Related
I am new to Functional programming.
The challenge I have is regarding the mental map of how a binary search tree works in Haskell.
In other programs (C,C++) we have something called root. We store it in a variable. We insert elements into it and do balancing etc..
The program takes a break does other things (may be process user inputs, create threads) and then figures out it needs to insert a new element in the already created tree. It knows the root (stored as a variable) and invokes the insert function with the root and the new value.
So far so good in other languages. But how do I mimic such a thing in Haskell, i.e.
I see functions implementing converting a list to a Binary Tree, inserting a value etc.. That's all good
I want this functionality to be part of a bigger program and so i need to know what the root is so that i can use it to insert it again. Is that possible? If so how?
Note: Is it not possible at all because data structures are immutable and so we cannot use the root at all to insert something. in such a case how is the above situation handled in Haskell?
It all happens in the same way, really, except that instead of mutating the existing tree variable we derive a new tree from it and remember that new tree instead of the old one.
For example, a sketch in C++ of the process you describe might look like:
int main(void) {
Tree<string> root;
while (true) {
string next;
cin >> next;
if (next == "quit") exit(0);
root.insert(next);
doSomethingWith(root);
}
}
A variable, a read action, and loop with a mutate step. In haskell, we do the same thing, but using recursion for looping and a recursion variable instead of mutating a local.
main = loop Empty
where loop t = do
next <- getLine
when (next /= "quit") $ do
let t' = insert next t
doSomethingWith t'
loop t'
If you need doSomethingWith to be able to "mutate" t as well as read it, you can lift your program into State:
main = loop Empty
where loop t = do
next <- getLine
when (next /= "quit") $ do
loop (execState doSomethingWith (insert next t))
Writing an example with a BST would take too much time but I give you an analogous example using lists.
Let's invent a updateListN which updates the n-th element in a list.
updateListN :: Int -> a -> [a] -> [a]
updateListN i n l = take (i - 1) l ++ n : drop i l
Now for our program:
list = [1,2,3,4,5,6,7,8,9,10] -- The big data structure we might want to use multiple times
main = do
-- only for shows
print $ updateListN 3 30 list -- [1,2,30,4,5,6,7,8,9,10]
print $ updateListN 8 80 list -- [1,2,3,4,5,6,7,80,9,10]
-- now some illustrative complicated processing
let list' = foldr (\i l -> updateListN i (i*10) l) list list
-- list' = [10,20,30,40,50,60,70,80,90,100]
-- Our crazily complicated illustrative algorithm still needs `list`
print $ zipWith (-) list' list
-- [9,18,27,36,45,54,63,72,81,90]
See how we "updated" list but it was still available? Most data structures in Haskell are persistent, so updates are non-destructive. As long as we have a reference of the old data around we can use it.
As for your comment:
My program is trying the following a) Convert a list to a Binary Search Tree b) do some I/O operation c) Ask for a user input to insert a new value in the created Binary Search Tree d) Insert it into the already created list. This is what the program intends to do. Not sure how to get this done in Haskell (or) is am i stuck in the old mindset. Any ideas/hints welcome.
We can sketch a program:
data BST
readInt :: IO Int; readInt = undefined
toBST :: [Int] -> BST; toBST = undefined
printBST :: BST -> IO (); printBST = undefined
loop :: [Int] -> IO ()
loop list = do
int <- readInt
let newList = int : list
let bst = toBST newList
printBST bst
loop newList
main = loop []
"do balancing" ... "It knows the root" nope. After re-balancing the root is new. The function balance_bst must return the new root.
Same in Haskell, but also with insert_bst. It too will return the new root, and you will use that new root from that point forward.
Even if the new root's value is the same, in Haskell it's a new root, since one of its children has changed.
See ''How to "think functional"'' here.
Even in C++ (or other imperative languages), it would usually be considered a poor idea to have a single global variable holding the root of the binary search tree.
Instead code that needs access to a tree should normally be parameterised on the particular tree it operates on. That's a fancy way of saying: it should be a function/method/procedure that takes the tree as an argument.
So if you're doing that, then it doesn't take much imagination to figure out how several different sections of code (or one section, on several occasions) could get access to different versions of an immutable tree. Instead of passing the same tree to each of these functions (with modifications in between), you just pass a different tree each time.
It's only a little more work to imagine what your code needs to do to "modify" an immutable tree. Obviously you won't produce a new version of the tree by directly mutating it, you'll instead produce a new value (probably by calling methods on the class implementing the tree for you, but if necessary by manually assembling new nodes yourself), and then you'll return it so your caller can pass it on - by returning it to its own caller, by giving it to another function, or even calling you again.
Putting that all together, you can have your whole program manipulate (successive versions of) this binary tree without ever having it stored in a global variable that is "the" tree. An early function (possibly even main) creates the first version of the tree, passes it to the first thing that uses it, gets back a new version of the tree and passes it to the next user, and so on. And each user of the tree can call other subfunctions as needed, with possibly many of new versions of the tree produced internally before it gets returned to the top level.
Note that I haven't actually described any special features of Haskell here. You can do all of this in just about any programming language, including C++. This is what people mean when they say that learning other types of programming makes them better programmers even in imperative languages they already knew. You can see that your habits of thought are drastically more limited than they need to be; you could not imagine how you could deal with a structure "changing" over the course of your program without having a single variable holding a structure that is mutated, when in fact that is just a small part of the tools that even C++ gives you for approaching the problem. If you can only imagine this one way of dealing with it then you'll never notice when other ways would be more helpful.
Haskell also has a variety of tools it can bring to this problem that are less common in imperative languages, such as (but not limited to):
Using the State monad to automate and hide much of the boilerplate of passing around successive versions of the tree.
Function arguments allow a function to be given an unknown "tree-consumer" function, to which it can give a tree, without any one place both having the tree and knowing which function it's passing it to.
Lazy evaluation sometimes negates the need to even have successive versions of the tree; if the modifications are expanding branches of the tree as you discover they are needed (like a move-tree for a game, say), then you could alternatively generate "the whole tree" up front even if it's infinite, and rely on lazy evaluation to limit how much work is done generating the tree to exactly the amount you need to look at.
Haskell does in fact have mutable variables, it just doesn't have functions that can access mutable variables without exposing in their type that they might have side effects. So if you really want to structure your program exactly as you would in C++ you can; it just won't really "feel like" you're writing Haskell, won't help you learn Haskell properly, and won't allow you to benefit from many of the useful features of Haskell's type system.
I want to make a hash of sets. Well, SetHashes, since they need to be mutable.
In fact, I would like to initialize my Hash with multiple identical copies of the same SetHash.
I have an array containing the keys for the new hash: #keys
And I have my SetHash already initialized in a scalar variable: $set
I'm looking for a clean way to initialize the hash.
This works:
my %hash = ({ $_ => $set.clone } for #keys);
(The parens are needed for precedence; without them, the assignment to %hash is part of the body of the for loop. I could change it to a non-postfix for loop or make any of several other minor changes to get the same result in a slightly different way, but that's not what I'm interested in here.)
Instead, I was kind of hoping I could use one of Raku's nifty hyper-operators, maybe like this:
my %hash = #keys »=>» $set;
That expression works a treat when $set is a simple string or number, but a SetHash?
Array >>=>>> SetHash can never work reliably: order of keys in SetHash is indeterminate
Good to know, but I don't want it to hyper over the RHS, in any order. That's why I used the right-pointing version of the hyperop: so it would instead replicate the RHS as needed to match it up to the LHS. In this sort of expression, is there any way to say "Yo, Raku, treat this as a scalar. No, really."?
I tried an explicit Scalar wrapper (which would make the values harder to get at, but it was an experiment):
my %map = #keys »=>» $($set,)
And that got me this message:
Lists on either side of non-dwimmy hyperop of infix:«=>» are not of the same length while recursing
left: 1 elements, right: 4 elements
So it has apparently recursed into the list on the left and found a single key and is trying to map it to a set on the right which has 4 elements. Which is what I want - the key mapped to the set. But instead it's mapping it to the elements of the set, and the hyperoperator is pointing the wrong way for that combination of sizes.
So why is it recursing on the right at all? I thought a Scalar container would prevent that. The documentation says it prevents flattening; how is this recursion not flattening? What's the distinction being drawn?
The error message says the version of the hyperoperator I'm using is "non-dwimmy", which may explain why it's not in fact doing what I mean, but is there maybe an even-less-dwimmy version that lets me be even more explicit? I still haven't gotten my brain aligned well enough with the way Raku works for it to be able to tell WIM reliably.
I'm looking for a clean way to initialize the hash.
One idiomatic option:
my %hash = #keys X=> $set;
See X metaoperator.
The documentation says ... a Scalar container ... prevents flattening; how is this recursion not flattening? What's the distinction being drawn?
A cat is an animal, but an animal is not necessarily a cat. Flattening may act recursively, but some operations that act recursively don't flatten. Recursive flattening stops if it sees a Scalar. But hyperoperation isn't flattening. I get where you're coming from, but this is not the real problem, or at least not a solution.
I had thought that hyperoperation had two tests controlling recursing:
Is it hyperoperating a nodal operation (eg .elems)? If so, just apply it like a parallel shallow map (so don't recurse). (The current doc quite strongly implies that nodal can only be usefully applied to a method, and only a List one (or augmentation thereof) rather than any routine that might get hyperoperated. That is much more restrictive than I was expecting, and I'm sceptical of its truth.)
Otherwise, is a value Iterable? If so, then recurse into that value. In general the value of a Scalar automatically behaves as the value it contains, and that applies here. So Scalars won't help.
A SetHash doesn't do the Iterable role. So I think this refusal to hyperoperate with it is something else.
I just searched the source and that yields two matches in the current Rakudo source, both in the Hyper module, with this one being the specific one we're dealing with:
multi method infix(List:D \left, Associative:D \right) {
die "{left.^name} $.name {right.^name} can never work reliably..."
}
For some reason hyperoperation explicitly rejects use of Associatives on either the right or left when coupled with the other side being a List value.
Having pursued the "blame" (tracking who made what changes) I arrived at the commit "Die on Associative <<op>> Iterable" which says:
This can never work due to the random order of keys in the Associative.
This used to die before, but with a very LTA error about a Pair.new()
not finding a suitable candidate.
Perhaps this behaviour could be refined so that the determining factor is, first, whether an operand does the Iterable role, and then if it does, and is Associative, it dies, but if it isn't, it's accepted as a single item?
A search for "can never work reliably" in GH/rakudo/rakudo issues yields zero matches.
Maybe file an issue? (Update I filed "RFC: Allow use of hyperoperators with an Associative that does not do Iterable role instead of dying with "can never work reliably".)
For now we need to find some other technique to stop a non-Iterable Associative being rejected. Here I use a Capture literal:
my %hash = #keys »=>» \($set);
This yields: {a => \(SetHash.new("b","a","c")), b => \(SetHash.new("b","a","c")), ....
Adding a custom op unwraps en passant:
sub infix:« my=> » ($lhs, $rhs) { $lhs => $rhs[0] }
my %hash = #keys »my=>» \($set);
This yields the desired outcome: {a => SetHash(a b c), b => SetHash(a b c), ....
my %hash = ({ $_ => $set.clone } for #keys);
(The parens seem to be needed so it can tell that the curlies are a block instead of a Hash literal...)
No. That particular code in curlies is a Block regardless of whether it's in parens or not.
More generally, Raku code of the form {...} in term position is almost always a Block.
For an explanation of when a {...} sequence is a Hash, and how to force it to be one, see my answer to the Raku SO Is that a Hash or a Block?.
Without the parens you've written this:
my %hash = { block of code } for #keys
which attempts to iterate #keys, running the code my %hash = { block of code } for each iteration. The code fails because you can't assign a block of code to a hash.
Putting parens around the ({ block of code } for #keys) part completely alters the meaning of the code.
Now it runs the block of code for each iteration. And it concatenates the result of each run into a list of results, each of which is a Pair generated by the code $_ => $set.clone. Then, when the for iteration has completed, that resulting list of pairs is assigned, once, to my %hash.
I'm making a script that sorts the depth for my objects by prioritizing the y variable, but then afterwards checks to see if the objects that are touching each other have a higher depth the further to the right they are, but for some reason the last part isn't working.
Here's the code:
ds_grid_sort(_dg,1,true);
_yy = 0;
repeat _inst_num
{
_inst = _dg[# 0, _yy];
with _inst
{
with other
{
if (x > _inst.x and y = _inst.y)
{
_inst.depth = depth + building_space;
}
}
}
_yy++;
}
I've identified that the problem is that nothing comes out as true when the game checks the y = _inst.y part of the _inst statement, but that doesn't make any sense seeing how they're all at the same y coordinate. Could someone please tell me what I'm doing wrong?
As Steven mentioned, it's good practice to use double equal signs for comparisons (y == _inst.y) and a single equals sign for assignments (_yy = 0;), but GML doesn't care if you use a single equals sign for comparison, so it won't be causing your issue. Though it does matter in pretty much every other language besides GML.
From what I understand, the issue seems to be your use of other. When you use the code with other, it doesn't iterate through all other objects, it only grabs one instance. You can test this by running this code and seeing how many debug messages it shows:
...
with other
{
show_debug_message("X: "+string(x)+"; Y: "+string(y));
...
You could use with all. That will iterate through all objects or with object, where object is either an object or parent object. That will iterate through all instances of that object. However, neither of these functions check whether the objects overlap (it's just going to iterate over all of them), so you'll have to check for collisions. You could do something like this:
...
with all
{
if place_meeting(x, y, other)
{
if (x > _inst.x and y = _inst.y)
{
_inst.depth = depth + building_space;
}
}
...
I don't know what the rest of your code looks like, but there might be an easier way to achieve your goal. Is it possible to initially set the depth based on both the x and y variables? Something such as depth = -x-y;? For people not as familiar with GameMaker, objects with a smaller depth value are drawn above objects with higher depth values; that is why I propose setting the depth to be -x-y. Below is what a view of that grid would look like (first row and column are x and y variables; the other numbers would be the depth of an object at that position):
Having one equation that everything operates on will also make it so that if you have anything moving (such as a player), you can easily and efficiently update their depth to be able to display them correctly relative to all the other objects.
I think it should be y == _inst.y.
But I'm not sure as GML tends to accept such formatting.
It's a better practise to use == to check if they're equal when using conditions.
I'm trying to understand this algorithm the DFA minimization algorithm at http://www.cs.umd.edu/class/fall2009/cmsc330/lectures/discussion2.pdf where it says:
while until there is no change in the table contents:
For each pair of states (p,q) and each character a in the alphabet:
if Distinct(p,q) is empty and Distinct(δ(p,a), δ(q,a)) is not empty:
set distinct(p,q) to be x
The bit I don't understand is "Distinct(δ(p,a), δ(q,a))" I think I understand the transition function where δ(p,a) = whatever state is reached from p with input a. but with the following DFA:
http://i.stack.imgur.com/arZ8O.png
resulting in this table:
imgur.com/Vg38ZDN.png
shouldn't (c,b) also be marked as an x since distinct(δ(b,0), δ(c,0)) is not empty (d) ?
Distinct(δ(p,a), δ(q,a)) will only be non-empty if δ(p,a) and δ(q,a) are distinct. In your example, δ(b,0) and δ(c,0) are both d. Distinct(d, d) is empty since it doesn't make sense for d to be distinct with itself. Since Distinct(d, d) is empty, we don't mark Distinct(c, b).
In general, Distinct(p, p) where p is a state will always be empty. Better yet, we don't consider it because it doesn't make sense.
I want to check whether a value is equal to 1. Is there any difference in the following lines of code
Evaluated value == 1
1 == evaluated value
in terms of the compiler execution
In most languages it's the same thing.
People often do 1 == evaluated value because 1 is not an lvalue. Meaning that you can't accidentally do an assignment.
Example:
if(x = 6)//bug, but no compiling error
{
}
Instead you could force a compiling error instead of a bug:
if(6 = x)//compiling error
{
}
Now if x is not of int type, and you're using something like C++, then the user could have created an operator==(int) override which takes this question to a new meaning. The 6 == x wouldn't compile in that case but the x == 6 would.
It depends on the programming language.
In Ruby, Smalltalk, Self, Newspeak, Ioke and many other single-dispatch object-oriented programming languages, a == b is actually a message send. In Ruby, for example, it is equivalent to a.==(b). What this means, is that when you write a == b, then the method == in the class of a is executed, but when you write b == a, then the method in the class of b is executed. So, it's obviously not the same thing:
class A; def ==(other) false end; end
class B; def ==(other) true end; end
a, b = A.new, B.new
p a == b # => false
p b == a # => true
No, but the latter syntax will give you a compiler error if you accidentally type
if (1 = evaluatedValue)
Note that today any decent compiler will warn you if you write
if (evaluatedValue = 1)
so it is mostly relevant for historical reasons.
Depends on the language.
In Prolog or Erlang, == is written = and is a unification rather than an assignment (you're asserting that the values are equal, rather then testing that they are equal or forcing them to be equal), so you can use it for an assertion if the left hand side is a constant, as explained here.
So X = 3 would unify the variable X and the value 3, whereas 3 = X would attempt to unify the constant 3 with the current value of X, and be equivalent of assert(x==3) in imperative languages.
It's the same thing
In general, it hardly matters whether you use,
Evaluated value == 1 OR 1 == evaluated value.
Use whichever appears more readable to you. I prefer if(Evaluated value == 1) because it looks more readable to me.
And again, I'd like to quote a well known scenario of string comparison in java.
Consider a String str which you have to compare with say another string "SomeString".
str = getValueFromSomeRoutine();
Now at runtime, you are not sure if str would be NULL. So to avoid exception you'll write
if(str!=NULL)
{
if(str.equals("SomeString")
{
//do stuff
}
}
to avoid the outer null check you could just write
if ("SomeString".equals(str))
{
//do stuff
}
Though this is less readable which again depends on the context, this saves you an extra if.
For this and similar questions can I suggest you find out for yourself by writing a little code, running it through your compiler and viewing the emitted asembler output.
For example, for the GNU compilers, you do this with the -S flag. For the VS compilers, the most convenient route is to run your test program in the debugger and then use the assembeler debugger view.
Sometimes in C++ they do different things, if the evaluated value is a user type and operator== is defined. Badly.
But that's very rarely the reason anyone would choose one way around over the other: if operator== is not commutative/symmetric, including if the type of the value has a conversion from int, then you have A Problem that probably wants fixing rather than working around. Brian R. Bondy's answer, and others, are probably on the mark for why anyone worries about it in practice.
But the fact remains that even if operator== is commutative, the compiler might not do exactly the same thing in each case. It will (by definition) return the same result, but it might do things in a slightly different order, or whatever.
if value == 1
if 1 == value
Is exactly the same, but if you accidentally do
if value = 1
if 1 = value
The first one will work while the 2nd one will produce an error.
They are the same. Some people prefer putting the 1 first, to void accidentally falling into the trap of typing
evaluated value = 1
which could be painful if the value on the left hand side is assignable. This is a common "defensive" pattern in C, for instance.
In C languages it's common to put the constant or magic number first so that if you forget one of the "=" of the equality check (==) then the compiler won't interpret this as an assignment.
In java, you cannot do an assignment within a boolean expression, and so for Java, it is irrelevant which order the equality operands are written in; The compiler should flag an error anyway.