I had (confidently) deduced the 2x2 boxes in R1-4 C9-12 wrongly (exactly the opposite of the actual solution). And the grid allowed me to continue a long way, before getting a contradiction in the 11-16 range. In these columns, R1-2 and R3-4 had the same set of inequality signs, and this induced the mistake. And this error took a lot of time to "undo".

I'm really surprised by my position in this contest. My printer gobbled up 10 randomly placed inequality signs, so I first had to slowly scan the PDF and place them (and keep the PDF open all the time just in case I went wrong). Then for some odd reason, I used triangles to mark the 13579 group when I usually use circles, and when it came to filling in this group later I realized that I'd made some of the triangles too small and spent a few more minutes erasing the smaller triangles (while keeping an eye on the PDF to make sure I didn't take away more inequality markings). I wouldn't have defeated Endo, but its nice to know I might have had a 'marathon bonus' on this puzzle if external factors hadn't played a part in it the way they did.

The break-in was really cool and while I expected some more solvers to see it quickly I did expect there to be many DNFs which is a good reason to not have it in the Marathon. But as a fun contest, well this was a lot of fun :)

Password removed and solutions appended.

Thanks Ivan for a great puzzle (and also for the neat write up). I'm not sure how you felt when we "rejected" this one from Marathon, but now I'm very glad we did that.

Thanks also to Branko for testing this.

Congratulations for double-podium-weekend.

Spent a rainy weekend periodically fiddling with this puzzle, trying to find a break in.. I marked the 2x2 region divisions, found and marked with various colors the 4- and 3-chains, but eventually put it on hold until the end of the contest, figuring I'd find the hint I needed from the thread comments. I think I would have eventually realized what each 2x2 quartet belonging to a single leg of the number map implies for categorizing and placing the quartets, but my thought process never really came near the direction required to realize the parity break-in.

However, getting stumped was not discouraging at all, I knew from looking at the puzzle (and from Ivan's past puzzles) that it would eventually be enjoyable, and I patiently waited out the contest duration. Sure enough, by about bullet 3 or 4 of Ivan's great break-in explanation, I was off to the races. And was happy to find that all the cell-coloring work I had done earlier was not wasted, and very useful during the solve process. As expected, an extremely fun solve after getting over that first hurdle. Thank you Ivan!

I believe there is a somewhat simpler way to determine that the L-blocks must all contain only odd numbers.

If one looks at any row (or column) of 2x2 blocks, there are 6 blocks each containing 4 values and 2 L-blocks each containing 3 values (when the 10s are excluded). In these blocks we need to place 10 odds, 8 evens and 12 tens with the restriction that each block can only contain values of the same type. The only way the odds can be accommodated is by using one 4-block and both L-blocks. The remaining groups fit comfortably in 2 and 3 4-blocks respectively.

I still can't understand how anyone can solve this puzzle in 41 minutes! :)