Penrose Sudoku 
[Sep. 19th, 200804:20 am]




Comments: 
Am I the only person struggling to se the row/column equivalents?
This is quite beautiful but I'm having difficulty with making out the row/column equivalents as well.
I did find it difficult for a moment. It looks like they go through opposite edges of the spaces. All rows/columns are 8 spaces long, so "this number must go in one of these spaces" logic works with them, provided you can keep track of them. Now then, as for actually *solving* it...
Yeah, ok. I can see it now.
Long time no see, btw
Has anyone solved this? I've convinced myself the 3's can't all be placed...
From: (Anonymous) 20080920 08:00 pm (UTC)
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Wow  my first impression is this is possibly the most beautiful sudoku variant I've come across  can't wait to give it a go!
Tom.C
Its a beautiful design, but this particular example is rather tough as most absolutely minimal puzzles are. I did get to an answer, but it required forward checking of a couple possible placements of some numbers.
Indeed. I debated as to whether the first public example of this puzzle should be a tough "absolute minimal" puzzle or a casual warmup.
From: (Anonymous) 20080922 11:40 am (UTC)
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Haha well it's a welcome novelty to have variants where you have to spend hours working out all the tricks and subtleties of a design, rather than already have them in your solving arsenal as second nature  or indeed to have them gently served up to you on a plate.
The upshot of it so far? I've placed 2 numbers, but I'm getting a feel for the logic required.
Tom.C
From: (Anonymous) 20080923 08:40 pm (UTC)
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This is a most incredibile sudoku I have ever seen! Please, can you explain the rules and publish the solution? Nikola Zivanovic
The rules are pretty much same as a normal sudoku. Fill in the cells with numbers 18 so that every region contains all numbers 18, every row contains all numbers 18, and all columns contain all numbers 18.
The only catch is that "rows" and "columns" zigzag a bit using 5way symmetry.
A good way to think about it goes like this. Imagine trying to figure out what a "row" is in a normal sudoku. You start with a cell on an edge. Imaging "entering" that edge and exiting out the other edge parallel to that; you'll get the next cell in the row. Now continue across and exit across the next edge parallel to that, and you get the next cell in the row. Eventually, after crossing a series of celledgecelledge (where all the edges you cross are parallel), you'll exit out the other side. That's what a "row" is.
From: (Anonymous) 20080925 09:06 pm (UTC)
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Thanks! I understand now. Nikola Zivanovic
I spent a fair amount of time on this one, managed to place an 8 and a 3, and then gave up. I'd be interested in a set of logic explaining how one can go about solving this one completely.
Well, I can tell you how I found an answer, which was easier than finding that it was the answer. First, let's standardize some notation as on this diagram. There are 10 bands (what would be rows/columns in a 2d sudoku) labeled A to I. All cells can be described as some pair of the letters A through I with no AB, CD, EF, GH, IJ for a total of 40 cells. Look at HJ (I choose this cell in part because it interacts with all the empty cells but one in the region on the right). Let's say HJ is not 4 or 5, but some other X. Then DF is also X. Checking these shows IB is also X. As are EG and AC. You'll see that all of these cells are in the same relative position in their groupings (and it turns out if any two of those positions are a single number, I'm pretty sure they all must be). Now, none of the X's overlap with a number yet, so X must be 1. Once you've filled in those places with ones, it is MUCH easier to make progress on this sphere. For example, an 8 in AE or EJ now forces an 8 in BD (which can be worked around the diagram to force all other 8s). The placement of an X in DF happens to force a 7 in AH which is also valuable to work from. Working forward, you should be able to find an answer once all those 1's go in. So, to know this is THE answer, we must show HJ is not a 4 or 5. Well, the easy part is the 5. If HJ is a 5, then BI is a 5, but then none of the cells in the lowerleft region can be 5. So HJ cannot be 5. Proving HJ is not 4 is nowhere near as simple and I can eventually do it but only with a couple chains of eliminations from a bifurcated guess. I welcome any good descriptions of how to eliminate 4 from HJ, but I really liked HJ as the key cell to work in on this puzzle.
Fascinating. If HJ is 4 then CA must be 2 or 6, and I assume that both of those cases can be checked easily, but that is a bit of backtracking.
From: (Anonymous) 20080930 08:02 pm (UTC)
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And after AC is 2 or 6 that means EG must be 4 or 7, although that doesn't seem to help all that much.
From: computrix 20090222 05:37 am (UTC)
Penrose Sudoku  has anyone solved it?  (Link)

I just recently found this puzzle  and have been working on it. I have gotten close  but still have 1 error. I posted my result on my journal page  just joined to find out more about this. Has anyone solved the puzzle? Just want to know if there is a solution...
From: (Anonymous) 20091115 05:31 am (UTC)
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Lol ok,, definatly time to visit sudoku tips (http://sudokutips.us) for this baby I think. I've never seen anything like this before, it's well.. it's absolutely facinating. I think im going to make a big pot of cofee and get down to it. Ohh does this look fun.
From: (Anonymous) 20091115 05:32 am (UTC)
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Lol ok,, definatly time to visit http://sudokutips.us leav minding I've never seen anything like this before, it;s facinating. I think im going to amke a big pot of cofee and get down to it.  

