The FSRS (Free Spaced Repetition Scheduler) algorithm is based on a variant of the DSR (Difficulty, Stability, Retrievability) model, which is used to predict memory states.

Here is a visualizer to preview the interval with specific parameters and review history: Anki FSRS Visualizer (open-spaced-repetition.github.io)

In case you find this article difficult to understand, perhaps you will like this more: https://expertium.github.io/Algorithm.html.

• $R$: Retrievability (probability of recall)
• $S$: Stability (interval when R=90%)
  • $S^\prime_r$: new stability after recall
  • $S^\prime_f$: new stability after forgetting
• $D$: Difficulty ( $D \in [1, 10]$ )
• $G$: Grade (rating at Anki):
  • $1$: again
  • $2$: hard
  • $3$: good
  • $4$: easy

[0.40255, 1.18385, 3.173, 15.69105, 7.1949, 0.5345, 1.4604, 0.0046, 1.54575, 0.1192, 1.01925, 1.9395, 0.11, 0.29605, 2.2698, 0.2315, 2.9898, 0.51655, 0.6621]

The stability after a same-day review:

$$S^\prime(S,G) = S \cdot e^{w_{17} \cdot (G - 3 + w_{18})}$$

The initial difficulty after the first rating:

$$D_0(G) = w_4 - e^{w_5 \cdot (G - 1)} + 1,$$

where $w_4 = D_0(1)$, i.e., the initial difficulty when the first rating is Again.

Linear Damping for the new difficulty after review:

$$\Delta D(G) = - w_6 \cdot (G - 3)$$

$$D^\prime = D + \Delta D \cdot \cfrac{10 - D}{9}$$

In FSRS 5, $D_0(4)$ is the target of mean reversion. Earlier versions had $D_0(3)$ as the target.

$$D^{\prime\prime} = w_7 \cdot D_0(4) + (1 - w_7) \cdot D^\prime.$$

The other formulas are the same as FSRS 4.5.

[0.4872, 1.4003, 3.7145, 13.8206, 5.1618, 1.2298, 0.8975, 0.031, 1.6474, 0.1367, 1.0461, 2.1072, 0.0793, 0.3246, 1.587, 0.2272, 2.8755]

The formula of forgetting curve is changed in this update.

The retrievability after $t$ days since the last review:

$$R(t,S) = \left(1 + FACTOR \cdot \cfrac{t}{S}\right)^{DECAY},$$

where $R(t,S)=0.9$ when $t=S$.

The next interval can be calculated by solving for t in the above equation after putting the request retention in place of R:

$$I(r,S) = \cfrac{S}{FACTOR} \cdot \left(r^\cfrac{1}{DECAY} - 1\right),$$

where $I(r,S)=S$ when $r=0.9$.

In FSRS v4, $DECAY=-1$ and $FACTOR=\cfrac{1}{9}$

In FSRS-4.5, $DECAY=-0.5$ and $FACTOR=\cfrac{19}{81}$

The new forgetting curve drops sharply before $S$ and flatly after $S$.

[0.4, 0.6, 2.4, 5.8, 4.93, 0.94, 0.86, 0.01, 1.49, 0.14, 0.94, 2.18, 0.05, 0.34, 1.26, 0.29, 2.61]

The $w_i$ denotes w[i].

The initial stability after the first rating:

$$S_0(G) = w_{G-1}.$$

For example, $S_0(1)=w_0$ is the initial stability when the first rating is again. When the first rating is easy, the initial stability is $S_0(4)=w_3$.

The initial difficulty after the first rating:

$$D_0(G) = w_4 - (G-3) \cdot w_5,$$

where the $D_0(3)=w_4$ when the first rating is good.

The new difficulty after review:

$$D^\prime(D,G) = w_7 \cdot D_0(3) +(1 - w_7) \cdot (D - w_6 \cdot (G - 3)).$$

It will calculate the new difficulty with $D^\prime = D - w_6 \cdot (G - 3)$, and then apply the mean reversion $w_7 \cdot D_0(3) + (1 - w_7) \cdot D^\prime$ to avoid "ease hell".

The retrievability after $t$ days since the last review:

$$R(t,S) = \left(1 + \cfrac{t}{9 \cdot S}\right)^{-1},$$

where $R(t,S)=0.9$ when $t=S$.

The next interval can be calculated by solving for t in the above equation after putting the request retention in place of R:

$$I(r,S) = 9 \cdot S \cdot \left(\cfrac{1}{r} - 1\right),$$

where $I(r,S)=S$ when $r=0.9$.

The new stability after a successful review (the user pressed "Hard", "Good" or "Easy"):

$$S^\prime_r(D,S,R,G) = S\cdot(e^{w_8} \cdot (11-D) \cdot S^{-w_9} \cdot (e^{w_{10}\cdot(1-R)}-1) \cdot w_{15}(\textrm{if G = 2}) \cdot w_{16}(\textrm{if G = 4}) + 1).$$

Let $SInc$ (the increase in stability) denotes $\cfrac{S^\prime_r(D,S,R,G)}{S}$, which is equivalent to Anki's factor.

1. The larger the value of D, the smaller the SInc value. This means that the increase in memory stability for difficult material is smaller than for easy material.
2. The larger the value of S, the smaller the SInc value. This means that the higher the stability of the memory, the harder it becomes to make the memory even more stable.
3. The smaller the value of R, the larger the SInc value. This means that the best time to review your material is when you almost forgot it (provided that you succeeded in recalling it).
4. The value of SInc is always greater than or equal to 1 if the review was successful.

In FSRS, a delay in reviewing (i.e., overdue reviews) affects the next interval as follows:

As the delay increases, retention (R) decreases. If the review was successful, the subsequent stability (S) would be higher, according to point 3 above. However, instead of increasing linearly with the delay like the SM-2/Anki algorithm, the subsequent stability converges to an upper limit, which depends on your FSRS parameters.

You can modify them in this playground: https://www.geogebra.org/calculator/ahqmqjvx.

The stability after forgetting (i.e., post-lapse stability):

$$S^\prime_f(D,S,R) = w_{11}\cdot D^{-w_{12}}\cdot ((S+1)^{w_{13}} - 1)\cdot e^{w_{14}\cdot(1-R)}.$$

For example, if $D=2$ and $R=0.9$, with default parameters, $S^\prime_f(S=100) = 2\cdot 2^{-0.2} \cdot ((100+1)^{0.2}-1) \cdot e^{1(1-0.9)} \approx 3$ and $S^\prime_f(S=1) \approx 0.3$.

[0.9605, 1.7234, 4.8527, -1.1917, -1.2956, 0.0573, 1.7352, -0.1673, 1.065, 1.8907, -0.3832, 0.5867, 1.0721]

The $w_i$ denotes w[i].

The initial stability after the first rating:

$$S_0(G) = w_0 + (G-1) \cdot w_1,$$

where the $S_0(1)=w_0$ is the initial stability when the first rating is again. When the first rating is easy, the initial stability is $S_0(4)=w_0 + 3 \cdot w_1$.

The initial difficulty after the first rating:

$$D_0(G) = w_2 + (G-3) \cdot w_3,$$

where the $D_0(3)=w_2$ when the first rating is good.

The new difficulty after review:

$$D^\prime(D,G) = w_5 \cdot D_0(3) +(1 - w_5) \cdot (D + w_4 \cdot (G - 3)).$$

It will calculate the new difficulty with $D^\prime = D + w_4 \cdot (G - 3)$, and then apply the mean reversion $w_5 \cdot D_0(3) + (1 - w_5) \cdot D^\prime$ to avoid "ease hell".

The retrievability of $t$ days since the last review:

$$R(t,S) = 0.9^{\frac{t}{S}},$$

where $R(t,S)=0.9$ when $t=S$.

The next interval can be calculated by solving for t in the above equation after putting the request retention in place of R.

$$I(r,S) = S \cdot \cfrac{\ln(r)}{\ln(0.9)},$$

where $I(r,S)=S$ when $r=0.9$.

Note: the intervals after Hard and Easy ratings are calculated differently. The interval after Easy rating will multiply easyBonus. The interval after Hard rating is lastInterval multiplied hardInterval.

The new stability after recall:

$$S^\prime_r(D,S,R) = S\cdot(e^{w_6}\cdot (11-D)\cdot S^{w_7}\cdot(e^{w_8\cdot(1-R)}-1)+1).$$

Let $SInc$ (the increase in stability) denotes $\cfrac{S^\prime_r(D,S,R)}{S}$, which is equivalent to Anki's factor..

1. The larger the value of D, the smaller the SInc value. This means that the increase in memory stability for difficult material is smaller than for easy material.
2. The larger the value of S, the smaller the SInc value. This means that the higher the stability of the memory, the harder it becomes to make the memory even more stable.
3. The smaller the value of R, the larger the SInc value. This means that the best time to review your material is when you almost forgot it (provided that you succeeded in recalling it).
4. The value of SInc is always greater than or equal to 1 if the review was successful.

The following 3D visualization could help understand.

In FSRS, a delay in reviewing (i.e., overdue reviews) affects the next interval as follows:

As the delay increases, retention (R) decreases. If the review was successful, the subsequent stability (S) would be higher, according to point 3 above. However, instead of increasing linearly with the delay like the SM-2/Anki algorithm, the subsequent stability converges to an upper limit, which depends on your FSRS parameters.

The stability after forgetting (i.e., post-lapse stability):

$$S^\prime_f(D,S,R) = w_9\cdot D^{w_{10}}\cdot S^{w_{11}}\cdot e^{w_{12}\cdot(1-R)}.$$

For example, if $D=2$ and $R=0.9$, with default parameters, $S^\prime_f(S=100) = 2\cdot 2^{-0.2} \cdot 100^{0.2} \cdot e^{1(1-0.9)} \approx 5$ and $S^\prime_f(S=1) \approx 2$.

You can play the function in post-lapse stability - GeoGebra.

If you find this introduction helpful, I'd be grateful if you could give it a star: