Counting Linear Regions

The question "how many" has motivated me in what little math I have done. For example, how many tetrahedron overlaps are there? That led me to the question of how many equivalent linear spaces there are. I used a recursive formula to define linear space in the first entry on this blog. Now I will present the closed form for f(d,b), the number of nonempty regions in a linear space of d dimensions and b boundaries. For example,
f(3,b) = 1 + b + ∑_x=0..b−1x + ∑_y=0..b−1∑_x=0..y−1x
By definition,
f(d,b) =
f(d,b−1) + f(d−1,b−1) =
f(d,b−2) + f(d−1,b−2) + f(d−1,b−1) =
f(d,0) + f(d−1,0) + f(d−1,1) + f(d−1,2) + ... + f(d−1,b−1) =
1 + ∑_x=0..b−1f(d−1,x) =
1 + ∑_y=0..b−1(1+∑_x=0..y−1f(d−2,x)) =
1 + b + ∑_y=0..b−1∑_x=0..y−1f(d−2,x) =
Removing subscripts for brevity,
1 + b + ∑∑f(d−2,x) =
∑⁰1 + ∑¹1 + ∑²(1+∑f(d−3,x)) =
∑⁰1 + ∑¹1 + ∑²1+∑³f(d−3,x) =
∑⁰1 + ∑¹1 + ... + ∑^d−11 + ∑^df(0,x) =
∑⁰ + ∑¹ + ... + ∑^d−1 + ∑^d =
∑_{x>−1,x<d+1,x<b+1}∑^x
I have always liked figurate numbers. For example, the triangular numbers are 1, 3, 6, 10, 15, .... Consider points on a page. Each line has one more point than the last. Then the total number of points is ∑², the triangular number P₂(b−1). Now make a stack of pages with successive triangular numbers on them. The total number of points in the stack is ∑³, the tetrahedral number P₃(b−2). In general, ∑^d is the polytopic number P_d(b−d+1). The diagonals of Pascal's triangle of binomial coefficients are the polytopic numbers. Thus, ∑^d = P_d(b−d+1) = ^(b+1)!/_d!(b−d+1)!, and f(d,b) = ∑_{x>−1,x<d+1,x<b+1}^(b+1)!/_x!(b−x+1)!.

Drawing Abstractions

This post shall contain observations of interest more to computer programmers than to mathematics enthusiasts.

My work experience is in creating models of hardware designs, and using the models to verify the designs by comparing the outputs of the two in response to various inputs. The difference between the models I write and the designs I verify is that the designs are synthesizable into real hardware. The models I write to verify the synthesizable designs contain more architectural than microarchitectural detail. The difference between architecture and microarchitecture is that the end user of the hardware does not have direct control over or direct access to the microarchitectural details. Corroboration between the model and design is merely anecdotal evidence of the correctness of the design. However, this sort of anecdotal verification is often essential to verify a hardware design.

The program described in the last post is a procedure to find mathematical abstractions. I can think of it as a model of some of the mathematical structures governed by theorems and definitions given in previous posts. Even though I would doubt the correctness of the theorems if I found it impossible to model the intended structures, my goal in modelling the structures is not to verify the theorems. In fact, success in modeling the structures would be only anecdotal support of the theorems. My goal in modeling the structures is to satisfy the curiosity which helped to motivate me to produce the theorems. In that way, I could more easily move on to related topics.

The challenge of modeling mathematical structures differs from the challenge of modeling hardware designs. Hardware designs are more concrete than the mathematical structures I am modeling. I have an intuitive sense of what a portion of the model must do in order to model the corresponding portion of the design. However, in my simplex overlap model, the sidednesses are represented by 1s and 0s, and sometimes I lack intuition as to whether some intermediate 1s and 0s are correct. This intuition about the correctness of intermediate results is essential to debugging a computer program. Other than by direct inspection of and reasoning about the program source code, the only way I know to debug a program is by taking it one step at a time, and checking the results after each step. For me, the direct inspection technique works better when it is used in conjunction with the intermediate result technique.

Often, the intermediate results in my simplex overlap program are the sidedness relations of linear spaces. At least for fewer than 4 dimensions, I do have an intuitive notion of what a linear space is. I can represent such a linear space by a drawing. Thus, to check that a sidedness relation does represent a linear space, I attempt to draw lines and points which satisfy the sidedness relation. Unfortunately, converting the sidedness relation to a drawing requires a lot of trial and error. This leads me to desire a program which could draw the lines and points automatically, given any linear sidedness relation. If the program failed on a particular sidedness relation, when it succeeded in much less time on equally complex sidedness relations, I would suspect that the sidedness relation it failed on was not in fact linear. In that way, I could detect faulty intermediate results in the simplex overlap program.

Another advantage of having drawings of intermediate linear space results in the simplex overlap program is that I could use those drawings to check other intermediate results which are not linear spaces. For example, the inside of a triangle overlap would be obvious from its drawing.

Yet another advantage of a program to convert the sidedness relations of a linear space into a drawing is that when I do find all of the tetrahedron overlaps, then I will be able to more easily publish drawings of them. I suspect the actual drawings will be more inspirational and convincing than a simple count of their number.

Now I will describe the program which converts sidedness relations to drawings. It starts out with random coordinates for the specified number of points and boundaries. The coordinates of a boundary are the coordinates of a point on the boundary, together with the coordinates of a unit normal to the boundary. Given a set of coordinates, the program calculates a score by taking the sum of the signed perpendicular distances of the points from the boundaries. The program attempts to increase this score by randomly changing one coordinate at a time. The program reverts the change if it does not improve the score. As the score improves, the way the coordinates are changed changes from pure random to a sort of Brownian motion.

I wonder whether this program to convert sidedness relations to coordinates would be easier to debug than the simplex overlap program it is intended to help debug. I suspect it would be easier to debug because coordinates are less abstract than sidedness relations. Coordinates are concrete in that they can be displayed on a computer screen in a graphical way. To me, graphics are less abstract than symbols.

Overlap Program Description

By the Continuity Theorem (see previous posts), the migration operation spans all linear sidedness spaces, starting from any one. Similarly, migrations are sufficient to find all simplex overlaps of a given dimension. I need only check each migration from each overlap found so far, and find if it is equivalent to one found already. To simplify the equivalence check, I will impose an ordering on equivalent sections and remember only the first in order. The program to do this is described in the following.

Start with a single entry list. Finding this initial overlap example is the most difficult part of the program. Create the example overlap by modifying an overlap with one fewer boundary than desired. First choose a boundary. Then choose a boundary in the section by it. In the one dimensional case, choose a region and divide it by a new boundary. In the next dimension up, divide regions by a new boundary close to the chosen one, passing through the new boundary in the previous dimension. Continue up to the original dimension.

One question is whether the above method could be used to find all overlaps without using migrations. This would be less efficient than using migrations as below, because the above method would find all possible sidedness relations. The migration method below finds all overlaps while going through only some of the possible sidedness relations.

Now fill out the list with a representative from each overlap equivalence class. For each overlap in the growing list, for each possible migration, consider the resultant overlap. If the overlap is not equivalent to any on the list, then add it to the end of the list.

Determine equivalence by keeping a list of sections which compare as the least of all permutations and inversions. By definition, the equivalence class is all permutations and inversions. A one to one mapping is a permutation. An inversion is a reversal of all bits in the sidedness relation indexed by a boundary.

First Principles

This entry shall contain some philosophization. Philosophization, sidedness, get it?

What are some of the advantages and disadvantages of deriving mathematical results from first principles, instead of basing new results on published results? Both techniques are useful. I found it easier to get the information I needed to design a program to count tetrahedron overlaps by assuming points do not lie on lines instead of assuming they always lie on lines as in (widely published) incidence geometry. I do not doubt someone more familiar with incidence geometry would find it easier to use than I do. Both techniques, using first principles and using published results, encourage insight and interrelationships between different parts of mathematics. I already tied sidedness geometry to linear spaces, and I can imagine eventually learning enough about incidence geometry to find how sidedness and incidence geometries relate. The inspiration for trying to relate incidence and sidedness is that they are so obviously similar. So far these are advantages for both the first principle technique and the published result technique.

The difference between first principles and published results is that a hypothetical person who is intelligent, but hitherto illiterate, could understand an argument from first principles, but the same hypothetical person could never understand an argument based on published results. I am much closer to such an hypothetical person than a professional mathematician, so I admit a bias toward first principles. Perhaps the general public is closer to that hypothetical, intelligent but hitherto illiterate, person than a professional mathematician, and for that reason my bias is mitigated. However, I do find myself peculiarly deficient of memory when reason will suffice. For some reason, I associate this peculiarity of mine with creativity. Thus, in my mind, I consider first principles to be more creative than I imagine reasoning from published results would be.

Mathematics is all about abstraction, generality, and precision. That is, start with nothing, end up with everything, and make no mistakes along the way. With credentials like that, why don't we have more mathematicians in the white house? Both creativity and abstraction are, to borrow a phrase from Marvin Minsky, suitcase words, meaning many different things. Since I do not have a superior understanding of the complexities of abstraction, I must rely on my intuition that first principles are similar to abstraction. Since even I was able make a connection from first principles to linear space, I must assume connections to linear space are not the most general. However, my fantasy is that the number of overlaps is always prime. Now that would be a promising path from nothing to everything! As to precision, I draw from personal experience. In my profession, we discourage designers from verifying their own designs; we prefer for the verification to be done by someone else. Similarly, the correctness of my, and presumably other people's, proofs would be assisted by the attention of others. Since results derived from published results are more likely to be reviewed by others, a result from first principles has less chance for such review. While first principles are more abstract than published results, published results are more well reviewed than first principles, so neither has superior advantages.

Overlap Examples

Here are the 11 triangle overlaps. The top left one differs from the disjoint one by a single migration. I call this kind of migration, where a point jumps across a boundary, a point migration.



Here are three tetrahedron overlaps. The right two differ from the disjoint one by single migrations. The middle one is a point migration since it is a point jumping across a boundary. The rightmost one is a segment migration since it is a segment jumping across another segment.

Tetrahedron Overlaps

Linear Space Equivalence

Two linear spaces are equivalent iff they are isomorphic to the same finite set of nonparallel noncoincidental linear surfaces. Two finite sets of nonparallel noncoincidental linear surfaces are equivalent iff they are isomorphic to the same linear space. A finite set of nonparallel noncoincidental linear surfaces are said to be equivalent to a linear space iff they are isomorphic.

Migration

In space Sⁿ = (B,P,S,n), regions R₁ and R₂ are neighbors with respect to b₀∈B iff R₁(b) = R₂(b) and R₁(b₀) ≠ R₂(b₀), for b∈B₀, where B = B₀∪{b₀}. In space Sⁿ = (B,P,S,n), region R₂ is the opposite of region R₁ iff for b∈B, R₁(b) ≠ R₂(b) ⇔ R₁ has a nonempty neighbor with respect to b. Two spaces S₁ⁿ = (B,P₁,S,n) and S₂ⁿ = (B,P₂,S,n) are migrations of each other iff for each region R in S₁ⁿ and S₂ⁿ, one or both of the following is true. R is empty in S₁ⁿ ⇔ R is empty in S₂ⁿ. Or, R is empty ⇔ the opposite of R is not empty.

Example

Regions A, B, and C have empty opposites. Region A is the opposite of region E. Region E is the opposite of region D.

Migration Theorem

If S₁ⁿ and S₂ⁿ are migrations of each other then, S₁ⁿ is linear ⇔ S₂ⁿ is linear.

Without loss of generality, assume S₁ⁿ is linear, R is empty in S₂ⁿ,
and its opposite is not empty in S₂ⁿ. Suppose S₃ⁿ is any subspace of S₂ⁿ. If R is not a region of S₃ⁿ, then it is linear because its identical subspace in S₁ⁿ is linear. If R is a region of S₃ⁿ, then so is its opposite, so the number of nonempty regions in S₃ⁿ is unaltered, so it is linear, and the theorem is proved.

Continuity Theorem

There exists continuous L:R→{m linear surfaces in Rⁿ}, such that L(r) is nonparallel and nonconicidental for r ≠ r₀∈R, only if there exist r₁, r₂∈R, and linear spaces S₁ⁿ and S₂ⁿ, such that r₁<r₀<r₂, L(r₁)≡S₁ⁿ, L(r₂)≡S₂ⁿ, and S₁ⁿ and S₂ⁿ are migrations of each other.

As a set of linear surfaces changes continuously, its regions cannot suddenly become empty. Before becoming empty, a region must become infinitesimally small. As a region becomes infinitesimally small, so does its opposite. Thus, as a region becomes empty, its opposite becomes nonempty.

Simplex Overlap

A simplex overlap is two disjoint sets of boundaries, each of which is a simplex. The inside of a simplex overlap Sⁿ = (B,P,S,n) is a space S₁ⁿ = (B,P₁,S,n), such that P₁ is the union of the insides of the simplices. Two simplex overlaps, (M₁,M₂), and (N₁,N₂), where M₁, M₂, N₁, N₂ are the simplices of the overlaps, are equivalent iff there exists one to one mapping g:M₁∪M₂→N₁∪N₂, such that g(M₁) = N_i for some i∈{1,2}, and for each b∈M₁∪M₂ the section of the inside of (M₁,M₂) by b is equivalent to the section of the inside of (N₁,N₂) by g(b). There are 11 equivalence classes of triangle overlap. How many equivalence classes of tetrahedron overlap are there?

Sidedness Geometry

Linear Space

Take points and boundaries as undefined. Define a sidedness relation S:B×P→{0,1}, where B is a finite set of boundaries and P is a finite set of points. Two points p₀, p₁ are said to be on the same side of a boundary b iff S(b,p₀) = S(b,p₁), and they are said to be opposite b iff S(b,p₀) ≠ S(b,p₁). Define a space as Sⁿ = (B,P,S,n), where S is a sidedness relation on B×P, and integer n>0 is called the dimension of the space. Define a region as R:B→{0,1}. Note that there are 2^m regions in a space with m boundaries. A point p∈P is said to be in region R iff S(b,p) = R(b) for each b∈B. A region is said to be empty iff no points are in it. Define a subspace of Sⁿ = (B,P,S,n) to be S₀ⁿ = (B₀,P,S₀,n) such that B₀⊂B, and S₀(b,p) = S(b,p) for each b∈B₀, and p∈P. A space Sⁿ is said to be linear iff in each subspace the number of nonempty regions is equal to f(n,m), where m is the number of boundaries in the subspace, f(1,m) = m+1, f(n,0) = 1, and f(n+1,m+1) = f(n+1,m)+f(n,m). A sidedness relation S:B×P→{0,1}, where B is a finite set of linear surfaces in Euclidean space Rⁿ and P⊂Rⁿ is a finite set of points not on the surfaces in B, is given by n_b∙(p-p_b)>0, where n_b is a normal to b∈B, p_b is a point on b, and p∈P. A set of n linear surfaces in Rⁿ determines an intersection point. A set of linear surfaces in Rⁿ is associated with a set of intersection points determined by all subsets of n linear surfaces. A finite set of linear surfaces in Rⁿ is noncoincidental iff no two surfaces in it are parallel, no n+1 surfaces in it share a point, and no linear surface goes through more than n intersection points. A space Sⁿ is isomorphic with a finite set of noncoincidental linear surfaces in Rⁿ iff there exist one to one mappings g:B→B, h:P→P, and s_b:{0,1}→{0,1}, such that S(b,p) = s_b(S(g(b),h(p))) for each b∈B and p∈P.

Linear Space Theorem

A finite set of noncoincidental linear surfaces in Rⁿ exists if and only if isomorphic linear space Sⁿ exists.
First I will prove that a finite set of linear surfaces in Rⁿ exists only if isomorphic linear space Sⁿ exists. Each point in R divides one of a finite set of regions spanning R, so the number of regions in R is f(1,m) = m+1, where m is the number of points dividing R. Thus, the "only if" part of the theorem is true for n = 1. There exist any number of points in Rⁿ on either side of a linear surface, so f(n,1) = 2, for all n. Thus, the "only if" part of the theorem is true for m = 1. Suppose the "only if" part of the theorem is true for n < i. Also suppose that if n = i, then the "only if" part of the theorem is true for m < j. I must show that the "only if" part of the theorem is true for n = i and m = j. Let L be a set of j-1 noncoincidental linear surfaces in Rⁱ, and let l be a linear surface in Rⁱ, noncoincidental with L. By supposition, L divides Rⁱ into f(i,j-1) regions. The intersections of l with the surfaces in L are isomorphic with linear surfaces in R^i-1. By supposition, l divides f(i-1,j-1) of the regions formed by L in Rⁱ. Therefore, there are f(i,j-1)+f(i-1,j-1) = f(i,j) regions formed by L∪{l} in Rⁱ, and the "only if" part of the theorem is proved. The "if" part I will prove below after proving some supportive theorems.

Section

A boundary b is said to extend S₀ⁿ = (B₀,P₀,S₀,n) to S₁ⁿ = (B₁,P₁,S₁,n) iff S₀ⁿ is a subspace of S₁ⁿ, and B₁ = B₀∪{b}. If b extends S₀ⁿ = (B₀,P₀,S₀,n) to S₁ⁿ, then b divides region R₀ in S₀ⁿ into regions R_1a and R_1b in S₁ⁿ iff R_1a and R_1b are not empty, R_1a ≠ R_1b, and R_1a(b₀) = R_1b(b₀) = R₀(b₀) for each b₀∈B₀. If b extends S₀ⁿ = (B₀,P₀,S₀,n), then S₁^n-1 is the section of S₀ⁿ by b iff S₁^n-1 is isomorphic with S₁ⁿ = (B₀,P⊂P₀,S₀,n), where p∈P iff p is in a region divided by b.

Section Theorem

If S^n-1 is a section of S₀ⁿ by b, b extends S₀ⁿ to S₁ⁿ, S₀ⁿ is linear, and S₁ⁿ is linear, then S^n-1 is linear.
The number of regions in S^n-1 is the number of regions divided by b. The number of regions divided by b is the number of regions in S₀ⁿ subtracted from the number of regions in S₁ⁿ. Since S₀ⁿ and S₁ⁿ are linear, the number of regions in S^n-1 is f(n,m+1)-f(n,m) = (f(n,m)+f(n-1,m))-f(n,m) = f(n-1,m). The same reasoning applies to each subspace of S^n-1, because each subspace of S^n-1 is the section of a subspace of S₀ⁿ by b. Therefore, S^n-1 is linear.

Antisection Theorem

If S^n-1 is a section of S₀ⁿ by b, b extends S₀ⁿ to S₁ⁿ, S^n-1 is linear, and S₀ⁿ is linear, then S₁ⁿ is linear.
The number of regions in S^n-1 is the number of regions divided by b. The number of regions in S₁ⁿ is the number of regions divided by b added to the number of regions in S₀ⁿ. Since S^n-1 and S₀ⁿ are linear, the number of regions in S₁ⁿ is f(n-1,m)+f(n,m) = f(n,m+1). The same reasoning applies to each subspace of S₁ⁿ, because each subspace of S₁ⁿ is the extension of a subspace of S₀ⁿ by b. Therefore, S₁ⁿ is linear.

Simplex

A simplex in linear space Sⁿ is any n+1 boundaries in Sⁿ. Note that because n linear surfaces are linearly independent in Rⁿ, and the "only if" part of the Linear Space Theorem is true, f(n,n) = 2ⁿ. By definition of linear space, f(1,2) = 2²-1, and if f(n-1,n) = 2ⁿ-1, then f(n,n+1) = f(n,n)+f(n-1,n) = 2ⁿ+2ⁿ-1 = 2ⁿ⁺¹-1, so the number of nonempty regions in a simplex is 2ⁿ⁺¹-1. Consider the empty region of a simplex in space Sⁿ. Reverse any one sidedness of the empty region, and call the resultant region a vertex region of the simplex. Since there are n+1 boundaries in a simplex, there are n+1 vertex regions in it. Reverse all of the sidednesses of the empty region and call the resultant region the inside of the simplex. Since there is only one empty region in a simplex, there is only one inside in it.

Simplex Theorem

There exists no boundary dividing all vertex regions of a simplex.
Suppose there does exist boundary b dividing all n+1 vertex regions of simplex b₁, b₂, ... b_n+1. Consider the section of b₁, b₂, ... b_n by b. That section is a simplex with vertices sectioned from the vertices opposite b₁, b₂, ... b_n. Since the vertex region opposite b_n+1 is opposite one of b₁, b₂, ... b_n from each of the sectioned vertex regions, it is not sectioned to any one region of the sectioned simplex. This contradicts the supposition of b, so no such b exists, and the theorem is proved.

Independent Boundary Theorem

If Sⁿ extended by b₀ is linear, and Sⁿ extended by b₁ is linear, then there exists a linear superspace of Sⁿ extended by b₀ and Sⁿ extended by b₁.
(This proof attempt is wrong. I am working on a correction.) For S¹ it is true because b₀ and b₁ divide different regions, so points on one or the other side of b₀ in the region it divides will all be on one side of b₁. Suppose it is true for S^n-1. I must prove it for Sⁿ. Suppose Sⁿ is not linear if it is extended by both b₀ and b₁ together, but Sⁿ is linear if it is extended by b₀ or b₁ individually. Discard boundaries from Sⁿ until the reduced Sⁿ is linear when extended by both b₀ and b₁. This is possible because b₀ and b₁ alone in Sⁿ is a linear space. Let b₂ be the last boundary discarded from Sⁿ. Consider the section by b₂, S^n-1. By the Section Theorem, S^n-1 is linear because Sⁿ is linear and Sⁿ extended by b₂ is linear. Similarly, S^n-1 extended by the section of b₀, or by the section of b₁, is linear. Thus, by supposition, S^n-1 extended by the sections of both b₀ and b₁ is linear. By the Antisection Theorem, since Sⁿ extended by b₀ and b₁ is linear, and S^n-1 extended by the sections of b₀ and b₁ is linear, Sⁿ extended by b₀, b₁, and b₂ is linear. This contradicts the supposition, so the theorem is proved.

"If" Part of Linear Space Theorem

If linear space Sⁿ exists, then a finite set of linear surfaces in Rⁿ isomorphic with Sⁿ exist.
(This proof attempt, in particular the struckout text, is wrong. I am working on a correction.) Suppose no R₀ⁿ exits with boundaries and sidednesses isomorphic to some linear space S₀ⁿ. Let S₁ⁿ be boundaries of S₀ⁿ which are isomorphic with the boundaries and sidednesses of some R₁ⁿ. Until impossible, extend S₁ⁿ by boundaries from S₀ⁿ such that S₁ⁿ is still isomorphic with an extension of R₁ⁿ. By supposition, there exists boundary b₀ in S₀ⁿ such that b₀ is not in S₁ⁿ. Let R₁, R₂, ... R_n+1 be the n+1 regions of S₁ⁿ divided by b₀ such that no b₀ exists which divides those regions in R₁ⁿ. Let p₁ and q₁ be in R₁, opposite b₀, and similarly for (p₂,q₂), ... (p_n+1,q_n+1). Construct b₁, b₂, ... b_n+1 such that p₁ and q₁ are in one vertex region of the simplex formed by b₁, b₂, ... b_n+1, and (p₂,q₂), ... (p_n+1,q_n+1) are in the other vertex regions. The prior construction is possible because, given n+1 points in Rⁿ a simplex can be constructed with those points as vertices; a point can be chosen in each of R₁, R₂, ... R_n+1; the vertex region of each point at least partially overlaps the region the point is in; and without loss of generality, we may assume p_i and q_i are in the portion of R_i which overlaps the vertex region. Extend R₁ⁿ by b₁, b₂, ... b_n+1 to R₂ⁿ. Since R₂ⁿ obeys the "only if" part of the Linear Space Theorem, S₁ⁿ can be similarly extended to S₂ⁿ such that S₂ⁿ is isomorphic to R₂ⁿ. Since S₁ⁿ can be extended by b₀, and b₀ divides different regions from b₁, b₂, ... b_n+1, S₂ⁿ can be extended by b₀ to S₃ⁿ because of the Independent Boundary Theorem. Now, b₀ goes through each of the vertex regions of b₁, b₂, ... b_n+1 in S₃ⁿ. This violates the Simplex Theorem. Therefore the supposition was false, and there does exist R₀ⁿ with boundaries and sidednesses isomorphic with S₀ⁿ.