Almost Locked Sets
Posted: Mon Oct 02, 2006 5:10 pm
I'm still trying to understand the "XZ rule" thoroughly.
A. r1c1 = 4 ==> r2c1 = 6 ==> r2c8 = 4 ==> r4c7 = 4 ==> r1c7 = 1
B. r1c1 = 4 ==> r1c3 = 1
So clearly r1c1 <> 4, and we must have r1c1 = 6.
Since I'm trying to gain a better understanding of the "ALS XZ rule", I went back and tried to cast this exclusion in those terms. This is what I found.
So the only candidates for {r2c1, r2c2, r8c1} are v=3469, the only candidates for {r1c3, r1c7, r3c3} are v=1345, and "3" is restricted common. Since r1c1 can "see" all the instances of "4" in both sets, the rule says I can exclude "4" from r1c1. Everything seems dandy.
But then I got to thinking about the proof that lies behind the "ALS XZ rule". In this instance it doesn't really apply, because I can complete the two ALS's as follows.
If r1c1 = 4 I can set r8c1 = 9, r2c1 = 6, r2c2 = 9
Also, r1c3 = 1, r3c3 = 3, r1c7 = 5
and no contradiction is yet apparent, because the "locked common" value, "3", can "escape" into r7c2. I have to extend the argument farther, along the lines identified as a "DIC" above, before the contradiction is forced.
I'm thinking the reason the proof fails here is that the definition
So I was working the "nightmare" puzzle for Monday, 2 October, 2006. Some fairly basic stuff got me to this point, where I applied a "Double Implication Chain".Solving Guide wrote:An almost locked set consists of N cells with exactly N+1 candidates.
Code: Select all
*-----------------------------------------------------------*
| 46* 7 14B | 9 25 3 | 145A 8 2456 |
| 469A 39 5 | 278 1 78 | 34 46A 27 |
| 8 2 13 | 4 57 6 | 9 15 1357 |
|-------------------+-------------------+-------------------|
| 57 14 2 | 36 36 1457 | 145A 9 8 |
| 3 14 6 | 15 8 9 | 7 2 145 |
| 57 8 9 | 1257 27 1457 | 6 3 145 |
|-------------------+-------------------+-------------------|
| 1 39 8 | 56 4 2 | 35 7 569 |
| 49 5 7 | 1368 36 18 | 2 146 1469 |
| 2 6 34 | 157 9 157 | 8 145 1345 |
*-----------------------------------------------------------*B. r1c1 = 4 ==> r1c3 = 1
So clearly r1c1 <> 4, and we must have r1c1 = 6.
Since I'm trying to gain a better understanding of the "ALS XZ rule", I went back and tried to cast this exclusion in those terms. This is what I found.
Code: Select all
*-----------------------------------------------------------*
| 46* 7 14+ | 9 25 3 | 145+ 8 2456 |
| 469- 39- 5 | 278 1 78 | 34 46 27 |
| 8 2 13+ | 4 57 6 | 9 15 1357 |
|-------------------+-------------------+-------------------|
| 57 14 2 | 36 36 1457 | 145 9 8 |
| 3 14 6 | 15 8 9 | 7 2 145 |
| 57 8 9 | 1257 27 1457 | 6 3 145 |
|-------------------+-------------------+-------------------|
| 1 39 8 | 56 4 2 | 35 7 569 |
| 49- 5 7 | 1368 36 18 | 2 146 1469 |
| 2 6 34 | 157 9 157 | 8 145 1345 |
*-----------------------------------------------------------*But then I got to thinking about the proof that lies behind the "ALS XZ rule". In this instance it doesn't really apply, because I can complete the two ALS's as follows.
If r1c1 = 4 I can set r8c1 = 9, r2c1 = 6, r2c2 = 9
Also, r1c3 = 1, r3c3 = 3, r1c7 = 5
and no contradiction is yet apparent, because the "locked common" value, "3", can "escape" into r7c2. I have to extend the argument farther, along the lines identified as a "DIC" above, before the contradiction is forced.
I'm thinking the reason the proof fails here is that the definition
is not quite restrictive enough. I'm not quite sure how it should be phrased, but it seems that cases like this -- where the two instances of candidate "9" in the set marked with a minus sign can't "see" each other -- need to be excluded somehow. dcbAn almost locked set consists of N cells with exactly N+1 candidates