03 Heuristic Search

How to Relax Informally

Informal definition of Relaxation

You have a problem, \(\mathcal{P}\), whose perfect heuristic \(h^*\) you wish to estimate.
You define a simpler problem, \(\mathcal{P}^{\prime}\), whose perfect heuristic \(h^{\prime *}\) can be used to estimate \(h^*\).
You define a transformation, \(r\), that simplifies instances from \(\mathcal{P}\) into instances \(\mathcal{P}^{\prime}\).
Given \(\Pi \in \mathcal{P}\), you estimate heuristic \(h^*(\Pi)\) by computing heuristic \(h^{\prime *}(r(\Pi))\).

Relaxation

Relaxation means simplifying a problem by taking the solution to a simpler problem as the heuristic estimate for the solution to the actual problem.

Relaxation in Route-Finding

Distance from Bucharest to

How do we derive straight-line distances in route-finding by relaxation?:

Problem \(\mathcal{P}\)?: Route finding.
Simpler problem \(\mathcal{P}^{\prime}\)?:
Perfect heuristic \(h^{\prime *}\) for \(\mathcal{P}^{\prime}\)?:
Relaxation Transformation \(r\)?:

Relaxation in the 8-Puzzle

Action: move(X,Y)
Pre: blank(X), adjacent(X,Y)
Add: blank(X)
Del: blank(Y)

Relaxation in the 8-Puzzle?

A perfect heuristic \(h^*\) for \(\mathcal{P}\): Actions = A tile can move from square \(X\) to square \(Y\) if \(X\) is adjacent to \(Y\) and \(Y\) is blank.

How do we derive a heuristic which specifies Manhattan (city-block) distance?:

How do we derive a heuristic which counts the number of misplaced tiles?:

“Goal-Counting” Relaxation for Logistics in Australia?

Propositions \(P\): \(\mathit{at}(x)\) for \(x \in \{\mathit{Sy}, \mathit{Ad}, \mathit{Br}, \mathit{Pe}, \mathit{Da}\}\); \(\mathit{v}(x)\) for \(x \in \{\mathit{Sy}, \mathit{Ad}, \mathit{Br}, \mathit{Pe}, \mathit{Da}\}\).
Actions \(a \in A\): \(\mathit{drive}(x,y)\) where \(x,y\) have a road; \(\mathit{pre}_a = \{\mathit{at}(x)\}\), \(\mathit{add}_a = \{\mathit{at}(y), \mathit{v}(y)\}\), \(\mathit{del}_a = \{\mathit{at}(x)\}\).
Initial state \(I\): \(\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\).
Goal \(G\): \(\mathit{at}(\mathit{Sy}), \mathit{v}(x)\) for all \(x\).

Let’s act as if we can achieve each goal directly:

Problem \(\mathcal{P}\): All STRIPS planning tasks; Simpler problem \(\mathcal{P}^{\prime}\): All STRIPS planning tasks with empty preconditions and deletes; Perfect heuristic \(h^{\prime *}\) for \(\mathcal{P}^{\prime}\): Optimal plan cost \((= h^*)\).

Transformation \(r\) ?:

Heuristic value here?:

Notes:

Optimal STRIPS planning with empty preconditions and deletes is still \(\mathit{NP}\)-hard! (Reduction from MINIMUM COVER, of goal set by add lists.)
Relaxation approximates the perfect heuristic \(h^{\prime *}\) for \(\mathcal{P}^{\prime}\).

How to Relax Formally

Relaxation: Definition for Planning

Definition: Relaxation

Let \(\mathcal{P}\) be a class of planning problems and \(h^*(\Pi)\) denote the optimal plan cost of problem \(\Pi \in \mathcal{P}\).

A relaxation is a triple \(\mathcal{R} = (\mathcal{P}', r, h'^*)\) where

\(\mathcal{P}'\) is another class of planning problems,
\(r : \mathcal{P} \to \mathcal{P}'\) is a transformation between the problems, and
\(h'^*(\Pi')\) is the optimal plan cost of \(\Pi' \in \mathcal{P}'\), such that

for all \(\Pi \in \mathcal{P}\) \(\,\,{\color{blue}h^{\prime *}(r(\Pi)) \le h^*(\Pi)}\), and the heuristic induced by \(\mathcal{R}\) is \({\color{blue}\,h^{\mathcal R}(\Pi) = h^{\prime *}(r(\Pi))}\).

The relaxation is:

Native if \(\mathcal P' \subseteq \mathcal P\) and \(h^{\prime *}(\Pi') = h^*(\Pi')\) for all \(\Pi' \in \mathcal P'\);
Efficiently constructible if there exists a polynomial-time algorithm that, given \(\Pi \in \mathcal{P}\), computes \(r(\Pi)\);
Efficiently computable if there exists a polynomial-time algorithm that, given \(\Pi' \in \mathcal{P}^{\prime}\), computes \(h^{\prime *}(\Pi')\).

The steps involved:

You have a problem, \(\mathcal{P}\), whose perfect heuristic \(h^*\) you wish to estimate.
You define a simpler problem, \(\mathcal{P}^{\prime}\), whose perfect heuristic \(h^*prime\) can be used to admissiblyestimate \(h^*\).
You define a transformation, \(r\), from \(\mathcal{P}\) into \(\mathcal{P}^{\prime}\).
Given \(\Pi \in \mathcal{P}\), you estimate \(h^*(\Pi)\) by \(h^{\prime *}(r(\Pi))\).
Hence goal counting just approximates \(h^{\prime *}\) by number-of-false-goals.

Relaxations: Illustration (Example: Route-finding)

Problem \(\mathcal{P}\): Route finding

Simpler problem \(\mathcal{P}^{\prime}\):

Perfect heuristic \(h^{\prime *}\) for \(\mathcal{P}^{\prime}\):

Transformation \(r\)?:

Native Relaxations: Illustration (Example: “Goal-counting”)

Problem \(\mathcal{P}\): All STRIPS planning tasks.

Simpler problem \(\mathcal{P}^{\prime}\): All STRIPS planning tasks with empty preconditions and deletes

Perfect heuristic \(h^{\prime *}\) for \(\mathcal{P}^{\prime}\): Optimal plan cost \(= h^{*}\)

Transformation \(r\)?:

Relaxation in Route-Finding: Properties?

Distance from Bucharest to

Assume relaxation \(\mathcal{R} = (\mathcal{P}^{\prime}, r, h^{\prime *})\): You are pretending to be a bird!

Native?:

Efficiently constructible?:

Efficiently computable?:

Relaxation in the 8-Puzzle: Properties?

Relaxation \(\mathcal{R} = (\mathcal{P}^{\prime}, r, h^{\prime *})\): Use more generous actions rule to obtain Manhattan distance.

Native?:

Efficiently constructible?:

Efficiently computable?:

What can we do with relaxations?

What if \(\mathcal{R}\) is not efficiently constructible?

Either (a) approximate \(r\), or (b) design \(r\) in a way so that it will typically be feasible, or (c) just live with it and hope for the best.
The vast majority of known relaxations (in planning) are efficiently constructible.

What if \(\mathcal{R}\) is not efficiently computable?

Either (a) approximate \(h^{\prime *}\), or (b) design \(h^{\prime *}\) in a way so that it will typically be feasible, or (c) just live with it and hope for the best.
Many known relaxations (in planning) are efficiently computable, some aren’t. The latter use (a); (b) and (c) are not used anywhere right now.

“Goal-Counting” Relaxation: Properties?

Propositions \(P\): \(\mathit{at}(x)\) for \(x \in \{\mathit{Sy}, \mathit{Ad}, \mathit{Br}, \mathit{Pe}, \mathit{Da}\}\); \(\mathit{v}(x)\) for \(x \in \{\mathit{Sy}, \mathit{Ad}, \mathit{Br}, \mathit{Pe}, \mathit{Da}\}\).
Actions \(a \in A\): \(\mathit{drive}(x,y)\) where \(x,y\) have a road; \(\mathit{pre}_a = \{\mathit{at}(x)\}\), \(\mathit{add}_a = \{\mathit{at}(y), \mathit{v}(y)\}\), \(\mathit{del}_a = \{\mathit{at}(x)\}\).
Initial state \(I\): \(\mathit{at}(\mathit{Sy}), \mathit{v}(\mathit{Sy})\).
Goal \(G\): \(\mathit{at}(\mathit{Sy}), \mathit{v}(x)\) for all \(x\).

Relaxation \(\mathcal{R} = (\mathcal{P}^{\prime}, r, h^{\prime *})\): Remove preconditions and deletes, then use \(h^*\).

Native?:

Efficiently constructible?:

Efficiently computable?:

What approach, (a), (b) or (c), do we take if not construtible and/or computable?:

How to Relax During Search

How to Relax During Search: Diagram

Using a relaxation \(\mathcal{R} = (\mathcal{P}^{\prime}, r,h^{\prime *})\) during search:

\(\Pi_s\): is problem \(\Pi\) with initial state replaced by \(s\)

i.e., \(\Pi = (F,A,c,I,G)\) changed to \((F,A,c,s,G)\) (initial sate \(s\))

This is the task of finding a plan for search state \(s\)

We will be using this notation in the course

How to Relax During Search: Logistics Problem

Initial state \(I\): \(AC\)
- \(AC\) is the state encoding used here, that denotes that truck is at \(A\) and package is at \(C\)
Goal \(G\): \(AD\)
- The goal is for the truck to take package to D from C, then return to A
Actions \(A\): \(drXY\) (drive from \(X\) to \(Y\)), \(loX\) (load package \(X\)), \(ulX\) (unload \(X\))
- Each action also has corresponding \(pre\), \(add\) & \(del\)

How to Relax During Search: Goal-Counting?

How to Relax During Search: Ignoring Deletes?

Conclusion

Robot Question

Question

Say we have a robot with one gripper, two rooms \(A\) and \(B\), and \(n\) balls we must transport.

The actions available are \(moveXY\), \(pickB\) and \(dropB\); say \(h=\) “the number of balls not yet in room \(B\)”.
Can \(h\) be derived as \(h^{\mathcal{R}}\) for a relaxation \(\mathcal{R}\)?

(A): No

(C): Sure, every admissible \(h\) can be derived via a relaxation.

(B): Yes, just drop the deletes

(D): I’d rather relax at the beach

Summary

Relaxation is a method to compute heuristic functions.
Given a problem \(\mathcal P\) we want to solve, we define a relaxed problem \(\mathcal P'\).
- The heuristic is obtained by mapping into \(\mathcal P'\) and taking the solution to this simpler problem as the estimate.
Relaxations can be native, efficiently constructible, and/or efficiently computable.
- None of these is a strict requirement for usefulness.
During search, the relaxation is used only inside heuristic computation for each state.
- It does not affect the rest of the search (This can be confusing, especially for native relaxations such as ignoring deletes.)

Goal Counting: Limitations

The goal-counting approximation, which is essentially \(h=\) “count the number of goals currently not true” is a very uninformative heuristic function:

The range of heuristic values is small \((0 \dots |G|)\).
We can transform any planning task into an equivalent one where \(h(s) = 1\) for all non-goal states \(s\).

How?:

Ignores almost all structure: heuristic value does not depend on the actions at all!

Is \(h\) safe/goal-aware/admissible/consistent?:

We will see how to compute better heuristic functions in the next Module.