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Guest
Guest
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 Posted: Fri Sep 16, 2005 7:39 pm Post subject: Problems from archives |
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Apologies for the old problems but I'm trying to improve my understanding of sudoku but cannot understand the hint to the foll two puzzles from the archives:
17th March
xxx 91x xxx
3xx x2x x1x
175 xx8 629
x52 xxx 496
6xx 5x9 237
xxx x6x 185
5x6 8xx 94x
x8x x5x xx2
xxx x94 xxx
the hint is 1 in r5c3.Why?
Second problem is more recent 7th August
8x1 xx4 7xx
xx5 7xx x38
6x7 9x8 45x
5x9 xxx xxx
x7x x9x x15
xxx xx5 3x9
x52 1x7 xx4
78x xx9 xxx
xx6 8xx 5xx
the hint is 1 in r4c6.
Thanks for any help |
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chobans
Joined: 21 Aug 2005
Posts: 39
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| Posted: Fri Sep 16, 2005 9:38 pm Post subject: Re: Problems from archives |
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| Guest wrote: | Second problem is more recent 7th August
8x1 xx4 7xx
xx5 7xx x38
6x7 9x8 45x
5x9 xxx xxx
x7x x9x x15
xxx xx5 3x9
x52 1x7 xx4
78x xx9 xxx
xx6 8xx 5xx
the hint is 1 in r4c6. |
In box 1(r1c1-r3c3), 3 can ONLY go in column 2. So we can eliminate 3 in other cells in column 2. This leaves r4c2 with possibilities of 1,2,4 or 6.
In box 4(r4c1-r6c3), 3 can ONLY go in row 5. So we can eliminate 3 in other cells in row 5. This leaves r5c4 with 2,4,6 and r5c6 with 2,6.
Box 5(r4c4-r6c6): r5c4 - {2,4,6}, r5c6 - {2,6}, r6c4 - {2,4,6}.
That makes it a naked triple. So we can eliminate 2,4,6 from other cells in box 5. This leaves r4c4 with ONLY possibility of 3. Then you get 1 in r4c6. |
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chobans
Joined: 21 Aug 2005
Posts: 39
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| Posted: Fri Sep 16, 2005 11:33 pm Post subject: Re: Problems from archives |
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| Guest wrote: | 17th March
xxx 91x xxx
3xx x2x x1x
175 xx8 629
x52 xxx 496
6xx 5x9 237
xxx x6x 185
5x6 8xx 94x
x8x x5x xx2
xxx x94 xxx
the hint is 1 in r5c3.Why? |
I had to think a bit about this one.
In row 7, 7 can ONLY go in box 8(r7c4-r9c6). So you can eliminate 7 from the other cells in box 8. Both r8c4 and r8c6 becomes 1,3,6 after this.
Row 8: r8c4 - {1,3,6}, r8c6 - {1,3,6}, r8c7 - {3,7}, r8c8 - {6,7}
Yes, it's a naked quadruple. So you can eliminate 1,3,6 and 7 from other cells in row 8. r8c3 changes from {1,3,4,7,9} to {4,9}.
Column 3: r1c3 - {4,8}, r2c3 - {4,8,9}, r8c3 - {4,9}
That makes it a naked triple, so you can eliminate 4,8 and 9 from other cells in column 3. So this reduces r5c3 from {1,4,8} to 1. |
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Guest
Guest
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| Posted: Sun Sep 18, 2005 7:14 am Post subject: |
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| Thanks so much. I was puzzling over those for a long time. |
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