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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1


Volume 13 (2015)

A greedy algorithm for interval greedoids

Hua Mao
Published Online: 2018-03-31 | DOI: https://doi.org/10.1515/math-2018-0026


We show that the greedy algorithm provided in this paper works for interval greedoids with positive weights under some conditions, and also characterize an exchangeable system to be an interval greedoid with the assistance of the greedy algorithm.

Keywords: Interval greedoid; Exchangeable system; Greedy algorithm; Positive weight

MSC 2010: 90C27; 05B35

1 Introduction and Preliminaries

Let 𝓕 (called feasible sets) be a set system on E (i.e. a non-empty family 𝓕 ⊆ 2E where 2E is the set of all subsets of a finite set E). We can suppose ∪𝓕 = E in this paper, since E ∖ ∪𝓕 ≠ ∅ will bring xE ∖ ∪𝓕 to own nothing in 𝓕 which is not interesting for studying. Actually, such supposition is also done in [1].

Let ω : E → ℝ+ be a weighting on E. Abbreviating ω(X)=xXω(x), especially ω(∅) = 0, we want to find an A ∈ 𝓕 satisfying ω(A) = max{ω(X)∣ X ∈𝓕}. We call this problem (𝓕, ω). An element of 𝓕 is optimal if it has the maximal weight. The greedy algorithm for (E, 𝓕) attempts to solve the above problem. In fact, Helman et al. [2] point that obtaining an exact characterization of the class of problems for which the greedy algorithm returns an optimal solution has been an open problem. The process of greedy algorithm (cf.[3, p.14]) is as follows.

  1. Set X = ∅.

  2. Set T = {xEXXx ∈ 𝓕},

    If T = ∅, stop;

    If T ≠ ∅, choose xT such that ω(x) ≥ ω(y) for all yT.

  3. Set X = Xx and go to (2).

Björner et al. indicate [1] that greedoids were invented around 1980 by Korte and Lovász. The relative definitions to greedoids are reviewed as follows.

Definition 1.1

([1, 3]). Let 𝓕 be a set system on E.

  1. A greedoid is a pair (E, 𝓕), where 𝓕 satisfies the following conditions:

    • (G1)

      For every non-empty X ∈ 𝓕, there is an xX such that X ∖ {x} ∈ 𝓕. (accessible)

    • (G2)

      For X, Y ∈𝓕 such that |X| > |Y|, there is an xXY such that Y ∪ {x} ∈ 𝓕. (exchangeable)

  2. A greedoid (E, 𝓕) has the interval property (or to be an interval greedoid) if ABC, A, B, C ∈ 𝓕 and xEC, then A ∪ {x} ∈𝓕 and C ∪ {x} ∈ 𝓕 imply B ∪ {x} ∈ 𝓕.

  3. A maximal element in (𝓕, ⊆) is called a basis.

  4. A loop in (E, 𝓕) is an element xE that is contained in no basis.

  5. A language 𝓛 over E is a non-empty set 𝓛 ⊆ E* (the free monoid of all words over the alphabet E) of words over the alphabet E; it is called simple if every word in 𝓛 is simple (i.e. it does not contain any letter more than once).

A greedoid language over E is a pair (E, 𝓛), where 𝓛 is a simple language 𝓛 ⊆ Es (the set of simple words in E*) satisfying the following conditions:

  • (L1)

    If α = β γ and α ∈ 𝓛, then β ∈ 𝓛.   (hereditary)

  • (L2)

    If α, β ∈ 𝓛 and |α| > |β|, then α contains a letter x such that β x ∈𝓛.   (exchange).

Björner et al. indicate [1] that greedoids were originally developed to give a unified approach to the optimality of various greedy algorithms known in combinatorial optimization. Such algorithms can be loosely characterized as having locally optimal strategy and no backtracking. Nowadays, researchers provide different greedy algorithms to characterize the different kinds of greedoids (cf. [1, 2, 3, 4]). Helman et al. [2] characterize greedy structures. [2, Theorem 1] is the best and main result in [2]. That is, let (E, 𝓕) be an accessible set system. Then (E, 𝓕) has an optimal greedy basis for every positive weighted linear function if and only if (E, 𝓕) is a matroid embedding (cf. [2, Definition 2], i.e. A matroid embedding is an accessible set system which is extensible, closure-congruent, and the hereditary closure of which is a matroid). Below [2, Definition 2], Helman et al. say that (S, 𝓒1) in [2, Example 1] is a matroid embedding, yet not a greedoid. Combining [2, Example 2] and the definition of a matroid embedding, (S, 𝓒2) in [2, Example 2] is a greedoid, yet not a matroid embedding. In other words, [2, Theorem 1] does not characterize a greedoid structure with greedy algorithm. Korte et al. say [4, p.358, Theorem 14.7]: Let (E, 𝓕) be a greedoid. The greedy algorithm finds a set F ∈ 𝓕 of maximum weight for each modular weight function if and only if (E, 𝓕) has the so-called strong exchange axiom (see [2], [3, p.160], [4, p.358], or say: For all A ∈ 𝓕, B is a maximal in 𝓕 and AB. If xEB with A ∪ {x} ∈ 𝓕, then there exists a yBA such that A ∪ {y} ∈ 𝓕 and (B ∖ {y}) ∪ {x} ∈ 𝓕). Actually, Korte et al. [3, p.160, Theorem 2.2] is the same result as [4, p.358, Theorem 14.7]. However, from [3], it is easily seen that a greedoid can not be ensured to satisfy the strong exchange axiom. Hence, we may be asserted that [3, p.160, Theorem 2.2] or [4, p.358, Theorem 14.7] is not a characterization for all of greedoids with greedy algorithm, but only a characterization for a part of class of greedoids. In addition, among the known characterizations relative to greedoids with greedy algorithm, we think [1, Theorem 8.5.2] (the same as [3, p.157, Theorem 1.4]) to be better, that is: suppose (E, 𝓛) is a simple hereditary language, then (E, 𝓛) is a greedoid if and only if greedy algorithm gives an optimal solution for every compatible objective function on 𝓛. In the characterizations using Definition 1.1 and the greedy algorithms for a greedoid (E, 𝓕) proved in [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4], 𝓕 must be hereditary (i.e. XY, Y ∈ 𝓕 ⇒ X ∈ 𝓕).

Now returning to our question: under what conditions on a greedoid, can every linear function be optimized by the greedy algorithm? Up to now, we do not find an answer for all of greedoids. Though we do not find out the solution to the open problem for all of greedoids, using the research methods in [1, 2, 3, 4] for reference, we can pay our attention to some special class of greedoids to look for the answer. By [1, 3], an interval greedoid (E, 𝓕0) does not ask 𝓕0 to be hereditary or satisfy strong exchange axiom. The authors describe [1] that the ‘interval property’ characterizes a very large class of greedoids and interval greedoids behave better than general greedoids in many respects. In some types of study, the interval property has to be assumed to obtain meaningful results [1, 3, 5, 6]. Hence, this paper will focus on interval greedoids in hope to find the answer for the open problem.

We may find from Definition 1.1 that for a greedoid (E, 𝓕), if 𝓕 is hereditary, then (E, 𝓕) is interval. In addition, it is necessary to generalize the results in [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4] for interval greedoids. This is done in this paper.

Lemma 1.2

Let 𝓕 ⊆ 2E and xE be a loop. Then 𝓕 is a set system on E ∖ {x}.


Suppose that a loop x is contained in a X ∈ 𝓕. Then, there is a basis BX satisfying XBX according to Definition 1.1(3). This follows xBX, a contrary to the loop of x. Therefore, we demonstrate that 𝓕 is defined on E ∖ {x}.

By Lemma 1.2, this paper only considers the set systems with no loops.

Lemma 1.3

([3, p.47]). For a given set system 𝓕 on E, the property (G2) holds if and only if for any AE, all bases of A have the same cardinality.

According to Lemma 1.3, we can state that in a set system such that (G2), then XE is a basis of AE if and only if X ∈ 𝓕, X ⊆ A satisfies |X|=maxYF,YA|Y|.

2 Main results

We give some notions for a set system (E, 𝓕):

  1. 𝓕(k) = {X ∈ 𝓕 ∣ |X| = k};

  2. 𝓕|A = {XXA, X ∈ 𝓕} for any A ∈ 𝓕;

  3. n = maxXF|X|;

  4. Let ω : E → ℝ+ be a positive weight function (i.e. ω(x) > 0 for any xE). For XE, define ωX: X → ℝ+ as ωX(x) = ω(x) for any xX.

We know that, generally, the solution of the greedy algorithm in Section 1 is not optimal. The already existing greedy algorithms for greedoids (see [1, 3, 4]) are satisfied (or say, characterized) by some different classes of greedoids. In order to search out a characterization of a type of greedy algorithms for some class of interval greedoids, we provide a type of greedy algorithm (i.e. Algorithm 1) as follows. After that, we will demonstrate under what conditions for a set system, Algorithm 1 has an optimal solution. We also find under what conditions Algorithm 1 characterizes an interval greedoid.

Algorithm 1

(Interval Greedy Algorithm). Input: 𝓕, a set system on E; ω: E → ℝ+, a positive weight function; n, maxXF |X|.

Output: S, the greedy solution.

  1. Set S = ∅, j = 0.

  2. If j < n − 1, then go to 3.

    If j = n − 1, then go to 4.

    If jn, then S: = S, stop.

  3. Set Dj+1 = {ethere exist Sj+1 ∈ 𝓕(j+1) and SSj+1 such that eESj+1 and Sj+1 ∪ {e} ∈ 𝓕}, and Gj = {eESS ∪ {e} ∈ 𝓕}.

    • 3.1

      If Dj+1Gj ≠ ∅, then choose ej+1Dj+1Gj such that ω(ej+1) = maxeDj+1Gj ω(e), and set S: = S ∪ {ej+1}, j: = j+1, go to {2}.

    • 3.2

      If Dj+1Gj = ∅, then S: = S, and j: = j+1, go to 2.

  4. Set Gj = {eESS ∪ {e} ∈ 𝓕}.

  • 4.1

    If Gj ≠ ∅, then choose ej+1Gj such that ω(ej+1) = maxeGj ω(e), and set S := S ∪ {ej+1}, j := j+1, go to 2

  • 4.2

    If Gj = ∅, then S := S and j: = j+1, go to 2.

We say the greedy algorithm works if ω(S) ≥ ω(A) for ∀ A ∈ 𝓕. In the process of Algorithm 1, we can use St+1 to stand for the solution when the cyclic variable j is tn − 1.

Example 2.1

Let E1 = {a1, a2, a3, a4} and 𝓕1 = {∅, {a1}, {a4}, {a1, a2}, {a1, a4}, {a1, a2, a3}, {a1, a3, a4}, {a1, a2, a3, a4}}. We can easily check that 𝓕1 satisfies (G1) and (G2) in Definition 1.1 (1).

Let A = ∅, B = {a1, a2} and C = {a1, a2, a3}. We easily find ABC. For a4E1C, we obtain A ∪ {a4} = {a4} ∈ 𝓕1, C ∪ {a4} = {a1, a2, a3, a4} ∈ 𝓕1 and B ∪ {a4} = {a1, a2, a4} ∉ 𝓕1. Using Definition 1.1 (2), (E1, 𝓕1) is not an interval greedoid.

Define ω1 : E1 → ℝ+ as ω1(a1) = 5, ω1(a2) = 4, ω1(a3) = 3, ω1(a4) = 2. Then, we can demonstrate that {a1, a2, a3, a4} is an optimal set. Applying Algorithm 1 on (𝓕1, ω1), we look for the solution S of Algorithm 1 as follows: There is n = 4.

When j = 0, there are S0 = ∅, D1 = {a1, a2, a4}, G0 = {a1, a4}, and so S1 = {a1}.

When j = 1, there are D2 = {a3}, G1 = {a2, a4}, and so D2G1 = ∅. Thus, we attain S2 = S1 = {a1}.

When j = 2, there are D3 = {a2, a4}, G2 = {a2, a4}, and so S3 = {a1, a2}.

When j = 3, there is j = n − 1. We find G3 = {a3}. Hence, there is S4 = {a1, a2, a3}.

When j = 4, there is jn. So, we obtain S = S4.

Actually, ω1({a1, a2, a3}) = 12 < 14 = ω1({a1, a2, a3, a4}) indicates that S is not optimal.

Remark 2.2

After analyzing Example 2.1, we attain some properties as follows.

  1. If we hope Algorithm 1 to work for (E, 𝓕), then (E, 𝓕) should be an interval greedoid.

  2. There are {a1}, {a4} ∈ 𝓕1 and {a2}, {a3} ∉ 𝓕1 holds. Hence, we can ask {x} ∈ 𝓕 for any xE if we hope Algorithm 1 to work for a set system 𝓕 with any positive weight function ω.

Example 2.3

Let E2 = {a1, a2, a3, a4} and 𝓕2 = {∅, {a1}, {a4}, {a1, a2}, {a1, a4}, {a1, a2, a3}, {a1, a3, a4}}. We easily check up 𝓕2 to satisfy (G1) and (G2) and the interval property. Hence, (E2, 𝓕2) is an interval greedoid.

Define ω2 : E2 → ℝ+ as ω2(a1) = 5, ω2(a2) = 4, ω2(a3) = 3, ω2(a4) = 2. Then, we can demonstrate that {a1, a2, a3} is an optimal set. Applying Algorithm 1 on (𝓕2, ω2), we look for the solution S of Algorithm 1 as follows: There is n = 3.

When j = 0, there are S0 = ∅, D1 = {a1, a2, a4}, G0 = {a1, a4}, and so S1 = {a1}.

When j = 1, there are D2 = {a3}, G1 = {a2, a4}, and so D2G1 = ∅. Thus, we attain S2 = S1 = {a1}.

When j = 2, there is j = n − 1. We find G2 = {a2, a4}. Hence, there is S3 = {a1, a2}.

When j = 3, there is jn. So, we obtain S = S3.

Actually, ω2({a1, a2, a3}) = 12 > 9 = ω2({a1, a2}) indicates that S is not optimal.

Remark 2.4

In Example 2.3, B1 = {a1, a2, a3} and B2 = {a1, a3, a4} are two bases of 𝓕2 such that B1 ∖ {a3} = {a1, a2},B2 ∖ {a3} = {a1, a4} ∈ 𝓕2. But ω2(a1) > ω2(a2) > ω2(a3) > ω2(a4) and 𝓕(2) = {{a1, a2}, {a1, a4}} imply that (E2,𝓕2) does not satisfy the following condition:

(G3) Let X, Y be bases in 𝓕 and S = Y ∖ {y0} ∈ 𝓕 for some y0Y. Then there is x0XS such that S ∪ {x0} ∈ 𝓕 and ω(x0) = maxxXω(x).

Hence, we should ask if 𝓕 with ω satisfy (G3) if we hope Algorithm 1 to work for (𝓕,ω) though (E, 𝓕) is an interval greedoid.

Lemma 2.5

Let 𝓕 ⊆ 2E with ∅ ∈ 𝓕 and ω : E → ℝ+ be a positive weight function.

  1. If 𝓕 satisfies (G2), then there is 𝓕(k) ≠ ∅ for any k = 0, 1, …, n.

  2. An optimal set in (𝓕,ω) is a basis.


  1. Using Lemma 1.3 and Definition 1.1 (3), all of bases in 𝓕 have the cardinality n. Let B ∈ 𝓕 be a basis. From Björner et al. [1, 8.2.A], we know that ∅ ∈ 𝓕 and (G2) together define greedoids as well as (G1) and (G2). Considering (G1) on B, we may easily obtain 𝓕(k) ≠ ∅, (k = 0, 1, …, n).

  2. Since an optimal set S satisfies ω(S) ≥ ω(B) for any basis B of 𝓕. If S is not a basis, then SBS holds for some basis BS according to Definition 1.1(3). Thus, there is ω(S) < ω(BS) since ω is positive, a contradiction with ω(S) ≥ ω(B). Hence S is a basis.


Theorem 2.6

Let 𝓕 ⊆ 2E satisfy ∅ ∈ 𝓕 and {x} ∈ 𝓕 for any xE. Let ω : E → ℝ+ be a positive weight function. If (E, 𝓕) is an interval greedoid satisfying the condition (G3), then Algorithm 1 works for (𝓕,ω).


  • Step 1

    Let 𝓕k = {XX ∈ 𝓕, |X| ≤ k} (k ≥ 1). We prove the following statements.

    (st1) ∅ ∈ 𝓕k,{x} ∈ 𝓕k for any xE.

    (st2) (E, 𝓕k} is an interval greedoid.

    {x} ∈ 𝓕 and |{x}| = 1 ≤ k for any xE imply {x} ∈ 𝓕k for any xE. ∅ ∈ 𝓕 and | ∅ | = 0 ≤ k together means ∅ ∈ 𝓕k. Hence, we can say that 𝓕k is a set system with no loops. According to 𝓕k ⊆ 𝓕 and 𝓕 satisfying both of (G1) and (G2), we easily obtain that 𝓕k satisfies both of (G1) and (G2).

    Let ABC, A, B, C ∈ 𝓕k and aEkC satisfy A ∪ {a} ∈ 𝓕k and C ∪ {a} ∈ 𝓕k. Using 𝓕k ⊆ 𝓕 and the interval property of 𝓕, we obtain B ∪ {a} ∈ 𝓕. C ∪ {a} ∈ 𝓕k follows | C ∪ {a}| ≤ k. Combining with B ∪ {a} ⊆ C ∪ {a}, we decide | B ∪ {a}| ≤ k. Hence, there is B ∪ {a} ∈ 𝓕k.

    Therefore, (E, 𝓕k) is an interval greedoid.

  • Step 2

    We will prove that Algorithm 1 works for (𝓕,ω) by induction on n.

    If n = 0. This means 𝓕 = {∅. Hence, the needed result follows.

    If n = 1. By Lemma 2.5 (1) and Definition 1.1 (2), there are 𝓕(0) ≠ ∅ and 𝓕(1) ≠ ∅.

    Then, in the process of Algorithm 1, when j = 0 = 1 − 1 = n − 1, according to S0 = ∅, there is G0 = {eE ∖ ∅ ∣ ∅ ∪ {e} ∈ 𝓕} = E since {x} ∈ 𝓕 for any xE. Hence, G0 ≠ ∅ holds. Choose ω(e0+1) = ω(e1) = maxeG0ω(e), and put S1 = ∅ ∪ {e1} = {e1} and j : = 0 + 1 = 1. When j = 1, then j ≥ 1 = n follows the process of Algorithm 1 to stop. Therefore, the solution of Algorithm 1 is optimal. That is to say, Algorithm 1 works for (𝓕,ω).

    Suppose that if n ≤ m − 1, then the needed result is correct. Now, let n = m.

    Since (E, 𝓕) is an interval greedoid, 𝓕 satisfies (G1) and (G2). Combining Lemma 2.5, there is m = |B| for any basis B of 𝓕. Utilizing Lemma 2.5 (1), we obtain 𝓕(k) ≠ ∅ where k = 0, 1, …, m.

    Let S be the solution of Algorithm 1 for (𝓕,ω). During the process of Algorithm 1, when j < m − 1, according to the interval property, ∅ ∪ {e0} ∈ 𝓕 for any e0E, ∅ ⊆ SjSj+1, Sj+1 ∪ {e} ∈ 𝓕 for any eDj+1, and the definitions of Dj+1 and Gj, there is Sj ∪ {e} ∈ 𝓕 for any eDj+1, and so Dj+1Gj ≠ ∅. Considering Lemma 2.5 (1), we arrive at 𝓕(m) ≠ ∅. So, there is Gm−1 = {eESm−1Sm−1 ∪ {e} ∈ 𝓕} ≠ ∅. Therefore, we can demonstrate that S is a basis. Hence, |S| = m holds.

    Since S is accessible according to the process of Algorithm 1 for the interval greedoid (E, 𝓕) and ω, there is S = Sm. Using Step 1 and the inductive supposition, Algorithm 1 works for (E, 𝓕m−1). That is to say, the solution Sm−1 of Algorithm 1 for (E, 𝓕m−1) satisfies ω(Sm−1) ≥ ω(X) for any X ∈ 𝓕m−1. Combining the process of Algorithm 1, we confirm Sm−1 = Sm−1.

    Considering Gm−1 = {eESm−1Sm−1 ∪ {e} ∈ 𝓕} and 𝓕(m) ≠ ∅, we obtain Sm = Sm−1 ∪ {em} where ω(em) = maxeGm1ω(e). Moreover, Sm is the solution of Algorithm 1 for (𝓕,ω).

    Let B be a basis of 𝓕. We easily find |B| = |Sm−1| + 1. Thus, Sm−1 ∪ {b} ∈ 𝓕 holds for some bBSm−1 according to (G2) satisfied by 𝓕. Thus, Sm−1 ∪ {b} is a basis of 𝓕. Using (G3), there is Sm−1 ∪ {b0} ∈ 𝓕 where ω(b0) = maxxBS¯m1ω(x).

    On the other hand, for any xB, if B ∖ {x} ∈ 𝓕, then ω(B ∖ {x}) ≤ ω(Sm−1) = ω(Sm−1) in view of the inductive supposition. Since (B ∖ {x}) ∪ {x} is a basis, there is ω((B ∖ {x}) ∪ {x}) ≤ ω(Sm−1 ∪ {b0}) ≤ ω(Sm−1 ∪ {em}) = ω(Sm) in virtue of (G3) and the process of searching Sm.

    Therefore, ω(Sm) ≥ ω(B) holds for any basis B in 𝓕. Furthermore, ω(Sm) ≥ ω(X) is correct for any X ∈ 𝓕 since X must be contained in a basis and ω is positive. Thus, Sm is optimal.

    Summing up, Algorithm 1 works for (𝓕,ω).


Remark 2.7

Example 2.1 shows that (E1, 𝓕1) is not an interval greedoid. In addition, for {a1} ⊂ {a1, a2} and a3E1 ∖ {a1, a2}, there are {a1, a2} ∪ {a3} = {a1, a2, a3} ∈ 𝓕1 and {a1} ∪ {a3} = {a1, a3} ∉ 𝓕1. That is to say, 𝓕 does not satisfy the semi-interval property (that is, if YZ, Y, Z ∈ 𝓕, aEZ, then Z ∪ {a} ∈ 𝓕 ⇒ Y ∪ {a} ∈ 𝓕)}.

It is more interesting that the converse of Theorem 2.6 is also true under some pre-conditions.

Theorem 2.8

Let 𝓕 be a set system on E with ∅ ∈ 𝓕. If for any positive weight function ω : E → ℝ+, there are the following statements:

  • (s1)

    Algorithm 1 works for (𝓕|A, ωA) for any A ∈ 𝓕.

  • (s2)

    𝓕 has semi-interval property.

  • (s3)

    𝓕 satisfies (G2).

Then, (E, 𝓕) is an interval greedoid.


  • Step 1

    To prove : 𝓕 is accessible.

    Let A ∈ 𝓕. Since A is the basis of 𝓕|A and ω is positive, there is ω(A) = maxXF|A ω(X). Let SA be the solution of Algorithm 1 for (𝓕|A, ωA). Consider the process of Algorithm 1 and A ∈ 𝓕, we can decide SAA. Hence, we obtain ωA(SA) ≤ ωA(A) since ωA is positive. Furthermore, since ωA is positive, we find SAAωA(SA) < ωA(A). By (s1), we confirm ωA(SA) = maxXF|A ωA(X). Summing up the above results, we may follow SA = A.

    From ∅ ∈ 𝓕 and the process of Algorithm 1 for (𝓕|A, ωA), we may assert that SA is accessible. We will prove this assertion as follows.

    Let m = |SA|, that is, m = |A| since SA = A.

    If m = 0. Then SA = ∅. So, SA is accessible.

    If m = 1. Then A = {a} and S0 = ∅.

    When j = 0 = m − 1 = 1 − 1. There is Gj = G0 = {a}. Hence, Gj ≠ ∅ and ω(e0+1) = ω(a) and S1 = ∅ ∪ {a} = {a}.

    When j = 0 + 1 = 1 = m, then stop, and SA = {a}. We easily obtain SA ∖ {a} = ∅ ∈ 𝓕. Hence, SA is accessible.

    Suppose that if mk − 1, then SA is accessible. Now, let m = k > 1.

    If for every j < k − 1, there is Dj+1Gj ≠ ∅ in the process of Algorithm 1, then according to the definitions of Dj+1 and Gj, there are S0 = ∅ ⊂ S1S2 ⊂ … ⊂ Sj+1 and Si+1 = Si ∪ {ei+1} for some ei+1ASi, (i = 0, …, j + 1), we can state that Sj is accessible (j = 0, …, k − 1).

    If for some j0 < k − 1, there is Dj0+1Gj0 = ∅ in the process of Algorithm 1, then Sj0+1 = Sj0. This follows |Sj0+1| < j0+1. Furthermore, we obtain |Sk| < k according to the process of Algorithm 1. Thus, we attain SA (that is Sk) satisfying |SA| < k = |A| = |SA|, a contradiction. Hence, for every j < k − 1, there is Dj+1Gj ≠ ∅.

    So, Sj is accessible (j = 0, 1, …, k − 1).

    Let j = k − 1. If Gk−1 ≠ ∅, then we obtain Sk = Sk−1 ∪ {ek}. Thus, we may easily find Sk to be accessible since Skek−1 = Sk−1 ∈ 𝓕 and the above discussion.

    If Gk−1 = ∅, then Sk = Sk−1. Thus, there is |Sk| = |Sk−1| = k − 1. On the other hand, Sk = Sm = SA = A hints |Sk| = |A| = m = k. This is a contradiction to |Sk| = k − 1. Therefore, we confirm Gk ≠ ∅.

    Adding up the above discussion with the induction, we can state that SA, that is A, is accessible.

    According to the arbitrariness of A, we attain that 𝓕 is accessible.

  • Step 2

    To prove : 𝓕 satisfies the interval property.

    Let X, Y, Z ∈ 𝓕 satisfy XYZ. Let aEZ satisfy X ∪ {a}, Z ∪ {a} ∈ 𝓕. Using the statement (s2), there is Y ∪ {a} ∈ 𝓕. Therefore, 𝓕 satisfies the interval property.

  • Step 3

    𝓕 is exchangeable according to the statement (s3).

  • Step 4

    Combining Steps 1, 2 and 3 with Definition 1.1, (E, 𝓕) is an interval greedoid.


Example 2.9

Let E3 = {a1, a2, a3, a4} and 𝓕3 = {∅, {a1}, {a4}, {a1, a2}, {a3, a4}}. There is |{a3, a4}| = 2 = |{a1}| + 1, but no element a ∈ {a3, a4} ∖ {a1} satisfies {a1} ∪ {a} ∈ 𝓕3. Define ω3 : E3 → ℝ+ as ω3(a1) = 5, ω3(a2) = 1, ω3(a3) = ω3(a4) = 4. The solution of Algorithm 1 for (𝓕3,ω3) is {a1}. But, ω3({a1}) = 5 < ω3({a3, a4}) = 8 implies Algorithm 1 not to work for (𝓕3,ω3).

Example 2.9 shows that if we want Algorithm 1 to work for (𝓕, ω), then 𝓕 should satisfy (G2). Combining Theorems 2.6 and 2.8, we give the following characterization for interval greedoids.

Theorem 2.10

Let 𝓕 be an exchangable set system on E with ∅ ∈ 𝓕 and {x} ∈ 𝓕 for any xE. 𝓕A satisfies (G3) for any A ∈ 𝓕. Then 𝓕 is the set of feasible sets of an interval greedoid on E if and only if for all positive weight functions ω : E → ℝ+, 𝓕 satisfies the statements (s1) and (s2).


  • (⇒)

    We easily prove (A, 𝓕|A) to be an interval greedoid for any A ∈ 𝓕. Combining Theorem 2.1 and 𝓕|A satisfying (G3), we obtain the correctness of (s1). ∅, {x} ∈ 𝓕 and the interval of 𝓕 follow the correctness of (s2).

  • (⇐)

    Using Theorem 2.8, all of needed results are straightforward.


Next, we will compare our results with some known results for greedoids.

  • (I)

    To compare our results with [1, Theorem 8.5.2] (or say [3, p.157, Theorem 1.4]).

  1. In [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4], the authors give a kind of greedy algorithm to characterize a greedoid (E, 𝓕13), where 𝓕13 is asked to be hereditary. In other words, if a greedoid (E, 𝓕) does not satisfy the hereditary property for 𝓕, then the characterizations with greedy algorithms in [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4] will not be successful.

  2. Let 𝓕 be a set system on E satisfying the hereditary. Then, we easily find that 𝓕 has the following properties:

  • ∅, {x} ∈ 𝓕 holds for any xE; 𝓕 satisfies the condition (G3).

  • Let YZ, Y, Z ∈ 𝓕 and aEZ. If Z ∪ {a} ∈ 𝓕 is correct, then Y ∪ {a} ∈ 𝓕 holds according to Y ∪ {a} ⊆ Z ∪ {a} and the hereditary property of 𝓕. This implies that every hereditary sets system satisfies the semi-interval property.

  • Considered items (1) and (2), we know that the characterization for greedoids with the greedy algorithm provided in [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4] are really effective only for some of interval greedoids and not for the other kinds of greedoids.

  • Evidently, the given conditions in Theorems 2.6, 2.8 and 2.10 do not ask 𝓕 to be hereditary. Combining the above three items, we can say that for a hereditary set system 𝓕, Theorem 2.6 and Theorem 2.8 are satisfied by much more greedoids than that in [1, Theorem 8.5.2] (or say, [3, p.157, Theorem 1.4]) respectively.

    Moreover, the characterization (i.e. Theorem 2.3) proposed in this paper for interval greedoids generalize the results in [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4] respectively.

    Therefore, Algorithm 1 generalizes the greedy algorithm for [1, Theorem 8.5.2] and [3, p.157, Theorem 1.4].

  • (II)

    To compare our results with [4, p.358, Theorem 14.7].

    It is well known that a greedoid is perhaps not satisfying the strong exchange axiom. In other words, not every greedoid has strong exchange axiom, though any greedoid is exchangeable. We also know that an interval greedoid can not be ensured to satisfy strong exchange axiom. Thus, we can state that Theorem 2.10 is a characterization of a greedy algorithm for some class of interval greedoids. [4, p.358, Theorem 14.7] can not substitute for the results in this paper. Therefore, Algorithm 1 is a new algorithm and not covered by the algorithm for [4, p.358, Theorem 14.7].

    More generalized characterization for greedoids with greedy algorithms will be studied in the future. We also hope to give more answers to the open problem stated in Section 1.


This research is supported by NNSF of China (61572011) and NSF of Hebei Province (A2018201117).


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About the article

Received: 2016-12-31

Accepted: 2018-02-08

Published Online: 2018-03-31

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 260–267, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0026.

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© 2018 Mao, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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