dailysudoku.com

geoff1946 · Joined: 23 Sep 2005 Posts: 1

Dipping into the archives, both my wife and I got stuck on this one
-3- 2-- 781
--8 --- 3-4
-74 -3- 96-
--- 18- 6-7
-17 6-9 -38
8-- 375 149
-41 -6- 89-
7-- --- 41-
98- 413 -7-

Notice there are only two options for placement of 7's
r2c4 r2c6 r7c4 and r7c6

This site's hint programme suggests r2c4 should be a 7 but how do you logically justify that decision?
Geoff & June

David Bryant · Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

It took too long to figure this one out, so I'm not going to spell out the whole position. The key to getting a "7" in r2c4 lies in r1c6. Let me explain.

The only possible values in r1c6 are {4, 6}. The only (apparent) possibilities at r2c4 are {5, 7, 9}. Let's consider what would happen if we were to place a "4" at r1c6. Well, we would have to place a "7" at r2c4, because the {5, 9} pair would then appear in r1c5 & r2c5, leaving a "7" as the only possibility at r2c4.

But what if we put a "6" in r1c6? Then the following moves are forced:

1. r4c6 = 4 (because there are only two spots where a "4" can fit in column 6)
2. r5c5 = 2 (because of the {2, 4} pair in the middle center 3x3 box)
3. r5c7 = 5 (because of the {2, 5} pair in the middle right 3x3 box)
4. r4c8 = 2 (for the same reason)
5. r2c8 = 5 (because of the {2, 5} pair in column 8)
6. r2c5 = 9 (only {5, 9} can fit in this spot)
7. r2c4 = 7 (the alternative values 5 & 9 have been eliminated)

So whichever way one fills in r1c6, a "7" must appear at r2c4, qed.

Samgj, the author of these puzzles, seems to like this particular sort of formation -- it came up in another one of his puzzles just a few days ago.

Happy solving! dcb

chobans · Joined: 21 Aug 2005 Posts: 39

I'll try to explain it with elimination method.

At your position, these are the logical steps I took.

In column 1, 6 has to be in box 1(r1c1-r3c3) so you can eliminate 6 from other cells in box 1. So r2c2 becomes {2,5,9}

r2c2-{2,5,9}, r2c4-{5,7,9}, r2c5-{5,9}, r2c8-{2,5}

Yes, your typical naked quadruple. So you can eliminate 2,5,7 and 9 from other cells in row 2.

Since you eliminated the possibility of 7 at r2c6, the ONLY place the 7 can go in row 2 is at r2c4.

Guest · Posted: Fri Sep 23, 2005 5:59 pm Post subject:

David
Thank you for taking the time to reply. Brilliant piece of logic. My wife thinks it's trial and error, however, but I like to think it's considering two possible options which both place a 7 in its correct cell.
Thanks again.
Geoff & June - Durham England.

David Bryant · Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Thanks for the kind words, Geoff. Chobans' answer is actually shorter, and simpler. That seems to be a tendency on this board -- I see a long complex chain of inference, and Chobans sees a simpler, more direct road to the same conclusion.

June will probably like Chobans' method better. dcb Smile

Guest · Posted: Sat Sep 24, 2005 7:59 am Post subject:

Thank you Choban.
Geoff and June

Guest · Posted: Sat Sep 24, 2005 8:54 am Post subject:

There's another way to see this. (In box 1, 6 has to be in column1). In row2, columns 2, 4, 5 and 8 can't have 1 or 6. So that leaves positions 1 and 6 for this pair. Now there's only one position for the 7, in column 4.

someone_somewhere · Joined: 07 Aug 2005 Posts: 275 Location: Munich

you can summarize it:
2 not in r4c1, it is in r6c2 or r6c3 (Row on 3x3 Block interaction)
2 not in r4c2, it is in r6c2 or r6c3 (Row on 3x3 Block interaction)
2 not in r4c3, it is in r6c2 or r6c3 (Row on 3x3 Block interaction)
2 not in r5c1, it is in r6c2 or r6c3 (Row on 3x3 Block interaction)
6 not in r1c3, it is in r1c1 or r2c1 (Column on 3x3 Block interaction)
6 not in r2c2, it is in r1c1 or r2c1 (Column on 3x3 Block interaction)
2 not in r2c1, Hidden Pair 1 6 in r2c1 and r2c6 (in Row)
5 not in r2c1, Hidden Pair 1 6 in r2c1 and r2c6 (in Row)
7 not in r2c6, Hidden Pair 1 6 in r2c1 and r2c6 (in Row)
7 in r2c4 7 in r7c6 - Unique Horizontal

see u,