I feel this round can use all the practice necessary. The Joker rule is something new, as far as I'm aware, and seems to be a bit tricky. Basically the Joker only follows the standard Sudoku rules, but ignores any variant in the puzzle. This is because the Joker digit can assume whatever digit it needs to be for the puzzle to work, even if it means being different digits in the same cell.

The Battenburg Sudoku was one of my own inventions. The reason I picked this type first is because it has both a positive and a negative rule in effect as all possible markings are given. The "All possible markings" rule is a bit odd with the Joker. The way I understand it, in the given solution there are no places where any extra markings HAVE TO be placed. Even if there are markings that could be placed if the Joker assumes the correct parity, they don't have to be placed as the Joker can assume an incorrect parity. I do hope my interpretation is correct, but it seems to conform with the given example in the Instruction Booklet. If it isn't, my apologies for providing an incorrect puzzle. It has happened before. It should still be fun to solve.

I think the puzzle worked out pretty well. The opening should properly show the way this variant interacts with the Joker. Once you've determined the value of the Joker, the solve really takes off. It's probably a harder puzzle because you have to keep reminding yourself of the Joker rule and thread carefully through the puzzle.

[Edit: Puzzle image fixed.]

Rules for Sudoku

In this Sudoku, everywhere a 2x2 square forms a Checkerboard pattern of Odd and Even digits a Battenburg marking is given. All possible Battenburg markings have been given.

Additionally, one digit in the puzzle is a Joker. This Joker digit can be both Odd or Even. The same Joker cell may assume different parities for different 2x2 areas.

..44:01..That was a brutal puzzle and finding the Joker didn't help much.The solve was equally hard even after that, but, what it did do is alleviate my fear of the Joker Round .So a big thanks to you Bram !!

ReplyDeleteFirst, is *every occurrence* of one digit a Joker, or just one cell? Second, and more importantly, when you say "All possible Battenburg markings have been given," does the Joker cooperate? I'm having trouble in the lower left.

ReplyDeleteOK. It makes more sense, and certainly works, if all nine occurrences of the mystery digit are Jokers.

DeleteAs for the second, and more pressing, question: no. That is, if a 2x2 area with a Joker might get a Battenburg marking if the Joker takes one parity, then at the constructor's discretion, the Joker may take the other parity and the area won't get the mark.

Oh wait. You actually say that in the post. Sorry.

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This puzzle is impossible, because there is no odd or even number that can be placed in the center of the bottom left 3x3 square, while satisfying the conditions. Unless you assume that this should be a joker, (I do not understand this concept of a joker). However, even then it is also impossible to place a number in the center of the top right 3x3 square. An odd value there would result in two missing marks (with the 1 and 3), and an odd value would mean that we need to place 6 odd numbers and 3 even in this box.

ReplyDelete