Reconstructing the degree sequence of a sparse graph from a partial deck

The deck of a graph $G$ is the multiset of cards $\{G-v:v\in V(G)\}$. Myrvold (1992) showed that the degree sequence of a graph on $n\geq7$ vertices can be reconstructed from any deck missing one card. We prove that the degree sequence of a graph with average degree $d$ can reconstructed from any deck missing $O(n/d^3)$ cards. In particular, in the case of graphs that can be embedded on a fixed surface (e.g. planar graphs), the degree sequence can be reconstructed even when a linear number of the cards are missing.

The Reconstruction Conjecture remains open, even for basic classes of graphs such as planar graphs and graphs of bounded maximum degree. There have been numerous surveys of partial results and related problems, for instance [1,3,4,17]. For a more detailed introduction, we refer to [13].
One closely related problem is that of determining which graph parameters are reconstructible in the sense that they are determined by the deck. For instance, given a full deck of cards, one can reconstruct the number of edges m by summing over the edges present in all of the cards and dividing by n − 2, where n is the number of vertices. The degree sequence can then be easily deduced as well. Other reconstructible parameters include connectedness, planarity, the number of Hamiltonian cycles and the chromatic polynomial (all of these are discussed in [13] together with further examples).
Some of these parameters can be reconstructed even when a subset of the deck is missing: • Bowler, Brown, Fenner and Myrvold [6] showed that any n 2 + 2 cards suffice to determine whether the graph is connected.
• Groenland, Guggiari and Scott [9] proved that the number of edges can be reconstructed from any n − 1 20 √ n cards. This improves on the work of Myrvold [15] and the work of Brown and Fenner [7], who proved the bounds n − 1 and n − 2 respectively.
The main contribution of the present paper is a proof that the degree sequence of graphs with bounded average degree can be reconstructed even with a linear number of missing cards. Theorem 2. Let G have n ≥ 3 vertices and average degree bounded by some d ∈ N. Then the degree sequence of G can be reconstructed from its deck when at most n 10 4 d 3 of the cards are missing. Since graphs that can be embedded on a fixed surface have bounded average degree, the following corollary is immediate.
Corollary 3. For any surface S, there is an ε > 0 such that for any n-vertex graph G embeddable on S, the degree sequence of G can be reconstructed from any collection of at least (1 − ε)n cards.
In particular, our results hold for planar graphs. A graph class C is recognisable if no graph G ∈ C has the same deck of cards as any graph H ∈ C. Planar graphs are of particular interest as they were shown to be recognisable by Bilinksi, Kwan and Yu in [2], but are still not known to be reconstructible in general. Partial results include the reconstructibility of outerplanar graphs [8], maximal planar graphs [12], and certain 5-connected planar graphs [2].
To reconstruct the degree sequence, we will first reconstruct the number of edges from the partial deck.
Theorem 4. Let G be a graph on n ≥ 3 vertices with average degree at most d for some d ∈ N. Then the number of edges in G can be reconstructed from any deck missing at most n 4d+6 − d − 5 cards. The proof of Theorem 4 is contained in Section 2, in which we also give a more general result for counting cliques in G. Section 3 is devoted to proving Theorem 2. We discuss the tightness of our results and conclude with some open problems in Section 4.

Notation
Throughout, we shall let G be the original graph on n vertices to be reconstructed. We write G i = G − v i and order the vertices as v 1 , . . . , v n such that G 1 , . . . , G n−k are the cards that have been given from the deck D(G).
For any graph H, let d t (H) be the number of vertices of degree t in H. That is, be the number of vertices of degree less than t in H.
We use the notation [n] = {1, . . . , n}, and write [a, b] to denote the set of integers between a and b inclusive.

Edge and clique counts
In order to reconstruct the degree sequence, we first reconstruct the number of edges. We note the following simple fact.
Observation 5. A graph with average degree bounded by d ∈ N has at least n 2 vertices of degree at most 2d and at least n d+1 vertices of degree at most d. Given enough cards, this will allow us to assume that the card with the most edges corresponds to a low-degree vertex, and therefore has small "error". Another important feature is that the property of having small average degree is recognisable from the partial deck in the following sense. Lemma 6. Let G be a graph on n ≥ 8 vertices with average degree d. From any deck of G missing k ≤ n 4 cards, we can reconstruct a quantity d that satisfies Proof. By [9, Lemma 2.1], we can calculate from the cards G 1 , . . . , G n−k an estimate m for the number of edges m that satisfies for n ≥ 8. Hence d = 2 m n satisfies the claimed inequality. Lemma 6 allows us to assume that the average degree d is known up to a potential error of 1. This approximation will be used to prove Theorem 4, but once we have that result and enough cards for it to apply, the average degree can then be computed exactly. Thus, when we come to reconstructing the degree sequence we will be able to assume that d is known.
Proof of Theorem 4. Let k be the number of missing cards and let the partial deck of cards consist of G 1 , . . . , G n−k with G i = G − v i . After possibly reordering the cards, we may assume that |E( . We may assume that n ≥ 2(d + 6)(2d + 3) ≥ 36, else there would be no cards missing. So by Lemma 6, we can recognise from the partial deck that G has average degree at most d + 1. Then Observation 5 and the conditions on k together imply that for each t ∈ [0, n]. The upper bound comes from the observation that v 1 has at most d + 1 neighbours. Only these vertices can drop degree from t to t − 1 when v 1 is deleted, and hence be counted in d <t (G 1 ) but not in d <t (G). For the lower bound, since the degree of a vertex in G 1 is at most the degree of the corresponding vertex in G, the only possible loss comes from the possibility that the deleted vertex v 1 may have had degree at most t. Applying Observation 5 first and then the fact that The additional plus one accounts for the fact that v 1 could have degree at most 2(d + 1). It follows that there must be some t ∈ {0, . . . , 2(d + 1)} such that where the last inequality holds by our assumptions on k. Let us choose t to be the smallest integer satisfying (2), noting that this is determined by G 1 and does not depend on any other information about G. Our next goal is to find a card corresponding to a vertex with degree exactly t.
by the bounds in (1) and (2). Since we are missing at most k cards from our deck, the (j + 2)nd card is certainly within the first j + k + 2 cards in the whole deck. Hence, we also have the reverse inequality d G (v j+2 ) ≤ t. This proves our claim that d G (v j+2 ) = t. Since we can compute j and t, and we have the card G j+2 , the number of edges may now be reconstructed by the formula |E(G)| = |E(G j+2 )| + t.
The preceding proof extends easily to reconstructing clique counts from an incomplete deck by replacing d t (H) and d <t (H) with analogous notions in terms of "clique degree". Namely, for each fixed r ∈ N, let c(v) be the number of r-cliques which contain the vertex v. Then let the number of vertices v for which c(v) = t (similarly, c(v) < t) be denoted by c t (H) (c <t (H)). Any vertex of degree at most 2(d + 1) is in it at most 2(d+1) r−1 cliques, and therefore By the pigeonhole principle, there is some t ∈ {0, . . . , 2(d + 1)} such that Again, only the neighbours of v 1 may appear in less cliques in G 1 than in G, and so c <t Since the number of r-cliques in G j+2 is the number of r-cliques in G minus c(v j+2 ), it suffices to choose k to guarantee c(v j+2 ) = t for j = c <t (G 1 ). We obtain the following result.

Degree sequence reconstruction
Once we know the number of edges m in a graph G, deducing its degree sequence from the complete deck is a simple matter of subtracting from m the number of edges seen in each card. Losing one card G i does not pose a problem as the missing degree is given by . However, as soon as we are missing just two cards it is no longer known whether the degree sequence can still be reconstructed. The main result in this section shows that for every graph G on n ≥ 3 vertices with average degree at most d for some d ∈ N, the degree sequence of G can be reconstructed from its deck when at most n 10 4 d 3 of the cards are missing. Since the degree sequence can be reconstructed if no cards are missing, we may assume that n ≥ 10 4 d 3 , which implies that the number of missing cards is at most n 4d+6 − d − 5. Applying Theorem 4 then allows us to reconstruct the number of edges in G. This means that we can determine d G (v i ) for all vertices corresponding to cards in our partial deck, as well as the average degree of G.
The total number of occurrences of degree t vertices across all of the cards is n i=1 d t (G i ). At the same time, each vertex v of degree t in G still has degree t in n − (t + 1) cards, namely all those In order to guess d t from d t+1 or vice versa, we first obtain a good estimate on the left-hand side of the equation above.
Lemma 8. Suppose we know the number of edges of G and that the average degree of G is at most d ∈ N. Moreover, assume that k < n 10 3 d 2 is the number of missing cards and n > 10 4 d 3 is the number of vertices. Then for every t ∈ [0, n], we can reconstruct an estimate s t from the given cards such that We again label the given cards G 1 , . . . , G n−k so that |E(G 1 )| ≥ · · · ≥ |E(G n−k )|. Let the missing cards be G n−k+1 , . . . , G n ordered arbitrarily, and let into three sets I 1 , I 2 , and I 3 defined as follows: We assume that the vertex numbering of G 1 is inherited from G for the sake of the argument, but we do not exploit that these labels are present on our given card. In particular, we do not access the set I 1 , only the multiset , so we can write [n] = I 1 ⊔ I 2 ⊔ I 3 as a disjoint union. Moreover, note that 1 ∈ I 2 .
For each j = 1, 2, 3, we estimate i∈I j d t (G i ). Recall that we know the number of edges of G and hence the degrees of v 1 , . . . , v n−k . This is enough to reconstruct the set I 2 , and to read off d t (G i ) for each v i ∈ I 2 by examining of the relevant card. Therefore, we can determine i∈I 2 d t (G i ) exactly.
We estimate i∈I 1 d t (G i ) by i∈I 1 d t (G 1 −v i ), and we now bound the error in this estimation. The vertex v 1 has degree at most d in G by Observation 5, and so, for each i ∈ [2, n], the vertex v 1 has degree at most d in the graph G i . It follows that for all i ∈ [2, n], Since at most n 100d vertices can have degree greater than 100d 2 in G and hence also in G 1 , we find that Finally, we can express I 3 as the union In this form, we see that all vertices v i with i ∈ I 3 have degree at most 100d 2 +1 in G. The first set in the union has cardinality at most k and the second has cardinality at most d G (v 1 ) ≤ d (since all such vertices must be adjacent to v 1 ).
so we can estimate i∈I 3 d t (G i ) by |I 3 |d t (G 1 ). Note that we can reconstruct which is reconstructible from our partial deck.
Using (4) and (5), the margin of error | n i=1 d t (G i ) − s t | is then given by and this is less than n 8 for k < n 10 3 d 2 and n > 10 4 d 3 . We now deduce the proof of the main result.
Proof of Theorem 2. Following the discussion at the start of this section, we may assume that n ≥ 10 4 d 4 , that we have already reconstructed the number of edges in G, and we have therefore determined the best possible upper bound d ∈ N on the average degree. In particular, for every t ∈ [0, n], Lemma 8 provides an estimate and, estimating n i=1 d t (G i ) by s t , we obtain the following estimate for d t (G) case. Replacing K 1,p with K 2,p in this example produces a planar graph with a linear number of common cards but a different number of edges. These examples can be generalised to graph classes with a larger (constant) average degree d as well. Indeed, consider adding disjoint copies of the same (3p + 1)-vertex graph H to both G 1 and G 2 . The resulting graphs will still have about 2 3 × 1 2 = 1 3 of their cards in common, and we can create the desired average degree by choosing the density of H.
Let c(G, H) denote the number of cards that G and H have in common, and let c(n) := max{c(G, H) : G, H distinct graphs on n vertices}. The graph reconstruction conjecture states that c(n) ≤ n − 1 for n ≥ 3. The examples above are variations on constructions by Bowler, Brown and Fenner [5] which lead to a bound c(n) ≥ ( 2 3 + o(1))n. The authors of [5] conjecture that the bound is tight and also propose a characterisation of the extremal graphs.
A good first step towards Conjecture 9 would be to determine whether c(n) ≥ (1 − o(1))n. A positive answer would disprove Conjecture 9, whereas a negative answer would prove the Reconstruction Conjecture in a strong form 1 . We remark that the answer to the equivalent question in the 'small' cards set-up has been answered. Let s(G) denote the smallest ℓ for which the multiset D ℓ (G) of ℓ-vertex induced subgraphs of G determines G, and let s(n) = max{s(G) : G graph on n vertices}. Nýdl [16] proved that s(n) ≥ (1 − o(1))n by constructing, for any ε > 0, two non-isomorphic graphs on n vertices with the same set of ℓ-vertex subgraphs for all ℓ ≤ (1 − ε)n.
Groenland, Guggiari and Scott [9] conjectured that the degree sequence of a graph can be reconstructed from a deck of cards with a constant number of cards missing (for n sufficiently large). It follows from Theorem 2 that the conjecture holds for graphs where the average degree is at most c k n 1 3 (for some c k depending only on k), but it is not yet known to hold for general graphs and we repeat it below.
Conjecture 10 (Groenland, Guggiari and Scott [9]). Fix k ∈ N and let n be sufficiently large. For any graph G on n vertices, the degree sequence of G is reconstructible from any n − k cards.