java - 来自 SuperMemo (SM-2) 的间隔重复算法

Question

为了在 Android 中制作一个词汇练习应用程序，我想实现 SuperMemo (SM-2) algorithm在 java 。这是间隔重复软件的流行选择，据我所知，Anki 甚至采用了它。给出的源代码示例here由于缺乏代码格式并且是用 Delphi 编写的，因此(对我而言)很难理解。

SuperMemo 的作者 states :

1. Split the knowledge into smallest possible items.
2. With all items associate an E-Factor equal to 2.5.
3. Repeat items using the following intervals: I(1):=1
   I(2):=6
   for n>2: I(n):=I(n-1)*EF
   where:
   I(n) - inter-repetition interval after the n-th repetition (in days),
   EF - E-Factor of a given item
   If interval is a fraction, round it up to the nearest integer.
4. After each repetition assess the quality of repetition response in 0-5 grade scale: 5 - perfect response
   4 - correct response after a hesitation
   3 - correct response recalled with serious difficulty
   2 - incorrect response; where the correct one seemed easy to recall
   1 - incorrect response; the correct one remembered
   0 - complete blackout.
5. After each repetition modify the E-Factor of the recently repeated item according to the formula:
   EF':=EF+(0.1-(5-q)*(0.08+(5-q)*0.02))
   where:
   EF' - new value of the E-Factor,
   EF - old value of the E-Factor,
   q - quality of the response in the 0-5 grade scale.
   If EF is less than 1.3 then let EF be 1.3.
6. If the quality response was lower than 3 then start repetitions for the item from the beginning without changing the E-Factor (i.e. use intervals I(1), I(2) etc. as if the item was memorized anew).
7. After each repetition session of a given day repeat again all items that scored below four in the quality assessment. Continue the repetitions until all of these items score at least four.

以下是关于 Stack Overflow 的一些相关(但不同)问题:

• What is the spaced repetition algorithm to generate the day intervals?
• Open Source implementation of a Spaced Repetition Algorithm in Java
• Spaced repetition (SRS) for learning

你如何在 Java 中实现它？

(我最近一直在研究这个问题，我想我有答案了，所以我将其作为问答对提交，以帮助其他人做同样的事情。)

Answer 1

SuperMemo 算法

以下是我们在实现 SuperMemo (SM-2) algorithm 时将处理的一些术语的 spaced repetition .

• 重复次数 - 这是用户看到抽认卡的次数。 0表示他们还没有研究过，1表示他们是第一次，以此类推。它在某些文档中也称为 n。
• 质量 - 也称为评估质量。这就是抽认卡的难度(由用户定义)。比例从 0 到 5。
• easiness - 这也称为 easiness factor 或 EFactor 或 EF。它是用于增加间隔重复中的“空间”的乘数。范围从 1.3 到 2.5。
• 间隔 - 这是重复之间的时间长度(以天为单位)。它是间隔重复的“空间”。
• nextPractice - 这是 date/time抽认卡何时再次复习。

默认值

int repetitions = 0;
int interval = 1;
float easiness = 2.5;

代码

我找到了 this Python implementation比 SuperMemo example source code 更容易理解，所以我或多或少地遵循了这一点。

private void calculateSuperMemo2Algorithm(FlashCard card, int quality) {

    if (quality < 0 || quality > 5) {
        // throw error here or ensure elsewhere that quality is always within 0-5
    }

    // retrieve the stored values (default values if new cards)
    int repetitions = card.getRepetitions();
    float easiness = card.getEasinessFactor();
    int interval = card.getInterval();

    // easiness factor
    easiness = (float) Math.max(1.3, easiness + 0.1 - (5.0 - quality) * (0.08 + (5.0 - quality) * 0.02));

    // repetitions
    if (quality < 3) {
        repetitions = 0;
    } else {
        repetitions += 1;
    }

    // interval
    if (repetitions <= 1) {
        interval = 1;
    } else if (repetitions == 2) {
        interval = 6;
    } else {
        interval = Math.round(interval * easiness);
    }

    // next practice 
    int millisecondsInDay = 60 * 60 * 24 * 1000;
    long now = System.currentTimeMillis();
    long nextPracticeDate = now + millisecondsInDay*interval;

    // Store the nextPracticeDate in the database
    // ...
}

注意事项

• 上面的代码没有设置easiness的上限。应该是 2.5 吗？文档和源代码似乎相互矛盾。
• 文档还表明，如果质量评估低于 3，则不应更新简单系数，但这似乎与源代码相矛盾。在我看来，更新它也更有意义(只要它保持在 1.3 以上)。不管怎样，我每次都会更新它。
• Anki 源代码是 here .不过，这是一个大项目，我还没有深入挖掘到看到他们的算法版本。
• This post讨论 SM-2 的一些问题以及这些问题的解决方案。