The New Sudoku Players' Forum

by **ArkieTech** » Wed Jun 25, 2014 10:33 pm

Code: Select all: *-----------* |...|219|.3.| |3..|48.|2..| |.8.|.63|...| |---+---+---| |2.5|...|.83| |816|.3.|429| |43.|...|5.7| |---+---+---| |...|85.|.4.| |..8|.74|..1| |.4.|692|...| *-----------*

Play/Print this puzzle online

by **pjb** » Wed Jun 25, 2014 10:49 pm

Still staying with DPs:

Code: Select all: 67 #56 4 | 2 1 9 |b78 3 #568 3 #569 17 | 4 8 57 | 2 157-9 #56 17-9 8 2 | 57 6 3 |a79 1579 4 ---------------------+----------------------+--------------------- 2 7 5 | 9 4 6 | 1 8 3 8 1 6 | 57 3 57 | 4 2 9 4 3 9 | 1 2 8 | 5 6 7 ---------------------+----------------------+--------------------- 679 69 37 | 8 5 1 | 3679 4 2 569 2 8 | 3 7 4 | 69 59 1 157 4 137 | 6 9 2 | 378 57 58

(9=7)r3c7 - (7=8)r1c7 - (8=9)UR:r12c29 => -9 r2c8, r3c1; stte

Phil

by **SteveG48** » Thu Jun 26, 2014 12:53 am

Code: Select all: *-----------------------------------------------------------* |a67 56 4 | 2 1 9 |b78 3 568 | | 3 569 f1-7 | 4 8 57 | 2 1579 56 | |e179 8 2 | 57 6 3 | 79 1579 4 | *-------------------+-------------------+-------------------| | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | *-------------------+-------------------+-------------------| | 679 69 37 | 8 5 1 | 3679 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | |d157 4 137 | 6 9 2 |c378 d57 d58 | *-----------------------------------------------------------*

(7)r1c1 = (7-8)r1c7 = r9c7 - (8=157)r9c189 - (1)r3c1 = (1)r2c3 => -7 r2c3 ; stte

by **Leren** » Thu Jun 26, 2014 1:06 am

Code: Select all: *--------------------------------------------------------------* |e67 56 4 | 2 1 9 |d78 3 568 | | 3 569 1-7 | 4 8 57 | 2 1579 56 | | 179 8 2 | 57 6 3 | 79 1579 4 | |--------------------+--------------------+--------------------| | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | |--------------------+--------------------+--------------------| | 69-7 69 a37 | 8 5 1 |b3679 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | | 15-7 4 137 | 6 9 2 |c378 57 58 | *--------------------------------------------------------------*

(7=3) r7c3 - r7c7 = (3-8) r9c7 = (8-7) r1c7 = (7) r1c1 = - 7 r2c3, r79c1; stte

Leren

by **7b53** » Thu Jun 26, 2014 3:07 am

Code: Select all: *-----------------------------------------------------------* | 67 56 4 | 2 1 9 | 78 3 568 | | 3 569 17 | 4 8 57 | 2 1579 56 | | 179 8 2 | 57 6 3 | 79 1579 4 | *-------------------+-------------------+-------------------| | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | *-------------------+-------------------+-------------------| | 679 69 37 | 8 5 1 | 3679 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | | 157 4 137 | 6 9 2 | 378 57 58 | *-----------------------------------------------------------*

any solution for UR(37)r79c37....

by **ArkieTech** » Thu Jun 26, 2014 3:08 am

Code: Select all: *-----------------------------------------------------------* |b67 b56 4 | 2 1 9 |c78 3 b568 | | 3 569 17 | 4 8 57 | 2 1579 56 | | 179 8 2 | 57 6 3 |c79 1579 4 | |-------------------+-------------------+-------------------| | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | |-------------------+-------------------+-------------------| |a679 a69 37 | 8 5 1 | 37-69 4 2 | | 59-6 2 8 | 3 7 4 |c69 59 1 | | 157 4 137 | 6 9 2 | 378 57 58 | *-----------------------------------------------------------* (69=7)r7c12-(7=8)r1c129-(8=69)r138c7 => -69r7c7, -6r8c1

by **daj95376** » Thu Jun 26, 2014 3:30 am

_

I'm sure Mike Barker has a designation for Phil's UR, but I'll just use internal strong links to force DPs.

Code: Select all: + ------------------------------------------------------- SL on <5> | +------- SL on <6> v v +--------------------------------------------------------------+ | 67 *56 4 | 2 1 9 | 78 3 *56+8 |<- SL on <5> | 3 *56+9 17 | 4 8 57 | 2 1579 *56 |<- SL on <6> | 179 8 2 | 57 6 3 | 79 1579 4 | |--------------------+--------------------+--------------------| | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | |--------------------+--------------------+--------------------| | 679 69 37 | 8 5 1 | 3679 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | | 157 4 137 | 6 9 2 | 378 57 58 | +--------------------------------------------------------------+ # 41 eliminations remain =5r1c9 ; =6r2c9 =6r1c2 ; =5r2c2 ; DP \ => - (56=8) r1c9 =6r1c9 ; =5r1c2 =5r2c9 ; =6r2c2 ; DP / -or- =5r2c2 ; =6r2c9 =6r1c2 ; =5r1c9 ; DP \ => - (56=9) r2c2 =6r2c2 ; =5r1c2 =5r2c9 ; =6r1c9 ; DP /

A simpler linkage.

Code: Select all: (8-56)r1c9 = (56)r1c12 - (56=9)r2c2 => =8r1c9 then =9r2c2 (9-56)r2c2 = (56)r1c12 - (56=8)r1c9 => =9r2c2 then =8r1c9

Since one value must be true, both values are true.

Note: I don't have anything significant for the <37> UR other than -7r9c37. Externals force -7r2c8.

[Edit: made "simpler linkage" even simpler.]
_

by **blue** » Thu Jun 26, 2014 7:35 am

7b53 wrote:any solution for UR(37)r79c37....

There's this:

Code: Select all: +-------------------+-----------+-------------------+ | 6-7 56 4 | 2 1 9 | (78) 3 568 | | 3 569 (17) | 4 8 57 | 2 1579 56 | | 179 8 2 | 57 6 3 | 79 1579 4 | +-------------------+-----------+-------------------+ | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | +-------------------+-----------+-------------------+ | 69(7) 69 (37) | 8 5 1 | 69(37) 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | | 157 4 (137) | 6 9 2 | (378) 57 58 | +-------------------+-----------+-------------------+ UR<37>r79c37 || 7r7c1 ------------- 7r1c1 || 1r9c3 - (1=7)r2c3 - 7r1c1 || 8r9c7 - (8=7)r1c7 - 7r1c1 => r1c1<>7; stte

Another take on the same thing: (added)

Code: Select all: 1r9c3 - (1=7)r2c3 ------------------ || \ (NP: 37r9c37) -UR- (HP: 37r7c37) = 7r7c1 - 7r1c1 => r1c1<>7; stte || / 8r9c7 - (8=7)r1c7 ------------------

by **daj95376** » Thu Jun 26, 2014 3:34 pm

blue wrote:
Code: Select all
UR<37>r79c37 || 7r7c1 ------------- 7r1c1 || 1r9c3 - (1=7)r2c3 - 7r1c1 || 8r9c7 - (8=7)r1c7 - 7r1c1 => r1c1<>7; stte

Hmmm! A mix of external and internal constraints on a UR. That's something I would have never tried. However, since it appears that your objective is to provide a scenario that accounts for <7> being eliminated from the UR, your second stream can also be an external.

Code: Select all: UR<37>r79c37 || 7r7c1 ------------- 7r1c1 || 7r2c3 ------------- 7r1c1 || 8r9c7 - (8=7)r1c7 - 7r1c1 => r1c1<>7; stte

Although the following is accurate in the general sense of a chain, it bothers me in the specific case where the NP has a <7> in r9c3. Then, the conclusion of =7r7c1 seems to be a contradiction.

blue wrote:Another take on the same thing: (added)

Code: Select all
1r9c3 - (1=7)r2c3 ------------------ || \ (NP: 37r9c37) -UR- (HP: 37r7c37) = 7r7c1 - 7r1c1 => r1c1<>7; stte || / 8r9c7 - (8=7)r1c7 ------------------

Nice catch !!!

_

by **7b53** » Thu Jun 26, 2014 9:20 pm

blue wrote:There's this:

Code: Select all
+-------------------+-----------+-------------------+ | 6-7 56 4 | 2 1 9 | (78) 3 568 | | 3 569 (17) | 4 8 57 | 2 1579 56 | | 179 8 2 | 57 6 3 | 79 1579 4 | +-------------------+-----------+-------------------+ | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | +-------------------+-----------+-------------------+ | 69(7) 69 (37) | 8 5 1 | 69(37) 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | | 157 4 (137) | 6 9 2 | (378) 57 58 | +-------------------+-----------+-------------------+ UR<37>r79c37 || 7r7c1 ------------- 7r1c1 || 1r9c3 - (1=7)r2c3 - 7r1c1 || 8r9c7 - (8=7)r1c7 - 7r1c1 => r1c1<>7; stte

so all three scenarios are pointing r1c1<>7
guess the first one { 7r7c1 represent (69)r7c7 }

then perhaps just that one cell 7r1c1 is enough to prove the DP(37)r79c37
7r1c1 - r2c3 = (37)r79c3
7r1c1 - r7c1 = (37)r7c37
7r1c1 - (7=8)r1c7 - (8=37)r9c7

not only 7r1c1 = DP(37)r79c37 , also 7r1c1 = DP(56)r12c29

by **tlanglet** » Fri Jun 27, 2014 4:14 am

Another view of the AUR(56)r12c29 is

AUR(56)r12c29[8r1c9=9r2c2]-(9=17)r3c1,r2c3-(7=65)r1c12 => r1c9<>56

Ted

by **blue** » Fri Jun 27, 2014 5:13 am

Here's a nice one, using externals:

AUR(56)r12c29[6r1c1=6r7c2] => -6r78c1,r1c2; stte

by **Sudtyro2** » Fri Jun 27, 2014 8:29 pm

Code: Select all: *-----------------------------------------------------------* | 67 *56 4 | 2 1 9 | 78 3 *568 | | 3 *569 17 | 4 8 57 | 2 1579 *56 | | 179 8 2 | 57 6 3 | 79 1579 4 | *-------------------+-------------------+-------------------| | 2 7 5 | 9 4 6 | 1 8 3 | | 8 1 6 | 57 3 57 | 4 2 9 | | 4 3 9 | 1 2 8 | 5 6 7 | *-------------------+-------------------+-------------------| | 679 69 37 | 8 5 1 | 3679 4 2 | | 569 2 8 | 3 7 4 | 69 59 1 | | 157 4 137 | 6 9 2 | 378 57 58 | *-----------------------------------------------------------*

blue wrote:Here's a nice one, using externals:
AUR(56)r12c29[6r1c1=6r7c2] => -6r78c1,r1c2; stte

Hi blue,
Sorry, slow learner on externals, once again! Two questions:
1. How is your 6r1c1=6r7c2 "external" strong inference derived?
2. Ted's ADP External Inferences posting requires a look at all UR digits. What about the UR 5-digit, which has multiple externals?

SteveC

by **blue** » Fri Jun 27, 2014 9:21 pm

Sudtyro2 wrote:1. How is your 6r1c1=6r7c2 "external" strong inference derived?
2. Ted's ADP External Inferences posting requires a look at all UR digits. What about the UR 5-digit, which has multiple externals?

If 6r1c1 is false, there's a <56> locked pair in r1c29, that eliminates 8r1c9.
If 6r7c2 is false, there's a <56> locked pair in r12c2, that eliminates 9r2c2.
If both are false, the combination produces the bare UR pattern.

Following Davids suggestion here, the best I could come up with, was to use one of these AIC's.

6r7c2 = (HP: <56>r12c2) - 9r2c2 = (NP: <56>r2c29) -UR- (HP: <56>r1c29) = 6r1c1
6r1c1 = (HP: <56>r1c29) - 8r1c9 = (NP: <56>r12c9) -UR- (HP: <56>r12c2) = 6r7c2.

If something "smaller" would work, I couldn't see it.
[ David, Danny ? ... any ideas ? ]

by **daj95376** » Sat Jun 28, 2014 12:01 am

_

Essentially, 6r1c1 works as an external in [b1] and [r1] for the DP.

Code: Select all: 6r1c1 = { HP(56-8)r1c29 + HP(56-9)r12c2 } ; DP => =6 r1c1

Given the above scenario, maybe Ted will consider expanding how externals can be used.

_

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