Related
The .split_off method on std::collections::LinkedList is described as having a O(n) time complexity. From the (docs):
pub fn split_off(&mut self, at: usize) -> LinkedList<T>
Splits the list into two at the given index. Returns everything after the given index, including the index.
This operation should compute in O(n) time.
Why not O(1)?
I know that linked lists are not trivial in Rust. There are several resources going into the how's and why's like this book and this article among several others, but I haven't got the chance to dive into those or the standard library's source code yet.
Is there a concise explanation about the extra work needed when splitting a linked list in (safe) Rust?
Is this the only way? And if not why was this implementation chosen?
The method LinkedList::split_off(&mut self, at: usize) first has to traverse the list from the start (or the end) to the position at, which takes O(min(at, n - at)) time. The actual split off is a constant time operation (as you said). And since this min() expression is confusing, we just replace it by n which is legal. Thus: O(n).
Why was the method designed like that? The problem goes deeper than this particular method: most of the LinkedList API in the standard library is not really useful.
Due to its cache unfriendliness, a linked list is often a bad choice to store sequential data. But linked lists have a few nice properties which make them the best data structure for a few, rare situations. These nice properties include:
Inserting an element in the middle in O(1), if you already have a pointer to that position
Removing an element from the middle in O(1), if you already have a pointer to that position
Splitting the list into two lists at an arbitrary position in O(1), if you already have a pointer to that position
Notice anything? The linked list is designed for situations where you already have a pointer to the position that you want to do stuff at.
Rust's LinkedList, like many others, just store a pointer to the start and end. To have a pointer to an element inside the linked list, you need something like an Iterator. In our case, that's IterMut. An iterator over a collection can function like a pointer to a specific element and can be advanced carefully (i.e. not with a for loop). And in fact, there is IterMut::insert_next which allows you to insert an element in the middle of the list in O(1). Hurray!
But this method is unstable. And methods to remove the current element or to split the list off at that position are missing. Why? Because of the vicious circle that is:
LinkedList lacks almost all features that make linked lists useful at all
Thus (nearly) everyone recommends not to use it
Thus (nearly) no one uses LinkedList
Thus (nearly) no one cares about improving it
Goto 1
Please note that are a few brave souls occasionally trying to improve the situations. There is the tracking issue about insert_next, where people argue that Iterator might be the wrong concept to perform these O(1) operations and that we want something like a "cursor" instead. And here someone suggested a bunch of methods to be added to IterMut (including cut!).
Now someone just has to write a nice RFC and someone needs to implement it. Maybe then LinkedList won't be nearly useless anymore.
Edit 2018-10-25: someone did write an RFC. Let's hope for the best!
Edit 2019-02-21: the RFC was accepted! Tracking issue.
Maybe I'm misunderstanding your question, but in a linked list, the links of each node have to be followed to proceed to the next node. If you want to get to the third node, you start at the first, follow its link to the second, then finally arrive at the third.
This traversal's complexity is proportional to the target node index n because n nodes are processed/traversed, so it's a linear O(n) operation, not a constant time O(1) operation. The part where the list is "split off" is of course constant time, but the overall split operation's complexity is dominated by the dominant term O(n) incurred by getting to the split-off point node before the split can even be made.
One way in which it could be O(1) would be if a pointer existed to the node after which the list is split off, but that is different from specifying a target node index. Alternatively, an index could be kept mapping the node index to the corresponding node pointer, but it would be extra space and processing overhead in keeping the index updated in sync with list operations.
pub fn split_off(&mut self, at: usize) -> LinkedList<T>
Splits the list into two at the given index. Returns everything after the given index, including the index.
This operation should compute in O(n) time.
The documentation is either:
unclear, if n is supposed to be the index,
pessimistic, if n is supposed to be the length of the list (the usual meaning).
The proper complexity, as can be seen in the implementation, is O(min(at, n - at)) (whichever is smaller). Since at must be smaller than n, the documentation is correct that O(n) is a bound on the complexity (reached for at = n / 2), however such a large bound is unhelpful.
That is, the fact that list.split_off(5) takes the same time if list.len() is 10 or 1,000,000 is quite important!
As to why this complexity, this is an inherent consequence of the structure of doubly-linked list. There is no O(1) indexing operation in a linked-list, after all. The operation implemented in C, C++, C#, D, F#, ... would have the exact same complexity.
Note: I encourage you to write a pseudo-code implementation of a linked-list with the split_off operation; you'll realize this is the best you can get without altering the data-structure to be something else.
Let's set the context/limitations:
A linked-list consists of Node objects.
Nodes only have a reference to their next node.
A reference to the list is only a reference to the head Node object.
No preprocessing or indexing has been done on the linked-list other than construction (there are no other references to internal nodes or statistics collected, i.e. length).
The last node in the list has a null reference for its next node.
Below is some code for my proposed solution.
Node cursor = head;
Node middle = head;
while (cursor != null) {
cursor = cursor.next;
if (cursor != null) {
cursor = cursor.next;
middle = middle.next;
}
}
return middle;
Without changing the linked-list architecture (not switching to a doubly-linked list or storing a length variable), is there a more efficient way to find the middle element of singly-linked list?
Note: When this method finds the middle of an even number of nodes, it always finds the left middle. This is ideal as it gives you access to both, but if a more efficient method will always find the right middle, that's fine, too.
No, there is no more efficient way, given the information you have available to you.
Think about it in terms of transitions from one node to the next. You have to perform N transitions to work out the list length. Then you have to perform N/2 transitions to find the middle.
Whether you do this as a full scan followed by a half scan based on the discovered length, or whether you run the cursor (at twice speed) and middle (at normal speed) pointers in parallel is not relevant here, the total number of transitions remains the same.
The only way to make this faster would be to introduce extra information to the data structure which you've discounted but, for the sake of completeness, I'll include it here. Examples would be:
making it a doubly-linked list with head and tail pointers, so you could find it in N transitions by "squeezing" in from both ends to the middle. That doubles the storage requirements for pointers however so may not be suitable.
having a skip list with each node pointing to both it's "child" and its "grandchild". This would speed up the cursor transitions resulting in only about N in total (that's N/2 for each of cursor and middle). Like the previous point, there's an extra pointer per node required for this.
maintaining the length of the list separately so you could find the middle in N/2 transitions.
same as the previous point but caching the middle node for added speed under certain circumstances.
That last point bears some extra examination. Like many optimisations, you can trade space for time and the caching shows one way to do it.
First, maintain the length of the list and a pointer to the middle node. The length is initially zero and the middle pointer is initially set to null.
If you're ever asked for the middle node when the length is zero, just return null. That makes sense because the list is empty.
Otherwise, if you're asked for the middle node and the pointer is null, it must be because you haven't cached the value yet.
In that case, calculate it using the length (N/2 transitions) and then store that pointer for later, before returning it.
As an aside, there's a special case here when adding to the end of the list, something that's common enough to warrant special code.
When adding to the end when the length is going from an even number to an odd number, just set middle to middle->next rather than setting it back to null.
This will save a recalculation and works because you (a) have the next pointers and (b) you can work out how the middle "index" (one-based and selecting the left of a pair as per your original question) changes given the length:
Length Middle(one-based)
------ -----------------
0 none
1 1
2 1
3 2
4 2
5 3
: :
This caching means, provided the list doesn't change (or only changes at the end), the next time you need the middle element, it will be near instantaneous.
If you ever delete a node from the list (or insert somewhere other than the end), set the middle pointer back to null. It will then be recalculated (and re-cached) the next time it's needed.
So, for a minimal extra storage requirement, you can gain quite a bit of speed, especially in situations where the middle element is needed more often than the list is changed.
I have a process that lots of threads make links to in order to associate themselves with this process. I made the system this way so that at any time I can easily see who is associated to the central process without having to also keep track of any lists inside of the application itself. I can simply do process_info(self(),links) and erlang keeps track of which processes are still alive or not, etc....
At least, I thought so, until I found out that the list being returned on this thread isn't accurate at this moment:
% P is my central pid
(node#host)212> P.
<0.803.0>
% Get all the links to it
(node#host)213> {links,L} = process_info(P,links).
{links,[<0.29179.155>,<0.6492.250>,<0.29990.293>|...]}
% Counting out all the links
(node#host)214> length(L).
31154
% Filtering out all of the dead processes, put them in Lf
(node#host)215> Lf = lists:filter(fun(Pid) -> try process_info(Pid,links) of {links,_} -> true; _ -> false catch _:_ -> false end end, L).
[<0.29179.155>,<0.6492.250>,<0.29990.293>,<0.23619.530>|...]
% Lf is only about half the size, half the linked processes are dead!
(node#host)216> length(Lf).
15654
% Proof that the links haven't changed in the interim
(node#host)217> {links,L} = process_info(P,links).
{links,[<0.29179.155>,<0.6492.250>,<0.29990.293>|...]}
The only thing I can think of that would cause this is network connectivity issues, since some of the links may come from threads on a node on another machine. Is that a possible explanation for this? And is there a way to make the thread clean up its link list?
This was my fault. It's actually a problem with how I was checking if processes are dead. process_info doesn't work on threads on remote nodes. D'oh!
HOW I WISH I HAD PHRASED MY QUESTION TO BEGIN WITH
Take a table with 26 keys, a-z and let them have integer values.
Create a process, Ouch, that does two things over and over again
In one transaction, write random values for a, b, and c such that those values always sum to 10
In another transaction, read the values for a, b and c and complain if their values do not sum to 10
If you spin-up even a few of these processes you will see that very quickly a, b and c are in a state where their values do not sum to 10. I believe there is no way to ask mnesia to "lock these 3 records before starting the writes (or reads)", one can only have mnesia lock the records as it gets to them (so to speak) which allows for the set of records' values to violate my "must sum to 10" constraint.
If I am right, solutions to this problem include
lock the entire table before writing (or reading) the set of 3 records -- I hate to lock whole table for 3 recs,
Create a process that keeps track of who is reading or writing which keys and protects bulk operations from anyone else writing or reading until the operation is completed. Of course I would have to make sure that all processes made use of this... crap, I guess this means writing my own AccessMod as the fourth parameter to activity/4 which seems like a non-trivial exercise
Some other thing that I am not smart enough to figure out.
thoughts?
Ok, I'm an ambitious Erlang newbee, so sorry if this is a dumb question, but
I am building an application-specific, in-memory distributed cache and I need to be able to write sets of Key, Value pairs in one transaction and also retrieve sets of values in one transaction. In other words I need to
1) Write 40 key,value pairs into the cache and ensure that no one else can read or write any of these 40 keys during this multi-key write operation; and,
2) Read 40 keys in one operation and get back 40 values knowing that all 40 values have been unchanged from the moment that this read operation started until it ended.
The only way I can think of doing this is to lock the entire table at the beginning of the fetch_keylist([ListOfKeys]) or at the beginning of the write_keylist([KeyValuePairs], but I don't want to do this because I have many processes simultaneously doing their own multi_key reads and writes and I don't want to lock the entire table any time any process needs to read/write a relatively small subset of records.
Help?
Trying to be more clear: I do not think this is just about using vanilla transactions
I think I am asking a more subtle question than this. Imagine that I have a process that, within a transaction, iterates through 10 records, locking them as it goes. Now imagine this process starts but before it iterates to the 3rd record ANOTHER process updates the 3rd record. This will be just fine as far as transactions go because the first process hadn't locked the 3rd record (yet) and the OTHER process modified it and released it before the first process got to it. What I want is to be guaranteed that once my first process starts that no other process can touch the 10 records until the first process is done with them.
PROBLEM SOLVED - I'M AN IDIOT... I guess...
Thank you all for your patients, especially Hynek -Pichi- Vychodil!
I prepared my test code to show the problem, and I could in fact reproduce the problem. I then simplified the code for readability and the problem went away. I was not able to again reproduce the problem. This is both embarrassing and mysterious to me since I had this problem for days. Also mnesia never complained that I was executing operations outside of a transaction and I have no dirty transactions anywhere in my code, I have no idea how I was able to introduce this bug into my code!
I have pounded the notion of Isolation into my head and will not doubt that it exists again.
Thanks for the education.
Actually, turns out the problem was using try/catch around mnesia operations within a transaction. See here for more.
Mnesia transaction will do exactly this thing for you. It is what is transaction for unless you do dirty operations. So just place your write and read operations to one transaction a mnesia will do rest. All operations in one transaction is done as one atomic operation. Mnesia transaction isolation level is what is sometimes known as "serializable" i.e. strongest isolation level.
Edit:
It seems you missed one important point about concurrent processes in Erlang. (To be fair it is not only true in Erlang but in any truly concurrent environment and when someone arguing else it is not really concurrent environment.) You can't distinguish which action happen first and which happen second unless you do some synchronization. Only way you can do this synchronization is using message passing. You have guaranteed only one thing about messages in Erlang, ordering of messages sent from one process to other process. It means when you send two messages M1 and M2 from process A to process B they arrives in same order. But if you send message M1 from A to B and message M2 from C to B they can arrive in any order. Simply because how you can even tell which message you sent first? It is even worse if you send message M1 from A to B and then M2 from A to C and when M2 arrives to C send M3 from C to B you don't have guarantied that M1 arrives to B before M3. Even it will happen in one VM in current implementation. But you can't rely on it because it is not guaranteed and can change even in next version of VM just due message passing implementation between different schedulers.
It illustrates problems of event ordering in concurrent processes. Now back to the mnesia transaction. Mnesia transaction have to be side effect free fun. It means there may not be any message sending outside from transaction. So you can't tell which transaction starts first and when starts. Only thing you can tell if transaction succeed and they order you can only determine by its effect. When you consider this your subtle clarification makes no sense. One transaction will read all keys in atomic operation even it is implemented as reading one key by one in transaction implementation and your write operation will be also performed as atomic operation. You can't tell if write to 4th key in second transaction was happen after you read 1st key in first transaction because there it is not observable from outside. Both transaction will be performed in particular order as separate atomic operation. From outside point of view all keys will be read in same point of time and it is work of mnesia to force it. If you send message from inside of transaction you violate mnesia transaction property and you can't be surprised it will behave strange. To be concrete, this message can be send many times.
Edit2:
If you spin-up even a few of these processes you will see that very
quickly a, b and c are in a state where their values do not sum to 10.
I'm curious why you think it would happen or you tested it? Show me your test case and I will show mine:
-module(transactions).
-export([start/2, sum/0, write/0]).
start(W, R) ->
mnesia:start(),
{atomic, ok} = mnesia:create_table(test, [{ram_copies,[node()]}]),
F = fun() ->
ok = mnesia:write({test, a, 10}),
[ ok = mnesia:write({test, X, 0}) || X <-
[b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z]],
ok
end,
{atomic, ok} = mnesia:transaction(F),
F2 = fun() ->
S = self(),
erlang:send_after(1000, S, show),
[ spawn_link(fun() -> writer(S) end) || _ <- lists:seq(1,W) ],
[ spawn_link(fun() -> reader(S) end) || _ <- lists:seq(1,R) ],
collect(0,0)
end,
spawn(F2).
collect(R, W) ->
receive
read -> collect(R+1, W);
write -> collect(R, W+1);
show ->
erlang:send_after(1000, self(), show),
io:format("R: ~p, W: ~p~n", [R,W]),
collect(R, W)
end.
keys() ->
element(random:uniform(6),
{[a,b,c],[a,c,b],[b,a,c],[b,c,a],[c,a,b],[c,b,a]}).
sum() ->
F = fun() ->
lists:sum([X || K<-keys(), {test, _, X} <- mnesia:read(test, K)])
end,
{atomic, S} = mnesia:transaction(F),
S.
write() ->
F = fun() ->
[A, B ] = L = [ random:uniform(10) || _ <- [1,2] ],
[ok = mnesia:write({test, K, V}) || {K, V} <- lists:zip(keys(),
[10-A-B|L])],
ok
end,
{atomic, ok} = mnesia:transaction(F),
ok.
reader(P) ->
case sum() of
10 ->
P ! read,
reader(P);
_ ->
io:format("ERROR!!!~n",[]),
exit(error)
end.
writer(P) ->
ok = write(),
P ! write,
writer(P).
If it would not work it would be really serious problem. There are serious applications including payment systems which rely on it. If you have test case which shows it is broken, please report bug at erlang-bugs#erlang.org
Have you tried mnesia Events ? You can have the reader subscribe to mnesia's Table Events especially write events so as not to interrupt the process doing the writing. In this way, mnesia just keeps sending a copy of what has been written in real-time to the other process which checks what the values are at any one time. take a look at this:
subscriber()->
mnesia:subscribe({table,YOUR_TABLE_NAME,simple}),
%% OR mnesia:subscribe({table,YOUR_TABLE_NAME,detailed}),
wait_events().
wait_events()->
receive
%% For simple events
{mnesia_table_event,{write, NewRecord, ActivityId}} ->
%% Analyse the written record as you wish
wait_events();
%% For detailed events
{mnesia_table_event,{write, YOUR_TABLE, NewRecord, [OldRecords], ActivityId}} ->
%% Analyse the written record as you wish
wait_events();
_Any -> wait_events()
end.
Now you spawn your analyser as a process like this:
spawn(?MODULE,subscriber,[]).
This makes the whole process to run without any process being blocked, mnesia needs not lock any tabel or record because now what you have is a writer process and an analyser process. The whole thing will run in real-time. Remember that there are many other events that you can make use of if you wish by pattern matching them in the subscriber wait_events() receive body.
Its possible to build a heavy duty gen_server or complete application intended for reception and analysis of all your mnesia events. Its usually better to have one capable subscriber than many failing event subscribers. If i have understood you question well, this unblocking solution fits your requirements.
mnesia:read/3 with write locks seems to be suffient.
Mnesia's transaction is implemented by read-write lock and locks are well-formed (holding lock untill the end of transaction). So the isolation level is serializable.
The granularity of locks are per record as long as you access by primary key.
A lot of what I'm reading says that removing an internal element in a doubly linked list (DLL) is O(1); but why is this the case?
I understand why it's O(n) for SLLs; traverse the list O(n) and remove O(1) but don't you still need to traverse the list in a DLL to find the element?
For a doubly linked list, it's constant time to remove an element once you know where it is.
For a singly linked list, it's constant time to remove an element once you know where it and its predecessor are.
Since that link you point to shows a singly linked list removal as O(n) and a doubly linked one as O(1), it's certain that's once you already know where the element is that you want to remove, but not anything else.
In that case, for a doubly linked list, you can just use the prev and next pointers to remove it, giving you O(1). Ignoring the edge cases where you're at the head or tail, that means something like:
corpse->prev->next = corpse->next
corpse->next->prev = corpse->prev
free (corpse)
However, in a singly linked list where you only know the node you want deleted, you can't use corpse->prev to get the one preceding it because there is no prev link.
You have to instead find the previous item by traversing the list from the head, looking for one which has a next of the element you want to remove. That will take O(n), after which it's once again O(1) for the actual removal, such as (again, ignoring the edge cases for simplicity):
lefty = head
while lefty->next != corpse:
lefty = lefty-> next
lefty->next = corpse->next
free (corpse)
That's why the two complexities are different in that article.
As an aside, there are optimisations in a singly-linked list which can make the deletion O(n) (the deletion being effectively O(1) once you've found the item you want to delete, and the previous item). In code terms, that goes something like:
# Delete a node, returns true if found, otherwise false.
def deleteItem(key):
# Special cases (empty list and deleting head).
if head == null: return false
if head.data == key:
curr = head
head = head.next
free curr
return true
# Search non-head part of list (so prev always exists).
prev = head
curr = head.next
while curr != null:
if curr.data == key:
# Found it so delete (using prev).
prev.next = curr.next
free curr
return true
# Advance to next item.
prev = curr
curr = curr.next
# Not found, so fail.
return false
As it's stated where your link points to:
The cost for changing an internal element is based on already having a pointer to it, if you need to find the element first, the cost for retrieving the element is also taken.
So, for both DLL and SLL linear search is O(n), and removal via pointer is O(1).
The complexity of removal in DLL is O(1).
It can also be O(1) in SLL if provided pointer to preceding element and not to the element itself.
This complexity is assuming you know where the element is.
I.e. the operation signature is akin to remove_element(list* l, link* e)
Searching for the element is O(n) in both cases.
#Matuku: You are correct.
I humbly disagree with most answers here trying to justify how delete operation for DLL is O(1). It's not.
Let me explain.
Why are we considering the scenario that we 'would' have the pointer to the node that is being deleted? LinkedLists (Singly/Doubly) are traversed linearly, that's their definition. They have pointers only to the head/tail. How can we suddenly have a pointer to some node in between? That defeats the purpose of this data structure. And going by that assumption, if I have a DLL list of say 1 million nodes, then do I also have to maintain 1 million pointers (let's call them access pointers) pointing to each of those nodes so that I can delete them in O(1)? So how would I store those 1 millions access pointers? And how do I know which access pointer points to the correct data/node that I want to delete?
Can we have a real world example where we 'have' the pointer to the data that has to be deleted 100% of the time?
And if you know the exact location/pointer/reference of/to the node to be deleted, why to even use a LinkedList? Just use array! That's what arrays are for - direct access to what you want!
By assuming that you have direct access to any node you want in DLL is going against the whole idea of LinkedList as a conceptual Data Structure. So I agree with OP, he's correct. I will stick with this - Doubly LinkedLists cannot have O(1) for deleting any node. You still need to start either from head or tail, which brings it down to O(n).
" If " we have the pointer to the node to be deleted say X, then of course it's O(1) because we have pointers to the next and prev node we can delete X. But that big if is imaginary, not real.
We cannot play with the definition of the sacred Data Structure called LinkedLists for some weird assumptions we may have from time to time.