Stable priority queue

[Aaron W. LaFramboise]

Hello,

I understand that an easy way to make the standard std::priority_q ueue
stable is by including an integer stamp with each node that is
incremented each time a new node is pushed into the queue. However, no
matter what reasonably-sized type I use for the stamp, eventually the
stamp will 'wrap around' and possibly cause incorrect ordering of
elements. For my application, which queues elements very quickly and
runs for an indefinite amount of time, this scenario is a real concern,
and will eventually cause incorrect results.

Is there any easy and painless way of correcting this order-stamp
wraparound problem?
Aaron W. LaFramboise

[Dietmar Kuehl]

Claudio Puviani wrote:

"Dietmar Kuehl" <di***********@ yahoo.com> wrote

Use a 128-bit integer (eg. made out of four longs): I
doubt that this one will overflow on any currently available
machine before the sun (note, not the SUN) cheases to exist.
Even if you account for faster processors you should be
on the safe side.
Gaa! Do people no longer do basic back-of-the envelope calculations?

I had to study mathematics because I'm notoriously bad at
calculating things and mathematics is the only more or less
technical topic where you don't need to do calculations - you
only prove that they can be done. With 128-bit I'm definitely
on the safe side...
If a 5GHz computer had an integer increment that took only one CPU cycle
and did NOTHING but increment a 64-bit int 5 billion times per second
24-hours a day without interruption, it would take 117 _YEARS_ before you
ran out of numbers!!! If you throw in the obligatory jump to implement the
loop and made the jump also take a single CPU cycle, the duration leaps to
a whopping 234 years.
Well, this computation seems reasonable (even though I'm not fit
to verifying it). Actually, the OP mentioned that this approach
does not work for him and all I wanted to do in this area is to
show that it is just a question of how many bits you spend to
make the approach work.
The most generous thing I can say about using a 128-bit integer for this
purpose is that it's absurdly over-engineered. There's a lot worse that
could be said about the idea.

To me, the difference of using 128-bits or 64-bits is not really
there: yes, I need twice as much memory but who cares? The template
class for doing simple arithmetics does not care how many (unsigned)
longs it takes for its representation and there is no portable
64-bit integer. Over-engineered? I'd say I was too lazy to do a more
precise computation.
--
<mailto:di***** ******@yahoo.co m> <http://www.dietmar-kuehl.de/>
<http://www.contendix.c om> - Software Development & Consulting

[Siemel Naran]

"Dietmar Kuehl" <di***********@ yahoo.com> wrote in message
news:c4upai$2mi v4i$1@ID-

Siemel Naran wrote:

Then maybe we can use the protected member 'c'.

'std::priority_ queue' actually just uses 'std::make_heap ', etc. Anyway,
accessing the member directly is no good: the internal organization of
the heap essentially makes immediate stability without some tool
impossible.

I don't understand what you mean. The protected member 'c' is already has
the data organized in the exact manner. Thus

void MyPriorityQueue ::write() const {
std::ofstream file("MyPriorit yQueue.dat");
file << c;
};

void MyPriorityQueue ::read() const {
std::ifstream file("MyPriorit yQueue.dat");
file >> c;
container_type c2 = c;
std::make_heap( c2.begin(), c2.end());
assert(std::equ al(c2.begin(), c2.end(), c.begin());
};

How do they achieve O(lg(N)) for both push and pop (ie. how does the heap work?).

The essential idea of several heap structure is to maintain the heap
property: the heap is viewed as a tree and each tree has its minimum
element in the root (reverse the comparison and it has the maximum
element in the root; this does not really matter as long as you stick
with one definition - I will use the minimum). In the container stored
in the 'std::priority_ queue' a binary tree is implicitly stored: the
element with index '0' is the root. The children of a node with index
'n' are the nodes with '2 * (n + 1) - 1' and '2 * (n + 1)'. The parent
of the node 'n' is the node '(n - 1) / 2' (I guess, I got the numbers
wrong somewhere; just draw a picture and number the nodes and things
should be relatively easy to work out). In any case, the tree is fully
balanced.

Thanks for jogging my memory. I implemented it once. It was an array of
fixed length (2^n)-1, and was essentially a perfectly balanced binary tree
stored in a regular C array.

Now out of curiosity, how much faster is a heap implemented through
std::priority_q ueue<T, std::vector<T>> using make_heap, over one implemented
through std::set<T>.

A related question. Have you or anyone measured the time improvement of
inserting N elements into a std::vector<T> using reserve(N) and
push_back(const value_type&) and calling strd::sort, versus inserting N
elements into a std::multimap<T > by calling insert(const value_type&)?

Inserting into a heap is pretty simple: you add a node to the lowest
level and exchange it with its parent if the parent has a bigger key.
This far, the representation of the tree does not at all matter and
other heaps, eg. a Fibonacci-heap, use a different structure than an
array to hold a heap although they also maintain the heap property.
Since the heap in the array is balanced, insertion takes at most
O(log n) steps.

Extracting from a heap, is also simple: you simply exchange the root
with the last element and bubble down the root again: check which of
the child nodes is the smaller one and exchange the node with this
child until the node is smaller than both children. This again takes
at most O(log n) steps.

This form of heap is called a "d-heap" in its general form where "d"
is the arity of the tree: the same approach works with non-binary
trees, too, although the computation of child and parent nodes changes
slightly. When I compared different priority queues, I found that
"d == 5" had the best performance for integer values stored in the
heap. An even bigger 'd' is probably better if the nodes itself become
bigger. Of course, this was based on a few tests with my own
implementation and probably does not hold up to a scientific analysis.
What does this term "arity" mean? Perhaps the heap I implemented has arity
= 2 because each parent has 2 children? What do you mean by the "nodes
itself become bigger"? Is it that sizeof(T) where T is the object stored in
the heap gets bigger, or that the number of nodes increases?
Other forms of priority queues provide additional capabilities, eg.
ability to change the key for the priority fast. Of some interest in
this direction are Fibonacci-heaps which can do changes of the keys
in such a form that it becomes amortized constant, at least when using
it for Dijkstra's algorithms for finding shortest paths (I don't
remember the details; it is possible that it is amortized constant in
general applications, too).

Thanks for the info. Changing a key in O(1) time is fascinating.

[Siemel Naran]

"Dietmar Kuehl" <di***********@ yahoo.com> wrote in message
news:c4upai$2mi v4i$1@ID-

Siemel Naran wrote:

Then maybe we can use the protected member 'c'.

'std::priority_ queue' actually just uses 'std::make_heap ', etc. Anyway,
accessing the member directly is no good: the internal organization of
the heap essentially makes immediate stability without some tool
impossible.

I don't understand what you mean. The protected member 'c' is already has
the data organized in the exact manner. Thus

void MyPriorityQueue ::write() const {
std::ofstream file("MyPriorit yQueue.dat");
file << c;
};

void MyPriorityQueue ::read() const {
std::ifstream file("MyPriorit yQueue.dat");
file >> c;
container_type c2 = c;
std::make_heap( c2.begin(), c2.end());
assert(std::equ al(c2.begin(), c2.end(), c.begin());
};

How do they achieve O(lg(N)) for both push and pop (ie. how does the heap work?).

The essential idea of several heap structure is to maintain the heap
property: the heap is viewed as a tree and each tree has its minimum
element in the root (reverse the comparison and it has the maximum
element in the root; this does not really matter as long as you stick
with one definition - I will use the minimum). In the container stored
in the 'std::priority_ queue' a binary tree is implicitly stored: the
element with index '0' is the root. The children of a node with index
'n' are the nodes with '2 * (n + 1) - 1' and '2 * (n + 1)'. The parent
of the node 'n' is the node '(n - 1) / 2' (I guess, I got the numbers
wrong somewhere; just draw a picture and number the nodes and things
should be relatively easy to work out). In any case, the tree is fully
balanced.

Thanks for jogging my memory. I implemented it once. It was an array of
fixed length (2^n)-1, and was essentially a perfectly balanced binary tree
stored in a regular C array.

Now out of curiosity, how much faster is a heap implemented through
std::priority_q ueue<T, std::vector<T>> using make_heap, over one implemented
through std::set<T>.

A related question. Have you or anyone measured the time improvement of
inserting N elements into a std::vector<T> using reserve(N) and
push_back(const value_type&) and calling strd::sort, versus inserting N
elements into a std::multimap<T > by calling insert(const value_type&)?

Inserting into a heap is pretty simple: you add a node to the lowest
level and exchange it with its parent if the parent has a bigger key.
This far, the representation of the tree does not at all matter and
other heaps, eg. a Fibonacci-heap, use a different structure than an
array to hold a heap although they also maintain the heap property.
Since the heap in the array is balanced, insertion takes at most
O(log n) steps.

Extracting from a heap, is also simple: you simply exchange the root
with the last element and bubble down the root again: check which of
the child nodes is the smaller one and exchange the node with this
child until the node is smaller than both children. This again takes
at most O(log n) steps.

This form of heap is called a "d-heap" in its general form where "d"
is the arity of the tree: the same approach works with non-binary
trees, too, although the computation of child and parent nodes changes
slightly. When I compared different priority queues, I found that
"d == 5" had the best performance for integer values stored in the
heap. An even bigger 'd' is probably better if the nodes itself become
bigger. Of course, this was based on a few tests with my own
implementation and probably does not hold up to a scientific analysis.
What does this term "arity" mean? Perhaps the heap I implemented has arity
= 2 because each parent has 2 children? What do you mean by the "nodes
itself become bigger"? Is it that sizeof(T) where T is the object stored in
the heap gets bigger, or that the number of nodes increases?
Other forms of priority queues provide additional capabilities, eg.
ability to change the key for the priority fast. Of some interest in
this direction are Fibonacci-heaps which can do changes of the keys
in such a form that it becomes amortized constant, at least when using
it for Dijkstra's algorithms for finding shortest paths (I don't
remember the details; it is possible that it is amortized constant in
general applications, too).

Thanks for the info. Changing a key in O(1) time is fascinating.

[Siemel Naran]

"Claudio Puviani" <pu*****@hotmai l.com> wrote in message news:0izcc.2608 0

Gaa! Do people no longer do basic back-of-the envelope calculations?

If a 5GHz computer had an integer increment that took only one CPU cycle and did NOTHING but increment a 64-bit int 5 billion times per second 24-hours a day without interruption, it would take 117 _YEARS_ before you ran out of
numbers!!! If you throw in the obligatory jump to implement the loop and
made the jump also take a single CPU cycle, the duration leaps to a whopping 234 years.
How do you get 117 years? In a 64 integer there are 2^64 = 1.84... * 10^19
possible numbers. In one year we can perform 86,400 seconds * 5 billion
increments/second = 4.32 * 10^14 increments. I believe 1 billion = 10^9.
Then 2^64 / 4.32e14 = 42700 years. What am I doing wrong in my calculation?

For 32 bit integers, I get 2^32 / 4.32e14 = 9.94e-6 years which is less than
one day.
So now the question is how to get a 64 bit integer. On some platforms, this
will be int, on others long, and on others long long. BTW, is long long a
standard type? My compiler Borland 6 accepts long long but not long long
long.

I thought people proposed adding int32, int64 to the standard. Is there any
update on this?

We could also use std::bitset<64> , but it doesn't have an operator++ or
operator+.

We could also use the sizeof trick though, which is a cool use of template
template parameters.

const int CHAR_BITS=8;
template <class T>
struct Eq16
{
static const bool result=((sizeof (T)*CHAR_BITS)= =16);
};
void test_find_typel ist(ostream& out)
{
typedef typelist<int, typelist<short, typelist<char> > > BasicTypes;
typedef find_typelist<E q16, BasicTypes>::re sult int16;
out << sizeof(int16) << '\n'; // print 2
}

where

template <class T, class U=void>
struct typelist
{
typedef T car;
typedef U cdr;
};

template <template <class> class Predicate, class list>
struct typelist_find
{
typedef typename list::car car;
typedef typename list::cdr cdr;

typedef typename
template_if<
Predicate<car>: :result,
car,
typename find_typelist<P redicate,cdr>:: result :: result result;
};

template <template <class> class Predicate>
struct typelist_find<P redicate,void>
{
typedef void result;
};


The most generous thing I can say about using a 128-bit integer for this
purpose is that it's absurdly over-engineered. There's a lot worse that
could be said about the idea.

Such as?

Anyway, quantum computers of the future may indeed be much faster. But as a
practical matter I'd just use long long.

[Siemel Naran]

"Claudio Puviani" <pu*****@hotmai l.com> wrote in message news:0izcc.2608 0

Gaa! Do people no longer do basic back-of-the envelope calculations?

If a 5GHz computer had an integer increment that took only one CPU cycle and did NOTHING but increment a 64-bit int 5 billion times per second 24-hours a day without interruption, it would take 117 _YEARS_ before you ran out of
numbers!!! If you throw in the obligatory jump to implement the loop and
made the jump also take a single CPU cycle, the duration leaps to a whopping 234 years.
How do you get 117 years? In a 64 integer there are 2^64 = 1.84... * 10^19
possible numbers. In one year we can perform 86,400 seconds * 5 billion
increments/second = 4.32 * 10^14 increments. I believe 1 billion = 10^9.
Then 2^64 / 4.32e14 = 42700 years. What am I doing wrong in my calculation?

For 32 bit integers, I get 2^32 / 4.32e14 = 9.94e-6 years which is less than
one day.
So now the question is how to get a 64 bit integer. On some platforms, this
will be int, on others long, and on others long long. BTW, is long long a
standard type? My compiler Borland 6 accepts long long but not long long
long.

I thought people proposed adding int32, int64 to the standard. Is there any
update on this?

We could also use std::bitset<64> , but it doesn't have an operator++ or
operator+.

We could also use the sizeof trick though, which is a cool use of template
template parameters.

const int CHAR_BITS=8;
template <class T>
struct Eq16
{
static const bool result=((sizeof (T)*CHAR_BITS)= =16);
};
void test_find_typel ist(ostream& out)
{
typedef typelist<int, typelist<short, typelist<char> > > BasicTypes;
typedef find_typelist<E q16, BasicTypes>::re sult int16;
out << sizeof(int16) << '\n'; // print 2
}

where

template <class T, class U=void>
struct typelist
{
typedef T car;
typedef U cdr;
};

template <template <class> class Predicate, class list>
struct typelist_find
{
typedef typename list::car car;
typedef typename list::cdr cdr;

typedef typename
template_if<
Predicate<car>: :result,
car,
typename find_typelist<P redicate,cdr>:: result :: result result;
};

template <template <class> class Predicate>
struct typelist_find<P redicate,void>
{
typedef void result;
};


The most generous thing I can say about using a 128-bit integer for this
purpose is that it's absurdly over-engineered. There's a lot worse that
could be said about the idea.

Such as?

Anyway, quantum computers of the future may indeed be much faster. But as a
practical matter I'd just use long long.

[Rob Williscroft]

Siemel Naran wrote in
news:Fc******** ************@bg tnsc05-news.ops.worldn et.att.net:

How do you get 117 years? In a 64 integer there are 2^64 = 1.84... *
10^19 possible numbers. In one year we can perform 86,400 seconds * 5
billion increments/second = 4.32 * 10^14 increments. I believe 1
billion = 10^9. Then 2^64 / 4.32e14 = 42700 years. What am I doing
wrong in my calculation?

For 32 bit integers, I get 2^32 / 4.32e14 = 9.94e-6 years which is
less than one day.

86,400 == 24 * 60 * 60 i.e. the No. of seconds in 1 day *not* 1 year.

I put 'days in a year' into google and got:

1 year = 365.242199 days, following there calculator link I did:

42700 / 365.242199 and got:

42700 / 365.242199 = 116.908726

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Rob Williscroft]

Siemel Naran wrote in
news:Fc******** ************@bg tnsc05-news.ops.worldn et.att.net:

How do you get 117 years? In a 64 integer there are 2^64 = 1.84... *
10^19 possible numbers. In one year we can perform 86,400 seconds * 5
billion increments/second = 4.32 * 10^14 increments. I believe 1
billion = 10^9. Then 2^64 / 4.32e14 = 42700 years. What am I doing
wrong in my calculation?

For 32 bit integers, I get 2^32 / 4.32e14 = 9.94e-6 years which is
less than one day.

86,400 == 24 * 60 * 60 i.e. the No. of seconds in 1 day *not* 1 year.

I put 'days in a year' into google and got:

1 year = 365.242199 days, following there calculator link I did:

42700 / 365.242199 and got:

42700 / 365.242199 = 116.908726

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Buster]

Siemel Naran wrote:

So now the question is how to get a 64 bit integer. On some platforms, this
will be int, on others long, and on others long long. BTW, is long long a
standard type? My compiler Borland 6 accepts long long but not long long
long.

#include <boost/cstdint.hpp>

uint64_t n;

--
Regards,
Buster.

[Buster]

Siemel Naran wrote:

So now the question is how to get a 64 bit integer. On some platforms, this
will be int, on others long, and on others long long. BTW, is long long a
standard type? My compiler Borland 6 accepts long long but not long long
long.

#include <boost/cstdint.hpp>

uint64_t n;

--
Regards,
Buster.

[Dietmar Kuehl]

"Siemel Naran" <Si*********@RE MOVE.att.net> wrote:

Siemel Naran wrote:

Then maybe we can use the protected member 'c'.

'std::priority_ queue' actually just uses 'std::make_heap ', etc. Anyway,
accessing the member directly is no good: the internal organization of
the heap essentially makes immediate stability without some tool
impossible.

I don't understand what you mean. The protected member 'c' is already has
the data organized in the exact manner.

The problem is, that you cannot get stable behavior of the priority queue
through simply manipulating the member 'c' in an 'std::priority_ queue'.
The way elements in a priority queue simple does not provide any useful
form of stability - unless you somehow encode the order in the key used
for comparison.
Thanks for jogging my memory. I implemented it once. It was an array of
fixed length (2^n)-1, and was essentially a perfectly balanced binary tree
stored in a regular C array.
I implemented a bunch of heap variations, including two array based version
(one allowing key modification, the other not allowing it; the one allowing
key modifications is a *lot* slower!). The implementation should be available
from <http://www.dietmar-kuehl.de/cxxrt/heaps.tar.gz>.
Now out of curiosity, how much faster is a heap implemented through
std::priority_q ueue<T, std::vector<T>> using make_heap, over one implemented
through std::set<T>.
It depends somewhat on 'T': if 'T' is expensive to copy or swap, the vector
version will probably eventually become slower with increasing costs. For
'T' being something like 'int' the vector version is something like ten
times faster than typical node based approaches, including sets. At least,
this is what I seem to remember from running very naive tests.
A related question. Have you or anyone measured the time improvement of
inserting N elements into a std::vector<T> using reserve(N) and
push_back(const value_type&) and calling strd::sort, versus inserting N
elements into a std::multimap<T > by calling insert(const value_type&)?
I don't know for the inserts themselves but if the set of elements is fixed,
using an array with binary search is *much* faster than using 'std::map' or
'std::multimap' (again, I don't have current figures...). Likewise, a hash
(see eg. the library TR's 'std::tr1::unor dered_map' and
'std::tr1::unor dered_multimap' ) is much faster. Of course, these figure also
depend somewhat on the data types.

This form of heap is called a "d-heap" in its general form where "d"
is the arity of the tree: the same approach works with non-binary
trees, too, although the computation of child and parent nodes changes
slightly. When I compared different priority queues, I found that
"d == 5" had the best performance for integer values stored in the
heap. An even bigger 'd' is probably better if the nodes itself become
bigger. Of course, this was based on a few tests with my own
implementation and probably does not hold up to a scientific analysis.

What does this term "arity" mean? Perhaps the heap I implemented has arity
= 2 because each parent has 2 children?

Yes, this is what I mean with "arity": the [maximum] number of children of a
node. Of course, in a d-heap only leaves and at most one other node have less
than 'd' children.
What do you mean by the "nodes itself become bigger"? Is it that sizeof(T)
where T is the object stored in the heap gets bigger, or that the number of
nodes increases?
I mean 'sizeof(T)', ie. the data stored in a node: with bigger 'd' the number
of necessary moves when modifying the heap becomes smaller. If the type 'T'
is "bigger" or at least more expensive to move, the cost of the moves
are more dominant than the costs of other operations.

[relating to Fibonacci-heaps]
Thanks for the info. Changing a key in O(1) time is fascinating.

It is not O(1), it is *amortized* O(1): occasionally a higher costs is
incurred but if this cost can be averaged over multiple operations and
then becomes O(1) again. Of course, changing the key involves reordering
the corresponding node such that the other performance guarantees still
hold. Just the actual change of the key is trivially O(1) but doing the
reordering in amortized O(1) is somewhat tricky.
--
<mailto:di***** ******@yahoo.co m> <http://www.dietmar-kuehl.de/>
<www.contendix. com> - Software Development & Consulting