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To start with we need to look for small compartments which have several
numbers in them. The 3/1 at the top implies a 2 since 2 is the gap between
3 and 1. Likewise 7/8/9 in the first column.
In the center is a 4 in a two-cell compartment. It's complement can be 3 or 5
but there is a 5 in the black cell in the center row, so the answer is 3.
Lastly, the right-most column has a 7 which could lead to a combination
from 5/6/7 to 7/8/9. To match with the horizontal two-cell compartment
that has the 6 we need 5 or 7. 5 is the only possibility.
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Having made a start we can add 6 to the right-most column since that's the
gap between 5 and 7.
We can add 2 into the central horizontal compartment to
make a straight of 2/1/3.
On the left hand side 6 is implied for the two-cell compartment that has 7
since 8 is used up vertically in the second column.
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Next the 7 can be filled in the small cluster on the right.
At this point a little thought is required. Take the four-cell compartment
in the second-to-last column that ends in 7 and 6. The possibilities range
from 4/5/6/7 to 6/7/8/9. However the 5 in that column removes the lower
possibilities, so we have to fit 8 and 9 in the remaining cells. 8 is used
but in the top row so 9 must start the compartment there.
The placement of 5 might seem a little strange. But consider what has been eliminated
on the row and column. 1/3/4/6/7/9 have been used up leaving 2/5/8. One of these
numbers must compliment the cell to the left to make a two-cell straight. Because
7 and 9 have been used up in the row 8 is what we call a 'stranded digit'. It can't
be part of a straight in the row so it can be dropped from consideration. What about the 2? A 2 would
require a 1 or 3 in the vertical straight but 1 is used up. A 2/3 straight won't leave enough numbers in the rest of the row (4 and 5 remain), so 5 it is.
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Now we can fill in the 8.
Where the 7 has been placed is another process of elimination. Looking at what's
been used up we find 2 and 7 are possible, but vertically, the straight contains
a 5, so 2 is too far away from 5 to be used, since the straight contains 3 cells.
With the 7 in place the 2 on the bottom row comes into place.
We can see that only 2
and 9 remain as possibles, but 9 is rules out because of the bottom row. The straight
is a long one - eight digits, but it has a 1 as a clue. So 9 is off the end of the
set of possibilities. 2 is inserted.
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Now we can attack the section on the lower right. The 1 clue in the last column
is very useful as it means the bottom right corner can only be a 2 or a 3. But the 2
has been place in the row. So 3 goes in. This gives us the 2 in the final column.
Because so many numbers have been used we can place the 4.
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The remainder of that section falls into place, the order of which I've
indicated with arrows.
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In the center the 3 is placed as it plugs the gap 2/?/4.
We can also plug the gap 5/?/7 in the 6th column with 6.
The 4 in the 4/5 straight is deduced from the fact that 6 is a clue in the central
column. To match 5, 4 must be placed.
The 2 in the top left corner is found from the fact that only 1 and 2 remain
as possibilities in that cell. But as the top row is an 8-straight and we have
placed 9, 1 can't fit.
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This gives us 1 in the first column, near the top since 4 is used in the third row.
With very few possibles left the 3 can be placed in the top row and this gives
us the 5 as well.
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We finish the top left section placing 2, 3 and 4 and we complete the top row.
Now, to get to get to the rest of the board we need to look at the remaining cells
carefully. Usually its simply a question of finding a congested cell with many
numbers in the row and columns. The 7 can be placed with certainty since its the
only remaining number.
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The two-cell straight starting with 8 in the 7th row can be completed with 9, having
placed the 7. The arrow shows this.
We have a 1 to place in the 3d column.
That 7 placement also allows us to place the 9 above it in that column, since
7/9 were the only chances.
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We are snow-balling to a conclusion now. The 3 is placed since 2 in the compartment below means 132 is the only fit. The 8 is placed since 9 has been placed in the central
straight.
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Four straights can have their final cells filled now. All for simple
reasons. We have four remaining cell and I leave it to the reader to mentally
fill those in.
I hope this was a useful walk though, and best of luck solving Str8ts.
Andrew Stuart
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