DAJ: SK Loops don't use pure AIC logic and are a special case that works only when a closed loop is formed.

To use your diagram:

- Code: Select all
` +--------------------------------------------------------------------------------+`

| *1 48+7 34578 | 3567 3689 5678 | 3489 39+6 *2 |

| 38+2 9 38+7 | 4 12368 12678 | 38+1 5 38+6 |

| 23458 48+2 *6 | 1235 12389 1258 | *7 39+1 3489 |

|--------------------------+--------------------------+--------------------------|

| 2468 5 1478 | 9 1246 3 | 128 1267 678 |

| 234689 12468 13489 | 126 7 1246 | 123589 12369 35689 |

| 2369 1267 1379 | 8 5 126 | 1239 4 3679 |

|--------------------------+--------------------------+--------------------------|

| *7 48+1 14589 | 1235 12348 12458 | *6 39+2 3459 |

| 45+6 3 45+1 | 12567 1246 9 | 45+2 8 45+7 |

| 45689 48+6 *2 | 3567 3468 45678 | 3459 39+7 *1 |

+--------------------------------------------------------------------------------+

(48=27)r13c2 - (27=38)r2c13 - (38=16)r2c79 - (16=39)r13c8 -

(39=27)r79c8 - (27=45)r8c79 - (45=16)r8c13 - (16=48)r79c2 - loop =>

r2c5<>38; r5c2<>48; r5c8<>39; r7c3<>1; r3c1<>2; r8c5<>4; r8c4<>5; r9c1<>6; r1c3<>7; r2c6<>8

To take the eliminations made by the second weak link in your diagram:

Specifiying the truths needed to consider a node true can be written two ways as the loop is followed

(27#2=38#1)r2c13 - (38#2=16#1)r2c79 or (27#1=38#2)r2c13 - (38#1=16#2)r2c79

One of these will eliminate (38)r2c5 and (8)r2c6 whichever way the counts are alternated when both the digits in one of the pairs are true.

But there is another possibility namely that every term has one truth. Now it still works because the truth in the repeating pair (38) can't be for the same digit in r2c13 as it is in r2c79.

Steve K's find was therefore brilliant, but you can't hold its notation up as as an example of a valid AIC.

Now does that ring the bells for you? It should do.

PS Just in case: (27#2) will be false when one or both digits are false, (38#1) will be true when one or both digits are true.