Hi I am writing an algorithm in algorithm design language for a mock test, for adding an element to a priority-queue.
Now I plan to do this by using the general method of inserting an element into a regular queue at the back and then sorting it into correct place by comparing it (according to priority value) to the element that is positioned before it in the priority-queue.
This is the algorithm I have so far (only for adding items to the priority-queue):
element- element to be added to the queue.
queue[ ]- priority queue element is being added to.
n- the size of the priority queue.
tail- the last element in the priority queue.
procedure AddQ (IN element, INOUT queue[ ], IN n, INOUT tail)
if tail= n then
print (“Queue is full”)
else {
tail← tail+ 1
queue (tail) ← element
}
end
What I am not sure of is should I have the section which prints that the Queue is full. Am I right is thinking that priority queues have a pointer in the tail element pointing to null to say that there are not elements left?
If so does this mean that priority-queues cannot have empty space (like a standard array for example)for new elements as the tail in the priority-queue points to null, saying no more elements are in the queue? Or am I taking the wrong approach here?
I am confused as to how this works, if someone could clarify I would be very grateful! Thankyou.
Priority queues in java are dynamic , they can have a minimum number of elements that needs to be in them however the maximum number of elements you put in them is left up to you.If you want to go through a queue to figure out what the last element is or to figure out where the last element is at any point ,you'll need an iterator object that'll help you do just that.
Related
This one's kind of an open ended design question I'm afraid.
Anyway: I have a big two-dimensional array of stuff. This array is mutable, and is accessed by a bunch of threads. For now I've just been dealing with this as a Arc<Mutex<Vec<Vec<--owned stuff-->>>>, which has been fine.
The problem is that stuff is about to grow considerably in size, and I'll want to start holding references rather than complete structures. I could do this by inverting everything and going to Vec<Vec<Arc<Mutex>>, but I feel like that would be a ton of overhead, especially because each thread would need a complete copy of the grid rather than a single Arc/Mutex.
What I want to do is have this be an array of references, but somehow communicate that the items being referenced all live long enough according to a single top-level Arc or something similar. Is that possible?
As an aside, is Vec even the correct data type for this? For the grid in particular I really want a large, fixed-size block of memory that will live for the entire length of the program once it's initialized, and has a lot of reference locality (along either dimension.) Is there something else/more specialized I should be using?
EDIT:Giving some more specifics on my code (away from home so this is rough):
What I want:
Outer scope initializes a bunch of Ts and somehow collectively ensures they live long enough (that's the hard part)
Outer scope initializes a grid :Something<Vec<Vec<&T>>> that stores references to the Ts
Outer scope creates a bunch of threads and passes grid to them
Threads dive in and out of some sort of (problable RW) lock on grid, reading the Tsand changing the &Ts in the process.
What I have:
Outer thread creates a grid: Arc<RwLock<Vector<Vector<T>>>>
Arc::clone(& grid)s are passed to individual threads
Read-heavy threads mostly share the lock and sometimes kick each other out for the writes.
The only problem with this is that the grid is storing actual Ts which might be problematically large. (Don't worry too much about the RwLock/thread exclusivity stuff, I think it's perpendicular to the question unless something about it jumps out at you.)
What I don't want to do:
Top level creates a bunch of Arc<Mutex<T>> for individual T
Top level creates a `grid : Vec<Vec<Arc<Mutex>>> and passes it to threads
The problem with that is that I worry about the size of Arc/Mutex on every grid element (I've been going up to 2000x2000 so far and may go larger). Also while the threads would lock each other out less (only if they're actually looking at the same square), they'd have to pick up and drop locks way more as they explore the array, and I think that would be worse than my current RwLock implementation.
Let me start of by your "aside" question, as I feel it's the one that can be answered:
As an aside, is Vec even the correct data type for this? For the grid in particular I really want a large, fixed-size block of memory that will live for the entire length of the program once it's initialized, and has a lot of reference locality (along either dimension.) Is there something else/more specialized I should be using?
The documenation of std::vec::Vec specifies that the layout is essentially a pointer with size information. That means that any Vec<Vec<T>> is a pointer to a densely packed array of pointers to densely packed arrays of Ts. So if block of memory means a contiguous block to you, then no, Vec<Vec<T>> cannot give that you. If that is part of your requirements, you'd have to deal with a datatype (let's call it Grid) that is basically a (pointer, n_rows, n_columns) and define for yourself if the layout should be row-first or column-first.
The next part is that if you want different threads to mutate e.g. columns/rows of your grid at the same time, Arc<Mutex<Grid>> won't cut it, but you already figured that out. You should get clarity whether you can split your problem such that each thread can only operate on rows OR columns. Remember that if any thread holds a &mut Row, no other thread must hold a &mut Column: There will be an overlapping element, and it will be very easy for you to create a data races. If you can assign a static range of of rows to a thread (e.g. thread 1 processes rows 1-3, thread 2 processes row 3-6, etc.), that should make your life considerably easier. To get into "row-wise" processing if it doesn't arise naturally from the problem, you might consider breaking it into e.g. a row-wise step, where all threads operate on rows only, and then a column-wise step, possibly repeating those.
Speculative starting point
I would suggest that your main thread holds the Grid struct which will almost inevitably be implemented with some unsafe methods, e.g. get_row(usize), get_row_mut(usize) if you can split your problem into rows/colmns or get(usize, usize) and get(usize, usize) if you can't. I cannot tell you what exactly these should return, but they might even be custom references to Grid, which:
can only be obtained when the usual borrowing rules are fulfilled (e.g. by blocking the thread until any other GridRefMut is dropped)
implement Drop such that you don't create a deadlock
Every thread holds a Arc<Grid>, and can draw cells/rows/columns for reading/mutating out of the grid as needed, while the grid itself keeps book of references being created and dropped.
The downside of this approach is that you basically implement a runtime borrow-checker yourself. It's tedious and probably error-prone. You should browse crates.io before you do that, but your problem sounds specific enough that you might not find a fitting solution, let alone one that's sufficiently documented.
I'm reading how the probabilistic data structure count-min-sketch is used in finding the top k elements in a data stream. But I cannot seem to wrap my head around the step where we maintain a heap to get our final answer.
The problem:
We have a stream of items [B, C, A, B, C, A, C, A, A, ...]. We are asked to find out the top k most frequently appearing
items.
My understanding is that, this can be done using micro-batching, in which we accumulate N items before we start doing some real work.
The hashmap+heap approach is easy enough for me to understand. We traverse the micro-batch and build a frequency map (e.g. {B:34, D: 65, C: 9, A:84, ...}) by counting the elements. Then we maintain a min-heap of size k by traversing the frequency map, adding to and evicting from the heap with each [item]:[freq] as needed. Straightforward enough and nothing fancy.
Now with CMS+heap, instead of a hashmap, we have this probabilistic lossy 2D array, which we build by traversing the micro-batch. The question is: how do we maintain our min-heap of size k given this CMS?
The CMS only contains a bunch of numbers, not the original items. Unless I also keep a set of unique elements from the micro-batch, there is no way for me to know which items I need to build my heap against at the end. But if I do, doesn't that defeat the purpose of using CMS to save memory space?
I also considered building the heap in real-time when we traverse the list. With each item coming in, we can quickly update the CMS and get the cumulative frequency of that item at that point. But the fact that this frequency number is cumulative does not help me much. For example, with the example stream above, we would get [B:1, C:1, A:1, B:2, C:2, A:2, C:3, A:3, A:4, ...]. If we use the same logic to update our min-heap, we would get incorrect answers (with duplicates).
I'm definitely missing something here. Please help me understand.
Keep a hashmap of size k, key is id, value is Item(id, count)
Keep a minheap of size k with Item
As events coming in, update the count-min 2d array, get the min, update Item in the hashmap, bubble up/bubble down the heap to recalculate the order of the Item. If heap size > k, poll min Item out and remove id from hashmap as well
Below explanation comes from a comment from this Youtube video:
We need to store the keys, but only K of them (or a bit more). Not all.
When every key comes, we do the following:
Add it to the count-min sketch.
Get key count from the count-min sketch.
Check if the current key is in the heap. If it presents in the heap, we update its count value there. If it not present in the heap, we check if heap is already full. If not full, we add this key to the heap. If heap is full, we check the minimal heap element and compare its value with the current key count value. At this point we may remove the minimal element and add the current key (if current key count > minimal element value).
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.
In Swift 3 Collection indices have to conform to Comparable instead of Equatable.
Full story can be read here swift-evolution/0065.
Here's a relevant quote:
Usually an index can be represented with one or two Ints that
efficiently encode the path to the element from the root of a data
structure. Since one is free to choose the encoding of the “path”, we
think it is possible to choose it in such a way that indices are
cheaply comparable. That has been the case for all of the indices
required to implement the standard library, and a few others we
investigated while researching this change.
In my implementation of a custom linked list collection a node (pointing to a successor) is the opaque index type. However, given two instances, it is not possible to tell if one precedes another without risking traversal of a significant part of the chain.
I'm curious, how would you implement Comparable for a linked list index with O(1) complexity?
The only idea that I currently have is to somehow count steps while advancing the index, storing it within the index type as a property and then comparing those values.
Serious downside of this solution is that indices must be invalidated when mutating the collection. And while that seems reasonable for arrays, I do not want to break that huge benefit linked lists have - they do not invalidate indices of unchanged nodes.
EDIT:
It can be done at the cost of two additional integers as collection properties assuming that single linked list implements front insert, front remove and back append. Any meddling around in the middle would anyway break O(1) complexity requirement.
Here's my take on it.
a) I introduced one private integer type property to my custom Index type: depth.
b) I introduced two private integer type properties to the collection: startDepth and endDepth, which both default to zero for an empty list.
Each front insert decrements the startDepth.
Each front remove increments the startDepth.
Each back append increments the endDepth.
Thus all indices startIndex..<endIndex have a reflecting integer range startDepth..<endDepth.
c) Whenever collection vends an index either by startIndex or endIndex it will inherit its corresponding depth value from the collection. When collection is asked to advance the index by invoking index(_ after:) I will simply initialize a new Index instance with incremented depth value (depth += 1).
Conforming to Comparable boils down to comparing left-hand side depth value to the right-hand side one.
Note that because I expand the integer range from both sides as well, all the depth values for the middle indices remain unchanged (thus are not invalidated).
Conclusion:
Traded benefit of O(1) index comparisons at the cost of minor increase in memory footprint and few integer increments and decrements. I expect index lifetime to be short and number of collections relatively small.
If anyone has a better solution I'd gladly take a look at it!
I may have another solution. If you use floats instead of integers, you can gain kind of O(1) insertion-in-the-middle performance if you set the sortIndex of the inserted node to a value between the predecessor and the successor's sortIndex. This would require to store (and update) the predecessor's sortIndex on your nodes (I imagine this should not be to hard since it is only changed on insertion or removal and it can always be propagated 'up').
In your index(after:) method you need to query the successor node, but since you use your node as index, that is be straightforward.
One caveat is the finite precision of floating points, so if on insertion you the distance between the two sort indices are two small, you need to reindex at least part of the list. Since you said you only expect small scale, I would just go through the hole list and use the position for that.
This approach has all the benefits of your own, with the added benefit of good performance on insertion in the middle.
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.