spaced-repetition.py

dict = {
    "problem_id": [
        [2.5],  # Current effort factor, defaults to 2.5 for new items
        [0],  # previous spacings, start with a 0 so new terms come quickly
    ]
}

# Using https://www.supermemo.com/en/archives1990-2015/english/ol/sm2
# Effort factor is calculated each time you answer correctly, and helps determine how long till the next interval
# I = which iteration this is, increments by 1 every time you get it right
# q is quality of answer, where 5 = perfect, 4 = correct w/hesitation, 3 = correct w/difficulty,
#   2 = wrong but quickly remember when shown, 1 = wrong and recognize correct answer, 0 = no recollection
def get_interval(problem_id, q):
    (EFs, spacings) = dict[problem_id]

    # An effort factor needs to be calculated, regardless of whether they answered it right or wrong
    # EF':= EF+(0.1-(5-q)*(0.08+(5-q)*0.02))
    new_EF = EFs[-1] + (0.1 - (5 - q) * (0.08 + (5 - q) * 0.02))
    # We want to lock it between 1.3 and 2.5, just to make sure it's not going to be too often or too rare
    if new_EF < 1.3:
        new_EF = 1.3
    if new_EF > 2.5:
        new_EF = 2.5
    EFs.append(new_EF)

    # If they got it wrong, you want to drop spacings all the way back no matter what
    if q < 3:
        spacings = 1
        return

    # Otherwise, you follow the formula
    # I(1):=1
    # I(2):=6
    # I(n>2):=I(n-1)*EF
    i = len(spacings)
    if i == 1:
        spacings.append(1)
    elif i == 2:
        spacings.append(6)
    else:
        spacings.append(spacings[-1] * new_EF)