After the simple eliminations are completed in a Sudoku, we sometimes get to a point where the next step isn't obvious. The Pincer is a pattern that may help eliminate one more candidate and get past the impasse. A Chain is an 'extended' pincer and is a bit harder to detect. But once you get adept at finding Pincers and Chains, they become invaluable in solving Sudokus. Others may have called this pattern the 'XY Wing'.

To get started, perform the simple eliminations and mark each cell with possible remaining candidates. Now the goal is to eliminate more candidates. For the basic Pincer, find a sequence of 3 cells, each with only 2 candidates, that have the pattern AB - BC - CA. Further, the cells containing AB and BC should be in the same zone, and BC and CA should be in a zone. 'Zone' means the same row, column or macro cell. If you find such a pattern, then other cells that are in the same zone as both AB and CA cannot have the digit A as a candidate. This is because one of either AB or CA will be the digit A. Think of AB and CA as the tips of the Pincer and they squeeze out any A that is in its grips. Applying this technique to cells with 3 candidates is possible, but is more tricky.

Here is a simple example.

The cells circled in green force the purple cell to NOT
be a 4. So it is 5. Look at the green chain: The candidate 4 is eliminated from the purple cell Click on the puzzle to magnify it.

*b9* - *c8* - *c6*
(or 48 - 58 - 45). Note that the cell *b4* (containing 45) is
controlled by both 45 and 48. *b4*
because it is common to 48 and 45.

In the above example, the green cells *b9* and *c6*
(48 and 45) are both 'controlled' by the green cell *c8* (58). That cell,
called the pivot, can either be a 5 or an 8. If it is 5, then *c6* will
be 4. If it is 8 then *b9* will be 4. Either way, one of the pincer tips
will be a 4. The purple cell marked 45 is controlled by both 45 and 48.
Therefore, the purple cell *b4* cannot be a 4.

After this elimination, the rest of the puzzle can be rapidly solved.

The trick is to find these Pincers. Firstly, all the cells that are part of a pincer are those with exactly 2 candidates. Start with one of them, say AB. Look for a BC that is in the same zone. Then look for a CA. If one is found, check if AC and CA control some other common cell, and if that cell has A as a candidate.

Another simple example:

The cell in purple It becomes 29 and along with the other 29 in

*f4* cannot be a 4.*e4*,
it forces the red 24 in *d4* to be a 4.

Another example, with a rare double yield...

The cells circled green The chain is: 89 - 58 - 59.

*a2*, *c3* and *d3*
force the cells circled in purple *d2* & *e2* to be NOT 9. So
they become 5 and 2.

Now a game where the Pincer is applied twice. First:

Cells circled in green force the cells circled in purple
to be NOT 5. So purple 35 in The chain is: 75 - 37 - 35

*e8* becomes 3.

Soon, we get to this situation. The green cells force the
purple cell in Alternatively, in this case, the chain could start with

*g3* to be NOT 4. It becomes 6. The chain is: 46 - 63
- 34*c1*
(36), then *i1* (34) and *g3* (46). Now the 6 in the green
circled *a3* (46) is eliminated. Either way...

Chains:

Chains are like longer pincers. A chain is defined as a sequence of connected pairs where the first and last cell have a common candidate and they control a different common cell. Consider a chain like: AB - BC - CD - DA. If the two end cells AB and DA can both see a target cell, then the target cannot have an A as a candidate. This is because if the first cell is A, then the target cannot be A. If the first cell is B, then the last cell in the chain will be an A, and again the target cannot be A. The first cell can only be A or B. This is the same logic used while explaining pincers.

Here is a simple chain which is easy to recognize because it has 2 reflecting pairs. In the following, the chain is like: AB - BC - CB - BA. The end points of the chain control a cell that contains A, which is then eliminated. (AKA XY Chain) View the chain as: 28 - 68 - 68 - 28.

Here we have a 68 in That means the 28 in the bottom row Note that there are the two blue 28 could not be used.

*a5* and 28 in *b6*
circled in red. We are looking for a 62, but we find another 68 & 28
in the last row (*a8* & *e8*). The cell in purple *b8* cannot
be a 2.*e8* must be a
2.

Here is another chain: AB - BC - CD - DA. It eliminates A from cells common to AB and DA.

Start with Then we notice the 3 purples starting at

*a2* (green 59) -> *i2* (red 58)
-> *i8* (red 48) -> *h7* (red 49). The chain is: 59 ->
58 -> 48 -> 49. So, *a7* (red 569) cannot be a 9. It becomes a
56.*a7* 56
-> 68 -> 58. This eliminates the two 5s in the bottom row (*b9*
& *c9*, circled yellow)

Sometimes we find a chain while searching for Pincers. In Pincers, when we find a potential AB and BC, we search for a suitable cell with candidates CA. Often we don't find it and continue our search. Sometimes we then find another pair AD and DC which also needs a CA. When we note two separate pairs that need the same missing cell, we can try to join them together. One cell of each pair needs to be share a zone. The previous example had a common missing link, both 59 & 58, and 49 & 48 needed a 98. There is no cell with 98 as candidates, but the 2 pairs fit together.

Here is another example of such a chain: AB - BC - CD - DA. The chain is in green and the affected cell is in red. The two pairs 69 & 46 and 48 & 89 both needed a 49.

The chain is: 69 - 46 - 48 - 89 ( Think of this as 2 pairs that both need a 94.

*f3* - *f4* -
*f8* - *d7*). The cell *d3* circled in red cannot be 9.

And here is another...

*i1*.

In the next piece above, we see Green 78 - Green 17 and we
search for a 18. It is just not there. Then we see Blue 13 - Blue 28 and note
that it too needs an 18. We see that the 17 and 13 share a zone, and *walla!*,
we have a chain that knocks out the 8 from *i1* (red 58). The chain is 78
- 17 - 13 - 38, and the game is over.

Often, there are puzzles where many cells have the same candidate pairs. It helps to form chains out of them. I like to mark them with slanted slashes. Again, this is only for like pairs of candidates. When a reflected chain is long, we can eliminate candidates if a cell sees both left & right slashes for the same candidates, it cannot contain those candidate numbers.

The following puzzle has several cells marked 26. Starting
at *c3*, we mark it with a upward slant. The matching cells *b2* and
*c8* get an downward slant. We find that *c3* and *i9* have
opposite slant. The common cell *i3* cannot contain a 2 or a 6 and must
be a 5. The rest is easy.

Mark chains with matching pairs with slashes,
alternating left and right slant, like / and \. Note two 26s circled in red. They have opposite slant markings.
Any cell common to them cannot have a 2 or a 6.

The next puzzle, by coincidence, also has chains of 26.

The force is strong in this puzzle, but you need to be
persistent. The cells *e3*, *e7* and *d8* circled in red are
clearly connected. If *e3* is 2 then *e7* must be 6 and *d8*
must be 2. Similarly if *e3* is 6, *d8* will be equal to *e3*
and *e7* will be opposite. Marked with right or left slashes (shown in
purple). Cells with the same slant will be equal.

Next, note that there is also a 26 in *c1*. In this
case it is not too hard to connect it with the chain. If *c1* is a 6,
then *c7* will be a 2 and so *e7* will be a 6. But if *c1* is a
2, *c7* is 7, *c4* is 6, *d4* is 7, *d8* is 8, *d1* is
4, *d5* is 2, *d8* is 6 and so *e7* will be 2. In both cases *c1*
will be the same as *e7* and so it should be marked like *e7*. That
was a bit messy, but here comes the reward.

Let's look for pincers. Looking at column 3, *c1* and *c7*
(26 and 27) form a pincer that eliminates 2 in any other cell common to them.
Well, there are no target cells containing a 2. However, we just showed that *c1*
is equivalent to *e7*. So *c7* and *e7* cannot allow any 2s.
This eliminates 2 in *a7*, *b7* and *h7*, the cells marked with
a red underline. The puzzle falls apart easily after this.

The lesson here is that it often helps to find and mark related cells with slashes. They are easy to find and can help form reflected chains.

Finding chains is not easy. Here is one way to do it. But
first, to set the stage, here is a long chain: 18 -> 28 -> 28 -> 38
-> 23 -> 21 (*a5* -> *e5* -> *e9* -> *d8*
-> *f7* -> *b7*). The 2 ends 18 & 21 eliminate candidate 1
from *b5* and *b6*, which forces *i6* to be 1 and then
everything else falls into place. Later, I found a shorter chain: 18 -> 28
-> 23 -> 23 -> 12, with the same effect.

BTW, not all pincers result in useful eliminations. There
is a pincer 69 -> 16 -> 19 that eliminates the 9 in *e1* below, but
nothing exciting came out of that.

This is going to be a brute force search with a way to avoid
dead ends. Say we start with 18 in *a5*. Find another cell in the same
zone with two candidates, one of which is common to the current cell. Cell *e5*
has a 28, where 8 is common to 18 and 28. Drop the 8 and look for 12 starting
from *e5*. If there is no 12, then look for a cell with a common
candidate. The 23 in *f5* has a 2 that is common to 12 and 23. Drop the
2 and look for the remaining candidate pair: 13. There is no 13, but there is
a 23 in *f7*. Drop the 3 that is common and look for the remaining
candidate pair: 12. Wow, there is one at *b7* which completes the chain.
Next, verify that the start and end cells in the chain have a common
candidate. Start cell 18 and 12 share the candidate 1. Then check if there
are any cells common to the start and end cells. Yup, cells *b5* & *b6*
are common and they both have 1 as a candidate. Eliminate the 1 from those
cells and move on.

Most of the examples here are from (c) Daily Sudoku. I thank them for many relaxing hours.

*May 2020*