
S2-->G0

14. A sequence plan structure
NIL
finish
On(A,B)
Clear(F2)
On(B,C)
Clear(A) 5 4
Move(A,B,F2)
Move(B,A,C)
On(A,F2)
Clear(B)
On(B,A)
Clear(C)
On(C,F3)
Clear(F1) 1 2 3
Move(A,B,F2)
Move(B,C,A)
Move(C,F1,F3)
On(A,B)
Clear(F2)
On(B,C)
Clear(A)
On(C,F1)
Clear(F3)
start T
Plan sequence: Move(A,B,F2), Move(B,C,A),Move(C,F1,F3),
Move(B,A,C), Move(A,F2,B)

15. Partially ordered plans
部分的に
ordered
順序に整理した
With respect to
…に関して
To block world problem,
1->2->3->4->5 are totally ordered plans.
Let us tale another example for the shoe-and-socks problem below
An total order plan
Partial order plan
Start
Start
Start
Right Sock
Left Sock
RightSock
LeftShoeOn, RightShoeOn
Right Shoe
End
LeftSockOn
RightSockOn
Left Shoe
RightShoe
Left Sock
In a partial order plan, some steps are ordered with respect to each other and other steps are unordered.
Left Shoe
LeftShoeOn
RightShoeOn
Finish
Finish

16. ? Exercises Ex. (optional)
Write a recursive STRIPS algorithm for block world problem in Pseudo code.