UR+2B/2SL

Advanced methods and approaches for solving Sudoku puzzles

Re: UR+2B/2SL

Hi,

When I coded them, I was not aware of the zoo thread
and also not of the Mike Barker 'May 16 2006' post which ronk recently linked as addendum.

When I submit yor mentioned examples on UR+3C/2SL (= hidden UR acccording to hodoku; = hidden UR of type 1 according to scanraid)
to my code, I get:
Code: Select all
`000000008000008630090130400047300000000510090000000000620800010078000005000625800 #1singlesUR+3C/2SL (HUR1) (79)[14|66] cancel 9[66]orig/ha M-wing  WeakLinkInDiscontLoop-7 2[64]-2[69]=1[69]-1[63]=1[23]-2[23]=2[24]-2[64] cancel 2[64]singles002090000004607000090004300708000009040020800050030100000500000006008900000000037 #2singlesUR+3x/1SL yb  (46)[95|77] cancel 6[77]orig/ha M-wing  WeakLinkInDiscontLoop-7 1[81]-1[88]=2[88]-2[97]=2[96]-1[96]=1[91]-1[81] cancel 1[81]singles000030040000206030102090000000003890003000027054000000500087200001600000400000005 #3singlesUR+3C/2SL (HUR1) (68)[12|51] cancel 8[51]orig/ha M-wing  WeakLinkInDiscontLoop-7 7[11]-7[21]=9[21]-9[61]=9[64]-7[64]=7[14]-7[11] cancel 7[11]singles201000008000003000000000062603007000052094700700300100400006051900500040007010000 #4singles, line-block, naked triplet in row 1UR+3X (27)[75|24]  cancel 9[3][4]pointing pair : 9 in block 2 aligned in row 1 : cancel 9[12]UR+3x/1SL yb  (27)[75|24] cancel 7[24]xy-chain        WeakLinkInDiscontLoop-13 7[74]-7[34]=8[34]-8[54]=6[54]-6[59]=3[59]-3[99]=9[99]-9[77]=2[77]-2[75]=7[75]-7[74] cancel 7[74]singles`

Yes, the UR+3C/2SL is not the bottleneck of the puzzles. In two of the cases, easier UR methods (methods with higher preference; higher application priority)
apply before the UR+3C/2SL got its chance.

When I switch off the UR+3x/1SL in #2, I get
UR+3C/2SL (HUR1) (46)[95|77] cancel 6[77]

When I switch off UR+3X for #4, I get
UR+3x/1SL yb (27)[75|24] cancel 7[24]
When I switch off UR+3x/1SL for #4, I get finally the
UR+3C/2SL (HUR1) (27)[75|24] cancel 7[24]

surbier
surbier
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Re: UR+2B/2SL

daj95376 wrote:When I reached these puzzles:

--- UR+3C/2SL: both strong links share a node, do not include the bivalue cell and have equal labels => "b" can be removed from "abZ"

Code: Select all
` ab     abX         |        a|     a   |abY-----abZ`

I was able to duplicate the UR pattern, but I wasn't able to get it to crack the puzzles. Do you get any better results?

I believe Mike was applying these two patterns simultaneously.

Code: Select all
` ab     abX         |        a|     a   |abY-----abZRemoves "b" from the "abZ" cell ab     abX |b| |   aabY-----abZRemoves "b" from the "abX" cell`

In the 2nd illustration, note that a "strong link corner" may exist at the "abX" cell too ... for a total of three exclusions.
ronk
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Re: UR+2B/2SL

Thanks surbier for confirming that UR+3C/2SL is insufficient to crack these puzzles.

Thanks ronk for supplying the companion UR pattern.

ronk wrote:I believe Mike was applying these two patterns simultaneously.
...
In the 2nd illustration, note that a "strong link corner" may exist at the "abX" cell too ... for a total of three exclusions.

I found four eliminations in the first three puzzles corresponding to:

Code: Select all
` ab-----abX |   b   |b|      a| |   a   |abY-----abZRemoves "a" from "ab" and "b" from "abX","abY","abZ"`

The fourth puzzle uses your second pattern once.

[Edit: specifically listed eliminations.]
Last edited by daj95376 on Mon Apr 11, 2011 7:02 pm, edited 1 time in total.
daj95376
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Re: UR+2B/2SL

daj95376 wrote:I found four eliminations in the first three puzzles corresponding to:

Code: Select all
` ab-----abX |   b   |b|      a| |   a   |abY-----abZRemoves "a" from "ab" and "b" from "abZ" ... a Naked Single on "b" completes the eliminations`

The fourth puzzle uses your second pattern once.

The "b" in "ab" becomes a hidden single, so IMO that 4th elimination shouldn't be counted as part of the underlying UR move.
ronk
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Re: UR+2B/2SL

ronk wrote:
daj95376 wrote:I found four eliminations in the first three puzzles corresponding to:

Code: Select all
` ab-----abX |   b   |b|      a| |   a   |abY-----abZRemoves "a" from "ab" and "b" from "abZ" ... a Naked Single on "b" completes the eliminations`

The fourth puzzle uses your second pattern once.

The "b" in "ab" becomes a hidden single, so IMO that 4th elimination shouldn't be counted as part of the underlying UR move.

I should have simply listed all of the eliminations directly attributable to the UR pattern.

Code: Select all
`Removes "a" from "ab" and "b" from "abX","abY","abZ"`

The elimination of "a" from "ab" can be derived based solely on the UR pattern.

Code: Select all
`"ab"=a => { "abX"=b and "abY"=b } either of which => "abZ"=a ... unacceptable!!!`
daj95376
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Re: UR+2B/2SL

daj95376 wrote:The elimination of "a" from "ab" can be derived based solely on the UR pattern.

Code: Select all
`"ab"=a => { "abX"=b and "abY"=b } either of which => "abZ"=a ... unacceptable!!!`

A hazard of using induction instead of deduction is that one doesn't pause to count the strong links. Don't know about the "3x" part, but you're now talking about a "UR+3x/3SL" ... instead of an overlay of "UR+3x/2SL"s.
ronk
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Re: UR+2B/2SL

ronk wrote:A hazard of using induction instead of deduction is that one doesn't pause to count the strong links. Don't know about the "3x" part, but you're now talking about a "UR+3x/3SL" ... instead of an overlay of "UR+3x/2SL"s.

That's why I didn't try to give the pattern a name! It was only suppose to show that combining a UR+3C/2SL with 2x of your second pattern (UR+3N/2SL) results in the three eliminations from the individual patterns ... and ... one additional elimination based on examining the combination as a whole.

From a practical standpoint, it's sufficient to perform the three patterns individually and let the fourth elimination fall by way of a Hidden Single. I was just impressed that "the combination" provided directly for the elimination in "a" -- as found by my solver -- and not as a subsequent Hidden Single elimination.
daj95376
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OT: UR+2B/2SL

daj95376 wrote:From a practical standpoint, ...

... you could solve them this way:
Code: Select all
` *----------------------------------* | 47  3  46  | 79 5  .  | 1  2  8  | | 17  5  12  |*27 4  8  | 6  3  9  | | 8   9  26  | 1  3  26 | 4  5  7  | |------------+----------+----------| | 19  4  7   | 3  6 x29 | 5  8  12 | | 2   8  3   | 5  1  4  | 7  9  6  | | 5   6  19  | .  8  .  | 3  4  12 | |------------+----------+----------| | 6   2  5   | 8  7  3  | 9  1  4  | | 3   7  8   | 4  9  1  | 2  6  5  | | 49  1  49  | 6  2  5  | 8  7  3  | *----------------------------------*`

DP 79 in r16c46, either r2c4 must be 7 or r4c6 must be 9. Both directly imply r1c4=9.
(When i get a number, i normally dont look for other UR eliminations, though
r2c4=7->r6c6=7 and r4c6=9 both imply r6c6<>9.)
Code: Select all
` *----------------------------------* | 47  3  46  | 9  5  67 | 1  2  8  | | 17  5  12  | 27 4  8  | 6  3  9  | | 8   9  26  | 1  3  6  | 4  5  7  | |------------+----------+----------| | 19  4  7   | 3  6  29 | 5  8  12 | | 2   8  3   | 5  1  4  | 7  9  6  | | 5   6  19  | 27 8  #  | 3  4  12 | |------------+----------+----------| | 6   2  5   | 8  7  3  | 9  1  4  | | 3   7  8   | 4  9  1  | 2  6  5  | | 49  1  49  | 6  2  5  | 8  7  3  | *----------------------------------*`

Now here you have a good chance to see that a 9 in r6c6 leads to the deadly pattern too.
eleven

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Re: UR+2B/2SL

Free Press April 22, 2011 (and posted by Keith elsewhere)

Is this UR pattern among MB's UR patterns?

Code: Select all
` (*) UR cell (#) bivalue support cell *r8c5=1 SL[ *r7c5=8 *r8c9=8 ] r7c9<>8   ||                  ||   ||                #r5c9=3   r7c9<>3 #r7c6=2                       r7c9<>2                              *r7c9=1 =>  r8c5<>1 +------------------------------------------------------------------------+ |  8      7      69     |  1      236     4      |  2356   2369   235    | |  59     356    2      |  7      36      8      |  1      369    4      | |  146    136    146    |  236    5       9      |  7      8      23     | |-----------------------+------------------------+-----------------------| |  3      1258   189    |  4      79      6      |  258    127    12578  | |  46     68     7      |  25     12      125    |  9      34    #38     | |  59     125    149    |  8      79      3      |  25     1247   6      | |-----------------------+------------------------+-----------------------| |  7      4      5      |  9     *18+236 #12     |  2368   1236  *18+23  | |  2      9      3      |  56    *18+6    15     |  4      167   *18+7   | |  16     168    168    |  23     4       7      |  23     5      9      | +------------------------------------------------------------------------+ # 87 eliminations remain`

TIA !!!
daj95376
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Re: UR+2B/2SL

daj95376 wrote:Is this UR pattern among MB's UR patterns?

I didn't make a close check, but I don't recall any of Mike's patterns having two "support cells" that didn't "see" each other. Here's a "ttt-like diagram" for that.
Code: Select all
`UR(18)r78c59 ||(6)r8c5 ||(236-8)r7c5 = (8)r8c5 ||(7-8)r8c9 = (8)r8c5 ||(2)r7c9 - (2=1)r7c6 ||(3)r7c9 - (3=8)r5c9 - (8)r8c9 = (8)r8c5  ==> r8c5<>1`
ronk
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Re: UR+2B/2SL

Thanks Ron for the deductive perspective on the elimination. For some reason, I'm drawn to the Dark Side when it comes to UR eliminations. I love making an invalid assignment and watching the domino effect produce a deadly pattern. The main drawback to this approach is the possible creation of secondary contradictions while striving for the DP.

If I were to tackle this UR manually, then I wouldn't look for a common elimination resulting from so many non-UR candidates in the pattern. I'd have noticed the two strong links on <8> and immediately tested r8c5=1 to see if it would force r7c9=1. Thus, the reason for the format I chose.

Regards, Danny
daj95376
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Re: UR+2B/2SL

daj95376 wrote:... I wouldn't look for a common elimination resulting from so many non-UR candidates in the pattern. I'd have noticed the two strong links on <8> and immediately tested r8c5=1 to see if it would force r7c9=1.

I agree on the "so many non-UR candidates", but that was the only practical way to preserve your two bivalued "helper cells." With too many internal non-UR candidates, I normally look for the external "DP busters." In this case:

Code: Select all
`UR(18)r78c59 ||(1)r78c6 ||(8)r7c7 - (8)r7c5 = (8)r8c5 ||(1-7)r4c9 = (7-8)r8c9 = (8)r8c5  ==> r8c5<>1`
ronk
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Re: UR+2B/2SL

In addition to my post from April 11 (if not outdated):

When applying the mentioned URs simultaneously to examples 1, and 3 , there remain only singles.
For #4, when applying the latter two URs simultaneously (keeping r2c4<>7 on hold), it also solves
down to singles.

Code: Select all
`201000008000003000000000062603007000052094700700300100400006051900500040007010000 #4+--------------------------+--------------------------+--------------------------+|   2     469       1      | 69      456     59       |   3       7       8      ||   8     67      459      |#267    #267       3      | 459       1     459      ||   3     479     459      |978      478       1      | 459       6       2      |+--------------------------+--------------------------+--------------------------+|   6     489       3      |   1     258       7      | 4589    289     459      ||   1       5       2      | 68        9       4      |   7     38      36       ||   7     489     49       |   3     2568    25       |   1     289     4569     |+--------------------------+--------------------------+--------------------------+|   4       3       8      |#279    #27        6      | 29        5       1      ||   9       1       6      |   5       3     28       | 28        4       7      ||   5       2       7      |   4       1     89       |   6     389     39       |+--------------------------+--------------------------+--------------------------+27+6  -2- 27+6|227+9  -7- 27UR+3X         (27)[r7c5|r2c4] r3c4<>9  (supporting cell (69)[r1c4] )pointing pair                 r1c2<>9UR+3x/1SL yb  (27)[r7c5|r2c4] r2c4<>7  (supporting cell (67)[r2c2] )UR+3N/2SL     (27)[r7c5|r2c4] r2c5<>7  (or a UR+3U/2SL)`
surbier
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Re: UR+2B/2SL

Yes, the individual eliminations can be found using two of Mike Barker's patterns.

A forcing network perspective that might be used manually.

Code: Select all
` Either (one of) r2c45=6 and/or r7c4=9 r2c45=6 ->                     r2c2=7  =>  r2c45<>7 r7c4 =9 -> r1c4=6 -> r5c4=8 -> r3c4=7  =>  r2c45<>7 +--------------------------------------------------------------+ |  2     469   1     |  69    456   59    |  3     7     8     | |  8     67    459   | *27+6 *27+6  3     |  459   1     459   | |  3     479   459   |  789   478   1     |  459   6     2     | |--------------------+--------------------+--------------------| |  6     489   3     |  1     258   7     |  4589  289   459   | |  1     5     2     |  68    9     4     |  7     38    36    | |  7     489   49    |  3     2568  25    |  1     289   4569  | |--------------------+--------------------+--------------------| |  4     3     8     | *27+9 *27    6     |  29    5     1     | |  9     1     6     |  5     3     28    |  28    4     7     | |  5     2     7     |  4     1     89    |  6     389   39    | +--------------------------------------------------------------+ # 63 eliminations remain -or- (7=6)r2c2 - (6)r2c45 =UR= (9)r7c4 - (9=687)r153c4  =>  r2c45<>7`

Since I have a difficult time following most of Mike Barker's explanations, I have no idea if he has a scenario for this approach.

Too bad ronk dropped the idea of examining (and explaining) Mike Barker's patterns. I could see one thread where the patterns are presented and discussed ... and a separate thread -- for reference purposes -- that posts just the consensus drawn from those discussions. (What should have been done for the Ultimate Fish Guide.)
daj95376
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Re: UR+2B/2SL

You've gone over to the dark side again daj!

(8a)r3c5 = (8)r3c4 - (8=96)r15c4 - (9b=27)r7c45 -[UR]- (27)r2c45 = (78c)r3c45
=> [ac] r3c5 <> 4, [bc] r3c4 <> 9 making (4)r1c5 single and (78)r2c45 a naked pair.
David P Bird
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