dailysudoku.com

[Paladin]

I could use some help with this puzzle. Although www.sudokusolver.co.uk indicates that this puzzles is only rated at 1252 (which means that it shouldn't be that hard to solve if you can apply an advanced solving technique), I don't seem to be able to get anywhere. I can't seem to find anything that will allow me to solve more than four (4) cells.

The start grid, and my four cell answers, are as follows:

[Marty R.]

You're not alone; I solved the same four cells and hit a brick wall.

[Ruud]

I am not always in the mood to provide clues for my daily nightmares, but this one deserves a little extra attention.

A series of new techniques has been developed by a couple of guys at the sudoku players forum. These techniques were way over our simple heads until Myth Jellies came up with the Filet-O-Fish theory. This was both brilliant and simple. From this, a couple of techniques have been developed, and one of those I implemented in SudoCue.

This is it:

Finned-X-Wing.

Try to find a situation in the sudoku (there is one in the nightmare, right after you placed the 4th digit) that looks like an X-Wing, but it has one additional candidate in the defining rows (or columns) in the same box as one of the 4 cells that form the X-Wing.

When there are candidates in that same box, that will be eliminated by the X-Wing, they will also be eliminated by the extra candidate that must be true if the X-Wing does not exist.

Try to find it. This is a one-time-only hint from me. The finned X-Wing will not have it's last appearance in the daily nightmares series, so get used to it!

In case you wonder: there are 2 other techniques that I recently implemented. Those are BUG and XYZ-Wing. Better learn about them too.

Ruud.

P.S. Has anyone of you solved the clueless sudoku yet?

[David Bryant]

Here -- I'll give more of a hint than Ruud did!

This puzzle isn't particularly elegant -- it's just a hard grind all the way through. Here's a (general) description of the way I solved it.

1. Resolved top center 3x3 box into {2, 7}, {3, 5}, and {1, 4} pairs.
2. Found {1, 4, 9} triplet in row 3, setting r3c1 = 6.
3. Found hidden pair {1, 3} in r7c1 & r9c1, setting r8c1 = 8.
4. The "4" in column 1 lies in middle left 3x3 box.
5. r5c4 = 8 (unique horizontal); r7c5 = 8 (unique vertical)
6. Coloring on "5" reveals pair {7, 9} in r2c1.
7. Coloring on "6" reveals {2, 4, 9} in r8c9.
8. XY-Wing from r2c1 eliminates "4" at r6c8; r6c1 = 4 (unique horizontal).
9. The "7" in row 6 lies in middle center 3x3 box.
10. {5, 9} pair revealed in column 4 -- r9c4 = 7.
11. {3, 5} pair revealed in column 5 -- r7c5 = 8.
12. r8c7 = 5 (unique horizontal), revealing {1, 2} pair in r7c7 & r9c7.
13. Triplet {4, 7, 9} lies in row 2.

After these steps and the associated eliminations the puzzle looks like this.

[TKiel]

After doings the 13 items in Davids list my candidate grid looked like this

[keith]

After the position given by David, nothing exotic is needed to solve the puzzle.

What I do not understand, and I have been staring at this thing for hours: What is the "coloring" logic that says R2C1 is not <5>, and that R8C9 is not <6>?

In other words, how do you proceed from the following position without using chains?

[TKiel]

R5c1(A) is conjugate with r5c6(a) which is conjugate with r1c6(A) which is conjugate with r2c5(a). R2c1 shares a row with r2c5(a) and a column with r5c1(A), one of which must be the value 5. Therefore it can't be.

R6c5(A) is conjugate with r8c5(a). R5c6(B) is conjugate with r5c9(b). A & B share a group so both can't be true, therefore either/both a & b are true. R8c9 shares a row with 'a' and a column with 'b', therefore can't be 6.

[David Bryant]

[keith]

Tracy,

Sorry, but I still don't see the first one.


ON <5> in R2C1:

>> R5c1(A) is conjugate with r5c6(a)
I agree

>> which is conjugate with r1c6(A)
no, there is a possible <5> in r7c6

>> which is conjugate with r2c5(a)
I agree

>> R2c1 shares a row with r2c5(a) and a column with r5c1(A),
>> one of which must be the value 5. Therefore it can't be.
I would agree, if there were only two possible 5's in c6.


ON <6> in R8C9:

>> R6c5(A) is conjugate with r8c5(a)
I agree

>> R5c6(B) is conjugate with r6c9(b)
I think you mean r5c9(b). If so, I agree.

>> A & B share a group so both can't be true
I agree. But, both can be false, because r6c6 is possibly 6

>> therefore either/both a & b are true. R8c9 shares a row
>> with 'a' and a column with 'b', therefore can't be 6.
I agree. You have taught me something new!


I missed the <6> because I was looking for a single conjugate chain. I still do not understand the <5>.

Thank you,

Keith

[TKiel]

Keith,

I made an error in my post explaining the conjugate 5's. It's actually exactly like the 6's. R5c1(A) and r5c6(a) are conjugates. R1c6(B) and r2c5(b) are conjugates. A & B share a group and can't both be true, so either/both a & b have to be true. R2c1 shares a row with r2c5(b) and a column with r5c1(a), so it can't be true either.

[keith]

Tracy and David,

Thank you. These colors, chains and cycles are coming into focus for me.

Keith

[Orsi_AC]

I've solved this puzzle using a single technique:
After you solve the fourth cell you will find in the row 6 cols 5 and 6 (3,6,7). As (6,7) appears together in the same box you can eliminate the number 3 from these cells and the numbers 6 and 7 from the others cells in the same row.
Good look.

[TKiel]

Orsi AC,

I'm not aware of a technique that allows one to make those exclusions based on that logic. It sounds like you are combining a hidden pair (two values that appear in only two cells) and a naked pair (two cells containing only the same two values) but 6 & 7 both appear in numerous cells in both the row and box in which they are grouped. A hidden pair allows the exclusion of all non pair values from those two cells and a naked pair allows the exclusion of the paired values from all other cells in the group (row, column, box). You've made exclusions both within & outside the two cells.

[guest]

I found this puzzle to be very straightforward without having to go 'fishing' or 'painting'. Basic logic only was needed.

[David Bryant]