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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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 Posted: Tue May 25, 2010 6:16 am Post subject: two coffee squiggly |
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From today's paper:
Very few clues, no house has more than three clues.
Enjoy !
Walkthru:
| *** SPOILER *** wrote: |
Since the puzzle starts out with so few givens, I thought I'd give a short tour how to use the Law of Leftovers (LoL) in the early stages.
There are two noteable examples of the Law of Leftovers
LoL_A: the "leftovers" below the imaginary line between rows 5/6 (i.e. cells r6c4 and r6c8) must be the same as the leftovers above that line (cells r45c9). Since r6c4=7, we know that one of r56c9 must be 7.
LoLB: similarly, if one puts the divide between columns 3 and 4, the leftovers in r7c23 must be the same as r2c45 (note that both leftover groups already contain a "4")
Start with row 7
There is a single "4" in r7c8
Naked single "3" r7c7 (because of LoL_A, "7" in this box must be in r45c9)
Single "5" (row) r7c4 (because of LoL_B, since r2c45<>"5", r7c23 cannot be 5)
both r2c45 and r7c23 must contain the same two candidates out of {1,6,7}
Single "9" (box) r6c7, thus
triple {2,6,7} col 9
naked pair 35 r89c9
np 14 r12c9
in column 7:
naked single "1" r3c7
single "8" (col) r8c7
np 57 r24c7
this leaves a triple {3,4,9} in r1c456 and solves
r1c9=1 and r2c9=4
A few singles:
single "7" (row) r1c8
single "9" (col) r9c8
single "1" (box) r7c6
which leaves a naked pair 67 in r7c23 and - because of LoLB - also in r2c45. That solves r2c4=6, r2c5=7 and the puzzle
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Tue May 25, 2010 8:53 am Post subject: |
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| The Law of Leftovers. Genius. |
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