# Introduction to combinatorics in Sage#

This thematic tutorial is a translation by Hugh Thomas of the
combinatorics chapter written by Nicolas M. Thiéry in the book “Calcul
Mathématique avec Sage” [CMS2012]. It covers mainly the treatment in
`Sage`

of the following combinatorial problems: enumeration (how
many elements are there in a set \(S\)?), listing (generate all the
elements of \(S\), or iterate through them), and random selection
(choosing an element at random from a set \(S\) according to a given
distribution, for example the uniform distribution). These questions
arise naturally in the calculation of probabilities (what is the
probability in poker of obtaining a straight or a four-of-a-kind of
aces?), in statistical physics, and also in computer algebra (the number
of elements in a finite field), or in the analysis of
algorithms. Combinatorics covers a much wider domain (partial orders,
representation theory, …) for which we only give a few pointers
towards the possibilities offered by `Sage`

.

Todo

Add link to some thematic tutorial on graphs

A characteristic of computational combinatorics is the profusion of types of objects and sets that one wants to manipulate. It would be impossible to describe them all or, a fortiori, to implement them all. After some Initial examples, this chapter illustrates the underlying method: supplying the basic building blocks to describe common combinatorial sets Common enumerated sets, tools for combining them to construct new examples Constructions, and generic algorithms for solving uniformly a large class of problems Generic algorithms.

This is a domain in which `Sage`

has much more extensive capabilities
than most computer algebra systems, and it is rapidly expanding; at the
same time, it is still quite new, and has many unnecessary limitations
and incoherences.

## Initial examples#

### Poker and probability#

We begin by solving a classic problem: enumerating certain combinations of cards in the game of poker, in order to deduce their probability.

A card in a poker deck is characterized by a suit (hearts, diamonds, spades, or clubs) and a value (2, 3, …, 10, jack, queen, king, ace). The game is played with a full deck, which consists of the Cartesian product of the set of suits and the set of values:

We construct these examples in `Sage`

:

```
sage: Suits = Set(["Hearts", "Diamonds", "Spades", "Clubs"])
sage: Values = Set([2, 3, 4, 5, 6, 7, 8, 9, 10,
....: "Jack", "Queen", "King", "Ace"])
sage: Cards = cartesian_product([Values, Suits])
```

There are \(4\) suits and \(13\) possible values, and therefore \(4\times 13=52\) cards in the poker deck:

```
sage: Suits.cardinality()
4
sage: Values.cardinality()
13
sage: Cards.cardinality()
52
```

Draw a card at random:

```
sage: Cards.random_element() # random
(6, 'Clubs')
```

Now we can define a set of cards:

```
sage: Set([Cards.random_element(), Cards.random_element()]) # random
{(2, 'Hearts'), (4, 'Spades')}
```

This problem should eventually disappear: it is planned to change the implementation of Cartesian products so that their elements are immutable by default.

Returning to our main topic, we will be considering a simplified version
of poker, in which each player directly draws five cards, which form his
*hand*. The cards are all distinct and the order in which they are drawn
is irrelevant; a hand is therefore a subset of size \(5\) of the
set of cards. To draw a hand at random, we first construct the set of
all possible hands, and then we ask for a randomly chosen element:

```
sage: Hands = Subsets(Cards, 5)
sage: Hands.random_element() # random
{(4, 'Hearts', 4), (9, 'Diamonds'), (8, 'Spades'),
(9, 'Clubs'), (7, 'Hearts')}
```

The total number of hands is given by the number of subsets of size \(5\) of a set of size \(52\), which is given by the binomial coefficient \(\binom{52}{5}\):

```
sage: binomial(52,5)
2598960
```

One can also ignore the method of calculation, and simply ask for the size of the set of hands:

```
sage: Hands.cardinality()
2598960
```

The strength of a poker hand depends on the particular combination of
cards present. One such combination is the *flush*; this is a hand all
of whose cards have the same suit. (In principle, straight flushes
should be excluded; this will be the goal of an exercise given below.)
Such a hand is therefore characterized by the choice of five values from
among the thirteen possibilities, and the choice of one of four suits.
We will construct the set of all flushes, so as to determine how many
there are:

```
sage: Flushes = cartesian_product([Subsets(Values, 5), Suits])
sage: Flushes.cardinality()
5148
```

The probability of obtaining a flush when drawing a hand at random is therefore:

```
sage: Flushes.cardinality() / Hands.cardinality()
33/16660
```

or about two in a thousand:

```
sage: 1000.0 * Flushes.cardinality() / Hands.cardinality()
1.98079231692677
```

We will now attempt a little numerical simulation. The following function tests whether a given hand is a flush or not:

```
sage: def is_flush(hand):
....: return len(set(suit for (val, suit) in hand)) == 1
```

We now draw 10000 hands at random, and count the number of flushes obtained (this takes about 10 seconds):

```
sage: n = 10000
sage: nflush = 0
sage: for i in range(n): # long time
....: hand = Hands.random_element()
....: if is_flush(hand):
....: nflush += 1
sage: n, nflush # random
(10000, 18)
```

Exercises

A hand containing four cards of the same value is called a *four
of a kind*. Construct the set of four of a kind hands (Hint: use
`Arrangements`

to choose a pair of distinct values at random,
then choose a suit for the first value). Calculate the number of
four of a kind hand, list them, and then determine the probability
of obtaining a four of a kind when drawing a hand at random.

A hand all of whose cards have the same suit, and whose values are
consecutive, is called a *straight flush* rather than a *flush*.
Count the number of straight flushes, and then deduce the correct
probability of obtaining a flush when drawing a hand at random.

Calculate the probability of each of the poker hands (see Wikipedia article Poker_hands), and compare them with the results of simulations.

### Enumeration of trees using generating functions#

In this section, we discuss the example of complete binary trees, and illustrate in this context many techniques of enumeration in which formal power series play a natural role. These techniques are quite general, and can be applied whenever the combinatorial objects in question admit a recursive definition (grammar) (see Species, decomposable combinatorial classes for an automated treatment). The goal is not a formal presentation of these methods; the calculations are rigorous, but most of the justifications will be skipped.

A *complete binary tree* is either a leaf \(\mathrm{L}\), or a
node to which two complete binary trees are attached (see
Figure: The five complete binary trees with four leaves).

Exercise: enumeration of binary trees

Find by hand all the complete binary trees with \(n=1, 2, 3, 4, 5\)
leaves (see Exercise: complete binary tree iterator to find them using `Sage`

).

Our goal is to determine the number \(c_n\) of complete binary trees with \(n\) leaves (in this section, except when explicitly stated otherwise, “trees” always means complete binary trees). This is a typical situation in which one is not only interested in a single set, but in a family of sets, typically parameterized by \(n\in \NN\).

According to the solution of Exercise: enumeration of binary trees, the first terms are given by \(c_1,\dots,c_5=1,1,2,5,14\). The simple fact of knowing these few numbers is already very valuable. In fact, this permits research in a gold mine of information: the Online Encyclopedia of Integer Sequences, commonly called “Sloane”, the name of its principal author, which contains more than 190000 sequences of integers:

```
sage: oeis([1,1,2,5,14]) # optional -- internet
0: A000108: Catalan numbers: ...
1: ...
2: ...
```

The result suggests that the trees are counted by one of the most famous sequences, the Catalan numbers. Looking through the references supplied by the Encyclopedia, we see that this is really the case: the few numbers above form a digital fingerprint of our objects, which enable us to find, in a few seconds, a precise result from within an abundant literature.

Our next goal is to recover this result using `Sage`

. Let
\(C_n\) be the set of trees with \(n\) leaves, so that
\(c_n=|C_n|\); by convention, we will define
\(C_0=\emptyset\) and \(c_0=0\). The set of all trees is
then the disjoint union of the sets \(C_n\):

Having named the set \(C\) of all trees, we can translate the recursive definition of trees into a set-theoretic equation:

In words: a tree \(t\) (which is by definition in \(C\)) is either a leaf (so in \(\{\mathrm{L}\}\)) or a node to which two trees \(t_1\) and \(t_2\) have been attached, and which we can therefore identify with the pair \((t_1,t_2)\) (in the Cartesian product \(C\times C\)).

The founding idea of algebraic combinatorics, introduced by Euler in
a letter to Goldbach of 1751 to treat a similar problem , is to
manipulate all the numbers \(c_n\) simultaneously, by encoding them
as coefficients in a formal power series, called the *generating
function* of the \(c_n\)’s:

where \(z\) is a formal variable (which means that we do not have to worry about questions of convergence). The beauty of this idea is that set-theoretic operations \((A\uplus B\), \(A\times B)\) translate naturally into algebraic operations on the corresponding series (\(A(z)+B(z)\), \(A(z)\cdot B(z)\)), in such a way that the set-theoretic equation satisfied by \(C\) can be translated directly into an algebraic equation satisfied by \(C(z)\):

Now we can solve this equation with `Sage`

. In order to do so, we
introduce two variables, \(C\) and \(z\), and we define the
equation:

```
sage: C, z = var('C,z')
sage: sys = [ C == z + C*C ]
```

There are two solutions, which happen to have closed forms:

```
sage: sol = solve(sys, C, solution_dict=True); sol
[{C: -1/2*sqrt(-4*z + 1) + 1/2}, {C: 1/2*sqrt(-4*z + 1) + 1/2}]
sage: s0 = sol[0][C]; s1 = sol[1][C]
```

and whose Taylor series begin as follows:

```
sage: s0.series(z, 6)
1*z + 1*z^2 + 2*z^3 + 5*z^4 + 14*z^5 + Order(z^6)
sage: s1.series(z, 6)
1 + (-1)*z + (-1)*z^2 + (-2)*z^3 + (-5)*z^4 + (-14)*z^5
+ Order(z^6)
```

The second solution is clearly aberrant, while the first one gives the expected coefficients. Therefore, we set:

```
sage: C = s0
```

We can now calculate the next terms:

```
sage: C.series(z, 11)
1*z + 1*z^2 + 2*z^3 + 5*z^4 + 14*z^5 + 42*z^6 +
132*z^7 + 429*z^8 + 1430*z^9 + 4862*z^10 + Order(z^11)
```

or calculate, more or less instantaneously, the 100-th coefficient:

```
sage: C.series(z, 101).coefficient(z,100)
227508830794229349661819540395688853956041682601541047340
```

It is unfortunate to have to recalculate everything if at some point we
wanted the 101-st coefficient. Lazy power series (see
`sage.combinat.species.series`

) come into their own here, in that
one can define them from a system of equations without solving it, and,
in particular, without needing a closed form for the answer. We begin by
defining the ring of lazy power series:

```
sage: L.<z> = LazyPowerSeriesRing(QQ)
```

Then we create a “free” power series, which we name, and which we then define by a recursive equation:

```
sage: C = L()
sage: C._name = 'C'
sage: C.define( z + C * C )
```

```
sage: [C.coefficient(i) for i in range(11)]
[0, 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862]
```

At any point, one can ask for any coefficient without having to redefine \(C\):

```
sage: C.coefficient(100)
227508830794229349661819540395688853956041682601541047340
sage: C.coefficient(200)
129013158064429114001222907669676675134349530552728882499810851598901419013348319045534580850847735528275750122188940
```

We now return to the closed form of \(C(z)\):

```
sage: z = var('z')
sage: C = s0; C
-1/2*sqrt(-4*z + 1) + 1/2
```

The \(n\)-th coefficient in the Taylor series for \(C(z)\) being given by \(\frac{1}{n!} C(z)^{(n)}(0)\), we look at the successive derivatives \(C(z)^{(n)}(z)\):

```
sage: derivative(C, z, 1)
1/sqrt(-4*z + 1)
sage: derivative(C, z, 2)
2/(-4*z + 1)^(3/2)
sage: derivative(C, z, 3)
12/(-4*z + 1)^(5/2)
```

This suggests the existence of a simple explicit formula, which we will now seek. The following small function returns \(d_n=n! \, c_n\):

```
sage: def d(n): return derivative(s0, n).subs(z=0)
```

Taking successive quotients:

```
sage: [ (d(n+1) / d(n)) for n in range(1,17) ]
[2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62]
```

we observe that \(d_n\) satisfies the recurrence relation \(d_{n+1}=(4n-2)d_n\), from which we deduce that \(c_n\) satisfies the recurrence relation \(c_{n+1}=\frac{(4n-2)}{n+1}c_n\). Simplifying, we find that \(c_n\) is the \((n-1)\)-th Catalan number:

We check this:

```
sage: n = var('n')
sage: c = 1/n*binomial(2*(n-1),n-1)
sage: [c.subs(n=k) for k in range(1, 11)]
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862]
sage: [catalan_number(k-1) for k in range(1, 11)]
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862]
```

We can now calculate coefficients much further; here we calculate \(c_{100000}\) which has more than \(60000\) digits:

```
sage: cc = c(n = 100000)
```

This takes a couple of seconds:

```
sage: %time cc = c(100000) # not tested
CPU times: user 2.34 s, sys: 0.00 s, total: 2.34 s
Wall time: 2.34 s
sage: ZZ(cc).ndigits()
60198
```

The methods which we have used generalize to all recursively defined
objects: the system of set-theoretic equations can be translated into a
system of equations on the generating function, which enables the
recursive calculation of its coefficients. If the set-theoretic
equations are simple enough (for example, if they only involve Cartesian
products and disjoint unions), the equation for \(C(z)\) is
algebraic. This equation has, in general, no closed-form solution.
However, using *confinement*, one can deduce a *linear* differential
equation which \(C(z)\) satisfies. This differential equation, in
turn, can be translated into a recurrence relation of fixed length on
its coefficients \(c_n\). In this case, the series is called
*D-finite*. After the initial calculation of this recurrence relation,
the calculation of coefficients is very fast. All these steps are purely
algorithmic, and it is planned to port into `Sage`

the implementations
which exist in `Maple`

(the `gfun`

and `combstruct`

packages) or
`MuPAD-Combinat`

(the `decomposableObjects`

library).

For the moment, we illustrate this general procedure in the case of complete binary trees. The generating function \(C(z)\) is a solution to an algebraic equation \(P(z,C(z)) = 0\), where \(P=P(x,y)\) is a polynomial with coefficients in \(\QQ\). In the present case, \(P=y^2-y+x\). We formally differentiate this equation with respect to \(z\):

```
sage: x, y, z = var('x, y, z')
sage: P = function('P')(x, y)
sage: C = function('C')(z)
sage: equation = P(x=z, y=C) == 0
sage: diff(equation, z)
diff(C(z), z)*D[1](P)(z, C(z)) + D[0](P)(z, C(z)) == 0
```

or, in a more readable format,

From this we deduce:

In the case of complete binary trees, this gives:

```
sage: P = y^2 - y + x
sage: Px = diff(P, x); Py = diff(P, y)
sage: - Px / Py
-1/(2*y - 1)
```

Recall that \(P(z, C(z))=0\). Thus, we can calculate this fraction
mod \(P\) and, in this way, express the derivative of
\(C(z)\) as a *polynomial in* \(C(z)\) *with coefficients in*
\(\QQ(z)\). In order to achieve this, we construct the quotient
ring \(R= \QQ(x)[y]/ (P)\):

```
sage: Qx = QQ['x'].fraction_field()
sage: Qxy = Qx['y']
sage: R = Qxy.quo(P); R
Univariate Quotient Polynomial Ring in ybar
over Fraction Field of Univariate Polynomial Ring in x
over Rational Field with modulus y^2 - y + x
```

Note: `ybar`

is the name of the variable \(y\) in the quotient ring.

Todo

add link to some tutorial on quotient rings

We continue the calculation of this fraction in \(R\):

```
sage: fraction = - R(Px) / R(Py); fraction
(1/2/(x - 1/4))*ybar - 1/4/(x - 1/4)
```

Note

The following variant does not work yet:

```
sage: fraction = R( - Px / Py ); fraction # todo: not implemented
Traceback (most recent call last):
...
TypeError: denominator must be a unit
```

We lift the result to \(\QQ(x)[y]\) and then substitute \(z\) and \(C(z)\) to obtain an expression for \(\frac{d}{dz}C(z)\):

```
sage: fraction = fraction.lift(); fraction
(1/2/(x - 1/4))*y - 1/4/(x - 1/4)
sage: fraction(x=z, y=C)
2*C(z)/(4*z - 1) - 1/(4*z - 1)
```

or, more legibly,

In this simple case, we can directly deduce from this expression a linear differential equation with coefficients in \(\QQ[z]\):

```
sage: equadiff = diff(C,z) == fraction(x=z, y=C)
sage: equadiff
diff(C(z), z) == 2*C(z)/(4*z - 1) - 1/(4*z - 1)
sage: equadiff = equadiff.simplify_rational()
sage: equadiff = equadiff * equadiff.rhs().denominator()
sage: equadiff = equadiff - equadiff.rhs()
sage: equadiff
(4*z - 1)*diff(C(z), z) - 2*C(z) + 1 == 0
```

or, more legibly,

It is trivial to verify this equation on the closed form:

```
sage: Cf = sage.symbolic.function_factory.function('C')
sage: equadiff.substitute_function(Cf, s0.function(z))
(4*z - 1)/sqrt(-4*z + 1) + sqrt(-4*z + 1) == 0
sage: bool(equadiff.substitute_function(Cf, s0.function(z)))
True
```

In the general case, one continues to calculate successive derivatives
of \(C(z)\). These derivatives are *confined* in the quotient ring
\(\QQ(z)[C]/(P)\) which is of finite dimension \(\deg P\)
over \(\QQ(z)\). Therefore, one will eventually find a linear
relation among the first \(\deg P\) derivatives of \(C(z)\).
Putting it over a single denominator, we obtain a linear
differential equation of degree \(\leq \deg P\) with coefficients
in \(\QQ[z]\). By extracting the coefficient of \(z^n\) in
the differential equation, we obtain the desired recurrence relation on
the coefficients; in this case we recover the relation we had already
found, based on the closed form:

After fixing the correct initial conditions, it becomes possible to calculate the coefficients of \(C(z)\) recursively:

```
sage: def C(n): return 1 if n <= 1 else (4*n-6)/n * C(n-1)
sage: [ C(i) for i in range(10) ]
[1, 1, 1, 2, 5, 14, 42, 132, 429, 1430]
```

If \(n\) is too large for the explicit calculation of \(c_n\), a sequence asymptotically equivalent to the sequence of coefficients \(c_n\) may be sought. Here again, there are generic techniques. The central tool is complex analysis, specifically, the study of the generating function around its singularities. In the present instance, the singularity is at \(z_0=1/4\) and one would obtain \(c_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}}\).

#### Summary#

We see here a general phenomenon of computer algebra: the best *data
structure* to describe a complicated mathematical object (a real number,
a sequence, a formal power series, a function, a set) is often an
equation defining the object (or a system of equations, typically with
some initial conditions). Attempting to find a closed-form
solution to this equation is not necessarily of interest: on the one
hand, such a closed form rarely exists (e.g., the problem of
solving a polynomial by radicals), and on the other hand, the equation,
in itself, contains all the necessary information to calculate
algorithmically the properties of the object under consideration (e.g.,
a numerical approximation, the initial terms or elements, an asymptotic
equivalent), or to calculate with the object itself (e.g., performing
arithmetic on power series). Therefore, instead of solving the equation,
we look for the equation describing the object which is best suited to
the problem we want to solve.

As we saw in our example, confinement (for example, in a finite dimensional vector space) is a fundamental tool for studying such equations. This notion of confinement is widely applicable in elimination techniques (linear algebra, Gröbner bases, and their algebro-differential generalizations). The same tool is central in algorithms for automatic summation and automatic verification of identities (Gosper’s algorithm, Zeilberger’s algorithm, and their generalizations; see also Exercise: alternating sign matrices).

Todo

add link to some tutorial on summation

All these techniques and their many generalizations are at the heart of
very active topics of research: automatic combinatorics and analytic
combinatorics, with major applications in the analysis of algorithms. It is
likely, and desirable, that they will be progressively implemented in
`Sage`

.

## Common enumerated sets#

### First example: the subsets of a set#

Fix a set \(E\) of size \(n\) and consider the subsets of
\(E\) of size \(k\). We know that these subsets are counted
by the binomial coefficients \(\binom n k\). We can therefore
calculate the number of subsets of size \(k=2\) of
\(E=\{1,2,3,4\}\) with the function `binomial`

:

```
sage: binomial(4, 2)
6
```

Alternatively, we can *construct* the set \(\mathcal P_2(E)\) of
all the subsets of size \(2\) of \(E\), then ask its
cardinality:

```
sage: S = Subsets([1,2,3,4], 2)
sage: S.cardinality()
6
```

Once `S`

has been constructed, we can also obtain the list of its
elements:

```
sage: S.list()
[{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}]
```

or select an element at random:

```
sage: S.random_element() # random
{1, 4}
```

More precisely, the object `S`

models the set
\(\mathcal P_2(E)\) equipped with a fixed order (here,
lexicographic order). It is therefore possible to ask for its
\(5\)-th element, keeping in mind that, as with `Python`

lists, the first
element is numbered zero:

```
sage: S.unrank(4)
{2, 4}
```

As a shortcut, in this setting, one can also use the notation:

```
sage: S[4]
{2, 4}
```

but this should be used with care because some sets have a natural indexing other than by \((0, 1, \dots)\).

Conversely, one can calculate the position of an object in this order:

```
sage: s = S([2,4]); s
{2, 4}
sage: S.rank(s)
4
```

Note that `S`

is *not* the list of its elements. One can, for example,
model the set \(\mathcal P(\mathcal P(\mathcal P(E)))\) and
calculate its cardinality (\(2^{2^{2^4}}\)):

```
sage: E = Set([1,2,3,4])
sage: S = Subsets(Subsets(Subsets(E))); S
Subsets of Subsets of Subsets of {1, 2, 3, 4}
sage: n = S.cardinality(); n
2003529930406846464979072351560255750447825475569751419265016973...
```

which is roughly \(2\cdot 10^{19728}\):

```
sage: n.ndigits()
19729
```

or ask for its \(237102124\)-th element:

```
sage: S.unrank(237102123) # random print output
{{{2, 4}, {1, 4}, {}, {1, 3, 4}, {1, 2, 4}, {4}, {2, 3}, {1, 3}, {2}},
{{1, 3}, {2, 4}, {1, 2, 4}, {}, {3, 4}}}
```

It would be physically impossible to construct explicitly all the elements of \(S\), as there are many more of them than there are particles in the universe (estimated at \(10^{82}\)).

Remark: it would be natural in `Python`

to use `len(S)`

to ask for the
cardinality of `S`

. This is not possible because `Python`

requires that the
result of `len`

be an integer of type `int`

; this could cause
overflows, and would not permit the return of {Infinity} for infinite
sets:

```
sage: len(S)
Traceback (most recent call last):
...
OverflowError: cannot fit 'int' into an index-sized integer
```

### Partitions of integers#

We now consider another classic problem: given a positive integer \(n\), in how many ways can it be written in the form of a sum \(n=i_1+i_2+\dots+i_\ell\), where \(i_1,\dots,i_\ell\) are positive integers? There are two cases to distinguish:

the order of the elements in the sum is not important, in which case we call \((i_1,\dots,i_\ell)\) a

*partition*of \(n\);the order of the elements in the sum is important, in which case we call \((i_1,\dots,i_\ell)\) a

*composition*of \(n\).

We will begin with the partitions of \(n=5\); as before, we begin by constructing the set of these partitions:

```
sage: P5 = Partitions(5); P5
Partitions of the integer 5
```

then we ask for its cardinality:

```
sage: P5.cardinality()
7
```

We look at these \(7\) partitions; the order being irrelevant, the entries are ordered, by convention, in decreasing order.

```
sage: P5.list()
[[5], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1],
[1, 1, 1, 1, 1]]
```

The calculation of the number of partitions uses the Rademacher
formula (Wikipedia article Partition_(number_theory)), implemented in `C`

and highly optimized, which makes it very fast:

```
sage: Partitions(100000).cardinality()
27493510569775696512677516320986352688173429315980054758203125984302147328114964173055050741660736621590157844774296248940493063070200461792764493033510116079342457190155718943509725312466108452006369558934464248716828789832182345009262853831404597021307130674510624419227311238999702284408609370935531629697851569569892196108480158600569421098519
```

Partitions of integers are combinatorial objects naturally equipped with many operations. They are therefore returned as objects that are richer than simple lists:

```
sage: P7 = Partitions(7)
sage: p = P7.unrank(5); p
[4, 2, 1]
sage: type(p)
<class 'sage.combinat.partition.Partitions_n_with_category.element_class'>
```

For example, they can be represented graphically by a Ferrers diagram:

```
sage: print(p.ferrers_diagram())
****
**
*
```

We leave it to the user to explore by introspection the available operations.

Note that we can also construct a partition directly by:

```
sage: Partition([4,2,1])
[4, 2, 1]
```

or:

```
sage: P7([4,2,1])
[4, 2, 1]
```

If one wants to restrict the possible values of the parts
\(i_1,\dots,i_\ell\) of the partition as, for example, when giving
change, one can use `WeightedIntegerVectors`

. For example, the
following calculation:

```
sage: WeightedIntegerVectors(8, [2,3,5]).list()
[[0, 1, 1], [1, 2, 0], [4, 0, 0]]
```

shows that to make 8 dollars using 2, 3, and 5 dollar bills, one can use a 3 and a 5 dollar bill, or a 2 and two 3 dollar bills, or four 2 dollar bills.

Compositions of integers are manipulated the same way:

```
sage: C5 = Compositions(5); C5
Compositions of 5
sage: C5.cardinality()
16
sage: C5.list()
[[1, 1, 1, 1, 1], [1, 1, 1, 2], [1, 1, 2, 1], [1, 1, 3],
[1, 2, 1, 1], [1, 2, 2], [1, 3, 1], [1, 4], [2, 1, 1, 1],
[2, 1, 2], [2, 2, 1], [2, 3], [3, 1, 1], [3, 2], [4, 1], [5]]
```

The number \(16\) above seems significant and suggests the existence of a formula. We look at the number of compositions of \(n\) ranging from \(0\) to \(9\):

```
sage: [ Compositions(n).cardinality() for n in range(10) ]
[1, 1, 2, 4, 8, 16, 32, 64, 128, 256]
```

Similarly, if we consider the number of compositions of \(5\) by length, we find a line of Pascal’s triangle:

```
sage: x = var('x')
sage: sum( x^len(c) for c in C5 )
x^5 + 4*x^4 + 6*x^3 + 4*x^2 + x
```

The above example uses a functionality which we have not seen yet:
`C5`

being iterable, it can be used like a list in a `for`

loop or
a comprehension (Set comprehension and iterators).

Prove the formulas suggested by the above examples for the number of compositions of \(n\) and the number of compositions of \(n\) of length \(k\); investigate by introspection whether

`Sage`

uses these formulas for calculating cardinalities.

### Some other finite enumerated sets#

Essentially, the principle is the same for all the finite sets with
which one wants to do combinatorics in `Sage`

; begin by constructing
an object which models this set, and then supply appropriate methods,
following a uniform interface 1. We now give a few more typical
examples.

Intervals of integers:

```
sage: C = IntegerRange(3, 21, 2); C
{3, 5, ..., 19}
sage: C.cardinality()
9
sage: C.list()
[3, 5, 7, 9, 11, 13, 15, 17, 19]
```

Permutations:

```
sage: C = Permutations(4); C
Standard permutations of 4
sage: C.cardinality()
24
sage: C.list()
[[1, 2, 3, 4], [1, 2, 4, 3], [1, 3, 2, 4], [1, 3, 4, 2],
[1, 4, 2, 3], [1, 4, 3, 2], [2, 1, 3, 4], [2, 1, 4, 3],
[2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1],
[3, 1, 2, 4], [3, 1, 4, 2], [3, 2, 1, 4], [3, 2, 4, 1],
[3, 4, 1, 2], [3, 4, 2, 1], [4, 1, 2, 3], [4, 1, 3, 2],
[4, 2, 1, 3], [4, 2, 3, 1], [4, 3, 1, 2], [4, 3, 2, 1]]
```

Set partitions:

```
sage: C = SetPartitions(["a", "b", "c"])
sage: C # random print output
Set partitions of {'a', 'c', 'b'}
sage: C.cardinality()
5
sage: C.list()
[{{'a', 'b', 'c'}},
{{'a', 'b'}, {'c'}},
{{'a', 'c'}, {'b'}},
{{'a'}, {'b', 'c'}},
{{'a'}, {'b'}, {'c'}}]
```

Partial orders on a set of \(8\) elements, up to isomorphism:

```
sage: C = Posets(8); C
Posets containing 8 elements
sage: C.cardinality()
16999
```

```
sage: C.unrank(20).plot()
Graphics object consisting of 20 graphics primitives
```

One can iterate through all graphs up to isomorphism. For example, there are 34 simple graphs with 5 vertices:

```
sage: len(list(graphs(5)))
34
```

Here are those with at most \(4\) edges:

```
sage: up_to_four_edges = list(graphs(5, lambda G: G.size() <= 4))
sage: pretty_print(*up_to_four_edges)
```

However, the *set* `C`

of these graphs is not yet available in
`Sage`

; as a result, the following commands are not yet
implemented:

```
sage: C = Graphs(5) # todo: not implemented
sage: C.cardinality() # todo: not implemented
34
sage: Graphs(19).cardinality() # todo: not implemented
24637809253125004524383007491432768
sage: Graphs(19).random_element() # todo: not implemented
Graph on 19 vertices
```

What we have seen so far also applies, in principle, to finite algebraic structures like the dihedral groups:

```
sage: G = DihedralGroup(4); G
Dihedral group of order 8 as a permutation group
sage: G.cardinality()
8
sage: G.list()
[(), (1,3)(2,4), (1,4,3,2), (1,2,3,4), (2,4), (1,3), (1,4)(2,3), (1,2)(3,4)]
```

or the algebra of \(2\times 2\) matrices over the finite field \(\ZZ/2\ZZ\):

```
sage: C = MatrixSpace(GF(2), 2)
sage: C.list()
[
[0 0] [1 0] [0 1] [0 0] [0 0] [1 1] [1 0] [1 0] [0 1] [0 1]
[0 0], [0 0], [0 0], [1 0], [0 1], [0 0], [1 0], [0 1], [1 0], [0 1],
[0 0] [1 1] [1 1] [1 0] [0 1] [1 1]
[1 1], [1 0], [0 1], [1 1], [1 1], [1 1]
]
sage: C.cardinality()
16
```

Exercise

List all the monomials of degree \(5\) in three variables (see
`IntegerVectors`

). Manipulate the ordered set partitions
`OrderedSetPartitions`

and standard tableaux
(`StandardTableaux`

).

Exercise

List the alternating sign matrices of size \(3\), \(4\),
and \(5\) (`AlternatingSignMatrices`

), and try to guess the
definition. The discovery and proof of the formula for the
enumeration of these matrices (see the method `cardinality`

),
motivated by calculations of determinants in physics, is quite a
story. In particular, the first proof, given by Zeilberger in 1992
was automatically produced by a computer program. It was 84 pages long,
and required nearly a hundred people to verify it.

Exercise

Calculate by hand the number of vectors in \((\ZZ/2\ZZ)^5\), and
the number of matrices in \(GL_3(\ZZ/2\ZZ)\) (that is to say,
the number of invertible \(3\times 3\) matrices with
coefficients in \(\ZZ/2\ZZ\)). Verify your answer with `Sage`

.
Generalize to \(GL_n(\ZZ/q\ZZ)\).

### Set comprehension and iterators#

We will now show some of the possibilities offered by `Python`

for
constructing (and iterating through) sets, with a notation that is
flexible and close to usual mathematical usage, and in particular the
benefits this yields in combinatorics.

We begin by constructing the finite set \(\{i^2\ \|\ i \in \{1,3,7\}\}\):

```
sage: [ i^2 for i in [1, 3, 7] ]
[1, 9, 49]
```

and then the same set, but with \(i\) running from \(1\) to \(9\):

```
sage: [ i^2 for i in range(1,10) ]
[1, 4, 9, 16, 25, 36, 49, 64, 81]
```

A construction of this form in `Python`

is called *set comprehension*.
A clause can be added to keep only those elements with \(i\) prime:

```
sage: [ i^2 for i in range(1,10) if is_prime(i) ]
[4, 9, 25, 49]
```

Combining more than one set comprehension, it is possible to construct the set \(\{(i,j) \ | \ 1\leq j < i <5\}\):

```
sage: [ (i,j) for i in range(1,6) for j in range(1,i) ]
[(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3),
(5, 1), (5, 2), (5, 3), (5, 4)]
```

or to produce Pascal’s triangle:

```
sage: [[binomial(n,i) for i in range(n+1)] for n in range(10)]
[[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1],
[1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1],
[1, 7, 21, 35, 35, 21, 7, 1],
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]
```

The execution of a set comprehension is accomplished in two steps; first
an *iterator* is constructed, and then a list is filled with the
elements successively produced by the iterator. Technically, an
*iterator* is an object with a method `next`

which returns a new value
each time it is called, until it is exhausted. For example, the
following iterator `it`

:

```
sage: it = (binomial(3, i) for i in range(4))
```

returns successively the binomial coefficients \(\binom 3 i\) with \(i=0,1,2,3\):

```
sage: next(it)
1
sage: next(it)
3
sage: next(it)
3
sage: next(it)
1
```

When the iterator is finally exhausted, an exception is raised:

```
sage: next(it)
Traceback (most recent call last):
...
StopIteration
```

More generally, an *iterable* is a `Python`

object `L`

(a list,
a set, …) over whose elements it is possible to iterate. Technically,
the iterator is constructed by `iter(L)`

. In practice, the commands
`iter`

and `next`

are used very rarely, since `for`

loops and list
comprehensions provide a much pleasanter syntax:

```
sage: for s in Subsets(3):
....: print(s)
{}
{1}
{2}
{3}
{1, 2}
{1, 3}
{2, 3}
{1, 2, 3}
```

```
sage: [ s.cardinality() for s in Subsets(3) ]
[0, 1, 1, 1, 2, 2, 2, 3]
```

What is the point of an iterator? Consider the following example:

```
sage: sum( [ binomial(8, i) for i in range(9) ] )
256
```

When it is executed, a list of \(9\) elements is constructed, and
then it is passed as an argument to `sum`

to add them up. If, on the
other hand, the iterator is passed directly to `sum`

(note the absence
of square brackets):

```
sage: sum( binomial(8, i) for i in range(9) )
256
```

the function `sum`

receives the iterator directly, and can
short-circuit the construction of the intermediate list. If there are a
large number of elements, this avoids allocating a large quantity of
memory to fill a list which will be immediately destroyed 2.

Most functions that take a list of elements as input will also accept an iterator (or an iterable) instead. To begin with, one can obtain the list (or the tuple) of elements of an iterator as follows:

```
sage: list(binomial(8, i) for i in range(9))
[1, 8, 28, 56, 70, 56, 28, 8, 1]
sage: tuple(binomial(8, i) for i in range(9))
(1, 8, 28, 56, 70, 56, 28, 8, 1)
```

We now consider the functions `all`

and `any`

which denote
respectively the \(n\)-ary *and* and *or*:

```
sage: all([True, True, True, True])
True
sage: all([True, False, True, True])
False
sage: any([False, False, False, False])
False
sage: any([False, False, True, False])
True
```

The following example verifies that all primes from \(3\) to \(99\) are odd:

```
sage: all( is_odd(p) for p in range(3,100) if is_prime(p) )
True
```

A *Mersenne prime* is a prime of the form \(2^p -1\). We verify
that, for \(p<1000\), if \(2^p-1\) is prime, then
\(p\) is also prime:

```
sage: def mersenne(p): return 2^p -1
sage: [ is_prime(p)
....: for p in range(1000) if is_prime(mersenne(p)) ]
[True, True, True, True, True, True, True, True, True, True,
True, True, True, True]
```

Is the converse true?

Exercise

Try the two following commands and explain the considerable difference in the length of the calculations:

```
sage: all( is_prime(mersenne(p))
....: for p in range(1000) if is_prime(p) )
False
sage: all( [ is_prime(mersenne(p))
....: for p in range(1000) if is_prime(p)] )
False
```

We now try to find the smallest counter-example. In order to do this, we
use the `Sage`

function `exists`

:

```
sage: exists( (p for p in range(1000) if is_prime(p)),
....: lambda p: not is_prime(mersenne(p)) )
(True, 11)
```

Alternatively, we could construct an iterator on the counter-examples:

```
sage: counter_examples = (p for p in range(1000)
....: if is_prime(p) and not is_prime(mersenne(p)))
sage: next(counter_examples)
11
sage: next(counter_examples)
23
```

Exercise

What do the following commands do?

```
sage: cubes = [t**3 for t in range(-999,1000)]
sage: exists([(x,y) for x in cubes for y in cubes], # long time (3s, 2012)
....: lambda x_y: x_y[0] + x_y[1] == 218)
(True, (-125, 343))
sage: exists(((x,y) for x in cubes for y in cubes), # long time (2s, 2012)
....: lambda x_y: x_y[0] + x_y[1] == 218)
(True, (-125, 343))
```

Which of the last two is more economical in terms of time? In terms of memory? By how much?

Exercise

Try each of the following commands, and explain its result. If possible, hide the result first and try to guess it before launching the command.

Todo

hide the results by default

Warning

it will be necessary to interrupt the execution of some of the commands

```
sage: x = var('x')
sage: sum( x^len(s) for s in Subsets(8) )
x^8 + 8*x^7 + 28*x^6 + 56*x^5 + 70*x^4 + 56*x^3 + 28*x^2 + 8*x + 1
```

```
sage: sum( x^p.length() for p in Permutations(3) )
x^3 + 2*x^2 + 2*x + 1
```

```
sage: factor(sum( x^p.length() for p in Permutations(3) ))
(x^2 + x + 1)*(x + 1)
```

```
sage: P = Permutations(5)
sage: all( p in P for p in P )
True
```

```
sage: for p in GL(2, 2): print(p); print("")
[1 0]
[0 1]
[0 1]
[1 0]
[0 1]
[1 1]
[1 1]
[0 1]
[1 1]
[1 0]
[1 0]
[1 1]
```

```
sage: for p in Partitions(3): print(p) # not tested
[3]
[2, 1]
[1, 1, 1]
...
```

```
sage: for p in Partitions(): print(p) # not tested
[]
[1]
[2]
[1, 1]
[3]
...
```

```
sage: for p in Primes(): print(p) # not tested
2
3
5
7
...
```

```
sage: exists( Primes(), lambda p: not is_prime(mersenne(p)) )
(True, 11)
```

```
sage: counter_examples = (p for p in Primes()
....: if not is_prime(mersenne(p)))
sage: for p in counter_examples: print(p) # not tested
11
23
29
37
41
43
47
...
```

#### Operations on iterators#

`Python`

provides numerous tools for manipulating iterators; most of them
are in the `itertools`

library, which can be imported by:

```
sage: import itertools
```

We will demonstrate some applications, taking as a starting point the permutations of \(3\):

```
sage: list(Permutations(3))
[[1, 2, 3], [1, 3, 2], [2, 1, 3],
[2, 3, 1], [3, 1, 2], [3, 2, 1]]
```

We can list the elements of a set by numbering them:

```
sage: list(enumerate(Permutations(3)))
[(0, [1, 2, 3]), (1, [1, 3, 2]), (2, [2, 1, 3]),
(3, [2, 3, 1]), (4, [3, 1, 2]), (5, [3, 2, 1])]
```

or select only the elements in positions 2, 3, and 4 (analogue of
`l[1:4]`

):

```
sage: import itertools
sage: list(itertools.islice(Permutations(3), int(1), int(4)))
[[1, 3, 2], [2, 1, 3], [2, 3, 1]]
```

To apply a function to all the elements, one can do:

```
sage: [z.cycle_type() for z in Permutations(3)]
[[1, 1, 1], [2, 1], [2, 1], [3], [3], [2, 1]]
```

and similarly to select the elements satisfying a certain condition:

```
sage: [z for z in Permutations(3) if z.has_pattern([1,2])]
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2]]
```

#### Implementation of new iterators#

It is easy to construct new iterators, using the keyword `yield`

instead of `return`

in a function:

```
sage: def f(n):
....: for i in range(n):
....: yield i
```

After the `yield`

, execution is not halted, but only suspended, ready
to be continued from the same point. The result of the function is
therefore an iterator over the successive values returned by `yield`

:

```
sage: g = f(4)
sage: next(g)
0
sage: next(g)
1
sage: next(g)
2
sage: next(g)
3
```

```
sage: next(g)
Traceback (most recent call last):
...
StopIteration
```

The function could be used as follows:

```
sage: [ x for x in f(5) ]
[0, 1, 2, 3, 4]
```

This model of computation, called *continuation*, is very useful in
combinatorics, especially when combined with recursion. Here is how to
generate all words of a given length on a given alphabet:

```
sage: def words(alphabet,l):
....: if l == 0:
....: yield []
....: else:
....: for word in words(alphabet, l-1):
....: for l in alphabet:
....: yield word + [l]
sage: [ w for w in words(['a','b'], 3) ]
[['a', 'a', 'a'], ['a', 'a', 'b'], ['a', 'b', 'a'],
['a', 'b', 'b'], ['b', 'a', 'a'], ['b', 'a', 'b'],
['b', 'b', 'a'], ['b', 'b', 'b']]
```

These words can then be counted by:

```
sage: sum(1 for w in words(['a','b','c','d'], 10))
1048576
```

Counting the words one by one is clearly not an efficient method in this case, since the formula \(n^\ell\) is also available; note, though, that this is not the stupidest possible approach - it does, at least, avoid constructing the entire list in memory.

We now consider Dyck words, which are well-parenthesized words in the letters “\((\)” and “\()\)”. The function below generates all the Dyck words of a given length (where the length is the number of pairs of parentheses), using the recursive definition which says that a Dyck word is either empty or of the form \((w_1)w_2\) where \(w_1\) and \(w_2\) are Dyck words:

```
sage: def dyck_words(l):
....: if l==0:
....: yield ''
....: else:
....: for k in range(l):
....: for w1 in dyck_words(k):
....: for w2 in dyck_words(l-k-1):
....: yield '('+w1+')'+w2
```

Here are all the Dyck words of length \(4\):

```
sage: list(dyck_words(4))
['()()()()', '()()(())', '()(())()', '()(()())', '()((()))',
'(())()()', '(())(())', '(()())()', '((()))()', '(()()())',
'(()(()))', '((())())', '((()()))', '(((())))']
```

Counting them, we recover a well-known sequence:

```
sage: [ sum(1 for w in dyck_words(l)) for l in range(10) ]
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862]
```

Exercise: complete binary tree iterator

Construct an iterator on the set \(C_n\) of complete binary trees with \(n\) leaves (see Enumeration of trees using generating functions).

Hint: `Sage`

4.8.2 does not yet have a native data structure to
represent complete binary trees. One simple way to represent them is
to define a formal variable `Leaf`

for the leaves and a formal
2-ary function `Node`

:

```
sage: var('Leaf')
Leaf
sage: function('Node', nargs=2)
Node
```

The second tree in Figure: The five complete binary trees with four leaves can be represented by the expression:

```
sage: tr = Node(Node(Leaf, Node(Leaf, Leaf)), Leaf)
```

## Constructions#

We will now see how to construct new sets starting from these building blocks. In fact, we have already begun to do this with the construction of \(\mathcal P(\mathcal P(\mathcal P(\{1,2,3,4\})))\) in the previous section, and to construct the example of sets of cards in Initial examples.

Consider a large Cartesian product:

```
sage: C = cartesian_product([Compositions(8), Permutations(20)]); C
The Cartesian product of (Compositions of 8, Standard permutations of 20)
sage: C.cardinality()
311411457046609920000
```

Clearly, it is impractical to construct the list of all the elements of this Cartesian product! And, in the following example, \(H\) is equipped with the usual combinatorial operations and also its structure as a product group:

```
sage: G = DihedralGroup(4)
sage: H = cartesian_product([G,G])
sage: H in Groups()
True
sage: H.an_element()
((1,3), (1,3))
sage: t = H([G.gen(0), G.gen(0)])
sage: t
((1,2,3,4), (1,2,3,4))
sage: t*t
((1,3)(2,4), (1,3)(2,4))
```

We now construct the union of two existing disjoint sets:

```
sage: C = DisjointUnionEnumeratedSets(
....: [ Compositions(4), Permutations(3)] )
sage: C
Disjoint union of Family (Compositions of 4,
Standard permutations of 3)
sage: C.cardinality()
14
sage: C.list()
[[1, 1, 1, 1], [1, 1, 2], [1, 2, 1], [1, 3], [2, 1, 1], [2, 2],
[3, 1], [4], [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1],
[3, 1, 2], [3, 2, 1]]
```

It is also possible to take the union of more than two disjoint sets, or even an infinite number of them. We will now construct the set of all permutations, viewed as the union of the sets \(P_n\) of permutations of size \(n\). We begin by constructing the infinite family \(F=(P_n)_{n\in N}\):

```
sage: F = Family(NonNegativeIntegers(), Permutations); F
Lazy family (<class 'sage.combinat.permutation.Permutations'>(i))_{i in Non negative integers}
sage: F.keys()
Non negative integers
sage: F[1000]
Standard permutations of 1000
```

Now we can construct the disjoint union \(\bigcup_{n\in \NN}P_n\):

```
sage: U = DisjointUnionEnumeratedSets(F); U
Disjoint union of
Lazy family (<class 'sage.combinat.permutation.Permutations'>(i))_{i in Non negative integers}
```

It is an infinite set:

```
sage: U.cardinality()
+Infinity
```

which doesn’t prohibit iteration through its elements, though it will be necessary to interrupt it at some point:

```
sage: for p in U: # not tested
....: print(p)
[]
[1]
[1, 2]
[2, 1]
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
...
```

Note: the above set could also have been constructed directly with:

```
sage: U = Permutations(); U
Standard permutations
```

### Summary#

`Sage`

provides a library of common enumerated sets, which can be
combined by standard constructions, giving a toolbox that is flexible
(but which could still be expanded). It is also possible to add new
building blocks to `Sage`

with a few lines (see the code in
`FiniteEnumeratedSets().example()`

). This is made possible by the
uniformity of the interfaces and the fact that `Sage`

is based on an
object-oriented language. Also, very large or even infinite sets can
be manipulated thanks to lazy evaluation strategies (iterators, etc.).

There is no magic to any of this: under the hood, `Sage`

applies the
usual rules (for example, that the cardinality of \(E\times E\) is
\(|E|^2\)); the added value comes from the capacity to manipulate
complicated constructions. The situation is comparable to `Sage`

’s
implementation of differential calculus: `Sage`

applies the usual
rules for differentiation of functions and their compositions, where
the added value comes from the possibility of manipulating complicated
formulas. In this sense, `Sage`

implements a *calculus* of finite
enumerated sets.

## Generic algorithms#

### Lexicographic generation of lists of integers#

Among the classic enumerated sets, especially in algebraic combinatorics, a certain number are composed of lists of integers of fixed sum, such as partitions, compositions, or integer vectors. These examples can also have supplementary constraints added to them. Here are some examples. We start with the integer vectors with sum \(10\) and length \(3\), with parts bounded below by \(2\), \(4\) and \(2\) respectively:

```
sage: IntegerVectors(10, 3, min_part=2, max_part=5,
....: inner=[2, 4, 2]).list()
[[4, 4, 2], [3, 5, 2], [3, 4, 3], [2, 5, 3], [2, 4, 4]]
```

The compositions of \(5\) with each part at most \(3\), and with length \(2\) or \(3\):

```
sage: Compositions(5, max_part=3,
....: min_length=2, max_length=3).list()
[[3, 2], [3, 1, 1], [2, 3], [2, 2, 1], [2, 1, 2], [1, 3, 1],
[1, 2, 2], [1, 1, 3]]
```

The strictly decreasing partitions of \(5\):

```
sage: Partitions(5, max_slope=-1).list()
[[5], [4, 1], [3, 2]]
```

These sets share the same underlying algorithmic structure, implemented
in the more general (and slightly more cumbersome) class
`IntegerListsLex`

. This class models sets of vectors
\((\ell_0,\dots,\ell_k)\) of non-negative integers, with
constraints on the sum and the length, and bounds on the parts and on
the consecutive differences between the parts. Here are some more
examples:

```
sage: IntegerListsLex(10, length=3,
....: min_part=2, max_part=5,
....: floor=[2, 4, 2]).list()
[[4, 4, 2], [3, 5, 2], [3, 4, 3], [2, 5, 3], [2, 4, 4]]
sage: IntegerListsLex(5, min_part=1, max_part=3,
....: min_length=2, max_length=3).list()
[[3, 2], [3, 1, 1], [2, 3], [2, 2, 1], [2, 1, 2],
[1, 3, 1], [1, 2, 2], [1, 1, 3]]
sage: IntegerListsLex(5, min_part=1, max_slope=-1).list()
[[5], [4, 1], [3, 2]]
sage: list(Compositions(5, max_length=2))
[[5], [4, 1], [3, 2], [2, 3], [1, 4]]
sage: list(IntegerListsLex(5, max_length=2, min_part=1))
[[5], [4, 1], [3, 2], [2, 3], [1, 4]]
```

The point of the model of `IntegerListsLex`

is in the compromise
between generality and efficiency. The main algorithm permits
iteration through the elements of such a set \(S\) in reverse
lexicographic order with a good complexity in most practical use
cases. Roughly speaking, the time needed to iterate through all the
elements of \(S\) is proportional to the number of elements, where the
proportion factor is controlled by the length \(l\) of the longest
element of \(S\). In addition, the memory usage is also controlled by
\(l\), which is to say negligible in practice.

This algorithm is based on a very general principle for traversing a
decision tree, called *branch and bound*: at the top level, we run
through all the possible choices for \(\ell_0\); for each of these
choices, we run through all the possible choices for \(\ell_1\),
and so on. Mathematically speaking, we have put the structure of a
prefix tree on the elements of \(S\): a node of the tree at depth
\(k\) corresponds to a prefix \(\ell_0,\dots,\ell_k\) of one
(or more) elements of \(S\) (see Figure: The prefix tree of the partitions of 5.).

The usual problem with this type of approach is to avoid bad decisions which lead to leaving the prefix tree and exploring dead branches; this is particularly problematic because the growth of the number of elements is usually exponential in the depth. It turns out that the constraints listed above are simple enough to be able to reasonably predict when a sequence \(\ell_0,\dots,\ell_k\) is a prefix of some element \(S\). Hence, most dead branches can be pruned.

### Integer points in polytopes#

Although the algorithm for iteration in `IntegerListsLex`

is
efficient, its counting algorithm is naive: it just iterates over all
the elements.

There is an alternative approach to treating this problem: modelling the
desired lists of integers as the set of integer points of a polytope,
that is to say, the set of solutions with integer coordinates of a
system of linear inequalities. This is a very general context in which
there exist advanced counting algorithms (e.g. Barvinok), which are
implemented in libraries like `LattE`

. Iteration does not pose a hard problem
in principle. However, there are two limitations that justify the
existence of `IntegerListsLex`

. The first is theoretical: lattice
points in a polytope only allow modelling of problems of a fixed
dimension (length). The second is practical: at the moment only the
library `PALP`

has a `Sage`

interface, and though it offers multiple
capabilities for the study of polytopes, in the present application it
only produces a list of lattice points, without providing either an
iterator or non-naive counting:

```
sage: A = random_matrix(ZZ, 6, 3, x=7)
sage: L = LatticePolytope(A.rows())
sage: L.points() # random
M(4, 1, 0),
M(0, 3, 5),
M(2, 2, 3),
M(6, 1, 3),
M(1, 3, 6),
M(6, 2, 3),
M(3, 2, 4),
M(3, 2, 3),
M(4, 2, 4),
M(4, 2, 3),
M(5, 2, 3)
in 3-d lattice M
sage: L.npoints() # random
11
```

This polytope can be visualized in 3D with `L.plot3d()`

(see
Figure: The polytope L and its integer points, in cross-eyed stereographic perspective.).

### Species, decomposable combinatorial classes#

In Enumeration of trees using generating functions, we showed how to use the recursive
definition of binary trees to count them efficiently using generating
functions. The techniques we used there are very general, and apply
whenever the sets involved can be defined recursively (depending on
who you ask, such a set is called a *decomposable combinatorial class*
or, roughly speaking, a *combinatorial species*). This includes all
the types of trees, but also permutations, compositions, functional
graphs, etc.

Here, we illustrate just a few examples using the `Sage`

library on
combinatorial species:

```
sage: from sage.combinat.species.library import *
sage: o = var('o')
```

We begin by redefining the complete binary trees; to do so, we stipulate the recurrence relation directly on the sets:

```
sage: BT = CombinatorialSpecies()
sage: Leaf = SingletonSpecies()
sage: BT.define( Leaf + (BT*BT) )
```

Now we can construct the set of trees with five nodes, list them, count them…:

```
sage: BT5 = BT.isotypes([o]*5)
sage: BT5.cardinality()
14
sage: BT5.list()
[o*(o*(o*(o*o))), o*(o*((o*o)*o)), o*((o*o)*(o*o)),
o*((o*(o*o))*o), o*(((o*o)*o)*o), (o*o)*(o*(o*o)),
(o*o)*((o*o)*o), (o*(o*o))*(o*o), ((o*o)*o)*(o*o),
(o*(o*(o*o)))*o, (o*((o*o)*o))*o, ((o*o)*(o*o))*o,
((o*(o*o))*o)*o, (((o*o)*o)*o)*o]
```

The trees are constructed using a generic recursive structure; the
display is therefore not wonderful. To do better, it would be necessary
to provide `Sage`

with a more specialized data structure with the
desired display capabilities.

We recover the generating function for the Catalan numbers:

```
sage: g = BT.isotype_generating_series(); g
x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + O(x^6)
```

which is returned in the form of a lazy power series:

```
sage: g[100]
227508830794229349661819540395688853956041682601541047340
```

We finish with the Fibonacci words, which are binary words without two consecutive “\(1\)”s. They admit a natural recursive definition:

```
sage: Eps = EmptySetSpecies()
sage: Z0 = SingletonSpecies()
sage: Z1 = Eps*SingletonSpecies()
sage: FW = CombinatorialSpecies()
sage: FW.define(Eps + Z0*FW + Z1*Eps + Z1*Z0*FW)
```

The Fibonacci sequence is easily recognized here, hence the name:

```
sage: L = FW.isotype_generating_series().coefficients(15); L
[1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987]
```

```
sage: oeis(L) # optional -- internet
0: A000045: Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
1: ...
2: ...
```

This is an immediate consequence of the recurrence relation. One can also generate immediately all the Fibonacci words of a given length, with the same limitations resulting from the generic display.

```
sage: FW3 = FW.isotypes([o]*3)
sage: FW3.list()
[o*(o*(o*{})), o*(o*(({}*o)*{})), o*((({}*o)*o)*{}),
(({}*o)*o)*(o*{}), (({}*o)*o)*(({}*o)*{})]
```

### Graphs up to isomorphism#

We saw in Some other finite enumerated sets that `Sage`

could generate
graphs and partial orders up to isomorphism. We will now describe the
underlying algorithm, which is the same in both cases, and covers a
substantially wider class of problems.

We begin by recalling some notions. A graph \(G=(V,E)\) is a set \(V\) of vertices and a set \(E\) of edges connecting these vertices; an edge is described by a pair \(\{u,v\}\) of distinct vertices of \(V\). Such a graph is called labelled; its vertices are typically numbered by considering \(V=\{1,2,3,4,5\}\).

In many problems, the labels on the vertices play no role. Typically a chemist wants to study all the possible molecules with a given composition, for example the alkanes with \(n=8\) atoms of carbon and \(2n+2=18\) atoms of hydrogen. He therefore wants to find all the graphs consisting of \(8\) vertices with \(4\) neighbours, and \(18\) vertices with a single neighbour. The different carbon atoms, however, are all considered to be identical, and the same for the hydrogen atoms. The problem of our chemist is not imaginary; this type of application is actually at the origin of an important part of the research in graph theory on isomorphism problems.

Working by hand on a small graph it is possible, as in the example of
Some other finite enumerated sets, to make a drawing, erase the labels, and
“forget” the geometrical information about the location of the
vertices in the plane. However, to represent a graph in a computer
program, it is necessary to introduce labels on the vertices so as to
be able to describe how the edges connect them together. To compensate
for the extra information which we have introduced, we then say that
two labelled graphs \(g_1\) and \(g_2\) are *isomorphic* if there is a
bijection from the vertices of \(g_1\) to those of \(g_2\), which maps
bijectively the edges of \(g_1\) to those of \(g_2\); an *unlabelled
graph* is then an equivalence class of labelled graphs.

In general, testing if two labelled graphs are isomorphic is expensive.
However, the number of graphs, even unlabelled, grows very
rapidly. Nonetheless, it is possible to list unlabelled graphs very efficiently
considering their number. For example, the program `Nauty`

can list the
\(12005168\) simple graphs with \(10\) vertices in
\(20\) seconds.

As in Lexicographic generation of lists of integers, the general principle of the algorithm is to organize the objects to be enumerated into a tree that one traverses.

For this, in each equivalence class of labelled graphs (that is to say, for each unlabelled graph) one fixes a convenient canonical representative. The following are the fundamental operations:

Testing whether a labelled graph is canonical

Calculating the canonical representative of a labelled graph

These unavoidable operations remain expensive; one therefore tries to minimize the number of calls to them.

The canonical representatives are chosen in such a way that, for each
canonical labelled graph \(G\), there is a canonical choice of an edge
whose removal produces a canonical graph again, which is called the
father of \(G\). This property implies that it is possible to organize
the set of canonical representatives as a tree: at the root, the graph
with no edges; below it, its unique child, the graph with one edge;
then the graphs with two edges, and so on. The set of children of a
graph \(G\) can be constructed by *augmentation*, adding an edge in all
the possible ways to \(G\), and then selecting, from among those graphs,
the ones that are still canonical 3. Recursively, one obtains all
the canonical graphs.

In what sense is this algorithm generic? Consider for example planar graphs (graphs which can be drawn in the plane without edges crossing): by removing an edge from a planar graph, one obtains another planar graph; so planar graphs form a subtree of the previous tree. To generate them, exactly the same algorithm can be used, selecting only the children which are planar:

```
sage: [len(list(graphs(n, property = lambda G: G.is_planar())))
....: for n in range(7)]
[1, 1, 2, 4, 11, 33, 142]
```

In a similar fashion, one can generate any family of graphs closed under deletion of an edge, and in particular any family characterized by a forbidden subgraph. This includes for example forests (graphs without cycles), bipartite graphs (graphs without odd cycles), etc. This can be applied to generate:

partial orders, via the bijection with Hasse diagrams which are oriented graphs without cycles and without edges implied by the transitivity of the order relation;

lattices (not implemented in

`Sage`

), via the bijection with the meet semi-lattice obtained by deleting the maximal vertex; in this case an augmentation by vertices rather than by edges is used.

REFERENCES:

- CMS2012
Alexandre Casamayou, Nathann Cohen, Guillaume Connan, Thierry Dumont, Laurent Fousse, François Maltey, Matthias Meulien, Marc Mezzarobba, Clément Pernet, Nicolas M. Thiéry, Paul Zimmermann

Calcul Mathématique avec Sagehttps://www.sagemath.org/sagebook/french.html

- 1
Or at least that should be the case; there are still many corners to clean up.

- 2
Technical detail:

`range`

returns an iterator on \(\{0,\dots,8\}\) while`range`

returns the corresponding list. Starting in`Python`

3.0,`range`

will behave like`range`

, and`range`

will no longer be needed.- 3
In practice, an efficient implementation would exploit the symmetries of \(G\), i.e., its automorphism group, to reduce the number of children to explore, and to reduce the cost of each test of canonicity.