Havel–Hakimi algorithm

The Havel–Hakimi algorithm is an algorithm in graph theory solving the graph realization problem. That is, it answers the following question: Given a finite list of nonnegative integers in non-increasing order, is there a simple graph such that its degree sequence is exactly this list? A simple graph contains no double edges or loops.^[1] The degree sequence is a list of numbers in nonincreasing order indicating the number of edges incident to each vertex in the graph.^[2] If a simple graph exists for exactly the given degree sequence, the list of integers is called graphic. The Havel-Hakimi algorithm constructs a special solution if a simple graph for the given degree sequence exists, or proves that one cannot find a positive answer. This construction is based on a recursive algorithm. The algorithm was published by Havel (1955), and later by Hakimi (1962).

The algorithm

The Havel-Hakimi algorithm is based on the following theorem.

Let ${\displaystyle A=(s,t_{1},...,t_{s},d_{1},...,d_{n})}$  be a finite list of nonnegative integers that is nonincreasing. Let ${\displaystyle A'=(t_{1}-1,...,t_{s}-1,d_{1},...,d_{n})}$  be a second finite list of nonnegative integers that is rearranged to be nonincreasing. List ${\displaystyle A}$  is graphic if and only if list ${\displaystyle A'}$  is graphic.

If the given list ${\displaystyle A}$  is graphic, then the theorem will be applied at most ${\displaystyle n-1}$  times setting in each further step ${\displaystyle A:=A'}$ . Note that it can be necessary to sort this list again. This process ends when the whole list ${\displaystyle A'}$  consists of zeros. Let ${\displaystyle G}$  be a simple graph with the degree sequence ${\displaystyle A}$ : Let the vertex ${\displaystyle S}$  have degree ${\displaystyle s}$ ; let the vertices ${\displaystyle T_{1},...,T_{s}}$  have respective degrees ${\displaystyle t_{1},...,t_{s}}$ ; let the vertices ${\displaystyle D_{1},...,D_{n}}$  have respective degrees ${\displaystyle d_{1},...,d_{n}}$ . In each step of the algorithm, one constructs the edges of a graph with vertices ${\displaystyle T_{1},...,T_{s}}$ —i.e., if it is possible to reduce the list ${\displaystyle A}$  to ${\displaystyle A'}$ , then we add edges ${\displaystyle \{S,T_{1}\},\{S,T_{2}\},\cdots ,\{S,T_{s}\}}$ . When the list ${\displaystyle A}$  cannot be reduced to a list ${\displaystyle A'}$  of nonnegative integers in any step of this approach, the theorem proves that the list ${\displaystyle A}$  from the beginning is not graphic.

Proof

The following is a summary based on the proof of the Havel-Hakimi algorithm in Invitation to Combinatorics (Shahriari 2022).

To prove the Havel-Hakimi algorithm always works, assume that ${\displaystyle A'}$  is graphic, and there exists a simple graph ${\displaystyle G'}$  with the degree sequence ${\displaystyle A'=(t_{1}-1,...,t_{s}-1,d_{1},...,d_{n})}$ . Then we add a new vertex ${\displaystyle v}$  adjacent to the ${\displaystyle s}$  vertices with degrees ${\displaystyle t_{1}-1,...,t_{s}-1}$  to obtain the degree sequence ${\displaystyle A}$ .

To prove the other direction, assume that ${\displaystyle A}$  is graphic, and there exists a simple graph ${\displaystyle G}$  with the degree sequence ${\displaystyle A=(s,t_{1},...,t_{s},d_{1},...,d_{n})}$  and vertices ${\displaystyle S,T_{1},...,T_{s},D_{1},...,D_{n}}$ . We do not know which ${\displaystyle s}$  vertices are adjacent to ${\displaystyle S}$ , so we have two possible cases.

In the first case, ${\displaystyle S}$  is adjacent to the vertices ${\displaystyle T_{1},...,T_{s}}$  in ${\displaystyle G}$ . In this case, we remove ${\displaystyle S}$  with all its incident edges to obtain the degree sequence ${\displaystyle A'}$ .

In the second case, ${\displaystyle S}$  is not adjacent to some vertex ${\displaystyle T_{i}}$  for some ${\displaystyle 1\leq i\leq s}$  in ${\displaystyle G}$ . Then we can change the graph ${\displaystyle G}$  so that ${\displaystyle S}$  is adjacent to ${\displaystyle T_{i}}$  while maintaining the same degree sequence ${\displaystyle A}$ . Since ${\displaystyle S}$  has degree ${\displaystyle s}$ , the vertex ${\displaystyle S}$  must be adjacent to some vertex ${\displaystyle D_{j}}$  in ${\displaystyle G}$  for ${\displaystyle 1\leq j\leq n}$ : Let the degree of ${\displaystyle D_{j}}$  be ${\displaystyle d_{j}}$ . We know ${\displaystyle t_{i}\geq d_{j}}$ , as the degree sequence ${\displaystyle A}$  is in non-increasing order.

Since ${\displaystyle t_{i}\geq d_{j}}$ , we have two possibilities: Either ${\displaystyle t_{i}=d_{j}}$ , or ${\displaystyle t_{i}>d_{j}}$ . If ${\displaystyle t_{i}=d_{j}}$ , then by switching the places of the vertices ${\displaystyle T_{i}}$  and ${\displaystyle D_{j}}$ , we can adjust ${\displaystyle G}$  so that ${\displaystyle S}$  is adjacent to ${\displaystyle T_{i}}$  instead of ${\displaystyle D_{j}.}$  If ${\displaystyle t_{i}>d_{j}}$ , then since ${\displaystyle T_{i}}$  is adjacent to more vertices than ${\displaystyle D_{j}}$ , let another vertex ${\displaystyle W}$  be adjacent to ${\displaystyle T_{i}}$  and not ${\displaystyle D_{j}}$ . Then we can adjust ${\displaystyle G}$  by removing the edges ${\displaystyle \left\{S,D_{j}\right\}}$  and ${\displaystyle \left\{T_{i},W\right\}}$ , and adding the edges ${\displaystyle \left\{S,T_{i}\right\}}$  and ${\displaystyle \left\{W,D_{j}\right\}}$ . This modification preserves the degree sequence of ${\displaystyle G}$ , but the vertex ${\displaystyle S}$  is now adjacent to ${\displaystyle T_{i}}$  instead of ${\displaystyle D_{j}}$ . In this way, any vertex not connected to ${\displaystyle S}$  can be adjusted accordingly so that ${\displaystyle S}$  is adjacent to ${\displaystyle T_{i}}$  while maintaining the original degree sequence ${\displaystyle A}$  of ${\displaystyle G}$ . Thus any vertex not connected to ${\displaystyle S}$  can be connected to ${\displaystyle S}$  using the above method, and then we have the first case once more, through which we can obtain the degree sequence ${\displaystyle A'}$ . Hence, ${\displaystyle A}$  is graphic if and only if ${\displaystyle A'}$  is also graphic.

The time complexity of the algorithm is ${\displaystyle O(n^{2})}$ .

Examples

Let ${\displaystyle 6,3,3,3,3,2,2,2,2,1,1}$  be a nonincreasing, finite degree sequence of nonnegative integers. To test whether this degree sequence is graphic, we apply the Havel-Hakimi algorithm:

First, we remove the vertex with the highest degree — in this case, ${\displaystyle 6}$  —  and all its incident edges to get ${\displaystyle 2,2,2,2,1,1,2,2,1,1}$  (assuming the vertex with highest degree is adjacent to the ${\displaystyle 6}$  vertices with next highest degree). We rearrange this sequence in nonincreasing order to get ${\displaystyle 2,2,2,2,2,2,1,1,1,1}$ . We repeat the process, removing the vertex with the next highest degree to get ${\displaystyle 1,1,2,2,2,1,1,1,1}$  and rearranging to get ${\displaystyle 2,2,2,1,1,1,1,1,1}$ . We continue this removal to get ${\displaystyle 1,1,1,1,1,1,1,1}$ , and then ${\displaystyle 0,0,0,0,0,0,0,0}$ . This sequence is clearly graphic, as it is the simple graph of ${\displaystyle 8}$  isolated vertices.

To show an example of a non-graphic sequence, let ${\displaystyle 6,5,5,4,3,2,1}$  be a nonincreasing, finite degree sequence of nonnegative integers. Applying the algorithm, we first remove the degree ${\displaystyle 6}$  vertex and all its incident edges to get ${\displaystyle 4,4,3,2,1,0}$ . Already, we know this degree sequence is not graphic, since it claims to have ${\displaystyle 6}$  vertices with one vertex not adjacent to any of the other vertices; thus, the maximum degree of the other vertices is ${\displaystyle 4}$ . This means that two of the vertices are connected to all the other vertices with the exception of the isolated one, so the minimum degree of each vertex should be ${\displaystyle 2}$ ; however, the sequence claims to have a vertex with degree ${\displaystyle 1}$ . Thus, the sequence is not graphic.

For the sake of the algorithm, if we were to reiterate the process, we would get ${\displaystyle 3,2,1,0,0}$  which is yet more clearly not graphic. One vertex claims to have a degree of ${\displaystyle 3}$ , and yet only two other vertices have neighbors. Thus the sequence cannot be graphic.

See also

• Erdős–Gallai theorem

Notes

1. From Shahriari (2022, p. 48): "Definition 2.17 (Graphs & Subgraphs). A simple graph (or just a graph) G is a pair of sets (V, E) where V is a nonempty set called the set of vertices of G, and E is a (possibly empty) set of unordered pairs of distinct elements of V. The set E is called the set of edges of G. If the number of vertices of G is finite, then G is a finite graph (or a finite simple graph)."
2. From Shahriari (2022, p. 355): "Definition 10.6 (The degree sequence of a graph; Graphic sequences). The degree sequence of a graph is the list of the degrees of its vertices in non-increasing order. A non-increasing sequence of non-negative integers is called graphic, if there exists a simple graph whose degree sequence is precisely that sequence.”

References

• Havel, Václav (1955), "A remark on the existence of finite graphs", Časopis pro pěstování matematiky (in Czech), 80: 477–480
• Hakimi, S. L. (1962), "On realizability of a set of integers as degrees of the vertices of a linear graph. I", Journal of the Society for Industrial and Applied Mathematics, 10: 496–506, MR 0148049.
• Shahriari, Shahriar (2022), Invitation to Combinatorics, Cambridge U. Press.
• West, Douglas B. (2001). Introduction to graph theory. Second Edition. Prentice Hall, 2001. 45-46.