P: n/a

Hi,
I am trying to mark consective numbers in a data set and get a count as
to how many consecutive numbers exist in the a given line of data.
Here is an example line:
3, 5, 7, 9, 10, 13, 14
The relevant part of the script which currently checks for a
consecutive value works in this fashion:
N.B. $cc is the consective counter and $one thru $seven are my result
numbers.
if ($two == $one + 1) {
$cc++;
}
if ($three == $two + 1) {
$cc++;
}
if ($four == $three + 1) {
$cc++;
}
if ($five == $four + 1) {
$cc++;
}
if ($six == $five + 1) {
$cc++;
}
if ($seven == $six + 1) {
$cc++;
}
Now this is all find and dandy, ugly I know but it gets the job done,
at least in the above example, it denotes the line as having two sets
of consective numbers.
My problem results when I have a line like the following:
3, 5, 7, 8, 9, 13, 15
I would like the script to tell me that there are 3 numbers in a row,
but it tells me there are two sets of consecutive numbers; 7,8 and 8,9.
At the same time though, I want it to still report the number of sets
of consective numbers.
Like I said I know it's ugly but if someone can offer up some guiding
light as to the error of my ways, I would appreciate the help.
TIA,
Adam  
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Theres no logical way that 3, 5, 7, 9, 10, 13, 14 can return 2, while
3, 5, 7, 8, 9, 13, 15 returns 3. If the second one is said to have 3
consecutive numbers, then the first one has 4. If the first one is
said to have 2, then the second one also has 2. It depends on how you
look at it, but I can't see any logical way of having the first one
return 2 and the second one return 3.  
P: n/a

Theres no logical way that 3, 5, 7, 9, 10, 13, 14 can return 2, while
3, 5, 7, 8, 9, 13, 15 returns 3. If the second one is said to have 3
consecutive numbers, then the first one has 4. If the first one is
said to have 2, then the second one also has 2. It depends on how you
look at it, but I can't see any logical way of having the first one
return 2 and the second one return 3.  
P: n/a

DJ Craig wrote: Theres no logical way that 3, 5, 7, 9, 10, 13, 14 can return 2, while 3, 5, 7, 8, 9, 13, 15 returns 3. If the second one is said to have 3 consecutive numbers, then the first one has 4. If the first one is said to have 2, then the second one also has 2. It depends on how you look at it, but I can't see any logical way of having the first one return 2 and the second one return 3.
I can.
In the first one, 9 and 10 are consecutive. So are 13 and 14  but they
are not consecutive with 9 and 10. So you have 2 sets of consecutive
numbers, the longer of which is 2.
In the second set, 7, 8 and 9 are consecutive, but there are no other
consecutive numbers. The only set is 3 elements long.

==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp. js*******@attglobal.net
==================  
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DJ Craig wrote: Theres no logical way that 3, 5, 7, 9, 10, 13, 14 can return 2, while 3, 5, 7, 8, 9, 13, 15 returns 3. If the second one is said to have 3 consecutive numbers, then the first one has 4. If the first one is said to have 2, then the second one also has 2. It depends on how you look at it, but I can't see any logical way of having the first one return 2 and the second one return 3.
I can.
In the first one, 9 and 10 are consecutive. So are 13 and 14  but they
are not consecutive with 9 and 10. So you have 2 sets of consecutive
numbers, the longer of which is 2.
In the second set, 7, 8 and 9 are consecutive, but there are no other
consecutive numbers. The only set is 3 elements long.

==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp. js*******@attglobal.net
==================  
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Adam wrote:
<snip likely homework assignment>
Hmmm, now what business problem are you trying to solve here? None? This
is perhaps something more abstract? Perhaps a homework assignment?

Kenneth Downs
Secure Data Software, Inc.
(Ken)nneth@(Sec)ure(Dat)a(.com)  
P: n/a

Adam wrote:
<snip likely homework assignment>
Hmmm, now what business problem are you trying to solve here? None? This
is perhaps something more abstract? Perhaps a homework assignment?

Kenneth Downs
Secure Data Software, Inc.
(Ken)nneth@(Sec)ure(Dat)a(.com)  
P: n/a

In article <11**********************@g44g2000cwa.googlegroups .com>, do*******@hotmail.com says... Hi,
I am trying to mark consective numbers in a data set and get a count as to how many consecutive numbers exist in the a given line of data.
Here is an example line:
3, 5, 7, 9, 10, 13, 14
The relevant part of the script which currently checks for a consecutive value works in this fashion:
N.B. $cc is the consective counter and $one thru $seven are my result numbers.
if ($two == $one + 1) { $cc++; } if ($three == $two + 1) { $cc++; } if ($four == $three + 1) { $cc++; } if ($five == $four + 1) { $cc++; } if ($six == $five + 1) { $cc++; } if ($seven == $six + 1) { $cc++; }
Now this is all find and dandy, ugly I know but it gets the job done, at least in the above example, it denotes the line as having two sets of consective numbers.
My problem results when I have a line like the following:
3, 5, 7, 8, 9, 13, 15
I would like the script to tell me that there are 3 numbers in a row, but it tells me there are two sets of consecutive numbers; 7,8 and 8,9. At the same time though, I want it to still report the number of sets of consective numbers.
Like I said I know it's ugly but if someone can offer up some guiding light as to the error of my ways, I would appreciate the help.
Try this for a giggle
================================================== ================
<?php
//$arr = array(3, 5, 7, 8, 9, 13, 15);
//var_dump($arr);
$arr = array(1);
chkArray($arr);
$arr = array(1,2);
chkArray($arr);
$arr = array(1,3,2);
chkArray($arr);
$arr = array(1,2,3);
chkArray($arr);
$arr = array(1,2,3,4);
chkArray($arr);
$arr = array(1,2,9,3,4,5);
chkArray($arr);
// ===============================================
function chkArray($arr) {
$consecutiveNumberPairCount = 0;
$longestConsecutiveRun = 0;
$currentLongestConsecutiveRun = 0;
$maxIndex = count($arr)  1;
foreach($arr as $index => $value) {
//print "$index : $value\n";
if ($index < $maxIndex) {
//print '$arr[$index] ' . $arr[$index] . '  $arr[$index+1]
' . $arr[$index+1] . "\n";
if (($arr[$index] + 1) == $arr[$index+1]) {
$consecutiveNumberPairCount =
$consecutiveNumberPairCount + 1;
}
}
if ($index > 0) {
//print '$arr[$index] ' . $arr[$index] . '  $arr[$index+1]
' . $arr[$index+1] . "\n";
if (($arr[$index] 1 ) == $arr[$index1]) {
if ($currentLongestConsecutiveRun == 0) {
$currentLongestConsecutiveRun = 2;
}
else {
$currentLongestConsecutiveRun =
$currentLongestConsecutiveRun + 1;
}
if ($currentLongestConsecutiveRun >
$longestConsecutiveRun) {
$longestConsecutiveRun =
$currentLongestConsecutiveRun;
}
}
else {
//print 'current longest ' .
$currentLongestConsecutiveRun . ' : longest ' . $longestConsecutiveRun .
"\n";
$currentLongestConsecutiveRun = 0;
}
}
}
print "pairs = $consecutiveNumberPairCount\n";
print "longest consecutive = $longestConsecutiveRun\n";
}
?>
================================================== ===================  
P: n/a

Here's a little function that does what you want. It returns an array
of lengths of consecutive sequences. For [3, 5, 7, 8, 9, 13, 15] it
yields [3]. For [3, 5, 7, 9, 10, 13, 14], you get [2, 2]. I'll leave it
to you to figure out how it works ;)
function seq($a) {
$b = range($a[count($a)  1], $a[1]);
$c = array_map(create_function('$a,$b', 'return $a + $b;'), $a, $b);
$d = array_count_values($c);
$e = array_filter($d, create_function('$n', 'return $n != 1;'));
return array_values($e);
}
print_r(seq(array(3, 5, 7, 8, 9, 13, 15)));
print_r(seq(array(3, 5, 7, 9, 10, 13, 14)));   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 2671
 replies: 8
 date asked: Jul 17 '05
