Supermemo 2 Algorithm, Unobscured (Python 3)

repetition_algorithm.ipynb

supermemo_2.py

def supermemo_2(x: [int], a=6.0, b=-0.8, c=0.28, d=0.02, assumed_score=2.5, min_score=1.3, theta=1.0) -> float:
    """
    Returns the number of days until seeing a problem again based on the 
    history of answers x to the problem, where the meaning of x is:
    x == 0: Incorrect, Hardest
    x == 1: Incorrect, Hard
    x == 2: Incorrect, Medium
    x == 3: Correct, Medium
    x == 4: Correct, Easy
    x == 5: Correct, Easiest
    @param x The history of answers in the above scoring.
    @param theta When larger, the delays for correct answers will increase.
    """
    assert all(0 <= x_i <= 5 for x_i in x)
    correct = [x_i >= 3 for x_i in x]
    # If you got the last question incorrect, just return 1
    if not correct[-1]:
        return 1.0
    
    # Calculate the latest consecutive answer streak
    r = 0
    for c_i in reversed(correct):
        if c_i:
            r+=1
        else:
            break
    
    return a*(max(min_score, assumed_score + sum(b+c*x_i+d*x_i*x_i for x_i in x)))**(theta*r)

supermemo_yavascript.js

function daysTillNextTestAlgorithm(recent, x, a = 6.0, b = -0.8, c = 0.28, d = 0.02, theta = 0.2) {
  if (recent < 4) {
    return 1
  }

  const history = [recent, ...x]

  // Calculate latest correctness streak
  let streak = 0
  for (let i = 0; i < history.length; i++) {
    if (history[i] > 3) {
      streak++
    } else {
      break
    }
  }

  // Sum up the history
  const historySum = history.reduce(
    (prev, val) => prev + (b + (c * val) + (d * val * val)),
    0
  )

  return a * Math.pow(Math.max(1.3, 2.5 + historySum), theta * streak)
}