Optimal Packing of High- Precision Rectangles By Eric Huang & Richard E. Korf 25 th AAAI Conference, 2011 Florida Institute of Technology CSE 5694 Robotics.

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Optimal Packing of High-Precision

RectanglesBy Eric Huang & Richard E. Korf

25th AAAI Conference, 2011

Florida Institute of TechnologyCSE 5694 Robotics & AIMindaugas Beliauskas

Problem Overview

Rectangle-packing problem – finding smallest enclosing rectangle that can contain a given rectangles without overlap

Why Is It Important?

Many practical applications Mechanical prospective

Loading a set of rectangular objects on a pallet without stacking them

Cutting stock – requires least material to be unused

Computer science prospective Scheduling and allocating

contiguous memory addresses to programs, where width of rectangle – time required to run it, height – memory it needs

Time, s

Mem

ory

, G

b

Sub ProblemMinimal Bounding Box

Problem

Finding the smallest rectangle that can contain rectangles of sizes:

Such a problem – minimal bounding box problem Finds a bounding box of least area that can contain a

given set of rectangles

In other words, here number and dimensions of boxes are known

Sub ProblemContainment Problem

Packaging of given rectangles in a given bounding box of rectangles of sizes:

Models rectangles positions over possible positions in the bounding box

In other words, the enclosing rectangle is known

Algorithm that solves minimal bounding box problem calls algorithm that solves containment problem as a subroutine

Unfeasible Problem

1968 Meir and Moser proposed a problem:

Finding the smallest square that can contain an infinite series of rectangles of sizes Rectangles cannot overlap Rectangles are unoriented – can be rotated 90° Unit square has exactly enough area for all rectangles

No space can be wasted Suggest infeasible task

Solution Strategy

Minimal bounding box problem Boxes of all sizes are generated and tested in non-

decreasing order of area until all feasible solutions of smallest area are found Lower bound – sum of all rectangle areas Upper bound – setting the height of the tallest box, and

placing rectangles to the first available position on the right

Containment problem All possible locations tested for each rectangle up to N

Minimum Bounding Box Problem

Generating all subset sums prior to searching – width and the height will be a subset sum of rectangles dimensions Unoriented problem - all widths and heights need to be

considered together

Excluding infeasible pairs of dimensions Example: for N = 4 consists four boxes of 60x30, 30x20, 20x15

and 15x12. Bounding box width of 57, requires 30 + 15 + 12, can’t have height of 47, because this combination requires rotating 12x15 box. Two rectangles cannot be simultaneously vertically and horizontally Bounding box of size 57 x 47 is then eliminated

Learning From Infeasible Attempts

Minimal bounding box problem calls containment problem When placing rectangles, the algorithm will run into

infeasible solution – the dimension of bounding box needs to be increased

Too small increment may cause to to the same partial solutions we already explored

Example

of oriented

boxes:

Containment Problem

We prune the sums of rectangles, when they overlap or exceeds the limits

Assigning x-coordinates dynamically Place rectangle on the left side of bounding box For the assigned rectangles coordinates, add

coordinate of unassigned rectangle (width or height) into each set and insert the new sums back

Build a set of subset sums

Once possible x-coordinates are assigned, the y-coordinates need to be assigned

Containment Problem

Previously used solution: Perfect Packing Transformation – creation of 1x1 rectangles in addition to placed existing rectangles to check for all empty left space Results in # of original rectangles + # of new 1x1

rectangles For N = 15, requires creating 1.5 billion of 1x1

rectangles, because the problem is scaled up by the least common multiple This requires too much memory and time

Suggested solution: Widening Existing Rectangles & Turning Empty Space Into Large Rectangles

Widening Existing Rectangles

Once x coordinates are assigned, we have to assign y coordinates Partial solution

example:

For any assignment of y-coordinates the space right of 40x10 rectangle must be always empty, hence it is replaced with 60x10 rectangle. Same procedure for 10x20 and 20x10 rectangles

Method used to generate possible rectangles’ y-coordinates

Turning Empty Space Into Large Rectangles

Now, instead of creating 300 1x1 rectangles, the new method suggests creating 10 30x1 for single hash marks and 20 30x1 for double-hashed empty spaces, instead of 900 rectangles

Method used to fit unplaced rectangles

Fitting Unplaced Rectangles

Finally, after packing transformation, we assign y-coordinates by asking which rectangle can be placed in a given empty corner We limit the y-coordinate to subset sums of height in the original

instance For an empty corner, we disallow assigning any rectangle that is not in a

subset sum, because only rectangles to create perfect packing can be assigned

We generate subset sums for every x-coordinate solution

Result is far fewer values in the possible solution set, because we already picked the rectangles’ orientation

Experimental Algorithm Results

Column 1 – size of the problem

Column 2 – method when problem scaled completely in integers

Column 3 – testing only those bounding boxes, whose widths and heights are subset sums of rectangles widths and heights

Column 4 – rejecting mutually exclusive bounding boxes widths and heights

Column 5 – using information of infeasible attempts

Experimental Algorithm Results

LCM – least common multiple

Packer needs to compare minimal bounding box At N=12, the number of

boxes to be compared exceeds 32bit integer

Most of optimal solutions have width of 1/2

Experimental Algorithm Results

Comparison to different algorithms of computing times

Empty Space – precomputes all subset sums prior searching over x-coordinates

Dynamic – dynamically computes subset sums

HK11 – added ability to learn and prune unfeasible subtrees

Solution of Unfeasible Problem

Recall the infeasible problem mentioned in the beginning

Main infeasibility – no space can be wasted, since the series of rectangles equals unit square

A packing solution for rectangles in the unit square of N = 50000 is presented on the right!

Conclusion

New benchmark consisting of instances with rectangles

Techniques presented Dynamically using subset sets to limit the number of possible positions

(Widening Existing Rectangles and Turning Empty Space Into Large Rectangles)

Rules to filter out subsets (Rejecting pairs of mutually exclusive subset sums)

Methods to learn from infeasible trees (Recalls infeasible result if noticed)

Solves problems problems up to two orders of magnitude faster

Solved infinite series of rectangles in the unit square, proving that such a packing exists

Thank you!

Time for questions!