The odd Hadwiger's conjecture is "almost'' decidable

The odd Hadwiger's conjecture, made by Gerads and Seymour in early 1990s, is an analogue of the famous Hadwiger's conjecture. It says that every graph with no odd $Kt$-minor is $(t-1)$-colorable. This conjecture is known to be true for $t \leq 5$, but the cases $t \geq 5$ are wide open. So far, the most general result says that every graph with no odd $Kt$-minor is $O(t \sqrt{\log t})$-colorable. In this paper, we tackle this conjecture from an algorithmic view, and show the following: For a given graph $G$ and any fixed $t$, there is a polynomial time algorithm to output one of the following: \begin{enumerate} \item a $(t-1)$-coloring of $G$, or \item an odd $K{t}$-minor of $G$, or \item after making all "reductions" to $G$, the resulting graph $H$ (which is an odd minor of $G$ and which has no reductions) has a tree-decomposition $(T, Y)$ such that torso of each bag $Yt$ is either \begin{itemize} \item of size at most $f1(t) \log n$ for some function $f1$ of $t$, or \item a graph that has a vertex $X$ of order at most $f2(t)$ for some function $f2$ of $t$ such that $Yt-X$ is bipartite. Moreover, degree of $t$ in $T$ is at most $f3(t)$ for some function $f_3$ of $t$. \end{itemize} \end{enumerate} Let us observe that the last odd minor $H$ is indeed a minimal counterexample to the odd Hadwiger's conjecture for the case $t$. So our result says that a minimal counterexample satisfies the lsat conclusion.