Sudoku Players' Forum (clone)

lunababy_moonchild wrote:

afjt wrote:I'm not sure how Animator ascertained the set of 9 numbers before he started solving !

Perhaps he'll explain

I will... in a few days... (I'm not going to post it yet since that might spoil the puzzle for many people...)

PaulIQ164 wrote:

Bigtone53 wrote:Could PaulIQ confirm the original link? It comes up on my machine as an unknown URL.

Thanks

Well, the link works fine for me. But you could try going to the image file directly, which is at: http://www.apathysketchpad.com/images/uploads/doublekiller.png

I should mention that the website seems to be playing up at the moment, so you might not be able to access the puzzle. I can always stick it on ImageShack if this continues to be a problem.

I couldn't get into the URL this morning but it's back on now. Thanks Paul for lending your brother's puzzle to the forum - he sounds cool. Is he as intelligent, romantic and good looking as you?

I found the first 6 clues by innies and outies then the last 3 from box 2 combinations and that led to box 3 and so on down the yellow brick road. Nice puzzle. I'd like to know how Animator figured out the clues before solving too.

The requested URL could not be retrieved

--------------------------------------------------------------------------------

While trying to retrieve the URL: http://www.apathysketchpad.com/images/uploads/doublekiller.png

The following error was encountered:

Unable to determine IP address from host name for unknown server name
This means that:
The cache was not able to resolve the hostname presented in the URL.
Check if the address is correct.

Sorry Paul but I am not there yet. Let me try another machine.

I've stuck the puzzle on imageshack since the website seems to keep going down:


(click on it for the fullsize picture)

I've stuck the puzzle on imageshack since the website seems to keep going down

Thanks Paul. An interesting diversion.

Great innovative puzzle!

Even before starting, I would say you could sum up all the cages in the grid and divide the sum by 9 to obtain the sum of the set of 9 numbers used... And then you know 1,2,4 must be among them from the 7-cage (the "numbers can't be duplicated within cages" rule still sticks, right?). I think that's more or less how Animator managed to find out all 9 numbers before starting... Hope I didn't spoil too much... But I'd say it'd be great fun also to start the solving process without knowing the set first...

Would you consider to post your link in djape's blog? There're heaps of killer fans (what some called as "killerholics" there!

udosuk wrote:Even before starting, I would say you could sum up all the cages in the grid and divide the sum by 9 to obtain the sum of the set of 9 numbers used... And then you know 1,2,4 must be among them from the 7-cage (the "numbers can't be duplicated within cages" rule still sticks, right?). I think that's more or less how Animator managed to find out all 9 numbers before starting

Perhaps the crux to this is the definition of "starting the puzzle". I would say that I had already started once I had pencilled in (even implicitly) the 1,2,4 into the 7-cage.

Incidentally, I obtained the cage-total by summing the whole 9-cages rather than the box 2 and 5 short-cut used by others. I guess it could be argued that if you apply the whole-grid sum rule you haven't yet started the puzzle, whereas using the a 2-cage sum you have !

About afjt's comment regarding "the crux", a bit of yes and no...

The "no" part is that it should not be counted as "pencilling in" when you determine 1,2,4 must be inside the 7-cage. So, first you collect all the cage sums and sum them together. In the process you must have come across this sum of 7 and a bit of logical thinking will make you sure that 1,2,4 must be there. I wouldn't say that constitutes "starting solving the puzzle".

The "yes" part is that the aforementioned argument must be based on 1 assumption: ONLY POSITIVE INTEGERS are used... But of course the mighty creaters PaulIQ164 & bro didn't say anything specific, there were every possibilities that ZERO could be one of the entries, even NEGATIVE INTEGERS and heaven forbids, FRACTIONS... So it was not right for me to assume that 1,2,4 must be in the 7-cage. Of course if you do it the proper way by first working out the 7 innies/outies and then focus on nonet-2 you will quite easily determine all 9 entries and it would become obvious that the 7-cage is indeed {124}...

BTW I'm posting the link to djape's blog cause I think it is a sin to not letting "the other side of the killer community" know about this marvellous puzzle... The original creators will be properly credited of course!

When I solved it (and I had nothing to do with its creation by the way) I assumed all the numbers were positive integers. If you don't make that assumption, I guess you no longer know that the 7 has to be 1+2+4. But thinking about it, the '45-rule' (not that it's necessarily 45 any more) still works. I'm not sure how many of the other tactics we use solving Killers are/aren't dependent on that assumtion whan solving this sort of puzzle (which has, by the way, been dubbed a 'Face Ripper Su Doku').

This is a very interesting variant.

Obviously, we also have to figure out what numbers are used.

Try to solve it by logic, not "'cheating" with a computer.

For a regular Killer Sudoku, numbers 1..9 add up to 45. We can deduce the equivalent of the 45 for this grid. The whole grid is made of 9 such "45" groups. For regular killer this gives 9*45=405. If we add up all the cages of this grid, we get 585 (equiv. of 405). Divided by 9, this gives 65 (equiv. of 45) which is the sum of all numbers used in the grid.

Now, that we know the sum of all numbers used in the grid, we can use innies/outies using 65 instead of 45. We will then get some of the numbers used in the grid.

In Nonet 2, R3C6 = 65-(12+24+12+8) = 9 -> R4C6 = 2.
...

That's a nice one indeed

Animator wrote:

lunababy_moonchild wrote:

afjt wrote:I'm not sure how Animator ascertained the set of 9 numbers before he started solving !

Perhaps he'll explain

I will... in a few days... (I'm not going to post it yet since that might spoil the puzzle for many people...)

I think a week is long enough to wait..!

Ok, here are two possible ways to find them. (Initially I used the long version.)

(small note: feel free to see this as 'solving' (either in whole or in parts). It all depends on how you define solving.)

• Figure out the total that has to go in each box/row/column. (= the sum of all numbers).
  
  The idea (used in this post): Look for a unit (box in this case). The only requirement: all the cages need to be fixed inside that unit.
  
  In this grid that is impossible. So let's take the next best thing: look for 2 units in which the cages are fixed inside those.
  And this exists. Box 2 and Box 5: All the cells of the cages in those two boxes are in box 2 or box 5.
  
  Now, take the sum of all the cages. This equals the sum of all the numbers of 2 units.
  
  Divide it by two and you find the sum of one unit.
  
  (12 + 12 + 24 + 8 + (11) + 18 + 18 + 17 + 10) / 2 = 130 / 2 = 65.
  
  (A more general apporach would be to sum all cages and devide by 9)
• Use the total.
  
  Some cells can now be filled in by using the calculated sum per unit.
  Look for a unit (a box in this case) in which all cells but one are part of a cage in that unit.
  
  Box 2, box 3 and box 5 fit this description.
  
  Box 2: The sum of the cages in the box is 56 (8 + 12 + 12 + 24). We know that the box-total is 65. So the one remaining cell has to be 9.
  
  Results: r3c6 = 9; r3c9 = 2; r4c6 = 2
• Maximum number:
  
  At some point we are going to need to check what combinations are possible. It is easier to do that if we first calculate the maximum number.
  
  Take a look at box 2, 3, 5. (We look only at those three because in those boxes there are no 'shared' cages.)
  
  Write down the highest cage-sum for each box: Box 2: 24; Box 3: 32; Box 5: 18
  
  Now select the lowest number. This is the 18 from Box 5.
  Look at the cage that has this sum. It uses 2 cells.
  
  Assumption: each number is bigger then zero. (read: only positive numbers and not zero).
  The minimum and maximum number we need to get to a sum of 18 (with 2 cells): 1 and 17.
  
  If a combination includes a number bigger then 17 then it simply is impossible to fill in Box 5.
• Choose the unit of which you write a list of possibilities
  
  Again the question: what unit?
  
  • Box 2:
    It has two cages that need the number 12, both of these cages use two cells.
    With what numbers can we represent 12?:
    
    • 1 + 11
    • 2 + 10
    • 3 + 9 (We can ignore this combination! Box 2 already has the number 9)
    • 4 + 8
    • 5 + 7
  • Box 5:It also has two cages that need the same number (and all of these cages use the same amount of cells).
    
    • 1 + 17
    • 2 + 16 (ignore this one: Box 5 already has the number 2)
    • 3 + 15
    • 4 + 14
    • 5 + 13
    • 6 + 12
    • 7 + 11
    • 8 + 9
  Bottomline: For box 2 you need to select 2 ways out of 4 to write 12. For box 5 it's 2 ways out of 7 to write 18. (I assume that Box 2 is faster so I used that)
• Write down the combinations.
  
  
  • 24 = (7 + 17), (8 + 16), (10 + 14), (11 + 13)
  • 12 = (1 + 11), (2 + 10), ( 4 + 8)
  • 8 = (1 + 7), (2 + 6), ( 3 + 5)
  Possible shortcut 1: If you look at row 3 then you see that it already contains the number 2. The cage that holds the number 8 is also on row 3. This means that it cannot include the number. So the 2 + 6 combination is invalid.
  Possible shortcut 2: The final list of numbers need to include the number 2. There is only one combination that has the number 2 so this one has to be used no matter what. (That brings it down to select 1 way out of 3 to write 12).
• Combine them:
  
  If you choose (7 + 17) for 24:
  
  Only 3 options remain for the number 12: (1 + 11), (2 + 10), (4 + 8). ((5 + 7) is no longer possible since 24 uses the 7)
  
  If we choose (1 + 11) and (2 + 10):
  
  Only one option remains for the number 8: (3 + 5).
  
  Combination A: 1, 2, 3, 5, 7, 9, 10, 11, 17. (The 9 is already in box 2. That's why it is added in the combination)
• Repeat.
  
  Possible combinations for box 2
  
  
  • 24 = (7 + 17)
    • A: 1 2 3 5 7 9 10 11 17
    • B: 1 2 4 6 7 8 9 11 17
    • C: 1 3 4 5 7 8 9 11 17
    • D: 2 3 4 5 7 8 9 10 17
  • 24 = (8 + 16)
    • E: 1 2 3 5 8 9 10 11 16
    • F: 1 2 5 6 7 8 9 11 16
  • 24 = (10 + 14)
    • G: 1 2 4 6 8 9 10 11 14
    • H: 1 3 4 5 8 9 10 11 14
    • I: 1 2 5 6 7 9 10 11 14
    • J: 2 4 5 6 7 8 9 10 14
  • 24 = (11 + 13)
    • K: 1 2 4 7 8 9 10 11 13
    • L: 2 3 4 5 8 9 10 11 13
    • M: 2 4 5 6 7 8 9 11 13
  Combination C and H can be ignored. (C nor H has the number 2. But it's already filled in in two cells so you know it is needed.)
• Continue
  
  There are two possiblities: either you use some shortcuts and are done or you use a longer version.
  
  
  • Shortcuts:
    Apply shortcut 1: each combination with a 6 is invalid.
    Apply shortcut 2: each combination that does not have the number 2 and 10 is invalid.
    
    Current list of combinations:
    • A: 1 2 3 5 7 9 10 11 17
    • D: 2 3 4 5 7 8 9 10 17
    • E: 1 2 3 5 8 9 10 11 16
    • K: 1 2 4 7 8 9 10 11 13
    • L: 2 3 4 5 8 9 10 11 13
    So now it's time for shortcut 3: there is a cage in the grid that has the sum 7 and it uses 3 cells. This means that this cage holds the numbers 1, 2 and 4.
    Conclusion: all combinations that do not have 1, 2 and 4 are invalid.
    
    This leaves only one possible combination: K = 1, 2, 4, 7, 8, 9, 10, 11, 13
  • Long version:
    • 'Try' each combination in another unit
      
      Try in box 5: (remember box 5 already has the number 2 so we cannot use that in additions)
      
      
      • A: invalid. (twice 18, but no 17)
      • B: valid. (18 = (1 + 17), 18 = (7 + 11), 17 = (8 + 9), 10 = (4 + 6))
      • D: invalid. (the combination includes the number 17. The only sum we can make with that is 18. But the combination does not include 1.)
      • E: invalid. (only one sum of 18)
      • F: invalid. (only one sum of 18)
      • G: valid. (18 = (4 + 14), 18 = (8 + 10), 17 = (6 + 11), 10 = (1 + 9))
      • I: invalid. (14 is included. Possible sums: 17 and 18 but there is no 3 or 4)
      • J: invalid. (twice 18, but no 17)
      • K: valid. (18 = (7 + 11), 18 = (8 + 10), 17 = (4 + 13), 10 = (1 + 9)).
      • L: invalid. (11 is included. Possible sums: 17 and 18 but there is no 6 or 7)
      • M: valid. (18 = (5 + 13), 18 = (7 + 11), 17 = (8 + 9), 10 = (4 + 6))
    • Repeat on another unit
      
      Only 4 combinations remaining...
      To find the correct combination we repeat the previous step but for another unit.
      
      The easiest: column 6. We limit the search to number 18 and 24. (note that 2 and 9 are already filled in)
      
      
      • B: 24 = ( 7 + 17). 18 is not possible ==> invalid.
      • G: 24 = (10 + 14). No 18 ==> invalid.
      • K: 24 = (11 + 13), 18 = (8 + 10). ==> valid.
      • M: 24 = (11 + 13). No 18 => invalid
    • The numbers: combination K = 1, 2, 4, 7, 8, 9, 10, 11, 13

This was a fun puzzle. It's amazing how you have to shift gears to do it. I had to create new addition facts (even for the 2-cages!). Once I determined the numbers, it helped to list them in order to keep track of them. Was it hard to create?

Creating it should be as hard (or maybe as easy) as creating a normal Killer.


All you need to do is change some numbers in the solution and re-calculate the cage-sums.