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Question

Going through graph theory, i found out that , if a graph $G$ is $k\,\text{regular}$ then in its complement $G^{'}$, all the vertex will be having a common degree $n-k-1$ i.e $n-k-1 \text{regular graph}$ .

Doubt

If i take $G$ with $4$ vertices and is $2$ regular graph , then its complement should be $4-2-1 =1\text{regular graph}$,But $G^{'}$ is actually $2$ regular graph.

Why is this contradiction?

Please help

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It is a $1$-regular graph indeed. The complement of a square graph is a graph consisting of two diagonals crossing one another each of whose vertex has degree $1$.

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