Stuck on this puzzle

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Stuck on this puzzle

Postby Terri » Tue Aug 16, 2005 4:11 pm

I have been staring at this puzzle for days, and I can't figure out what to do to next. It was in the Friday, August 12th Boston Globe. Here is the puzzle so far:

* 5 * | * 1 * | * 2 *
1 * * | * * * | 5 * 9
* * * | 5 * 7 | * * *

6 7 * | 1 9 5 | * 8 2
* 1 5 | * * * |7 * *
8 2 * | 7 3 * | * 1 5

5 * * | 4 7 1 | * * *
4 * * | * 5 * | * * 3
* 9 * | * 8 * | * 5 *

I can see the naked pair in row 4 and the naked triple in column 5. I also see that the 8 and 2 must go in the middle row of box 5, but I am missing something! What step do you suggest I should try next?
Terri
 
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Postby The Central Scrutinizer » Wed Aug 17, 2005 3:36 am

Terri, this is my first time giving advice, so I hope my explanation isn't too muddled.

{Edited to remove a missed possibility}

One thing that can be done is in the center row (row 5):

{3,9} {1} {5} {2,6,8} {2,4,6} {2,4,6,8} {7} {3,4,6,9} {4,6}

Row 5 has a hidden pair of 3 9 that can only be found in cell 1 and 8, so you can eliminate the 4 & 6 from the choices in cell 8.


There's lots of neat things going on in this puzzle.

Try the middle 3x3 box on the bottom (box #8):

It eliminates down to looking like this


Code: Select all
{4}        {7}    {1}
{2,6,9}    {5}    {2,6,9}
{2,3,6}    {1}    {2,3,6}



The only two places that the 3 can reside in this box is in the bottom row (row 9)...

...which looks like this:

{2,3,7} {9} {1,2,3,6,7} {2,3,6} {8} {2,3,6} {1,2,4} {5} {1,4,7,6}

But you know the 3 for row 9 is in either column 4 or 6, so you can eliminate it from c1 & c3.

{2,7} {9} {1,2,6,7} {2,3,6} {8} {2,3,6} {1,2,4} {5} {1,4,7,6}

Looking back at box #8, you can do the same thing for the 9, which must be in row 8. You can remove all possibilities of the 9 from row8 c7-8.

Doing this hopefully gets you going!

{Note}Annette's post below is right. My initial post about Box#6 omitted a possibility of another 6, so I had to edit it. The other stuff rings true.

I'm very sorry.
Last edited by The Central Scrutinizer on Sat Aug 20, 2005 10:55 am, edited 1 time in total.
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Postby Anette » Wed Aug 17, 2005 8:02 pm

Scrutinizer, how did you eliminate 6 from R6C7?
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Postby The Central Scrutinizer » Wed Aug 17, 2005 9:44 pm

Annette, that's a really good question and, at this point, I'm not sure. It's possible the first thing I talked about above wasn't the first thing I did. I'll have to look at the pencil smeared tattered piece of paper I did this on.

Gosh, if I gave bad advice or anything based on faulty logic or omission, I'm truly sorry.:( I was trying to be methodical and keep track of my steps.

The other stuff I said pans out. I think.
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Postby Anette » Wed Aug 17, 2005 10:02 pm

I still had the 6 in that cell, but managed to eliminate the 4 in R5C8 based on the hidden pair

After that, I could elimintate the 4's in box 3 in C7 and C9 based on the fact that all candidate 4's in C8 now were in box 3.
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Postby Rickh1 » Fri Aug 19, 2005 11:00 pm

Scrutinizer Wrote:
Row 5 has a hidden pair of 3 9 that can only be found in cell 1 and 8, so you can eliminate the 4 & 6 from the choices in cell 8.


What is the name of the technique you used in order to come up with this deduction? I'm brand new this week and trying to learn.
Thanks, Rick
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Postby The Central Scrutinizer » Sun Aug 21, 2005 12:16 pm

Rick, technique names, I'm not very good at. I'm only a 2 week veteran myself. I think this is what's called a hidden pair, and I can at least try and explain the logic behind it.

That row in question:

{3,9} {1} {5} {2,6,8} {2,4,6} {2,4,6,8} {7} {3,4,6,9} {4,6}

How many places can you put the number 3 ? (two, only cells one and eight)


How many places can you put the number 9 ? (only two: cells one and eight)

You have two numbers (3 & 9) that you know both go in the same two cells. Knowing which number goes where isn't important. What is important is knowing that since those two numbers must go in those two cells, than no other numbers can possibly be in those two cells. Any other number possibilities in those two cells are just camoflauge and can be eliminated:

{3,9} {1} {5} {2,6,8} {2,4,6} {2,4,6,8} {7} {3,4,6,9} {4,6}

Take cell eight for instance. Let's say you put the 6 there. Then you put the 9 in cell one. Now there's nowhere to possibly put the 3.

If you put the 4 in cell eight, then throw the 3 into cell one, again, you have nowhere the 9 can go.

When you're newbies like we are, I have no problem justifying working through a logic step in this fashion. I think the good puzzlers here call that 'Forcing Chains,' or--horror of horrors--'Trial and Error.'

But being new, I have to sometimes prove to myself that a piece of logic works. I can read as many of the descriptions of these complicated fancy moves until my eyes glaze over, but there is nothing better than seeing it in action. I guess it's from my time in Missouri: Show me, seeing is believing.

Now, I've got Saturday's Sun-Times puzzle to work on. It's the 'Hard' one for the week, and last Saturday's took me a few hours. Fun fun fun!
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Postby silverpie » Tue Aug 30, 2005 7:51 pm

The Central Scrutinizer wrote:Rick, technique names, I'm not very good at. I'm only a 2 week veteran myself. I think this is what's called a hidden pair, and I can at least try and explain the logic behind it.

That row in question:

{3,9} {1} {5} {2,6,8} {2,4,6} {2,4,6,8} {7} {3,4,6,9} {4,6}

How many places can you put the number 3 ? (two, only cells one and eight)

How many places can you put the number 9 ? (only two: cells one and eight)

You have two numbers (3 & 9) that you know both go in the same two cells. Knowing which number goes where isn't important. What is important is knowing that since those two numbers must go in those two cells, than no other numbers can possibly be in those two cells. Any other number possibilities in those two cells are just camoflauge and can be eliminated:

{3,9} {1} {5} {2,6,8} {2,4,6} {2,4,6,8} {7} {3,4,6,9} {4,6}

Take cell eight for instance. Let's say you put the 6 there. Then you put the 9 in cell one. Now there's nowhere to possibly put the 3.

If you put the 4 in cell eight, then throw the 3 into cell one, again, you have nowhere the 9 can go.


Ah yes, that is indeed a hidden pair. It's also a naked quad (2468 are the only possibilities for cells four/five/six/nine, so none of the other cells can have any of those). In fact, any time you have a hidden, the opposite cells form a naked something.
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Postby native » Thu Sep 08, 2005 11:01 am

356 914 827
147 328 569
289 567 341

673 195 482
915 842 736
824 736 915

532 471 698
468 259 173
791 683 254
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Postby hwnag99 » Sun Sep 11, 2005 10:34 pm

Hi, this is another newbee. I played with this puzzle for whole Saturday afternoow. Guessed 4 times and got it solved. But latter I found that it seems we are not supposed to GUESS, we should use LOGIC to solve the Su Doku puzzle. Is this true ? If it is, can some expert show me how to get the first number out ? ( any one will do ) My level is, after all the effort, I cannot put donw any number onto any cell. I start to guess 3 at R4C3. Any instruction, tip, rule will be highly appreciated.

Thanks in advance.
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Postby victor » Sun Sep 25, 2005 6:31 pm

Saw this while glancing thro' old problems.
(a) One simple way to see the 39 pair in row 5 is to note that they CAN'T go in columns 4,5,6 or 9.
(b) Look at the 3 right hand boxes. We knew that in the bottom one (box 9) 4 had to be in columns 7 or 9. In the middle one (box 6), we now know that the same thing's true about 4s. SO: in the top box (box 3), 4 must come in the middle column (c8). We also know that 4 has to be in the middle column in box 1. SO: in box 2 it's in row 1, and that fixes it as r1c6.
(c) Etc.!
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