Discrete Applied Mathematics ( ) –

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Discrete Applied Mathematics

journal homepage: www.elsevier.com/locate/dam

On the linear description of the Huffman trees polytope✩

Jean-Francois Maurras a, Thanh Hai Nguyen b,∗, Viet Hung Nguyen c

a Laboratoire d’Informatique Fondamentale de Marseille, Université dAix Marseille, Faculté des Sciences de Luminy, 163, Avenue de Luminy, 13288 MarseilleCedex 8, Franceb CEA LIST, Laboratoire des fondements des Systèmes Temps Réels Embarqués, Centre de Saclay, Nano-Innov, 91191 Gif-Sur-Yvette Cedex, Francec Laboratoire d’Informatique de Paris 6, Université de Paris 6, 4 place Jussieu 75005 Paris, France

a r t i c l e i n f o

Article history:Received 20 July 2010Received in revised form 8 May 2012Accepted 17 May 2012Available online xxxx

Keywords:Polyhedral combinatoricsCombinatorial optimizationCombinatorial inequalitiesBinary treesPolytopesHyperplaneHuffman codingHuffman treeFibonacci

a b s t r a c t

The Huffman tree is a well known concept in data compression discovered by DavidHuffman in 1952 [7]. A Huffman tree is a binary tree and represents the most efficientbinary code for a given alphabetwith respect to a distribution of frequency of its characters.In this paper, we associate any Huffman tree of n leaves with the point in Qn havingas coordinates the length of the paths from the root to every leaf from the left to right.We then study the convex hull, that we call Huffmanhedron, of those points associatedwith all the possible Huffman trees of n leaves. First, we present some basic properties ofHuffmanhedron, especially, about its dimensions and its extreme vertices. Next we givea partial linear description of Huffmanhedron which includes in particular a completecharacterization of the facet defining inequalities with nonnegative coefficients that aretight at a vertex corresponding to some maximum height Huffman tree (i.e. a Huffmantree of depth n − 1). The latter contains a family of facet defining inequalities in whichcoefficients follow in some way the law of the Fibonacci sequence. This result shows thatthe number of facets of Huffmanhedron is at least a factorial of n and consequently showsthat the facial structure of Huffmanhedron is rather complex. This is quite in contrast withthe fact that using the algorithm of Huffman described in [7], one can minimize any linearobjective function over theHuffmanhedron inO(n log n) time.We also give two proceduresfor lifting and facet composition allowing us to derive facet-defining inequalities from theones in lower dimensions.

© 2012 Elsevier B.V. All rights reserved.

1. Introduction

In 1952, David Huffman discovered the concept of the Huffman tree which offers the most efficient binary code for thecharacters of an alphabet Λ = {c1, c2, c3, . . . , cn} in the context of a language using Λ as its alphabet. The Huffman tree isa binary tree of n leaves labeled by the characters in Λ. It can be built in O(n log n) time by a greedy algorithm also given byHuffman based on the frequency of the characters in the language in question. The Huffman tree is optimal in the sense thatit minimizes the linear function

ni=1 fili, where fi denotes the frequency of ci and li is the length (in terms of the number

of edges) of the path from the root to the leaf with label ci. To every Huffman tree of n leaves, we associate any Huffmantree with n leaves with the point x = (x1, x2, . . . , xn)T ∈ Qn where xi = li. Such a point is called a Huffman point. In thispaper, we propose to study the convex hull, denoted by HPn, of the Huffman points in Qn. For our convenience, we call thispolyhedronHuffmanhedron. When n = 1 and 2, there is only one possible Huffman tree, thus the Huffmanhedron is a unique

✩ This research was partly supported by the ANR grant BLAN06-1-138894 (project OPTICOMB).∗ Corresponding author. Tel.: +33 684850376; fax: +33 169082082.

E-mail addresses: Jean-Francois.Maurras@lif.univ-mrs.fr (J.-F. Maurras), ThanhHai.Nguyen@cea.fr (T.H. Nguyen), Hung.Nguyen@lip6.fr (V.H. Nguyen).

0166-218X/$ – see front matter© 2012 Elsevier B.V. All rights reserved.doi:10.1016/j.dam.2012.05.004

2 J.-F. Maurras et al. / Discrete Applied Mathematics ( ) –

Fig. 1. All the possible Huffman trees for Λ = {a, b, c, d}.

Huffman point that is (0) for n = 1 and (1, 1) for n = 2. When n = 3, there are three Huffman points which are (1, 2, 2),(2, 1, 2), (2, 2, 1), the Huffmanhedron is thus a triangle in Q3. In this dimension, the Huffmanhedron clearly has an emptyinterior. Therefore, we will focus to study the Huffmanhedron from the 4-dimensional space.

In the following, we give an example of all the possible Huffman points in dimension 4. Let us consider an alphabet of4 letters Λ = {a, b, c, d}. With all possible frequencies for each character, we can generate 13 Huffman trees as shown inFig. 1.The first 12 trees which correspond to the permutations of 4 characters a, b, c , d, give us the following 12 Huffman pointsin Q4: (3, 3, 2, 1), (3, 3, 1, 2), (3, 2, 3, 1), (3, 2, 1, 3), (3, 1, 3, 2), (3, 1, 2, 3), (2, 3, 3, 1), (2, 3, 1, 3), (2, 1, 3, 3), (1, 3, 3, 2),(1, 3, 2, 3), (1, 2, 3, 3), respectively. The point (2, 2, 2, 2) associated to the 13th tree remains stable under permutations ofits leaves. Note that the first 12 Huffman points lie on the hyperplane defined by

n=4i=1 xi = 9 = n(n+1)

2 − 1 (where n = 4).The Huffmanhedron was introduced by Maurras and Edmonds in one of their discussions in 1974. They were motivated

by the fact that optimizing a linear function over HPn is easy by the Huffman algorithm. The simplicity of this algorithmcould suggest that HPn has a ‘‘nice’’ facial structure and it is hopeful to find a complete linear description of HPn. They wouldlike for a start to investigate HPn at least for small dimensions. But the project was abandoned because the structure of HPnseemed to be quite complex evenwhen nwas small, and at that time, computerswere not powerful enough to enumerate allthe facet defining inequalities of HPn for small n. With the current computer technology and using some software like Lrs [2],Porta [4] and Cdd [5], we have been able to obtain a complete facial description of HPn up to dimension 8. For example, indimension 8 where the are 41,245 Huffman points, the Huffmanhedron has 102,691 facets and it takes about 4 months toenumerate them all with Lrs software on an Intel(R) Core(TM)2 Quad CPU Q6600 2.40 GHz computer with 2.00G of RAM.

Due to the construction of Huffman points, a related polyhedron to HPn is the well-known combinatorial permutahedron,Πn−1, which was introduced by Schoute [12] in 1911. The latter is defined on an n-element set N = {n, . . . , 1}. With eachpermutation π of N , we associate an incident vector x(π) = (π(n), . . . , π(1)) ∈ Rn. The permutahedron is the polytope

Πn−1 = conv{x(π), π is a permutation of N}.This geometric object is an (n − 1)-dimensional polytope in Rn with n! vertices and 2n

− 2 facets. Independently, severalauthors (cf., e.g., Rado [11], Balas [3], Gaiha and Gupta [6] and Young [16]) studied the permutahedron and derived acharacterization of Πn−1 via the following linear inequalities

i∈S

π(i) ≥ f (S), S ⊆ N,

i∈N

π(i) = f (N), where f (S) =|S| + 1

2

Arnim et al. [1], and Schulz [13,14], among others, gave several interesting results on the permutahedron of series–parallelposets and its generalization of a poset.

J.-F. Maurras et al. / Discrete Applied Mathematics ( ) – 3

In this paper, we first prove some basic properties of the Huffmanhedron, especially, about its dimension and itsextreme vertices. Next we give a partial linear description of the Huffmanhedron which includes in particular a completecharacterization of the facet defining inequalities with nonnegative coefficients that are tight at a vertex correspondingto some maximum height Huffman tree (i.e. a Huffman tree of depth n − 1). The latter contains a family of facet defininginequalities inwhich coefficients follow in someway the law of the Fibonacci sequence. This result shows that the number offacets of theHuffmanhedron is at least a factorial of n and consequently shows that the facial structure of theHuffmanhedronis rather complex. This is quite contrast with the fact that using the algorithm of Huffman described in [7], one canminimizeany linear objective function over the Huffmanhedron in O(n log n) time. We also give two procedures of lifting and facetcomposition allowing us to derive facet defining inequalities from the ones in lower dimensions.

2. Basic properties of the Huffmanhedron

Given a Huffman point x, such that

x1 ≥ x2 ≥ · · · ≥ xn−1 ≥ xn,

we can see that a vector a ∈ Rn minimizing

f (x) =n

i=1

aixi,

obviously has the coefficients in nondecreasing order from a1 to an, i.e.

a1 ≤ a2 ≤ · · · ≤ an−1 ≤ an.

Hence, any facet-defining inequality aTx ≥ a0 for HPn such that aTx = a0, should have the coefficients of the left hand sidein nondecreasing order. Moreover, by the construction of Huffman points explained in the previous section, we can obtainother Huffman points by simply permuting the coordinates of x. In the same way, we also obtain facet defining inequalitiesfor HPn by simply permuting some coefficients of the facet defining inequalities aTx ≥ a0 with nondecreasing coefficients.

Remark 2.1. If the inequality I ≡n

i=1 aixi ≥ a0 defines a facet for HPn, then the following inequality

I ′ ≡a1 · · · ai−1 aj ai+1 · · · aj−1 ai aj+1 · · · an

x ≥ a0,

for some pair i < j, ai = aj, also defines a facet for HPn.

For example, I ≡ x1 + x2 + x3 + 2x4 ≥ 10 is a facet defining inequality with nondecreasing coefficients for HP4, andlet EI = {(3, 3, 2, 1), (3, 2, 3, 1), (2, 3, 3, 1), (2, 2, 2, 2)} is the set of the Huffman points satisfying I at equality. Let us callthese pointsHuffman tight points with respect to (w.r.t) I . Let us consider I ′ ≡ 2x1+x2+x3+x4 ≥ 10, which is obtained fromI by permuting the 1st and 4th coefficients of I . The set EI ′ which can be obtained in the same way by exchanging the 1stand 4th coordinates of the points in EI , i.e. EI ′ = (1, 3, 3, 2), (1, 3, 2, 3), (1, 2, 3, 3), (2, 2, 2, 2), clearly is the set of Huffmantight points w.r.t I ′. By Remark 2.1, I ′ thus defines a facet for HP4.

Consequently, from now on we only focus on characterizing facet defining inequalities aTx ≥ a0 with nondecreasingcoefficients of a.

In the sequel, we introduce several terminologies which will be used throughout this paper. We use HVn ⊂ Qn to denotethe set of all the Huffman points in n-dimensional space. Let xnmax = (n − 1, n − 1, n − 2, . . . , 2, 1) be the Huffman pointcorresponding to the leftmost maximum height n-tree which is the Huffman tree of n leaves having maximum height. LetT nmax denote the latter. Given a Huffman point x, we use the notation {(x1, . . . , xi−1, ⟨xi, xi+1, . . . , xk⟩, xk+1, . . . , xn)}, to

denote the set of the Huffman points obtained from x by replacing the sequence of coordinates ⟨xi, xi+1, . . . , xk⟩ by any ofits permutations. For example, the set EI in the above example can be shortly rewritten as EI = {(⟨3, 3, 2⟩, 1), (2, 2, 2, 2)}.

In the following, we characterize several simple properties of the Huffman point which will be useful later.

Property 2.1. 1. For every x ∈ HVn, there are at least two coordinates of x equal tomax{xi,∀i = 1, . . . , n}.2. For every x ∈ HVn, with n ≥ 3 and let i, j be two indices such that xi = xj, we have

(x1, . . . , xi−1, xi − 1, xi+1, . . . , xj−1, xj+1, . . . , xn) ∈ HVn−1.

3. For every x ∈ HVn−1, with n ≥ 3, for some i > 0, we have

(x1, . . . , xi−1, xi + 1, xi + 1, xi+1, . . . , xn−1) ∈ HVn.

We investigate some basic properties on the Huffmanhedron. We first determine the dimension of HPn. Then we provethat Huffman’s algorithm is an optimization algorithm over HPn and all the Huffman points are the extreme points, i.e. thevertices, of HPn.

Proposition 2.1. For n ≥ 4, HPn is full dimensional.

4 J.-F. Maurras et al. / Discrete Applied Mathematics ( ) –

Proof. Note that all the points which are obtained by permuting coordinates of xnmax, belong to the hyperplane H definedby the rank equality

ni=1 xi =

n(n+1)2 − 1 and H is unique in this sense. By Property 2.1-3, generating from xn−1max in HVn−1,

we obtain a Huffman point x0 in HVn, x0 = (n− 2, n− 2, n− 3, . . . , 3, 2, 2, 2). Note that this point clearly does not lie onH . Hence, the claim follows. �

Note that Huffman’s algorithm which was originally established in [7] works with nonnegative weights correspondingto the occurrence frequencies of characters. We show in the following that the algorithm still works even in the presence ofnegativeweights. Before looking to it, let us recall Huffman’s algorithm: Let C be a set of n characters and that each characterc in C has a defined frequency f [c], and let Q be the priority queue of C where the characters c are in nondecreasing orderin the function of their frequency f [c].

Algorithm 2.1 (Huffman’s Algorithm).

Input :– Q ← COutput :– the root of the tree.while |Q | > 1

do allocate a new node {k, h}, whereleft({k, h}) ← k ←Extract_Min(Q)right({k, h}) ← h←Extract_Min(Q \ {k})f [{k, h}] ← f [k] + f [h]Insert (Q, {k,h})

end doend whilereturn Extract_Min(Q) ◃ return the root of the tree.

Proposition 2.2. Huffman’s algorithm is an optimization algorithm on HPn.

Proof. Given a linear function f with f (x) =n

i=1 aixi, where a ∈ Qn, a = 0 and x ∈ HVn, as explained above, we only needto prove Proposition 2.2 for the case when f (x) contains negative coefficients. Then, we first need to prove the followinglemma:

Lemma 2.1 ([9]). There is an optimal tree T ∗ in which two coefficients having the lowest value correspond to two leaves havingsame father node in T ∗.

Proof. It is similar to the proof of Lemma 4.31 in [9]. For more details, the interested reader is referred to [9] to p. 171. �

Let E be the set of the coefficient values of f (x) and let ak, ah be two lowest-value coefficients. Let us suppose that T ∗ isan optimal tree for E. From Lemma 2.1, the two leaves k and h having weights, respectively, ak and ah are siblings in T ∗. LetT ′ be a tree which is derived from T ∗ by deleting two leaves h, k and replacing them by a leaf {k, h} having weight ak,h equalto ak + ah. Fig. 2 is followed from p. 172 of [9] to illustrate this construction.

Lemma 2.2. T ∗ is an optimal tree for E if and only if T ′ is an optimal tree for E ′ = E \ {ak, ah}

ak,h.

Proof. Let x∗ be the incident Huffman point to T ∗. Let us denote OPT (E, T ∗) be the value obtained by applying x∗ on f (x),i.e. OPT (E, T ∗) =

ni=1 aix

∗

i . Using a similar notation for T ′, we get OPT (E ′, T ′) =n

i=1 aix′

i . As ak and ah are the two lowestvalue in E, then, from Lemma 2.1 we obtain that x∗k = x∗h . So, it implies that

OPT (E, T ∗) =n

i=1,i=h,k

aix∗i + akx∗k + ahx∗h

=

ni=1,i=h,k

aix∗i + ak,h(x∗k − 1)+ ak + ah

= OPT (E ′, T ′)+ ak + ah.

If T ′ is not an optimal tree for E ′, then the equation OPT (E, T ∗) = OPT (E ′, T ′)+ ak + ah allows us to obviously deducethat T ∗ is not an optimal tree for E, and conversely. Indeed, suppose that T ′ is an optimal tree for E ′, it is clear thatOPT (E ′, T ′) < OPT (E ′, T ′). Therefore, OPT (E, T ∗) > OPT (E ′, T ′) + ak + ah, which implies that T ∗ is not an optimal treefor E. Thus, the claim follows. Note that this proof is always valid even if ak and ah are negative values. �

By Lemma 2.2, we then obtained a new set E ′ where |E ′| = |E| − 1. This progressive procedure will finally provide acoefficient set containing either all non-negative coefficients or only one negative coefficient. For the first case, according

J.-F. Maurras et al. / Discrete Applied Mathematics ( ) – 5

Fig. 2. There is an optimal solution in which the two lowest-value coefficients label sibling leaves; deleting them and labeling their parent with a newcoefficient having the combined value yields an instance with a smaller coefficient set.

Fig. 3. Merging a′1 with a′2 first and then with a′3 .

Fig. 4. Merging a′3 with a′2 first and then with a′1 .

to the Huffman’s algorithm, the tree obtained is clearly an optimal tree for them. Thus the initial tree obtained by thisconstruction is also an optimal tree for E.

Note that for each time we replace two lowest-coefficient values by a new coefficient value, then we need n− 1 steps toreduce the initial set to a set having exactly one coefficient. Thus, for the second case, we should go back to the step where acoefficient set only contains there coefficients. In this step, we may prove that an optimal tree for these 3 coefficient valuescan be found here. Remark that in this case, there are only two following trees in 3-dimensional space, one of them is clearlyan optimal tree, but which one is an optimal tree?

Suppose that E ′ contains {a′1, a′

2, a′

3}, where a′1 ≤ a′2 ≤ a′3. For the tree in Fig. 3, we get that 2a′1 + 2a′2 + a′3, and for thetree in Fig. 4, we have that a′1 + 2a′2 + 2a′3. It is easy to see that the left tree allows us to obtain an optimal cost. Thus thisone is an optimal tree for E ′. According to Lemma 2.2, the claim follows. �

Theorem 2.1. Each Huffman point x ∈ HVn is a vertex of HPn.

Proof. By recurrence on n. In 3-dimensional space, we only have 3 Huffman points (2, 2, 1), (2, 1, 2) and (1, 2, 2). HP3clearly is a triangle having 3 vertices as these points. Suppose that Theorem 2.1 is verified in Qn, i.e. for each point x0 in HVn,we can find an hyperplane H0

≡n

i=1 a0i xi = b0 which exposes x0, i.e., H0 is valid for HPn and only contains x0. Assume

that the coordinates of x0 are in a non-increasing order. Thus, the coefficients of H0 are in ascending order. By Property 2.1,generating x1 in HVn+1 from x0 by replacing x0i with x0i +1 and x0i +1, in this proof we choose the first coordinate x01, we thusobtain: x1 = (x01 + 1, x01 + 1, x02, x

03, . . . , x

0n). Suppose that without loss of generality, a01 = 0. So, we consider the following

hyperplane: H1≡n+1

i=1 a1i xi = b10 where a11 = a12 = a01, and a1i = 2a0i−1, ∀i > 2 and b10 = 2b0. Then, applying Huffman’salgorithm to the coefficient values a1i , for the first step of this algorithm, there is a unique choice that is merging a11 witha12, which yields a new coefficient value equal to 2a01. From this step through the end, this algorithm will continue with themultiple coefficient values of H0. By the assumption, we thus get a unique optimal tree to which the point x0 is associated.Finally, the point x1 is a unique Huffman point which is exposed by H1. Hence, the claim follows. �

In the sequel, a Huffman point of the Huffmanhedron is called a Huffman vertex and a tight Huffman point w.r.tsome inequality, is called a tight Huffman vertex w.r.t such an inequality. Before looking to the partial description of theHuffmanhedron, we need to introduce e-property that is especially useful for the proof of facetness.

6 J.-F. Maurras et al. / Discrete Applied Mathematics ( ) –

a cb d

Fig. 5. Fig. 5(a)–(d) show e-property which is satisfied for the 2th, 3th, 4th, 5th index, respectively.

Definition 2.1 (e-Property). Let E be a vertex subset of HVn and let K be an increasing index subset of N = {1, 2, . . . , n}. Esatisfies the e-property for K if and only if for each k ∈ K , there exist two vertices x, y in E, such that:

– xk = yk,– for all j > k ∈ K , xj = yj.

For example, let us consider E = {(⟨4, 4, 3⟩, 2, 1), (⟨3, 3, 2⟩, 2, 2), (⟨3, 3, 3⟩, 3, 1)}. Recall that the pair ‘⟨, ⟩’ indicates thepossibility of permutation over all the coordinates between this pair, which allows us to generate other vertices in E. Notethat E satisfies e-property for the index set {2, 3, 4, 5}. The proof for this can be illustrated in Fig. 5.

Proposition 2.3. Let x0 be a point in HVn and let E be a subset of Huffman points, for some pair i < j:

E = {(x01, x02, . . . , x

0i−1, ⟨x

0i , . . . , x

0j ⟩, x

0j+1, . . . , x

0n)}.

If there is a lowest index k, where i < k < j, such that x0i = x0k , then E satisfies e-property for {i+ 1, . . . , j}.

Proof. Note that a point generated by exchanging two coordinates between the ith and jth indices also belongs to E. Thus,from x0, by permuting two coordinates x0i and x0k we obtain xk in E. Similarly, from xk, by exchanging two coordinates xki andxki+1, we deduce the point xi+1 in E. Using these two points xk and xi+1, it obviously implies that e-property of E holds forthe index i+ 1. By the same way, generating the point xt from xk by switching xki with a coordinate xkt , where i < t ≤ j, twoobtained points xk and xt also allow E to verify e-property for the index t . �

3. A partial description for the Huffmanhedron

In this section, we present a partial description for the Huffmanhedron. We first describe two classes of trivial facets andrank facet defining inequalities. Then we give a complete characterization of the facet defining inequalities with nonnegativecoefficients that are tight at a vertex corresponding to some maximum height Huffman tree. Finally, we provide twoprocedures of lifting and facet composition and show that applying them to the facet defining inequalities we know for theHuffmanhedron gives all the facets that contain at least a vertex associated with some maximum height Huffman tree.

3.1. Trivial facets and rank facets

Proposition 3.1 (Trivial Facets).1. For n ≥ 5, the inequality ITi ≡ xi ≥ 1 defines a facet for HPn.2. For n ≥ 6, the inequality ITi,j ≡ xi + xj ≥ 3, for all 1 ≤ i < j ≤ n, defines a facet for HPn.

Proof. In every Huffman tree of n leaves, there is at most one leaf of depth 1 and the others are of depth at least 2. Hence,the validity of ITi and ITi,j are verified.

Let us consider the inequality ITi , when i = n. It means that all the Huffman vertices having the last coordinate equal to 1are tight vertices w.r.t ITn . Suppose that all tight vertices w.r.t ITn also satisfy another valid inequality I ≡

ni=1 cixi ≥ c0 at

equality. So picking a vertex x0 = (x01, . . . , x0i , . . . , x

0j , . . . , x

0n−1, 1) such that x0i = x0j , generating from x0 by switching x0i ,

x0j , we then get x1 = (x01, . . . , x0j , . . . , x

0i , . . . , x

0n−1, 1). Rewriting I with the coordinates of these two vertices, subtracting

coordinate by coordinate of these two equalities, finally we deduce that (ci − cj)(x0i − x0j ) = 0, which implies that ci = cj.Repeating this procedure for another index, we will obtain ci = cj for all i, j < n. Moreover, there are always two tightpoints such that sum of the first n− 1 coordinates of these two vertices are different. Therefore, ci = 0 for all i < n. Whichconcludes that ITn defines a facet for HPn.

Let us consider the last inequality ITi,j , when i = n− 1 and j = n. Using the above argument similarly, we will also obtainthe facetness of ITn−1,n . �

Proposition 3.2 (Rank Facets).

1. The inequality IH ≡ 1Tx ≤ n(n+1)2 − 1 defines a facet of HPn.

2. For n = 2k+ c with k ∈ N+ and 0 < c < 2k, the inequality IL ≡ 1Tx ≥ 2c(k+ 1)+ (n− 2c)k defines a facet for HPn.

Proof. Note that the sum of the coordinates of a given Huffman point x is maximumwhen x is associated to somemaximumheight tree, this sum is equal to n(n+1)

2 −1. In addition, this sum isminimumwhen x corresponds to some equilibrated tree, i.e.in which there are no two leaves such that the difference of its depths strictly is greater than 1. Indeed, Huffman’s algorithm

J.-F. Maurras et al. / Discrete Applied Mathematics ( ) – 7

applied tominimize the linear function f where f = 1will clearly give an equilibrated tree as an optimal tree. It is obviouslyverified that the sum of all the lengths of the root-to-leaf paths in this tree is equal to 2c(k + 1) + (n − 2c)k. Hence, thevalidity of IH and IL are justified.

Let us suppose that theHuffman tight verticesw.r.t IH satisfy another valid inequality at equality I ′ ≡n

i=1 cixi = c0. Notethat all the coefficients of IH are equal, then switching two different coordinates x0i , x

0j of a tight vertex x

0 will provide anothertight vertex w.r.t IH . Therefore, rewriting I ′ for the coordinates of these two tight vertices, then subtracting coordinate bycoordinate the two equalities, we finally obtain the following equality (ci − cj)(x0i − x0j ) = 0. This implies that ci = cj.Repeating this procedure for other indices, we deduce that all the coefficients of I ′ are equal. Hence, we get the contradiction.

We may use the same argument for proving IH to show the facetness of IL. �

3.2. Characterization of the vertices associated with maximum height Huffman trees

Given a Huffman point x, let us call the facet defining inequalities characterizing x those that are tight at x. In this sectionwe described all the facet defining inequalities with nonnegative coefficients defining xnmax. By Remark 2.1, we obtain allfacet defining inequalities with nonnegative coefficients characterizing the Huffman points which are permutations of thecoordinates of xnmax.

Given a polyhedron P ∈ Rn+which is the dominant of the convex hull of a set of points. Let u be an extreme point of P

and a ∈ Rn be a vector. Suppose that we know a system In ofm linear inequalities

In =

c11a1 + c12a2 + · · · + c1nan ≥ 0

c21a1 + c22a2 + · · · + c2nan ≥ 0. . .cm1 a1 + cm2 a2 + · · · + cmn an ≥ 0

that express the necessary and sufficient conditions for aTx to get optimal at u over P . Thus it is clear that a belongs tothe cone Cn generated by the inequalities in In, i.e. a is a conic combination of the extreme rays of Cn. Let F = {f1x ≥1, f2x ≥ 1, . . . , fpx ≥ 1} be the set of all the facet defining inequalities for P characterizing u, i.e. fiu = 1 and fi ≥ 0 for alli = 1, . . . , p. We can see that optimizing a function aTx over F with a satisfying In will accept u as an optimal solution.In the following, we present two standard basic properties in polyhedral theory.

Lemma 3.1. A linear function aTx gets optimal at u over P if and only if a is conic combination of f1, f2, . . . , fp.

Corollary 3.1. The extreme rays of Cn are in one-to-one correspondence with the facets fi.

We can now use Corollary 3.1 to find the facet defining inequalities characterizing xnmax. We first characterize the systemIn that express the necessary and sufficient conditions so that aTx gets optimal at xnmax over HPn.

Theorem 3.1. The point xnmax minimizes a linear functionn

i=1 aixi over HPn if and only if a is super-Fibonacci, which meansthat the coefficients ai satisfy the following system of inequalities

In =−a1 − · · · − ai−2 + ai ≥ 0, for all 3 ≤ i ≤ n,−ai−1 + ai ≥ 0, for all 3 ≤ i ≤ n.

Proof. ‘‘⇒’’. The ith leaf of T nmax from left to right is associated to the weight ai. Note that in the depth n − 1 of this tree,

two leaves having weight a1 and a2 aremerged before joining with the leaf having weight a3. Hence, according to Huffman’salgorithm, the following conditions must hold: a1, a2 ≤ a3 and a1 + a2 ≤ a4. Similarly for the other depths in this tree, weeasily deduce that T n

max is an optimal tree for a if a is super-Fibonacci.‘‘⇐’’. By recurrence on n. When n = 3, we have a3 ≥ max(a2, a1). Hence, from Huffman’s algorithm, we clearly obtain

T 3max as an optimal tree for them. Suppose that we obtain T n−2

max as an optimal tree for a1, a2, . . . , an−2. In the n-dimensionalspace, as a is super-Fibonacci, we get that an ≥ max(an−1,

n−2i=1 ai). According to Huffman’s algorithm, an−1 is merged with

T n−2max to provide T n−1

max , after that, an is merged with T n−1max . Hence, we obtain T n

max as an optimal tree for f . �

Remark 3.1. Theorem 3.1 also applies to the dominant of HPn (dom(HPn)). In this case all the components of a arenonnegative.

We focus now to characterize the extreme rays of Cn, the cone defined by the inequalities of In. By Corollary 3.1 andRemark 3.1, we can then derive all the facet defining inequalities with nonnegative coefficients characterizing xnmax.

Let us see the case when n = 3. In the 3-dimensional space, the system I3 is given as:

I3 =

a1, a2 ≥ 0−a2 + a3 ≥ 0−a1 + a3 ≥ 0.

8 J.-F. Maurras et al. / Discrete Applied Mathematics ( ) –

Note that in this dimension, there are only 3 Huffman points (2, 2, 1), (2, 1, 2), (1, 2, 2). Moreover, these points lie on thehyperplane defined by

3i=1 xi = 5. It means that a1 = (1, 1, 1)T . By the fact that these three Huffman points all belong to

the hyperplane defined by3

i=1 xi = 5, we can always set a coefficient ai to a fixed positive value. Let us then fix a3 = 1.Hence, we can get that a2 = (0, 1, 1)T and a3 = (1, 0, 1)T . These extreme rays ai ≥ 0 for C3 correspond to the facet defininginequalities (ai)T x ≥ ai0 for dom(HPn) with ai0 is the optimal value while applying Huffman’s algorithm to minimize (ai)Tx.Consequently, we obtain the following facet defining inequalities characterizing the point (2, 2, 1) of HP3 as:

x2 + x3 ≥ 3x1 + x3 ≥ 3x1 + x2 + x3 = 5.

When n = 4, the system I4 is given as follows:

I4 =

a1, a2 ≥ 0−a2 + a3 ≥ 0−a1 + a3 ≥ 0−a3 + a4 ≥ 0−a1 − a2 + a4 ≥ 0.

Let us consider the sub-system:

I4=

−a3 + a4 ≥ 0−a1 − a2 + a4 ≥ 0a4 ≥ 0.

If either a4 = a3 and a4 > a2 + a1 or a4 > a3 and a4 = a2 + a1, then letting a4 = max(a3, a2 + a1), we continue to solveI4 with the inequalities of I3. As the case when n = 3, we get (a1, a2, a3)T ∈ {(1, 1, 1)T , (0, 1, 1)T , (1, 0, 1)T }. Therefore, weobtain a1 = (1, 1, 1, 2)T , a2 = (0, 1, 1, 1)Ta3 = (1, 0, 1, 1)T .

If a4 > a3 and a4 > a2 + a1, then fixing a4 = 1, we deduce that a4 = (0, 0, 0, 1)T .If the two first inequalities of I

4are saturated, i.e, a4 = a2+a1 = a3, fixing a4 = 1, thenwe also obtain a1 = (0, 1, 1, 1)T ,

a2 = (1, 0, 1, 1)T .From these extreme rays, by the same reasoning as we have done for the case when n = 3, we derive the facet defining

inequalities characterizing the point x4max:

x4 ≥ 1,x2 + x3 + x4 ≥ 6,x1 + x3 + x4 ≥ 6,x1 + x2 + x3 + 2x4 ≥ 10.

Suppose that we know all the extreme rays of Cn−2 and Cn−1. We will show that all the extreme rays for Cn can becompletely derived from those of Cn−2 and Cn−1.

In =

a1, a2 ≥ 0,−ai−1 + ai ≥ 0, ∀i = 3, . . . , n−a1 − a2 − · · · − ai−2 + ai ≥ 0 ∀i = 3, . . . , n.

Let us consider the sub-system of inequalities

In=

−an−1 + an ≥ 0−a1 − a2 − · · · − an−2 + an ≥ 0.

If none of these inequalities of Inis tight, then we get an = 1 and ai = 0 for all i < n. (This is a trivial facet ITn , see

Proposition 3.1).If exactly one of these inequalities is tight, by choosing an = max(an−1,

n−2i=1 ai), we continue to solve In with the

inequalities in In−1. Note that in the casewhere the components of a are positive, the Fibonacci numbers clearly satisfy theseconditions. Hence, we obtain an extreme ray of Cn: a = (α1, . . . , αn)

T , where α1 = α2 = α3 = 1 and αi = αi−1 + αi−2,∀i ≥ 4. This result is reformulated as follows, the optimal value while applying the Huffman’s algorithm to minimize thelinear function αTx will be found and proved in the next section:

Theorem 3.2 (Fibonacci Facets). Given a vector α ∈ Zn such that α1 = α2 = α3 = 1 and αi = αi−1 + αi−2, ∀i ≥ 4. Theinequality

fibon ≡

ni=1

αixi ≥ Fn+3 − 3,

where Fn+3 is the (n+ 3)th Fibonacci number [15], defines a facet for HPn, for all n ≥ 3.

J.-F. Maurras et al. / Discrete Applied Mathematics ( ) – 9

Note that in this case, we also obtain a trivial facet ITn−1,n . In fact, this facet defining inequality is generated from thetrivial facet defining inequality ITn−1 for HPn−1. Moreover, when the vector a contains k components equal to zero, then then − k positive components of a are the ones of an extreme ray of Cn−k. From this observation, a may be generated fromthe zero-lifting procedure. This result is generalized in the theorem below, the proof for it will be presented in the nextsection:

Theorem 3.3 (Zero-Lifting). Given I0 ≡n

i=1 aixi ≥ a0 a facet defining inequality for HPn, with a1 = · · · = ad and d ≥ 1, andad < ai ≤ ai+1, for all i > d. Suppose that EI0 , the subset of tight vertices w.r.t I0, satisfies e-property for {d+ 1, d+ 2, . . . , n}.Then in Qn+1, the inequalities

Lift1 ≡

0 a1 a2 · · · an−1 anx ≥ a0 + a1 (1)

defines facet for HPn+1.

For the last case, when an = (a1 + a2 + · · · + an−2) = an−1, then we continue to solve In with the inequalities in In−2. Notethat if the vector a only contains the positive components, then it could be obtained by applying composition procedureswiththe inputs being two Fibonacci facets of HPk and HPh. This application will be discussed later in this section. We introducethe facet composition results, the proof for this will be presented in the next section:

Theorem 3.4 (Facets Composition). Let In ≡n

i=1 aixi ≥ a0, be a facet defining inequality for HPn with a1 = · · · = ak, wherek ≥ 3, and ak < ai ≤ ai+1, for all i > k. Suppose that EIn , the subset of the tight vertices w.r.t In, satisfies e-property for{k+ 1, k+ 2, . . . , n}.

Let Im ≡m

i=1 bixi ≥ b0 be a facet defining inequality for HPm with b1 = · · · = bl, where l ≥ 3, and bl < bi ≤ bi+1, for alli > l. Suppose that EIm , the subset of the tight vertices w.r.t Im, satisfies e-property for {l+1, l+2, . . . ,m}. Then the inequalities

In,m ≡ b1n

i=1

aixi +

n

i=1

ai

mj=2

bixn+i−1 ≥ b1a0 + b0n

i=1

ai,

Im,n ≡ a1mi=1

bixi +

mi=1

bi

n

j=2

aixm+i−1 ≥ a1b0 + a0mi=1

bi,

define facets for HPn+m−1.

By Remark 3.1, all the inequalities that we derived from the extreme rays of Cn define facets for dom(HPn). However, someof them do not define facets for HPn but only (n− 2)-faces. These inequalities are presented in the following remark:

Remark 3.2. 1. The facet defining inequality x4 ≥ 1 for dom(HP4) is a valid inequality for HP4. Indeed, that inequality isobtained from the convex combination of two facet defining inequalities x1+x2+x3+2x4 ≥ 10 and−x1−x2−x3−x4 ≥−9.

2. Some facet defining inequalities for dom(HPn), which are recursively generated from the facet defining inequalityx2+x3 ≥ 3 for dom(HP3) are (n−2)-faces of HPn. For example, the facet defining inequality for dom(HP5), x2+x3+2x4+2x5 ≥ 13 that is generated in the casewhere a5 = a4 =

3i=1 ai, is simply a valid inequality for HP5. In fact, this inequality

is obtained from the convex combination of two facet defining inequalities for HP5, x1 + x2 + x3 + 3x4 + 3x5 ≥ 20 and−x1 + x2 + x3 + x4 + x5 ≥ 6, obtained by Porta [4]. Recursively, some inequalities generated through the facet defininginequality for dom(HP5), x2 + x3 + 2x4 + 2x5 ≥ 13, are (n− 2)-faces of HPn.

3. The facet defining inequalities for dom(HPn) generated from the Zero-lifting procedure which only contain twocoefficients equal to 0 are the (n − 2)-face of HPn. In fact, the two first coordinates of any tight Huffman vertex w.r.tsuch an inequality are always equal, and are equal to the maximum value coordinate, as Property 2.1. Consequently,there do not exist n linearly independent points.

In order to interpret the corresponding facet defining inequalities for dom(HPn) characterizing xnmax, we introduce somenotations through the following example which allows us to enumerate all these facets:

Let fibon−2 ≡n−2

i=1 αixi ≥ Fn+1 − 3 be the Fibonacci facet defining inequality for HPn−2 and let fibo3 ≡3

i=1 xi ≥ 5be the Fibonacci facet defining inequality for HP3, applying the facet composition operation to fibon−2 and fibo3, weobtain the inequality :

α1 α2 · · · αn−2

n−2i=1

αi

n−2i=1

αi

≥ Fn+1 − 3+ 5

n−2i=1

αi

defining a facet for HPn.

10 J.-F. Maurras et al. / Discrete Applied Mathematics ( ) –

Note that the facets which are generated from two Fibonacci facets for HPp and HPq, have exactly two equal coefficients,except for the three first coefficients. In the above example, we have the (n− 1)th and nth coefficients that are equal ton−2

i=1 αi. We call two equal coefficients 1-leap, and these facets Fibonacci 1-leap facets. By our convenience, the Fibonaccifacet is called a Fibonacci 0-leap facet. Generally, applying a Fibonacci k-leaps facet for HPm with a Fibonacci l-leaps facet forHPn−m+1, we obtain two Fibonacci (k+ l+ 1)-leaps facets for HPn. Moreover, using the zero-lifting procedure to apply to allthese facets, we also generate facets which contain xnmax.

Let us use Ff ,m,{i1,...,ik} ≡n

i=1 aixi ≥ a0 to denote a facet defining inequality for dom(HPn), where m ∈ N denotes thenumber of zeros added at the beginning of the left hand side and the set {i1, . . . , ik} is a set of k indices where there are kleaps in this inequality. The notation Ff ,m,{i1,...,ik} unifies all the notations of facet defining inequalities generated from theFibonacci inequalities. In particular,

1. whenm = 0 and {i1, . . . , ik} = ∅, Ff ,0,∅ denotes the Fibonacci facet defining inequality.2. when m = 0 and K = {i1, . . . , ik} = ∅, Ff ,0,{i1,...,ik} denotes the Fibonacci-k leaps facet defining inequality where

aij = aij+1 for all ij ∈ K .3. when m = 0 and {i1, . . . , ik} = ∅, Ff ,m,∅ denotes the m-Zero facet defining inequality where the first m coefficients are

equal to 0.4. whenm = 0 and K = {i1, . . . , ik} = ∅, Ff ,m,{i1,...,ik} denotes the m-Zero Lifting Fibonacci-k leaps inequalities.

Using this notation, we established the following result:

Theorem 3.5. The following facet defining inequalities for dom(HPn)

ITn ≡ xn ≥ 1,ITn−1,n ≡ xn−1 + xn ≥ 3,Ff ,m,{i1,...,ik} ≡ a1x1 + · · · + an−2xn−2 + an−1xn−1 + anxn ≥ a0,

characterize xnmax.

3.3. Proofs of Theorems 3.2–3.4

In this section, we prove the theorems described in the above section. Before looking into this, we introduce a technicaltool allowing to prove facetness of a valid inequality I for HPn. Consider a valid linear inequality I for HPn, I defines a facetfor HPn if the set of Huffman points that are tight w.r.t I do not satisfy a valid inequality I ′ ≡ cT x ≥ c0 at equality. Thus, letus suppose the contrary. Adding an appropriate linear combination of I to I ′, as HPn is full-dimensional with n ≥ 4, thenwe can fix one coefficient of I ′ to 0. For more details of this argument, the interesting reader is referred to [10]. By usingparticular properties of the tight Huffman points w.r.t I satisfying cT x ≥ c0 at equality, we may derive that ci = c0 = 0.

Lemma 3.2. Let I ≡ aTx ≥ b0 be a valid inequality for HPn and let EI ⊂ HVn be a set of the tight vertices w.r.t I. If EI satisfiese-property for an index subset K ⊆ N = {1, 2, . . . , n} and moreover the coefficients ai are pairwise equal for all i ∈ N \ K , thenI defines a facet for HPn.

Proof. Suppose that I does not define facet for HPn, then all the tight vertices in EI w.r.t I may also satisfy at equality anothervalid inequality for HPn, let I ′ ≡ cTx ≥ c0 with c = 0, be such an inequality. Note that for all i in N \ K , the coefficientsai are pairwise equal, as the assumption. Therefore, the vertex x1 which is obtained from a vertex x0 in EI by exchangingtwo different coordinates x0i and x0j with i, j ∈ N \ K , is also a tight vertex w.r.t I . Furthermore, these vertices are also tightvertices w.r.t I ′. Writing I ′ with these two vertices and subtracting coordinate by coordinate the two obtained equalities, weget that ci = cj. The reasoning is continued for another index in N \ K , we finally obtain that ci = cj for i, j ∈ N \ K .

Pick a coefficient cj of I ′ with j ∈ N \ K , as the vertices in EI satisfy both I and I ′ at equality, without loss of generality wecan fix cj at 0. Since ci = cj for i, j ∈ N \ K , which implies that ci = 0 for all i ∈ N \ K . Let k be the smallest index in K , asthe assumption, EI satisfies e-property for K , then there are two vertices x, y in EI such that (see Definition 2.1): xk = yk and∀j ∈ K , j > k, xj = yj. So, rewriting I ′ by all the coordinates of x and y, and subtracting coordinate by coordinate these twoequalities, we get that ck(xk − yk) = 0, thus ck = 0. Repeating this procedure for a subset K \ {k}, we will obtain ci = 0 forall i in K . Hence, the claim follows. �

3.4. The proof of Theorem 3.2

To prove the validity of the Fibonacci inequality, it is enough to show the following result:

Lemma 3.3.

(n− 1)α1 + (n− 1)α2 + (n− 2)α3 + · · · + 2αn−1 + αn = Fn+3 − 3.

J.-F. Maurras et al. / Discrete Applied Mathematics ( ) – 11

Proof. Let us process by recurrence on n. When n = 3, we obviously obtain that 2α1+2α2+α3 = 5 = F6−3.When n = d,we get:

−α1 +

di=1

(d− i+ 1)αi = −α1 +

d−1i=1

(d− 1− i+ 1)αi +

di=1

αi

= Fd+2 − 3+d

i=1

αi.

As the Fibonacci sequence starting from F0 = 0 and F1 = 1, we then deduce the equality αi = Fi−1, for all i > 1. Hence, weget that

di=1 αi = αd+2 = Fd+1. In addition, we have Fd+1 + Fd+2 = Fd+3. Thus the claim follows. �

From Huffman’s algorithm, we obtain that the vertex xnmax is an optimal point for the linear function f (x) =n

i=1 αixi.From Lemma 3.3, the validity of fibon is concluded.

3.5. The proof of Theorem 3.3

Let us consider the following linear function

f (x) =0 a1 a2 · · · an−1 an

x

where x ∈ HVn+1. FromHuffman’s algorithm,wewill get the optimal value of f onHPn. In the first step, Huffman’s algorithmchooses two first coefficient values of f to merge. Note that from this step through the end, this algorithm will clearlybe applied on the coefficient values of I0. Let T0 be an optimal tree for the coefficient values of I0. Hence, according toLemma 2.2, an optimal tree for the ones of f is easily obtained by branching a 2-tree on the leftmost leaf of T0, i.e theleaf has weight equal to a1. Let T1 be this tree. Let x0 be a vertex associated to T0, then the vertex corresponding to T1 is:x1 = (x01 + 1, x01 + 1, x02, . . . , x

0n−1, x

0n). Comparing the optimal value of f given by x1 with the second member of I0, the

validity of Lift1 is proved.Let ELift1 be a subset of tight vertices w.r.t Lift1. From the above construction, we easily establish the following inclusion:

ELift1 ⊇x ∈ HVn+1 | ∀x0 ∈ EI0 , x =

x01 + 1, ⟨x01 + 1, x02, . . . , x

0d⟩, x

0d+1, . . . , x

0n

. (2)

Note that EI0 satisfies e-property for K = {d + 1, d + 2, d + 3, . . . , n}, as a hypothesis, then taking a pair of vertices in EI0which satisfy e-property for an index k in K , wemay prove that the extensions of these vertices in ELift1 by the link (2), allowto verify e-property for an index k+ 1. Using this argument for all indices in K \ {k}, we may show that e-property of ELift1holds for {d+ 2, d+ 3, . . . , n+ 1}. We now proceed to prove that ELift1 also satisfy e-property for the first index. For this,it is enough to choose a vertex x1 ∈ EI0 such that x11 = x1i , with i ≤ d. Exchanging these two coordinates, we get a vertexx2 which also belongs to EI0 . Applying the link (2) for x1 and x2, we obtain respectively two tight vertices w.r.t ELift1 , weeasily prove that these two vertices satisfy e-property for {1}. Hence, all the conditions of Lemma 3.2 are satisfied, the claimfollows.

3.6. The proof of Theorem 3.4

Let a′ denote the left hand side vector of In,m. Using Huffman’s algorithm, we will get an optimal value for a′. Supposethat the first n − 1 steps of this optimization algorithm yield a tree, let us call it Tn. Note that the first n coefficients of a′are exactly the multiples of the ones of In. Thus, a Huffman vertex corresponding to Tn is clearly a tight vertex w.r.t In. Inparticular, a root of Tn has weight equal to b1

ni=1 ai. Consequently, from the nth step, Huffman’s algorithm continues to be

applied on the m remaining weights which are the coefficient multiples of Im: {b1n

i=1 ai, b2n

i=1 ai, . . . , bmn

i=1 ai}. LetTm be a subtree which is provided by Huffman’s algorithm applied on these values. It is easy to verify that a Huffman vertexassociated to Tm is a tight vertex w.r.t Im. Hence, the tree which is obtained by branching Tn on the leftmost leaf of Tm havingweight equal to b1

ni=1 ai (see Fig. 6) is an optimal tree for a′. Let us call Tn,m this tree.

Letx1 = (x11, . . . , x1n)be theHuffmanpoint corresponding to Tn and letx2 = (x21, . . . , x

2m)be theHuffmanpoint associated

to Tm. Thus, the Huffman point associated to Tn,m is:

x0 = (x11 + x21, x12 + x21, . . . , x

1n + x21, x

22, . . . , x

2m).

Verifying the optimal value of a′ given by x0 with the second member of In,m, we obtain the validity of In,m. Using the sameargument, we obtain also the validity of Im,n.

Let EIn,m be a set of tight vertex w.r.t In,m. Due to the above construction we have

EIn,m ⊇x ∈ HVn+m−1|∀x1 ∈ EIn ,∀x

2∈ EIm ,

x =⟨x11 + x21, . . . , x

1k + x21⟩, . . . , x

1n + x21, ⟨x

22, . . . , x

2l ⟩, . . . , x

2m

. (3)

12 J.-F. Maurras et al. / Discrete Applied Mathematics ( ) –

Fig. 6. Compositions of an optimal tree Tn,m .

Note that the coefficients between the 1st and the lth indices of Im are pairwise equal. Hence, applying Proposition 2.3,we deduce that e-property of EIm holds for {2, 3, . . . , l}. From the hypothesis, we obtain that EIm satisfies e-property for{2, 3, . . . ,m}.

In the sequel, we will show that EIn,m satisfies e-property for {k + 1, k + 2, . . . , n + m − 2, n + m − 1}. We first provethat EIn,m satisfies e-property for Kk+1,n = {k+ 1, k+ 2, . . . , n}. For this, we take two vertices in EIn that satisfy e-propertyfor an index i in Kk+1,n and merge themwith an any vertex in EIm by the link (3) to generate two vertices in EIn,m . It is easy toverify that these vertices also respect e-property for the same index i in Kk+1,n. This reasoning will be repeated for anotherindex in Kk+1,n.

Using the above argument, we similarly obtain that EIn,m satisfies e-property for {n+1, n+2, . . . , n+m−2, n+m−1}.As the k first coefficients of In,m are pairwise equal and EIn,m satisfies e-property for {k+1, k+2, . . . , n+m−2, n+m−1},

all the conditions of Lemma 3.2 are satisfied. Thus the claim follows.

4. Concluding remarks

In this paper, we have studied the facial structure of the Huffmanhedron. We have seen that although Huffman’salgorithm optimizes linear functions over this polyhedron in O(n log n), the facial structure of the Huffmanhedron is rathercomplex. Indeed, the number of Fibonacci facets togetherwith the ones derived from the lifting and composition is a factorialin terms of n. However, Kaibel and Pashkovich [8] have recently given extended formulations for the Huffmanhedronwith O(n log n) variables and inequalities. This very interesting result makes the question of obtaining a complete lineardescription in the original space even more challenging.

Acknowledgments

Wewould like to thank the anonymous referees for their helpful comments and suggestions for the improvement of thepaper. A special thanks to the second referee for suggesting to us the correspondence between the extreme rays of the coneCn and the facet defining inequalities characterizing the point xnmax.

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