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by **ArkieTech** » Mon Apr 13, 2015 11:19 pm

Code: Select all: *-----------* |5..|2..|..1| |..9|...|7..| |.43|..7|62.| |---+---+---| |...|..3|8..| |.6.|1.8|.9.| |..7|4..|...| |---+---+---| |.86|..4|35.| |..1|...|4..| |7..|9..|..8| *-----------*

by **pjb** » Mon Apr 13, 2015 11:38 pm

Code: Select all: 5 7 8 | 2 cD34 6 | 9 C34 1 6 2 9 | 35 345 1 | 7 8 34 1 4 3 | 8 9 7 | 6 2 5 ------------------------+----------------------+--------------------- 49 19 *25 | 56 *a256 3 | 8 B14 7 34 6 *25 | 1 7 8 |*25 9 34 8 13 7 | 4 *25 9 |*A125 136 26 ------------------------+----------------------+--------------------- 29 8 6 | 7 1 4 | 3 5 29 239 359 1 | 36 8 25 | 4 7 269 7 35 4 | 9 b36 25 | 12 16 8

Using DP (25) r4c35, r5c37, r6c57 => r1c5=4; stte
(6)r4c5 - (6=3)r9c5 - (3=4)r1c5
(1)r6c7 - (1=4)r4c8 - (4=3)r1c8 - (3=4)r1c5

Alternatively, using DP (29)r78c19:
(3)r5c9 = (3-6)r6c8 = r6c9 - (6=3)UR -3 r5c1; stte
Phil

by **SteveG48** » Tue Apr 14, 2015 12:24 am

Code: Select all: *--------------------------------------------------* | 5 7 8 | 2 4-3 6 | 9 a34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | *----------------+----------------+----------------| | 49 19 25 | 56 256 3 | 8 a14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 13 7 | 4 25 9 | 125 136 26 | *----------------+----------------+----------------| | 29 8 6 | 7 1 4 | 3 5 29 | | 239 359 1 | 36 8 25 | 4 7 269 | | 7 35 4 | 9 b36 25 | 12 a16 8 | *--------------------------------------------------*

(3=146)r149c8 - (6=3)r9c5 => -3 r1c5 ; stte

by **Marty R.** » Tue Apr 14, 2015 1:55 am

Code: Select all: +------------+-----------+-------------+ | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | +------------+-----------+-------------+ | 49 19 25 | 56 256 3 | 8 14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 13 7 | 4 256 9 | 125 136 26 | +------------+-----------+-------------+ | 29 8 6 | 7 1 4 | 3 5 29 | | 239 359 1 | 36 8 25 | 4 7 269 | | 7 35 4 | 9 36 25 | 12 16 8 | +------------+-----------+-------------+

Play this puzzle online at the Daily Sudoku site

(1=3)r6c2-(3=621)r9c257=>r6c7<>1

by **Ngisa** » Tue Apr 14, 2015 8:38 am

Code: Select all: +------------+-----------+-------------+ | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | +------------+-----------+-------------+ | 49 19 25 | 56 256 3 | 8 c14 267 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 a13 7 | 4 256 9 | b125 b136 26 | +------------+-----------+-------------+ | 29 8 6 | 7 1 4 | 3 5 29 | | 239 359 1 | 36 8 25 | 4 7 269 | | 7 5-3 4 | 9 e36 25 | 12 d16 8 | +------------+-----------+-------------+

(3=1)r6c2-r6c78=r4c8-(1=6)r9c8-(6=3)r9c5 => -3r9c2; stte

by **Leren** » Tue Apr 14, 2015 11:41 am

Code: Select all: *--------------------------------------------------------------* | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | |--------------------+--------------------+--------------------| | 49 19 25 | 56 256 3 | 8 14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 d13 7 | 4 25 9 | 125 e13-6 26 | |--------------------+--------------------+--------------------| | 29 8 6 | 7 1 4 | 3 5 29 | | 239 359 1 | 36 8 25 | 4 7 269 | | 7 c35 4 | 9 b36 25 | 12 a16 8 | *--------------------------------------------------------------*

(6) r9c8 - (6-3) r9c5 = r9c2 - r6c2 = (3) r6c8 => - 6 r6c8; stte

Leren

by **Sudtyro2** » Tue Apr 14, 2015 8:35 pm

Code: Select all: *--------------------------------------------------------------* | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | |--------------------+--------------------+--------------------| |*9-4 19 25 | 56 256 3 | 8 14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 13 7 | 4 25 9 | 125 136 *26 | |--------------------+--------------------+--------------------| | 29 8 6 | 7 1 4 | 3 5 29 | | 29+3 359 1 | 36 8 25 | 4 7 29+6 | | 7 35 4 | 9 36 25 | 12 16 8 | *--------------------------------------------------------------*

Phil's ADP(29)r78c19, but again trying “externals.”
Column externals(*) needed are 9r4c1 and 2r6c9, giving (notation?):
ADP[9r4c1 = 2r6c9] – (2=154)r56c7,r4c8 => -4r4c1; stte

SteveC

by **Marty R.** » Wed Apr 15, 2015 3:51 pm

Sudtyro2 wrote:
Code: Select all
*--------------------------------------------------------------* | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | |--------------------+--------------------+--------------------| |*9-4 19 25 | 56 256 3 | 8 14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 13 7 | 4 25 9 | 125 136 *26 | |--------------------+--------------------+--------------------| | 29 8 6 | 7 1 4 | 3 5 29 | | 29+3 359 1 | 36 8 25 | 4 7 29+6 | | 7 35 4 | 9 36 25 | 12 16 8 | *--------------------------------------------------------------*

Phil's ADP(29)r78c19, but again trying “externals.”
Column externals(*) needed are 9r4c1 and 2r6c9, giving (notation?):
ADP[9r4c1 = 2r6c9] – (2=154)r56c7,r4c8 => -4r4c1; stte

SteveC

I wish I knew more about externals. I thought both deadly candidates have to be covered. There doesn't seem to be coverage for the 2 in r8c1 or the 9 in r8c9. That makes me wonder if the strong inference should be: (9r8c2=2r8c6).

Where's Ted when he's needed? :lol:

by **Sudtyro2** » Wed Apr 15, 2015 6:18 pm

Really good questions, Marty, that I asked myself (several times). Here's what I finally got from the first three examples in Ted's ADP treatise:

Ted says (in his Example 1) that you must select any combination of inferences that cover all eight AUR digits. But, in his Example 2, he makes use of houses that are EMPTY of externals to do the covering. To me, this simply means that two UR digits must form a locked pair in that house. So, for the AUR above, c1 (empty of external 2s) covers the locked UR 2s, 2r89c1, while c9 similarly covers the locked UR 9s, 9r78c9. That then leaves the two externals to cover the remaining four UR digits, which in turn gives you the needed strong-inference link between the two externals.

So, just to illustrate further...

Code: Select all: *--------------------------------------------------------------* | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | |--------------------+--------------------+--------------------| | 9-4 19 25 | 56 256 3 | 8 14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 13 7 | 4 25 9 | 125 136 26 | |--------------------+--------------------+--------------------| | 29 8 6 | 7 1 4 | 3 5 29 | | 29+3 *359 1 | 36 8 25 | 4 7 29+6 | | 7 35 4 | 9 36 25 |*12 16 8 | *--------------------------------------------------------------*

Suppose we pick houses b7 and b9 to cover those remaining four UR digits. B7 has external 9r8c2 to cover the 9s, and b9 has external 2r9c7 to cover the 2s. The chain is a bit longer, but you get the same elimination:
9r4c1 = 9r4c2 – AUR[9r8c2 = 2r9c7] - (2=154)r56c7,r4c8 => -4r4c1; stte

The same method was also applied to Phil's 6-cell ADP(25) in the same grid. I'll spare you the details, but I needed only column externals, 5r2c5 and 2r9c7. This gives:
ADP[5r2c5 = 2r9c7] – (2=154)r56c7,r4c8 – 4r1c8 = 4r2c9 => -4r2c5; stte.

SteveC

by **Marty R.** » Wed Apr 15, 2015 7:48 pm

Re: April 14, 2015

Postby Sudtyro2 » Wed Apr 15, 2015 1:18 pm
Really good questions, Marty, that I asked myself (several times). Here's what I finally got from the first three examples in Ted's ADP treatise:

Ted says (in his Example 1) that you must select any combination of inferences that cover all eight AUR digits.

I'm sure I read that treatise at one time, but probably a long time ago. I don't understand the need to cover eight digits, or all four cells of the rectangle. Why cover the bivalue cells? If you cover the polyvalue cells, it's impossible to leave a Deadly Pattern.

by **Sudtyro2** » Wed Apr 15, 2015 9:25 pm

Code: Select all: *--------------------------------------------------------------* | 5 7 8 | 2 34 6 | 9 34 1 | | 6 2 9 | 35 345 1 | 7 8 34 | | 1 4 3 | 8 9 7 | 6 2 5 | |--------------------+--------------------+--------------------| |*9-4 19 25 | 56 256 3 | 8 14 7 | | 34 6 25 | 1 7 8 | 25 9 34 | | 8 13 7 | 4 25 9 | 125 136 *26 | |--------------------+--------------------+--------------------| | 29 8 6 | 7 1 4 | 3 5 29 | | 29+3 359 1 | 36 8 25 | 4 7 29+6 | | 7 35 4 | 9 36 25 | 12 16 8 | *--------------------------------------------------------------*

Marty,
Back to the original AUR(29)r78c19, grid and posting, I always have to remind myself that the key issue is that (from internals) either 3r8c1 or 6r8c9 MUST be true. Otherwise, you've got a real UR to deal with, and the puzzle is invalid.

So, looking only at c1 in the above grid, the 2r78c1 pair is locked (one must be true), so for internal 3r8c1 to be true we have to kill both 9s in r78c1, which happens if the external 9r4c1 is true. OR, looking at c9, the 9r89C9 pair is locked (one must be true), so for internal 6r8c9 to be true we have to kill both 2s in r78c9, which happens if the external 2r6c9 is true. Bottom line is that at least one of two externals must be true.

Ted has figured out a universal Rx to account for many ADPs in general using only the externals involved.

SteveC

by **pjb** » Wed Apr 15, 2015 11:16 pm

From my simplistic viewpoint, if 9 is false in r4c1, then r78c1 becomes a hidden pair, and r8c1 = 29. If also 2 is false in r6c9, then r78c9 also becomes a hidden pair, and r8c9 =29, ie the DP is realized. So if r4c1<> 9, r6c9=2 and vice versa.

Phil