tag:blogger.com,1999:blog-8283950071005526162.post5139867281938062090..comments2023-12-09T13:25:54.553+01:00Comments on Para's Puzzle Site: WSC Practice #4: Round 13 - Puzzle 9: Joker Battenburg SudokuParahttp://www.blogger.com/profile/02367804879917808922noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-8283950071005526162.post-68022044831072856242020-05-11T12:34:08.548+02:002020-05-11T12:34:08.548+02:00This puzzle is impossible, because there is no odd...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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8283950071005526162.post-79481378076233490702019-10-14T17:47:32.868+02:002019-10-14T17:47:32.868+02:00Generally I do not learn article on blogs, but I w...Generally I do not learn article on blogs, but I would like to say that this write-up very forced me to take a look at and do so! Your writing style has been amazed me. Thank you, very great article.<br /><br /><br /><br /><a href="https://www3.go123movies.io/" rel="nofollow">https://www3.go123movies.io/</a><br />Live Sportshttps://www.blogger.com/profile/15046287190667256966noreply@blogger.comtag:blogger.com,1999:blog-8283950071005526162.post-30164931441644866242017-11-29T20:57:59.291+01:002017-11-29T20:57:59.291+01:00Oh wait. You actually say that in the post. Sorry....Oh wait. You actually say that in the post. Sorry.Anonymoushttps://www.blogger.com/profile/05316761520404488684noreply@blogger.comtag:blogger.com,1999:blog-8283950071005526162.post-61861805649176632272017-11-29T20:54:31.737+01:002017-11-29T20:54:31.737+01:00OK. It makes more sense, and certainly works, if a...OK. It makes more sense, and certainly works, if all nine occurrences of the mystery digit are Jokers.<br /><br />As 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.Anonymoushttps://www.blogger.com/profile/05316761520404488684noreply@blogger.comtag:blogger.com,1999:blog-8283950071005526162.post-56132027959685307942017-11-27T21:32:11.949+01:002017-11-27T21:32:11.949+01:00First, is *every occurrence* of one digit a Joker,...First, 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.Bryce Herdtnoreply@blogger.comtag:blogger.com,1999:blog-8283950071005526162.post-34869921884007494252017-10-09T15:56:59.355+02:002017-10-09T15:56:59.355+02:00..44:01..That was a brutal puzzle and finding the .....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 !!Anonymousnoreply@blogger.com