## ah..stuck again. heh

Advanced methods and approaches for solving Sudoku puzzles

### ah..stuck again. heh

I'm trying to use the "Reducing Candidates" on Simes site but, not sure i'm getting it. Here's the row of candidates.

everything in [] are already set.

(2,4,6)(4,6)[8](2,6,7)[3][9][1](4,5,7)(2,4,5)

Now..4 and 6 show up in the first and second group. But 2 and 4 show up in the first and last group. And then there's 2 and 6 in the first and 3rd group! Argh!! Re-reading the page now but, if someone would'nt mind hinting at what i'm doing wrong..

I'll post the entire thing if it helps but, i'm trying to figure out the method used on that site so.. any tipe here would be great.
dcj1978

Posts: 17
Joined: 25 May 2005

Based on that information you can not remove any of the candidates.
Animator

Posts: 469
Joined: 08 April 2005

Ok, so the fact that I can't find anything that fits what i'm reading on the site doesn't mean I'm really missing something... because there's nothing there.

Reading too much into it I guess

Thanks
dcj1978

Posts: 17
Joined: 25 May 2005

### "ah...stuck again, heh". Author dcj1978

Animator wrote:Based on that information you can not remove any of the candidates.

I have a similar problem as author dcj1978 in relation to "hidden pairs"and this is an example. When candidates (2,4,6) occupy the first cell in a row and (4,6) occupy the second cell in the same row, why can't candidate 2 be removed from the first cell which then leaves a pair (4,6) in these two cells? I hope I'm replying in the proper way this time.
Regards Bonsai Cec
Cec

Posts: 1039
Joined: 16 June 2005

(That sure is a better way to reply :) )

In order to have a hidden pair, you need to find two cells that are the only candidates for two numbers.

If you want a hidden 4-6 pair then the numbers 4 and 6 should occur only in two cells.

In this example the number 2 occurs 3 times, the numbers 4 occurs 3 times and the number 6 also occurs 3 times. This makes a hidden pair of 4 and 6 impossible.

(Ofcourse you could start looking for a hidden triplet (since that requires 3 cells) but that one does not exists either, and you shouldn't be looking for it until you understand hidden pairs and normal/naked pairs)
Animator

Posts: 469
Joined: 08 April 2005

### ah...stuck again. heh

Thanks Animator for your explanation of "hidden pairs". Also relieved that I replied the correct way.
Regards, Bonsai Cec
Cec

Posts: 1039
Joined: 16 June 2005

What's the wrong way? Italics? Deutsch?

(After further browsing) Aaahh, I see.
Karyobin

Posts: 396
Joined: 18 June 2005

Return to Advanced solving techniques