An Alternative Construction to the Transitive Closure of a Directed Graph

One must add arrows which are forced by transitivity to form the transitive closure of a directed graph. We introduce a construction of a transitive directed graph which is formed by adding vertices instead of arrows and which preserves the transitive relationships formed by distinct vertices in the original directed graph. Our construction does not apply to all directed graphs.

A figure showing a reflexive directed graph D will only display D * and will not show the loops. Thus we assume the directed graphs in Figure 1 are both reflexive.
In Example 1.2 we show directed graph (b) is a compression of directed graph (a). The effect on the directed graphs is to map the two middle vertices of (a) to the middle vertex of (b).

Balanced and Stable Directed Graphs
there is an arrow from w to y if and only if there is an arrow from x to z.
A reflexive directed graph is balanced if and only if does not contain an induced subgraph isomorphic to either (a) or (b) in Figure 2 or either of the directed graphs in Figure 3. are reflexive directed graphs.
1. If D 1 is balanced then D 2 is balanced.
2. If D 1 is stable then D 2 is stable.
3. If D 1 is preordered then D 2 is preordered.

Clasps and Soloists
The next definition helps us identify vertices where the transitive relation fails.
Definition 3.1 Suppose D is a reflexive directed graph and x ∈ V (D).
and there is no arrow from w to y.

We say
and there is no arrow from w to y.
3. An unlocked clasp is a clasp which is not locked.
Directed graph (d) in Figure 2 contains a locked clasp determined by the vertex in the lower right corner. We are now able to state our main Theorem.  Figure 3. This is reminiscent of Kuratowki's characterization of planar graphs (see [2]). Theorem 3.2 is not a complete classification since both directed graphs in Figure 3 are compressions of preordered directed graphs. But both directed graphs in Figure 3 contain directed cycles so we can give a complete classification if D * is acyclic.
Then D is the compression of a preordered directed graph if and only if D does not contain an induced subgraph isomorphic to one of the directed graphs in Figure 2.
rs, sr ∈ A (D) then r and s are said to be paired in V (D). We say an element There is a locked clasp in D 2 if and only if there is a locked clasp in D.
3. If D contains a locked clasp then D is not the compression of a preordered directed graph.
wx, xy ∈ A (D), and there is no arrow from w to y. Suppose there exists z ∈ V (D) \ {x} such that x is paired with z. Applying the balance property to w, x, z, x and to x, z, x, y yields wz, zy ∈ A (D). This gives w = y, y = z, and w = z since there is no arrow from w to y. Applying the stable property to w, x, z, y gives wy ∈ A (D), which is a contradiction.
and there is no arrow from w to y in D 2 .
Properties (1) and (3) and there is no arrow from θ (w) to θ (y) in D. Therefore θ (x) is a locked clasp and there is no arrow from w 1 to , and there is no arrow from u 2 to y 2 in D 2 . Therefore x 2 is a locked clasp in D 2 .
(3) Part 3 follows immediately from part 2.   The proof of Theorem 3.2 is a constructive algorithm described in section 4. In each iteration of the algorithm we construct a preordered directed graph with one more vertex and define a compression. The algorithm stops when we arrive at a preordered directed graph and the desired compression is obtained by composition.
We finish this section with an example which covers the steps and constructions given in the proof of Theorem 3.2. The directed graphs in Figure 4 are stable. We may identify (i) as a compression of (ii) by mapping 2 and t 1 to 2. We may also identify (ii) as a compression of (iii) by mapping 4 and t 2 to 4.  Step 1 Y 1 = {4, 6} and A 1 is empty.
Step 4 We go back to step 1 with x 2 = 4 since 4 is the only clasp in D 2 .
Step 2 Use construction B with a 2 = 3, b 2 = t 1 , and y 2 = 6. Let D 3 be the . This gives (iii) in Figure 4.

Proof of Theorem 3.2
If D is the compression of a preordered directed graph then every clasp in D is unlocked by part 3 of Lemma 3.6. We assume D is stable and every clasp in D is unlocked and prove D is the compression of a preorder. We set D 1 = D, and describe an algorithm to construct stable directed graphs D 1 , . . . , D m such that D i is a compression of D i+1 for each i < m. In the last iteration D m is preordered and the desired compression map is obtained by composition.
Assume D 1 is not preordered and fix a clasp x 1 ∈ V (D 1 ). In the first iteration of our algorithm we have i = 1.
Step 1 The sets Y i and A i are defined below.
Note that Y i is nonempty since x i is a clasp.
Step 2 Construction A.
Construction B.
Step 3 Before moving on we prove θ i is a compression. Routine calculations show part (1) of Definition 1.1 hold and there is a well-defined map θ * i : The only condition left is part 2 of Definition 1. We check every possible case and make repeated use of the fact uv ∈ σ i if and only In cases 2, 3, and 4 we have x i = d r for some r ∈ {1, 2, 3}. We will show either (d 1 , d 2 , d 3 ) ∈ Trans (D i ) so that θ i (d j ) = d j for j = 1, 2, 3 or the desired transitive triple is obtained by replacing d r with t i so that θ i (d j ) = d j for j = r, and θ i (t i ) = d r .
We split cases between construction B and construction A when necessary. Note that x i is a soloist by part 1 of Lemma 3.6.
Check case 2 for construction B. We have On the other hand if clasp, which is a contradiction. We are left with Check case 3 for construction B.
is a locked clasp. This is a contradiction. We are left with Check case 4 for construction A. Suppose d j ∈ B i for j = 2 or j = 3 of Lemma 3.7. This gives cd 2 ∈ T i if and only if cd Step 4 If D i+1 is preordered then the algorithm stops and the compression from D i+1 to D is determined by composition. Otherwise fix a clasp x i+1 ∈ V (D i ) and go back to step 1 with i replaced by i + 1.
To study the algorithm we consider a given iteration i. Then A (D i+1 ) is stable by Theorem 2.2 and D i+1 contains no unlocked clasps by part 2 of Lemma 3.6.
This means we may repeat the algorithm as often as necessary. We must prove the algorithm stops eventually.
In Assume there is not an arrow formed by x i and any other vertex of D i+1 after using construction A. There exist b, z ∈ V (D i ) such that bx i ∈ A (D i ), x i z ∈ A (D i ), and bz / ∈ A (D i ) since x i is a clasp. Then bt i ∈ A (D i+1 ) and t i z ∈ A (D i+1 ) by assumption so b ∈ A i and z ∈ B i . Since b ∈ A i there exists y ∈ V (D i ) such that (b, x i , y) ∈ Trans (D i ). There must also exist a ∈ V (D i ) such that ax i ∈ A (D i ) and ay / ∈ A (D i ) since y ∈ Y i . This gives at i ∈ A (D i+1 ) by assumption so a ∈ A i .
Thus a, b ∈ A i satisfy the conditions in step 2 for construction B. This contradicts our assumption that we used construction A, so there is an arrow formed by x i and another vertex of D i+1 .
In construction B we have b i y i ∈ A (D i ) so b i z / ∈ T i for all z ∈ V (D i ) \ {x i }. This gives b i x i ∈ σ i and b i x i ∈ A (D i+1 ) so there is an arrow formed by x i with another vertex of D i+1 .