555
(Eingestellt am 2. August 2022, 01:01 Uhr von Fool on Hill)
Puzzle rules
Normal Sudoku rules apply.
Digits may not repeat in cages. The three given 5s must be greater than the values in orthogonally adjacent cells. V separates cells with a sum of 5 and X separates cells with a sum of 10. Black dots separate values with a ratio 1:2. Orthogonally adjacent digits cannot be consecutive (even on a dot or in an XV pair or in a maximum cell). Cells a knight's move apart in chess must be different. Not all dots, Xs or Vs are given.
For a significantly harder puzzle replace the three given 5s with 6s - I have now published that as: Number of the beast - to make it possible to record solutions to both
Here is the Cracking the Cryptic link: 555
Here it is on F-Puzzles: 555
Here are some hints on a couple of tricky steps which might help the solve (highlight to see text)
Geometric observations
The knight’s move constraint and the interaction of cages in the boxes is very constraining. Letter the cages from A to J (omitting I) starting at the top end (head) following the path of the cage to the tail. Number the cells in the related box from 1 to 9 top left to bottom right.
Observe first that A and J are different and are forced into cells 3 and 7, and then because B and H are different from each other and also from A and J they are forced into cells 2 and 8. Then the cells above and below the box which are not in the cage can be marked (top) 4DJ and (bottom) 6AF. The corollary of this is that candidates in the box can be used to restrict candidates in the three cells above and in the three cells below, and the candidates above and below restrict the content of the boxes. This geometry is useful throughout.
It is also surprising that the cell marked 6AF sees cells BCDEGHJ which all see each other, and this can be a source of unusual triples etc and other eliminations where a value is forced into a subset of these cells. (A similar observation applies to the cell marked 4DJ)
Easy starting steps
Fill in central column clues 28/4/1/37/37 and 123 options around the given 5s. r6c7 is 5. There are some easy eliminations. The existence of various 123 options and the geometry ought to suggest that colouring could be useful.
A tricky but useful observation to break in
On the eliminations it is easy to miss that r7c7 cannot be 3.
Note now that the 5 in r6c7 takes one of the positions in box 6 allocated to cells A and J, so r3c8 must be in r4c9 and must also appear around the 5 in box 9, hence in r6c7 where it cannot be a 3. Call this cell yellow – yellow is 1 or 2.
Now where does yellow go in box 5? Not in r4 or r5 c46. And this eliminates yellow from r6c3.
Now ask where yellow goes in the cage in c123. This gives the value of yellow and breaks open the puzzle.
A tricky but helpful elimination later in the solve
Later in the puzzle, quite a bit of rows 6 and 7 can be completed by taking care over eliminations. This restricts the values of cells in c123 in r67 and – see the geometric observations – gives the value of r5c2. After that the solve is standard.
Lösungscode: Column 5
Kommentare
am 6. August 2022, 00:40 Uhr von Fool on Hill
Note publication of the companion puzzle: Number of the beast with given 6s instead of 5s
am 2. August 2022, 06:17 Uhr von Markyboy
It's not so difficult that you get totally stuck, but it makes you think and has a very nice flow. Challenging until the end. Very nice to solve.
am 1. August 2022, 16:12 Uhr von Fool on Hill
Here is a variant with an obvious theme. This also solves with three given 6s replacing the three given 5s which would be harder as the maximum cells are less effective. However, I liked the theme. This is anti-knight and non-consecutive (overriding other clues) with other clues standard.
