Skip to main content

Is there a more elegant and efficient way to write brainf*** style loops in Mathematica?


I'm having some fun writing a brainf*** interpreter in Mathematica. See this wiki article for more info. It works nicely but I'd like to find an elegant, preferably functional, way to handle the looping structures.


This is a simple example of what I'm hoping to do with the solution I currently have in place. The variable ptr is the position of a pointer in the program. The variable paren is used to keep track of the brackets, it increments on "[" and decrements on "]" so it should be zero when I find the proper closing bracket.


ptr = 1;
paren = 1;
lst = {"[", "[", "-", "]", ">", ">", ">", ".", "<", "<", "<", "]",
">", ">", ">", ">", ">", ">", "."};


While[paren > 0, ptr++;
Switch[lst[[ptr]], "[", paren++, "]", paren--, _, Null]]

Which tells me the closing "]" is located at position 12 in lst.


In[287]:= ptr

Out[287]= 12

Is there a more elegant and/or efficient way to do this?



Answer




My solution isn't precisely what was asked for: it is the complete parser. At the moment, it is strictly a parser, but the functionality is all ready to be added.


Clear[bfinterpreter, bfeval, movf, movb, add1,  sub1, write, read, loop, 
stored, loopdepth, rls]
rls = {">" -> movf, "<" -> movb, "+" -> add1,
"-" -> sub1, "." -> write, "," -> read,
"]" :> With[{mode = loopdepth--}, loop[stored[mode]]]
};

bfeval[a : (">" | "<" | "+" | "-" | "." | "," | "]")] :=
With[{val = a /. rls},

If[loopdepth == 0,
val[ptr],
AppendTo[stored[loopdepth], val]; ## &[]]
]

bfeval["["] := (stored[++loopdepth] = {}; ## &[])
bfeval[_] := (## &[])

bfinterpreter[code_String] :=
Block[{loopdepth = 0, stored, ptr,

movf, movb, add1, sub1, write, read},
bfeval /@ StringSplit[code, ""]
];

A user would access this by passing bfinterpreter as String of BF commands. The string is split into individual characters via StringSplit[code, ""], and then bfeval is mapped over the resulting list. Scan would be better for the full implementation, but I used Map here to show that the parser works. In particular, using this on the OPs supplied BF code we get


(* spaces added for demo purposes *)
bfinterpreter["[[-]>> > . <<<]>>> >>>."]
(*
=> {loop[{ (* [ *)
loop[{sub1}], (* [-] *)

movf, (* > *)
movf, (* > *)
movf, (* > *)
write, (* . *)
movb, (* < *)
movb, (* < *)
movb (* < *)
}
][ptr], (* ] *)
movf[ptr], (* > *)

movf[ptr], (* > *)
movf[ptr], (* > *)
movf[ptr], (* > *)
movf[ptr], (* > *)
movf[ptr], (* > *)
write[ptr] (* . *)
}
*)

As you've probably figured out, all the magic happens in bfeval. The first form accepts all BF characters {">", "<", "+", "-", ".", ",", "]"}, but the open loop, "[". It works by applying a list of replacement rules to the supplied character and either storing it, if we're within a loop, or evaluating it immediately. The close loop rule, though, has added functionality in that it needs to both reduce loop counter, loopdepth, and return the stored commands for the loop, hence the use of RuleDelayed (:>).



The second form of bfeval simply increments the loop counter, and returns ##&[] to ensure the final list doesn't contain Nulls. And, the final form is the catch all for every other type of character.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...