2. Degree Sequences. 2.1 Degree Sequences

Degree Sequences

The concept of degrees in graphs has provided a framework for the study of various struc- tural properties of graphs and has therefore attracted the attention of many graph theorists.

Here we deliberate on the various criteria for a non-decreasing sequence of non-negative integers to be a degree sequence of some graph.

2.1 Degree Sequences

Letd_i,1≤ i ≤ n, be the degrees of the verticesv_iof a graph in any order. The sequence[di]ⁿ₁ is called the degree sequence of the graph. The non-negative sequence[di]ⁿ₁is called the degree sequence of the graph if it is the degree sequence of some graph, and the graph is said to realise the sequence.

The set of distinct non-negative integers occurring in a degree sequence of a graph is called its degree set. A set of non-negative integers is called a degree set if it is the degree set of some graph, and the graph is said to realise the degree set.

Two graphs with the same degree sequence are said to be degree equivalent. In the graph of Figure 2.1(a), the degree sequence isD= [1, 2, 3, 3, 3, 4]orD= [1 2 3³4]and its degree set is{1, 2, 3, 4}, while the degree sequence of the graph in Figure 2.1(b) is[1, 1, 2, 3, 3]

and its degree set is{1, 2, 3}.

Fig. 2.1

If the degree sequence is arranged as the non-decreasing positive sequenced₁ⁿ¹, d₂ⁿ², . . . dⁿ_k^k, (d₁< d2< . . . < dk), the sequencen₁, n2, . . ., nkis called the frequency sequence of the graph.

The two necessary conditions implied by Theorem 1.1 and Theorem 1.12 are not suffi- cient to ensure that a non-negative sequence is a degree sequence of a graph. To see this, consider the sequence[1, 2, 3, 4, . . ., 4, n − 1, n − 1]. The sum of the degrees is clearly even and∆ = n − 1. However, this is not a degree sequence, since there are two vertices with degreen−1, and this requires that each of the two vertices is joined to all the other vertices, and therefore^δ ≥ 2. But the minimum number in the sequence is 1.

A degree sequence is perfect if no two of its elements are equal, that is, if the frequency sequence is1, 1, . . ., 1. A degree sequence is quasi-perfect if exactly two of its elements are same.

Definition: Let D= [di]ⁿ₁ be a non-negative sequence and k be any integer 1≤ k ≤ n. Let D⁰= [d_i⁰]ⁿ₁ be the sequence obtained fromD by settingd_k= 0 andd⁰_i= di− 1for the d_k largest elements of D other than d_k. Let H_k be the graph obtained on the vertex set V = {v1, v2, . . ., vn}by joiningv_k to thed_kvertices corresponding to thed_kelements used to obtainD⁰. This operation of gettingD⁰andH_kis called laying offd_kandD⁰is called the residual sequence, andH_kthe subgraph obtained by laying offd_k.

Example LetD= [2, 2, 3, 3, 4, 4]. Taked3= 0. ThenD⁰= [2, 2, 0, 2, 3, 3]. The subgraph H_kin this case is shown in Figure 2.2.

Fig. 2.2

2.2 Criteria for Degree Sequences

Havel [112] and Hakimi [99] independently obtained recursive necessary and sufficient conditions for a degree sequence, in terms of laying off a largest integer in the sequence.

Wang and Kleitman [261] proved the necessary and sufficient conditions for arbitrary layoffs.

Theorem 2.1 A non-negative sequence is a degree sequence if and only if the residual sequence obtained by laying off any non-zero element of the sequence is a degree sequence.

Proof

Sufficiency Let the non-negative sequence be[di]ⁿ₁. Supposed_k is the non-zero element laid off and the residual sequence[d_i⁰]ⁿ₁is a degree sequence. Then there exists a graphG⁰

realising[d_i⁰]ⁿ₁ in whichv_k has degree zero and somed_k vertices, sayv_ij, 1≤ j ≤ dk have degreed_ij− 1. Now, by joiningv_k to these vertices we get a graphGwith degree sequence [di]ⁿ₁. (Observe that the subgraph obtained by such joining is precisely the subgraphH_k obtained by laying offd_k).

Necessity We are given that there is a graph realisingD= [di]ⁿ₁. Letd_kbe the element to be laid off. First, we claim there is a graph realisingDin whichv_k is adjacent to all the vertices in the setS ofd_k largest elements ofD− {d_k}. If not, let Gbe a graph realising Dsuch thatv_k is adjacent to the maximum possible number of vertices in S. Then there is a vertexv_i inS to whichv_k is not adjacent and hence a vertexv_j outsideS to whichv_k is adjacent (sinced(v_k) = |S|). By definition ofS, dj≤ di. Therefore there is a vertexv_hin V− {v_k}adjacent tov_i, but not adjacent tov_j. Note thatv_hmay be inS(Fig. 2.3).

Fig. 2.3

Construct a graphHfromGby deleting the edgesv_jv_kandv_hv_iand adding the edgesv_jv_h andv_iv_k. This operation does not change the degree sequence. ThusH is a graph realising the given sequence, in which one more vertex, namelyv_i ofSis adjacent tov_k, than inG. This contradicts the choice ofGand establishes the claim.

To complete the proof, ifGis a graph realising the given sequence and in whichv_k is adjacent to all vertices of S, let G⁰= G − vk. Then G⁰ has the residual degree sequence

obtained by laying offd_k. q

Definition: Let the subgraphHon the verticesvi, vj, vr, vsof a multigraphGcontain the edgesv_iv_j andv_rv_s. The operation of deleting these edges and introducing a pair of new edgesv_iv_sandv_jv_r, orv_iv_randv_jv_sis called an elementary degree preserving transformation (EDT), or simple exchange, or 2-switching, or elementary degree-invariant transformation.

Remarks

1. The result of an EDT is clearly a degree equivalent multigraph.

2. If an EDT is applied to a graph, the result will be a graph only if the latter pair of edges (vivs andvjvr), or (vivr andvjvs) does not exist inG.

Theorem 2.2 (Havel, Hakimi) The non-negative integer sequenceD= [di]ⁿ₁is graphic if and only ifD⁰is graphic, whereD⁰is the sequence (havingn− 1elements) obtained from Dby deleting its largest element^∆and subtracting 1 from its^∆next largest elements.

Proof

Sufficiency Let^D= [di]ⁿ₁ be the non-negative sequence withd₁≥ d2≥ . . . ≥ dn. LetG⁰ be the graph realising the sequenceD⁰. We add a new vertex adjacent to vertices inG⁰having degreesd₂− 1, . . ., d_∆+1− 1. Thosed_iare the^∆largest elements ofDafter^∆itself. (But the numbersd₂− 1, . . ., d_∆+1− 1need not be the^∆largest elements inD⁰).

Necessity LetGbe a graph realisingD= [di]ⁿ₁, d1≥ d2≥ . . . ≥ dn. We produce a graphG⁰ realisingD⁰, whereD⁰is the sequence obtained fromDby deleting the largest entryd₁and subtracting 1 fromd₁next largest entries.

Letwbe a vertex of degreed₁inGandN(w)be the set of vertices which are adjacent to w. LetSbe the set ofd₁number of vertices inGhaving the desired degreesd₂, . . ., dd1+1.

IfN(w) = S, we can deletewto obtainG⁰. Otherwise, some vertex ofS is missing from N(w). In this case, we modifyGto increase|N(w) ∩ S|without changing the degree of any vertex. Since|N(w) ∩ S|can increase at mostd1times, repeating this procedure converts an arbitrary graphGthat realisesD, into a graphG^∗that realisesD, and hasN(w) = S. From G^∗, we then deletewto obtain the desired graphG⁰realisingD⁰.

If N(w) 6= S, let x∈ S and z∈ S/ , so that wz is an edge and wx is not an edge, since d(w) = d1= |S|. By this choice ofS, d(x) ≥ d(z)(Fig. 2.4).

Fig. 2.4

We would like to add wxand delete wzwithout changing their respective degrees. It suffices to find a vertexyoutsideT = {x, z, w}such thatyxis an edge, whileyzis not. If such ayexists, then we also delete xyand add zy. Letq be the number of copies of the edgexz(0 or 1). Nowxhasd(x) − qneighbours outsideT, andzhasd(z) − 1 − qneighbours outsideT. Sinced(x) ≥ d(z), the desiredyoutsideT exists and we can perform the EDT (elementary degree preserving transformation or 2-switch). q

Algorithm: The above recursive conditions give an algorithm to check whether a non- negative sequence is a degree sequence and if so to construct a graph realising it.

The algorithm starts with an empty graph on vertex setV= {v1, v2, . . ., vn}and at thekth iteration generates a subgraphH_kofGby deleting (laying off) a vertex of maximum degree in the residual sequence at that stage. If the given sequence is a degree sequence, we end up with a null degree sequence (i.e., for eachi,d_i= 0) and the graph realising the original sequence is simply the sum of the subgraphsH_j. If not, at some stage, one of the elements of the residual sequence becomes negative, and the algorithm reports non-realisability of the sequence.

An obvious modification of the algorithm, obtained by choosing an arbitrary vertex of positive degree, gives the Wang-Kleitman algorithm for generating a graph with a given degree sequence.

Remarks

1. There can be many non-isomorphic graphs with the same degree sequence. The smallest example is the pair shown in Figure 2.5 on five vertices with the degree sequence[2, 2, 2, 1, 1].

Fig. 2.5

The problem of generating all non-isomorphic graphs of given order and size in- volves the problem of graph isomorphism for which a good algorithm is not yet known. So also is the problem of generating all non-isomorphic graphs with given degree sequence. In fact, even the problem of finding the number of non-isomorphic graphs with given order and size, or with given degree sequence (and several other problems of similar nature) has not been satisfactorily solved.

2. The Wang-Kleitman algorithm is certainly more general than the Havel-Hakimi algo- rithm, as it can generate more number of non-isomorphic graphs with a given degree sequence, because of the arbitrariness of the laid-off vertex. For example, not all the five non-isomorphic graphs with the degree sequence[3, 3, 2, 2, 1, 1]can be gener- ated by the Havel-Hakimi algorithm unlike the Wang-Kleitman algorithm.

3. Even the Wang-Kleithman algorithm cannot always generate all graphs with a given degree sequence. For example, the graphGwith degree sequence[3, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1]shown in Figure 2.6, cannot be generated by this algorithm. For

a. if we lay off a 3, it has to be laid off against the other 3’s and will generate a graph in which a vertex with degree 3 is adjacent to three other vertices with degree 3,

b. if we lay off a 2 it will generate a graph with a vertex of degree 2 adjacent to two vertices of degree 3,

c. if we lay off a one it will generate a graph in which a vertex of degree one is adjacent to a vertex of degree 3. None of these cases is realised in the given graphG.

Fig. 2.6

However, there are other methods of generating all graphs realising a degree se- quenceDfrom any one graph realisingDbased on a theorem by Hakimi [98]. But those will also be inefficient unless some efficient isomorphism testing is developed.

4. The graphs in Figure 2.5 show that the same degree sequence may be realised by a connected as well as a disconnected graph. Such degree sequences are called poten- tially connected, where as a degree sequenceDsuch that every graph realisingDis connected is called a forcibly connected degree sequence.

Definition: IfPis a graph property, andD= [di]ⁿ₁is a degree sequence, thenDis said to be potentially-P, if at least one graph realisingDis aP-graph, and it is said to be forcibly-P if every graph realising it is aP-graph.

Theorem 2.3 (Hakimi) If G₁ and G₂ are degree equivalent graphs, then one can be obtained from the other by a finite sequence of EDTs.

Proof SuperimposeG₁andG₂such that each vertex ofG₂coincides with a vertex ofG₁ with the same degree. Imagine the edges ofG₁are coloured blue and the edges ofG₂are coloured red. Then in the superimposed multigraphH, the number of blue edges incident equals the number of red edges incident at every vertex. We refer to this as blue-red parity.

If there is a blue edgev_iv_jand a red edgev_iv_jinH, we call it a blue-red parallel pair.

LetKbe the graph obtained fromHby deleting all such parallel pairs. ThenKis the null graph if and only ifG1andG2are label-isomorphic inHand hence originally isomorphic.

If this is not the case, we show that we can create more parallel pairs by a sequence of EDTs and delete them till the final resultant graph is null. This will prove the theorem.

LetB andRdenote the sets of blue and red edges in K. Ifvivj∈ B, we show that we can produce a parallel pair atvivj, so that the pair can be deleted. This would establish the claim made above.

Now, by construction, there is a blue-red degree parity at every vertex ofK. So there are red edgesv_iv_k, vjv_r inK. Ifv_k6= vr (Fig. 2.7(a)) an EDT inG₂switching the red edges to v_iv_j, vkv_rproduces a blue-red parallel atv_iv_j.

Fig. 2.7

Ifv_k= vr, again by degree parity, atv_kthere are at least two blue edges. Letv_kvsbe one such blue edge. Thenvs is distinct from bothvi andvj, for otherwise, there is a blue-red parallel pairv_iv_korv_jv_r. Then there is another red edgev_sv_t, vtdistinct fromv_iorv_j.

Letv_t6= vi. The two subcasesv_t= vjandv_t6= vjare shown in Figure 2.7(b) and (c). In the case of (b), one EDT ofG₂switchingv_iv_kandv_sv_t to positionsv_iv_jandv_sv_kproduces a blue- red pair atv_iv_jandv_kv_s. In the case of (c), one EDT ofG₂switchingv_iv_kandv_tv_sto positions v_sv_kandv_tv_i produces a blue-red parallel pair atv_kv_s(which can be deleted). Another EDT ofG₂switching the blue-red pairv_tv_iandv_jv_kto positionsv_iv_jandv_sv_kproduces a blue-red pairv_iv_j.

Since in both cases we get a blue-red pair atv_iv_j position, our claim is established and

the proof of the theorem is complete. q

Remarks In the related context of a (0, 1) matrixA(that is, a matrixAwhose elements are 0’s or 1’s), Ryser [227] defined an interchange as a transformation of the elements of Athat changes a minor of typeA1= 1

0 0 1



into a minor of the typeA1= 0 1

1 0



, or vice versa and proved an interchange theorem which can be interpreted as EDT theorem for bipartite graphs and digraphs.

The next result is a combinatorial characterisation of degree sequences, due to Erdos and Gallai [73]. Several proofs of the criterion exist; the first proof given here is due to Choudam [58] and the second one is due to Tripathi et al [246].

Theorem 2.4 (Erdos-Gallai) A non-increasing sequence[di]ⁿ₁of non-negative integers is a degree sequence if and only ifD= [di]ⁿ₁is even and the inequality

k

∑

i=1

d_i≤ k(k − 1) +

n

∑

i=k+1

min(di, k) (2.4.1)

is satisfied for each integerk, 1 ≤ k ≤ n. Proof

Necessity Evidently ∑ⁿ

i=1

d_iis even. LetU denote the subset of vertices with thekhighest degrees inD. Then the sums= ∑^k

i=1

d_i can be split ass₁+ s2, where s₁ is the contribution tosfrom edges joining vertices inU, each edge contributing 2 to the sum, and s₂is the contribution tosfrom the edges between vertices inUandU(whereU= V −U), each edge contributing 1 to the sum (Fig. 2.8).

s₁ is clearly bounded above by the degree sum of a complete graph onk-vertices, i.e., k(k − 1). Also, each vertexv_iofUcan be joined to at most min(di, k)vertices ofU, so that s₂is bounded above by ∑ⁿ

i=k+1

min(di, k). Together, we get (2.4.1).

Fig. 2.8

Sufficiency We induct on the sums= ∑ⁿ

i=1

diand use the obvious inequality

min(a, b) − 1 ≤ min(a − 1, b), (2.4.2)

for positive integersaandb.

For s= 2, clearly K2∪ (n − 2)K1 realises the only sequence[1, 1, 0, 0, . . . 0]or[1²0ⁿ⁻²] satisfying the conditions (2.4.1).

As induction hypothesis, let all non-increasing sequences of non-negative integers with even sum at mosts− 2and satisfying (2.4.1) be degree sequences.

LetD= [di]ⁿ₁be a sequence with sumsand satisfying (2.4.1). We produce a new non- increasing sequenceD⁰of non-negative integers by subtracting one each from two positive terms of Dand verify that D⁰ satisfies the hypothesis of the theorem. Since the trailing

zeros in the non-increasing sequences of non-negative integers do not essentially affect the argument, there is no loss of generality in assuming that d_n> 0, and we assume this to simplify the expression.

To defineD⁰, lett be the smallest integer (≥ 1)such that d_t> dt+1. That is, letD be d₁= d2= . . . = dt> dt+1≥ dt+2≥ . . . ≥ dn> 0.

IfDis regular (that is,d_i= d > 0, for alli) then lett ben− 1.

Thend_i⁰=







d_i, f or 1≤ i ≤ t − 1 and t + 1 ≤ i ≤ n − 1 , d_t− 1, f or i= t ,

d_n− 1, f or i= n .

Clearly, D⁰ is a non-increasing sequence of non-negative integers and ∑ⁿ

i=1

d_i⁰= s − 2 is even.

We verify thatD⁰satisfies (2.4.1) by considering several cases depending on the relative position ofkand the magnitudes ofd_kandd_n.

Case I Letk= n. Therefore, ∑^k

i=1

d⁰_i= ∑^k

i=1

d_i− 2 ≤ n(n − 1) − 2 < n(n − 1) =RHS of (2.4.1) forD⁰.

Case II Lett≤ k ≤ n − 1. Then ∑^k

i=1

d_i⁰= ∑^k

i=1

d_i− 1 ≤ k(k − 1) + ∑ⁿ

i=k+1

min(di, k) − 1(sinceDsatisfies (2.4.1))

= k(k − 1) + ⁿ⁻¹∑

i=k+1

min(d_i⁰, k) + min(dn, k) − 1

≤ k(k − 1) + ⁿ⁻¹∑

i=k+1min(d_i⁰, k) + min(dn− 1, k) by (2.4.2)

= k(k − 1) + ⁿ⁻¹∑

i=k+1

min(d_i⁰, k) + min(d_n⁰, k)

Therefore, ∑^k

i=1

d_i⁰≤ k(k − 1) + ∑ⁿ

i=k+1

min(d_i⁰, k). Case III Letk≤ t − 1.

Subcase III.1 Assumed_k≤ k − 1. Then ∑^k

i=1

d_i⁰= kdk≤ k(k − 1) ≤ k(k − 1) + ∑ⁿ

i=k+1min(d_i⁰, k), since the second term is non-negative.

Subcase III.2 Everyd_j= k, 1 ≤ j ≤ k. We first observe thatd_k+2+ . . . + dn≥ 2.

This is obvious ifk+ 2 ≤ n − 1, becaused_n> 0givesd_n≥ 1andd_n−1≥ 1. Whenk+ 2 = n, we havek= n − 2. Ask≤ t − 1, t ≥ k + 1 = n − 2 + 1 = n − 1. Sincet> n − 1is not possible, t= n − 1.

The sequenceDis[n −2, n −2, . . ., n −2, dn], or[(n −2)ⁿ⁻¹d_n]. Thens= (n −1) (n −2)+dn. Sincesis even,d_nis even and henced_n≥ 2. Thus,d_k+2+ . . .+ dn≥ 2.

Therefore,d_k+2+ . . . + dn− 2 ≥ 0. Now,

k

∑

i=1

d_i⁰ = ∑^k

i=1

d_i= k.k = k²= k²− k + k

= k²− k + dk+1, (because k ≤ t − 1, and d1= . . . = dt−1= dt, so ifd_t−1= k, thend_t= k, and ifd_k= k, d_k+1= k).

Thus, ∑^k

i=1

d_i⁰≤ k²− k + dk+1+ (dk+2+ . . . + dn− 2) = k(k − 1) + ∑ⁿ

i=k+1min(di, k) − 2,

(because min(dk+1, k) = dk+1, min(dk+2, k) = k = dk+2, . . .,min(dt, k) = k = dt, . . .,min (dt+1, k) = dt+1(asd_t+1< dt= k),. . .,min(dn, k) = dn(asdn< dt= k)).

Hence, ∑^k

i=1

d_i⁰≤ k(k − 1) + ∑ⁿ

i=k+1

min(di, k) + min(dt, k) + min(dn, k) − 2

i6=t, n

= k(k − 1) + ∑ⁿ

i=k+1min(d⁰_i, k) + min(d_t⁰+ 1, k) + min(d_n⁰+ 1, k) − 2

i6=t, n

≤ k(k − 1) + ∑ⁿ

i=k+1

min(d⁰_i, k) + min(d_t⁰, k) + 1 + min(d_n⁰, k) + 1 − 2

i6=t, n

= k(k − 1) + ∑ⁿ

i=k+1

min(d⁰_i, k).

Subcase III.3 Letd_k≥ k + 1. i. Letd_n≥ k + 1.

Then ∑^k

i=1

d⁰_i= ∑^k

i=1

d_i≤ k(k − 1) + ∑ⁿ

i=k+1

min(di, k) (sinceDsatisfies (2.4.1))

= k(k − 1) + ∑ⁿ

k+1

min(di, k) + min(dt, k) + min(dn, k)

i6=t, n

= k(k − 1) + ∑ⁿ

k+1

min(d_i⁰, k) + min(dt− 1, k) + min(dn− 1, k),

i6=t, n

(because min(dt, k) = min(dt− 1, k) = k, min(dn, k) = min(dn− 1, k) = k, as d_t ≥ k + 1, dn≥ k + 1implies thatd_t− 1 ≥ k, dn− 1 ≥ k).

So, ∑^k

i=1

d_i⁰≤ k(k − 1) + ∑ⁿ

k+1

min(d⁰_i, k) + min(d_t⁰, k) + min(d_n⁰, k)

i6=t, n

= k(k − 1) + ∑ⁿ

i=k+1min(d_i⁰, k).

ii. Let d_n≤ k and letr be the smallest integer such thatd_t+r+1≤ k. We verify that in (2.4.1),Dcan not attain equality for such a choice ofk. For, with equality, we have

k

∑

i=1

d_i= kdk= k(k − 1) +^t+r∑

k+1min(di, k) + ∑ⁿ

t+r+1min(di, k)

= k(k − 1) + (t + r − k)k + ∑ⁿ

t+r+1

d_i,

becausemin(di, k)=







k, f or i= k + 1, ..., t + r as di≥ k + 1, di, f or i= t + r + 1, ..., n as di≤ k .

So,kd_k= k(t + r − 1) + ∑^k

t+r+1

d_i.

Then

k+1

∑

i=1

d_i= (k + 1)dk= (k + 1) (

(t + r − 1) +1 k

n

∑

t+r+1

d_i )

, (usingd_kfrom above)

= (k + 1)(t + r − 1) +k+ 1 k

n

∑

t+r+1

d_i> (k + 1)(t + r − 1) +

n

∑

t+r+1

d_i

= (k + 1)k − (k + 1)k + (k + 1)(t + r − 1) +

n

∑

t+r+1

d_i

= (k + 1)k + (k + 1)(t + r − k − 1) +

n

∑

t+r+1

d_i= (k + 1)k +

t+r

∑

k+1

(k + 1) +

n

∑

t+r+1

d_i

= (k + 1)k +

n

∑

t+r+1

min(di, k + 1),

because min(di, k + 1) = k + 1fori= k + 1, . . ., t + r,and min(di, k + 1) = di, fori= t + r + 1, . . ., n. So,

k+1

∑

i=1

d_i> k(k + 1) +

n

∑

k+1

min(di, k + 1).

Therefore,

k+1

∑

i=1

d_i> k(k + 1) + (k + 1) +

n

∑

k+2

min(di, k + 1),

which is a contradiction to (2.4.1), forDfork+ 1. HenceDhas strict inequality fork. Therefore,

k

∑

i=1

d_i⁰=

k

∑

i=1

di< k(k − 1) +

n

∑

k+1

min(di, k).

Thus,

k

∑

i=1

d⁰_i=

k

∑

i=1

d_i≤ k(k − 1) +

n

∑

k+1

min(di, k) − 1

= k(k − 1) +

n−1

∑

i=k+1

min(di, k) + min (dt, k) + min (dn, k) − 1

i6=t

≤ k(k − 1) +

n−1

∑

i=k+1

min(d_i⁰, k) + min (dt− 1, k) + min (dn− 1, k),

i6=t

asmin(dn, k) − 1 ≤ min (dn− 1, k), min(dt, k) = k(sinced_t ≥ k + 1),min(dt− 1, k) = k (sinced_t− 1 ≥ k).

Therefore,

k

∑

i=1

d_i⁰≤ k(k − 1) +

n

∑

k+1

min(d_i⁰, k). Hence in all casesD⁰satisfies (2.4.1).

Therefore by induction hypothesis, there is a graphG⁰realisingD⁰. Ifv_tv_n∈ E(G/ ⁰), then G⁰+ vtv_ngives a realisationGofD. Ifv_tv_n∈ E(G⁰), sinced(vt|G⁰) = dt− 1 ≤ n − 2, there is a vertexvr such thatvrvt ∈ E(G/ ⁰). Also, sinced(vr|G⁰) > d(vn|G⁰), there is a vertexvs such thatvsvn∈ E(G/ ⁰). Making an EDT exchanging the edge pairvtvn,vrvsfor the edge pairvtvr, vsvn, we get a realisationG⁰⁰ofD⁰withvtvn∈ E(G/ ⁰⁰). ThenG⁰⁰+ vtvnrealisesD.

Second Proof of Sufficiency (Tripathi et al.) Let a subrealisation of a non-increasing sequence[d1, d1, . . ., dn]be a graph with verticesv₁, v1, . . ., vnsuch thatd(vi) = difor1≤ i ≤ n, whered(vi)denotes the degree ofv_i. Given a sequence[d1, d1, . . ., dn]with an even sum that satisfies (2.4.1), we construct a realisation through successive subrealisations. The initial subrealisation hasnvertices and no edges.

In a subrealisation, the critical index r is the largest index such thatd(vi) = di for 1≤ i< r. Initially,r= 1unless the sequence is all 0, in which case the process is complete.

Whiler≤ n, we obtain a new subrealisation with smaller deficiencyd_r− d(vr)at vertexv_r while not changing the degree of any vertex v_i withi< r (the degree sequence increases lexicograpically). The process can only stop when the subrealisation ofd.

LetS= {vr+1, . . ., vn}. We maintain the condition thatS is an independent set, which cer- tainly holds initially. Writeu_i↔ vjwhenv_iv_j∈ E(G); otherwise,v_i6↔ vj

Case 0 vr6↔ vifor some vertexv_isuch thatd(vi) < di. Add the edgeu_rv_i.

Case 1 vr6↔ vi for someiwithi< r. Sinced(vi) = di≥ dr> d(vr), there existsu∈ N(ui) − (N(vr) ∪ {vr}), where N(z) = {y : z ↔ y}. Ifdr− d(vr) ≥ 2, then replaceuvi with{uvr, vivr}. Ifd_r− d(vr) = 1, then since ∑d_i− ∑ d(vi)is even there is an index kwith k> rsuch that d(vk) < dk. Case 0 applies unlessv_r↔ vk; replace{vrv_k, uvi}with{uvr, viu_r}.

Case 2 v1, . . ., vr−1∈ N(vr)andd(vk) 6= min{r, dk}for somekwithk> r. In a subrealisation, d(vk) ≤ dk. Since Sis independent,d(vk) ≤ r. Henced(vk) < min{r, dk}, and case 0 applies unlessu_k↔ vr. Sinced(vk) < r,there existsiwithi< rsuch thatu_k6↔ vi. Sinced(vi) > d(vr), there existsu∈ N(vi) − (N(vr) ∪ {ur}). Replaceuv_iwith{uvr, viv_k}.

Case 3 v1, . . ., vr−16∈ N(vr)and v_i↔ vi for some i and j with i< j < r. Case 1 applies unlessv_i, vj∈ N(vr). Sinced(vi) ≥ d(vi) > d(vr), there existsu∈ N(vi) − (N(vr) ∪ {vr}) and w∈ N(vj)−(N(vr)∪ {vr})(possiblyu= w). Sinceu, w 6∈ N(vr), Case 1 applies unlessu, w ∈ S. Replace{viv_j, uvr}with{uvr, v, vr}.

If none of these case apply, thenv₁, . . ., vr are pairwise adjacent, andd(v_k) = min{r, d_k} fork> r. SinceSis independent,∑^r_i=1d(vi) = r(r − 1) + ∑ⁿ_k=r+1min{r, d_k}. By (2.4.1),∑^r_i=1d₁ is bounded by the right side. Hence we have already eliminated the deficiency at vertexr.

Increaserby 1 and continue. q

Tripathi and Vijay [245] have shown that the Erdos-Gallai condition characterising graphi- cal degree sequences of lengthnneeds to be checked only for as manykas there are distinct terms in the sequence and not for allk, 1 ≤ k ≤ n.

2.3 Degree Set of a Graph

The set of distinct non-negative integers occurring in a degree sequence of a graph is called its degree set. For example, let the degree sequence beD= [2, 2, 3, 3, 4, 4], then degree set is{2, 3, 4}. A set of distinct non-negative integers is called a degree set if it is the degree set of some graph and the graph is said to realise the degree set.

Let S= {d1, d2, . . ., dk} be the set of distinct non-negative integers. Clearly, S is the degree set as the graph

G= Kd1+1∪ Kd2+1∪ . . .∪ Kd_k+1,