User:Y5nw

y5nw is a player on the server.

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Notes

Note: This section is more or less written based on my own knowledge. Feel free to suggest corrections in my talk page.

Event chance calculation

For a one-shot event that fires with a probability of $p$ checked at an interval of $T$ (e.g. tree growth), the expected amount of time $t$ it takes for the event to fire is $E(t)=\frac{T}{p}$.

Note that, in some definition tables in Minetest (e.g. ABMs), the probability is specified using the reciprocal of $p$.

Calculations: Let $q:=1-p$ be the probability of the event not firing at the next check.

$\begin{array}{rl}E(t)& =\sum _{n=1}^{\mathrm{\infty }}(q^{n-1}(1-q))(Tn)=(1-q)T\sum _{n=1}^{\mathrm{\infty }}{q}^{n-1}n=(1-q)T\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^n{q}^{n-1}\\ & =(1-q)T\sum _{n=1}^{\mathrm{\infty }}\sum _{m=n}^{\mathrm{\infty }}{q}^{m-1}=\frac{(1-q)T}{q}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=n}^{\mathrm{\infty }}{q}^m\\ & =\frac{(1-q)T}{q}\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=0}^{\mathrm{\infty }}{q}^m-\sum _{m=0}^{n-1}{q}^m)=\frac{(1-q)T}{q}\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{1-q}-\frac{1-q^n}{1-q})=\frac{(1-q)T}{q}\sum _{n=1}^{\mathrm{\infty }}\frac{q^n}{1-q}\\ & =T\sum _{n=0}^{\mathrm{\infty }}{q}^n=T\cdot \frac{1}{1-q}=\frac{T}{p}\end{array}$

(Abandoned) Idea for prefix-notation syntax

For my railtour mod I considered using a simple prefix-notation syntax for filter expressions (although IMO filters have a low priority at the time of writing). A somewhat adequate prefix syntax can be made to sufficiently resemble infix notation used for mathematics. Consider the following example:

x >= 5

As an infix expression, this is parsed to something similar to $(\ge )(x,5)$ with $\ge :\mathbb{ℝ}\times \mathbb{ℝ}\to \mathbb{𝔹}$.

However, given

$\begin{array}{rl}F& :X\to \{\{Y\to X\}\to Y\},x\mapsto F_x\\ F_x& :\{X\to Y\}\to Y,f\mapsto f(x),x\in X\\ C& :\{X\times Y\to Z\}\to \{X\to \{Y\to Z\}\},f\mapsto C_f\\ C_f& :X\to \{Y\to Z\},C_f(x)(y):=f(x,y)\end{array}$

(i.e. $C$ effectively performs function currying.)

The above expression can be parsed into $F_x(C_{\ge })(5)$. It can be checked that

$\begin{array}{rl}& \phantom{=}{F}_x(C_{\ge })(5)\\ & =C_{\ge }(x)(5)\\ & =(\ge )(x,5)\end{array}$

It should be noted that this approach reaches its limitations when multiple "infix" operators are used, as in x >= y + 1, which is parsed into $F_x(C_{\ge })(y)(C_{+})(1)$. Evaluating this gives $(\ge )(x,F_y)(C_{+})(1)$ which does not have a sensible result.

The above issue can be avoided (albeit counterintuitively) by adding parentheses where appropriate, i.e. with x >= (y + 1) However, with the use of parentheses, a Lisp-based syntax à la (>= x (+ y 1)) appears more attractive as it also involves less function-related abstractions (which avoids issues arising from such abstractions; compare x >= 5 and 5 <= x with the custom prefix notation)