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6 5 1

5 1 3

2 2 must be one of those given as the next available number 5 would be too big. This means that the only available number to add to the 4 is 5.

2 9 5 2 must be one of those given as the next available number 5 would be two big. This means that the only available number to add to the 4 is 5.

26 9 5 The only number that can be combined with the 1 and 2 is 6 meaning that there must be a 9 in the end circle

9 26 5 The only number that can be combined with the 1 and 2 is 6 meaning that there must be a 9 in the end circle

9 26 5 We can now place a 9 in the top middle box through normal sudoku rules and the fact that a 9 can’t lie on one of the lines.

9 26 5 We can now place a 9 in the top middle box through normal sudoku rules and the fact that a 9 can’t lie on one of the lines.

9 26 5 Using similar rules of elimination we can find possible 9’s in the top right and bottom middle boxes.

9 26 5 Using similar rules of elimination we can find possible 9’s in the top right and bottom middle boxes.

9 26 5 Using similar rules of elimination we can find possible 9’s in the top right and bottom middle boxes.

9 26 5 Using similar rules of elimination we can find possible 9’s in the top right and bottom middle boxes.

9 26 5 The remaining two cells in the middle must be a 7 and an 8. The one on the left cannot be a 7 because the remaining two cells on the arrow would have to add up to 4 and the only way to do this is with a 1 and a 3 but we already have a 1 and a 3 on the central vertical line so the circle on the left must be an 8 and the remaining circle a 7.

9 26 5 13 The remaining two cells in the middle must be a 7 and an 8. The one on the left cannot be a 7 because the remaining two cells on the arrow would have to add up to 4 and the only way to do this is with a 1 and a 3 but we already have a 1 and a 3 on the central vertical line so the circle on the left must be an 8 and the remaining circle a 7.

9 26 8 5 The remaining two cells in the middle must be a 7 and an 8. The one on the left cannot be a 7 because the remaining two cells on the arrow would have to add up to 4 and the only way to do this is with a 1 and a 3 but we already have a 1 and a 3 on the central vertical line so the circle on the left must be an 8 and the remaining circle a 7.

9 26 8 5 7 The remaining two cells in the middle must be a 7 and an 8. The one on the left cannot be a 7 because the remaining two cells on the arrow would have to add up to 4 and the only way to do this is with a 1 and a 3 but we already have a 1 and a 3 on the central vertical line so the circle on the left must be an 8 and the remaining circle a 7.

9 26 8 5 7 The remaining two cells in the 8 arrow must add up to 5 which can be made up of a 1,4 or a 2,3. Clearly the 1,4 won’t fit...

9 26 8 5 7 14 The remaining two cells in the 8 arrow must add up to 5 which can be made up of a 1,4 or a 2,3. Clearly the 1,4 won’t fit... It clashes with the central vertical again so we must have a 2,3 in these cells, the 3 being at the end so it does not clash with the central vertical.

9 26 8 5 7 2 3 The remaining two cells in the 8 arrow must add up to 5 which can be made up of a 1,4 or a 2,3. Clearly the 1,4 won’t fit...

9 26 8 5 7 2 3 Let’s examine the top central box now. We have some information about 2..... Let’s look....

9 26 8 5 7 2 3 2 appears in two central columns which means the 2 in the top middle box must be on the right side. It cannot appear in the circle so must be in either of the two cells at the top right.

9 2 26 8 5 7 3 2 appears in two central columns which means the 2 in the top middle box must be on the right side. It cannot appear in the circle so must be in either of the two cells at the top right.

9 2 26 8 5 7 3 We also have a 6 in the left vertical eliminating the left side of the top middle box. It also cannot be with one of the 2’s because it would make the arrow total greater than 9. Can there be a 6 in the central cell?

9 2 6 26 8 5 7 3 We also have a 6 in the left vertical eliminating the left side of the top middle box. It also cannot be with one of the 2’s because it would make the arrow total greater than 9. Can there be a 6 in the central cell?

13 9 2 6 26 8 5 7 3 Well, the arrowhead would need to be a 1 or a 3. (We already have a 2 in that vertical so it cannot be a 2)

13 9 2 6 79 26 8 5 7 3 This means that the circle would need to be a 7 or a 9 but we already have a 7 and a 9 in that vertical so the 6 cannot be in the central cell. There are only two possibilities for the 6 in the two lower right cells.

9 2 6 26 8 5 7 3 This means that the circle would need to be a 7 or a 9 but we already have a 7 and a 9 in that vertical so the 6 cannot be in the central cell. There are only two possibilities for the 6 in the two lower right cells.

9 2 6 26 8 5 7 3 These two cells are also the only place where an 8 would fit in. The 8 cannot be on the central cell either because the only value that the end circle could have would then be 9 and we already have a 9 in that vertical.

9 2 68 26 8 5 7 3 These two cells are also the only place where an 8 would fit in. The 8 cannot be on the central cell either because the only value that the end circle could have would then be 9 and we already have a 9 in that vertical.

9 2 68 26 8 5 7 3 In fact, looking at the right vertical, the only other place to put a 6 or an 8 is the bottom cell. Neither will fit below the 7 or where the 2’s are. This means that tehe cell below the 7 must be a 4 as a 4 cannot fit in where the 2s are.

9 2 68 26 8 5 7 4 3 In fact, looking at the right vertical, the only other place to put a 6 or an 8 is the bottom cell. Neither will fit below the 7 or where the 2’s are. This means that tehe cell below the 7 must be a 4 as a 4 cannot fit in where the 2s are.

9 2 68 26 8 5 7 4 3 Focussing on that vertical on the right, since the 6,8 must occupy the two cells shown, 1 can only fit in with the 2’s.

9 12 68 26 8 5 7 2 4 3 We can also complete the arrow coming down from the 7. The two remaining cells must total 3=!+2 and the 2 must be in the central row so it doesn’t clash with the 2 in the lower middle box.

9 12 68 26 8 5 7 2 4 1 3 We can also complete the arrow coming down from the 7. The two remaining cells must total 3=!+2 and the 2 must be in the central row so it doesn’t clash with the 2 in the lower middle box.

3 9 12 68 26 8 5 7 2 4 1 We now have some possibilities for a 3 in the top central box. It must be in the left vertical.

3 9 12 68 26 8 5 7 2 4 1 However it cannot be in the circle.

9 12 3 68 26 8 5 7 2 4 1 Because the two arrow cells must be 1 or 2 but 1,2 are already accounted for in the middle box so this restricts 3 to the top left cells.

3 9 12 68 26 8 5 7 2 4 1 Let’s take a look at the boxes on the left and see if we can fill in some more 9s...

3 9 12 68 26 8 5 7 2 4 1 Let’s take a look at the boxes on the left and see if we can fill in some more 9s... In the box on the bottom left, there can be no 9 on the right vertical and no 9 on the arrow cells. This leaves top middle and bottom middle as possibilities.

3 9 12 68 26 8 5 7 2 4 1 This allows us to eliminate the central vertical in the middle box as far as 9s are concerned.

3 9 12 68 26 8 5 7 2 4 1 The 9 in the fourth row forces the nine in the left central box Remembering that the 9 cannot lie on an arrow line it must be in the circle.

3 9 12 68 26 8 5 7 2 4 1 The 9 in the fourth row forces the nine in the left central box Remembering that the 9 cannot lie on an arrow line it must be in the circle.

3 9 12 68 26 8 5 7 2 4 1 We can also find some 9 possibilities in the bottom right box. There are only 2 possible options.

3 9 12 68 26 8 5 7 2 4 1 We can also find some 9 possibilities in the bottom right box. There are only 2 possible options.

3 9 12 7 68 26 8 5 2 4 1 Return now to the top central box. Can a 7 be fit in? Let’s examine the diagonal arrow. If there is a 7 here then the circle must be 8 and the other number on the arrow 1 but 1 has already been accounted for so 7 cannot be on the diagonal arrow. What about on the other arrowhead?

3 9 12 37 68 26 8 5 7 2 4 1 If there is a 7 on the arrowhead then the circle must be 8 or 9 but we already have an 8 in the left of centre vertical and a 9 in the third row so there can be no 7 here. The other cells are already claimed by 1268 leaving only the circle at the lower left for the 7.

3 9 12 37 7 68 26 8 5 2 4 1 This in turn gives us two options for 7 in the lower box.

3 9 12 7 68 26 8 5 2 4 1 If there is a 7 on the arrowhead then the circle must be 8 or 9 but we already have an 8 in the left of centre vertical and a 9 in the third row so there can be no 7 here. The other cells are already claimed by 1268 leaving only the circle at the lower left for the 7.

34 9 12 7 68 26 8 5 2 4 1 3 In the top centre box 4 must share its options with 3

34 9 12 5 7 68 26 8 2 4 1 3 This leaves the middle cell as the only option for 5.

34 9 12 5 7 68 26 8 2 4 1 3 Consider now the options for the long arrow coming down from the 9 at top centre...They are very limited! The first two cells at the top containing 12 must add up to 3 which means that the remaining three must add up to six. The only 3 numbers that add up to six are 1,2,3. The next column along has a two in it already so the two cells in that column must contain 12.

34 9 12 5 7 68 13 26 8 2 4 1 3 This means that the final cell with the arrowhead must contain a 2. Check that they all add up to 9. We can now also fill in the diagonal arrow in the top central box. The 5 can be added to 3 or 4 giving us the options of 8 or 9 in the circled cell.

3 9 12 34 5 7 68 8 13 26 2 4 1 8 is the only option for the circle meaning that 3 must be on the arrowhead. This also implies the 6 and the 4 in the top centre box.

3 9 12 4 5 7 6 8 13 26 2 1 68 This in turn implies the 3 next to the 4.

3 9 12 4 5 7 6 8 13 26 2 1 68 This now gives us the 6 in the lower middle box.

3 9 12 4 5 7 6 8 13 26 2 1 This now gives us the 6 in the lower middle box. The 8 is constrained to the same boxes as the 7s.

3 9 12 4 5 7 6 8 13 26 2 1 This now gives us the 6 in the lower middle box. The 8 is constrained to the same boxes as the 7s.

3 9 12 4 5 7 6 8 13 26 2 1 78 This now gives us the 6 in the lower middle box. The 8 is constrained to the same boxes as the 7s.

3 9 12 4 5 7 6 8 13 26 2 1 78 The arrowhead is 78. The only other option for the remaining arrow line is 1 or 5. 1 it is!

3 9 12 4 5 7 6 8 13 26 2 1 78 The arrowhead is 78. The only other option for the remaining arrow line is 1 or 5. 1 it is! The circle head must therefore contain an 8 or 9 but the options for 9 in that row are already accounted for so it must be an 8 in the circle

3 9 12 4 5 7 6 8 13 26 2 1 78 The arrowhead is 78. The only other option for the remaining arrow line is 1 or 5. 1 it is! The circle head must therefore contain an 8 or 9 but the options for 9 in that row are already accounted for so it must be an 8 in the circle

3 9 12 4 5 7 6 8 13 26 2 1 78 Which means that the arrowhead must be a 7

3 9 12 4 5 7 6 8 13 26 2 1 Which forces another 8 in the lower middle. These 8’s imply two possibilities in the box on the lower right, sharing their options with the 9s.

3 9 12 4 5 7 6 8 13 26 2 1 Which forces another 8 in the lower middle. These 8’s imply two possibilities in the box on the lower right, sharing their options with the 9s.

3 9 12 4 5 7 6 8 13 26 2 1 89 Which forces another 8 in the lower middle. These 8’s imply two possibilities in the box on the lower right, sharing their options with the 9s.

3 9 12 4 5 7 6 8 13 26 2 1 89 Which forces another 8 in the lower middle. These 8’s imply two possibilities in the box on the lower right, sharing their options with the 9s. The 9s in the lower middle must also share their options with the 5

3 9 12 4 5 7 6 8 13 26 2 59 1 89 Look for 9s now in the right middle box....

3 9 12 4 5 7 6 8 13 26 2 59 1 89 9 cannot lie on the arrowhead so that only gives us 2 options.

3 9 12 4 5 7 6 8 13 26 2 59 1 89 Note now that the only options for a 9 in the far right column are in the lower right box, rightmost central cell.

3 9 12 4 5 7 6 8 13 26 2 59 1 Which means that the middle cell must be an 8.

3 9 12 4 5 7 6 8 13 26 2 59 1 Let’s look for 7s in the lower left. There can’t be a 7 in the central row because we already have a 7 further along to the right.

3 9 12 4 5 7 6 8 13 26 2 59 1 Also, there can’t be a 7 on the lower left arrowhead because that means that there must be a 12 on the remaining arrow cell and we already have 12 in that row. The 7 can’t be on the other arrrowhead either because it would take the total beyond the circled 9 so the 7 options must be shared with the 9s.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 Also, there can’t be a 7 on the lower left arrowhead because that means that there must be a 12 on the remaining arrow cell and we already have 12 in that row. The 7 can’t be on the other arrrowhead either because it would take the total beyond the circled 9 so the 7 options must be shared with the 9s.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 The top 2 rows of the lower boxes already contain 12 so the only place to fit them is on the arrowheads.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 Because of the 3 on the central line the only place for the lower right 3 is on the top left cell.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 This gives us two options for 3 in the lower right box.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 Consider the options for the circle at the lower right. It cannot be 123 because they are accounted for already. If it was a 4 then it would need to consist of 3+1 but the 1 is already accounted for. If it were a 5 then it would need to be 3+2 but the 2 is already spoken for in the cell above. We have a six already in that row (on the left). We have an 8 or 9 already so there must be a 7 in the circle and the other number on the arrow must be a 4.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 34 Note that this leaves only one option for 5 in the lower row.

3 9 12 4 5 7 6 8 13 26 2 79 59 1 34 Which in turn gives the only option for 9 in the lower centre box and the 7 and 9 in the lower left box.

3 9 12 4 5 7 6 8 13 26 2 1 34 Can we place a 7 in the middle right box?

3 9 12 4 5 7 6 8 13 26 2 1 34 7 cannot be in the top arrowhead because it would bring the total in the top circle to more than 9. So there are only two options for 7

3 9 12 4 5 7 6 8 13 26 2 7? 1 34 But can we have a 7 on that arrowline?

3 9 12 4 5 7 6 8 13 26 2 1 34 It would have to be paired with a 1 or 2 on the arrowhead to give us an 8 or 9 in the circle but we already have an 8 on that vertical and if it were a 9 in the circle then there would need to be a 2 on the arrowhead but we already have a two – in the centre of the box. So 7 cannot be on the arrowhead meaning that 7 must be in the top right of the box.

3 9 12 4 5 7 6 8 13 26 2 1 34 This determines the 7 in the central left box.

3 9 12 4 5 7 6 8 13 26 2 1 34 We can now determine the 8 in the centre right box.

3 9 12 4 5 7 6 8 13 26 2 98 1 34 There are only two options.

3 9 12 4 5 7 6 8 13 26 2 98 8? 1 34 But can there be an 8 on the arrowhead?

3 9 12 4 5 7 6 8 13 26 2 1 34 It would need to be accompanied by a 1 on the arrow but we already have an arrow in the central row. We can now put in the 8 and the 9 in the centre right box.

3 9 12 4 5 7 6 8 13 26 2 1 34 This determines the 8 in the top right box. The 8 is too big to be on the arrow line so it must be in the top right cell.

3 9 12 8 4 5 7 6 13 26 2 1 34 The 9 in the top right box is also determined by the usual sudoku rules.

3 9 12 8 4 5 7 6 13 26 2 1 34 The 9 in the top right box is also determined by the usual sudoku rules.

3 9 12 8 4 5 7 6 13 26 2 1 34 5 can also be determined in the middle right box. 5 cannot be in the top arrowhead because the top circle must be 7 or less and the 5 would take the arrow total to more than 7. This means that the 5 must be on the other arrowhead.

3 9 12 8 4 5 7 6 13 26 2 1 34 5 can also be determined in the middle right box. 5 cannot be in the top arrowhead because the top circle must be 7 or less and the 5 would take the arrow total to more than 7. This means that the 5 must be on the other arrowhead.

3 9 12 8 4 5 7 6 13 26 2 1 34 The other number on the arrow must be a 4

3 9 12 8 4 5 7 6 13 26 2 1 34 This determines the position of the 3 on the central row.

3 9 12 8 4 5 7 6 13 26 2 1 34 And the 6

3 9 12 4 8 5 7 6 13 26 2 1 34 And now the 4 in the third column from the right.

3 9 12 4 8 5 7 6 13 26 2 1 34 Leaving only the 5 to be slotted in to that column.

3 9 12 4 8 5 7 6 13 26 2 1 34 And the 6 can be placed at the end of that row.

3 9 12 4 8 5 7 6 13 26 2 1 34 The 5 in the top right box won’t fit into the circle so must be in the lower centre.

3 9 12 4 67 8 5 7 6 13 26 2 1 34 Top right box: 67 must fit in to the top middle two cells.

3 9 12 4 67 8 5 7 6 13 26 2 1 34 Second column from the right, 4 must be at the bottom

3 9 12 4 67 8 5 7 6 13 26 2 1 Which also determines the 3

3 9 12 4 67 8 5 7 6 13 26 2 1 Far right column:12 can be fitted in

3 9 12 4 6 8 5 7 13 26 2 1 And so the 3,6,7 can be fixed in the 2nd column from the right.

7 3 9 12 4 6 8 5 13 26 2 1 Follow the sevens to the top left box.

7 25 3 9 12 4 6 8 5 13 26 2 1 The circled 7 is made up of 25

7 25 3 9 1 4 6 8 5 12 13 26 2 Which means that it must be a 1 next to the circled 9 in the top row

7 25 3 9 1 4 6 8 5 2 26 12 The 123s in the top right can now be determined.

7 25 3 9 1 4 6 8 5 2 26 12 In the third row, the 4 cannot lie on the arrow line.

7 25 3 9 1 4 6 8 5 2 26 12 Which gives us the 1 as well.

7 25 3 9 1 4 6 8 5 2 26 12 The circle at the top left must be a 6 or 8. 8 won’t fit.

7 25 3 9 1 4 6 8 5 2 26 12 Fill in the 8 and the 2 on the arrow line.

7 25 3 9 1 4 6 8 5 2 12 And now the 2 and 6 in the centre.

7 2 5 3 9 1 4 6 8 Easy to complete now!