How to zero out array in O(1)? - c++

Question

Is there an way to zero out an array in with time complexsity O(1)? It's obvious that this can be done by for-loop, memset. But their time complexity are not O(1).

Answer 1

Yes
However not any array. It takes an array that has been crafted for this to work.
template <typename T, size_t N>
class Array {
public:
Array(): generation(0) {}
void clear() {
// FIXME: deal with overflow
++generation;
}
T get(std::size_t i) const {
if (i >= N) { throw std::runtime_error("out of range"); }
TimedT const& t = data[i];
return t.second == generation ? t.first : T{};
}
void set(std::size_t i, T t) {
if (i >= N) { throw std::runtime_error("out of range"); }
data[i] = std::make_pair(t, generation);
}
private:
typedef std::pair<T, unsigned> TimedT;
TimedT data[N];
unsigned generation;
};
The principle is simple:
we define an epoch using the generation attribute
when an item is set, the epoch in which it has been set is recorded
only items of the current epoch can be seen
clearing is thus equivalent to incrementing the epoch counter
The method has two issues:
storage increase: for each item we store an epoch
generation counter overflow: there is something as a maximum number of epochs
The latter can be thwarted using a real big integer (uint64_t at the cost of more storage).
The former is a natural consequence, one possible solution is to use buckets to downplay the issue by having for example up to 64 items associated to a single counter and a bitmask identifying which are valid within this counter.
EDIT: just wanted to get back on the buckets idea.
The original solution has an overhead of 8 bytes (64 bits) per element (if already 8-bytes aligned). Depending on the elements stored it might or might not be a big deal.
If it is a big deal, the idea is to use buckets; of course like all trade-off it slows down access even more.
template <typename T>
class BucketArray {
public:
BucketArray(): generation(0), mask(0) {}
T get(std::size_t index, std::size_t gen) const {
assert(index < 64);
return gen == generation and (mask & (1 << index)) ?
data[index] : T{};
}
void set(std::size_t index, T t, std::size_t gen) {
assert(index < 64);
if (generation < gen) { mask = 0; generation = gen; }
mask |= (1 << index);
data[index] = t;
}
private:
std::uint64_t generation;
std::uint64_t mask;
T data[64];
};
Note that this small array of a fixed number of elements (we could actually template this and statically check it's inferior or equal to 64) only has 16 bytes of overhead. This means we have an overhead of 2 bits per element.
template <typename T, size_t N>
class Array {
typedef BucketArray<T> Bucket;
public:
Array(): generation(0) {}
void clear() { ++generation; }
T get(std::size_t i) const {
if (i >= N) { throw ... }
Bucket const& bucket = data[i / 64];
return bucket.get(i % 64, generation);
}
void set(std::size_t i, T t) {
if (i >= N) { throw ... }
Bucket& bucket = data[i / 64];
bucket.set(i % 64, t, generation);
}
private:
std::uint64_t generation;
Bucket data[N / 64 + 1];
};
We got the space overhead down by a factor of... 32. Now the array can even be used to store char for example, whereas before it would have been prohibitive. The cost is that access got slower, as we get a division and modulo (when we will get a standardized operation that returns both results in one shot ?).

Answer 2

You can't modify n locations in memory in less than O(n) (even if your hardware, for sufficiently small n, maybe allows a constant-time operation to zero certain nicely-aligned blocks of memory, like for example flash memory does).
However, if the object of the exercise is a bit of lateral thinking, then you can write a class representing a "sparse" array. The general idea of a sparse array is that you keep a collection (perhaps a map, although depending on usage that might not be all there is to it), and when you look up an index, if it's not in the underlying collection then you return 0.
If you can clear the underlying collection in O(1), then you can zero out your sparse array in O(1). Clearing a std::map isn't usually constant-time in the size of the map, because all those nodes need to be freed. But you could design a collection that can be cleared in O(1) by moving the whole tree over from "the contents of my map", to "a tree of nodes that I have reserved for future use". The disadvantage would just be that this "reserved" space is still allocated, a bit like what happens when a vector gets smaller.

Answer 3

It's certainly possible to zero out an array in O(1) as long as you accept a very large constant factor:
void zero_out_array_in_constant_time(void* a, size_t n)
{
char* p = (char*) a;
for (size_t i = 0; i < std::numeric_limits<size_t>::max(); ++i)
{
p[i % n] = 0;
}
}
This will always take the same number of steps, regardless of the size of the array, hence it's O(1).

Answer 4

No.
You can't visit every member of an N-element collection in anything less than O(N) time.
You might, as Mike Kwan has observed, shift the cost from run- to compile-time, but that doesn't alter the computational complexity of the operation.

Answer 5

It's clearly not possible to initialize an arbitrarily sized array in a fixed length of time. However, it is entirely possible to create an array-like ADT which amortizes the cost of initializing the array across its use. The usual construction for this takes upwards of 3x the storage, however. To whit:
template <typename T, size_t arr_size>
class NoInitArray
{
std::vector<T> storage;
// Note that 'lookup' and 'check' are not initialized, and may contain
// arbitrary garbage.
size_t lookup[arr_size];
size_t check[arr_size];
public:
T& operator[](size_t pos)
{
// find out where we actually stored the entry for (*this)[pos].
// This could be garbage.
size_t storage_loc=lookup[pos];
// Check to see that the storage_loc we found is valid
if (storage_loc < storage.size() && check[storage_loc] == pos)
{
// everything checks, return the reference.
return storage[storage_loc];
}
else
{
// storage hasn't yet been allocated/initialized for (*this)[pos].
// allocate storage:
storage_loc=storage.size();
storage.push_back(T());
// put entries in lookup and check so we can find
// the proper spot later:
lookup[pos]=storage_loc;
check[storage_loc]=pos;
// everything's set up, return appropriate reference:
return storage.back();
}
}
};
One could add a clear() member to empty the contents of such an array fairly easily if T is some type that doesn't require destruction, at least in concept.

Answer 6

I like Eli Bendersky's webpage http://eli.thegreenplace.net/2008/08/23/initializing-an-array-in-constant-time, with a solution that he attributes to the famous book Design and Analysis of Computer Algorithms by Aho, Hopcroft and Ullman. This is genuinely O(1) time complexity for initialization, rather than O(N). The space demands are O(N) additional storage, but allocating this space is also O(1), since the space is full of garbage. I enjoyed this for theoretical reasons, but I think it might also be of practical value for implementing some algorithms, if you need to repeatedly initialize a very large array, and each time you access only a relatively small number of positions in the array. Bendersky provides a C++ implementation of the algorithm.
A very pure theorist might start worrying about the fact that N needs O(log(N)) digits, but I have ignored that detail, which would presumably require looking carefully at the mathematical model of the computer. Volume 1 of The Art of Computer Programming probably gives Knuth's view of this issue.

Answer 7

It is impossible at runtime to zero out an array in O(1). This is intuitive given that there is no language mechanism which allows the value setting of arbitrary size blocks of memory in fixed time. The closest you can do is:
int blah[100] = {0};
That will allow the initialisation to happen at compiletime. At runtime, memset is generally the fastest but would be O(N). However, there are problems associated with using memset on particular array types.

Answer 8

While still O(N), implementations that map to hardware-assisted operations like clearing whole cache lines or memory pages can run at <1 cycle per word.
Actually, riffing on Steve Jessop's idea...
You can do it if you have hardware support to clear an arbitrarily large amount of memory all at the same time. If you posit an arbitrarily large array, then you can also posit an arbitrarily large memory with hardware parallelism so that a single reset pin will simultaneously clear every register at once. That line will have to be driven by an arbitrarily large logic gate (that dissipates arbitrarily large power), and the circuit traces will have to be arbitrarily short (to overcome R/C delay) (or superconducting), but these things are quite common in extradimensional spaces.

Answer 9

It is possible to do it with O(1) time, and even with O(1) extra space.
I explained the O(1) time solution David mentioned in another answer, but it uses 2n extra memory.
A better algorithm exists, which requires only 1 bit of extra memory.
See the Article I just wrote about that subject.
It also explains the algorithm David mentioned, some more, and the state-of-the-art algorithm today. It also features an implementation of the latter.
In a short explanation (as I'll just be repeating the article) it cleverly takes the algorithm presented in David's answer and does it all in-place, while using only a (very) little extra memory.