Actually, permutation is your friend... and there are tons of code samples
and algorithms out there. I'd start here.
http://www.google.com/search?hl=en&client=firefox-a&channel=s&rls=org.mozill
a%3Aen-US%3Aofficial&hs=mXk&q=permutation+coldfusion&btnG=Search
..:.:.:.:.:.:.:.:.:.:.:.
Bobby Hartsfie
Hey that's pretty smooth. I appreciate it!!
On Thu, Feb 14, 2008 at 1:27 PM, Jeff Price <[EMAIL PROTECTED]> wrote:
> Try this chunk of code. I leave it up to you to add in a remove duplicates
> feature. Check cflib.org for some handy functions to do that.
>
> NOTE: I found it easier to build a l
Try this chunk of code. I leave it up to you to add in a remove duplicates
feature. Check cflib.org for some handy functions to do that.
NOTE: I found it easier to build a list instead of an array simply because I
could recurse with ListAppend but if I recurse with ArrayAppend it starts
making
I'd make some changes to make it easier to read. First, instead of
wordarray[arraylen(wordarray)+1] = ...
I'd use arrayappend().
Second, if you aren't going to use ccArr, sbArr, and rcArr, remove them
from the function. They are just confusing.
If the length of remainingchars is 1, you don't
I thought I had it with this.. which is similiar to what you suggested Ben
if(len(remainingchars) eq 1) {
wordArray[ArrayLen(wordArray)+1] = strbase &
mid(remainingchars,1,1);
In pseudocode:
function f(sofar, more)
{
array strings
foreach letter in more
{
strings = f(foreach+letter, more-letter)
}
return strings
}
HTH.
--Ben Doom
Greg Morphis wrote:
> anyone else? I was hoping to do this in CF alone?
>
> Thanks
anyone else? I was hoping to do this in CF alone?
Thanks
On Wed, Feb 13, 2008 at 6:59 PM, Dawson, Michael <[EMAIL PROTECTED]> wrote:
> Thinking outside the box... You could use a database for this. Create a
> table that contains a single column. That column contains a record for each
> lette
Thinking outside the box... You could use a database for this. Create a table
that contains a single column. That column contains a record for each letter
of the alphabet. Then, do a cartesian join to join that table to itself, three
other times. Concatenate the fields and you should have e
This is what I have so far..
if(len(remainingchars) eq 1) {
wordArray[ArrayLen(wordArray)+1] = strbase &
mid(remainingchars,1,1);
} else {
for (j = 1; j lt len(remainingchars); j=j+1) {
What have you got so far?
Mark
On Feb 14, 2008 10:58 AM, Greg Morphis <[EMAIL PROTECTED]> wrote:
> yes, I know that and it's that part that I need help with which is why
> I posted here.
> Thanks
>
> On Feb 13, 2008 5:34 PM, Mark Mandel <[EMAIL PROTECTED]> wrote:
> > Recursion is your friend ;o)
yes, I know that and it's that part that I need help with which is why
I posted here.
Thanks
On Feb 13, 2008 5:34 PM, Mark Mandel <[EMAIL PROTECTED]> wrote:
> Recursion is your friend ;o)
>
> Mark
>
>
> On Feb 14, 2008 9:57 AM, Greg Morphis <[EMAIL PROTECTED]> wrote:
> > Given a string, e.g. "ABCD
Recursion is your friend ;o)
Mark
On Feb 14, 2008 9:57 AM, Greg Morphis <[EMAIL PROTECTED]> wrote:
> Given a string, e.g. "ABCD"
>
> I need to come up with all combinations of letters
> eg
> ABCD
> ABDC
> ACBD
> ACDB
>
>
> And exlude strings like '', 'AAAB' unless you pass a string with
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