Please refer to row 7 of this puzzle. Note that r7c3 is 2,5.....r7c4 is 2,5,6,9 ...r7c5 is 3,5,6,9 and r8c5 is 3,5,9, I thought I could eliminate 5 in r7c5 because it could "see" 5 in r7c3 as well as r7c4 and r8c5. It would not eliminate. I am fairly new to Sudoku and especially ALS. Could someone please help me see the error of my ways. Thank you.
*-----------*
|...|32.|.76|
|.78|..5|4..|
|...|...|.3.|
|---+---+---|
|.8.|...|295|
|..3|...|6..|
|629|...|.1.|
|---+---+---|
|.1.|...|...|
|..7|4..|16.|
|49.|.78|...|
*-----------*
*-----------*
|...|32.|.76|
|.78|..5|42.|
|...|...|.3.|
|---+---+---|
|.8.|...|295|
|..3|...|6..|
|629|...|.13|
|---+---+---|
|.1.|...|...|
|..7|4..|16.|
|496|178|352|
*-----------*
*--------------------------------------------------------------------*
| 159 45 145 | 3 2 149 | 589 7 6 |
| 139 7 8 | 69 169 5 | 4 2 19 |
| 1259 456 1245 | 6789 14689 14679 | 589 3 189 |
|----------------------+----------------------+----------------------|
| 17 8 14 | 67 1346 13467 | 2 9 5 |
| 157 45 3 | 2789 14589 12479 | 6 48 478 |
| 6 2 9 | 578 458 47 | 78 1 3 |
|----------------------+----------------------+----------------------|
| 2358 1 25 | 2569 3569 369 | 789 48 4789 |
| 2358 35 7 | 4 359 239 | 1 6 89 |
| 4 9 6 | 1 7 8 | 3 5 2 |
*--------------------------------------------------------------------*
Help with ALS problem
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• Page 1 of 1
Re: Help with ALS problem
by ronk » Fri Jun 29, 2007 7:09 pm
Gee wrote:Note that r7c3 is 2,5.....r7c4 is 2,5,6,9 ...r7c5 is 3,5,6,9 and r8c5 is 3,5,9, I thought I could eliminate 5 in r7c5 because it could "see" 5 in r7c3 as well as r7c4 and r8c5. It would not eliminate. I am fairly new to Sudoku and especially ALS.
- Code: Select all
*--------------------------------------------------------------------*
| 159 45 145 | 3 2 149 | 589 7 6 |
| 139 7 8 | 69 169 5 | 4 2 19 |
| 1259 456 1245 | 6789 14689 14679 | 589 3 189 |
|----------------------+----------------------+----------------------|
| 17 8 14 | 67 1346 13467 | 2 9 5 |
| 157 45 3 | 2789 14589 12479 | 6 48 478 |
| 6 2 9 | 578 458 47 | 78 1 3 |
|----------------------+----------------------+----------------------|
| 2358 1 *25 |*2569 *3569 369 | 789 48 4789 |
| 2358 35 7 | 4 *359 239 | 1 6 89 |
| 4 9 6 | 1 7 8 | 3 5 2 |
*--------------------------------------------------------------------*
There is no combination of the cells you list that make up an ALS, which is characterized by N+1 candidates in N cells. In your pencilmarks, the following are ALSs:
{r23c9} = {189}, 3 candidates in 2 cells
{r5c89} = {478}, 3 candidates in 2 cells
{r7c789} = {4789}, 4 candidates in 3 cells
Moreover, you need a minimum of two ALSs with two like candidates before an elimination is possible.
Recommend you use this forum's search feature using "xz-rule" as a keyword.
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by Steve R » Fri Jun 29, 2007 7:20 pm
You need two ALSs to make eliminations. What’s more they must have a restricted common candidate, that is a candidate for both sets of cells which can be entered into only one set.
I couldn’t see this set up in your grid, perhaps because I didn’t look for long enough. Instead I found one a little later on in the solution path.
After using the many singles available the grid becomes:
Take A = {r2c4}. One cell with two candidates, (69), so an ALS.
Take B = {r7c4, r8c6}. Two cells with three candidates, (269), so an ALS.
6 is a common candidate but cannot be entered in both sets because the fourth column would then have two 6s. Thus 6 is an rcc for A and B.
If some other common candidate were to be eliminated from both sets, 6 would be forced into both simply to make the number of candidates for each equal to the number of cells in it. But this is exactly what would happen if r1c6 contained a 9: the ALSs permit 9 to be eliminated from r1c6.
This is often called the xz-rule. When bennys developed the principle he used x for the restricted common candidate and z for the candidate which was eliminated. More formally, z may be eliminated from any cell which “sees” all the cells open to z in A and B combined.
Steve
I couldn’t see this set up in your grid, perhaps because I didn’t look for long enough. Instead I found one a little later on in the solution path.
After using the many singles available the grid becomes:
- Code: Select all
+------------------------------------+
| 159 4 15 | 3 2 19 | 8 7 6 |
| 3 7 8 | 69A 169 5 | 4 2 19 |
| 129 6 12 | 8 4 7 | 5 3 19 |
--------------------------------------
| 17 8 4 | 67 136 136 | 2 9 5 |
| 17 5 3 | 279 19 129 | 6 8 4 |
| 6 2 9 | 5 8 4 | 7 1 3 |
--------------------------------------
| 8 1 25 | 26B 356 36 | 9 4 7 |
| 25 3 7 | 4 59 29B | 1 6 8 |
| 4 9 6 | 1 7 8 | 3 5 2 |
+------------------------------------+
Take A = {r2c4}. One cell with two candidates, (69), so an ALS.
Take B = {r7c4, r8c6}. Two cells with three candidates, (269), so an ALS.
6 is a common candidate but cannot be entered in both sets because the fourth column would then have two 6s. Thus 6 is an rcc for A and B.
If some other common candidate were to be eliminated from both sets, 6 would be forced into both simply to make the number of candidates for each equal to the number of cells in it. But this is exactly what would happen if r1c6 contained a 9: the ALSs permit 9 to be eliminated from r1c6.
This is often called the xz-rule. When bennys developed the principle he used x for the restricted common candidate and z for the candidate which was eliminated. More formally, z may be eliminated from any cell which “sees” all the cells open to z in A and B combined.
Steve
- Steve R
- Posts: 74
- Joined: 03 April 2006
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