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

Sub Problem Minimal 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 Problem Containment 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 , 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 = 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!