Chapter 53: Research Topics in Combinatorial Structures

Which small examples suggest a meaningful structural conjecture?

Five objects appear in three different settings: permutations of three elements avoiding the pattern $231$, rooted plane binary trees with three internal vertices, and triangulations of a convex pentagon. Is this a numerical accident, an explicit correspondence, or the visible part of a stronger structural identity?

A permutation contains $231$ if some three entries, read from left to right, have relative order middle, largest, smallest. Thus the five avoiding permutations of length three are $123,132,213,312,321$. A descent is a position $i$ with $\pi_i\gt\pi_{i+1}$. Their descent counts are respectively $0,1,1,1,2$, giving the polynomial

$$D_3(t)=1+3t+t^2.$$

This records more than the total five: it refines the objects by a statistic. Its two roots are $(-3\pm\sqrt5)/2$, both negative and real. After dividing by five, it is the probability generating function of the descent count of a uniformly chosen avoider. Geometry will supply the same coefficient vector as the $h$-vector of a pentagon's dual boundary complex. Those are precise connections, but each needs its own definition and proof.

A useful investigation therefore asks more than “do the first counts match?” Specify the object families, the size parameter, the statistic, the probability law, and the proposed correspondence. A bijection preserving size proves count equality; a bijection preserving a statistic proves a stronger distributional identity. Matching totals alone proves neither.

The finite example and this question-building discipline are the chapter foundation. The extensions use analytic combinatorics, bijective and computer-assisted enumeration, tableaux and symmetric functions, and advanced algebraic combinatorics. The final investigation develops an established Catalan–Narayana connection as research training; it does not claim a new theorem from a short numerical table.

53.1 Asymptotic enumeration of complex structures

Optional extension; not required to complete the chapter foundation.

Let $a_n$ count a specified family at size $n$. An asymptotic equivalence $a_n\sim b_n$ means $a_n/b_n\to1$. A growth rate, such as $\lim a_n^{1/n}$, retains less information: it can miss polynomial factors, constants, periodicity, and different behavior on residue classes. Labeling and rooting conventions must be fixed before any comparison.

Solved Catalan example. A rooted plane full binary tree is either a leaf or an internal root with an ordered left and right subtree. If size counts internal vertices, its numbers satisfy $C_0=1$ and $C_n=\sum_{k=0}^{n-1}C_kC_{n-1-k}$. The ordinary generating function $C(z)=\sum_{n\ge0}C_nz^n$ therefore satisfies $C=1+zC^2$. The solution with constant term one is

$$C(z)=\frac{1-\sqrt{1-4z}}{2z},\qquad C_n=\frac1{n+1}\binom{2n}{n}.$$

The recurrence determines the coefficients uniquely; binomial expansion, or formal Lagrange inversion, gives the displayed formula. Stirling's factorial estimate then gives

$$C_n\sim\frac{4^n}{\sqrt\pi\,n^{3/2}},\qquad C_n^{1/n}\longrightarrow4.$$

For example, $C_4=14$ and $C_5=42$. A count ratio near four supports a growth estimate, but the exact recurrence establishes the family and the asymptotic proof identifies the power $n^{-3/2}$. Primary analytic framework: Flajolet and Sedgewick's Analytic Combinatorics. Proof scope: the combinatorial recurrence and formula are established here; Stirling's theorem is a prerequisite, and the general complex-analytic transfer theorem is omitted.

More complex families can grow on a different scale. For labeled partial orders on $[n]$, the classical Kleitman–Rothschild theorem implies $\log_2 p_n=n^2/4+o(n^2)$. That is much faster than a fixed exponential $c^n$. Omitted prerequisites: the structural classification and asymptotic counting argument. Small three-element posets below do not prove this large-size theorem.

Exercise. Do $4^n$ and $4^n/n^{3/2}$ have the same exponential growth rate? Hint: take logarithms and divide by $n$. Solution: yes, both have root limit four, but their ratio is $n^{3/2}$, so they are not asymptotically equivalent.

53.2 Random posets and order types

Optional extension; not required to complete the chapter foundation.

A partial order is reflexive, antisymmetric, and transitive. Its strict relation is an acyclic transitive relation between distinct elements. A linear extension lists the elements in an order compatible with every required comparison. To speak of a random poset, choose a distribution: uniform labeled posets, uniform isomorphism classes, or an intersection of random linear orders are different models.

Solved example. On three labeled elements there are nineteen partial orders. They fall into five isomorphism types: the antichain; one comparable pair plus an isolated element; a minimum below two incomparable elements; a maximum above two incomparable elements; and a chain. Their labeled multiplicities are $1,6,3,3,6$ respectively. The linear-extension counts for these types are $6,3,2,2,1$. The sum of multiplicities is nineteen, and the expected number of linear extensions under the uniform labeled law is $(6+18+6+6+6)/19=42/19$.

Sampling one of the five types uniformly would instead give expectation $(6+3+2+2+1)/5=14/5$. Forgetting labels from a uniform labeled sample does not yield a uniform type: the chain gets probability $6/19$ and the antichain $1/19$.

“Order type” can mean an order-isomorphism class; in geometry it has another meaning. The planar order type of labeled points in general position records the orientation sign of every ordered triple. For the square points $(0,0),(1,0),(1,1),(0,1)$ labeled in that order, each increasing-index triple has positive determinant. This orientation data records geometric relationships, not a poset's comparison table. Uniform planar order types are not defined merely by choosing coordinates independently without analyzing the induced law.

Kleitman and Rothschild supply the classical asymptotic poset context. Scope: our nineteen-order classification is finite and proved by the listed types; no claim about geometric order-type asymptotics is inferred from it.

Exercise. What is the probability of a chain under a uniform labeled three-element poset? Hint: count total orders. Solution: $6/19$, not $1/5$. Under a uniform unlabeled-type law the latter value is correct.

53.3 Unimodality, real-rootedness, and log-concavity

Optional extension; not required to complete the chapter foundation.

A nonnegative sequence is unimodal when it increases up to a peak and then decreases, allowing equalities. It is log-concave when $a_k^2\ge a_{k-1}a_{k+1}$ at interior indices. With no internal zeros, log-concavity implies unimodality: consecutive positive ratios $a_k/a_{k-1}$ are nonincreasing, so they can cross one at most once. A polynomial is real-rooted if all its complex roots are real.

For a polynomial with nonnegative coefficients and only real roots, Newton's inequalities give log-concavity, in fact a stronger inequality after binomial normalization. Omitted proof prerequisite: the general Newton-inequality argument. The converse implications fail: a unimodal sequence need not be log-concave, and a log-concave coefficient sequence need not have a real-rooted polynomial.

Solved example. The next descent polynomial in our Catalan family is

$$D_4(t)=1+6t+6t^2+t^3=(1+t)(t^2+5t+1).$$

Its roots are $-1$ and $(-5\pm\sqrt{21})/2$, all strictly negative. The coefficient sequence is symmetric and unimodal; log-concavity checks read $6^2\ge1\cdot6$ and $6^2\ge6\cdot1$. These are exact algebraic checks, rather than roots estimated from a plot.

The SVG shows the four coefficients and both interior inequalities. Beneath them, the factorization gives every root exactly; bar height is a visual aid, not a root certificate.

For a counterexample to the converse, $1+t+t^2$ has log-concave coefficients $1,1,1$, but discriminant minus three, so its two roots are nonreal. The unimodal sequence $1,2,5,5$ fails log-concavity at the second entry because $2^2\lt1\cdot5$. These finite examples keep the hierarchy precise.

The general Narayana polynomials are established real-rooted families; primary treatments include Liu and Wang and Chen, Yang, and Zhang. Proof mechanism: recurrence relations together with interlacing or real-root-preserving operations establish all degrees. That induction is omitted here. Checking roots through length seven is not its replacement.

Exercise. Does symmetry plus unimodality prove real-rootedness? Hint: revisit $1+t+t^2$. Solution: no; it has both properties and nonreal roots. State which property a proof actually establishes.

53.4 Unlabeled enumeration and symmetry

Optional extension; not required to complete the chapter foundation.

A finite group $G$ acts on labeled objects by relabeling. Unlabeled objects are its orbits. Orbit–stabilizer gives orbit size $|G|/|\operatorname{Aut}(X)|$; dividing the number of labeled objects by $|G|$ works only if every stabilizer is trivial. Burnside's lemma instead counts orbits by $|G|^{-1}\sum_{g\in G}|\operatorname{Fix}(g)|$.

Proof by double counting. Count pairs $(g,X)$ with $gX=X$. Summing over group elements gives the fixed-point sum. Within an orbit, the number of objects times their equal stabilizer sizes is $|G|$. Thus each orbit contributes exactly $|G|$, proving the formula.

Solved poset example. Let $S_3$ relabel the nineteen posets. The identity fixes nineteen. A transposition fixes exactly three: the antichain, the swapped pair jointly below the third element, and that pair jointly above it. A three-cycle fixes only the antichain. Consequently

$$\frac{19+3\cdot3+2\cdot1}{6}=5$$

unlabeled types, as listed before. A chain has no nonidentity order-preserving automorphism and six labelings. The V-shaped poset has an automorphism swapping its upper elements and three labelings. The antichain has all six relabelings as automorphisms and only one labeled relation.

Symmetry also changes sampling. Uniformly selecting an orbit and then a labeled representative is not the same law as uniformly selecting among all labeled objects. A weighted construction must account for orbit size or automorphism size, depending on its intended target. This is a practical issue in generating test objects and estimating frequencies, not merely an enumerative detail.

For a broader original framework, Schmitt's Incidence Hopf Algebras90105-8) organizes isomorphism classes and decompositions algebraically. Scope: the orbit count above follows from the supplied group action; no asymptotic assertion that almost all objects are asymmetric is needed.

Exercise. Why is $19/6$ not the unlabeled poset count? Hint: compare the antichain with a chain. Solution: their stabilizers differ, so orbit sizes differ. Burnside's fixed-point terms correct the naive division.

53.5 Permutation patterns and growth rates

Optional extension; not required to complete the chapter foundation.

For a fixed permutation $\tau$, an occurrence in $\pi$ is a subsequence whose entries have the same relative order as $\tau$. Positions need not be consecutive. Write $\operatorname{Av}_n(\tau)$ for the length-$n$ avoiders and $a_n(\tau)$ for their number. Pattern avoidance is preserved by taking subsequences and standardizing their values, forming a permutation class.

Solved Catalan decomposition. Write a $231$-avoider as $L\,n\,R$, with $k$ entries before its largest value. Every entry of $L$ must be smaller than every entry of $R$: otherwise an entry of $L$, then $n$, then a smaller entry of $R$ would form $231$. Thus $L$ uses exactly $1,\ldots,k$ and $R$ uses $k+1,\ldots,n-1$. Both standardized blocks avoid $231$. Conversely, blocks satisfying these conditions create no crossing occurrence, so the construction is reversible.

There are $a_k(231)a_{n-1-k}(231)$ choices at each split. The resulting recurrence is the Catalan recurrence, with the same base. Hence $a_n(231)=C_n$ and its growth rate is four. For length three the five objects in the foundation confirm the base example; the reversible decomposition proves all lengths.

The general Marcus–Tardos theorem establishes exponential upper bounds for avoiders of any fixed nontrivial permutation pattern. Proof sketch: bound the maximum number of ones in pattern-avoiding binary matrices, then use a block-compression counting argument. The full extremal-to-enumerative reduction is omitted. A universal statement of exponential boundedness does not determine each pattern's exact growth constant.

For $231$, direct sums of avoiders remain avoiders. Concatenating a length-$n$ avoider with a shifted length-$m$ avoider injects pairs into length $n+m$, so $a_{n+m}\ge a_na_m$. Superadditivity of logarithms and the exponential upper bound yield a finite root growth limit; in this case the exact Catalan formula already identifies it. This argument is stated for this class, not as an unproved claim that every permutation class has a growth limit.

Exercise. Does $2413$ contain $231$? Hint: inspect entries in positions one, two, three. Solution: $2,4,1$ have relative order $231$, so it does. The final entry does not remove that occurrence.

53.6 Extremal problems for forbidden permutation patterns

Optional extension; not required to complete the chapter foundation.

The matrix version asks how many ones can be placed while avoiding a fixed permutation matrix. Containment preserves the order of selected rows and columns, and requires ones at the pattern's one positions. Other selected entries may also be one; this is not induced-submatrix containment. The distinction matters even in two-by-two examples.

Solved exact extremal case. Avoid the two-by-two identity pattern in an $r\times c$ zero-one matrix. It is forbidden to have ones at $(i,j)$ and $(i',j')$ with $i\lt i'$ and $j\lt j'$. Cells on a diagonal with fixed $i-j$ are ordered in both coordinates, so at most one cell on each such diagonal may be one. There are $r+c-1$ diagonals. Therefore at most $r+c-1$ ones are possible.

The bound is attained by placing ones in the entire first row and entire first column, counting their intersection once. Any pair in different rows has the lower one's column equal to one, so it cannot create a strictly increasing-column pair. Thus the maximum is exactly $r+c-1$; on a three-by-three board it is five. One extremal matrix is

$$\begin{pmatrix}1&1&1\\ 1&0&0\\ 1&0&0\end{pmatrix}.$$

This extremal problem permits several ones per row. In the permutation problem avoiding $12$, each row and column has exactly one one and the only permutation is decreasing. Linear extremal density and permutation count measure different things.

The established Marcus–Tardos result gives $O(n)$ ones in an $n\times n$ matrix avoiding any fixed permutation matrix. Proof sketch and omitted prerequisites: compress a block partition while preserving avoidance, bound the number of blocks using many distinct rows or columns, and iterate the resulting linear recurrence. The constant depends on the forbidden pattern. The diagonal proof above treats only the two-entry pattern and does not extend by replacing two with an arbitrary pattern length.

Exercise. Does an all-one two-by-two matrix contain the identity pattern under this convention? Hint: extra ones are permitted. Solution: yes. Its top-left and bottom-right ones form the pattern, even though the off-diagonal entries are also one.

53.7 Random planar maps

Optional extension; not required to complete the chapter foundation.

A planar map is a connected graph embedded on an oriented sphere, considered up to orientation-preserving equivalence. It includes cyclic orders of edges around vertices; loops and multiple edges are allowed in the model used here. Rooting selects a directed edge, or dart. A uniformly random rooted map with $n$ edges is not a uniformly random labeled simple planar graph.

A useful bridge is from random trees to embedded structures: Section 23.6 on random trees introduces tree-based random models, while Section 26.10 on asymptotics explains how generating functions translate counting into growth laws. Those are optional background connections. A plane tree is a special embedded map, but general maps have cycles and faces, so their counts are different.

Solved one-edge example. There are two rooted map types with one edge. The link joins two vertices and has one face; its Euler check is $2-1+1=2$. The loop has one vertex and two faces; its check is $1-1+2=2$. A rooted plane tree with one edge accounts only for the link. This small distinction already prevents identifying the map sequence with Catalan numbers.

For these general rooted planar maps, Tutte's formula is

$$M_n=\frac{2\cdot3^n(2n)!}{n!(n+2)!}\quad(n\ge1).$$

It gives $M_1=2,M_2=9,M_3=54,M_4=378$. Primary source: Tutte's A Census of Planar Maps. The zero-edge convention can be assigned one degenerate rooted object, but the usual directed-edge root is meaningful only for positive edge count.

Rewrite the formula as $2\cdot3^n\binom{2n}{n}/[(n+1)(n+2)]$. Stirling then yields $M_n\sim(2/\sqrt\pi)12^n n^{-5/2}$. Both the exponential base and polynomial exponent differ from Catalan tree counts. Omitted prerequisites: the map decomposition giving Tutte's enumeration and metric scaling-limit theory. A small Euler check does not prove a random surface limit.

For exact computer checks, pair $2n$ labeled darts by a fixed involution $\alpha$ and encode vertex rotations by a permutation $\sigma$. Connectedness means the generated action is transitive. Faces are cycles of $\sigma\alpha$, and planarity requires $c(\sigma)-n+c(\sigma\alpha)=2$. With dart one as root, quotient by the fixed-pair relabelings fixing that dart. This provides a finite independent validation route through small $n$; it must retain embedding data, not just an adjacency matrix.

Exercise. Why can't an unrooted map count always be obtained by dividing $M_n$ by $2n$? Hint: consider automorphisms. Solution: different unrooted maps may have different dart orbits under automorphisms, so the number of inequivalent root choices varies.

53.8 Algebraic combinatorics of cluster structures

Optional extension; not required to complete the chapter foundation.

A cluster algebra organizes generators into overlapping clusters and replaces a generator through an exchange relation. Finite-type examples connect this algebraic mutation to combinatorial flips. The coefficient conventions matter; use coefficient-free type $A_2$ and independent variables $a,b$ below.

Solved five-variable example. Define

$$x_1=a,\quad x_2=b,\quad x_3=\frac{1+b}{a},\quad x_4=\frac{1+a+b}{ab},\quad x_5=\frac{1+a}{b}.$$

They satisfy $x_{i-1}x_{i+1}=1+x_i$ with indices read cyclically modulo five. Starting from $x_1,x_2$, the recurrence produces $x_3,x_4,x_5,x_6=a,x_7=b$. For instance $x_2x_4=(1+a+b)/a=1+x_3$; the last cyclic relation is $x_4x_1=(1+a+b)/b=1+x_5$. These are identities of rational functions, not numerical coincidences at one chosen input.

The clusters are the five adjacent pairs $\{x_1,x_2\},\{x_2,x_3\},\{x_3,x_4\},\{x_4,x_5\},\{x_5,x_1\}$. Label the pentagon vertices one through five in cyclic order and assign diagonals

$$d_1=13,\quad d_2=14,\quad d_3=24,\quad d_4=25,\quad d_5=35.$$

Adjacent diagonals in this list are compatible, and nonadjacent ones cross. Sending $x_i$ to $d_i$ identifies the five clusters with the five triangulations. The exchange graph has triangulations as vertices and diagonal flips as edges. The compatibility complex instead has individual diagonals as vertices and triangulations as edges; both happen to be five-cycles here, but their vertices represent different objects.

Primary theorem: Fomin and Zelevinsky's Finite Type Classification. Scope: our rational identities and explicit type-$A_2$ correspondence are verified directly. The classification for all finite types, general Laurent phenomenon, and positivity theorems require additional arguments and are not inferred from five variables.

Exercise. At $a=b=1$, list the five variable values and check one exchange relation. Hint: evaluate the formulas. Solution: $(1,1,2,3,2)$; for example $x_2x_4=3=1+x_3$. This numerical check complements, but does not replace, the rational-function proof.

53.9 Combinatorial Hopf algebras and categorification

Optional extension; not required to complete the chapter foundation.

A Hopf algebra combines a multiplication, a coproduct describing splitting, a unit, a counit, and an antipode satisfying compatibility axioms. Combinatorial examples often multiply by combining objects and coproduct by summing over decompositions. The splitting rules and coefficients must be specified; “all ways to split” is not a complete definition.

Solved polynomial model. Work over $\mathbb Q$ with the ordinary polynomial algebra $\mathbb Q[x]$. Define $\Delta(x)=x\otimes1+1\otimes x$ and extend multiplicatively. Set the counit to evaluation at zero and the antipode to $S(x)=-x$. Then

$$\Delta(x^2)=x^2\otimes1+2x\otimes x+1\otimes x^2.$$

The coefficient two counts the two choices of which labeled element goes to the first part of a two-element split. More generally $\Delta(x^n)=\sum_{k=0}^n\binom nk x^k\otimes x^{n-k}$. Coassociativity follows because splitting into three ordered parts gives the same multinomial coefficients whichever binary split is performed first. The antipode identity for $x^2$ reads $x^2-2x^2+x^2=0$, agreeing with its counit value. This is an explicit toy Hopf algebra, not the entire Hopf algebra of permutations or trees.

A simple form of decategorification maps a finite graded vector space $V=\bigoplus_kV_k$ to $\sum_k\dim(V_k)t^k$. Direct sums add these polynomials, while graded tensor products multiply them. If $V$ has one basis vector in degree zero and one in degree one, then $V\otimes V$ has dimensions $1,2,1$ and polynomial $(1+t)^2$. The extra dimension in the middle has a concrete basis: degree zero tensored with degree one, or the reverse.

Categorification seeks objects and maps whose algebraic shadows recover a given identity. Matching dimension sequences alone does not establish a categorification of every operation or prove that two categories are equivalent. One needs functors, maps, and compatibility statements, not just a new name for the coefficients.

Primary broader framework: Schmitt's Incidence Hopf Algebras90105-8). Omitted prerequisites: incidence coalgebras, species, representation-theoretic categories, and the coherence data for deeper categorifications. The polynomial splitting model above verifies its own elementary axioms directly.

Exercise. What is the coefficient of $x\otimes x^2$ in $\Delta(x^3)$? Hint: choose the singleton first part. Solution: three. Replacing it by one would use an unlabeled splitting convention incompatible with the coproduct just defined.

53.10 Connections among probability, algebra, geometry, and topology

Optional extension; not required to complete the chapter foundation.

Return to the refined Catalan family. If $D_n(t)=\sum_d a_{n,d}t^d$ counts $231$-avoiders by descents, then $D_n(1)=C_n$ and $D_n(t)/C_n$ is a probability generating function. Its derivative at one gives the expected statistic, and its second derivative gives the second factorial moment. Algebraic coefficients therefore encode probability laws once the sample space and normalization are fixed.

Solved probabilistic connection. For $n=3$, the descent law is $(1,3,1)/5$. Its mean is one and variance $2/5$. More generally, if $D_n$ has negative real roots $-r_1,\ldots,-r_{n-1}$ with $r_i\gt0$, its normalized factorization is

$$\frac{D_n(t)}{D_n(1)}=\prod_{i=1}^{n-1}\frac{r_i+t}{r_i+1}=\prod_{i=1}^{n-1}(1-p_i+p_it),\qquad p_i=\frac1{1+r_i}.$$

This is the generating function of a sum of independent Bernoulli variables. It gives an independent-variable representation of the descent distribution, not a proof that actual descent indicators at different permutation positions are independent. Consequently its variance is $\sum_ip_i(1-p_i)\le(n-1)/4$.

For $n=3$, $r_i=(3\pm\sqrt5)/2$ and the two Bernoulli probabilities are $(5\mp\sqrt5)/10$. Their variances sum to $2/5$, matching direct enumeration. For $n=4$, the root $-1$ supplies a probability $1/2$, and the remaining two factors supply complementary probabilities. Their sum gives mean $3/2$ and variance $15/28$.

Geometry supplies a different interpretation. The pentagon compatibility complex has five vertices, five edges, and no two-dimensional faces. Its $h$-polynomial is $1+3t+t^2$ by the defining face-to-$h$ transformation. For a hexagon, the compatibility complex has nine diagonal vertices, twenty-one compatible pairs, and fourteen triangulations. Thus

$$(1-t)^3+9t(1-t)^2+21t^2(1-t)+14t^3=1+6t+6t^2+t^3.$$

This matches $D_4(t)$. Established generalized-associahedron enumeration is treated in Chapoton and the finite-type framework of Fomin–Zelevinsky. Omitted prerequisites: the general type-$A$ $h$-vector theorem and broader geometric realization arguments. The displayed pentagon and hexagon counts can be checked exhaustively without those prerequisites.

Topology must keep boundary and filling separate. The pentagon compatibility complex is a circle: its boundary map has rank four, so ordinary Betti numbers are $\beta_0=1,\beta_1=1$. Filling that cycle by a two-cell gives a disk, with $\beta_0=1$ and higher Betti numbers zero. Neither has ordinary Betti numbers $1,3,1$. An $h$-vector, a statistic distribution, and a boundary homology vector are not interchangeable just because all are sequences of integers.

Exercise. Does the equality of $D_3(t)$ and the pentagon $h$-polynomial prove equality of homology groups? Hint: compare the sequences just calculated. Solution: no. The equality concerns a face transformation and a descent statistic; homology is computed from boundary maps, not from substituting a name for the coefficients.

Turn a cross-disciplinary observation into a precise research question

A defensible question begins with quantified objects and a falsifiable conclusion. For this chapter, ask: for a uniform $231$-avoiding permutation of length $n\ge1$, what is the exact descent polynomial, why is its coefficient sequence well behaved, and what probabilistic bound follows from real-rootedness? This replaces a vague numerical coincidence with a precise family, statistic, law, and implication.

The Catalan–Narayana count and general real-rootedness used here are established results, not newly claimed open problems. The primary sources cited above identify their scope. The optional extension below is a bounded computational exploration of changed hypotheses, not an assertion that every broader question is unresolved. Small examples can help rediscover a theorem or expose a missing assumption; a literature check is part of that work.

One application is average-case analysis of constrained ordering algorithms. A statistic such as descents measures local order changes, but an average is meaningful only under a stated input distribution. A known generating function gives exact finite moments and supports asymptotic study; sampling arbitrary permutations or unlabeled types would answer another problem. This connects asymptotic enumeration with probability and algebra while keeping the geometric and topological invariants distinct.