Joined: Wed Apr 16, 2008 1:16 am Posts: 1044 Location: Sydney, Australia
HATMAN wrote:
Where a cell value is shared between two cages it contributes half to each
Can't work this out. Do r1c2 and r2c1 share "half" (ie 6/2=3) with some other cells? Is r1c1 a singleton cage? Perhaps instead of half you mean the cages overlap the cells with the diagonal dashes. Do the cages have the total in the top left as per normal killer convention? Perhaps if you post the solution I could work it out.
Joined: Tue Jun 16, 2009 9:31 pm Posts: 282 Location: California, out of London
Don't understand all your questions Ed - but here are examples which clear things up (I hope).
Yes - every cage total is in the leftmost cell of the topmost row of the cage. (This may be a half-cell).
There is a cage in [r1c1, r1c2 (half), and r2c1 (half)] = 4. Possible solutions for that cage are: [124] (I.e., 1 + 2/2 + 4/2), , [142], [213], [231]
Similarly the cage r6c2 (half) + r6c3 = 5. Possible solutions are [24], [43], [62], [81].
Other example implications: a)The cells at r6c89,r7c789, r8c89 between them add up to 27+18 = 45. b) Whatever is in r6c8 can only be in r89c7 or r9c9 in n9. c) r9c7 (half) + r9c89 = 6. -> r9c7 must be even.
Joined: Tue Jun 16, 2009 9:31 pm Posts: 282 Location: California, out of London
Ed, - Sorry - I threw away the solution I came up with some time ago and no longer have it.
A couple more examples....
The halves add up to wholes. So the cage in [r1c1, r1c2 (half), and r2c1 (half)] = 4 could be [213] since 2 + 1/2 + 3/2 = 4. A cage with exactly one half-cell must have an even number in that half-cell. However, a cage with two half-cells could have 2 even numbers or 2 odd numbers in those half-cells.
You are right! The '6' in r9c8 should maybe have been in r9c7. The cages are delimited by the bold dashed lines so may be identified even when the cage total is perhaps not exactly where you expect it. The cage at the bottom right consists of r7c9(half), r8c9, r9c9. What is in r7c9 must be even.
The cells r9c56789 add up to 9+6 = 15 - must be {12345}. Since r7c9 must be even the possibilities for r9c56789 are [{35}2{14}] or [{25}4{13}]
It was a nice puzzle. It takes a while to get started, but once you break it open, it goes very smoothly.
Question that contains a spoiler, so I hid it.
Hidden Text:
Are the parity deductions intended as the opening deductions? I really liked those. I actually kept track of the parity of all cells through the whole solve.
Ed wrote:
Thanks wellbeback. I understand some of your examples but not others so there must be something I'm still missing.
Quote:
There is a cage in [r1c1, r1c2 (half), and r2c1 (half)] = 4.... [213], [231]
Don't get these. How can you halve 1 and 3?
Quote:
c) r9c7 (half) + r9c89 = 6. -> r9c7 must be even.
Isn't there a 6(2) at r9c89? How you can include r9c7 in the 6 total?
The easiest way may be to post the solution so I can work it out.
I think there easiest way to understand this, is to look at for example the 10 cage at R4C1 or the 8 cage at R5C1. They both consist of only half cells. So each cell is only counted as half the value of each cell. So as each cell counted as half the value adds to 10, you know that these 3 cells add to 20. Under the same reasoning the 4 cell in which the 8 cage lies add to 16.
Joined: Wed Apr 16, 2008 1:16 am Posts: 1044 Location: Sydney, Australia
Thanks to wellbeback and Para for the hints and explanations. I've looked at the puzzle a bit each day but just can't seem to get anything happening. I assume from HATMANS intro I should be able to make big chunks so must be missing something. Do you guys remember how you got started?
Joined: Tue Jun 16, 2009 9:31 pm Posts: 282 Location: California, out of London
This isn't how I actually got started the first time - but here's a possible start...
Hidden Text:
(a) 5-cage @ r6c2 has one half-cell -> r6c2 is even. Possibilities are [24], [43], [62], [81].
8-cage @ r5c1 has 4 half cells. Those four whole cells add to 16. Since r6c2 is even at least one of the other cells in the 8-cage must also be even. They can't all be even since 2+4+6+8 does not add to 16. (b) -> 8-cage consists of 2 even numbers and 2 odd numbers.
Innies - outies n7 -> half cell at r7c2 is 1 greater than half-cell at r6c1. -> Whole cell at r7c2 is 2 greater than whole cell at r6c1. (c) (Using (b)) -> r7c2 and r6c1 are odd, and r5c1 is even.
Since the two even numbers sum to at least 6 -> two odd number sum to at most 10. (d) -> Choices for [r6c1,r7c2] are [13] and [35]
10-cage at r4c1 consists of 3 half cells. Those whole cells must add to 20. (e) (using (c)) -> r5c1 from (468).
Also using (b) choices for [r6c1,r7c2] and possibilities for [r5c1,r6c2] and r6c3 are: [r6c1,r7c2] = [13] -> [r5c1,r6c2] = +12 = [84] -> r6c2 = 3. [r6c1,r7c2] = [35] -> [r5c1,r6c2] = +8 = [62] -> r6c2 = 4.
A little more work on the remainder of n4 will show that the latter is impossible.
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I liked it as much as them.