yes...it won't work if it overlaps...but again it was assumed that the @start time
would be in ascending order.
the example that you have cited...the start @start array will just need to be sorted
in ascending order.
my @start = (25,10, 5);
my @stop = (45,35,30);
# Sort the start /stop array to be in ascending order
foreach(0..$#start){
$sortHash{$start[$_]}{$stop[$_]} = ();
}
foreach (sort {$a <=> $b} keys %sortHash) {
foreach $inIndex (sort {$a <=> $b} keys %{$sortHash{$_}} ) {
push (@startA, $_);
push (@stopA, $inIndex);
}
}
# The rest will work fine!!!!!
# Assuming that the start and the stop array have the
# same number of elements and each start corresponds with
# each stop
$i = 0;
$hash{$startA[0]} = $stopA[0];
$refStart = $startA[0];
$refStop = $stopA[0];
for ($i = 1; $i <= $#startA; $i++) {
if ($refStop >= $startA[$i]) {
$refStop = $stopA[$i];
$hash{$refStart} = $stopA[$i];
} else {
$refStart = $startA[$i];
$refStop = $stopA[$i];
$hash{$refStart} = $stopA[$i];
}
}
foreach (sort { $a <=> $b } keys %hash) {
print $_, " ", $hash{$_}, "\n";
}
-----Original Message-----
From: Jonathan E. Paton [mailto:[EMAIL PROTECTED]]
Sent: Friday, May 31, 2002 2:18 PM
To: [EMAIL PROTECTED]
Subject: RE: union of times algorithm
--- "Shishir K. Singh" <[EMAIL PROTECTED]> wrote:
> Is there something wrong with this algo ??
Yes, it isn't valid perl. You didn't debug yours either,
did you? :P Anyway, yes there is a problem... I think.
What happens if your timeslices overlap? E.g.
my @start = (25, 10, 5);
my @stop = (40, 35, 30);
I don't think your algorithm produces the correct
results, and there is no provision to fix this... I'm
more in a algorithm writing mood rather than a algorithm
validating mood. Can you comment please?
Jonathan Paton
> ####################################################
> my @start = (3,4, 15, 23, 29, 34, 37);
> my @stop = (5,10,20, 29, 33, 36, 37);
>
> # Assuming that the start and the stop array have the
> # same number of elements and each start corresponds with
> # each stop
>
> my $i = 0;
> my $hash{$start[0]} = $stop[0];
> my $refStart = $start[0];
> my $refStop = $stop[0];
>
> for ($i = 1; $i <= $#start; $i++) {
> if ($refStop >= $start[$i]) {
> $refStop = $stop[$i];
> $hash{$refStart} = $stop[$i];
> } else {
> $refStart = $start[$i];
> $refStop = $stop[$i];
> $hash{$refStart} = $stop[$i];
> }
> }
>
> foreach (sort { $a <=> $b } keys %hash) {
> print $_, " ", $hash{$_}, "\n";
> }
> ######################################################
> -----Original Message-----
> From: Jonathan E. Paton [mailto:[EMAIL PROTECTED]]
> Sent: Friday, May 31, 2002 1:58 PM
> To: [EMAIL PROTECTED]
> Subject: RE: union of times algorithm
>
>
> --- [EMAIL PROTECTED] wrote:
> > What we have is as stated a union of times. So all I do is take the
> > timeslices and generate a hash element for each time. I don't check to see
> > if it is already there, but just set to one. Then I sort the hash down
> > numerically and total where current minus previous equal 1. Here is a shot:
> >
> > use Data::Dumper;
> >
> > my @timeslices = (
> > [4, 9],
> > [3, 6],
> > [2, 4],
> > [11,12],
> > [14,14],
> > [17,19]
> > );
> >
> > my %MyHashCounts = ();
> >
> > for(my $MyId=0;$MyId<=$#timeslices;$MyId++){
> > for($MyId1=$timeslices[$MyId][0];$MyId1<=$timeslices[$MyId][1];$MyId1++)
> > {
> > $MyHashCounts{$MyId1}=1;
> > }
> > }
> > print Dumper(\@timeslices);
> > print Dumper(\%MyHashCounts);
> >
> > my $MyPrev = -10;
> > my $MyTotal = 0;
> > foreach my $MyKeys (sort {$a <=> $b} keys %MyHashCounts) {
> > printf "%3d ", $MyKeys;
> > $MyTotal++ if ( ($MyKeys - $MyPrev) == 1 );
> > $MyPrev = $MyKeys;
> > }
> > printf "\n";
> > printf "Total: %4d\n", $MyTotal;
> >
> > Output:
> > 2 3 4 5 6 7 8 9 11 12 14 17 18 19
> > Total: 10
>
> Which is wrong, since the 14 sneaked in even though the
> contribution of [14,14] should be zero. Notice:
>
> 0 1 2
> 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
> > | [4, | 9], | |
> > | [3, 6], | | | | |
> > |[2,| 4], | | | | | | |
> > | | | | | | | | | |[11,
> > | | | | | | | | | | | 12],
> > | | | | | | | | | | | | |[14,
> > | | | | | | | | | | | | | 14],
> > | | | | | | | | | | | | | | | |[17,
> | | | | | | | | | | | | | | | | | |19]
> X X X X X X X X X X X = 11
>
> Looking at my algorithm, I can remove the use of the
> hash.
>
> The algorithm you are using is inferiour to the approach
> I tried (even though BOTH our algorithms are wrong). If
> I start using numbers in the region of Unix timestamps,
> (i.e. large) it is an impossible to satisfy memory/CPU hog.
>
> The algorithm I use is pictorally:
>
> XXXXXXXXXXXXXXXXXXX
> YYYYYYYYYYYYYYYYYYYYYYY
> ZZZZZZZZZZZZZZZZZZZZZZ
> ^ ^ ^
> start X ^ start Y end Z ^
> start Z end Y
> ^
> end X
>
> Sorting results in the following events:
>
> ^start X
> ^start Z
> ^end X
> ^start Y
> ^end Z
> ^ end Y
>
> And we can find the number of overlapping sections now:
> ACTIVE
> V
> 1 ^start X
> 12 ^start Z
> 12 ^end X
> 23 ^start Y
> 2 ^end Z
> 3 ^ end Y
>
> So, by sorting we can loop through, starting and stopping "active"
> timeslots. We need to do this in order, otherwise we will get
> incorrect results. I sort the second column to always get a start
> event before a stop event.
>
> Jonathan Paton
>
> =====
> $_=q|.,&@$$. ,.@$&@$. .&$$@. ,,$ ....!$_=$p.'&$@.',y'&$@' ...,';for(/\S+/g){
> !|.q| .$ .,@, ,$, .,.. @, ,$ ,,@ ...,,.!++$.<22?${'y'.$_}=chr$.+64:[$$=${'y'
> !|.q| ,@$@&.,. $$$&, ..@&&$,,, $., ....!.$_},$y.=($.=~/22\|26\|3(3\|7)/x?' '
> !|.q|. @ ., ,.&,,, , .$..&. .,$ .,,!.$$:"\l$$")]};$y=~/ (.*)/;warn"$1\n"
> !|.q|. $ .,. .,$$&&$...&., @.,.&@$@ ...|,map{-$|--?$r:$p.=$_}split'!';eval$r
>
> __________________________________________________
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=====
$_=q|.,&@$$. ,.@$&@$. .&$$@. ,,$ ....!$_=$p.'&$@.',y'&$@' ..,';for(/\S+/g){
!|.q| .$ .,@, ,$, .,.. @, ,$ ,,@ ..,,.!++$.<22?${'y'.$_}=chr$.+64:[$$=${'y'
!|.q| ,@$@&.,. $$$&, ..@&&$,,, $., ...!.$_},$y.=($.=~/22\|26\|3(3\|7)/x?' '
!|.q|. @ ., ,.&,,, , .$..&. .,$ .,,!.$$:"\l$$")]};$y=~/ (.*)/;warn"$1\n"
!|.q|. $ .,. .,$$&&$...&., @.,.&@$@ ..|,map{-$|--?$r:$p.=$_}split'!';eval$r
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