Slide 1

Arrow SudokuOr Sudokarrow if you read the Saturday iPurchase Arrow Sudoku magazine from Clarity Media here: http://www.puzzle-magazine.com/arrow-sudoku-magazine.php

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5131

222 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 5.6

29522 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 5.7

269526The only number that can be combined with the 1 and 2 is 6 meaning that there must be a 9 in the end circle8

9269526The only number that can be combined with the 1 and 2 is 6 meaning that there must be a 9 in the end circle9

9269526We can now place a 9 in the top middle box through normal sudoku rules and the fact that a 9 cant lie on one of the lines.10

99269526We can now place a 9 in the top middle box through normal sudoku rules and the fact that a 9 cant lie on one of the lines.11

99269526Using similar rules of elimination we can find possible 9s in the top right and bottom middle boxes.12

9999269526Using similar rules of elimination we can find possible 9s in the top right and bottom middle boxes.13

9999269526Using similar rules of elimination we can find possible 9s in the top right and bottom middle boxes.14

999926952699Using similar rules of elimination we can find possible 9s in the top right and bottom middle boxes.15

999926952699The 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 7.16

9999269526913139The 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 7.17

9999269852699The 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 7.18

99992698526799The 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 7.19

99992698526799The 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 wont fit...20

999926985267914149The 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 wont 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.21

9999269852679239The 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 wont fit...22

9999269852679239Lets examine the top central box now. We have some information about 2..... Lets look....23

99992698526792392 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.24

9229992698526792392 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.25

922999269852679239We 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 2s because it would make the arrow total greater than 9. Can there be a 6 in the central cell?26

9262999269852679239We 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 2s because it would make the arrow total greater than 9. Can there be a 6 in the central cell?27

139262999269852679239Well, 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)28

13926299979269852679239This 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.29

92299966269852679239This 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.30

92299966269852679239These 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.31

9229996868269852679239These 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.32

922999686826985267923968In 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 2s are. This means that tehe cell below the 7 must be a 4 as a 4 cannot fit in where the 2s are.33

9229996868269852679243968In 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 2s are. This means that tehe cell below the 7 must be a 4 as a 4 cannot fit in where the 2s are.34

9229996868269852679243968Focussing on that vertical on the right, since the 6,8 must occupy the two cells shown, 1 can only fit in with the 2s.35

912129996868269852679243968We 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 doesnt clash with the 2 in the lower middle box.36

91212999686826985267924132968We 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 doesnt clash with the 2 in the lower middle box.37

39123129993686826985267924132968We now have some possibilities for a 3 in the top central box. It must be in the left vertical.38

39123129993686826985267924132968However it cannot be in the circle.39

91212129993686826985267924132968Because 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.40

3912312999686826985267924132968Lets take a look at the boxes on the left and see if we can fill in some more 9s...41

3912312999686826985267924132968Lets 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.42

391231299968682698526799241329968This allows us to eliminate the central vertical in the middle box as far as 9s are concerned.43

391231299968682698526799241329968The 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.44

3912312999686826998526799241329968The 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.45

391231299968682699852679924132999968We can also find some 9 possibilities in the bottom right box. There are only 2 possible options.46

391231299968682699852679924132999968We can also find some 9 possibilities in the bottom right box. There are only 2 possible options.47

3912371299968682699852679924132999968Return now to the top central box. Can a 7 be fit in? Lets 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?

48

3912371299968682699852679924132999968If 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.

49

39123712999768682699852679924132999968This in turn gives us two options for 7 in the lower box.

50

391231299976868269985267992417329999768If 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.

51

34912341299976868269985267992417329999768In the top centre box 4 must share its options with 352

349123451299976868269985267992417329999768This leaves the middle cell as the only option for 5.53

349123451299976868269985267992417329999768Consider 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 12.54

34912345129997686813269139852267992417329999768This 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. 55

3912345129997688132691398522679924173299997688 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.56

39123451299976813269139852267992417329999768This in turn implies the 3 next to the 4.57

39123451299976813269139852267992417329999768This now gives us the 6 in the lower middle box.58

3912345129997681326913985226799241732999976This now gives us the 6 in the lower middle box. The 8 is constrained to the same boxes as the 7s.59

3912345129997681326913985226799241732999976This now gives us the 6 in the lower middle box. The 8 is constrained to the same boxes as the 7s.60

391234512999768132691398522679924178329999786This now gives us the 6 in the lower middle box. The 8 is constrained to the same boxes as the 7s.61

391234512999768132691398522679924178329999786The arrowhead is 78. The only other option for the remaining arrow line is 1 or 5. 1 it is!62

3912345129997681326913985226799241178329999786The 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 circle63

39123451299976813269139852267989241178329999786The 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 circle64

3912345129997681326913985226798924117329999786Which means that the arrowhead must be a 765

391234512999768132691398522679892411732999986Which forces another 8 in the lower middle. These 8s imply two possibilities in the box on the lower right, sharing their options with the 9s.66

391234512999768132691398522679892411732999986Which forces another 8 in the lower middle. These 8s imply two possibilities in the box on the lower right, sharing their options with the 9s.67

39123451299976813269139852267989241173289899986Which forces another 8 in the lower middle. These 8s imply two possibilities in the box on the lower right, sharing their options with the 9s.68

39123451299976813269139852267989241173289899986Which forces another 8 in the lower middle. These 8s 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 569

3912345129997681326913985226798592411732898995986Look for 9s now in the right middle box....70

39123451299976813269139852267985924117328989959869 cannot lie on the arrowhead so that only gives us 2 options.71

391234512999768132691398522679998592411732898995986Note now that the only options for a 9 in the far right column are in the lower right box, rightmost central cell.72

3912345129997681326913985226799985924117328995986Which means that the middle cell must be an 8.73

3912345129997681326913985226799985924117328995986Lets look for 7s in the lower left. There cant be a 7 in the central row because we already have a 7 further along to the right.74

39123451299976813269139852267999859241173289795986Also, there cant 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 cant 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.75

391234512999768132691398522679979859241173289795986Also, there cant 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 cant 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.76

3912345129997681326913985226799798592411732891279125986The top 2 rows of the lower boxes already contain 12 so the only place to fit them is on the arrowheads.77

39123451299976813269139852267993798592411732891279125986Because of the 3 on the central line the only place for the lower right 3 is on the top left cell.78

3912345129997681326913985226799379859241173289127912598633This gives us two options for 3 in the lower right box.79

3912345129997681326913985226799379859241173289127912598633Consider 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 4.80

391234512999768132691398522679937985924117328912791259867

3434Note that this leaves only one option for 5 in the lower row.81

39123451299976813269139852267993798592411732891279125867

3434Which in turn gives the only option for 9 in the lower centre box and the 7 and 9 in the lower left box.82

39123451299976813269139852267993789241173289129125867

3434Can we place a 7 in the middle right box?83

3912345129997681326913798527267993789241173289129125867

34347 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 784

3912345129997681326913798527?267993789241173289129125867

3434But can we have a 7 on that arrowline? 85

391234512999768132691379852267993789241173289129125867

3434It 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.86

391234512999768132691379852267993789241173289129125867

3434This determines the 7 in the central left box.87

3912345129997681326913797852267993789241173289129125867

3434We can now determine the 8 in the centre right box.88

391234512999768132691379785226798983789241173289129125867

3434There are only two options.89

39123451299976813269137978522679898?3789241173289129125867

3434But can there be an 8 on the arrowhead?90

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3434It 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.91

3912345129997681326913797852267893789241173289129125867

3434This 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.92

39128345129997681326913797852267893789241173289129125867

3434The 9 in the top right box is also determined by the usual sudoku rules.93

39128345129997681326913797852267893789241173289129125867

3434The 9 in the top right box is also determined by the usual sudoku rules.94

39128345129997681326913797852267893789241173289129125867

34345 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.95

391283451299976813269137978522678953789241173289129125867

34345 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.96

3912834512999768132691379785242678953789241173289129125867

3434The other number on the arrow must be a 497

39128345129997681326913793785242678953789241173289129125867

3434This determines the position of the 3 on the central row.98

391283451299976813269137937856242678953789241173289129125867

3434And the 699

3912483451299976813269137937856242678953789241173289129125867

3434And now the 4 in the third column from the right.100

39124834512999768132691379378562426789537892451173289129125867

3434Leaving only the 5 to be slotted in to that column.101

391248345129997681326913793785624267895378924516173289129125867

3434And the 6 can be placed at the end of that row.102

3912483451299976813526913793785624267895378924516173289129125867

3434The 5 in the top right box wont fit into the circle so must be in the lower centre.103

3912467834512967976813526913793785624267895378924516173289129125867

3434Top right box: 67 must fit in to the top middle two cells.104

3912467834512967976813526913793785624267895378924516173289129125867

434Second column from the right, 4 must be at the bottom105

3912467834512967976813526913793785624267895378924516173289129125867

43Which also determines the 3106

39124678345129671297681351226913793785624267895378924516173289129125867

43Far right column:12 can be fitted in107

3912468345129712976813512269133793785624267895378924516173289129125867

43And so the 3,6,7 can be fixed in the 2nd column from the right.108

73912468345129712976813512269133793785624267895378924516173289129125867

43Follow the sevens to the top left box.109

725253912468345129712976813512269133793785624267895378924516173289129125867

43The circled 7 is made up of 25110

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43Which means that it must be a 1 next to the circled 9 in the top row111

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43The 123s in the top right can now be determined.112

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43In the third row, the 4 cannot lie on the arrow line.113

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43Which gives us the 1 as well.114

725253914686345297141976835226913793785624267895378924516173289129125867

43The circle at the top left must be a 6 or 8. 8 wont fit.115

72525391468683452971419768352226913793785624267895378924516173289129125867

43Fill in the 8 and the 2 on the arrow line.116

725253914686834529714197683522691379378562427895378924516173289129125867

43And now the 2 and 6 in the centre.117

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43Easy to complete now!118