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I need a help with following:
flatten ([]) -> [];
flatten([H|T]) -> H ++ flatten(T).
Input List contain other lists with a different length
For example:
flatten([[1,2,3],[4,7],[9,9,9,9,9,9]]).
What is the time complexity of this function?
And why?
I got it to O(n) where n is a number of elements in the Input list.
For example:
flatten([[1,2,3],[4,7],[9,9,9,9,9,9]]) n=3
flatten([[1,2,3],[4,7],[9,9,9,9,9,9],[3,2,4],[1,4,6]]) n=5
Thanks for help.
First of all I'm not sure your code will work, at least not in the way standard library works. You could compare your function with lists:flatten/1 and maybe improve on your implementation. Try lists such as [a, [b, c]] and [[a], [b, [c]], [d]] as input and verify if you return what you expected.
Regarding complexity it is little tricky due to ++ operator and functional (immutable) nature of the language. All lists in Erlang are linked lists (not arrays like in C++), and you can not just add something to end of one without modifying it; before it was pointing to end of list, now you would like it to link to something else. And again, since it is not mutable language you have to make copy of whole list left of ++ operator, which increases complexity of this operator.
You could say that complexity of A ++ B is length(A), and it makes complexity of your function little bit greater. It would go something like length(FirstElement) + (lenght(FirstElement) + length(SecondElement)) + .... up to (without) last, which after some math magic could be simplified to (n -1) * 1/2 * k * k where n is number of elements, and k is average length of element. Or O(n^3).
If you are new to this it might seem little bit odd, but with some practice you can get hang of it. I would recommend going through few resources:
Good explanation of lists and how they are created
Documentation on list handling with DO and DO NOT parts
Short description of ++ operator myths and best practices
Chapter about recursion and tail-recursion with examples using ++ operator
I'm new to Fortran, and I want to parse a few lines to an array from
a big text file like below. After I read the complete file, I separated the
file into a few parts. And I want to parse specific lines into , L(i),
and write a few columns from lines. I tried to write a small part
of the program. I read all lines from the text file, but I don't know how
can I parse specific Line(i).
text file:
...
...
a b c d
1.2564E+2 0.2564E+2 1.25047E+2 3.2564E+1
1.2564E+2 0.2564E+2 1.25047E+2 3.2564E+1
1.2564E+2 0.2564E+2 1.25047E+2 3.2564E+1
1.2564E+2 0.2564E+2 1.25047E+2 3.2564E+1
.....
.....
character*256 Line(155)
integer ierr, n, i, s
real:: a, b, c, d
open(10,file='b.txt', status='old')
do i=1,155
read(10,'(a)',iostat=ierr) Line(i)
if (ierr /= 0) exit
end do
close(10)
You may read from a character variable using read:
do i = 1, 155
read(line(i), *) a, b, c, d
print*,'a =',a
print*,'b =',b
print*,'c =',c
print*,'d =',d
enddo
If you are not planning to use Line for anything else, you could have done that in the first place, reading a, b, c and d directly from your file.
I'm trying to teach myself Prolog. Below, I've written some code that I think should return all paths between nodes in an undirected graph... but it doesn't. I'm trying to understand why this particular code doesn't work (which I think differentiates this question from similar Prolog pathfinding posts). I'm running this in SWI-Prolog. Any clues?
% Define a directed graph (nodes may or may not be "room"s; edges are encoded by "leads_to" predicates).
room(kitchen).
room(living_room).
room(den).
room(stairs).
room(hall).
room(bathroom).
room(bedroom1).
room(bedroom2).
room(bedroom3).
room(studio).
leads_to(kitchen, living_room).
leads_to(living_room, stairs).
leads_to(living_room, den).
leads_to(stairs, hall).
leads_to(hall, bedroom1).
leads_to(hall, bedroom2).
leads_to(hall, bedroom3).
leads_to(hall, studio).
leads_to(living_room, outside). % Note "outside" is the only node that is not a "room"
leads_to(kitchen, outside).
% Define the indirection of the graph. This is what we'll work with.
neighbor(A,B) :- leads_to(A, B).
neighbor(A,B) :- leads_to(B, A).
Iff A --> B --> C --> D is a loop-free path, then
path(A, D, [B, C])
should be true. I.e., the third argument contains the intermediate nodes.
% Base Rule (R0)
path(X,Y,[]) :- neighbor(X,Y).
% Inductive Rule (R1)
path(X,Y,[Z|P]) :- not(X == Y), neighbor(X,Z), not(member(Z, P)), path(Z,Y,P).
Yet,
?- path(bedroom1, stairs, P).
is false. Why? Shouldn't we get a match to R1 with
X = bedroom1
Y = stairs
Z = hall
P = []
since,
?- neighbor(bedroom1, hall).
true.
?- not(member(hall, [])).
true.
?- path(hall, stairs, []).
true .
?
In fact, if I evaluate
?- path(A, B, P).
I get only the length-1 solutions.
Welcome to Prolog! The problem, essentially, is that when you get to not(member(Z, P)) in R1, P is still a pure variable, because the evaluation hasn't gotten to path(Z, Y, P) to define it yet. One of the surprising yet inspiring things about Prolog is that member(Ground, Var) will generate lists that contain Ground and unify them with Var:
?- member(a, X).
X = [a|_G890] ;
X = [_G889, a|_G893] ;
X = [_G889, _G892, a|_G896] .
This has the confusing side-effect that checking for a value in an uninstantiated list will always succeed, which is why not(member(Z, P)) will always fail, causing R1 to always fail. The fact that you get all the R0 solutions and none of the R1 solutions is a clue that something in R1 is causing it to always fail. After all, we know R0 works.
If you swap these two goals, you'll get the first result you want:
path(X,Y,[Z|P]) :- not(X == Y), neighbor(X,Z), path(Z,Y,P), not(member(Z, P)).
?- path(bedroom1, stairs, P).
P = [hall]
If you ask for another solution, you'll get a stack overflow. This is because after the change we're happily generating solutions with cycles as quickly as possible with path(Z,Y,P), only to discard them post-facto with not(member(Z, P)). (Incidentally, for a slight efficiency gain we can switch to memberchk/2 instead of member/2. Of course doing the wrong thing faster isn't much help. :)
I'd be inclined to convert this to a breadth-first search, which in Prolog would imply adding an "open set" argument to contain solutions you haven't tried yet, and at each node first trying something in the open set and then adding that node's possibilities to the end of the open set. When the open set is extinguished, you've tried every node you could get to. For some path finding problems it's a better solution than depth first search anyway. Another thing you could try is separating the path into a visited and future component, and only checking the visited component. As long as you aren't generating a cycle in the current step, you can be assured you aren't generating one at all, there's no need to worry about future steps.
The way you worded the question leads me to believe you don't want a complete solution, just a hint, so I think this is all you need. Let me know if that's not right.
so far I have the following working:
gen_phrase(S1,S3,Cr) :- noun_phrase(S1,S2,Cr1), verb_phrase(S2,S3,Cr2),
append([Cr1],[Cr2],Cr),add_rule(Cr).
question_phrase(S1,S5,Cr) :- ist(S1,S2),noun_phrase(S2,S3,Cr1),
noun_phrase(S3,S4,Cr2),
append([Cr1],[Cr2],Cr).
add_rule([X,Y]) :-
Fact =.. [Y, X],
assertz(Fact).
Given test run, code generates following:
1 ?- gen_phrase([the,comp456,is,a,computing,course],S3,Cr).
S3 = []
Cr = [comp456, computing_course].
add_rule(Cr) asserts existence of predicate computing_course(comp456).
Now what I would like to do is ask a question:
4 ?- question_phrase([is,the,comp456,a,computing,course],X,Cr).
Cr = [comp456, computing_course] .
What I need to do is extract computing_course and comp456, which I can do, then convert it into form accepted by prolog. This should look like Y(X) where Y = computing_course is a predicate and X = comp456 is atom. The result should be something similar to:
2 ?- computing_course(comp456).
true.
And later on for questions like "What are computing courses":
3 ?- computing_course(X).
X = comp456.
I thought about using assertz, however I still do not know how to call predicate once it is constructed. I am having hard time finding what steps need to be taken to accomplish this. (Using swi-prolog).
Edit: I have realized that there is a predicate call(). However I would like to construct something like this:
ask([X,Y]) :- call(Y(X)).
2 ?- gen_phrase([a,comp456,is,a,computing,course],S3,Cr).
S3 = [],
Cr = [comp456, computing_course]
4 ?- question_phrase([is,the,comp456,a,computing,course],X,Cr),ask(Cr).
ERROR: toplevel: Undefined procedure: ask/1 (DWIM could not correct goal)
It doesn't appear that such call() is syntactically correct. Would be good to know if this is at all possible and how.
call/N it's what you need (here N == 2):
ask([X,Y]) :- call(Y,X).
You could as well use something very similar to what you already use in add_rule/1:
ask([X,Y]) :- C =.. [Y,X], call(C).
The first form it's more efficient, and standardized also.
I'm writing a LSL to Lua translator, and I'm having all sorts of trouble implementing incrementing and decrementing operators. LSL has such things using the usual C like syntax (x++, x--, ++x, --x), but Lua does not. Just to avoid massive amounts of typing, I refer to these sorts of operators as "crements". In the below code, I'll use "..." to represent other parts of the expression.
... x += 1 ...
Wont work, coz Lua only has simple assignment.
... x = x + 1 ...
Wont work coz that's a statement, and Lua can't use statements in expressions. LSL can use crements in expressions.
function preIncrement(x) x = x + 1; return x; end
... preIncrement(x) ...
While it does provide the correct value in the expression, Lua is pass by value for numbers, so the original variable is not changed. If I could get this to actually change the variable, then all is good. Messing with the environment might not be such a good idea, dunno what scope x is. I think I'll investigate that next. The translator could output scope details.
Assuming the above function exists -
... x = preIncrement(x) ...
Wont work for the "it's a statement" reason.
Other solutions start to get really messy.
x = preIncrement(x)
... x ...
Works fine, except when the original LSL code is something like this -
while (doOneThing(x++))
{
doOtherThing(x);
}
Which becomes a whole can of worms. Using tables in the function -
function preIncrement(x) x[1] = x[1] + 1; return x[1]; end
temp = {x}
... preincrement(temp) ...
x = temp[1]
Is even messier, and has the same problems.
Starting to look like I might have to actually analyse the surrounding code instead of just doing simple translations to sort out what the correct way to implement any given crement will be. Anybody got any simple ideas?
I think to really do this properly you're going to have to do some more detailed analysis, and splitting of some expressions into multiple statements, although many can probably be translated pretty straight-forwardly.
Note that at least in C, you can delay post-increments/decrements to the next "sequence point", and put pre-increments/decrements before the previous sequence point; sequence points are only located in a few places: between statements, at "short-circuit operators" (&& and ||), etc. (more info here)
So it's fine to replace x = *y++ + z * f (); with { x = *y + z * f(); y = y + 1; }—the user isn't allowed to assume that y will be incremented before anything else in the statement, only that the value used in *y will be y before it's incremented. Similarly, x = *--y + z * f(); can be replaced with { y = y - 1; x = *y + z * f (); }
Lua is designed to be pretty much impervious to implementations of this sort of thing. It may be done as kind of a compiler/interpreter issue, since the interpreter can know that variables only change when a statement is executed.
There's no way to implement this kind of thing in Lua. Not in the general case. You could do it for global variables by passing a string to the increment function. But obviously it wouldn't work for locals, or for variables that are in a table that is itself global.
Lua doesn't want you to do it; it's best to find a way to work within the restriction. And that means code analysis.
Your proposed solution only will work when your Lua variables are all global. Unless this is something LSL also does, you will get trouble translating LSL programs that use variables called the same way in different places.
Lua is only able of modifying one lvalue per statement - tables being passed to functions are the only exception to this rule. You could use a local table to store all locals, and that would help you out with the pre-...-crements; they can be evaluated before the expression they are contained in is evauated. But the post-...-crements have to be evaluated later on, which is simply not possible in lua - at least not without some ugly code involving anonymous functions.
So you have one options: you must accept that some LSL statements will get translated to several Lua statements.
Say you have a LSL statement with increments like this:
f(integer x) {
integer y = x + x++;
return (y + ++y)
}
You can translate this to a Lua statement like this:
function f(x) {
local post_incremented_x = x + 1 -- extra statement 1 for post increment
local y = x + post_incremented_x
x = post_incremented_x -- extra statement 2 for post increment
local pre_incremented_y = y + 1
return y + pre_incremented_y
y = pre_incremented_y -- this line will never be executed
}
So you basically will have to add two statements per ..-crement used in your statements. For complex structures, that will mean calculating the order in which the expressions are evaluated.
For what is worth, I like with having post decrements and predecrements as individual statements in languages. But I consider it a flaw of the language when they can also be used as expressions. The syntactic sugar quickly becomes semantic diabetes.
After some research and thinking I've come up with an idea that might work.
For globals -
function preIncrement(x)
_G[x] = _G[x] + 1
return _G[x]
end
... preIncrement("x") ...
For locals and function parameters (which are locals to) I know at the time I'm parsing the crement that it is local, I can store four flags to tell me which of the four crements is being used in the variables AST structure. Then when it comes time to output the variables definition, I can output something like this -
local x;
function preIncrement_x() x = x + 1; return x; end
function postDecrement_x() local y = x; x = x - 1; return y; end
... preIncrement_x() ...
In most of your assessment of configurability to the code. You are trying to hard pass data types from one into another. And call it a 'translator'. And in all of this you miss regex and other pattern match capacities. Which are far more present in LUA than LSL. And since the LSL code is being passed into LUA. Try using them, along with other functions. Which would define the work more as a translator, than a hard pass.
Yes I know this was asked a while ago. Though, for other viewers of this topic. Never forget the environments you are working in. EVER. Use what they give you to the best ability you can.