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

𝓕^{(k)} = {*X* ∈ 𝓕 ∣ |*X*| = *k*};

𝓕|_{A} = {*X* ∣ *X* ⊆ *A*, *X* ∈ 𝓕} for any *A* ∈ 𝓕;

*n* =
$\begin{array}{}\underset{X\in \mathcal{F}}{max}\end{array}$|*X*|;

Let *ω* : *E* → ℝ^{+} be a *positive weight function* (i.e. *ω*(*x*) > 0 for any *x* ∈ *E*). For *X* ⊆ *E*, define *ω*_{X}: *X* → ℝ^{+} as *ω*_{X}(*x*) = *ω*(*x*) for any *x* ∈ *X*.

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*,
$\begin{array}{}\underset{X\in \mathcal{F}}{max}\end{array}$ |*X*|.

*Output*: *S*, *the greedy solution*.

*Set* *S* = ∅, *j* = 0.

*If j* < *n* − 1, *then go to* 3.

*If j* = *n* − 1, *then go to* 4.

*If j* ≥ *n*, *then* *S*: = *S*, *stop*.

*Set D*_{j+1} = {*e* ∣ *there exist S*_{j+1} ∈ 𝓕^{(j+1)} *and* *S* ⊂ *S*_{j+1} *such that e* ∈ *E* ∖ *S*_{j+1} *and S*_{j+1} ∪ {*e*} ∈ 𝓕}, *and G*_{j} = {*e* ∈ *E* ∖ *S* ∣ *S* ∪ {*e*} ∈ 𝓕}.

3.1

*If D*_{j+1} ∩ *G*_{j} ≠ ∅, *then choose e*_{j+1} ∈ *D*_{j+1} ∩ *G*_{j} such that ω(*e*_{j+1}) =
$\begin{array}{}\underset{e\in {D}_{j+1}\cap {G}_{j}}{max}\end{array}$ *ω*(*e*), *and set* *S*: = *S* ∪ {*e*_{j+1}}, *j*: = *j*+1, *go to* {2}.

3.2

*If D*_{j+1} ∩ *G*_{j} = ∅, *then* *S*: = *S*, *and j*: = *j*+1, *go to* 2.

*Set G*_{j} = {*e* ∈ *E* ∖ *S* ∣ *S* ∪ {*e*} ∈ 𝓕}.

4.1

*If G*_{j} ≠ ∅, *then choose e*_{j+1} ∈ *G*_{j} such that ω(*e*_{j+1}) =
$\begin{array}{}\underset{e\in {G}_{j}}{max}\end{array}$ *ω*(*e*), *and set* *S* := *S* ∪ {*e*_{j+1}}, *j* := *j*+1, *go to* 2

4.2

*If G*_{j} = ∅, *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 *S*_{t+1} to stand for the solution when the cyclic variable *j* is *t* ≤ *n* − 1.

#### Example 2.1

*Let E*_{1} = {*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}} *and* 𝓕_{1} = {∅, {*a*_{1}}, {*a*_{4}}, {*a*_{1}, *a*_{2}}, {*a*_{1}, *a*_{4}}, {*a*_{1}, *a*_{2}, *a*_{3}}, {*a*_{1}, *a*_{3}, *a*_{4}}, {*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}}}. *We can easily check that* 𝓕_{1} *satisfies (G1) and (G2) in Definition 1.1 (1)*.

Let *A* = ∅, *B* = {*a*_{1}, *a*_{2}} and *C* = {*a*_{1}, *a*_{2}, *a*_{3}}. We easily find *A* ⊂ *B* ⊂ *C*. For *a*_{4} ∈ *E*_{1} ∖ *C*, we obtain *A* ∪ {*a*_{4}} = {*a*_{4}} ∈ 𝓕_{1}, *C* ∪ {*a*_{4}} = {*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}} ∈ 𝓕_{1} and *B* ∪ {*a*_{4}} = {*a*_{1}, *a*_{2}, *a*_{4}} ∉ 𝓕_{1}. Using Definition 1.1 (2), (*E*_{1}, 𝓕_{1}) is not an interval greedoid.

Define *ω*_{1} : *E*_{1} → ℝ^{+} as *ω*_{1}(*a*_{1}) = 5, *ω*_{1}(*a*_{2}) = 4, *ω*_{1}(*a*_{3}) = 3, *ω*_{1}(*a*_{4}) = 2. Then, we can demonstrate that {*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}} is an optimal set. Applying Algorithm 1 on (𝓕_{1}, *ω*_{1}), we look for the solution *S* of Algorithm 1 as follows: There is *n* = 4.

When *j* = 0, there are *S*_{0} = ∅, *D*_{1} = {*a*_{1}, *a*_{2}, *a*_{4}}, *G*_{0} = {*a*_{1}, *a*_{4}}, and so *S*_{1} = {*a*_{1}}.

When *j* = 1, there are *D*_{2} = {*a*_{3}}, *G*_{1} = {*a*_{2}, *a*_{4}}, and so *D*_{2} ∩ *G*_{1} = ∅. Thus, we attain *S*_{2} = *S*_{1} = {*a*_{1}}.

When *j* = 2, there are *D*_{3} = {*a*_{2}, *a*_{4}}, *G*_{2} = {*a*_{2}, *a*_{4}}, and so *S*_{3} = {*a*_{1}, *a*_{2}}.

When *j* = 3, there is *j* = *n* − 1. We find *G*_{3} = {*a*_{3}}. Hence, there is *S*_{4} = {*a*_{1}, *a*_{2}, *a*_{3}}.

When *j* = 4, there is *j* ≥ *n*. So, we obtain *S* = *S*_{4}.

Actually, *ω*_{1}({*a*_{1}, *a*_{2}, *a*_{3}}) = 12 < 14 = *ω*_{1}({*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}}) indicates that *S* is not optimal.

#### Example 2.3

*Let E*_{2} = {*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}} *and* 𝓕_{2} = {∅, {*a*_{1}}, {*a*_{4}}, {*a*_{1}, *a*_{2}}, {*a*_{1}, *a*_{4}}, {*a*_{1}, *a*_{2}, *a*_{3}}, {*a*_{1}, *a*_{3}, *a*_{4}}}. *We easily check up* 𝓕_{2} *to satisfy (G1) and (G2) and the interval property*. *Hence*, (*E*_{2}, 𝓕_{2}) *is an interval greedoid*.

Define *ω*_{2} : *E*_{2} → ℝ^{+} as *ω*_{2}(*a*_{1}) = 5, *ω*_{2}(*a*_{2}) = 4, *ω*_{2}(*a*_{3}) = 3, *ω*_{2}(*a*_{4}) = 2. Then, we can demonstrate that {*a*_{1}, *a*_{2}, *a*_{3}} is an optimal set. Applying Algorithm 1 on (𝓕_{2}, *ω*_{2}), we look for the solution *S* of Algorithm 1 as follows: There is *n* = 3.

When *j* = 0, there are *S*_{0} = ∅, *D*_{1} = {*a*_{1}, *a*_{2}, *a*_{4}}, *G*_{0} = {*a*_{1}, *a*_{4}}, and so *S*_{1} = {*a*_{1}}.

When *j* = 1, there are *D*_{2} = {*a*_{3}}, *G*_{1} = {*a*_{2}, *a*_{4}}, and so *D*_{2}∩ *G*_{1} = ∅. Thus, we attain *S*_{2} = *S*_{1} = {*a*_{1}}.

When *j* = 2, there is *j* = *n* − 1. We find *G*_{2} = {*a*_{2}, *a*_{4}}. Hence, there is *S*_{3} = {*a*_{1}, *a*_{2}}.

When *j* = 3, there is *j* ≥ *n*. So, we obtain *S* = *S*_{3}.

Actually, *ω*_{2}({*a*_{1}, *a*_{2}, *a*_{3}}) = 12 > 9 = *ω*_{2}({*a*_{1}, *a*_{2}}) indicates that *S* is not optimal.

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 ⊆ 2^{E} *with* ∅ ∈ 𝓕 *and* *ω* : *E* → ℝ^{+} *be a positive weight function*.

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

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

#### Proof

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*).

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

□

#### Theorem 2.6

*Let* 𝓕 ⊆ 2^{E} *satisfy* ∅ ∈ 𝓕 *and* {*x*} ∈ 𝓕 *for any* *x* ∈ *E*. *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*).

#### Proof

Step 1

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

(st1) ∅ ∈ 𝓕_{k},{*x*} ∈ 𝓕_{k} for any *x* ∈ *E*.

(st2) (*E*, 𝓕_{k}} is an interval greedoid.

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

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

Therefore, (*E*, 𝓕_{k}) 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 *S*_{0} = ∅, there is *G*_{0} = {*e* ∈ *E* ∖ ∅ ∣ ∅ ∪ {*e*} ∈ 𝓕} = *E* since {*x*} ∈ 𝓕 for any *x* ∈ *E*. Hence, *G*_{0} ≠ ∅ holds. Choose *ω*(*e*_{0+1}) = *ω*(*e*_{1}) = $\begin{array}{}\underset{e\in {G}_{0}}{max}\omega (e),\end{array}$ and put *S*_{1} = ∅ ∪ {*e*_{1}} = {*e*_{1}} 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, ∅ ∪ {*e*_{0}} ∈ 𝓕 for any *e*_{0} ∈ *E*, ∅ ⊆ *S*_{j} ⊆ *S*_{j+1}, *S*_{j+1} ∪ {*e*} ∈ 𝓕 for any *e* ∈ *D*_{j+1}, and the definitions of *D*_{j+1} and *G*_{j}, there is *S*_{j} ∪ {*e*} ∈ 𝓕 for any *e* ∈ *D*_{j+1}, and so *D*_{j+1} ∩ *G*_{j} ≠ ∅. Considering Lemma 2.5 (1), we arrive at 𝓕^{(m)} ≠ ∅. So, there is *G*_{m−1} = {*e* ∈ *E* ∖ *S*_{m−1} ∣ *S*_{m−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* = *S*_{m}. Using Step 1 and the inductive supposition, Algorithm 1 works for (*E*, 𝓕_{m−1}). That is to say, the solution *S*^{m−1} of Algorithm 1 for (*E*, 𝓕_{m−1}) satisfies *ω*(*S*^{m−1}) ≥ *ω*(*X*) for any *X* ∈ 𝓕_{m−1}. Combining the process of Algorithm 1, we confirm *S*^{m−1} = *S*_{m−1}.

Considering *G*_{m−1} = {*e* ∈ *E* ∖ *S*_{m−1} ∣ *S*_{m−1} ∪ {*e*} ∈ 𝓕} and 𝓕^{(m)} ≠ ∅, we obtain *S*_{m} = *S*_{m−1} ∪ {*e*_{m}} where *ω*(*e*_{m}) = $\begin{array}{}\underset{e\in {G}_{m-1}}{max}\omega (e).\end{array}$ Moreover, *S*_{m} is the solution of Algorithm 1 for (𝓕,*ω*).

Let *B* be a basis of 𝓕. We easily find |*B*| = |*S*_{m−1}| + 1. Thus, *S*_{m−1} ∪ {*b*} ∈ 𝓕 holds for some *b* ∈ *B* ∖ *S*_{m−1} according to (G2) satisfied by 𝓕. Thus, *S*_{m−1} ∪ {*b*} is a basis of 𝓕. Using (G3), there is *S*_{m−1} ∪ {*b*_{0}} ∈ 𝓕 where *ω*(*b*_{0}) = $\begin{array}{}\underset{x\in B\setminus {\overline{S}}_{m-1}}{max}\omega (x).\end{array}$

On the other hand, for any *x* ∈ *B*, if *B* ∖ {*x*} ∈ 𝓕, then *ω*(*B* ∖ {*x*}) ≤ *ω*(*S*^{m−1}) = *ω*(*S*_{m−1}) in view of the inductive supposition. Since (*B* ∖ {*x*}) ∪ {*x*} is a basis, there is *ω*((*B* ∖ {*x*}) ∪ {*x*}) ≤ *ω*(*S*_{m−1} ∪ {*b*_{0}}) ≤ *ω*(*S*_{m−1} ∪ {*e*_{m}}) = *ω*(*S*_{m}) in virtue of (G3) and the process of searching *S*_{m}.

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

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

□

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*:

*Then*, (*E*, 𝓕) *is an interval greedoid*.

#### Proof

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*) = $\begin{array}{}\underset{X\in \mathcal{F}{|}_{A}}{max}\end{array}$ *ω*(*X*). Let *S*_{A} be the solution of Algorithm 1 for (𝓕|_{A}, *ω*_{A}). Consider the process of Algorithm 1 and *A* ∈ 𝓕, we can decide *S*_{A} ⊆ *A*. Hence, we obtain *ω*_{A}(*S*_{A}) ≤ *ω*_{A}(*A*) since *ω*_{A} is positive. Furthermore, since *ω*_{A} is positive, we find *S*_{A} ⊂ *A* ⇔ *ω*_{A}(*S*_{A}) < *ω*_{A}(*A*). By (s1), we confirm *ω*_{A}(*S*_{A}) = $\begin{array}{}\underset{X\in \mathcal{F}{|}_{A}}{max}\end{array}$ *ω*_{A}(*X*). Summing up the above results, we may follow *S*_{A} = *A*.

From ∅ ∈ 𝓕 and the process of Algorithm 1 for (𝓕|_{A}, *ω*_{A}), we may assert that *S*_{A} is accessible. We will prove this assertion as follows.

Let *m* = |*S*_{A}|, that is, *m* = |*A*| since *S*_{A} = *A*.

If *m* = 0. Then *S*_{A} = ∅. So, *S*_{A} is accessible.

If *m* = 1. Then *A* = {*a*} and *S*_{0} = ∅.

When *j* = 0 = *m* − 1 = 1 − 1. There is *G*_{j} = *G*_{0} = {*a*}. Hence, *G*_{j} ≠ ∅ and *ω*(*e*_{0+1}) = *ω*(*a*) and *S*_{1} = ∅ ∪ {*a*} = {*a*}.

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

Suppose that if *m* ≤ *k* − 1, then *S*_{A} is accessible. Now, let *m* = *k* > 1.

If for every *j* < *k* − 1, there is *D*_{j+1} ∩ *G*_{j} ≠ ∅ in the process of Algorithm 1, then according to the definitions of *D*_{j+1} and *G*_{j}, there are *S*_{0} = ∅ ⊂ *S*_{1} ⊂ *S*_{2} ⊂ … ⊂ *S*_{j+1} and *S*_{i+1} = *S*_{i} ∪ {*e*_{i+1}} for some *e*_{i+1} ∈ *A* ∖ *S*_{i}, (*i* = 0, …, *j* + 1), we can state that *S*_{j} is accessible (*j* = 0, …, *k* − 1).

If for some *j*_{0} < *k* − 1, there is *D*_{j0+1} ∩ *G*_{j0} = ∅ in the process of Algorithm 1, then *S*_{j0+1} = *S*_{j0}. This follows |*S*_{j0+1}| < *j*_{0}+1. Furthermore, we obtain |*S*_{k}| < *k* according to the process of Algorithm 1. Thus, we attain *S*_{A} (that is *S*_{k}) satisfying |*S*_{A}| < *k* = |*A*| = |*S*_{A}|, a contradiction. Hence, for every *j* < *k* − 1, there is *D*_{j+1} ∩ *G*_{j} ≠ ∅.

So, *S*_{j} is accessible (*j* = 0, 1, …, *k* − 1).

Let *j* = *k* − 1. If *G*_{k−1} ≠ ∅, then we obtain *S*_{k} = *S*_{k−1} ∪ {*e*_{k}}. Thus, we may easily find *S*_{k} to be accessible since *S*_{k} ∖ *e*_{k−1} = *S*_{k−1} ∈ 𝓕 and the above discussion.

If G_{k−1} = ∅, then *S*_{k} = *S*_{k−1}. Thus, there is |*S*_{k}| = |*S*_{k−1}| = *k* − 1. On the other hand, *S*_{k} = *S*_{m} = *S*_{A} = *A* hints |*S*_{k}| = |*A*| = *m* = *k*. This is a contradiction to |*S*_{k}| = *k* − 1. Therefore, we confirm *G*_{k} ≠ ∅.

Adding up the above discussion with the induction, we can state that *S*_{A}, 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 *X* ⊆ *Y* ⊆ *Z*. Let *a* ∈ *E* ∖ *Z* 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* *E*_{3} = {*a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}} *and* 𝓕_{3} = {∅, {*a*_{1}}, {*a*_{4}}, {*a*_{1}, *a*_{2}}, {*a*_{3}, *a*_{4}}}. *There is* |{*a*_{3}, *a*_{4}}| = 2 = |{*a*_{1}}| + 1, *but no element* *a* ∈ {*a*_{3}, *a*_{4}} ∖ {*a*_{1}} *satisfies* {*a*_{1}} ∪ {*a*} ∈ 𝓕_{3}. *Define* *ω*_{3} : *E*_{3} → ℝ^{+} *as* *ω*_{3}(*a*_{1}) = 5, *ω*_{3}(*a*_{2}) = 1, *ω*_{3}(*a*_{3}) = *ω*_{3}(*a*_{4}) = 4. *The solution of* *Algorithm 1* *for* (𝓕_{3},*ω*_{3}) *is* {*a*_{1}}. *But*, *ω*_{3}({*a*_{1}}) = 5 < *ω*_{3}({*a*_{3}, *a*_{4}}) = 8 *implies* *Algorithm 1* *not to work for* (𝓕_{3},*ω*_{3}).

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* *x* ∈ *E*. 𝓕_{A} *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)*.

#### Proof

(⇒)

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.

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*, 𝓕_{13}), where 𝓕_{13} 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.

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 *x* ∈ *E*; 𝓕 satisfies the condition (G3).

Let *Y* ⊆ *Z*, *Y*, *Z* ∈ 𝓕 and *a* ∈ *E* ∖ *Z*. 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.

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