Nice set of puzzles.
I was able to solve only 4 during the competition, but could not submit the fourth because I ran out of time.
Now I am taking my time to solve other puzzles.
Renban was very good although I have started from the R1 and R3 Renban groups. As you said "What is very helpful for this puzzle is to notice what digits have to be in a renban group, and what digits cannot be in a group", I did the same thing and puzzle was solved very smoothly.

Still more to solve.

Thanks Richard and LMI for wonderful test.

Regards,
Ravi

Let me first thank the author for providing the sudokus.
I'm quite new to these competitions, that might also be why I missed something, as I cannot see why there is a correct solution to one of the simpler sudokus: all odd/even
Let me explain, and hopefully one of you can prove me wrong :)

The grey fields either contain all even or all odd numbers. There are 5 odd numbers (1,3,5,7,9) and 4 even numbers (2,4,6,8) possible. Look at the rightmost column, there are 3 empty grey fields and even numbers 2 and 6 already present. That means that the 3 grey fields in this column cannot be filled with all even numbers (as there are only two left: 4 and 8). Hence, all grey fields should be odd-numbered.
Now look at the second row, there are only two non-grey empty fields, which are located in the top-middle 3x3 square. Number 4 is still missing in this row, hence it should be filled in somewhere in the middle 3, however there is already a 4 present in the top-middle 3x3 square. Hence, the grey fields cannot all be odd-numbered. A contradiction!

edit:
oops... I now see that the restriction on odd/even should be interpreted per 3x3 square... sorry

Edited by gjdv 2012-07-12 2:01 AM