I have a nested r-tree like datastructure in Python (list of lists). The key is a large number (about 10 digits). On each level there are about x number of items (eg:10) in the list. Then within each list, it recurses and has x items and so on. The height of the tree is h levels (eg: 5). Each level also has an indication of what range of keys it contains (like r-tree).
For a given key, I need to locate the corresponding entry in the tree. This can be trivially done by scanning through each level, check if the given key lies within the range. If so, then step into that layer and recurse till it reaches the leaf.
This can also be done by successively dividing the key by x and taking the quotient as list index.
So the question is, what is more effecient : walking through list sequentially (complexity = depth * x (eg:50)) or successively dividing the large number by x to get the actual list indices (complexity = h divisions (eg: 5 divisions)).
(ie) 50 range checks or 5 divisions ?
This needs to be scalable. So if this code is being accessed in cloud by very large number of users, what is efficient ? May be division is more expensive to perform at scale than range checks ?
You need to benchmark the code in somewhat realistic scenario.
The reason why it's so hard to say is that you are not just comparing division (by the way, modern compilers avoid divisions with a large number of tricks). On modern CPUs you have large caches so likely the list will fit into L2 or L3 which decreases the run-time dramatically. There's also the fancy vector/SIMD instructions that might be used to speed up all the checks in the linear case.
I would guess that going through the list sequentially will be faster, in addition the code will be simpler.
But don't take my word for it, take a real example and benchmark the two versions and pick based on the results. Especially if this is critical for your system's performance.
Related
People coming from other coding languages to python often ask how they should pre-allocate or initialize their list. This is especially true for people coming from Matlab where codes as
l = []
for i = 1:100
l(end+1) = 1;
end
returns a warning that explicitly suggest you to initialize the list.
There are several posts on SO explaining (and showing through tests) that list initialization isn't required in python. A good example with a fair bit of discussion is this one (but the list could be very long): Create a list with initial capacity in Python
The other day, however, while looking for operations complexity in python, I stumbled this sentence on the official python wiki:
the largest [cost for list operations] come from growing beyond the current allocation size (because everything must move),
This seems to suggest that indeed lists do have a pre-allocation size and that growing beyond that size cause the whole list to move.
This shacked a bit my foundations. Can list pre-allocation reduce the overall complexity (in terms of number of operations) of a code? If not, what does that sentence means?
EDIT:
Clearly my question regards the (very common) code:
container = ... #some iterable with 1 gazilion elements
new_list = []
for x in container:
... #do whatever you want with x
new_list.append(x) #or something computed using x
In this case the compiler cannot know how many items there are in container, so new_list could potentially require his allocated memory to change an incredible number of times if what is said in that sentence is true.
I know that this is different for list-comprehensions
Can list pre-allocation reduce the overall complexity (in terms of number of operations) of a code?
No, the overall time complexity of the code will be the same, because the time cost of reallocating the list is O(1) when amortised over all of the operations which increase the size of the list.
If not, what does that sentence means?
In principle, pre-allocating the list could reduce the running time by some constant factor, by avoiding multiple re-allocations. This doesn't mean the complexity is lower, but it may mean the code is faster in practice. If in doubt, benchmark or profile the relevant part of your code to compare the two options; in most circumstances it won't matter, and when it does, there are likely to be better alternatives anyway (e.g. NumPy arrays) for achieving the same goal.
new_list could potentially require his allocated memory to change an incredible number of times
List reallocation follows a geometric progression, so if the final length of the list is n then the list is reallocated only O(log n) times along the way; not an "incredible number of times". The way the maths works out, the average number of times each element gets copied to a new underlying array is a constant regardless of how large the list gets, hence the O(1) amortised cost of appending to the list.
I need a data structure to store positive (not necessarily integer) values. It must support the following two operations in sublinear time:
Add an element.
Remove the largest element.
Also, the largest key may scale as N^2, N being the number of elements. In principle, having O(N^2) space requirement wouldn't be a big problem, but if a more efficient option exists in terms of store, it would work better.
I am working in Python, so if such a data structure exists, it would be of help to have an implementation in this language.
There is no such data structure. For example, if there were, sorting would be worst-case linear time: add all N elements in O(N) time, then remove the largest element remaining N times, again in total O(N) time.
the best data structure you can choose for this operations is the heap: https://www.tutorialspoint.com/python_data_structure/python_heaps.htm#:~:text=Heap%20is%20a%20special%20tree,is%20called%20a%20max%20heap.
with this data structure both adding an element and removing the max are O(log(n)).
this is the most used data structure when you need a lot of operations on the max element, for example is commonly used to implement priority queues
Although constant time may be impossible, depending on your input constraints, you might consider a y-fast-trie, which has O(log log m) time operations and O(n) space, where m is the range, although they work with integers, taking advantage of the bit structure. One of the supported operations is next higher or lower element, which could let you keep track of the highest when the latter is removed.
I'm teaching myself data structures through this python book and I'd appreciate if someone can correct me if I'm wrong since a hash set seems to be extremely similar to a hash map.
Implementation:
A Hashset is a list [] or array where each index points to the head of a linkedlist
So some hash(some_item) --> key, and then list[key] and then add to the head of a LinkedList. This occurs in O(1) time
When removing a value from the linkedlist, in python we replace it with a placeholder because hashsets are not allowed to have Null/None values, correct?
When the list[] gets over a certain % of load/fullness, we copy it over to another list
Regarding Time Complexity Confusion:
So one question is, why is Average search/access O(1) if there can be a list of N items at the linkedlist at a given index?
Wouldnt the average case be the searchitem is in the middle of its indexed linkedlist so it should be O(n/2) -> O(n)?
Also, when removing an item, if we are replacing it with a placeholder value, isn't this considered a waste of memory if the placeholder is never used?
And finally, what is the difference between this and a HashMap other than HashMaps can have nulls? And HashMaps are key/value while Hashsets are just value?
For your first question - why is the average time complexity of a lookup O(1)? - this statement is in general only true if you have a good hash function. An ideal hash function is one that causes a nice spread on its elements. In particular, hash functions are usually chosen so that the probability that any two elements collide is low. Under this assumption, it's possible to formally prove that the expected number of elements to check is O(1). If you search online for "universal family of hash functions," you'll probably find some good proofs of this result.
As for using placeholders - there are several different ways to implement a hash table. The approach you're using is called "closed addressing" or "hashing with chaining," and in that approach there's little reason to use placeholders. However, other hashing strategies exist as well. One common family of approaches is called "open addressing" (the most famous of which is linear probing hashing), and in those setups placeholder elements are necessary to avoid false negative lookups. Searching online for more details on this will likely give you a good explanation about why.
As for how this differs from HashMap, the HashMap is just one possible implementation of a map abstraction backed by a hash table. Java's HashMap does support nulls, while other approaches don't.
The lookup time wouldn't be O(n) because not all items need to be searched, it also depends on the number of buckets. More buckets would decrease the probability of a collision and reduce the chain length.
The number of buckets can be kept as a constant factor of the number of entries by resizing the hash table as needed. Along with a hash function that evenly distributes the values, this keeps the expected chain length bounded, giving constant time lookups.
The hash tables used by hashmaps and hashsets are the same except they store different values. A hashset will contain references to a single value, and a hashmap will contain references to a key and a value. Hashsets can be implemented by delegating to a hashmap where the keys and values are the same.
A lot has been written here about open hash tables, but some fundamental points are missed.
Practical implementations generally have O(1) lookup and delete because they guarantee buckets won't contain more than a fixed number of items (the load factor). But this means they can only achieve amortized O(1) time for insert because the table needs to be reorganized periodically as it grows.
(Some may opt to reorganize on delete, also, to shrink the table when the load factor reaches some bottom threshold, gut this only affect space, not asymptotic run time.)
Reorganization means increasing (or decreasing) the number of buckets and re-assigning all elements into their new bucket locations. There are schemes, e.g. extensible hashing, to make this a bit cheaper. But in general it means touching each element in the table.
Reorganization, then, is O(n). How can insert be O(1) when any given one may incur this cost? The secret is amortization and the power of powers. When the table is grown, it must be grown by a factor greater than one, two being most common. If the table starts with 1 bucket and doubles each time the load factor reaches F, then the cost of N reorganizations is
F + 2F + 4F + 8F ... (2^(N-1))F = (2^N - 1)F
At this point the table contains (2^(N-1))F elements, the number in the table during the last reorganization. I.e. we have done (2^(N-1))F inserts, and the total cost of reorganization is as shown on the right. The interesting part is the average cost per element in the table (or insert, take your pick):
(2^N - 1)F 2^N
---------- ~= ------- = 2
(2^(N-1))F 2^(N-1)
That's where the amortized O(1) comes from.
One additional point is that for modern processors, linked lists aren't a great idea for the bucket lists. With 8-byte pointers, the overhead is meaningful. More importantly, heap-allocated nodes in a single list will almost never be contiguous in memory. Traversing such a list kills cache performance, which can slow things down by orders of magnitude.
Arrays (with an integer count for number of data-containing elements) are likely to work out better. If the load factor is small enough, just allocate an array equal in size to the load factor at the time the first element is inserted in the bucket. Otherwise, grow these element arrays by factors the same way as the bucket array! Everything will still amortize to O(1).
To delete an item from such a bucket, don't mark it deleted. Just copy the last array element to the location of the deleted one and decrement the element count. Of course this won't work if you allow external pointers into the hash buckets, but that's a bad idea anyway.
Input: A list of positive integers where one entry occurs exactly once, and all other entries occur exactly twice (for example [1,3,2,5,3,4,1,2,4])
Output: The unique entry (5 in the above example)
The following algorithm is supposed to be O(m) time and O(1) space where m is the size of the list.
def get_unique(intlist):
unique_val = 0
for int in intlist:
unique_val ^= int
return unique_val
My analysis: Given a list of length m there will be (m + 1)/2 unique positive integers in the input list, so that the smallest possible maximum integer in the list will be (m+1)/2. If we assume this best case, then when taking an XOR sum the variable unique_val will require ceiling(log((m+1)/2)) bits in memory, so I thought the space complexity should be at least O(log(m)).
Your analysis is certainly one correct answer, particularly in a language like Python which gracefully handles arbitrarily large numbers.
It's important to be clear about what you're trying to measure when thinking about space and time complexity. A reasonable assumption might be that the size of an integer is constant (e.g. you're using 64-bit integers). In that case, the space complexity is certainly O(1), but the time complexity is still O(m).
Now, you could also argue that using a fixed-size integer means you have a constant upper-bound on the size of m, so perhaps the time complexity is also O(1). But in most cases where you need to analyze the running time of this sort of algorithm, you're probably very interested in the difference between a list of length 10 and one of length 1 billion.
I'd say it's important to clarify and state your assumptions when analyzing space- and time-complexity. In this case, I would assume we have a fixed size integer and a value of m much smaller than the maximum integer value. In that case, O(1) space and O(m) time are probably the best answers.
EDIT (based on discussion in other answers)
Since all m gives you is a lower-bound no the maximum value in the list, you really can't provide a worst-case estimate of the space. I.e. a number in the list can be arbitrarily large. To have any reasonable answer as to the space complexity of this algorithm, you need to make some assumption about the maximum size of the input values.
The (space/time) complexity analysis is usually applied to algorithms on a higher level. While you can drop down to specific language implementation level, it may not be useful in all cases.
Your analysis is both right and possibly wrong. It's right for current cpython implementation where integers do not have a maximum value. It's ok if all your integers are relatively small and fit into the implementation-specific case of small numbers.
But it doesn't have to be valid for all other implementations of python. For example, you could have an optimizing implementation which figures out that intlist is not used again and instead of using unique_val, it reuses the space of the consumed list elements. (basically transforming this function into a space-optimized reduce call)
Then again, can we even talk about space complexity in a GC'd language with allocated integers? Your analysis of the complexity is wrong, because a ^= b will allocate new memory for big value b and the size of that depends on the system, architecture, python version, and luck.
Your original question is however "Why is the following algorithm O(1) space?". If you look at the algorithm itself and assume you have some arbitrary maximum integer limits, or your language can represent any number in a limited space, then the answer is yes. The algorithm itself with those conditions uses constant space.
The complexity of an algorithm is always dependent on the machine model (= platform) you use. E.g. we often say that multiplying and dividing IEEE floating point numbers is of run-time complexity O(1) - which is not always the case (e.g. on an 8086 processor without FPU).
For the above algorithm, the space complexity O(1) only holds as long as your input list has no element > 2147483647 (= sys.maxint). Usually, python stores integers as signed 32 bit values. For those datatypes, your processor has all relevant operations already implemented in hardware and it generally takes only a constant number of clock cycles (in most cases only one) to perform them (= run-time complexity O(1)) and only a constant number of memory addresses (only one) is occupied to store the result (= space complexity O(1)).
However, if your input exceeds 2147483647, python generally uses a software-implemented datatype to store these big integers. Operations on these are no longer in O(1) and they require more than constant O(1) space.
I am designing a software in Python and I was getting little curious about whether there is any time differences when popping out items from a dictionary of very small lengths and when popping out items from a dictionary of very large length or it is same in all cases.
You can easily answer this question for yourself using the timeit module. But the entire point of a dictionary is near-instant access to any desired element by key, so I would not expect to have a large difference between the two scenarios.
Check out this article on Python TimeComplexity:
The Average Case times listed for dict objects assume that the hash
function for the objects is sufficiently robust to make collisions
uncommon. The Average Case assumes the keys used in parameters are
selected uniformly at random from the set of all keys.
Note that there is a fast-path for dicts that (in practice) only deal
with str keys; this doesn't affect the algorithmic complexity, but it
can significantly affect the constant factors: how quickly a typical
program finishes.
According to this article, for a 'Get Item' operation, the average case is O(1), with a worse case of O(n). In other words, the worst case is that the time increases linearly with size. See Big O Notation on Wikipedia for more information.