sjs34 wrote:

Often lately I have not bothered with these puzzles as the exhaustive multiplication and addition possibilities that are required seem too tedious. In contrast, this week's on Nov 20 I found much more appealing and accessible. (Judging from the number of solvers, either others also found this to be true or it was just easier).

This week was a bit more straightforward than usual. I find that in general, in order to solve the Tuesday 9x9, I have to work out the different multiplication options and the sum of each of these options. This usually narrows it down to relatively few possibilities, although sometimes I also have to work out the product of each of the options for some of the addition cells in order to narrow it down further.

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michaele wrote:

Are you referring to puzzle A4V14? I have also solved this puzzle. Or have I lost track of which puzzle is being discussed.

oops, yes, a4v14 indeed, and I didn't bother to check if there had been new solutions since last time...

so it's FOUR (sorry)

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paulv66 wrote:

I finally cracked it!

THREE people solved it

(in fact:

correction: FOUR people (also

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pnm wrote:

paulv66 wrote:

I wonder is this a candidate for the most difficult book puzzle ever?

It could be.

So far 2 (TWO) people have solved it ...

I finally cracked it!

I wish I could say it was due to brilliant insight on my part, but it was more due to bloodymindedness and sheer perseverance. I don't know how many hours I spent on it, but I just kept trying to eliminate wrong options and I was stymied again and again until I realised that one of my early assumptions about the way two particular numbers were arranged in one of the rows was not necessarily correct.

Now I'm going to have to find something else to do with my spare time!

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paulv66 wrote:

I wonder is this a candidate for the most difficult book puzzle ever?

It could be.

So far 2 (TWO) people have solved it ...

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paulv66 wrote:

It's taken me 5 days to do the last two puzzles, so at this rate, it will another month before I complete the remaining 12.

That was way too optimistic of me! I'm almost there at this stage, but haven't managed to solve A4V14 yet. I know angelwhite mentioned that he had made several attempts at it without success and I see that bertoldo and agnostico are also 12 points shy of a full total, for which A4V14 seems to be the most likely culprit.

I wonder is this a candidate for the most difficult book puzzle ever?

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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

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sjs34 wrote:

The 9/6 was rated 51.2. Today's is 51.3 and is much easier.

In my opinion the 8x8 medium puzzle from 9/6 that has a rating of 51.3 is much easier than the 6/9 8x8 that had a rating of 51.2.

At first glance it might appear that being all multiplication will make it tricky, but the only solving techniques required to solve it are: Calculating factors for each cage, removing pencil marks for cells in the row and column of a solved cell, and solving a cell because a pencil mark appears only once in a row or column.

I would like to add that I don't say this to be complaining or argumentative. I have found this forum discussion interesting because I have been working on my own way of assessing the difficulty level of puzzles (so far with limited success).

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