The 9x9 puzzle from 2nd April 2013 has been described as "The world's hardest calcudoku", there is a very good step by step solving guide with excellent graphics in the forum created by clm.

I have been working on a way to assess the difficulty of calcudoku puzzles, for my method to work I need to find the easiest way to solve a puzzle, this is probably not very practical but I might find it useful when creating my own puzzles. I have been using this puzzle from 2013 and some of the difficult puzzles from the latest book to test my ideas. Below are the basic steps that I have for solving the puzzle,

I think my method is easier than the method provided by clm, there is only 1 use of parity checking, and adding cage values is used twice. There are 2 steps (12 & 16) that have long explanations of why a value is not valid, but these are not as complicated as they might first appear.

This is an old puzzle, but I hope it is still of interest to many of you, I have tried to be careful, but mistakes happen, if you see any mistakes let me know. If you would like a more complete step by step description or more images I am happy to provide more. I will work on adding notes to my screen shots so that they are more helpful.

"The World's Hardest Calcudoku" 2/4/2013

1. Solve single cell cages

2. Cage g8 = 1, 3

3. Add columns g,h & i: g4 + g6 = 10, g4 & g6 = 2,4,6,8

4. Column d: cage d7 = 2 & 8, remove 2 & 8 from the other cells in column d

(See Image 1)

5. With reference to cells d2, d3, d4 & d6: d1 and d9 can not be 6

(If d1 or d9 = 6 : cage d2(11+) = 4 & 7, d4=4,7,9 and d6=4,7,9. This would create 4 cells with only 3 possible values)

6. Recalculate cages d1(60x) and e8(378x)

(See Image 2)

7. Referring to cell f9: cage f2 can not be 2 & 6

8. f9 = 2 (only one cell in column d can have the value 2)

9. f5 = 6 (only one cell in column d can have the value 6)

(See Image 3)

10. f1 can not be 4. (in column d 8 appears only in cages f2(8-4=4) & f7(8+4=12), either cage f2 or f7 must be 4 & 8)

11. d1 & d5 can not be 4

(Based on parity check of columns a,b,c & d, cells d1 & d5 combined must be even,

this can be achieved by both cells being even, or both cells being odd,

however there is only one even value in the cells, so they must both have odd values)

(See Image 4)

12. e2 can not be 3: 3 would create an error in cages c4 & c6.

(If e2 = 3: cage b2=1,7, d9=3(only 3 in col. d, d5=5 (recalculate cage e4), cage d2=4, 7

d4 & d6=6, 9, this means either c4 or c6 = 7, error: 7 is already used in col. c (cage b2(6-))

13. Cage d1 must have a 3 in row 1: c1 can not be 3

14. Cage e8 must have a 3 in row 9: c9 can not be 3

15. c2 = 3, b2=9 (c2 is the only cell in column c that can be 3)

16. Cage d2 can not be 4, 7: this would create an error in cells d3 & f3:

(If cage d2=4,7: d9=3, recalculate cage e4:(e4=1,5,8 e5=1,8 e6=1,5,8),

recalculate cage d1:(d1=1, e1=3, f1=5, e2=4), d2=7, e3=4, f2=8, f3=4, error: both e3 and f3 =4)

17. Add all cages in rows 1,2 & 3: b3 + h3 = 9, 11 or 15: b3 = 5, 7 and h3 = 4

(note that b3=3 and h3=6 is not a valid option, although the sum is 9, it would result in both b4 and h4 having the same value of 5)

(also note that values 5 & 6 are not valid, the sum is 11, but cell d3= 5 or 6 so this is not a valid option)

18. All remaining cells can be solved using basic solving techniques.

(See Image 5)

Thank you for reading.

Image 1

Image 2

Image 3

Image 4

Image 5