bennys wrote:The only part that I agree is that you don't get the ALS method.
You disagree with everything else I suggested? The forcing chains I gave? The nice loop? That any bivalue cell is an Almost Locked Set? That there are dozens of Almost Locked Sets to choose from?
Carcul wrote:Hi Tso.
Tso wrote:I can't see how identifying these "almost locked sets" helps find a solution
In the case of this particular puzzle, it is very easy to find a solution without the need for ALSs, and is probably "harder" to find the "correct" combination of AlSs, because there are so many to choose from.
If this is the case, how is it different from describing this...
[12][23][31]|[..][..][..]|[..][..][..]
... as a continuous nice loop --> [r1c1]-2-[r1c2]-3-[r1c3]-1-[r1c1] --> that eliminates 1s, 2s and 3s from the rest of the row? Sure, its true, but all human solvers will find it easier to identify this as a simple naked triple.
Carcul wrote:However, in extremely difficult puzzles, we will not get very far with nice loops, and, in my opinion, it is in these cases where the ALSs can be very helpful. ... I guess one of the "secrets" is, when solving a puzzle, to keep track of some ALSs and then try to relate them with cells or/and with each other.
Regards, Carcul
Ok, but where are the algorithms? There are several methods of tracking down finding forcing chains that have been described in these forums. The ALS tactics do not seem to be a simple pattern, nor have any methods of *finding* them been described. And unlike forcing chains in which I must only hold two pieces of information in my head at once, ALS seems to require me to idenify and hold in my head, SET A, SET B, maybe SET C, D, x, y and z ... oh, wait, I see, it's so *obvious* to me now. How are you finding x, y and z?
Look at one of bennys original posts,
Almost locked rules (for now)bennys wrote:- Code: Select all
Almost locked rules
-------------------
ALS - almost locked set
For two sets A,B that have common candidate x we will say that x is restricted common if it can't be
in A and B in the same time.(being in A exclude it from B)
Almost locked sets xz rule
--------------------------
If
1)A,B ALS
2) x is restricted common
Then Any other common candidate (lets say z) cant appear outside of A and B if it can see all the z candidates in both A and B.
The reason?
If z appear then both A and B are locked but only one can get the x.
+-------------+-------------+-------------+
| 24 7 8 | 24 6 5 | 1 9 3 |
| 9 3 24 | 248 1 48 | 56 7 56 |
| 5 1 6 | 7 3 9 | 8 4 2 |
+-------------+-------------+-------------+
| 28 9 23 | 458 7 6 |#35 1 #45 |
|*17 6 5 | 3 9 14 | 2 8 #47 |
| 178 4 *13 | 58 2 18 | 357 6 9 |
+-------------+-------------+-------------+
| 6 5 7 | 1 4 2 | 9 3 8 |
| 14 2 14 | 9 8 3 | 67 5 67 |
| 3 8 9 | 6 5 7 | 4 2 1 |
+-------------+-------------+-------------+
A={R5C1,R6C3}
B={R4C7,R4C9,R5C9}
x=7
z=3
This seems like we're arbitrarily creating more specific patterns and adding complications where a general pattern already exists that is sufficient to explain these exclusions more easily. The 5 cells marked form 5/6 of a forcing chain that will exclude a 3 from both r4c3 and r6c7.
r6c3=1 => r5c1=7 => r5c9=4 => r4c9=5 => r4c7=3 => (r4c3<>3 AND r6c7<>3)
r6c3=3 => (r4c3<>3 AND r6c7<>3)
Therefore, (r4c3<>3 AND r6c7<>3)
Or, written as nice loops (I've marked the strong links, though, as this is a pure bivalue chain, no strong links are required):
[r6c3]-1-[r5c1]=7=[r5c9]=4=[r4c9]-5-[r4c7]=3=[r4c3]=3=[r6c3]
[r6c3]-1-[r5c1]=7=[r5c9]=4=[r4c9]-5-[r4c7]=3=[r6c7]=3=[r6c3]
Here's the partial grid with the extraneous removed:
- Code: Select all
+-------------+-------------+-------------+
| . . 2x3 | . . . |#35 . #45 |
|*17 . . | . . . | . . #47 |
| . . *13 | . . . |x357 . . |
+-------------+-------------+-------------+
Solving by ALS requires that I identify the cells marked # as an ALS and those with an * as another -- exclusions marked with 'x'. How these were chosen from among the other combinations using these same cells, I don't know. (Just in box 6, I count three 2-cell ALS and four 3-cell ALS -- and 3 one-cell ALS.)
But:
- Code: Select all
+-------------+-------------+-------------+
| . . 2x3 | . . 35 | . . 45 |
|*17 . . | . . . | . . 47 |
| . . *13 | . . x357 | . . . |
+-------------+-------------+-------------+
The exact same forcing chains / nice loops exist making the same exclusions for the same reasons --
but the arbitrary # ALS is gone! ... implying that the ALS tactic was coincidental.
Actually, each consecutive pair of cells in the chain, taken together form an ALS.
The ALS method of solving this puzzle requires me keep four bits of information in my head at once the sets A and B and the value and location of x and z.
