We consider the problem of completing a (0,-1)\documentclass[12pt]{minimal}
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\begin{document}$$(0,-1)$$\end{document}-matrix to an alternating sign matrix (ASM) by replacing some 0\documentclass[12pt]{minimal}
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\begin{document}$$0$$\end{document}s with -1\documentclass[12pt]{minimal}
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\begin{document}$$-1$$\end{document}s. An algorithm can be given to determine a completion or show that one does not exist. We are concerned primarily with bordered-permutation (0,-1)\documentclass[12pt]{minimal}
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\begin{document}$$(0,-1)$$\end{document} matrices, defined to be n×n\documentclass[12pt]{minimal}
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\begin{document}$$n\times n$$\end{document}(0,-1)\documentclass[12pt]{minimal}
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\begin{document}$$(0,-1)$$\end{document}-matrices with only 0\documentclass[12pt]{minimal}
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\begin{document}$$0$$\end{document}s in their first and last rows and columns where the -1\documentclass[12pt]{minimal}
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\begin{document}$$-1$$\end{document}s form an (n-2)×(n-2)\documentclass[12pt]{minimal}
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\begin{document}$$(n-2)\times (n-2)$$\end{document} permutation matrix. We show that any such matrix can be completed to an ASM and characterize those that have a unique completion.
