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A procedure by induction to get maximum spanning bipartite graph (same vertex set as $G$ and the maximum possible number of edges) from given graph $G$:

Given a planar graph $G$, we need two disjoint sets $V_1$ and $V_2$. We define $V_2=V \setminus V_1$ where $V = V(G)$. In the following, $H$ is the bipartite subgraph with $V(H)=V_1 \sqcup V_2$ and $E(H) = K_{V_1,V_2} \cap E(G)$ where $K_{V_1,V_2}$ is the complete bipartite graph on $V_1$, $V_2$.

  1. Start with first $i$ vertices of $V$ in vertex set $V_{01} \subseteq V$ and then $V_{02} = V \setminus V_{01}$

  2. In iteration $i$, let our current chosen vertex set be $V_{i1}$

  3. If $\exists x \in V_{i1}$ such that $d_{H}(x) \lt \frac12 d_{G}(x)$, then consider the bipartite spanning graph $H'$ with $V(H') = V_{i1}' \sqcup V_{i2}'$ where $V_{i1}' = V_{i1} \setminus x$, $V_{i2}'= V_{i2} \sqcup x$, $E(H') = K_{V_{i1}',V_{i2}'} \cap E(G) $ This makes $\vert E(H') \geq E(H) \vert$. We are basically moving vertex $x$ which satisfies above property from $V_{i1}$ to $V_{i2}$.

  4. Similarly do for $ V_{i2}$. If $\exists x \in V_{i2}$ such that $d_{H}(x) < \frac{1}{2} d_{G}(x) $, move vertex $x$ to $V_{i1}$. Go to step 2.

  5. Repeat until $(\forall x \in V(H))\, d_{H}(x) \geq \frac{1}{2} d_{G}(x)$

  6. After this $V_1 = V_{i1}$ and $V_2 = V_{i2}$. We are using the property that a spanning bipartite subgraph $H$ from $G$ is maximal in terms of the total number of edges only when $H$ with $V(H) = V_{1} \sqcup V_{2}$ satisfies the property that $$(\forall x \in V(G))\, d_{H}(x) \geq \frac{1}{2} d_{G}(x)$$

Analysis of this procedure:

  1. Could somebody help me analyze this algorithm?

  2. Can I assume I know the degrees of the vertices in the bipartite graph $H$, in the intermediate steps?

    2.1. If I can assume so, then each induction iteration has $2n$ steps. And, in the worst case, could there be $n!$ permutations of induction iteration (Any permutation of $i$ vertices is possible for the first vertex set). Is the worst-case upperbound $O(n!\times n)$?

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In answer to your comment:

By maximum spanning bipartite graph I mean same vertex set as given graph and the maximum number of edges possible. Is the complexity $O(n^3)$ correct for this algorithm? If so it would solvable in P time.

No, that problem, also known as Bipartite Edge Deletion, or Edge Bipartization, is:

Given a graph $G = (V, E)$ and an integer $k$, is it possible to delete at most $k$ edges $E'$ such that $G' = (V, E \setminus E')$ is bipartite,

which is NP-complete.

It happens to be "equivalent" to the "vertex deletion" version of the problem, which is also known as Odd Cycle Transversal.

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  1. We cannot analyze your algorithm since it is, as currently presented, not an algorithm to start with. In particular, you did not specify what is the starting value of $i$, in the very first iteration.

  2. Surely, when you have $V_{i1}$ and $V_{i2}=V\setminus V_{i1}$, it is easy to compute the degrees of the vertices in $H$.

    2.1. IMHO, even after you fix some starting value of $i$ for the first iteration, there cannot be any guarantee that your algorithm would stop. Since, swapping vertices between $V_{i1}$ and $V_{i2}$ does not necessarily decrease the number of "bad" vertices, i.e. those vertices $x$ for which $d_H(x)\le \frac12d_G(x)$.

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