(For an explanation of the notation used in the following logical deductions, please consult this thread).

The puzzle that I am talking about is the following one:

- Code: Select all
`*-----------------------*`

| . . . | . 7 . | 9 4 . |

| . . . | . 9 . | . . 5 |

| 3 . . | . . 5 | . 7 . |

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

| . . 7 | 4 . . | 1 . . |

| 4 6 3 | . . . | . . . |

| . . . | . . 7 | . 8 . |

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

| 8 . . | . . . | . . . |

| 7 . . | . . . | . 2 8 |

| . 5 . | 2 6 . | . . . |

*-----------------------*

which was considered until recently (October 2005?) the toughest known puzzle.

After the basic logic have been applied we arrive at the following grid:

- Code: Select all
`*------------------------------------------------------------*`

| 1256 12 1258 | 1368 7 1236 | 9 4 1236 |

| 126 7 1248 | 1368 9 12346 | 2368 136 5 |

| 3 1249 12489 | 168 1248 5 | 268 7 126 |

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

| 259 8 7 | 4 235 2369 | 1 3569 2369 |

| 4 6 3 | 1589 1258 129 | 257 59 279 |

| 1259 129 1259 | 3569 235 7 | 23456 8 23469 |

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

| 8 12349 12469 | 7 1345 1349 | 456 1569 1469 |

| 7 1349 1469 | 1359 1345 1349 | 456 2 8 |

| 19 5 149 | 2 6 8 | 347 139 13479 |

*------------------------------------------------------------*

From here, the solution of this puzzle that I have worked out makes use of the following steps:

1. [r8c3]{=6=[r7c3]=2=[r7c2](-2-[r6c2])-2-[r1c2]-1-[r6c2]}-9-[r3c3]=9=[r3c2]-9-[r6c2],

which means that, if r8c3=9 then r6c2 would be an empty cell. So, r8c3<>9.

2. [r5c4]{-8-[r5c5]=8=[r3c5](-8-[r3c7]=8=[r2c7]-8-[r2c3])=4=[r2c6]-4-[r2c3]}(-8-[r1c4]=8=[r1c3]=5=[r6c3]-5-[r6c4])-8-[r123c4](-3,6-[r6c4]-9-[r6c2])-1,3,6-[r1c6]-2-[r1c2](-1-[r6c2])-1-[r2c3]-2-[r7c3]=2=[r7c2]-2-[r6c2],

which means that r5c4=8 would make r6c2 an empty cell. So, r5c4<>8.

3. [r8c5](-3-[r8c46|r7c6])(-3-[r8c2]=3=[r7c2]=2=[r7c3]=6=[r8c3]-6-[r8c7])-3-[r46c5]-2,5-[r5c46]=(Almost Unique Rectangle:r5c46/r8c467)=2,5|4,5=[r8c467]-4-[r8c2|r9c1]-1,9-[r9c3]-4-[r2c3]=4=[r2c6]-4-[r78c6]-1,9-[r5c6],

which means that r8c5=3 would make r5c6 an empty cell. So, r8c5<>3.

4. In this step, I have made use of a concept that I call "Almost Nice Loops" (this step, and the concept of Almost Nice Loops, are explained in detail in this thread):

[r7c5](-3-[r7c6])(-3-[r7c2]=3=[r8c2])-3-[r46c5](-2,5-[r5c46]){-5-[r56c4]=5=[r8c4](-5-[r8c7])=9=[r8c6](-9-[r7c6])-9-[r5c6]-1-[r7c6]}-{Nice Loop: [r8c3]-4-[r8c5]=4=[r3c5]-4-[r2c6]=4=[r2c3]-4-[r8c3]}-4-[r8c3]-{Nice Loop: [r8c7]-4-[r8c5]=4=[r3c5]-4-[r2c6]=4=[r2c3]-4-[r9c3]=4=[r9c79]-4-[r8c7]}-4-[r8c7]-6-[r8c3]-1-[r8c5]-4-[r7c6],

which means that, if r7c5=3 then r7c6 would be an empty cell. So, r7c5<>3.

5. [r8c4]{-1-[r78c5]=(Almost Unique Rectangle: r78c5/r78c7)=1|6=[r78c7]-6-[r7c8]}(-1-[r5c4]=1=[r5c6]-1-[r12c6])-1-[r123c4](=1=[r3c5]-1-[r3c9])(-6-[r6c4]=6=[r4c6])-3,6-[r1c6]-2-[r1c2]-1-[r1c9]=1=[r2c8]=6=[r4c8]-6-[r4c6],

which means that, if r8c4=1 then r4c6 would be simultaneously "6" and not "6" - a contradiction. So, r8c4<>1.

6. [r78c6]{-1-[r78c5]=(Almost Unique Rectangle: r78c5/r78c7)=1|6=[r78c7]-6-[r7c8]}(-1-[r12c6])-1-[r5c6]=1=[r5c4]-1-[r123c4](-6-[r6c4]=6=[r4c6])(=1=[r3c5]-1-[r3c9])-3,6-[r1c6]-2-[r1c2]-1-[r1c9]=1=[r2c8]=6=[r4c8]-6-[r4c6],

and similarly we have r7c6/r8c6<>1.

7. [r8c5]=1=[r8c23](-1-[r9c3])-1-[r9c1]-9-[r9c3]-4-[r2c3]=4=[r2c6]-4-[r3c5]-2-[r46c5]-5-[r8c5], => r8c5<>5.

8. [r6c7]-5-[r8c7]=5=[r8c4]-5-[r5c4]=5=[r5c78]-5-[r6c7], => r6c7<>5.

9. [r5c7](-5-[r5c4|r4c8])-5-[r5c8](-9-[r9c8|r4c8])-9-[r5c4]-1-[r123c4]-6-[r6c4]=6=[r4c6]-6-[r4c8]-3-[r9c8]=3|4=[r9c3]-4-[r2c3]=4=[r2c6]-4-[r3c5]-2-[r46c5]-5-[r6c4]-9-[r6c123]=9=[r4c1]-9-[r9c1]-1-[r9c8],

from which we can conclude that r5c7=5 would make r9c8 an empty cell. So, r5c7<>5.

10. [r5c9](-9-[r5c46|r4c8])-9-[r5c8](-5-[r4c8])-5-[r5c4](-1-[r5c6]-2-[r1c6])-1-[r123c4](-3,6-[r1c6]-1-[r1c2]-2-[r6c2|r7c2])-6-[r6c4]=6=[r4c6]-6-[r4c8]-3-[r9c8]=3|4=[r9c3](-4-[r78c2])-4-[r2c3]=4=[r2c6]{-4-[r78c6]=(Almost Unique Rectangle: r78c2/r78c6)=4|1=[r78c2]-1-[r6c2]}-4-[r3c5]-2-[r46c5]-5-[r6c4]-9-[r6c2],

and r6c2=empty cell if r5c9=9, and so r5c9<>9.

11. [r45c6]-9-[r78c6]{(-4-[r7c5])-4-[r8c5]-1-[r7c5]-5-[r7c7]=5=[r8c7]=6=[r8c3]-6-[r7c3]}-4-[r2c6]=4=[r2c3](-4-[r7c3])-4-[r9c13](-1,9-[r7c3])-1,9-[r78c2]-(Unique Rectangle: r78c2/r78c6)-3,4-[r7c2]-2-[r7c3],

which means that if either of r4c6/r5c6 is "9" then r7c3 would be an empty cell. So, r4c6/r5c6<>9.

12. In this step I have used some Almost Locked Sets present in the grid:

[r2c3](-4-[r3c3|r3c2])(-4-[r9c13]-1,9-[r9c789]-4-[r78c7]-6-[r236c7])-4-[r2c6]=4=[r3c5]-4-[r78c5]-5-[r46c5]-2-[r4c6]-6-[r4c89]=6=[r6c9](-6-[r13c9]=6=[r2c8]-6-[r2c1|r1c2|r3c2]-1,2,9-[r3c3])=4=[r6c7]-4-[r9c7]-7-[r5c7]-2-[r3c7]-8-[r3c3],

which means that, if r2c3=4 then r3c3 would have no possible candidate. So, r2c3<>4.

13. Now we have a Type-1 Unique Rectangle in cells r78c2/r78c6 that enables us to deduce r7c2<>3,9.

14. [r7c3]{=2=[r7c2](-2-[r6c2])-2-[r1c2]-1-[r6c2]}-9-[r3c3]=9=[r3c2]-9-[r6c2],

and r6c2 would be an empty cell if r7c3=9. So, r7c3<>9.

15. [r7c23](-1-[r7c9])-1-[r7c5]-4-[r7c9]=(Almost Unique Rectangle: r4c89/r7c589)=1,4|3=[r4c9]-3-[r4c5]-5-[r4c1]-9-[r9c1]-1-[r7c23],

which implies r7c2/r7c3<>1.

16. [r3c4]=1=[r3c9]-1-[r7c9]=1=[r7c5]=4=[r8c5]-4-[r8c7]-6-[r3c7]-8-[r3c4],

which implies r3c4<>8 and that solves the puzzle.

Carcul