The New Sudoku Players' Forum

[baea808]

hi

okay, so, there's naked/hidden pairs and triplets (sets), and when they share 2 houses, they are also called "locked" sets (which allows eliminations in both houses instead of one).

so good so far, but!

all the cases of Locked Sets i've seen so far only produce those extra eliminations when the "previous" set was a NAKED Set.
i've never seen a Locked HIDDEN Set produce the extra eliminations on the second house, and i'm wondering if there's a reason for this. or if i'm just plain wrong.

if someone can think of an example to prove me wrong, i'd much appreciate it!!!

i have this following (INVALID) example...
i modified this puzzle by only removing a 9 from r1c4, so i could prove this point....
... but turns out that invalidated the whole puzzle (this position has now 3 solutions instead of 1):



as you can see, there is a HIDDEN 1-9 pair (in green) in box 1 AND c3 (meaning this is a LOCKED pair).
are the eliminations (in red, r78c3) from this pair legal?

my guess is yes, since they could also be achieved by the Pointing Pairs of 1 and 9, but it still feels a bit like cheating, since i had to use an invalid example (if it were valid, there would be no pair to begin with).

what are your thoughts on this? do you have a valid example of a Locked and Hidden Pair/Triplet that produces eliminations in both houses?

[creint]

Hidden sets: N digits locked in N cells in single constraint resulting in:
1. exclusions from other digits in cells
2. exclusions from target cells seeing those digits. But can also be found using locked singles.

In your example, 1,9 locked in box 1.
Locked singles will be first and excluding in 1,9 in col 3.
After that Hidden set 1,9 will remove 7 from r12c3.

Here an example:

Code: Select all

.------------------.-------------------.--------------------.
| 9     25    258  | 234   247    347  | 278  6       1     |
| 6     1     4    | 8     27     9    | 257  257*     3     |
| 78    237   2378 | 6     5      1    | 4    9       28    |
:------------------+-------------------+--------------------:
| 2     457   567  | 1     4678   478  | 568  3       9     |
| 457   3457  9    | 234   24678  3478 | 1    2458    24568 |
| 1     8     36   | 2349  2469   5    | 26   24      7     |
:------------------+-------------------+--------------------:
| 458   9     1    | 7     3      2    | 568  458     4568  |
| 4578  2457  2578 | 459   1489   6    | 3    124578*  2458  |
| 3     6     2578 | 45    148    48   | 9    124578*  2458  |
'------------------'-------------------'--------------------'

17 locked in box 9 column 8, but that 7 can be excluded in box 3 by locked singles using box 3 or box 9.
After the exclusions in r89c8 a locked single in box 9 is revealed.

[baea808]

In your example, 1,9 locked in box 1.
Locked singles will be first and excluding in 1,9 in col 3.
After that Hidden set 1,9 will remove 7 from r12c3.

thank you very much for your response!

so, a Hidden Set can never produce the extra eliminations from a Locked Set because these are the same eliminations as its Locked Singles?

[rjamil]

Actually, you can.

R. Jamil

[creint]

No. A single fully locked hidden set will become n locked singles. With almost hidden sets you can do some tricks but that is not the question.

[baea808]

Thank you both for your replies.

Indeed, for my purposes, it would have to be a "fully" locked set. But those are some interesting takes which I may use later on. Thank you again.

[baea808]

for documentation purposes, I think I have found a case where a Locked Hidden Pair produces extra eliminations:



Consider Row 8: digits 2-5 can only be inside Block 8: a locked hidden pair.
now, there is one elimination to make (candidate 5 in r7c4, circled in pink) .... which, of course, could also be made with locked singles (type 2 - claiming).

the picture is from the Pairs video from Sudoku Swami: https://www.youtube.com/watch?v=NBi6ivpq6_8

[StrmCkr]

from the post directly above this one .

its not producing extra eliminations they have always been valid

the problem lies with in the context of the solving engine you used to find them most of them are contextually constricted to apply the eliminations inside the sector and the cells used to apply the eliminations.
then the hierarchy applies the elimination from a simpler technique {box/line/reduction}. aka 1 fish, locked candidate{s}

now if the engine actually applies the 3 sectors of the found cells and tracks the digits in full
it would also include the box,row and col for line of sight eliminations and remove those outside the sector/cell.

this can add more time to the solving engine and slow it down so it is often skipped as blr is the clean up.

similar effects can be seen with many other techniques in pairs,triple, quad... etc
for a sector with N digits in N cells {naked} or N cells with N remaining digits {hidden}
if n digits of the set above all fall inside a 2nd sector they apply eliminations for those digits on that sector <- usually skipped for clean up
{or more simply for each digit of the set eliminate any cell that is visible to all cells containing that digit}

12 . 123| . . .| . . 13 ->>> eliminations for 123 on the row , and 2 in the box. most solvers will never show the 2 in the box and locked sets pick it up.

[kurbads]

Dude, you have 2 digits 1 and 9 (not 1–9) that can be only in one row in a block.

They can not be on the same row in another block.

That is not a technique. That is a basic principle of sudoku.

Furthermore, they are not hidden. They are very obvious.