Da Vinci Code Killer

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Da Vinci Code Killer

Postby mikejapan » Fri Feb 09, 2007 4:50 am

Crack the code, then complete the killer. It's a Zero X.

Image
Last edited by mikejapan on Tue Feb 13, 2007 4:29 am, edited 2 times in total.
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Postby mikejapan » Sun Feb 11, 2007 5:40 am

As this puzzle appears to be causing a lot of problems, here's a hint;

This is a killer but there are no cage totals. However, each cage has a number in it so that number must be related to the cage total. If you look at the two cell cage with a 5 in it, if the other number is a 9, the cage total would be 14. 1+4=5. The cages with 7 in them must be either 16/25/34 or 43. If there are only three cells, the cage must be 16.
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Postby Scott H » Mon Feb 12, 2007 7:27 am

Interesting enough with the code given. Since there are so many other possible codes, I question whether the code is even logically crackable.
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Postby mikejapan » Mon Feb 12, 2007 9:10 am

A killer must have cage totals. This one didn't have any. However, there was one number given in each cage. Therefore it is logical to assume that this given number must relate somehow to the cage total.
If you add the numbers 1/2/3/4/6/7/8/9 to the 5 in the 2 cell cage, the only logical number it can be is 9 because 14 is 1+4=5 - a definite connection. The other numbers have no logical connection.

(Well, I thought it was pretty logical and perfectly crackable)
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Postby mikejapan » Mon Feb 12, 2007 11:43 am

Scott H
Here's a killer just for you. I think this can be solved logically - with a little lateral thinking.
It's called "4 Calling Birds" and is one of twelve I did depicting the 12 days of Christmas (see Djape's site for all 12)

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Postby HATMAN » Wed Feb 14, 2007 12:11 pm

DVC1 start of solution (I seem to have a problem putting tiny text on this forum so see DJ's for the rest):
It has cages so it is fair to assume that it is a Killer of some form
No mention is made of using numbers other than 1-9 so it is fair to assume that this is a standard numbered Sudoku.
It is reasonable to assume that the cell values give a clue to the cage totals.
It is again reasonable to assume that the clue is for the cage containing the cell.
The most obvious assumption (especially to me) is some form of KiMo linkage.
Given the 8 at r3c1 we are limited to KiMo-10 or KiMo-9.

Consider the doublet at r34c4 with a 5 at r4c4 if it is a KiMo-10 then this implies r3c4 = 0, but we have already assumed 1-9 hence r3c4 = 9 and this is a KiMo-9 puzzle.
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Postby mikejapan » Wed Feb 14, 2007 4:18 pm

Way to go Hatman! I'm glad to see the code was crackable by someone.
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Postby HATMAN » Thu Feb 15, 2007 10:08 am

Mike

I was initially stumped by this one - it was only when Udosuk asked me to check his solution that I got what you were on about. So I then went through it in pedantic detail.

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