Jeff wrote:Neunmalneun wrote:Based on the uniqueness rule we know that r5c8=6 or one of the cells r9c8/r9c9=4. The combination of the BUG principle with the uniqueness rule seems to give a valid hint to r9c8=4 (which solves the puzzle easily)
I can understand the bit with the uniqueness rule. Could you explain how the BUG principle helps to conclude that r9c8=4?
Hi Jeff, since
Neunmalneun hasn't yet responded, let me take a crack at it.
There are nine non-BUG candidates, at least one of which must be TRUE to prevent a BUG grid. Three of them are r5c8=3, r9c8=4, and r9c9=3.
- Code: Select all
7 49 6 | 5 1 3 | 48 89+4 2
14 13+49 2 | 8 6 49 | 7 5 39+4
5 39+4 8 | 7 49 2 | 34 1 6
----------------+-----------------+-----------------
3 6 7 | 9 2 5 | 18+4 48 14
9 2 5 | 4 8 1 | 36 67+3 37
48 48 1 | 6 3 7 | 9 2 5
----------------+-----------------+-----------------
6 7 39 | 1 5 49 | 2 34 8
2 5 4 | 3 7 8 | 16 69 19
18 18 39 | 2 49 6 | 5 37+4 47+3
Coincidentally, these same three cells are part of an almost unique rectangle for which we have the AUR grid ...
- Code: Select all
7 49 6 | 5 1 3 | 48 48+9 2
14 1349 2 | 8 6 49 | 7 5 394
5 394 8 | 7 49 2 | 34 1 6
--------------------------------------------------
3 6 7 | 9 2 5 | 48+1 48 14
9 2 5 | 4 8 1 | 36 37+6 37
48 48 1 | 6 3 7 | 9 2 5
--------------------------------------------------
6 7 39 | 1 5 49 | 2 34 8
2 5 4 | 3 7 8 | 16 69 19
18 18 39 | 2 49 6 | 5 37+4 37+4
Since r9c8=4 is the only common placement preventing both the BUG and the AUR grid, Neunmalneun's deduction is the placement is valid. His deduction certainly seems logically correct to me.
[edit 1 & 4: corrected three typos in the above]
[edit 2 and 3: added and amended the following clarification]
[edit 5:
CORRECTION: That r9c8=4 was a correct placement was a "happy accident" ... the equivalent of a guess ... because the logic was faulty. Read on if you wish to know where the logic flaw occurred.]
In general, given that ...
- at least one member of set AUR must be true, and
- at least one member of set BUG must be true, and
- the intersection of sets AUR and BUG contain exactly one member
... then that single member of the intersection must be true.
For this example, specifically ...
AUR = {r5c8=6, r9c8=4, r9c9=4} (of which
at least one must be true), and
BUG = {r5c8=3, r9c8=4, r9c9=3, ..., ..., } (of which
at least one must be true)
The only set member common to (in the intersection of) sets AUR and BUG is r9c8=4. Therefore r9c8=4 must be true.
That set BUG also contains other unlisted members is irrelevant. The unlisted members of set BUG are for cells that are not common to set AUR, and would thus not be contained in the intersection anyway.
[edit 5:
CORRECTION: r9c8=4 is indeed the only set member common to the intersection of sets AUR and BUG, but taking the intersection of AUR and BUG is only valid if the criteria were ...
AUR = {r5c8=6, r9c8=4, r9c9=4} (of which
exactly one must be true), and
BUG = {r5c8=3, r9c8=4, r9c9=3, ..., ..., } (of which
exactly one must be true)
But the criteria is "at least one", not "exactly one", so both the method and conclusion were invalid. Just as likely or even more likely, the correct placements might have been r5c8=6 and/or r9c9=4 to fulfill the AUR set requirement, and one or more of the unlisted placements of the BUG set.]
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[edit: Corrections added to the original post 1-12-06]