Logic Masters Deutschland e.V.

Standard sudoku rules applies. Fill the grid with numbers from 1 to 5 with a gap of 0.5 so each digit occurs exactly once in every row, column and 3x3 box.

Circle means four digits who surround it fulfill two of the following three conditions, two odd integer, two even integer, two non-integer(fraction). in other words, digits around circle is either two odds and two evens, or two odds and two fraction, or two evens and two fractions.

All possible circles are given. Digits in the circle is the sum of the biggest digit and the smallest digit around the circle.

It’s not a good idea. To be honest, the type good is the basic binary group variant. So when i try to think of a ternary group, i had a test variant already. My first idea is involving negative. And the 0 problem arise. Even if i remove 0, there will be parity and negative, which is a quaternary group. Any quality of digits in integer i add will crisscross with the parity and separate digits into 4 groups. If i use 0 as one single group, then, it would not be a group. It could be easily be a single digit exclusion, not what i want. So i have to use a group that is not integer, which can be either fraction or complex number. Fraction is more intuitive, so i try it.

The result is not good. The problem is not rules writing or explaining, which can be easily fix by example or illustration, which i don’t bother to add because it’s not worthy. The problem is at the solver end. Pencil mark is really agonizing. There’re two ways to do it. One is multiply everything by two, and make everything integer again. Trouble is now the grouping is unintuitive. The other way is change fraction into another part attach to the number, for instance, color. Say, red is integer and green is fraction. Still strange, but easier to deal with clues.

To be honest, this puzzle is not hard. But the pencil mark part is hard. It’s not the experience i want when solving a puzzle. Juggling with pencil mark throughout the solve is toiling, even when the puzzle itself is easy. One evidence is that one cannot speed through any part of this puzzle, or he may risk of forgetting groups. At some point, ternary group solve some problem with the binary group, mainly negative restraint shows every cell’s property, a solve path so boring that no setter, at least here, wants. However, it seems that each variant has a self contained difficulty. I can only trade one difficulty into another one.