The New Sudoku Players' Forum

[David P Bird]

Code: Select all

+-----------------+-----------------+-------------------+
| 6   59    8     | 3      7      4 | 1      59     2   |
| 4   179   179   | 2      (16)   5 | 6789   789    3   |
| 3   2     157   | (16)   9      8 | (567)  4      57  |
+-----------------+-----------------+-------------------+
| 9   8     (157) | (15)   (135)  2 | 3-57   6      4   |
| 17  1567  3     | 14568  1456   9 | 2      578    578 |
| 2   4     56    | 568    (356)  7 | 358    1      9   |
+-----------------+-----------------+-------------------+
| 17  1379  1479  | 45     2      6 | 5789   35789  578 |
| 8   67    467   | 9      45     3 | (57)   2      1   |
| 5   39    2     | 7      8      1 | 4      39     6   |
+-----------------+-----------------+-------------------+

(7=153)r4c345 - (3=56#1)r6c5 - (56=1#1)r4c4,r2c5 - (1=6)r3c4 - (6=57)r38c7 => r4c7 <> 7

"In either direction?" In the right-to-left direction, only r2c5=6 true or r4c4=5 true does not yield r6c5=3 because it is a tri-valued cell.

Perhaps you're unfamiliar with my (56#1) notation. The argument is true when the the digits hold a single truth.

To walk through the links in the R to L direction:
if (1)r3c4 is true then r4c4,r2c5 can't hold a single truth for digit (1)
if (1#1)r4c4,r2c5 is false then (56)r4c4,r6c6 must be true (ie both of them)
if (56)r4c4,r6c6 is true then (56#1)r6c5 can't hold a single truth
if (56#1)r6c5 doesn't hold a truth then (3)r6c5 must be true
The test is to take any of these statements by itself and check that the inference must hold under all circumstances.

As a comment I think the split node version is actually easer to read than the AIC with an embedded XY-Wing that it ripped off:
(7=153)r4c345 - (3)r6c5 = XYWing(15)r4c4,(56)r6c5,(61)r2c5 - (1=6)r3c4 - (6=57)r38c7 => r4c7 <> 7

I notate it like this to allow the reader to step through the cells in the wing in order. The chain seems far easier to digest reading L to R than R to L.

(Sorry about the earlier typos)
[Edit another typo reported by DAJ]

[sultan vinegar]

To walk through the links in the R to L direction:
if (1)r3c4 is true then r4c4,r2c5 can't hold a single truth for digit (1)
if (1#1)r4c4,r2c5 is false then (56)r4c4,r6c6 must be true (ie both of them)
if (56)r4c4,r6c6 is true then (56#1)r6c5 can't hold a single truth
if (56#1)r6c5 doesn't hold a truth then (3)r6c5 must be true
The test is to take any of these statements by itself and check that the inference must hold under all circumstances.

Whilst I can read and follow your #1 notation, I think we need to be careful with it. Consider the second line in isolation, as required by an AIC.

if (1#1)r4c4,r2c5 is false then (56)r4c4,r6c6 must be true (ie both of them)

But if (1#1)r4c4,r2c5 is false, i.e. r4c4,r2c5 do not hold a single truth for (1), then there are two possibilities: r4c4,r2c5 hold no truths for (1); r4c4,r2c5 hold two truths for (1). If we have memory then we know that we mean the former case and not the latter. But AICs aren't allowed to have memory, so we need to tidy this up. How about:

R to L:
if (1)r3c4 is true then both of (1)r4c4, (1)r2c5 are false
if both of (1)r4c4, (1)r2c5 are false then both of (5)r4c4, (6)r6c6 are true
if both of (5)r4c4, (6)r6c6 are true then both of (5)r6c5, (6)r6c5 are false
if both of (5)r6c5, (6)r6c5 are false then (3)r6c5 is true.

L to R:
if (3)r6c5 is false then at least one of (5)r6c5, (6)r6c5 is true [OK, exactly one because it is located in a cell, but at least one in general]
if at least one of (5)r6c5, (6)r6c5 is true then at least one of (5)r4c4, (6)r2c5 is false
if at least one of (5)r4c4, (6)r2c5 is false then at least one of (1)r4c4, (1)r2c5 is true
if at least one of (1)r4c4, (1)r2c5 is true then (1)r3c4 is false.

[David P Bird]

But if (1#1)r4c4,r2c5 is false, i.e. r4c4,r2c5 do not hold a single truth for (1), then there are two possibilities: r4c4,r2c5 hold no truths for (1); r4c4,r2c5 hold two truths for (1).

Hi SV

Yes in previous discussions but not in this one I think I made it clear that the #1 symbol means 'at least one' not 'exactly one'.

As each post gathers on average 10 readers some of whom will benefit from the extra detail I tend to be quite definitive in my contributions. However when responding to a single contributor this can appear insulting, so I try to steer a middle line.

DPB

[sultan vinegar]

Another option for this example is to use AMSLS. These can 'linearise' the split node similarly to how ALS can 'linearise' certain branched/memory chains.

Code: Select all

+-----------------+-----------------+-------------------+
| 6   59    8     | 3      7      4 | 1      59     2   |
| 4   179   179   | 2      (16)   5 | 6789   789    3   |
| 3   2     157   | (16)   9      8 | (567)  4      57  |
+-----------------+-----------------+-------------------+
| 9   8     (157) | (15)   (135)  2 | 3-57   6      4   |
| 17  1567  3     | 14568  1456   9 | 2      578    578 |
| 2   4     56    | 568    (356)  7 | 358    1      9   |
+-----------------+-----------------+-------------------+
| 17  1379  1479  | 45     2      6 | 5789   35789  578 |
| 8   67    467   | 9      45     3 | (57)   2      1   |
| 5   39    2     | 7      8      1 | 4      39     6   |
+-----------------+-----------------+-------------------+

In SV's long-hand notation:
(57=3)ALS1357:r3c345 - (3=1)AMSLS1356:r4c4,r26c5 - (1=6)r3c4 - (6=57)ALS567:r38c7 => r4c7 <> 57.

In the more accepted short-hand notation:
(57=13)r3c345 - (3=156)r4c4,r26c5 - (1=6)r3c4 - (6=57)r38c7 => r4c7 <> 57.

It all depends on personal preferences of course. Each approach has advantages and disadvantages.

[David P Bird]

SV, you have a terminology mix-up.

Briefly and without going into further ramifications which I expect you'll enjoy exploring yourself:

To qualify as a Multi-Sector Naked Set with N digits in two houses X digits must be confined to the cells in one house and the remaining N-X digits must be confined to the cells of the other house.

Code: Select all

 *-----------------------*-----------------------*-----------------------*
 | (346)  (356)  <9>     | <8>    <7>    <1>     | (36)   245-3  245-36  |
 | 37-4   <8>    <2>     | <6>    3459   3459    | <1>    79     345     |
 | <1>    367-5  (45)    | 459    3459   <2>     | 79     <8>    3456    |
 *-----------------------*-----------------------*-----------------------*

(36)r1,(45)b1 MSNS:r1c127,r3c3 => r1c8 <> 3, r1c9 <> 36, r2c1 <> 4, r3c2 <> 5
For example this Sue de Coq pattern is a MSNS. Of the four cells two of them in r1 must hold (36) and two of them in b1 must hold (45). No internal chain is needed to prove these eliminations.

Although an XYWing locks digits in the cells of two houses, one of the digits isn't confined to either house within the cell set and must be absent from the overlap cell to enable an internal proving chain. It's not therefore a simple MSNS.

DPB

[sultan vinegar]

Yes, I agree that AMSLS is stretching it because reading from L to R not all of the remaining digits get locked. However, the grouped (1)'s that we need for the next link in the chain do get locked, so the logic works even though it's different to the ALS case where all of the remaining digits do get locked. Reading from R to L the remaining digits do all get locked, so there is no issue in that direction.

[David P Bird]

Picador SV come in now, your bull baiting time is up!