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Thus Ryser's Theorem can be interpreted as saying that any partial latin rectangle $R$ can be completed if and only if $R$ satisfies Hall's Condition for partial latin squares.\n  We define Hall's Condition for partial Sudoku squares and show that Hall's Condition for partial Sudoku squares gives a criterion for the completion of partial Sud"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1107.2634","kind":"arxiv","version":2},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.CO","submitted_at":"2011-07-13T19:24:14Z","cross_cats_sorted":[],"title_canon_sha256":"de64f3a9212835ceca685e928df0332302f26e065c6762f31519747af452259a","abstract_canon_sha256":"66e4889e165287d0ef8521d0f8082116b92bbcb9be4341973e3d3f84e8227129"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T04:18:15.886599Z","signature_b64":"mXf/l63E2qwx9C0C115RrnpC3SBOQ48oqfsDQMGjcu/zWn3qCyeVDZrUp9Ffs3V4q1KaYAvKobRTIwnvGGOZBw==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"d8e108338328448d97b432d006cc4f909212795ad0d27a309c48db86316a7b73","last_reissued_at":"2026-05-18T04:18:15.886190Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T04:18:15.886190Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"An analogue of Ryser's Theorem for partial Sudoku squares","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CO","authors_text":"A. 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