This Program tackles the following problem:
Imagine a set of points in 3D space. They represent the top right corner of a rectangular Box. This position vector defines our object, where the bottom left corner sits in the origin. It has a Volume denoted by:
V = x * e̅,x + y * e̅,y + z * e̅,z
We got a bunch of these scattered in space, and want to find another set of Boxes, which enclose all of these Points, while minimizing the volume difference between Box an Point:
Vᵦ,d = ∑ⁿ(Vᵦ - Vᵢ) , where Vᵦ = Volume of outer Box
Vᵢ = Volume of Point i
n = Number of Points in outer Box
Let's call it Death Volume! ⛧
Shown here is a 2D examplification with only one Point:
^
y |
|------------------B
|//////////////////|
|---------------P//| B: Box(x,y)
| |//| P: Point(x,y (but different ones))
| |//| /: Death Volume
| |//|
| |//|
| |//|
| |//|
| |//|
-------------------------->
x
We apply a classic divide-and-conqer strategy: a tree. More precisely, a 3-dimensional bipartion tree or short, kdTree. I'll just describe what we do here, you can read read the rest on wikipedia.
The tree takes a set of Points, an axis, a discriminant and a depth.
It splits the set of Points along the axis on the discriminant, which is a value exactly in the center between Zero and the value of Box in that dimension, respectively that value plus the offset from the previous split in this dimension. Then the axis is rotated, like so
axis = (axis + 1) % 3and the depth is decremented. Those values are passed to the new children Nodes, and the split is recursively applied, until the depth is Zero.
+-----+ +--------+-------(This is a split)
|Depth| | |
|-----| | |
| 0 | V 1 V
|-----| /--- ---\ /----+-------(These are Nodes;
| | --- --- /- | their Keys are
| 1 | 2 3 <-- | ordered in a breadth
|-----| /- -\ /- -\ - first walk)
| | -- -- -- -- -/
| 2 | 4 5 6 7 <-/
|-----| /-\ /-\ /-\ /-\
| | - - - -- - - -
| 3 | 8 9 10 11 12 13 14 15
| | ^
+-----+ |
|
|
+----(Those are Leaves)
We call the old space Parent and the new ones leftChild, if the Point- dimension in question was smaller than the discriminator and rightChild, if its Points had greater values in that dimension.
After the programm halts, the old Pointspace is partitioned into 2 ^ (depth) Subspaces.
If such a split makes sense in the context of death volume minimization, we have to do some (fairly easy) math:
To adress the 'quality' of a split, lets look at an example in 2D:
^
y |-----------B|-----------A
| | |
| | |
| | P |
| | |
| | |
| P' P'| P |
| | |
| P' | P|
| | |
| | |
-------------|------------>
d x
In the case the split is done, the total death volume is just the sum of those of the two spaces A and B, where
Vₐ,d + Vᵦ,d = ∑ᴺ(Vₐ - Vᵢ) + ∑ᴷ(Vᵦ - Vᵢ')
In the case of no split:
^
y |------------------------A
| |
| |
| P |
| |
| |
| P' P' P |
| |
| P' P|
| |
| |
-------------------------->
x
the death volume of the whole thing is
Vₐ,d|notB = ∑ᴺ(Vₐ - Vᵢ) + ∑ᴷ(Vₐ - Vᵢ')
If we subtract the two equations from another, we get:
Vₐ,d|notB – Vₐ,d + Vᵦ,d = ∑ᴺVₐ - ∑ᴺVᵢ + ∑ᴷVₐ - ∑ᴷ Vᵢ'
–(∑ᴺVₐ - ∑ᴺVᵢ + ∑ᴷVᵦ - ∑ᴷ Vᵢ')
= ∑ᴷ(Vₐ - Vᵦ)
This leaves us with a quantity i will call ∆V,d(A,B). As we will see, this is a excellent measure of a split.
The situation presented above corresponds exactly to parent- and children-Nodes in the kdTree, where A = Parent and B = leftChild. We can now assign to each Node a ∆V,d, where leftChild-Nodes get ∆V,d(Parent, leftChild) and rightChild- Nodes inherit those of their parents.
+---+
| |
| P | (x,y)
| |
+---+
|
|
+-------+-------+
|x<d x>d|
| |
+---+ +---+
| | | |
| L | (d,y) | R | (x,y)
| | | |
+---+ +---+
After all, their volume is identical, so they make no difference in terms of ∆V,d.
Now we can build the tree with a modified split function, which takes ∆V,d into consideration. We end up with a list of leaves, in which each has a 3D Point of their top right corner, and a ∆V,d - Value. Now we just compile a list of those having a positive ∆V,d, plus the largest non-empty Node and sort them in descending order.
Here are our best N Boxes, which also minimize the death volume.
Remember the 2D simplification from the beginning?
Here is what happens when i try to draw a three dimensional volume difference in ascii:
^ /----------------------+
? | /---///////////////////////|
| /--- //////////////////////////|
| +-------//////////////////////////|
| |\\\\\\\\\\\\\\\\\\\\\\\\\\-+/////|
| |\\\\\\\\\\\\\\\\\\\\\\\\\\-|/////|
| |\\\\\\\\\\\\\\\\\\\\\\-\\\\|/////|
| +--------|--#######+\\\\\\\\|/////|
| | | #######|\\\\\\\\|/////|
| | | #######|\\\\\\\\|/////|
| | | #######|\\\\\\\\|/////|
| | | #######|\\\\\\\\|/////|
| | | #######|\\\\\\\\|/////+
| | /-+----------\\\\\\\\+//////-
| | /---- |\\\\\\--////--
| | /-- |\\\\-////--
| +------------------+----+ --
+------------------------------------->
!
I h̗̞opͫ̍̅e͚͚ͭͤ y̜̝ͤ̆oͫ͊ụ̗̥̩̤̦͗͊̎̀̔̇ dó̜̪ͧn'̟͎̐̎t̾̔̄ m̘͙̝̞͓̘̼̬̩̖̗͓̬̱̈́̊͛͐̈͑̏ͯ̿̅̃̾ͩ́̚͝͠ͅin̨̥͚͕̙̼̥̖̱̘̭ͤ̐̇̓ͧ͑̊̎ͧ͒d̴̈́͋̅̿͋̈́ͩ̌͂̇̚
Just cd into the root and run it like
python3 run.pyThe csv format is not really standard, basically
Name,Box.x,Box.y,Box.z,Item.x,Item.y,Item.z,\n
Maybe when we start to care for proper .csv implementation this can change.
Here is an explanation of run.py.
from classes.tree import TreeControl
depth = 13
divCrit = 0.5 # tweaks the discriminant, normally set to 0.5, but you know, its your choice
startAxis = 0 # pretty much self explanatory
MaxNumPoints = None # cap of your points you want to partition
inPath = 'assets/1M_old.csv'
outPath = 'assets/kandidates.csv'
t = TreeControl()
t.getInitialItemBoxes(inPath, MaxNumPoints)
t.getInitialValues() # calculates initial total Volume, and Death Volume
t.initializeTree(depth, divCrit, startAxis)
p = [i[0] for i in t.itemBoxes]
t.tree.insert(p)
t.tree.grow()
t.isNumPointsConst() # sanity check
t.getBestNodes() # retrieves Nodes with positive ∆V,d. one can access them via TreeControl.bestNodes
t.writeNewBoxesCSV(outPath)
t.printInfo(extended=False,leaves=False ,bestN=30)- rewrite assignment of Points into the new Boxes
- pick bestN from NewBoxes
- Rotate / compare initial Placement of Hyperplane
- Draw Graphs!!