{"id":3157,"date":"2020-12-23T18:22:36","date_gmt":"2020-12-23T16:22:36","guid":{"rendered":"https:\/\/blog.atlant.is\/?p=3157"},"modified":"2021-02-21T10:33:29","modified_gmt":"2021-02-21T08:33:29","slug":"christmas-tree","status":"publish","type":"post","link":"https:\/\/blog.douzeb.is\/?p=3157","title":{"rendered":"Christmas tree"},"content":{"rendered":"\n
\"\"<\/a>
A puzzle by Joseph Nehme<\/a><\/figcaption><\/figure><\/div>\n\n\n\n

Mark Goodliffe and Simon Anthony from cracking the cryptic<\/a> are doing a fantastic job of making us discover the fascinating world of advanced sudoku.<\/p>\n\n\n\n

Here is a treat for the season: normal sudoku rules apply, and identical digits cannot be a chess knight’s move apart. The grey square marks an even digit. Any two cages that share an edge and whose totals have a difference of 1 have a white dot between them, and any two cells that share an edge whose totals have a ratio of 2:1 have a black dot between them; there is a negative constraint so two cages touching each other without a dot do not have totals with a difference of 1 or a ratio of 2:1<\/em>.<\/p>\n\n\n\n

Please try the puzzle here<\/a>.<\/p>\n\n\n\n\n\n\n\n

Possible solution<\/h2>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Cages A, B, F completely fill box 8, therefore add up to 45. The white dots tell that cage B and cage F must each total to cage A plus or minus 1. The only possibility for A, B and F together to total to a multiple of 3 is if they are x, x-1, x+1, in some order. And subsequently x=15.<\/p>\n\n\n\n

Therefore, cage A is 15.<\/p>\n\n\n\n

Cage B must be 14 or 16. If B were 16, then cage G would be at least 13, and cages C, D, E respectively at least 15, 14, 13, with a total of 42. Cell r7c7 could be no more than 2, making it impossible to fill cage G.<\/p>\n\n\n\n

Therefore, cage B is 14 and cage F is 16.<\/p>\n\n\n\n

Cage C must be 13 or 15. If C were 15, the C, D, E would again add up to 42 minimum, making it impossible to fill cage G.<\/p>\n\n\n\n

Therefore, cage C is 13. Similarly cage D is 12.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Cage H is either 14 or 16. If it were 16, it would contain the digits 7 and 9. Cage I would be either 17 or 15, thus containing the digits 9 and 8, 9 and 6, or 8 and 7. Impossible by sudoku.<\/p>\n\n\n\n

Therefore, cage H is 14, and cage J is 7 (two-cell cage J could not be 28).<\/p>\n\n\n\n

Cage K must be 8, with an absolute minimum of 15 for 5-cell J+K, comprised exactly of the digits 1, 2, 3, 4, 5.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now cage H must contain the digits 8 and 6. The digits 7 and 9 complete box 7.<\/p>\n\n\n\n

Now the possibilities for cage I are to be a 13-cage (with digits 9 and 4, or 7 and 6) or a 15-cage (with digits 9 and 6, or 7 and 8). By sudoku, only the first possibility is legit.<\/p>\n\n\n\n

Therefore, cage I is 13 with the digits 9 and 4.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

By sudoku, cage F cannot contain a 7 or a 9. It must contain a digit 8, otherwise it could not add up to 16.<\/p>\n\n\n\n

By sudoku, cell r7c7 in cage G cannot be 7, 8 or 9, and must therefore be 6 for cage G to reach its minimum value of 15.<\/p>\n\n\n\n

Conversely, cage F without a 6 now must contain the digits 8, 3, 5 in some order.<\/p>\n\n\n\n

Also, now that cage G contains the digits 6 and 9, the only possibility for cage C is the pair of digits 5 and 8.<\/p>\n\n\n\n

Cage D cannot contain the digits 5 and 8. The only possibility for a 12-cage is the pair of digits 3 and 9.<\/p>\n\n\n\n

The digit 3 no longer being available, we know that cell r7c9 must contain the digit 1, cage E is a 13-cage, and contains the digits 2, 4, 7.<\/p>\n\n\n\n

In turn this simplifies the situation of cages J and K.<\/p>\n\n\n\n

Finally, this further restricts the possibilities of digits for cages A and B.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

By knight move, cell r8c6 cannot be 9. The only possibility for 14-cage B is the triple of digits 7, 1, 6.<\/p>\n\n\n\n

The only possibility for 15-cage A is the triple of digits 9, 2, 4.<\/p>\n\n\n\n

By sudoku and knight move from cage H, digit 6 cannot be contained in cell r9c5. By knight move from cell r7c7, digit 6 cannot be contained in cell r9c6. Thus, the only place for digit 6 in cage B is in cell r8c6.<\/p>\n\n\n\n

Further sudoku and knight move considerations yield quite a few digits.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Cage L must be 15 or 17. If it were 17, cell r6c7 would have to contain a 9 (impossible by sudoku) or an 8 (impossible with the digit 8 one knight move away).<\/p>\n\n\n\n

Therefore, cage L is 15. It contains the pair of digits 8 and 7, the 8 above the 7.<\/p>\n\n\n\n

Digit 9 in box 5 is now restricted: it must belong to cage M. M being a 14-cage or a 16-cage, the possibilities are either the triple 9, 4, 1, or the triple 9, 6, 1.<\/p>\n\n\n\n

By sudoku and knight move, cell r6c5 must be 1. Then cell r5c5 must be 6. Finally cell r4c5 must be 9 and M is indeed a 16-cage.<\/p>\n\n\n\n

By sudoku, digit 4 must be in cell r4c6, and 2, 3, 5 complete box 5.<\/p>\n\n\n\n

By sudoku and knight move, cage O is restricted to either the pair of digits 3 and 2, or the pair of digits 3 and 5. Now is time to use the negative constraint (there is no dot between cage O and cage M) and eliminate the possibility of a 3 and 5 pair (adding up to 8, i.e. half of cage M).<\/p>\n\n\n\n

Therefore, cage O contains the pair of digits 3 and 2.<\/p>\n\n\n\n

N must be an 8-cage, containing either the triple of digits 1, 2, 5, or 1, 3, 4, which can be simplified by sudoku and knight move.<\/p>\n\n\n\n

By sudoku, cell r2c5 must contain 4, 5 or 8 and the rule set requires it to be even. It must be the digit 8 (there is already a digit 4 a knight move away).<\/p>\n\n\n\n

Digits 1 and 7 in cage B can be disambiguated.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

We fill the possibilities for the remaining cells in row 6.<\/p>\n\n\n\n

Then simplify by sudoku and knight move rules.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Do the same for row 5, for row 4, finally for box 2.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Disambiguate the pair 1 and 3 in cage K, and the pair 2 and 5 in cage J.<\/p>\n\n\n\n

The only possibility for cell r1c2 (naked single) is digit 8 – by sudoko and knight move with the nearby 6-7 pair in box 2.<\/p>\n\n\n\n

By sudoku, cell r3c8 must also contain an 8.<\/p>\n\n\n\n

The only place for a 9 in box 4 is r5c1. By sudoku and knight move, 9 in box 1 must be in cell r2c2.<\/p>\n\n\n\n

The 9 in box 3 must be in cell r3c9. This disambiguates the 3-9 pair in cage D.<\/p>\n\n\n\n

Fill in the possibilities for row 3. Simplify by sudoku and knight move rules.<\/p>\n\n\n\n

The 6-7 pair in column 2 looks at cell r5c3, which therefore must be a 3.<\/p>\n\n\n\n

Simplify by sudoku and knight move rules. Use the 4-7 pair in row 5.<\/p>\n\n\n\n

Digit 1 in box 1 must be in cell r2c3.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Fill in the possibilities for row 2. Simplify by sudoku and knight move rules.<\/p>\n\n\n\n

The 6-7 pair in column 2 looks at cell r2c1, making digit 6 impossible.<\/p>\n\n\n\n

This gives a 2-3-4-7 quadruple in row 2.<\/p>\n\n\n\n

Disambiguate the 6-7 pair in column 4, then the 6-7 pair in column 3.<\/p>\n\n\n\n

Continue with sudoku and knight move rules.<\/p>\n\n\n\n

Use the 1-3-4-6 quadruple in row 1.<\/p>\n\n\n\n

Box N is now complete with the triple of digits 1, 4, 3.<\/p>\n\n\n\n

Only place for digit 6 in row 3 is in cell r3c1.<\/p>\n\n\n\n

Only place for digit 6 in column 9 is in cell r1c9.<\/p>\n\n\n\n

Complete the puzzle.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Merry Christmas!<\/p>\n","protected":false},"excerpt":{"rendered":"

Mark Goodliffe and Simon Anthony from cracking the cryptic are doing a fantastic job of making us discover the fascinating world of advanced sudoku. Here is a treat for the season: normal sudoku rules apply, and identical digits cannot be a chess knight’s move apart. The grey square marks an even digit. Any two cages […]<\/p>\n","protected":false},"author":1,"featured_media":3158,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[16,5,67],"tags":[],"ppma_author":[76],"authors":[{"term_id":76,"user_id":1,"is_guest":0,"slug":"fred","display_name":"Fred","avatar_url":"https:\/\/secure.gravatar.com\/avatar\/ec0326d654fdf9f66e9eb42bb34a9bc4?s=96&d=mm&r=g","description":"","first_name":"","last_name":"","user_url":""}],"_links":{"self":[{"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/posts\/3157"}],"collection":[{"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=3157"}],"version-history":[{"count":8,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/posts\/3157\/revisions"}],"predecessor-version":[{"id":3271,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/posts\/3157\/revisions\/3271"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=\/wp\/v2\/media\/3158"}],"wp:attachment":[{"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3157"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3157"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3157"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/blog.douzeb.is\/index.php?rest_route=%2Fwp%2Fv2%2Fppma_author&post=3157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}