# Harder Killer than the Hardest

Updated on April 3, 2016

## Harder than the Hardest

I have come across a sudoku that is in my opinion much harder than the one at a website that claims to have the 10 hardest puzzles of all time.
http://www.conceptispuzzles.com/index.aspx?uri=info/article/424.

## My Sudoku Experience

I have solved the regular sudoku and the killer sudoku and have written hubs on both. Even developed a method to help in solving their killer.
http://hubpages.com/games-hobbies/Solving-Hardest-Killer-Sudoku

Then I looked for other hard killer sudokus to do. Found one I say is much harder.
The author has shown the solution because commentators have complained that there is no way to solve it without guessing. He solved it up to a point then left it up to you to solve the rest, which was much easier to finish. When I went back to show all the steps so I can write this hub, had a few problems with his explanation. Tried to solve it not using the obscure overlap method. Found a work around. Not much opportunity to use the method used solving the hardest killer on the 10 hardest site.

## Trouble With Solving

Since I had so much trouble with the author's explanation I put it in a solver.
http://www.sudokuwiki.org/killersudoku.htm It solved it to a point then ran out of strategies. Finally solved it using the author's suggestions and my own efforts.

## My Solving Conventions

I use the conventions: first number is the row, second is column. C means column, B means box, Cg/# means cage and number of cells, pr means pair, trip means triple No memorization of all those fishy words. Just common sense and ALS's mostly

## Solving by killersudokuonline author

3 at 6-2 innie 90-87 C1+C2, 3's can also be taken from the 21 Cg.
1,s needed in C7 B9, removes 1 from 1-7, leaving the single 3 B3, = 1 B2

1-3 pr R1, 7-8-9 trip R2 only combo that will work in a 24 Cg, 1-2-3 trip C9 It is
up to you to go around and find all the eliminations.

1 innie 4-8 C8+C9 = 89, also 1,s out of 31 Cg

3 B4, due to 1-2 pr C9
split cage rule rows 4 and 5 Forming new cages. 16/3 Cg B1, 12/2 Cg B2, 13/2 Cg B6 15/3 Cg B6
Using the combo chart you can remove numbers that don't fit.

## Combo Chart

Hubpages would not let me put a chart in this capsule. Said it didn't look like text. So I made a photo.

The chart is all the possibilities that can go into a certain number of
cells in a row, column, box or cage.

example: The newly formed 12/2 Cg at B4: Using the chart we find no 1,2 or 6 can be in a 12/2 cage

## More Solving

3-6, 4-5 in 9/2 Cg in R2 combo rule, 1-6, 2-5, 3-4 possibilities in 7/2 Cg in R3
2's out of 20/3 Cg,s 1's out of all 19/3 Cg's
1-1 pr B4, = 6 remved from 7 Cg/2 in R3 combo rule
New 18/5 Cg B4 only contain 1,2,4,5,6. 1,s remved from R7. Must be a 1 in R6
combo removals: 8/2 Cg R7, 12/2 Cg R8, 6/2 Cg R9
1 removed from 3-2 because all combos in the 19/5 Cg requires a 1
14/2 pseudo Cg in B3 combo reduced to 5-9 and 6-8
1 must be in 21/4 Cg B7 so only combos that contain 1 are concidered, no 2's in any
9 removed from 8-5 and 8-7 due to combo 13/3 rule
Out O strats.png
This is as far as the solver took the puzzle. The rest is done bya combination of my
own efforts and what I gleaned from the author's solving hints. Check all cages for
number combos that don't fit and remove those numbers
2 removed 3-2 pseudo 26/4 Cg in B1 combo rule.
5-8 removed from 5-8 and 5-9 it would make the 5-7 or 4-8 impossible at 5-1 or 5-2
7 and 4 can be remved from the 20/5 Cg for same reason
1 3-9 because two outties = 11 or 12 no 1 can be used leaving 1 at 3-9, 2 in 2-9
(no overlap method required)
no 6 in 2-1 or 2-2, no (pseudo Cg) 10/3 Cg can have 4-5-6 at 2-8 = 6-6 B3 = out 3 6's
ALS 1-5 5-6 = 5 out 2-8,
3-2 cannot be, 5 that would mean 7 Cg 3-4 leaving only 7-8-9 and no pseudo 12/2 Cg
same for 5's in 31/5 Cg and R3

Here is anexample of how the author has described the next step.
"The rule of 45 now allows us to deduce that x+y = 12.

We can now determine that r7c2 cannot be 5 as the 7-cage on r7 must be either 5+2
or 3+4. However, if r7c2 is a 5, then x+y must use either the 3 or the 4.

Now we can inspect the 20-cage on r6. Due to the 3 in r6c9, the only valid sums are
875, 974, and 965. However, 875 is not legal as:

r6c3 cannot be an 8. If r6c3=8, then r7c2=8 (by rule-of-necessity on the 28-cage)
which means that r7c7 must be a 9. But then there is no place for a 9 in r6."
impossible."

You will notice that he numbers rows from bottom up. I get lost alredy. I do just as
if I where reading normally. Don't even bother with the r or c. I have rewritten it to:

The rule of 45 now allows us to deduce that the pseudo Cg/2 = 12 (2 outties R3)

We can now determine that 4-2 cannot be 5 as the 7-cage on R3 must be either 5+2
or 3+4. However, if 3-2 is a 5, then pseudo Cg 12 must use either the 3 or the 4.?????
(must mean 3 or 4 out of 7 Cg which can't be done. very confusing) so, no 5 in 3-2

Now we can inspect the 20-cage on R4. Due to the 3 in 4-9, the only valid sums are
875, 974, and 965. However, 875 is not legal as:

4-3 cannot be an 8. If 4-3=8, then 3-2 also=8 (by rule-of-necessity on the 28-cage)
which means that 3-7 must be a 9. But then there is no place for a 9 in R4 impossible.?????
(haven't figured this one out) just say 5-7-8 in 20 Cg? Is he saying 3-7 is 9 or not?
4-3 cannot be either 5 or 7. If either was the case then the 28-cage would be impossible.
One use 5 in 4-3 there is 26794, if 7 there is none
Ignore whole paragraph for now.
Unless you can eliminate the 26794 you cannot use this paragraph???

I struggled on and off a few days to find alternate solutions and decided to continue on where the author left off. Will update if an easier solution is found.
Haven't determined if author has made any guesses. He claims not.

## Finishing Up

You can see right away that a 1 goes in 9-1. Remove a couple easy things I and the author have missed. Like the 8 st 8-2 and the 6,s R7 etc.
Other easies: 7 at 4-4 = 4 at 4-5, 3 single at 5-4
1-5 pr R4 leaves 6 1-4 = 5 1-5
289 combo R6 = 5-7 pr B6 forcing 2-3 1n 30 Cg.
looking at the combo chart, There is no 2 combo that will work leaving 3 B9 and the 7/2 Cg = 7 and 1 = 3 then 7 B1, 3 B8, 5 B8,
6-8 pr B8 = 1 B8. 2 = 9 B5 = 8 B5 = 6 B5 and 1 B5
8 then 9 B2, 8 then 2 B4, 8 then 5 B7
new 12/2 Cg R8 no 6 leaving 8 = 4 B9 also 6 B8
7, 9 then 4 B7 = 2 and6 B9, 9 B9 = 4 then 9 B6
2-6-9 leaves 7 out of 30 Cg B9 = 7 B9, 8 B9 = 7 then 5 B6 Last 4 and 7 B3.

## All Done

I had much more trouble solving this killer than the one on the 10 hardest list.
Would like to nominate it instead of one on their site other. Please leave comment when you have done both. Or if you have found a better solution. Please show how.

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