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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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 Posted: Mon Sep 13, 2010 7:09 pm Post subject: ADP External Inferences |
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Recently, more players on this forum have been using steps involving potential URs or Almost URs (AURs). These posts typically evaluate the extra or non-AUR digits in the AUR cells to determine if either pincer-like deletions or forced cell values are possible. However another approach is available using the AUR digits in the cells external to the AUR that share a house with the AUR cells. I have unsuccessfully searched the general Soduko community for a discussion thread on this topic. If anyone is aware of such a post, please yell out!
The following is my attempt to begin a discussion on using the external technique on all forms of "Almost Deadly Patterns" (ADPs). It represents my current understanding and is assuredly incorrect, incomplete, confusing, etc in many aspects; it definitely should not be taken as "gospel". These perspectives are derived from viewing posts on various Sudoku forums and from constructive feedback by several folks including Danny (DAJ95367), Luke(Luke451), Norm (storm_norm) and Ron (ronk). I have simply attempted to compile that assortment of information and add examples for this post.
Definitions
Internal, extra or naked: Refers to the non-ADP digits within the ADP cells.
External or hidden: Refers to the ADP digits that reside in non-ADP cells that "see" (share a house with) the ADP cells.
Principles
The overall approach to prevent the Deadly Pattern (ADP) is that at least one of the digits of the ADP must not be present in the final solution; that is, at least one ADP digit must be false. This can be achieved by working with the extra digits in the ADP cells or by working with ADP digits in cells external to the ADP that "see" the all the ADP cells. Here, we will consider using the external cells, but for clarity and reference, the examples with generally include both approaches.
The first External Principle is that the implication of ALL of the ADP digits must be considered since it is not possible to determine a priori what ADP digit(s) will be false.
The second External Principle is that only one house, (row, column or box) needs to be considered for each ADP digit within the ADP cells, but all external occurrences of the ADP digit in that house must be included in the inference analysis.
The third External Principle is that you may use different houses for each ADP digit.
The fourth External Principle is that an external candidate cell may be used for more that one ADP cell.
The fifth External Principle is that you may also use an internal, extra digit within an ADP instead of using an external inference. When used, an internal digit covers both ADP digits in that ADP cell.
In summary, all ADP digits must be considered and you may independently select any row, column, box or extra digit within a ADP cell for each ADP digit; mix & match at your discretion and to your advantage to gain maximum results.
Again, note that these guidelines apply to all ADPs including Type N URs, AURs, ABUG-Lites and AMUGs.
Examples
Most of the following examples are puzzles from this forum, "dailysudoku.com". They have been selected to demonstrate the use of external techniques in a variety of ways and to highlight the flexibility offered by this method.
Example 1
The following code, which is after completing basics, contains a Type 1 UR (58)r13c47, marked *.
| Code: |
*-----------------------------------------------------------*
| 9 7 6 |*58 34 1 |*3458#@3458 2 |
| 8 5 14 | 34 2 9 | 134 7 6 |
| 2 3 14 |*58 6 7 |*58 9 14 |
|-------------------+-------------------+-------------------|
| 1 6 5 | 2 7 34 | 9 34 8 |
| 3 4 9 | 1 8 5 | 6 2 7 |
| 7 8 2 | 9 34 6 |#1345 345 134 |
|-------------------+-------------------+-------------------|
| 46 9 38 | 346 1 2 | 7 348 5 |
| 46 2 7 | 346 5 348 |@348 1 9 |
| 5 1 38 | 7 9 348 | 2 6 34 |
*-----------------------------------------------------------* |
As is already well documented, this pattern allows for the deletion of the two AUR digits in the AUR cell containing the extra, free digit(s). In this case, r1c7<>58
However, consider an external analysis of this pattern to determine what deletions are possible.
Digit 5: only two non-AUR cells contain a 5 that share a house with an AUR cell: r1c8 shares a box/row house, and r6c7 shares a column house; both are marked #.
Digit 8: only two non-AUR cells contain an 8 that share a house with an AUR cell: r1c8 shares a box/row house, and r8c7 shares a column house; both are marked @.
As External Principle 3 states, we may select any combination of inferences that cover all eight AUR digits. First try using r1c8=5 and r8c7=8
If r1c8=5, then r1c7<>5
If r8c7=8, then r3c7=5, then r1c7<>5
Thus both inferences provide a common result; r1c7<>5
Now consider another combination of inferences: r6c7=5 and r8c7=8
If r6c7=5, then r3c7=8, then r1c7<>8
If r8c7=8, then r1c7<>8
Again, both inferences provide a common result; r1c7<>8
(Note: We may take both of these two deletions even though taking either one destroys the second because both are valid at this stage of activity.)
This simple example using external inferences also demonstrates that for a Type 1 UR, both AUR digits may be deleted from the AUR cell containing the extra, free non-AUR digits. In actuality, we would never perform an external analysis for a simple APD for which we already have processing guidelines/rules but it was hopefully a good introduction to the concept.
Example 2
| Code: |
*-----------------------------------------------------------*
|#37 4 #137 | 28 6 *2789 | 5 18 *79 |
| 9 17 5 | 38 #37 4 | 2 18 6 |
| 6 2 8 | 5 1 *79 |@49 3 *479 |
|-------------------+-------------------+-------------------|
| 28 3 4 | 126 5 12 | 69 7 @89 |
| 1 9 267 | 2468 47 #278 | 46 5 3 |
| 78 5 67 | 346 9 #37 | 1 2 48 |
|-------------------+-------------------+-------------------|
| 5 17 17 | 9 8 6 | 3 4 2 |
| 234 6 23 | 7 34 5 | 8 9 1 |
| 34 8 9 | 134 2 13 | 7 6 5 |
*-----------------------------------------------------------* |
The code above, which is after completing basics, contains a AUR 79 in r13c69, marked with an *.
Internal Analysis
The internal or extra, non AUR digits are the digits 2 & 8 in r1c6 and the digit 4 in r3c9. Thus to prevent the deadly pattern (DP) one or more of these three must to true. However, since we do not know which one, we must consider all three inferences and strive to find a common result.
If r1c6=2, then r1c4=8, then r2c4=3
If r1c6=8, then r2c4=3
If r3c9=4, then r3c7=9, then r3c6=7, then r6c6=3
This gives us two pincers on 3 in r2c4 & r6c6 which deletes the 3 in r6c4; r6c4<>3 which completes the puzzle.
Notationally, this could be written as
(2)r1c6-(2=8)r1c4-(8=3)r2c4
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(8)r1c6-(8=3)r2c4
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(4)r3c9-(4=9)r3c7-(9=7)r3c6-(7=3)r6c6; r6c4<>3
External Analysis
To use this technique, we need to look at all digit 7s and digit 9s that "see" (share a house) with the four AUR cells.
Digit 7: Instances of digit 7 that "see" any on the AUR cells are marked # in the grid above. Note that the 7s in r1c13 are row implications, r56c6 are column implications and r2c5 is a box implication. The houses to use for analysis is a mater of choice based on ease of use and the result it provides. However, it is normally best to look at all inferences for both digits before making choices.
Digit 9: Instances of digit 9 that "see" any of the AUR cells are marked @. Only two instances occur; r3c7 in row3/box3 and r4c9 in col9.
Now we have all the needed info to make some decision regarding what inferences we will use for each AUR digit. Consider the following:
For digit 7, use the house box2 which includes r2c5=7 for AUR cells r13c6, and use house box3, which is empty,for AUR cells r13c9. This results in r2c5=7 being the only implication needed to cover all four AUR digits 7.
For digit 9, use the house box2 which is empty for AUR cells r13c6, and for AUR cells r13c9 use the house box3 which only contains r3c7. Thus the only implication needed to cover all four AUR digit 9s is r3c7=9.
We now have only two inferences to resolve, one of which must be true: r2c5=7 and r3c7=9. One solution is the following.
If r2c5=7, then r1c6<>7
If r3c7=9, then r1c9=7, then r1c6<>7
Thus r1c6<>7, but this is not very productive. An alternate solution using the same inferences is
If r2c5=7, then r2c5<>3, then r2c4=3 due to the strong link on 3 in row2
If r3c7=9, then r3c6=7, then r6c6=3
This gives us two pincers on 3 in r2c4 & r6c6 which deletes the 3 in r6c4; r6c4<>3 which completes the puzzle.
As an AIC this could be notated as:
(3=7)r6c6-(7=9)r3c6-AUR79r13c69[(9)r3c7=(7-3)r2c5]=(3)r2c4; r6c4<>3
Each choice of inferences and solution paths may provide different results as noted above. (And you know which result you want so work those inferences to you advantage!)
Example 3
1. A 6-cell BUG-Lite+3 (14)r236c379 , marked *. This is the same puzzle as Example 1.
| Code: |
*-----------------------------------------------------------*
| 9 7 6 | 58 34 1 | 3458 3458 2 |
| 8 5 *14 | 34 2 9 |*134 7 6 |
| 2 3 *14 | 58 6 7 | 58 9 *14 |
|-------------------+-------------------+-------------------|
| 1 6 5 | 2 7 34 | 9 34 8 |
| 3 4 9 | 1 8 5 | 6 2 7 |
| 7 8 2 | 9 34 6 |*1345 345 *134 |
|-------------------+-------------------+-------------------|
| 46 9 38 | 346 1 2 | 7 348 5 |
| 46 2 7 | 346 5 348 | 348 1 9 |
| 5 1 38 | 7 9 348 | 2 6 34 |
*-----------------------------------------------------------* |
Internal analysis
To prevent the DP: r2c7=3, r6c7=3, r6c7=5 and/or r6c9=3
If r2c7=3, then r2c4=4, then r1c5<>4, then r6c5=4, then r6c79<>4
If r6c7=3, then r6c5=4, then r6c79<>4
If r6c7=5, then r6c8<>5, then r46c8=34, then r6c79<>4
If r6c9=3, then r6c5=4, then r6c79<>4
Notationally this analysis could look like the following:
(3)r2c7-(3=4)r2c4-(4)r1c5=(4)r6c5; r6c79<>4
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(3)r6c7-(3=4)r6c5; r6c79<>4
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(5)r6c7-(5=34)ls:r46c8; r6c79<>4
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(3)r6c9-(3=4)r6c5; r6c79<>4
External analysis
Digit 1: No external inferences to consider in box1, box3 & box6.
Digit 4: Box1 is empty, r2c4 for ADP cell r2c7, r1c78 for ADP cells r2c7 & r3c9 in box3, r6c5 & r46c8 for ADP cells r6c79 in box6, r18c7 sees ADP cells r26c7 and finally r9c9 sees r36c9.
This set of inferences consists entirely of many occurrences of one ADP digit, 4, which offers several possibilities. Here, consider using the house row for ADP cells r2c7 & r3c9 in box3 which only includes r2c4 since row3 is empty, and house box6 for ADP cells r6c79 in box 6 which includes the two cells r46c8. This set of inferences then consists of r2c4=4 and r46c8=4.
If r2c4=4, then r1c5<>4, then r6c5=4, then r6c79<>4
If r46c8=4, then r6c79<>4
As an AIC, this could be notated: ADP(14)r236c379[(4)r46c8=(4)r2c4]-(4)r1c5=(4)r6c5; r6c79<>4
Given this set of inferences and the solution paths used, we get the same result with either internal or external analysis. Are more results possible? Give it a try!
Example 4
Here is another example that also includes a mixed analysis using both internal and external inferences. This AUR (25)r78c25, marked *, was found and solved by "Marty R.", who deserves great credit for even finding it since all four AUR cells contain extra, free digits.
| Code: |
*--------------------------------------------------------------------*
| 45 8 3 | 7 569 69 | 12469 124 269 |
| 2 45 7 | 5689 1 689 | 469 69 3 |
| 6 1 9 | 4 3 2 | 7 5 8 |
|----------------------+----------------------+----------------------|
| 8 3 2 | 1 4 5 | 69 69 7 |
| 17 6 4 | 2 79 789 | 13 138 5 |
| 17 9 5 | 68 67 3 | 12 128 4 |
|----------------------+----------------------+----------------------|
| 3459 *245 8 | 569 *2569 469 | 3469 7 1 |
| 3459 *245 1 | 569 *25679 4679 | 8 34 69 |
| 49 7 6 | 3 8 1 | 5 24 29 |
*--------------------------------------------------------------------* |
Internal analysis
To prevent the DP: r78c2=4 and/or pseudocell r78c5=679 formed by combining r7c5=69 and r8c5=679
If r78c2=4, then r2c2=5
If r78c5=679, then form a naked triple set, NT(679), by joining r78c5=679 with r5c5=79 and r6c5=67. Then, the NT(679) deletes (69) from r1c5 which makes r1c5=5.
We now have two pincers on digit 5, r2c2=5 and r1c5=5, which deletes 5 from r1c1; r1c1<>5=4.
This could be notated as: (5=4)r2c2-AUR(25)r78c25[(4)r78c2=NT(679)r5678c5]-(69=5)r1c5; r1c1<>5=4
External Analysis
Digit 2: No external inferences
Digit 5: r2c2 and r78c1 cover AUR cells r78c2; r1c5 and r78c4 cover AUR cells r78c5
Working with single cells is frequently easiest, so try r2c2=5 and r1c5=5 which act as pincers to delete 5 from r1c1 (which is identical to the use of the same pincers for the internal analysis).
Notationally, this could be written as: AUR(25)r78c25[(5)r2c2= (5)r1c5]; r1c1<>5
Mixed Internal and External Analysis
We have already noted above all the inferences for both the internal and the external viewpoints. One combination of internal and external inferences is to use the internal inference r78c2=4 for AUR cells r78c2 which covers both digits 2 and 5. For AUR cells r78c5 use external inferences: no external inferences exist for digit 2 and r1c5=5 covers digit 5. Thus, we have only two inferences to consider: r78c2=4 and r1c5=5
If r78c2=4 then r2c2=5, which together with the second inference r1c5=5 form two pincers to delete 5 from r1c1; ric1<>5
In notational form: (5=4)r2c2-AUR(25)r78c25[(4)r78c2= (5)r1c5]; r1c1<>5
Ted
Edits
September 15, 2010: Deleted statement "this AUR is also categorized as a Type 6 UR" in Example 2 introduction
September 15, 2010: Rewrote Principle 2
September 15, 2010: Added to Principle 5 description
Last edited by tlanglet on Wed Sep 15, 2010 1:54 pm; edited 1 time in total |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 14, 2010 12:26 am Post subject: |
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An impressive effort!!!
This is how I (often) utilize internal/external values in a UR.
A 4-cell Deadly Pattern can occur when two values form overlapping X-Wings in the four cells. To prevent this occurrence, either a secondary value must be true in the UR containing the DP or else at least one of the primary values must be true in a cell where it would otherwise be eliminated by one of the X-Wings.
Your Example #1:
| Code: | +--------------------------------------------------------------+
| 9 7 6 | *58 34 1 | *58+34 3458 2 |
| 8 5 14 | 34 2 9 | 134 7 6 |
| 2 3 14 | *58 6 7 | *58 9 14 |
|--------------------+--------------------+--------------------|
| 1 6 5 | 2 7 34 | 9 34 8 |
| 3 4 9 | 1 8 5 | 6 2 7 |
| 7 8 2 | 9 34 6 | 1345 345 134 |
|--------------------+--------------------+--------------------|
| 46 9 38 | 346 1 2 | 7 348 5 |
| 46 2 7 | 346 5 348 | 348 1 9 |
| 5 1 38 | 7 9 348 | 2 6 34 |
+--------------------------------------------------------------+
# 43 eliminations remain
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If X-Wings exist in r13\c47 for <5> and <8>, then I get eliminations r6c7<>5 and r8c7<>8. To prevent the DP, I need:
| Code: | ( r6c7=5 | r8c7=8 ) to force r3c7=8|5 => r1c7<>58 q.e.d.
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If X-Wings exist in c47\r13 for <5> and <8>, then I get eliminations r1c8<>5 and r1c8<>8. To prevent the DP, I need:
| Code: | r1c8=5|8 to force r1c4=8|5 => r1c7<>58 q.e.d.
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No matter which pair of X-Wings I choose to block, the outcome is the same! |
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ttt
Joined: 06 Dec 2008
Posts: 42
Location: vietnam
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| Posted: Tue Sep 14, 2010 2:45 am Post subject: Re: ADP External Inferences |
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| tlanglet wrote: | Example 2
| Code: |
*-----------------------------------------------------------*
|#37 4 #137 | 28 6 *2789 | 5 18 *79 |
| 9 17 5 | 38 #37 4 | 2 18 6 |
| 6 2 8 | 5 1 *79 |@49 3 *479 |
|-------------------+-------------------+-------------------|
| 28 3 4 | 126 5 12 | 69 7 @89 |
| 1 9 267 | 2468 47 #278 | 46 5 3 |
| 78 5 67 | 346 9 #37 | 1 2 48 |
|-------------------+-------------------+-------------------|
| 5 17 17 | 9 8 6 | 3 4 2 |
| 234 6 23 | 7 34 5 | 8 9 1 |
| 34 8 9 | 134 2 13 | 7 6 5 |
*-----------------------------------------------------------* |
The code above, which is after completing basics, contains a AUR 79 in r13c69, marked with an * ; this AUR is also categorized as a Type 6 UR.
..................
As an AIC this could be notated as: (3=7)r6c6-(7=9)r3c6-AUR79r13c69[(9)r3c7=(7-3)r2c5]=(3)r2c4; r6c4<>3
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Hi Ted,
Some of my opinion on this:
1- (3=7)r6c6… => AUR(79)r13c69 is cracked, so we can’t use AUR in this case.
2- The easy way to start from r6c6 that is no need to use AUR: (3=7)r6c6-(7)r13c6=(7-3)r2c5=(3)r2c4
3- We can use AUR (external digits) for this one as: (3)r2c4=(3)r2c5-AUR(79)r13c69[(7)r2c5=(9)r3c7]-(4)r3c7=r5c7-r6c9=r6c4
BTW, another example on using AURs (external & internal digits  ) here.
ttt |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Sep 14, 2010 3:42 am Post subject: |
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| Kudos Ted for all the time and effort that must've been expended on this excellent post. I was completely oblivious to the external inferences, but hopefully no more. |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Tue Sep 14, 2010 11:34 am Post subject: Re: ADP External Inferences |
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| ttt wrote: | | tlanglet wrote: | As an AIC this could be notated as: (3=7)r6c6-(7=9)r3c6-AUR79r13c69[(9)r3c7=(7-3)r2c5]=(3)r2c4; r6c4<>3
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1- (3=7)r6c6… => AUR(79)r13c69 is cracked, so we can’t use AUR in this case. |
Not sure what you mean by "cracked", but I see nothing wrong with Ted's AIC. Would you please clarify? |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Sep 14, 2010 12:29 pm Post subject: |
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Danny: That is an interesting perspective. How often is this more limited view effective for general AURs?
ttt: I also do not understand your concern regarding the possible notation I offered for using the Example 2 AUR, nor how the alternate AIC you suggested differs. Is it the fact that the second step in the chain, (3=7)r6c6-(7=9)r3c6-AUR... effectively destroys the AUR?
Regarding your second comment, I realize that the AUR is not required to obtain the same deletion but it is sometimes hard to find a clean, simple solution that can not be duplicated in a different manner. In this case, my goal was to demonstrate use of the AUR.
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 14, 2010 2:58 pm Post subject: |
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| tlanglet wrote: | Danny: That is an interesting perspective. How often is this more limited view effective for general AURs?
ttt: I also do not understand your concern regarding the possible notation I offered for using the Example 2 AUR, nor how the alternate AIC you suggested differs. Is it the fact that the second step in the chain, (3=7)r6c6-(7=9)r3c6-AUR... effectively destroys the AUR?
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AFAIK, the overlapping X-Wing perspective works on all URs. It works on your examples #1, #2, and #4.
As for ttt's perspective, your first AIC SIS -- (3=7)r6c6 -- sets r6c6=7, thus preventing the possibility of the DP existing and being used later.
BTW: this isn't a UR Type 6.
| Code: | *-----------------------------------------------------------*
|#37 4 #137 | 28 6 *2789 | 5 18 *79 |
| 9 17 5 | 38 #37 4 | 2 18 6 |
| 6 2 8 | 5 1 *79 |@49 3 *479 |
|-------------------+-------------------+-------------------|
| 28 3 4 | 126 5 12 | 69 7 @89 |
| 1 9 267 | 2468 47 #278 | 46 5 3 |
| 78 5 67 | 346 9 #37 | 1 2 48 |
|-------------------+-------------------+-------------------|
| 5 17 17 | 9 8 6 | 3 4 2 |
| 234 6 23 | 7 34 5 | 8 9 1 |
| 34 8 9 | 134 2 13 | 7 6 5 |
*-----------------------------------------------------------*
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If you consider X-Wing c69\r13 for <7> and <9>, then one of r1c1=7, r1c3=7, or r3c7=9 must be true. However, we have:
(7)r1c13 - r2c2 = (7)r2c5
So the X-Wing results can be simplified to r2c5=7 or r3c7=9.
If you consider X-Wing r13\c69 for <7> and <9>, then one of r56c6=7 or r4c9=9 must be true. However, we have:
(7)r56c6 - r5c5 = (7)r2c5
(9)r4c9 - r4c7 = (9)r3c7
So the X-Wing results can (again) be simplified to r2c5=7 or r3c7=9.
Last edited by daj95376 on Tue Sep 14, 2010 4:00 pm; edited 1 time in total |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Sep 14, 2010 3:59 pm Post subject: |
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Danny: Thanks for the insight about how the AIC I formed impacted the AUR. I have never given that aspect much attention and have probably made the same error in the past.
Also, I just re-read Keith's gospel on UR Essentials and realized that a long time ago I forgot about the x-wing overlay requirement for a Type 6 UR. Terminology bites me again.
ttt: Thanks for the reference to a great example of mixed internal/external SIS for a ADP.
Ted |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Tue Sep 14, 2010 4:02 pm Post subject: |
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| daj95376 wrote: | If you consider X-Wing c69\r13 for <7> and <9>, then one of r1c1=7, r1c3=7, or r3c7=9 must be true. However, we have:
(7)r1c13 - r2c2 = (7)r2c5
So the X-Wing results can be simplified to r2c5=7 or r3c7=9. |
I think it is more accurate to view the "cover set" for the eight digit <7> and <9> candidates to be a "UR cover", not c69. And by choosing the base set to be 79b23 instead of 79r13, you get to (7)r2c5 = (9)r3c7 directly, without the need for a strong inference in r2. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 14, 2010 4:14 pm Post subject: |
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| ronk wrote: | | I think it is more accurate to view the "cover set" for the eight digit <7> and <9> candidates to be a "UR cover", not c69. And by choosing the base set to be 79b23 instead of 79r13, you get to (7)r2c5 = (9)r3c7 directly, without the need for a strong inference in r2. |
Hmmm ... okay!
In [b2] and [b3], either: r2c5=7, or r3c7=9, or the DP exists because Hidden Pairs are allowed to form for <79>. Clean!!! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Sep 14, 2010 4:41 pm Post subject: |
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Please bear with me, external inferences are totally new to me. This is where I've gotten on Danny's 9-13C after basics:
| Code: |
+----------------+------------+---------------+
| 3 2568 568 | 1 45 456 | 7 268 9 |
| 1268 268 4 | 28 9 7 | 5 2368 1236 |
| 12568 9 7 | 28 35 356 | 68 4 126 |
+----------------+------------+---------------+
| 7 258 58 | 3 6 9 | 18 1258 4 |
| 268 2468 39 | 5 7 1 | 368 2689 236 |
| 56 1 39 | 4 8 2 | 36 569 7 |
+----------------+------------+---------------+
| 9 3 56 | 7 145 45 | 2 16 8 |
| 568 568 1 | 9 2 35 | 4 7 36 |
| 4 7 2 | 6 13 8 | 9 13 5 |
+----------------+------------+---------------+
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Play this puzzle online at the Daily Sudoku site
I'm looking at the potential Type 6 on 45 in boxes 28. There are no external 4s, therefore, only 5s need be considered, right? And I can consider the 5s in any house? So I look at box 8 and see only one external 5. Does that mean that r8c6=5? Seems too simple. |
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ttt
Joined: 06 Dec 2008
Posts: 42
Location: vietnam
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| Posted: Tue Sep 14, 2010 4:59 pm Post subject: Re: ADP External Inferences |
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| ronk wrote: | | ttt wrote: | | tlanglet wrote: | As an AIC this could be notated as: (3=7)r6c6-(7=9)r3c6-AUR79r13c69[(9)r3c7=(7-3)r2c5]=(3)r2c4; r6c4<>3
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1- (3=7)r6c6… => AUR(79)r13c69 is cracked, so we can’t use AUR in this case. |
Not sure what you mean by "cracked", but I see nothing wrong with Ted's AIC. Would you please clarify? |
The first: I’m sorry by my poor English... 
For Ted’s AIC on his example No.2: AUR(79)r13c69 => on external digits: at least one of [(7)r2c5, (9)r3c7] must be true => at least one of [(7)r13c6, (9)r13c9] must be false. When (3=7)r6c6... meant: if r6c6<>3 then r6c6=7 then (7)r13c6 is false then UR(79)r13c69 is avoided... so, we can’t use AUR(79)r13c69 later on that chain.
I’m sorry, I can’t explain more... But, I feel that something is not clear on that Ted’s AIC.
Wow...! I had drunk too much bear and wine tonight, the real life is not easy as solving sudoku puzzles .
ttt |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Sep 14, 2010 7:23 pm Post subject: Re: ADP External Inferences |
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ttt,
| ttt wrote: |
For Ted’s AIC on his example No.2: AUR(79)r13c69 => on external digits: at least one of [(7)r2c5, (9)r3c7] must be true => at least one of [(7)r13c6, (9)r13c9] must be false. When (3=7)r6c6... meant: if r6c6<>3 then r6c6=7 then (7)r13c6 is false then UR(79)r13c69 is avoided... so, we can’t use AUR(79)r13c69 later on that chain. |
I wondered if my AIC had destroyed the AUR, and your response confirms that situation. I will hopefully be more aware of that condition in the future. I will also change Example 2 to used your AIC.
| ttt wrote: | | Wow....! I had drunk too much bear and wine tonight, the real life is not easy as solving sudoku puzzles[/color] . |
Life has many pleasure! Two of my favorites are beer/wine and sudoku, and both also have a down side such as when a beautiful, complicated step simply will not work due to one simple thing.
Ted |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Tue Sep 14, 2010 8:08 pm Post subject: Re: ADP External Inferences |
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| tlanglet wrote: | ttt,
| ttt wrote: | | For Ted’s AIC on his example No.2: AUR(79)r13c69 => on external digits: at least one of [(7)r2c5, (9)r3c7] must be true => at least one of [(7)r13c6, (9)r13c9] must be false. When (3=7)r6c6... meant: if r6c6<>3 then r6c6=7 then (7)r13c6 is false then UR(79)r13c69 is avoided... so, we can’t use AUR(79)r13c69 later on that chain. |
I wondered if my AIC had destroyed the AUR, and your response confirms that situation. |
But your AIC is valid nonetheless. Due to the AUR, at least one of r2c5=7 or r3c7=9 must be true. Now separately follow each inference stream.
| Code: | (7-3)r2c5 = (3)r2c4
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(9)r3c7 - (9=7)r3c6 - (7=3)r6c6 ==> r6c4<>3 |
That r3c6 is one of the AUR cells is of no consequence. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 14, 2010 8:49 pm Post subject: |
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Aaaaaagh!!!! We've finally had a breech between the two parallel Sudoku universes.
ttt lives in the network universe. He sees things in terms of downstream implications occurring.
I live mostly in the chain universe where one implication only affects the next implication. However, I will occasionally visit the alternate universe.
Ted's "chain" doesn't work in ttt's universe. However, it works in my universe as long as you bury your head in the sand about inserting the AUR logic.
Last edited by daj95376 on Wed Sep 15, 2010 1:23 pm; edited 2 times in total |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 14, 2010 9:29 pm Post subject: |
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| Marty R. wrote: | Please bear with me, external inferences are totally new to me. This is where I've gotten on Danny's 9-13C after basics:
| Code: |
+----------------+------------+---------------+
| 3 2568 568 | 1 45 456 | 7 268 9 |
| 1268 268 4 | 28 9 7 | 5 2368 1236 |
| 12568 9 7 | 28 35 356 | 68 4 126 |
+----------------+------------+---------------+
| 7 258 58 | 3 6 9 | 18 1258 4 |
| 268 2468 39 | 5 7 1 | 368 2689 236 |
| 56 1 39 | 4 8 2 | 36 569 7 |
+----------------+------------+---------------+
| 9 3 56 | 7 145 45 | 2 16 8 |
| 568 568 1 | 9 2 35 | 4 7 36 |
| 4 7 2 | 6 13 8 | 9 13 5 |
+----------------+------------+---------------+
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I'm looking at the potential Type 6 on 45 in boxes 28. There are no external 4s, therefore, only 5s need be considered, right? And I can consider the 5s in any house? So I look at box 8 and see only one external 5. Does that mean that r8c6=5? Seems too simple. |
First off, potential is incorrect because it actually is a UR Type 6 because of the X-Wing pattern for <4> -- which you noted!
Yes, only 5s need/can be considered for external prevention of the DP. However, your settling on one cell is too simplistic.
In all, there are six 5s that can directly prevent the DP:
r1c2=5, r1c3=5, r3c5=5, r3c6=5, r7c3=5, and r8c6=5
However, not all of them need be considered at once. If you recall that a UR is restricted to four cells that exist in only two rows, two columns, and two boxes, then you can use that in selecting cells to consider.
For the two rows of the UR, the cells to consider are: r1c23=5 and r7c3=5
For the two columns of the UR, the cells to consider are: r3c56=5 and r8c6=5
For the two boxes of the UR, the cells to consider are (again): r3c56=5 and r8c6=5
So, for example, you must use r8c6=5 and r3c56=5 together if you want to obtain an elimination -- like r1c6<>5. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Sep 14, 2010 9:51 pm Post subject: |
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Danny, thanks for the response, and yes, I misspoke when I used the word "potential."
As I noted, my elimination seemed too simple. However, I misinterpreted one of Ted's principles, but still don't know why I misinterpreted it.
"The second External Principle is that only one house, (row, column or box) needs to be considered for each DP digit, but all occurrences of the ADP digit in that house must be included in the inference analysis."
I thought I was following that when I used one house and accounted for all (just one) of the external candidates. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Sep 14, 2010 10:12 pm Post subject: |
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| Marty R. wrote: | Danny, thanks for the response, and yes, I misspoke when I used the word "potential."
As I noted, my elimination seemed too simple. However, I misinterpreted one of Ted's principles, but still don't know why I misinterpreted it.
"The second External Principle is that only one house, (row, column or box) needs to be considered for each DP digit, but all occurrences of the ADP digit in that house must be included in the inference analysis."
I thought I was following that when I used one house and accounted for all (just one) of the external candidates. |
I expect that Ted will be editing that Principle very soon!!!
===== ===== ===== ===== ===== FWIW
On the 9-13C grid that Marty presented, the UR Type 6 forces r1c5,r7c6=4. External 5s force r1c6<>5. Together, this leaves r1c6=6 to crack the puzzle. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Sep 14, 2010 11:23 pm Post subject: |
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| Marty R. wrote: | Danny, thanks for the response, and yes, I misspoke when I used the word "potential."
As I noted, my elimination seemed too simple. However, I misinterpreted one of Ted's principles, but still don't know why I misinterpreted it.
"The second External Principle is that only one house, (row, column or box) needs to be considered for each DP digit, but all occurrences of the ADP digit in that house must be included in the inference analysis."
I thought I was following that when I used one house and accounted for all (just one) of the external candidates. |
Marty: Principle 2 says that each ADP digit must be covered by one house. You indicated that you "accounted for all (just one) of the external candidates".
Danny has already listed all the external cells containing a digit 5 that share a house with each of the four ADP cells. So the various sets of inferences necessary to satisfy Principle 2 are:
r1c23 for r1c56 and r7c3 for r7c56; house row1 and row7
r1c23 for r1c56 and r3c56|r8c6 for r7c45; house row1 and col56
r1c23 for r1c56 and r8c6 for r7c56; house row1 and box8
r3c56|r8c6 for r1c56 and r7c3 for r7c56; col56 and row7
r3c56|r8c6 for r1c56 and r8c6 for r7c56; house col56 and col56 or box8 (Note that r8c6 is used twice per Principle 4, once for col 6 and once for box8)
To make things even more powerful (and confusing) it is possible to also use an extra, free digit thus another set could be
r3c5=5 for r1c5, r1c6=6 for r1c6 and r8c6=5 for r7c56; houses col1, free digit, and box8. For this set, the eliminations could be
If r3c5=5, then r1c5<>5
If r1c6=6 then, r3c6<>6, then r38c6=35, then r7c6<>5=4, then r7c5<>4 then r1c5=4<>5
If r8c6=5, then, r7c6=4, then r7c5<>4 then r1c5=4<>5
Does Principle 2 needed further clarification? If so, please suggest some wording or identify what is missing/confusing.
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Sep 15, 2010 12:26 am Post subject: |
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| Quote: | | Does Principle 2 needed further clarification? If so, please suggest some wording or identify what is missing/confusing. |
I certainly can't answer the question or make any suggestions since I'm back to square one, i.e., not understanding. I'll have to limit myself to internal techniques until I can digest the external. |
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