Sudoku

VoucherMan

I've just put this into a solver. Not a great one but it works through a step at a time so you can see how it solved it.

Claims this one wouldn't validate. There are numbers out of place.

googler

splishsplash wrote: »

Look at the first cube top left. The two missing numbers in the middle row MUST be 1 and 9, they can't be anything else. Therefore, 5 and 7 are the two missing from the first row, as there is a 5 on the top of the sixth column, 5 MUST go in the first box of the third row, and 7 MUST go in the top left box:
746
1?3
5?8

HTH

All you can say from the 5s in the top 3 rows is that a 5 must go in ONE of the two ?s in the third row, cols 1 & 2

5 and 7 CAN'T be the two missing from row 1, since there's already a 5 in row 1.....





It would help to have the original grid, rather than part-completed, but

Top R square - remaining two must be 1 and 2
Top Row - remining three must be 1, 2 and 7
Top L square - number in top row can't be 2 (already a 2 in that square), must be 1 or 7
Already have 7s in Columns 2 and 3, so
Must be 1 and 7 in L column of Top L square.
Therefore Column 2, top L square must be 5 and 9

5s are already in row 1 and 2, so Col 2, row 2 = 9, row 3 =5

9s in row 1 and 2, so 9 must be in row 3 of top Mid square
Already 9 in col 4, so 9 must be row 3, col 6

Remaining 3 in Top Mid square must be 1,2 and 7
Already a 2 and 7 in row 2, so row 2 col 5 = 1


Have to go now, solve rest later.

debitcardmayhem

There doesn't appear to be a logical move to my tired old eyes , it has to be trial (and error ) but the top left square(1) c1,r1 and c1,r3 one of which must be a 7. Only one of them is correct. You could try 1 or 9 in the c2r2 square as a starting point too.
I used T&E first and solved it then ran it into sudoku-savant and when checking it says "unique solution , involves guesswork (stinker)"

googler

googler wrote: »

Top R square - remaining two must be 1 and 2
Top Row - remining three must be 1, 2 and 7
Top L square - number in top row can't be 2 (already a 2 in that square), must be 1 or 7
Already have 7s in Columns 2 and 3, so
Must be 1 and 7 in L column of Top L square.
Therefore Column 2, top L square must be 5 and 9

5s are already in row 1 and 2, so Col 2, row 2 = 9, row 3 =5

9s in row 1 and 2, so 9 must be in row 3 of top Mid square
Already 9 in col 4, so 9 must be row 3, col 6

Remaining 3 in Top Mid square must be 1,2 and 7
Already a 2 and 7 in row 2, so row 2 col 5 = 1

With 1 and 7 in sq1, remaining in col1 are 4 and 5 to go in r5 and r9. r5 already has a 5, so c1r5 = 4, and c1r9 = 5

With 4s in r7 and r8, c3r9 must = 4

Row 5 remaining are 1 and 2, in c2 and c5. c5 already has 1, so c2r5 = 1, c5r5 = 2

With that 2 in place, top middle square needs 2 and 7 - c4r3 must be 2, with 7 in c5r1

From this, r1c9 = 2, c8r3 = 1, c1r1 = 1, c1r3 = 7

With 5s in c7 and c9, the 5 in square 9 must be in c8
With 2 in c9r1, and c8 full in square 6, the 2 in square 6 must be in c7 and the 2 in square 9 must be in c8; therefore, 2 and 5 in col8, r7&8, with 1 and 3 in col9, r7 and r9

With 2s in col 4 and 5, c6r7 must be 2, and c8r7 = 5, with c8r8 =2

This is where it falls apart, either with the OPs grid (and as suggested in post #12, where the solver says something is out of place) or my logic, because with 2 in c6r7 and c8r8, that prompts for a 2 in row 9 of square 7 - which is already full.

Original starting grid from the OP, please, rather than a part-completed one?

debitcardmayhem

VoucherMan wrote: »

I've just put this into a solver. Not a great one but it works through a step at a time so you can see how it solved it.

Claims this one wouldn't validate. There are numbers out of place.

It will validate look at my post 14

googler

debitcardmayhem wrote: »

It will validate look at my post 14

Well, since I can't get it to validate, and the other poster's solver at #12 agrees, could you be more specific?

debitcardmayhem

googler wrote: »

Well, since I can't get it to validate, and the other poster's solver at #12 agrees, could you be more specific?

I used sudoku-savant on Linux, it does state that it is a stinker (i.e. needs guesswork) however last Fridays Telegraph diabolical said nightmare (worse than stinker, needs guesswork), but I got that one fairly quickly(1/2 hour which is good for me on Friday) without trial and error. Work from the premise that 1,1 or 1,3 is a 7, one breaks down fairly quickly, the other well is the path to the solution.

Edit: I found it impossible yesterday since I tried it using his post and the ?'s all looked like 7's on my screen .

VoucherMan

googler wrote: »

Well, since I can't get it to validate, and the other poster's solver at #12 agrees, could you be more specific?

My bad! Early morning not fully awake I loaded a Killer Sudoku grid.

Redone on Sudoku Solver it works fine. Here's the puzzle in case anyone wants to go through it.

Took 30 steps to find the next number


?46¦ 3?5¦ 89?
2?3¦ 8?4¦ 765
??8¦ ?6?¦ 3?4
83?¦ ???¦ ?46
?27¦ 6?8¦ 539
96?¦??? ¦?78
6?9¦ ?8?¦ 4??
38?¦ 4?6¦ 9?7
?7?¦ 9?1¦ 68?

and another 44 to complete the puzzle.

DUTR

Wow, glad this generated so much interest and debate
Here is the original puzzle without my solution as in the original post




the difficulty rating was described as moderate, as said the challenging one was a doddle and completed in minutes.

debitcardmayhem

Still the same rating :cool: