Enforce S0(Again) < S0(Hard) < S0(Good) < S0(Easy)

[Expertium]

smooth_and_fill isn't strict enough. We need Again < Hard < Good < Easy, not just Again <= Hard <= Good <= Easy. This is so that the user is less likely to see the same interval if learning steps are removed.

[L-M-Sherlock]

I find out a case which may help us to clarify the problem:

rating	stability	count
again	0.47	65
hard	0.27	8587
good	1.13	15614
easy	100.0	4348

Intuitively, we need to decrease the stability of again. And the degree of decrement should depend on the sample size.

[Expertium]

I found a problem with the current way you estimate missing values of S0

import numpy as np

w1 = 3/5
w2 = 3/5

rating_stability = {1: 0.47, 2: 0.27, 3: 1.13, 4: 100.0}

def impute(rating_stability):
    S0_1 = np.power(rating_stability[2], 1 / w1) * np.power(rating_stability[3], 1 - 1 / w1)
    S0_2 = np.power(rating_stability[1], w1) * np.power(rating_stability[3], 1 - w1)
    S0_3 = np.power(rating_stability[2], 1 - w2) * np.power(rating_stability[4], w2)
    S0_4 = np.power(rating_stability[2], 1 - 1 / w2) * np.power(rating_stability[3], 1 / w2)
    new_rating_stability = {1: round(S0_1, 4), 2: round(S0_2, 4), 3: round(S0_3, 4), 4: round(S0_4, 4)}
    return new_rating_stability

new_rating_stability = impute(rating_stability)
print(new_rating_stability)

The result is {1: 0.104, 2: 0.6676, 3: 9.3874, 4: 2.9346}, which is non-monotonic. I had an idea related to using these estimates, but it seems that your method needs to be revised.