But let's leave the ad hominem out...

Please do pick up a book or two on algorithms and data structures; this is stuff anyone who is serious about programming should know.

Yes. Let's do that.

In a heap with 1,000 elements, you need at most 10 swaps. How much money will you bet on splice?

Quite a lot, were I a betting man! :)

#! perl -slw
use strict;
use Benchmark qw[ cmpthese ];

our $SIZE ||= 1000;
our $ITERS||= -5;

sub spliceIt{
    my @a = reverse 0 .. $SIZE;
    unshift @a, splice @a, $SIZE, 1; ## Update: changed push to unshif
+t!
#   print "@a";
}

sub heapIt {
    my @a = reverse 0 .. $SIZE;
    $a[ $SIZE ] = $SIZE + 1;
    moveUp( \@a, $SIZE );
#   print "@a";
}

sub moveUp {
    my( $ref, $l ) = @_;
    my $p = int $l /2;
    return if $p >= $l;
    @{ $ref }[ $p, $l ] = @{ $ref }[ $l, $p ];
    moveUp( $ref, $p );
}

print "Testing $SIZE items for $ITERS iterations";

cmpthese( $ITERS, {
    splice => \&spliceIt,
    heap   => \&heapIt,
});
__END__
P:\test>heaptest -ITERS=-5 -SIZE=100
Testing 100 items for -5 iterations
          Rate   heap splice
heap   19502/s     --   -37%
splice 30887/s    58%     --

P:\test>heaptest -ITERS=-5 -SIZE=1000
Testing 1000 items for -5 iterations
         Rate   heap splice
heap   3159/s     --    -4%
splice 3288/s     4%     --

P:\test>heaptest -ITERS=-5 -SIZE=10000
Testing 10000 items for -5 iterations
        Rate   heap splice
heap   327/s     --    -0%
splice 328/s     0%     --

P:\test>heaptest -ITERS=-5 -SIZE=20000
Testing 20000 items for -5 iterations
        Rate splice   heap
splice 159/s     --    -1%
heap   160/s     1%     --
[download]

From where you left off. A new item not currently in cache is called for, it is read from disk, the lowest item* (currently index 12) is replaced by the new item in the array** and the new item given a weight of 17.

 a) 0 [ 16 ] b) 0 [ 16 ] c) 0 [ 16 ] d) 0 [ 16 ] e) 0 * 17 ]
    1 [ 12 ]    1 [ 12 ]    1 [ 12 ]    1 * 17 ]    1 * 16 ]
    2 [ 13 ]    2 [ 13 ]    2 [ 13 ]    2 [ 13 ]    2 [ 13 ]
    3 [ 10 ]    3 [ 10 ]    3 * 17 ]    3 * 12 ]    3 [ 12 ]
    4 [  9 ]    4 [  9 ]    4 [  9 ]    4 [  9 ]    4 [  9 ]
    5 [ 11 ]    5 [ 11 ]    5 [ 11 ]    5 [ 11 ]    5 [ 11 ]
    6 [  7 ]    6 * 17 ]    6 * 10 ]    6 [ 10 ]    6 [ 10 ]
    7 [  6 ]    7 [  6 ]    7 [  6 ]    7 [  6 ]    7 [  6 ]
    8 [  5 ]    8 [  5 ]    8 [  5 ]    8 [  5 ]    8 [  5 ]
    9 [  4 ]    9 [  4 ]    9 [  4 ]    9 [  4 ]    9 [  4 ]
   10 [  3 ]   10 [  3 ]   10 [  3 ]   10 [  3 ]   10 [  3 ]
   11 [  8 ]   11 [  8 ]   11 [  8 ]   11 [  8 ]   11 [  8 ]
   12 * 17 ]   12 *  7 ]   12 [  7 ]   12 [  7 ]   12 [  7 ]
[download]

Now, another new item is called for, so I need to locate the lowest weighted item in the array. *How do I do this?

And another problem, when I need to locate one of these items that are moving around in this heap via it's key.

**How do I locate it?

Actually, it's just the original one. That of maintaining the linkage between the items in the array(heap) and their keys. No matter how long I "look at the pictures"--or read the text--at heaps, I do not see the mechanism by which the lowest weighted item in the heap is located (other than a linear search).

To re-state the requirements. I need to be able to:

1. Locate the highest weighted item.

   This is required to allow promotion of the lastest accessed item to the top in the classic LRU algorithm.

2. Locate the lowest weighted item.

   Also an LRU requirement(or any variation), as this is the one that will be discarded when the cache is full and a new element must be added.

3. Locate an item in the cache via it's key.

   As the items get moved around, that linkage *must* be maintained.

   Embedding the key within the item would require a linear search to locate it. The purpose of the exercise was to avoid a linear search.