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December 26 "hard" (rookie solver)

 
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ADBjester



Joined: 28 Dec 2005
Posts: 3

PostPosted: Wed Dec 28, 2005 6:13 am    Post subject: December 26 "hard" (rookie solver) Reply with quote

I am stumped early on in my very first "hard" puzzle. I had little trouble with the Medium, and Easy was... well, easy. But I'm very stuck on this one, and will not move past the hint given until I understand WHY this hint is correct.

I am here thus far:

036 708 000
700 000 900
500 000 370
000 126 754
004 807 200
070 043 800
017 000 092
009 000 000
000 609 480

The Draw tool's hint wants to place a 9 in row 1, column 5... and I just can't figure out why.

I have put in pencilmarks that narrow it down to (1,5,9)... and pencilmarks on EVERY other square in the puzzle. I've triple-checked the pencilmarks, and they're accurate -- I can't eliminate 1 or 5 in this square either vertically or horizontally, nor by logic that I can trace.

They say you don't ever have to "guess", so I'm reluctant to try the tactic of "OK, assume its a 5" and then continue on until I hit a paradox, thus eliminating the 5.

I've looked for associative eliminations in adjoining squares and columns, but just can't fathom why this is the next logical square, and why it must be a 9.

Any hints or help, please? TIA....

Jester
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ADBjester



Joined: 28 Dec 2005
Posts: 3

PostPosted: Wed Dec 28, 2005 6:34 am    Post subject: Never mind Reply with quote

Sneaky!
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Wed Dec 28, 2005 7:05 am    Post subject: December 26 "hard" (rookie solver) Reply with quote

Itís quite a tricky one.

First look at column 8. The 2 and the 4 are already in the bottom two boxes on the right, so 2 and 4 must occupy the top two places, that is they occupy r1c8 and r2 c8 in some order.

Now look at row1. The 6 and the 8 in that row are still missing from the top right-hand box. They cannot go in the top row so they must fill r2c9 and r3 c9.

The only entries unaccounted for in the top right-hand box are now 1 and 5. Again you do not know their exact positions but you do now know they lie in the top row.

Allowing for the 1 and the 5, the only remaining entries in the top row are 2, 4 and 9. So r1c5 must be 2, 4 or 9. As 2 and 4 are already placed in column 5, only 9 will serve.

Steve.
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