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# Packing

The packing principle says that if there is a matching between
two sets, *n* things in one set cannot have their partners
inside a set of size smaller than *n*.

## Two things cannot be where there is room for one

From two pairs in a box one concludes to the point of intersection:

since one box cannot have two 1's.

Or again, with two intersecting pairs for the same digit, but not in a
single box:

since otherwise two different values are forced for the point
opposite to the point of intersection.

## Three things cannot be where there is room for two

For example, suppose all possible positions for digit 3
in rows 2 and 5 are as shown.

Then position (6,7) cannot be 3, otherwise also position (5,3)
would be 3, and moreover one of positions (2,3) and (2,7),
so that columns 3 and 7 would have three digits 3.

**Terminology** People call this 'X-wing with fin'
('finned X-wing').

## Four things cannot be where there is room for three

After some work, the puzzle shown has proceeded to the given stage,
and the usual methods make no progress. Now look at the yellow cells.
They form an almost locked set, and if (1,6)2 then they form a
locked set and must have the digits 4,6,8. Then in the middle bottom
box digits 4,7,8 must fill the middle column. But also the digit 2
must be there, since it cannot be in the middle top box if (1,6)2.
Hence (1,6)!2, that is, (1,6)9.

## Four things cannot be where there is room for 3.5

Consider the following diagram, where only the candidates
for digit 8 are indicated (with a # mark). The goal is to
conclude that position (1,1) cannot have an 8.

One way to proceed is to assume that (1,1) has an 8, and to
count how many digits 8 are seen in the seven areas indicated
(four columns, two rows and one box). Since each of these areas
should contain a unique digit 8, the total should be 7.
Now each # mark is covered by at least two of these areas,
so is counted at least twice,
and that means that there is room for at most 3.5 digits 8
where we need 4. Therefore position (1,1) does not have an 8.

(In this case one can also follow the chain: if (1,1), then also
(4,2), then also (7,9), then also (2,3), then no further choice
is possible.)

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