Number of the beast
(Eingestellt am 5. August 2022, 13:00 Uhr von Fool on Hill)
Puzzle rules
Normal Sudoku rules apply.
Numbers in a cage may not repeat. Adjacent digits are not consecutive (this applies to the whole grid including dots, cages, maximum and XV clues - to clarify ambiguities noted in comments). Cells a knight's move apart must be different.
The black dot separates cells with a ratio 1:2. The X separates cells with a sum of 10 and the V separates cells with a sum of 5. There is no negative constraint on Kropki or XV pairs. The three maximum cells are greater than each of their orthogonal neighbours.
This is a companion puzzle to my puzzle 555 which has three given 5s and is rather easier
Here is the Cracking the Cryptic link: Number of the beast
Here it is on F-Puzzles: Number of the beast
Here are some notes and hints on a possible solve path - highlighting particular logic which I think is hard to discover, but makes the solve work. The geometric observations given in notes to 555 also apply to this puzzle. Highlight the text to see the hints.
Gains and losses
In contrast to 555 we in fact gain a 56 pair in r5c46. But we lose the fact that the options around the three given digits are restricted to three values. The options are 1234 with some easy eliminations, but no immediate colouring in prospect.
A key question – three values or four
If there are only three values around the given 6s, then r3c8=r4c9=r7c7= yellow = 1 or 2 and the solve continues similarly to the solve in 555.
But there is another option. Call r3c8=r4c9 yellow. It is possible that r7c7 is not yellow, in which case the three cells next to 6 in box 9 are all different from yellow, and with yellow they make up the digits 1-4. Which case applies?
A stress test
Now the geometric observations mean that r7c6 (and r3c4) are 347.
Suppose r7c7 is not yellow – make it blue. Then blue goes in r5c9 in box 6 (geometry) and is not 1 (row) or 2 because it is next to yellow =123. So blue is 4 and yellow is not 3. Also r3c6 becomes 7.
Now call r9c2 purple and r9c7 green to complete the set in column 9. R8c8 is purple and r8c6 is green and now green is not 3 because of the 37 pair on the X. So purple is 3, r4c4 is 3 (geometry) and r3c4 is 4 and there is no place for 4 in the cage in c123.
So r7c7 is yellow and the solve can proceed with yellow as 12 using the useful observation from 555 if necessary and checking the options for r4c8 to get more colouring in, and the positions of r8c8 in box 6.
One last trick
It is possible to show that r6c4 and r7c5 are 3 (noting positions for 2 in box 6 to resolve the 37 pair in col 5) and r6c8 and r7c9 are the same (green) 14, so that r6c1-3 and r7c1-4 are from 589 and the non-green digit from 1 and 4. Now where does r6c3 go in r7? Only in r7c4, so these two are equal and r5c2 sees all four of these values.
Now r5c2 cannot be 1 (row) or 4 (options for 3 in box 2) and cannot now be 589 or 2 or 3 or 6 and must therefore be 7.
Or a complete cheat
Solve 555 and note that the grid is such that 5 can be swapped with 6 throughout without breaking the rules. Then invoke uniqueness.
Lösungscode: Negative diagonal: 9 digits r1c1 to r9c9
Kommentare
am 16. September 2024, 10:48 Uhr von mezkur7
Sudokupad(w/sol): https://sudokupad.app/51bf1m0an6
am 18. August 2022, 10:48 Uhr von Fool on Hill
To clarify that non-consecutive applies to the whole grid
am 18. August 2022, 10:13 Uhr von Bankey
A clarification, pl. The constraint on adjacent cells not being consecutive applies to the whole grid or only to the cells covered by given cages and other given clues? It seems to be the former, but just wanted to be sure. Thanks. (The confusion arose because of the wording "even on other clues" which would be kind of superfluous if the constraint applies to the whole grid.)
I put "even on other clues" because someone asked me whether it applied on the dot and the V - which it does. I will clarify further.
@ Fool on Hill : thanks for your response. :)
am 9. August 2022, 19:50 Uhr von Fool on Hill
For those who have found this puzzle too tricky, I have added some hints.
am 9. August 2022, 18:39 Uhr von Fool on Hill
Have added some solving hints and notes to indicate how this can be done for those who have found it challenging.
am 5. August 2022, 21:42 Uhr von Fool on Hill
To clarify no negative constraint.
am 5. August 2022, 21:41 Uhr von Fool on Hill
There is no negative constraint on Kropki or XV.
am 5. August 2022, 20:42 Uhr von bansalsaab
Hi @foolonhill, is there a negative constraint on kropki, V and X?
am 4. August 2022, 10:30 Uhr von Fool on Hill
Not sure how difficult this is, but much more so than the companion puzzle 555. It is humanly possible, but might be harder than a 4*. I found it a real challenge.
It is easy to show that there are no solutions for given 7s and it is possible to show that there are no solutions with three 8s or three 9s (though I haven't got an easy way to do that by hand). My original goal was to leave the Maximum cells blank and say they all had to be the same. Making the three equal now implies that they are either 5 or 6.
This puzzle comes from disambiguating a rather constrained grid - for those who are into these things the solved grid has some interesting features which weren't used in the puzzle as I didn't need the implied negative constraints (I was constraining X and black Kropki options in an idiosyncratic way). Also compare the solution with the 555 solution to see another feature of the grid. Perhaps the grid - which I was pleased to discover - is more interesting than the puzzle?
