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.----------.---------------------.--------------------.
| 4 8 3 | 1 2 5 | 7 6 9 |
| 7 5 2 | 6 9 8 | 1 4 3 |
| 1 6 9 | 4 3 7 | 2 8 5 |
:----------+---------------------+--------------------:
| 8 2 4 | 39 7 6 | b39 5 1 |
| 3 7 16 | 259 145 124 | 4689 b29 2468 |
| 5 9 16 | 8 14 a12[3]4 | b346 7 246 |
:----------+---------------------+--------------------:
| 9 1 8 | 25 456 24 | 456 3 7 |
| 6 3 5 | 7 48-1 9 | 48 c2(1) 248 |
| 2 4 7 | 35 1568 a[1]3 | 5689 9-1 68 |
'----------'---------------------'--------------------'
(1,3)r96c6 = (392)b6p715 - (2=1)r8c8 => -1 r8c5,r9c8; stte
uncompressed, slightly different: Show (1=3)r9c6 - r6c6 = r6c7 - (3=9)r4c7 - r9c7 = (9)r9c8 => -1 r9c8; stte
Just for fun... As many probably know, the
Apollo Guidance Computers were implemented with just NOR gates. If they got us to the moon, they should probably work for sudoku too. Here's my (hidden) solution with NOR ( ':' ) gates only.
NORs only:
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(1r9c6:3r9c6) : ((3r9c6:3r9c6):(3r6c6:3r6c6)) : (3r6c6:3r6c7) : ((3r6c7:3r6c7):(3r4c7:3r4c7)) : (3r4c7:9r4c7) : ((9r4c7:9r4c7):(9r9c7:9r9c7)) : (9r9c7:9r9c8) => -1 r9c8; stte
Interestingly, if NOT gates are also allowed, it's exactly the same as the normal AIC form (as a product of sums) with both ORs and ANDs replaced with NORs.
ORs, ANDs, NOTs:
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(1r9c6|3r9c6) & (-3r9c6|-3r6c6) & (3r6c6|3r6c7) & (-3r6c7|-3r4c7) & (3r4c7|9r4c7) & (-9r4c7|-9r9c7) & (9r9c7|9r9c8) => -1 r9c8; stte
NORs, NOTs:
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(1r9c6:3r9c6) : (-3r9c6:-3r6c6) : (3r6c6:3r6c7) : (-3r6c7:-3r4c7) : (3r4c7:9r4c7) : (-9r4c7:-9r9c7) : (9r9c7:9r9c8) => -1 r9c8; stte
That's probably the simplest boolean form for an AIC! I wouldn't have guessed.