寻找产品最佳使用的算法

Question

I'm trying to solve a problem in Javascript allowing for optimal use of area with flooring products. I'll be either implementing this or looking for the algorithm in Javascript. The following is the problem:

Consider that I sell rolled carpets with width dimensions of 12, 13 and 15 ft. The length can be any size (because it's rolled).

Let's say I have 3 rooms:

• 25 x 10 (hallway)
• 30 x 30 (large room)
• 5 x 5 (closet)

How can I determine the minimal amount of product to use based on the available size options? I'd also like to ensure any wasted cuts are re-used. For larger rooms it's fine that there are seams to connect the rolled pieces. For example: in the hallway I can use a 15ft wide product and I'll have a roll remaining of 10 feet. I can use that to fill the closet.

The goal is to find the most optimal fit of product accounting for waste re-use.

Answer 1

For a large room of Width, and starting with our largest roll:

The integer, k15, of Width / 15ft gives the number of 15ft wide lengths.

Get Remaining Width = Width - 15ft x k15.

The integer, k13, of Remaining Width / 13ft gives the number of 13ft wide lengths.

Get next Remaining Width" = Remaining Width - 13ft x k13.

The integer, k12, of Remaining Width" / 12ft gives the number of 12ft wide lengths.

Assuming that you will never have a room that is an integer of any combination of these three sizes, you will always have offcuts.

I think, area optimization and polgon geometry are two extremly complex fields, but I hope I have gone part way into at least defining your problem.

Answer 2

For three carpet widths of 12ft, 13ft & 15ft:

Assuming three rooms, with carpet rolls infinitly long, the algorithm for determining the most efficient use is:

Give each room a dimension of: x0" & x1" for room one, y0" & y1" for room two, and z0" & z1" for room three.

Sort this list such that the smallest dimension of each room is written first, and the room with the largest 'smaller' dimension is also written first.

If x1" > x0", write it: (x0", x1") => (x0', x1') If y1" > y0", write it: (y0", y1") => (y0', y1')
If z1" > z0", write it: (z0", z1") => (z0', z1')

And:

If z0' > x0' > y0' then write => (x0, x1), (y0, y1) & (z0, z1)

Given this ordered set, begin with the largest room, X, with its smaller dimension as x0, the greater as x1.

Get the second largest room, Y.

We now need to find how we can lay the carpet in room X, such that the OFFCUT will fill room Y in ONE or more lengths.

We need to choose between our three widths, so that the 'offcut will be as wide as room Y'.

If x0 is 30, and y0 is 10, then 12ft + 13ft + 15ft gives us 30ft with 10ft spare to use in room Y.

The same carries to room Z.

Problem solved.