This example shows the kind of logic involved in using cage combinations in Killer Sudoku. There are many leaps of logic to be found in this type of puzzle, and this should also explain in more detail some of the results from the Killer Sudoku Solver. In this instance the solver returns

KILLER STEP (Hard) on D3: shape of length 3 with clue of 15+ can only be 1/2/3/4/5/6/8/9, removing 7

The question is - why does it eliminate the 7 in D3 and not D1?

Hovering over the 15 on the small board it lists the combinations. There are quite a few in this cage:

1/5/9

1/6/8

2/4/9

2/5/8

2/6/7

3/4/8

3/5/7

4/5/6

KILLER STEP (Hard) on D3: shape of length 3 with clue of 15+ can only be 1/2/3/4/5/6/8/9, removing 7

The question is - why does it eliminate the 7 in D3 and not D1?

Hovering over the 15 on the small board it lists the combinations. There are quite a few in this cage:

1/5/9

1/6/8

2/4/9

2/5/8

2/6/7

3/4/8

3/5/7

4/5/6

Now...

1/5/9 <-- no 7 used at all

1/6/8 <-- no 7 used at all

2/4/9 <-- no 7 used at all

2/5/8 <-- no 7 used at all

2/6/7

3/4/8 <-- no 7 used at all

3/5/7

4/5/6 <-- no 7 used at all

So if 7 is used (either D1 or D3) the combos are

3/5/7

2/6/7

But there is no 3 in the cage, so we are left with

2/6/7

Try and fit that in the cage - it can only go one way with the candidates that remain:

7/6/2 fits in D1,D2,D3

But there is no 2 and 6 in D1 and D2 if D3 is a 7

Voila...if - and its still and IF...7 is used the solution will be {7,6,2}, so the solver will remove 7 from D3.

Of course we don't know if 7 is actually used in the combo so that's as far as we can take it at this stage.

1/5/9 <-- no 7 used at all

1/6/8 <-- no 7 used at all

2/4/9 <-- no 7 used at all

2/5/8 <-- no 7 used at all

2/6/7

3/4/8 <-- no 7 used at all

3/5/7

4/5/6 <-- no 7 used at all

So if 7 is used (either D1 or D3) the combos are

3/5/7

2/6/7

But there is no 3 in the cage, so we are left with

2/6/7

Try and fit that in the cage - it can only go one way with the candidates that remain:

7/6/2 fits in D1,D2,D3

But there is no 2 and 6 in D1 and D2 if D3 is a 7

Voila...if - and its still and IF...7 is used the solution will be {7,6,2}, so the solver will remove 7 from D3.

Of course we don't know if 7 is actually used in the combo so that's as far as we can take it at this stage.

## Comments

Comments Talk## Saturday 1-Jun-2013

## ... by: PeteTy

i see we are only discussing the 15(3) cage8 in row 4, 3 in the box leaves only as possible

159 249 267 456

so Andrew's note

with 3 and 8 removed:

Now...

1/5/9 <-- no 7 used at all

---

2/4/9 <-- no 7 used at all

---

2/6/7

---

---

4/5/6 <-- no 7 used at all

becomes:

1/5/9 <-- no 7 used at all

2/4/9 <-- no 7 used at all

2/6/7 <-- a 7

4/5/6 <-- no 7 used at all

So if 7 is used (either D1 or D3) the combos are

2/6/7

cant be in col 3 since that is the only 2 in the cage

Now I understand the explanation....

doing a scan of each digit in the cage gives results that dont eliminate another digit.

I still dont understand why combinations with 3 and 8 are considered or mentioned

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as for the rest of the example picture

The cage F7 - F9 has a digit that cant exist but i cant resolve which digit can be removed since i don't know the cage sum

so box 3 and box 5 both have 2 cages where i cant figure the sum

----------------------

Perhaps if the picture just showed F1 to F4 and F1 to F4

i wouldnt worry so much about trying to solve the puzzle ... just worry about the cage with the example.