The New Sudoku Players' Forum

[creint]

Removing pencilmarks to make it harder.
A special case but hard to define what the restrictions are. Unique solution assumptions are dangerous until fully proven.

Can you still post your solution without using that 7?
Do you know any solvers that can solve this?

[Leren]

Hi creint,

Since you seem to not use uniqueness moves the best way forward is to re-define the original problem with the 7 as the only candidate in r6c3, reducing the total number of candidates to 604.

The only public Str8ts solvers I know about are the one on Andrew Stuart's site, and one I found here, but I haven't used it.

The solution with the 7 as a clue is relatively easy, by the standards of so called Extreme Stre8ts puzzles, but as with most Str8ts puzzles, takes a lot of moves. My solution takes 69 moves, none too difficult but not practical to post on the forum.

The solution with just the 3 removed is much harder. My solver (in humanistic mode) can't solve it, and I doubt whether any non-guessing solver can.

There is very little publicly available information on Str8ts humanistic solving methods. Apart from Andrew Stuart's explanations, the slideshow here is about the best I can do.

You can try and get as far as you can and then post a PM and I can suggest a way forward. As I said before you can remove 2 & 3 from the lower compartment in Column 1, which will solve a few cells.

Leren

[eleven]

Removing pencilmarks to make it harder.
A special case but hard to define what the restrictions are. Unique solution assumptions are dangerous until fully proven.

Isn't the proof obvious ?
If each digit in each cell can be replaced by the complement, every sequence keeps to be a sequence, and all normal sudoku rules are kept valid too to give a second solution for such a "symmetric" candidate grid.
No matter, how much candidates you have in the cell, as long as the complement of each digit is there too.

[yzfwsf]

For fun, I tried your examples. SudoRules allows to do this easily:

SKFR rate this sukaku at ER/EP/ED=8.9/1.2/1.2

[creint]

If you add an greater-than constraint then you don't need to remove a pencil mark. A lot of rules must apply and checked before this can be used, I don't know if this is an usable tactic for users.
My solver can solve it with easy tactics when 7 placed. Else it needs chains in chains, SE < 12.