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This paper aims made it clear that:
1) time, scheduling, resource
2) hierarchical planning
3) non-deterministic planning problem
Time, scheduling and resource
Classic planning can do:
1) what action
2) what order do
not do:
1) when the action occurs
2) How long
3) considering resource constraints
classic plan can not discuss the time and therefore can not solve the scheduling problem, we can not discuss resources constraints .
So how do you solve it?
First plan, after scheduling.
- What action do the planning stage
- In what order to do the planning stage
- When scheduling stage action occurs
- How long scheduling stage
- Consider scheduling resource constraints stage
The whole problem into a planning phase and a subsequent scheduling stage . During the planning phase election action, consider the order of constraints to meet the target problem. In the scheduling stage time information added to the plan to meet the resource constraints and deadlines .
Shop scheduling problem
Problem Description:
- Each action has a duration (DURATION) and a set of actions requested resource constraints .
- Solutions must provide a start time of each action, and must meet time constraints and resource constraints order
- Suppose the cost function is a completion time (from start to finish for a long time)
- Represented by the number of resource - e.g. Inspectors (2) represent two checker
Time Scheduling ignores resource constraints: critical path method
Partial Order plans through a path beginning from the start to the end of the finish.
Critical Path is the longest time span of the path, took a critical path delays the start time of the action plan will extend the time.
Each action has not on the critical path time window, this operation gives the earliest possible starting time and latest ES time the LS may start , defined as [ES, LS].
All actions together constitute the ES and LS scheduling problem.
LS-ES is called slack action. Each action on the critical path with no slack .
The following formula can be defined as the ES and LS and a contour thereof is calculated dynamic programming algorithm (operation of a given partial order, find the minimum scheduled duration, minimizing completion time). A and B are operation, A≺B A represents operation before B:
\begin{aligned}
& ES(Start) = 0 \\
&ES(B) = max_{A \prec B } {ES(A)+ Duration(A)} \\
&LS(Finish) = ES(Finish) \\
& LS(A) = min_{B \succ A}{LS(B) - Duration(A)} \\
\end{aligned}
Wherein, ES (B) is the maximum value of those immediate predecessor earliest completion time of the operation.
This process is repeated until each action is assigned a value ES.
There are resource constraints Time Scheduling
When the introduction of resource constraints, can not overlap with a resource use.
Hierarchical planning
Hierarchical decomposition - why the need to plan at a higher level of abstraction?
The answer: a lot of tasks into a single action; for large-scale problem, which is totally unrealistic
Primitive operation (PA, primitive Action) : can not be refined, provided with a standard - Effect Mode
rise operation (HLA, High Level Action) : Each HLA may have one or more refinement operation sequence, wherein each action can be an HLA or a PA.
Contains only a primitive action is called HLA refinement to achieve the HLA's .
HLA exactly one may realize; there may be more than one possible implementation:
- When a HLA happens when there is a realization that we can calculate from the premises and the effect of this implementation is a prerequisite and the effect of HLA, HLA then this in itself can be seen as a primitive action.
- When there is more than one HLA possible implementation, there are two options:
1) Search primitive solution: search an achievable work;
2) search for abstract solution: direct reasoning, no search implementation
High-level planning is HLA sequence, to achieve high-level planning is spliced to achieve each of the HLA HLA sequence.
Search primitive solution
Answer: In the current planning repeatedly selected a HLA, replacing it with its refinement, until the plan reaches the target.
For a non-hierarchical, b each state has available forward motion planner state space, the cost is O (b ^ d).
For a HTN (hierarchical task network) planner, assume a general refinement of the structure: every non-primitive action has r possible refinement, each refinement has k action. If the primitive actions d layer, the layers below the root is $log_kd$, the internal junction refinement points are $1+k+k^2+...+k^{log_k{d-1}}=(d-1)/(k-1)$. Each internal node having r possible refinement, thus constructed $r^{(d-1)/(k-1)} $a possible decomposition tree. R k large and small can result in substantial savings.
Search abstract solution
If you get high-level planning can be shown to achieve the target, then this is an abstract high-level planning solution. So we committed to this plan, every step of this plan will be refined. In this way, the search deflator get the order.
Each statement can achieve the goal of high-level planning is on "It has at least one implementation to achieve the goal of" significance can achieve our goals. This feature is known as HLA described downward refinement characteristics .
Reachable sets: Given a state s, HLA reaches a set denoted as h REACH (s, h), is an implementation of any of the HLA reachable state set.
Search senior planning to find a high-level planning that can reach the target set intersect.
HLA effect
when the HLA is only just a realization, we can calculate the effect of the realization of the premise and the HLA out of (and premise) effect (Exercise 11.3)
In many cases a HLA may have an infinite number of realization, we can only approximate results, which may produce any swing sets can reach
a HLA action h using two approximate:
optimistic description (optimistic description) $REACH^+ (s,h)$may exaggerate set reachable
pessimistic description (pessimistic description) $REACH^-(s,h)$may underestimate the reachable set.
REACH^-( s , h ) \subseteq REACH( s , h ) \subseteq REACH^+( s , h )
How accurate description of the target under test? (To determine whether a plan to achieve the target)
under precise description, a plan or work (it can reach the target set intersection) or does not work (it can reach the target set disjoint).
How to target test under similar description?
If pessimism can be set and reach goals set intersection, then this plan is to work, that is certainly solvable.
If optimistic to reach the target set disjoint sets, then this plan is not able to work.
There are intermediate states in the approximate Description: If you set optimistic goals and intersect, but do not intersect with the target set pessimism, then we can not determine whether the program is working. When this happens, how do?
When this happens, you can eliminate this uncertainty by further refinement of planning.
Non-deterministic planning problem
No planning for the sensor (sensorless planning) no observation environment;
for partially observable, non-deterministic environment contingency planning (contingency planning);
for the unknown environment online planning (online planning) and re planning (replanning).
Coloring problem
to consider this question: Given a chair and a table, the goal is to match them - have the same color. The initial state we have two cans of paint, but the paint color and furniture unknown. Only when the start of the table, as the eyes of Agent:
the Init (Object (the Table) ∧Object (Chair) ∧Can (. 1 C) ∧Can (C 2) ∧InView (the Table))
. Goal (Color (Chair, C) ∧Color (Table, c))
There are two operation modes: the tank cover removed from the pigment, paint the object using the open jar. Allow the premise and not effect the action list comprising variable variables (as it is in the case of partially observed; in the case of fully observable, which is not allowed)
- Action(RemoveLid(can),
PRECOND: Can(can)
EFFECT: Open(can)) - Action(Paint(x,can),
PRECOND: Object(x)∧ Can(can)
∧Color(can,c) ∧ Open(can)
EFFECT: Color(x,c))
Agent requires that the object is a (one at a time) into the line of sight of the operation mode:
the Action (the LookAt (X),
PRECOND: the InView (Y) ∧ - (X ≠ Y)
the EFFECT: the InView (X) ∧¬InView (Y))
To solve part of the problem can be observed, the actual action Agent, which will provide sensors perceive information, but when it is time to plan, it will need its sensor model. Chapter 4, this model is a given function, PERCEPT (s). For planning, we perceive mode (percept schema) to extend PDDL:
- Percept(Color(x,c),
PRECOND: Object(x) ∧ InView(x)) - Percept(Color(can,c),
PRECOND: Can(can) ∧ InView(can) ∧ Open(can))
For a fully observable environment for each stream (fluent) we will not have a prerequisite of modes of perception.
On the other hand, a sensor-Agent does not perceive pattern.
Sensorless planning
Contingency planning
Information sharing
Experiment download:
https://github.com/yyl424525/AI_Homework
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