{"id":495,"date":"2026-11-14T01:54:25","date_gmt":"2026-11-14T01:54:25","guid":{"rendered":"https:\/\/petermc.net\/blog\/?p=495"},"modified":"2026-11-14T01:54:25","modified_gmt":"2026-11-14T01:54:25","slug":"linear-recurrences-and-applications-to-sudoku","status":"publish","type":"post","link":"https:\/\/petermc.net\/blog\/2026\/11\/14\/linear-recurrences-and-applications-to-sudoku\/","title":{"rendered":"Linear recurrences and applications to sudoku"},"content":{"rendered":"

Warning: Spoilers for this puzzle<\/a> ahead.<\/p>\n

I am not a fan of sudoku puzzles.<\/p>\n

However there are sudoku variants that are interesting. Usually when they solve nothing like a sudoku.<\/p>\n

I want to share something that came up when I solved Lots in Space: Deconstructica by Marty Sears<\/a>.<\/p>\n

The puzzle has lines, where any three consecutive entries along a line have a different residue modulo 3 (which of course implies the sequence is periodic mod 3). And there are blue diamonds, where a digit on a blue diamond tells you how many distinct sums of adjacent pairs of numbers along that line there are.<\/p>\n

Now suppose there is a 3 in a blue diamond. Let \"a_1,a_2,\ldots\" be the entries along the line. The mod 3 condition, together with the fact that there are only three distinct sums of adjacent pairs tells you that \"a_i+a_{i+1}\" is a function of \"i \pmod{3}\" only. Thus we get a recurrence<\/p>\n

  <\/span>   <\/span>\"\[a_i+a_{i+1}=a_{i+3}+a_{i+4}\]\"<\/p>\n

The characteristic equation of this recurrence divides \"x^6-1\", so the solution is 6-periodic. I used this in my solve, but something tells me this was not an intended approach.<\/p>\n","protected":false},"excerpt":{"rendered":"

Warning: Spoilers for this puzzle ahead. I am not a fan of sudoku puzzles. However there are sudoku variants that are interesting. Usually when they solve nothing like a sudoku. I want to share something that came up when I … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"jetpack_post_was_ever_published":false},"categories":[1],"tags":[],"class_list":["post-495","post","type-post","status-publish","format-standard","hentry","category-uncategorised"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7V6a7-7Z","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/495","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/comments?post=495"}],"version-history":[{"count":2,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/495\/revisions"}],"predecessor-version":[{"id":497,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/495\/revisions\/497"}],"wp:attachment":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/media?parent=495"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/categories?post=495"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/tags?post=495"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}