The number of unordered edge pairs without common vertex in a graph G with 30 vertices and 105 edges is equal to 4822. What is the maximal possible value of the differences between degrees of vertices? (Azer Kerimov).
Solution
Suppose that G has n vertices and k edges and i-th vertex has a degree di. The number of unordered edge pairs without common vertex is equal to ((2k)−i=0∑n(2di))=2k(k−1)−21i=0∑ndi2+k=2k(k+1)−21i=0∑ndi2 Therefore, ∑i=030di2=1486. Let δi=di−7. Then ∑i=030δi=0 and 1486=∑i=030di2=∑i=03072+∑i=030δi2. Therefore, ∑i=030δi2=16.
Thus, ∣δi∣≤3 and di−dj≤5. An example with di=10,dj=5 can be constructed by perturbation of regular graph of degree 7. The answer is 5.
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