dailysudoku.com

[Leo P]

Having arrived at

528|391|647
493|726|xxx
176|845|xxx
---+ ---+---
xxx|653|7xx
6x7|489|xx2
xx5|172|x6x
---+ ---+---
x54|967|x31
x6x|xx4|97x
7x9|xx8|4x6

I completed the puzzle using trial and error. Surely there is something more elegant! What am I missing?

[Angel]

Hi Leo,

Did you notice the naked triple (2,3,8) and hidden pair (1,5) in column 7?
This should finish it off without trail and error.

[Leo P]

Hi Angel,

Thank you. I was certain I had overlooked something. It is much better than brute force.

Leo

[Ian]

This is the first VERY HARD Sudoku that I have solved without using HINTs or asking for help. I find the most difficult aspect of solving these is not so much the use of advanced techniques but recognising the patterns.

[David]

I am afraid I got as far as Leo (first post), but can not see how to reduce it down to the hidden pair / triplet. Could someone please explain ?

Thanks

[Angel]

Hi David,

In column 7 there are precisely 2 cells that can contain a 1 or a 5. Every other cell in this column already has a 1 and a 5 in its row. These 2 cells are r2c7 and r5c7. Since we have 2 cells and 2 values, we know that these 2 cells contain these numbers (we don't know in what order) and no others. This is what is called a hidden pair. The particular cell we are interested in is r5c7 -which we now know must contain 1 or 5. But we also know that r5c8 can only contain 1 or 5 (checking the rest of row 5 and column 8). Therefore the only place for a 3 in row 5 is now r5c2.

The other way of looking at this is to consider the possibilities for placing 2, 3 and 8 in column 7. Looking at the candidates for r3c7, r6c7 and r7c7 we get:
r3c7 2,3
r6c7 3,8
r7c7 2,8

That is we have 3 cells and 3 possible values, we don't know how they are distributed, but somehow these values must cover these three cells. The important thing to realise is that they cannot be in any other cell in this column. In particular 2,3 and 8 can be eliminated from r2c7 and r5c7. This in turn leads to the same conclusion that r5c7 must be either 1 or 5, and hence r5c2 must be 3. This is called a naked triplet.

It doesn't matter which of the hidden pair/naked triplet you spot, they lead to the same conclusion.

I hope this helps.

[David]

Thanks Angel. Embarrassingly obvious when you point it out, I will have to turn down my indie rock music down a bit - the waves must be pickling my synapses.

david