The New Sudoku Players' Forum

[dpletcher]

I’ve been doing Sudoku for about 3 months. I get the puzzles in the News Sentinel in Ft Wayne, IN and enjoy them a lot. Usually have no problems but have 2 HARD ones now that have me stumped. One I’ve been looking at for 3 weeks and need help with getting the next step. Here it is:

{6}{2}{3}|{7}{4}{8}| (5/9){1}(5/9)
{9}{8}{1}|{3}{2}{5}|{4}{7}{6}
{4}{5}{7}|{9}{1}{6}|{2}{3}{8}
(3/8)(3/9){6}|{4}(3/5){1}|{7}(2/8/9)(2/5/9)
{5}(3/7/9){4}|(2/6/8)(3/6/7)(2/3/7)|(3/6/8/9)(6/8/9){1}
(1/3/7/8)(1/3/7){2}|(6/8)(3/5/6/7){9}|(3/5/6/8)(6/8){4}
(2/3){6}{9}|{5}{8}(2/3)|{1}{4}{7}
(1/2/3/7)(1/3/7){8}|(2/6)(3/6/7/9){4}|(6/9){5}(2/9)
(2/7){4}{5}|{1}(6/7/9)(2/7)|(6/8/9)(2/6/8/9){3}

Can anyone give me a next step? I must be missing something or there is a technique that I’m not aware of.

[SteveF]

Have a close look at row 6, there is a naked pair that should allow some eliminations in that row.

That should allow you to enter a candidate in box 4.

[dpletcher]

OK, I see the (6/8) pairs in r6c4 & r6c8 so that eliminates the 6 & 8 from r6c7 but what abut the possible (2/6/8) in r5c4 and the (6/8/9) in r5c8.

[SteveF]

First the 6, 8 pair also means you can eliminate these values from r6c1 and r6c5 as well.

Then have a look at the possible candidates for r4c1 (in particular is there a candidate in box 4 that can only go in r4c1).

As a follow-up hint, there is also a naked pair in row 9.

[dpletcher]

Ah, yes. Thank you. I now have an 8 in r4c1 and I see the naked pair 2/7 in row 9; but I also have a naked pair 5/9 in row 1 and 2/3 in row 7. How does that help me? I'm missing a critical piece of logic here.

[Jeff]

The naked pairs in row 1 and row 7 won't allow you the eliminate any numbers, but the one in row 9 does. After 27 are eliminated from row 9, you will get a number similar to the '8' in box 4.

[dpletcher]

I'm still not seeing it. Even after eliminating for the 2/7 pair in row 9 I still have a 6/9 in col 5 and 6/8/9 in cols 7 and 8. What am I missing?

[Anette]

By eliminating 2 from R9C8, you only have one place left in column 8 where 2 can go.

[SteveF]

I'm not sure I see that arguement Anette, but following the elimination of 2 from r9c8 - where can a 2 go in box 9?

[Anette]

Well, you get both the 2 in box 9 and the 2 in column 8 by eliminating it from R9C8.

I guess it depends on your techniques, I always check rows and columns before boxes.

[dpletcher]

Ah yes, I see those now. I'll take it from there and see if I can't break this open soon. Thanks so much for your help.

[dpletcher]

Here I am again. I got that one solved in just a few minutes after our last post. Now here is another one that I am stuck on.

(4/5/6) (1/2/4/5/6) {8} | (1/4){9} (2/6) | {3} {7} (1/6)
(4/6) {9} {7} | {5} {3} (2/6) | (2/4/6/8) (1/2/4/6) (1/6/8)
{3} (1/2/4/6) (1/4/6) | {8} (1/4) {7} | (2/4/6) {5} {9}

{9} (1/4/6) (1/4/6) | {7} {8} (1/5) | (2/5/6) (2/6) {3}
{2} {3} {5} | {9} {6} {4} | {1} {8} {7}
{7} {8} (1/6) | (1/3) {2} (1/5/3) | (5/6) {9} {4}

{1} {7} {9} | {6} {5} (3/8) | (4/8) (3/4) {2}
(4/5/6/8) (4/5/6) {3} | {2} (1/4) {9} | {7} (1/4/6) (1/5/6/8)
(4/5/6/8) (4/5/6) {2} | (1/3/4) {7} (1/3/8) | {9} (1/3/4/6) (1/5/6/8)


What next step am I missing here?

[Ianac]

Look in box 3 , you can eliminate 1 in c1 r9 which should give you a good start. ( possibles locked to a row )

[Ianac]

Sorry got my grid wrong should be R1C9

[dpletcher]

Thank you so much. I got it finished in about 5 min after your tip. Now, please explain why we were able to eliminate the 1 in r1c9 when there was a possible 1 in several other cells in row 1. What is the logic behind the "possibles locked to a row" thing?