Tsumego: rating difficulty level

I don’t exactly know how but we might be able to (1) initialize the difficulty of a puzzle to a certain value and (2) let it slowly drift as players attempt the puzzle. Process 1 could be coarse but after Process 2 we can let it take finer grained values.

One idea is to treat puzzles’ difficulties and players’ puzzle-strengths (≠ Go-strengths) on equal footing, as probability distributions:

puzzle-dist-labeled

For example, a puzzle based on finding/knowing a clever trick may stay as a flatter curve like Puzzle A. On the other hand, the curve of a puzzle that’s less tricky but requires thorough reading may get tighter like Puzzle B.

Then I wonder if this interpretation is reasonable:
EncounterRateA : EncounterRateB = AreaA : AreaB (?)

In other words, is it reasonable to let the encounter rate of a puzzle for a player be proportional to the overlapping area?

I like the puzzle-strength idea, but he overlapping area is not a good idea. It means:
M: probability multiplier for encounter
P(s): Players “probability density” as a function of strength
A(s): Puzzles “probability density” as a function of strength
M= sum for s (min(A(s);P(s)))
If you want to use a similar formula, use:
M= sum for s (A(s)*P(s))
or
M= sum for s (sqrt(A(s))*sqrt(P(s)))

edit: edited some typos and notations

I understand this calculates overlapping area.

I still don’t see why it’s not a good idea looking at the 3 formulas but maybe my math knowledge isn’t enough.


sum for ssum over s
I think it’s clearer.

The problem with the overlapping area:
Imagine your figure, but with all Puzzle A, Puzzle B and Player X centered around the same s. Puzzle B will lose frequency, just because it’s deviation is smaller (some of it’s curve is way above the player curve). It’s strange to play a tsumego more rarely, just because it’s rank is more sure.

The second and third formula were just handwavy guesses, those would be the first I try (these kind of problems came up in my job a lot). If you’d like to calculate a good formula the way would be:

  • a model to get the winning ratio w from A(s) and P(s)
  • some requirements for a pairing system to get M from w.
2 Likes

(Emphases mine.)

That makes a lot of sense!
too-tall

Interesting insights.

1 Like

Here I put 4 problems (black to play in each of them) to illustrate how what you call now scale of problem and you called before complexity: quantity of stones needed to make the problem, is not really a good criteria

In my opinion, the two problems on the right are far more difficult as the ones on the left.

Agreed.
In general however I think that the number of stones is in an indicator for the complexity of a puzzle, but of course there will always be exceptions.
What do you suggest as a better criterium?

I think:
Forget about this one. It’s too restricted to some very particular cases when there is more to read because more different tactical situations, but this is too rare to be the core of the problems. And even I would say in my opinion that it’s more the reverse which is the case in general. Curiously the more stones, the less complicated.

Another one : you can combine wideness x deepness
(and eliminate the specific case of the first move which applies for each move, not only the first)

1 Like