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The following source doesn't compile because Stopover is unbound.
I'm new to Erlang, how can I rewrite it?
-module(distances).
-export([ path/2 ]).
path( madrid, paris ) ->
{ km, 1049 };
path( paris, moscou ) ->
{ km, 2482 };
path( moscou, berlin ) ->
{ km, 1603 };
path( From, To ) ->
path( From, Stopover ) + path( Stopover, To ).
The usage of this module maybe:
path( madrid, moscou ).
And the epected answer should be { km, 3531}.
The following source doesn't compile because Stopover is unbound.
I'm new to Erlang, how can I rewrite it?
Look at this code:
-module(a).
-compile(export_all).
do_stuff() ->
Stopover.
Here's what happens when I try to compile it:
a.erl:5: variable 'Stopover' is unbound
The variable Stopover was never assigned a value, so erlang has no idea what should be returned by the function do_stuff(). You are doing something similar here:
path( From, Stopover ) + path( Stopover, To ).
The variables From and To are parameter variables for the function path(), and when path() is called, e.g. path(madrid, moscow), then madrid will be assigned to the variable From, and moscow will be assigned to the variable To. Note, however, that nowhere do you assign any value to the variable Stopover.
You need to redefine path() to look like this:
path(From, To, Stopover) ->
Next, you should try to see if adding tuples actually works:
2> {km, 5} + {km, 3}.
** exception error: an error occurred when evaluating an arithmetic expression
in operator +/2
called as {km,5} + {km,3}
3>
Nope!
What you need to do is use pattern matching to extract the distance, an integer, from each tuple, then add the two integers:
{km, Distance1} = path( From, Stopover ),
... = path(Stopover, To),
{km, Distance1 + Distance2}.
This question is already answered by #7stud, and I was wondering how to implement such a path search in erlang. Here is a possible solution:
-module(distances).
-export([ path/2,getTowns/0,start/1, stop/0 ]).
path(From,To) ->
Paths = getPath(),
path(From,To,maps:get(orderedTuple(From,To), Paths, not_found),Paths).
% distanceServer in charge to keep the liste of known distances
% server interfaces
start(Towns) ->
{ok,List} = file:consult(Towns),
Paths = lists:foldl(fun({A,B,D},Acc) -> maps:put(orderedTuple(A,B), D, Acc) end,#{},List),
start(Paths,distance_server).
stop() ->
distance_server ! stop.
getTowns() ->
K = maps:keys(getPath()),
L = lists:usort(lists:flatten([[A,B] || {A,B} <- K])),
io:format("list of towns :~n~p~n~n",[L]).
getPath() ->
distance_server ! {getPath,self()},
receive
Path -> Path
end.
% server fuctions
start(Paths,Server) ->
Pid = spawn(fun() -> distanceServer(Paths) end),
register(Server, Pid).
distanceServer(Path) ->
receive
stop -> stop;
{getPath,From} ->
From ! Path,
distanceServer(Path)
end.
% Searching path
path(From,To,not_found,Paths) -> % if not in the known list, seach for the shortest path
{KM,P} = searchBestPath({0,[From]},To,maps:keys(Paths),{no_dist,no_path}),
case P of
no_path -> not_found;
_ -> {lists:reverse(P),KM}
end;
path(From,To,KM,_) -> % else give the result. Assumption: the known path contains always the best one.
{[From,To],KM}.
searchBestPath({N,[To|_]}=Path,To,_,{BestD,_}) when N < BestD -> Path; % keep the new path if it is better
searchBestPath({N,_},_,_,{BestD,_}=Best) when N >= BestD -> Best; % cancel search if the path so far is longer or equal to the best found
searchBestPath({D,[H|_]=PathSoFar},To,Remaining,Best) ->
Next = [remove(H,{A,B}) || {A,B} <- Remaining, (A =:= H) orelse (B =:= H)], % list of all possible next steps
case Next of
[] -> Best;
Next -> lists:foldl(
fun(X,Acc) ->
{_,ND} = path(H,X), % will always match
R = Remaining -- [orderedTuple(H,X)], % necessary to avoid possible infinite loop in the first search
searchBestPath({D+ND,[X|PathSoFar]},To,R,Acc) % evaluate path for all possible next steps
end,
Best,Next)
end.
% helpers
orderedTuple(A,B) when B > A -> {A,B};
orderedTuple(A,B) -> {B,A}.
remove(X,{X,B}) -> B;
remove(X,{A,X}) -> A.
it uses an external file to define the "known distances", I have used this one for test:
{paris,lyon,465}.
{lyon,marseille,314}.
{marseille,nice,198}.
{marseille,toulouse,404}.
{toulouse,bordeaux,244}.
{bordeaux,paris,568}.
{bordeaux,nantes,347}.
{nantes,paris,385}.
{paris,lille,225}.
{paris,strasbourg,491}.
{lille,strasbourg,525}.
{lille,bruxelles,120}.
{rennes,brest,244}.
{rennes,paris,351}.
{rennes,nantes,113}.
and the result in the shell:
1> c(distances).
{ok,distances}
2> distances:start("distances.txt").
true
3> distances:getTowns().
list of towns :
[bordeaux,brest,bruxelles,lille,lyon,marseille,nantes,nice,paris,rennes,
strasbourg,toulouse]
ok
4> distances:path(bordeaux,bruxelles).
{[bordeaux,paris,lille,bruxelles],913}
5> distances:path(nice,bruxelles).
{[nice,marseille,lyon,paris,lille,bruxelles],1322}
6> distances:path(moscou,paris).
not_found
7> distances:stop().
stop
8>
next step could be to increase the list of known distances each time a new request is done.
I am reading Programming Erlang, when I type these into erlang REPL:
perms([]) -> [[]];
perms(L) -> [[H|T] || H <- L, T <- perms(L--[H])].
* 1: syntax error before: '->'
I know I cannot define functions this way in shell, so I change it to:
2> Perms = fun([]) -> [[]];(L) -> [[H|T] || H <- L, T <- Perms(L--[H])] end.
* 1: variable 'Perms' is unbound
Does this mean I cannot define a recursive function within shell?
Since OTP 17.0 there are named funs:
Funs can now be given names
More details in README:
OTP-11537 Funs can now be a given a name. Thanks to to Richard O'Keefe
for the idea (EEP37) and to Anthony Ramine for the
implementation.
1> Perms = fun F([]) -> [[]];
F(L) -> [[H|T] || H <- L, T <- F(L--[H])]
end.
#Fun<erl_eval.30.54118792>
2> Perms([a,b,c]).
[[a,b,c],[a,c,b],[b,a,c],[b,c,a],[c,a,b],[c,b,a]]
In older releases, you have to be a little bit more clever but once you get it:
1> Perms = fun(List) ->
G = fun(_, []) -> [[]];
(F, L) -> [[H|T] || H <- L, T <- F(F, L--[H])]
end,
G(G, List)
end.
#Fun<erl_eval.30.54118792>
2> Perms([a,b,c]).
[[a,b,c],[a,c,b],[b,a,c],[b,c,a],[c,a,b],[c,b,a]]
Here's my code snippet.
%% test.erl
-export([count_characters/1]).
count_characters(Str) ->
count_characters(Str, #{}).
count_characters([H|T], #{H := N} = X) ->
count_characters(T, X#{H := N+1});
count_characters([H|T], X) ->
count_characters(T, X#{H => 1});
count_characters([], X) ->
X.
%% ErShell
1> c(test).
test.erl:19: illegal use of variable 'H' in map
test.erl:20: illegal use of variable 'H' in map
test.erl:20: variable 'N' is unbound
test.erl:22: illegal use of variable 'H' in map
error
I just don't know why it complains such error, since the following code just worked out fine:
%% test2.erl
birthday(#{age := N} = Person) ->
Person#{age := N+1}.
%% ErShell
1> c(test2).
2> test2:birthday(#{age => 333}).
#{age => 334}
Thanks in advance.
The reason is simple: map hasn't been fully implemented yet. Take a look at: http://learnyousomeerlang.com/maps
Also, you might think of alternative implementations, using the stuff that's already possible with maps:
count_characters(Str) -> count_characters(Str, #{}).
count_characters([H|T], Map) ->
N = maps:get(H, Map, 0),
count_characters(T, maps:put(H, N + 1, Map));
count_characters([], Map) -> Map.
As of today (2021), the feature is still not implemented.
However, for larger keys it is recommended to use the update function for reasons of efficiency:
count_characters([H|T], Map) ->
case N = maps:get(H, Map, 0) of
0 -> count_chars(T, Map#{H => 1});
_ -> count_chars(T, Map#{H := N+1}) % update function maps:udpate(...)
end;
Part of my program requires me to be able to randomly shuffle list elements. I need a function such that when i give it a list, it will pseudo-randomly re-arrange the elements in the list. A change in arrangement Must be visible at each call with the same list. My implementation seems to work just fine but i feel that its rather long and is increasing my code base and also, i have a feeling that it ain't the best solution for doing this. So i need a much shorter implementation. Here is my implementation:
-module(shuffle).
-export([list/1]).
-define(RAND(X),random:uniform(X)).
-define(TUPLE(Y,Z,E),erlang:make_tuple(Y,Z,E)).
list(L)->
Len = length(L),
Nums = lists:seq(1,Len),
tuple_to_list(?TUPLE(Len,[],shuffle(Nums,L,[]))).
shuffle([],_,Buffer)-> Buffer;
shuffle(Nums,[Head|Items],Buffer)->
{Pos,NewNums} = pick_position(Nums),
shuffle(NewNums,Items,[{Pos,Head}|Buffer]).
pick_position([N])-> {N,[]};
pick_position(Nos)->
T = lists:max(Nos),
pick(Nos,T).
pick(From,Max)->
random:seed(begin
(case random:seed(now()) of
undefined ->
NN = element(3,now()),
{?RAND(NN),?RAND(NN),?RAND(NN)};
Any -> Any
end)
end
),
T2 = random:uniform(Max),
case lists:member(T2,From) of
false -> pick(From,Max);
true -> {T2,From -- [T2]}
end.
On running it in shell:
F:\> erl
Eshell V5.8.4 (abort with ^G)
1> c(shuffle).
{ok,shuffle}
2> shuffle:list([a,b,c,d,e]).
[c,b,a,e,d]
3> shuffle:list([a,b,c,d,e]).
[e,c,b,d,a]
4> shuffle:list([a,b,c,d,e]).
[a,b,c,e,d]
5> shuffle:list([a,b,c,d,e]).
[b,c,a,d,e]
6> shuffle:list([a,b,c,d,e]).
[c,e,d,b,a]
I am motivated by the fact that in the STDLIB there is no such function. Somewhere in my game, i need to shuffle things up and also i need to find the best efficient solution to the problem, not just one that works.
Could some one help build a shorter version of the solution ? probably even more efficient ? Thank you
1> L = lists:seq(1,10).
[1,2,3,4,5,6,7,8,9,10]
Associate a random number R with each element X in L by making a list of tuples {R, X}. Sort this list and unpack the tuples to get a shuffled version of L.
1> [X||{_,X} <- lists:sort([ {random:uniform(), N} || N <- L])].
[1,6,2,10,5,7,9,3,8,4]
2>
Please note that karl's answer is much more concise and simple.
Here's a fairly simple solution, although not necessarily the most efficient:
-module(shuffle).
-export([list/1]).
list([]) -> [];
list([Elem]) -> [Elem];
list(List) -> list(List, length(List), []).
list([], 0, Result) ->
Result;
list(List, Len, Result) ->
{Elem, Rest} = nth_rest(random:uniform(Len), List),
list(Rest, Len - 1, [Elem|Result]).
nth_rest(N, List) -> nth_rest(N, List, []).
nth_rest(1, [E|List], Prefix) -> {E, Prefix ++ List};
nth_rest(N, [E|List], Prefix) -> nth_rest(N - 1, List, [E|Prefix]).
For example, one could probably do away with the ++ operation in nth_rest/3. You don't need to seed the random algorithm in every call to random. Seed it initially when you start your program, like so: random:seed(now()). If you seed it for every call to uniform/1 your results become skewed (try with [shuffle:list([1,2,3]) || _ <- lists:seq(1, 100)]).
-module(shuffle).
-compile(export_all).
shuffle(L) ->
shuffle(list_to_tuple(L), length(L)).
shuffle(T, 0)->
tuple_to_list(T);
shuffle(T, Len)->
Rand = random:uniform(Len),
A = element(Len, T),
B = element(Rand, T),
T1 = setelement(Len, T, B),
T2 = setelement(Rand, T1, A),
shuffle(T2, Len - 1).
main()->
shuffle(lists:seq(1, 10)).
This will be a bit faster than the above solution, listed here as do2 for timing comparison.
-module(shuffle).
-export([
time/1,
time2/1,
do/1,
do2/1
]).
time(N) ->
L = lists:seq(1,N),
{Time, _} = timer:tc(shuffle, do, [L]),
Time.
time2(N) ->
L = lists:seq(1,N),
{Time, _} = timer:tc(shuffle, do2, [L]),
Time.
do2(List) ->
[X||{_,X} <- lists:sort([ {rand:uniform(), N} || N <- List])].
do(List) ->
List2 = cut(List),
AccInit = {[],[],[],[],[]},
{L1,L2,L3,L4,L5} = lists:foldl(fun(E, Acc) ->
P = rand:uniform(5),
L = element(P, Acc),
setelement(P, Acc, [E|L])
end, AccInit, List2),
lists:flatten([L1,L2,L3,L4,L5]).
cut(List) ->
Rand=rand:uniform(length(List)),
{A,B}=lists:split(Rand, List),
B++A.
I am new to F# and was reading about tail recursive functions and was hoping someone could give me two different implementations of a function foo - one that is tail recursive and one that isn't so that I can better understand the principle.
Start with a simple task, like mapping items from 'a to 'b in a list. We want to write a function which has the signature
val map: ('a -> 'b) -> 'a list -> 'b list
Where
map (fun x -> x * 2) [1;2;3;4;5] == [2;4;6;8;10]
Start with non-tail recursive version:
let rec map f = function
| [] -> []
| x::xs -> f x::map f xs
This isn't tail recursive because function still has work to do after making the recursive call. :: is syntactic sugar for List.Cons(f x, map f xs).
The function's non-recursive nature might be a little more obvious if I re-wrote the last line as | x::xs -> let temp = map f xs; f x::temp -- obviously its doing work after the recursive call.
Use an accumulator variable to make it tail recursive:
let map f l =
let rec loop acc = function
| [] -> List.rev acc
| x::xs -> loop (f x::acc) xs
loop [] l
Here's we're building up a new list in a variable acc. Since the list gets built up in reverse, we need to reverse the output list before giving it back to the user.
If you're in for a little mind warp, you can use continuation passing to write the code more succinctly:
let map f l =
let rec loop cont = function
| [] -> cont []
| x::xs -> loop ( fun acc -> cont (f x::acc) ) xs
loop id l
Since the call to loop and cont are the last functions called with no additional work, they're tail-recursive.
This works because the continuation cont is captured by a new continuation, which in turn is captured by another, resulting in a sort of tree-like data structure as follows:
(fun acc -> (f 1)::acc)
((fun acc -> (f 2)::acc)
((fun acc -> (f 3)::acc)
((fun acc -> (f 4)::acc)
((fun acc -> (f 5)::acc)
(id [])))))
which builds up a list in-order without requiring you to reverse it.
For what its worth, start writing functions in non-tail recursive way, they're easier to read and work with.
If you have a big list to go through, use an accumulator variable.
If you can't find a way to use an accumulator in a convenient way and you don't have any other options at your disposal, use continuations. I personally consider non-trivial, heavy use of continuations hard to read.
An attempt at a shorter explanation than in the other examples:
let rec foo n =
match n with
| 0 -> 0
| _ -> 2 + foo (n-1)
let rec bar acc n =
match n with
| 0 -> acc
| _ -> bar (acc+2) (n-1)
Here, foo is not tail-recursive, because foo has to call foo recursively in order to evaluate 2+foo(n-1) and return it.
However, bar ís tail-recursive, because bar doesn't have to use the return value of the recursive call in order to return a value. It can just let the recursively called bar return its value immediately (without returning all the way up though the calling stack). The compiler sees this and optimized this by rewriting the recursion into a loop.
Changing the last line in bar into something like | _ -> 2 + (bar (acc+2) (n-1)) would again destroy the function being tail-recursive, since 2 + leads to an action that needs to be done after the recursive call is finished.
Here is a more obvious example, compare it to what you would normally do for a factorial.
let factorial n =
let rec fact n acc =
match n with
| 0 -> acc
| _ -> fact (n-1) (acc*n)
fact n 1
This one is a bit complex, but the idea is that you have an accumulator that keeps a running tally, rather than modifying the return value.
Additionally, this style of wrapping is usually a good idea, that way your caller doesn't need to worry about seeding the accumulator (note that fact is local to the function)
I'm learning F# too.
The following are non-tail recursive and tail recursive function to calculate the fibonacci numbers.
Non-tail recursive version
let rec fib = function
| n when n < 2 -> 1
| n -> fib(n-1) + fib(n-2);;
Tail recursive version
let fib n =
let rec tfib n1 n2 = function
| 0 -> n1
| n -> tfib n2 (n2 + n1) (n - 1)
tfib 0 1 n;;
Note: since the fibanacci number could grow really fast you could replace last line tfib 0 1 n to
tfib 0I 1I n to take advantage of Numerics.BigInteger Structure in F#
Also, when testing, don't forget that indirect tail recursion (tailcall) is turned off by default when compiling in Debug mode. This can cause tailcall recursion to overflow the stack in Debug mode but not in Release mode.