Rows and columns are divided into compartments of white squares. Look at the diagram on the right which has two 'straights' filled in and highlighted. The black cells separate the compartments. Compartments can be both vertical and horizontal.

Squares in compartments need to be filled in with single numbers. These must complete a 'straight'. A straight is a set of numbers with no gaps and in any order, such as [6,8,7] or [3,2,4,5], as shown on the diagram.

No single number can repeat in any row or column - same rule as Sudoku.

Clues in black cells remove that number as an option in that row and column, and are not part of any straight.

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Squares in compartments need to be filled in with single numbers. These must complete a 'straight'. A straight is a set of numbers with no gaps and in any order, such as [6,8,7] or [3,2,4,5], as shown on the diagram.

No single number can repeat in any row or column - same rule as Sudoku.

Clues in black cells remove that number as an option in that row and column, and are not part of any straight.

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Or look at the walkthrough below.

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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.

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.

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.

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.

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.

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.

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.

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.

Because so many numbers have been used we can place the 4.

The remainder of that section falls into place, the order of which I've
indicated with arrows.

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.

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.

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.

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.

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.

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.

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.

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.

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

I hope this was a useful walk though, and best of luck solving Str8ts.

Andrew Stuart

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