The New Sudoku Players' Forum

[SuDoKu_StUdeNt]

Hi everyone,
I'm looking for a sudoku puzzle that requires the hidden pairs technique in order to solve it. I was searching the net for one puzzle that requires hidden pairs, but it turned out that my "mastered" methods:

naked pairs
naked triplet
naked quads
locked candidates type 1 & 2

were enough to solve it without using hidden pairs. I was able to find puzzles that require hidden triplet and hidden quads ( = the methods mentioned before weren't enough), but not a single puzzle that require hidden pairs.
Is there such a puzzle? Is there a possibility that once you master the naked subset and locked candidates the hidden pairs technique is no longer necessary?

[Nick67]

When you spot one of the naked patterns,
there is a good chance that there is a hidden pattern
lurking in the other cells.

Here is an example row of 9 cells, 4 of
them solved and 5 containing candidates:

4 [2368] 9 [68] [123] 5 7 [18] [168]

Cells 4, 8, and 9 form a naked triple.
Inside cells 2 and 5 is a hidden pair.
If you spot the triple first, then you can
remove the 6 and 8 candidates from cell 2,
and the 1 candidate from cell 5 ... and
the hidden pair is gone (naked now), without
ever having been noticed.

So ... I am suggesting that it seems likely that
you've just been missing the hidding pairs,
in noticing and acting on the naked patterns.

How likely is the above example? Well, generalizing
a bit, consider any row with 4 solved cells. If there is a
naked triple in the row, the other 2 cells must either
form a naked pair or hold a hidden pair.

[r.e.s.]

Some of the puzzles in the archive at
http://www.palmsudoku.com/pages/s-o-t-d.php
might be what you're looking for. (E.g., check out the "Tricky" puzzles for Oct. 11, 12, 15, 19)

[tso]

I tried those four puzzles -- I solved them all without using hidden pairs. Though may be some cases where using a hidden pair is arguably the simplest "next step", I agree with SuDoKu_StUdeNt's conjecture that hidden pairs are never be required once once the other tactics mentioned are mastered.

[r.e.s.]

I tried those four puzzles -- I solved them all without using hidden pairs. Though may be some cases where using a hidden pair is arguably the simplest "next step", I agree with SuDoKu_StUdeNt's conjecture that hidden pairs are never be required once once the other tactics mentioned are mastered.

Interesting ... Is there a logical basis for the conjecture, other than lack of a counterexample? (Perhaps they are just quite rare?)

BTW, I was only trying to offer a convenient source of possibilities to check out, and was not claiming they would do so. In particular -- as I think you yourself pointed out recently -- the mentioned site nicely exhibits a sufficient set of solving methods for each puzzle.

[PaulIQ164]

Well, a hidden pair is always countered by a naked set comprising the remaining empty cells in the unit. So if you've already got naked singles, pairs, triplets and quads in your arsenal, you'd only run into problems if you had a hidden pair and a naked quintet, sextet, or septet, which I'd have to imagine are pretty rare, but possible surely?

[Lardarse]

A naked 5-group would eventually be findable, but you would need a full candidate breakdown do have any chance to see it.

[tso]

Interesting ... Is there a logical basis for the conjecture, other than lack of a counterexample? (Perhaps they are just quite rare?)

I have no logical basis. It's more of a challenge for someone to come up with a counterexample than a claim that one doesn't exist.

[Lummox JR]

If your search for naked sets only extends as high as quads, then you'll definitely need hidden pairs, triples, and quads for some puzzles. Each naked set has a complementary hidden set, of a size depending on the number of unfilled cells in the group.

[QBasicMac]

...not a single puzzle that require hidden pairs.
Is there such a puzzle?

I believe there are many. I see them all the time. Maybe I don't understand.

So I went looking for today's sudoku. Actually they already posted tomorrows, Oct 27th. OK, I will try that. I believe that r8c1/r9c1 are what you are looking for. Right? First random puzzle I tried.

Mac

Original Puzzle

Code: Select all

5-- --- ---
6-- --4 3--
13- 67- ---
4-- 5-- 2-1
--- 286 ---
2-5 --1 --6
--- -35 -48
--1 4-- --9
--- --- --2


Solution So Far

Code: Select all

5-- --3 --4
6-- -54 3-7
134 67- --5
463 597 281
-1- 286 453
285 341 --6
--- -35 -48
-51 4-- --9
-4- --- 5-2


Pencilmarks So Far

Code: Select all

-         279       2789      189       12        -         1689      1269      -       
-         29        289       189       -         -         -         129       -       
-         -         -         -         -         289       89        29        -       
-         -         -         -         -         -         -         -         -       
79        -         79        -         -         -         -         -         -       
-         -         -         -         -         -         79        79        -       
79        279       2679      179       -         -         167       -         -       
378       -         -         -         26        28        67        367       -       
3789      -         6789      1789      16        89        -         1367      -       


[QBasicMac]

Err, of course if you notice the naked 79's above, the hidden pair of 38's becomes naked, too.

Now the question is, does there exist a puzzle for which no further progress can be made on simpler rules and one must either find the hidden pair or else use more advanced techniques (hidden triples, x-wings, etc.)

I bet a zillion dollars the answer is "yes", but I'm too lazy to look some more. It's like looking for a puzzle where r3c5=3, r8c7=4 and r9c2=9. I'm sure one exists, but would get bored looking for it.

(Actually, on a bet, I would do it. Simple: Find any solved puzzle where those three cells are unique and then substitute all numbers on the grid. Voila)

Mac

[cho]

Now the question is, does there exist a puzzle for which no further progress can be made on simpler rules and one must either find the hidden pair or else use more advanced techniques (hidden triples, x-wings,etc.)

If a hidden pair exists then the remaining candidates in the row/column/box they are in must make a naked set (or single - I guess you could say {1}{123}{123} contains a hidden pair) sized to the number of remaining unique candidates. How could it not be? If there are five remaining candidates, there has to be exactly five cells for them to go in. Hence they are (or were) a naked quintuple. Of course the value of knowing this has diminished somewhat after you find the hidden pair. ;)

So if spotting the naked quint is considered simpler than the hidden pair, your answer is no. Both produce the same result. Hopefully I haven't slipped up on the terminology, I could sure use a zillion bucks.

cho