The New Sudoku Players' Forum

[Guest]

i am baffled can anybody explain this to me please?


cheers

Flipside

[Anette]

When you look at a row or column or box, where you have several cells still not filled, you look at which candidates that are possible for each cell.

If there are 3 cells that contain 3 values and ONLY these values, i.e {1,8}{1,3}{1,3,8}, then you know that these 3 values have to be in these cells and therefore they can be eliminated as candidates from all other cells in this row (column, box).

Note that not all three of them have to be in each cell, just as long as they are NOT in any of the other cells.

[simes]

My attempt is here

Simes

[Guest]

i must be really dumb annette/sime i still dont get it

this is the puzzle that i am working on and according to simple sudoku i have a naked triple, but i cant c the logic behind it

.5.|...|..7
4..|59.|.61
...|.2.|.85
---+---+---
.6.|3..|..4
1..|.7.|..3
8..|..4|.9.
---+---+---
52.|.6.|...
79.|.15|..6
6..|...|.1.


.5.|..1|..7
4..|59.|.61
.1.|.2.|.85
---+---+---
.65|38.|174
14.|.7.|853
8..|154|692
---+---+---
521|.6.|7..
79.|.15|..6
6..|...|51.


{239} {5} {2689} {468} {34} {1} {2349} {234} {7}
{4} {78} {278} {5} {9} {378} {23} {6} {1}
{39} {1} {679} {467} {2} {367} {349} {8} {5}
{29} {6} {5} {3} {8} {29} {1} {7} {4}
{1} {4} {29} {269} {7} {269} {8} {5} {3}
{8} {37} {37} {1} {5} {4} {6} {9} {2}
{5} {2} {1} {489} {6} {389} {7} {34} {89}
{7} {9} {348} {48} {1} {5} {234} {234} {6}
{6} {38} {348} {24789} {34} {23789} {5} {1} {89}

Thanks allot

For your help

Flip

[Moschopulus]

{6} {38} {348} {24789} {34} {23789} {5} {1} {89}

This is your last row. You can see {38},{348} and {34}.
This is a triple. So the digits 3,4 and 8 go in these three cells (you don't know which cells yet).
You can deduce that 3,4,8 cannot go in any other cells in this row.
So you can remove 4 and 8 from {24789}, you can remove 3 and 8 from {23789}, and you can remove 8 from {89}.

You still don't know which way 3,4,8 go into those three cells, until you do more work.

[Guest]

Moschopulus

i still dont get the logic of this,

how do u know that this is a triple, thats the bit its making my brain hurt :>

{38},{348} {34}.

thanks for your patience, still a noobie at this

Flip

[Anette]

There are 3 cells.

There are 3 values that can go in ONLY these cells.

Since they can go nowhere else, they have to go in these cells (but note, you can't tell which goes in which, you just use this information to do other eliminations).

Even though they can't go all of them in each of these 3 cells, they still form a triple since there are no other numbers that can go there.

[PaulIQ164]

There are 3 cells.

There are 3 values that can go in ONLY these cells.

This is the wrong way round surely? The whole point of the technique is that you're deducing that the three numbers can only go in these cells.

Look at it this way: you have three cells with {38},{348} {34} as candidates. Whatever way you try and fill these three in, you'll use the numbers 3, 4 and 8. So those numbers can't be anywhere else in the row.

Perhaps an even better way (certainly more obvious to my eye) is to look at the other three numbers that need filling in on that row, the 2, 7 and 9. They can't be in the first three cells of the row (because of the 2 7 and 9 already in that box), and they can't be in the middle cell of the row (because there are already a 2 7 and 9 in column 5). This only leaves three cells for the 2, 7 and 9 to go in. The cell on the far right (r9c9) can't have the 2 or the 7 in it, so it must be the 9.

[stuartn]

PaulIQn - you're using the triple to exclude candidates from other cells - which is exactly what was being got at in earlier posts. You're surely making the issue more confused for newcomers to this august forum.

stuartnIQ>n

[PaulIQ164]

I just thought it's generally easier to visualise concepts if they can be explained directly from the numbers already placed in the puzzle, rather than a bare list of candidates.

[Anette]

I was saying it wrong.

It should have been: In those 3 cells, only these 3 values can go, no other values. And since there are 3 values that have to go in those 3 cells, they have to occupy those cells and cna therefore go nowhere else.

[silverpie]

There are 3 cells.

There are 3 values that can go in ONLY these cells.

This is the wrong way round surely? The whole point of the technique is that you're deducing that the three numbers can only go in these cells.

Both are actually valid techniques. In the naked triple, you know that the cells forming the triple can only receive those three values, and you deduce that is where those three numbers go. If instead you have three numbers each of which is restricted to the three cells (or a subset), and thus deduce that those numbers are the only ones that can go there, you've used a hidden triplet.

[Rostis]

how do u know that this is a triple, thats the bit its making my brain hurt :>

{38},{348} {34}.

OK, how about this. Try to put one of these three numbers, say the 8, in a totally different cell. Then the left one is three, the right one four, and you have nothing left for the middle cell. You have to keep these three numbers for these three cells.

Did that help?

/Rostis

[Guest]

can i run somthing by you guys/girls

because the numbers 3 4 8 only happen in 3 boxes (cells), in that these three boxes (cells) have no other numbers other than 3 4 8

then u can safely exclude them from any other box in this row

is this a correct understanding?

and it is called a triple due to have 3 numbers or less or is it called a triple because you muct have three boxes (cells)

thanks everybody for you help

Flip

[stuartn]

Both - the 3 numbers can only occupy the 3 cells.