Stuck with a "lookup up" question

hello forum,
is there a function to search a column for a particular string, e.g. "*pic1*" (wildcards needed) and if there is a match than just return "pic1"? Somehow it sounds very simple - but right now I´m stuck with it. I tried "lookup" but could not get it to work that way.
Regards, Michael

Hi Michael,
I think you could build a formula to do what you are describing starting with COUNTIF(). It can count how many times "*pic1*" shows up in a column and return that number. You could then test that value and return "pic1" (or whatever) if a number value is not enough.
You could also use conditional highlighting on those cells to highlight the cells containing "pic1".
Wayne's request is a good one. You have an idea of how you want to solve a problem but it may not be the simplest or most direct way. As an example, many people come to the forum with a question about IF() and end up using LOOKUP().
What is the problem you are trying to address?
quinn

Similar Messages

App will not download to iPad when purchased from App Store - stuck with Cloud icon

Device : iPad Air 32gb wi-fi model (not cellular data)
OS version : 7.1.1 (up to date)
Location : UK
Problem : App will not download to iPad when purchased from App Store - stuck with Cloud icon
Last night I tried to download an app over wi-fi onto my iPad Air, but the app did not install.
I have plenty of space (the app is 24.1mb, I have 3.4gb free space) and there are no restrictions whatsoever on my iPad - no blocks on App store purchases, no age-related restrictions, no country restrictions etc.
Normally from the App Store page you would tap the price (in this case the app was free), enter your iTunes password, and the app would download.
Also a new icon appears on the home screen and a circle shows the progress of the download.
However this did not happen.
Instead I have the Cloud icon (the one that shows when you have bought an app but it is not currently installed on your device) and a message that says "You've already purchased this, so it will be downloaded now at no additional cost".
If I press OK I am then prompted to enter my iTunes password, the Cloud symbol briefly changes to a circle (and I mean for a split-second, like the blink of an eye) and then the Cloud symbol appears again.
The app does not download - in fact it does not even *attempt* to download.
I have tried this >20 times now and it always occurs excatly like this.
After Googling this problem (which seems common) and reading over a dozen threads on this forum plus Apple's own official support pages, this is what I've tried:
I can re-download other apps that I have previously purchased on this iPad, with no problems.
I purchased another app on my iPhone 5 and was then able to download onto my iPad with no problems.
I have reset my iPad >5 times - both by pressing/holding the home/power buttons, and also a full switch off and restart.
I have logged out of my iTunes account on the iPad, and logged back in.
The wi-fi is Virgin Media fibre optic broadband with a verified speed of 20 meg (confirmed by Speedtest) plus an 802.11-n Belkin router and my iPad will happily load web pages, stream videos on YouTube, download other apps etc.
I have reset/rebooted my internet connection
In an attempt to eliminate by broadband connection from the possible causes, I have switched off my router, linked my iPad to the hotspot feature on my iPhone and tried to download the app over cellular data, but still cannot download the app.
I have tried on 2 other wi-fi networks but still cannot download this one app.
My iTunes account has a cerdit balance from an iTunes card, but the app was free anyway
Nothing I try will download the app in question.
The app is iPad only, so I cannot install on my iPhone to try and kick-start the process.
I guess I could install the app on my computer and then sync the iPad, but that's *working around* the problem - it's not *fixing* the problem.
(Since getting the iPad I use almost never use my computer. And anyway what if I did not have a computer? Didn't iOS go "PC-free" with version 5?)
I would like to call Apple but my 90-day telephone support has expired (the iPad is around 6 months old) and I don't want to buy Apple Care for this one problem.
I can't make it to an Apple Store to speak to a Genius.
Bearing in mind the list of things I've already tried, what do I do next?
Thanks in advance for any support.

Hi there.
Thank you very much for your quick reply, it's really appreciated.
No need to apologize, I actually found the phrase "If your iPad ever craps the bed" amusing under the circumstances
I checked the list of installed apps like you suggested (Settings>General>Usage>Storage>Show all apps) but this problem app is not in the list. I have also swiped down on the home screen to search for it, but again it does not show.
While it has not let me download the app (and I just tried it again) it has helped me to eliminate a few things from the list of what to try, and I can definitely conclude that it has not downloaded.
I guess I need to fire up my PC and download it / sync my iPad.
Thanks again for your help

How do I access compacted mail on a Mac version 10.5.8 stuck with Thunderbird version 16.0.2 ? I am unable to access needed documentation for Taxes, etc

How do I access compacted mail on a Mac version 10.5.8 stuck with Thunderbird version 16.0.2 ? I am unable to access needed documentation for Taxes, etc

Where in the world did you read Deleting All My Mail? I have reread my answer several times and cannot seem to find that in there.
When you mark a message for deletion it does just that. It marks it for deletion and hides the message. It is not truly deleted until you compact.
My point is that unless you deleted your important mail they should still be there somewhere.
Your question was about compacting and thinking it put your messages in a nice little box for safe keeping and now you want to open it.
I have to agree that Thunderbird did a horrible job of naming this process because most are like you and do not understand what it actually does. I included the link to the explanation of what compacting means in the Thunderbird world.
Your messages may still be available to your but finding them has nothing to do with compacting.
Thunderbird does not archive anything unless you use that process by selecting the message and pressing a. You have the option to setup where you want that archive to go and what folders to make.
The All Mail folder is a goofy gmail arrangement. If you have problems with that then you need to address that with Google.
If you go into your gmail account on the gmail site do you see the messages you are hunting in the All Mail folder there?

Stuck with problems and no help

hi,
We are stuck with the following set of problems. Repeated entries to this forum have not fetced any response. If anybody can help us on any of these issues, please reply -
1. We are not able to define any custom validator. We made our own validator that extends CompareValidor, but when we try to apply this rule to an attribute in the validation tab, we get the error - "no properties defined". Do i need to do something else besides creating a new class that extends CompareValidator and then writing my own code in the vlaidate method.
2. While deploying our jspApplication on java web server, we are getting the message "error - null". The details are as mentioned in the Topic 'can anyone help!!!' that has been posted by Prabhdeep ().
3. I have two tables, Customer and address. There is no FK between them (there cannot be because address is shared between many entities). In the address, i have 2 fields, BpType and BpId. BpType will contain 'C' to identity that this address is for a customer and id will have the customer id. In jdev 3.1, i have created entity and view objects for the 2 tables. also, i have made a view link. i need to put the condition Address.bpid = customer.customeid and address.bptype = 'c'. but it is not letting me put the second part of the condition. the chk on id works fine. as soon as i add a chk on type and test the application module, then while insertion of address i get the following error - "set method for BpType cannot be resolved".
4. I have made a jsp appl based on bc4j. now, i need to put a check on any entity in such a manner that if the check fails, my jsp should ask the user that an exception is occuring and does he want to save even with the exceptional data. if he says yes, i want that the same entity should now save the data without giving an exception. How do i achieve this interaction between my entity and my jsp.
Thanks a lot for any help,
Aparna.

I will answer Question #3:
I'm not sure if I understand exactly what is
happening. At first I thought you meant that you couldn't even add the second part of the association clause. But, later it sounds like you were able to add the second part of the clause (namely "address.bptype =
'c'") and that you are experiencing some runtime error.
Assuming the latter (that it is a RT problem), I take it this
"...cannot be resolved" error is the error with code 27020.
Under normal error conditions, you shouldn't get this error. Something "unusual" (or unforeseen by us) is happening if you are getting this error. Anyway, if you wrote your own client code, you should wrap a try/catch block around it, pick out the detail exception, and see what's going
on. See the example code below:
try
... make call that causes the exception ...
catch(oracle.jbo.AttrSetValException ex)
// Print stack trace for the outer ex
ex.printStackTrace();
// Get the inner exception
Object[] details = ex.getDetails();
if (details.length > 0)
Exception innerEx = (Exception) details[0];
// Now, print the inner exception's
// stack.
innerEx.printStackTrace();
This will help shed some light on the exception and the cause of it.
John - JDeveloper Team

Stuck with updating AACS key / WinDVD 8

I have a W700 with BluRay drive. Occasionally I used it to watch BluRay movies using WinDVD 8. But when I tried this recently, WinDVD told me, that the AACS key has expired / been deacvtived and asked me, if I want to renew it. I followed all the instructions from Corel (turning off UAC, making IE the standard browser), but it is giving me an error "Server is unable to process your renewal request".
I did run ThinkVantage System Update, which updated the WinDVD version from B09.572C14.63616.00000T0000 to B09.581C14.64008.00000T0000, but still cannot renew the AACS key.
I found another update for WinDVD on Lenovo's drivers and software page: windvd88171u-a.exe. According to the description (http://support.lenovo.com/de_DE/downloads/detail.page?DocID=DS003904) it should be applicable for LOGID 64008, but the update fails saying it doesn't seem applicable.
So I'm stuck with renewing the AACS key / updating WinDVD 8. Any help is very much appreciated.
Thanks in advance.
Andy
Solved!
Go to Solution.

Hi, Andy
a
I did some research and found a forum thread from Corel's site. I wasn't able to find too much, but I did find something simple that might help. A user claims to have had the same issue as you, and it would only update the AACS key if he was already logged in on the Corel site. I'm not sure if this will solve your issue, but it could be something to try.
Good luck,
Adam
Did someone help you today? Press the star on the left to thank them with a Kudo!
If you find a post helpful and it answers your question, please mark it as an "Accepted Solution!" This will help the rest of the community with similar issues identify the verified solution and benefit from it.

Stuck with File Path

I am writing a very simple script that exports an image's cutline for my newsroom, i am getting stuck with the having it write the file to the correct path. When I run script as below it works, writing the file to my documents:
function writeCutline () {
                // Linefeed shizzle
            if ($.os.search(/windows/i) != -1)
                fileLineFeed = "windows";
            else
                fileLineFeed = "macintosh";
    var fileName = app.activeDocument.name
    var MetaInfo = app.activeDocument.info;
    MetaInfo = MetaInfo.caption;
    var fileOut = new File(Folder.myDocuments + '/'+ fileName.substring(0,fileName.length-4)+ '.txt');
    fileOut.linefeed = fileLineFeed;
    fileOut.open("w", "TEXT", "????");
    fileOut.writeln(MetaInfo);
    fileOut.close();
writeCutline();
What I want it to do is:
var fileOut = new File( '/c/cutlines/'+ fileName.substring(0,fileName.length-4)+ '.txt');
but this does not work. What do I have to do different?

Thanks for the help guys. I wasn't actually putting the files on C: it was actually going to a mapped network drive and for some reason last night the server was working incorrectly and that is where my problems where coming from. But that raises my next question, is there anyway to address the path with mapping the drive, so it would be a refence to the server. If i was browsing the path it would be TARHSHARE/PHOTOS/photos/DailyATS/Cutlines. With PHOTOS mapped to drive P the path is P:/Photos/Daily/ATS/Cutlines. So what I want to do is get it work without mapping the drive, so the script will work on all of the photographers accounts without mapping the drive.

Stuck with these methods

Hello everyone.
the reason of this post is that I am a bit stuck with this class where I need to implement two methods and it is not working so far.
The class in question is:
public class MyPuzzle
    public static final int GRIDSIZE = 5;
    private int[][] squares;
    private String[][] rowConstraints;
    private String[][] columnConstraints;
    public MyPuzzle()
        squares = new int[GRIDSIZE][GRIDSIZE];
        rowConstraints = new String[GRIDSIZE][GRIDSIZE - 1];
        columnConstraints = new String[GRIDSIZE][GRIDSIZE - 1];
        for (int row = 0; row < GRIDSIZE; row++) {
            for (int column = 0; column < GRIDSIZE - 1; column++) {
                rowConstraints[row][column] = " ";
        for (int column = 0; column < GRIDSIZE; column++) {
            for (int row = 0; row < GRIDSIZE - 1; row++) {
                columnConstraints[column][row] = " ";
    public void setSquare(int row, int column, int val)
        if ((1 <= val) && (val <= GRIDSIZE)) {
            squares[row][column] = val;
        else {
            System.out.println("Error - Illegal value " + val);
    public void setRowConstraint(int row, int col, String relation)
        if (relation.equals("<") || relation.equals(">")) {
            rowConstraints[row][col] = relation;
    public void setColumnConstraint(int col, int row, String relation)
        if (relation.equals("<") || relation.equals(">")) {
            columnConstraints[col][row] = relation;
    public void fillPuzzle()
       setColumnConstraint(0, 1, ">");
       setRowConstraint(4, 0, "<");
       setRowConstraint(4, 2, "<");
       setColumnConstraint(4, 3, "<");
       setColumnConstraint(4, 2, "<");
       setRowConstraint(3, 1, "<");
       setRowConstraint(1, 3, ">");
       setSquare(4, 0, 4);
    public void printPuzzle()
        for (int row = 0; row < GRIDSIZE - 1; row++) {
            drawRow(row);
            drawColumnConstraints(row);
        drawRow(GRIDSIZE - 1);
    private void printTopBottom()
        for (int col = 0; col < GRIDSIZE; col++) {
            System.out.print("---   ");
        System.out.println();
    private void drawColumnConstraints(int row)
        for (int col = 0; col < GRIDSIZE; col++) {
            String symbol = " ";
            if (columnConstraints[col][row].equals("<")) {
                symbol = "^";
            else if (columnConstraints[col][row].equals(">")) {
                symbol = "V";
            System.out.print(" " + symbol + "    ");
        System.out.println();
    private void drawRow(int row)
        printTopBottom();
        for (int col = 0; col < GRIDSIZE; col++) {
            String symbol;
            if (squares[row][col] > 0) {
                System.out.print("|" + squares[row][col] + "|");
            else {
                System.out.print("| |");
            if (col < GRIDSIZE - 1) {
                System.out.print(" " + rowConstraints[row][col] + " ");
        System.out.println();
        printTopBottom();
[/code]
The two methods I am trying to implement are:
-isLegal should return a boolean value that tells us whether the puzzle configuration is legal or not. That is, if there is a repeated value in any row or column, or if there is an incorrect value (the number 1024 for instance) anywhere, or if any of the constraints is violated isLegal should return false; otherwise it should return true.
-if the puzzle is not legal, getProblems should return a String describing all the things that are wrong with the configuration.
If anyone may help, I will be eternally gratefull.
Thanks.

Hello everyone.
the reason of this post is that I am a bit stuck with this class where I need to implement two methods and it is not working so far.
The class in question is:
public class MyPuzzle
public static final int GRIDSIZE = 5;
private int[][] squares;
private String[][] rowConstraints;
private String[][] columnConstraints;
public MyPuzzle()
squares = new int[GRIDSIZE][GRIDSIZE];
rowConstraints = new String[GRIDSIZE][GRIDSIZE - 1];
columnConstraints = new String[GRIDSIZE][GRIDSIZE - 1];
for (int row = 0; row < GRIDSIZE; row++) {
for (int column = 0; column < GRIDSIZE - 1; column++) {
rowConstraints[row][column] = " ";
for (int column = 0; column < GRIDSIZE; column++) {
for (int row = 0; row < GRIDSIZE - 1; row++) {
columnConstraints[column][row] = " ";
public void setSquare(int row, int column, int val)
if ((1 <= val) && (val <= GRIDSIZE)) {
squares[row][column] = val;
else {
System.out.println("Error - Illegal value " + val);
public void setRowConstraint(int row, int col, String relation)
if (relation.equals("<") || relation.equals(">")) {
rowConstraints[row][col] = relation;
public void setColumnConstraint(int col, int row, String relation)
if (relation.equals("<") || relation.equals(">")) {
columnConstraints[col][row] = relation;
public void fillPuzzle()
setColumnConstraint(0, 1, ">");
setRowConstraint(4, 0, "<");
setRowConstraint(4, 2, "<");
setColumnConstraint(4, 3, "<");
setColumnConstraint(4, 2, "<");
setRowConstraint(3, 1, "<");
setRowConstraint(1, 3, ">");
setSquare(4, 0, 4);
public void printPuzzle()
for (int row = 0; row < GRIDSIZE - 1; row++) {
drawRow(row);
drawColumnConstraints(row);
drawRow(GRIDSIZE - 1);
private void printTopBottom()
for (int col = 0; col < GRIDSIZE; col++) {
System.out.print("--- ");
System.out.println();
private void drawColumnConstraints(int row)
for (int col = 0; col < GRIDSIZE; col++) {
String symbol = " ";
if (columnConstraints[col][row].equals("<")) {
symbol = "^";
else if (columnConstraints[col][row].equals(">")) {
symbol = "V";
System.out.print(" " + symbol + " ");
System.out.println();
private void drawRow(int row)
printTopBottom();
for (int col = 0; col < GRIDSIZE; col++) {
String symbol;
if (squares[row][col] > 0) {
System.out.print("|" + squares[row][col] + "|");
else {
System.out.print("| |");
if (col < GRIDSIZE - 1) {
System.out.print(" " + rowConstraints[row][col] + " ");
System.out.println();
printTopBottom();
}The two methods I am trying to implement are:
-isLegal should return a boolean value that tells us whether the puzzle configuration is legal or not. That is, if there is a repeated value in any row or column, or if there is an incorrect value (the number 1024 for instance) anywhere, or if any of the constraints is violated isLegal should return false; otherwise it should return true.
-if the puzzle is not legal, getProblems should return a String describing all the things that are wrong with the configuration.
If anyone may help, I will be eternally gratefull.
Thanks.

Stuck with search query

I am trying to search the whole record if any of the word matches with existing data in the rows of columns.
First I was trying the following code:
SQL> select prod from prod where prod like '%chromium ore%'
2 /
PROD
chromium ore andconcentrates, ferro-chrome, silico-chrome
chromium ore andconcentrates, ferro-chrome, silica-chrome, metal
chromium ore andconcentrates, ferro-chrome, silica-chrome
chromium ore andconcentrates, ferro-chrome, silico-chrome
chromium ore andconcentrates, ferro-chrome, silico-chrome, metal
chromium ore andconcentrates, ferro-chrome, silico-chrome
6 rows selected.
SQL> select prod from prod where prod like '%what is chromium ore%';
no rows selected
SQL> select count(unique(co_id)) from prod where (prod like upper('%ferro%') or prod like lower('%ferro%') or prod like initcap('%ferro%'));
COUNT(UNIQUE(CO_ID))
                3153
SQL> select count(unique(co_id)) from prod where (prod like upper('%What is ferro') or prod like lower('%What is ferro') or prod like initcap('%What is ferro'));
COUNT(UNIQUE(CO_ID))
                0
SQL> select count(unique(co_id)) from prod where (prod like upper('%What is ferro%') or prod like lower('%What is ferro%') or prod like initcap('%What is ferro%'));
COUNT(UNIQUE(CO_ID))
                0
Then I advised to try "Oracle Text". I did the following steps
create user test identified by test;
grant connect, resource, dba, ctxapp to test;
SQL> GRANT ctxapp, dba TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_CLS TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_DDL TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_DOC TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_OUTPUT TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_QUERY TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_REPORT TO test;
Grant succeeded.
SQL> GRANT EXECUTE ON CTX_THES TO test;
Grant succeeded.
SQL> connect test/test
Connected.
SQL> CREATE TABLE test (id NUMBER PRIMARY KEY, text VARCHAR2(200));
Table created.
SQL>
SQL> INSERT INTO test VALUES(1, '<HTML>California is a state in the US.</HTML>');
1 row created.
SQL> INSERT INTO test VALUES(2, '<HTML>Paris is a city in France.</HTML>');
1 row created.
SQL> INSERT INTO test VALUES(3, '<HTML>France is in Europe.</HTML>');
1 row created.
SQL>
SQL> CREATE INDEX idx_test ON test(text)
2       INDEXTYPE IS CTXSYS.CONTEXT PARAMETERS
3       ('FILTER CTXSYS.NULL_FILTER SECTION GROUP CTXSYS.HTML_SECTION_GROUP');
Index created.
SQL> SELECT SCORE(1), id, text FROM test WHERE CONTAINS(text, 'paris', 1) > 0;
SCORE(1)         ID TEXT
         4          2 <HTML>Paris is a city in France.</HTML>
SQL> SELECT SCORE(1), id, text FROM test WHERE CONTAINS(text, 'Where is Paris', 1) > 0;
no rows selected
SQL> select text from test where (upper(text) like '%where is paris%' or lower(text) like '%where is paris%' or initcap(text) like '%where is paris%');
no rows selected Still I stuck with searching technique. Please help me
Thanks & best regards

Hi,
Whenever you have a question, it helps to post:
(1) The version of Oracle (and any other relevant software) you're using
(2) A little sample data (just enough to show what the problem is) from all the relevant tables
(3) The results you want from that data
(4) Your best attempt so far (formatted)
Executable SQL statements (like "CREATE TABLE AS ..." or "INSERT ..." statements) are best for (2).
If you can present your problem using commonly available tables (for example, tables in scott schema, or views in the data dictionary), then you can omit (2).
Formatted tabular output is okay for (3).
This does what you requested in Oracle 10 (and up):
SELECT prod
FROM     prod
WHERE     REGEXP_LIKE ( ' ' || prod || ' '
              , '([^[:alnum:]]'
                || REPLACE ( 'what is chromium ore'
                     , '[^[:alnum:]])|([^[:alnum:]]'
                || '[^[:alnum:]])'
             );If the string containing all the target words ('what is chromium ore' in the example above) contains some special characters (such as '.' or '*') the results might not be what you expect.

Lightroom 5.2 my clone tool is stuck with the scissors and won't clone a spot. How do I get rid of the scissors?   HELP!!!!!

My clone tool in Lr is stuck in the scissors mode. How do I get it back to normal clone function.
Spending hours with this FRUSTRATION.
CAN SOMEONE HELP ME?

DOWNLOADED 5.7 CAN'T POST CLIP BECAUSE WHEN i LET GO OF THE ALT KEY THEY DISAPPEAR.
      Fr5.7om: Bob Somrak <[email protected]>
To: Cat Anders <[email protected]>
Sent: Sunday, February 15, 2015 5:22 PM
Subject: lightroom 5.2 my clone tool is stuck with the scissors and won't clone a spot. How do I get rid of the scissors?   HELP!!!!!
lightroom 5.2 my clone tool is stuck with the scissors and won't clone a spot. How do I get rid of the scissors?   HELP!!!!!
created by Bob Somrak in Photoshop Lightroom - View the full discussionFirst thing to do is update to LR 5.7.1.     No use trying to solve a problem on an outdated version. BTW, could you post a screen clip of these SCISSORS? If the reply above answers your question, please take a moment to mark this answer as correct by visiting: https://forums.adobe.com/message/7197183#7197183 and clicking ‘Correct’ below the answer Replies to this message go to everyone subscribed to this thread, not directly to the person who posted the message. To post a reply, either reply to this email or visit the message page: Please note that the Adobe Forums do not accept email attachments. If you want to embed an image in your message please visit the thread in the forum and click the camera icon: https://forums.adobe.com/message/7197183#7197183 To unsubscribe from this thread, please visit the message page at , click "Following" at the top right, & "Stop Following" Start a new discussion in Photoshop Lightroom by email or at Adobe Community For more information about maintaining your forum email notifications please go to https://forums.adobe.com/thread/1516624.

I can download yosemite but then it will not install? I am stuck with the icon on the dock with the downward arrow in it but when I double click on it or try to open it nothing happens. Thanks.

I can download yosemite but then it will not install? I am stuck with the icon on the dock with the downward arrow in it but when I double click on it or try to open it nothing happens. Thanks.

This is a DUPLICATE request.
:See response in https://support.mozilla.com/questions/893523#answer-271175

Hi, I was about to reset my iphone3g when I accidentally selected erase all contents and setting. It was stucked with the iphone logo, so i tried to fix it by restoring all my datas in iTunes. I was able to successfully fix it but when I inserted my sim c

hi, I was about to reset my iphone3g when I accidentally selected erase all contents and setting. It was stucked with the iphone logo, so i tried to fix it by restoring all my datas in iTunes. I was able to successfully fix it but when I inserted my sim card it says that simcard not recognized. I tried all the solutions in the net but nothing seemed to work. What should I do? I'm located here in the Philippines. I hope you could help. Thanks!

I know. I was with my brother when we did this. He didnt clearly understood the question that poped-up. Regarding the simcard, after successfully restoring it on itunes my simcard won't work. I'm not able to send and receive calls and messages.
hi, I was about to reset my iphone3g when I accidentally selected erase all contents and setting. It was stucked with the iphone logo, so i tried to fix it by restoring all my datas in iTunes. I was able to successfully fix it but when I inserted my sim c

Still stuck with the same old producer consumer weight problem need help

Hello All,
This is the problem I am stuck with right now.
I have two array lists one producer array list and one consumer array list denoted by a and b
P1 P2 P3 P4 P5
5 6 7 8 9
C1 C2 C3 C4 C5
2 3 4 5 6
Now we find all those producer consumer pairs which satisfy the criteria Pi>=Ci
We have the following sets
(5,2)(6,2)(7,2),(8,2),(9,2)
(5,3)(6,3)(7,3),(8,3),(9,3)
(5,4)(6,4)(7,4),(8,4),(9,4)
(5,5)(6,5)(7,5),(8,5),(9,5)
(6,6)(7,6)(8,6),(9,6)
Let us done each of them with Si
so we have S1,S2,S3,S4,S5
we assign a third parameter called weight to each element in Si which has satisfied the condition Pi>=Ci;
so we we will have
(5,2,ai),(6,2,bi),(7,2,ci)....etc for S1
similarly for S2 and so on.
We need to find in each set Si the the pair which has the smallest weight.
if we have (5,2,3) and (6,2,4) then 5,2,3 should be chosen.We should make sure that there is only one pair in every set which is finally chosen on the basis of weight.
Suppose we get a pair (5,2,3) in S1 and (5,2,3) in S2 we should see that (5,2,3) is not used to compare to compare with any other elements in the same set S2,
Finally we should arrive at the best element pair in each set.They should be non repeating in other sets.
Given a problem
P0 P1 P2 P3 P4
9 5 2 2 8
6 5 4 5 3
C0 C1 C2 C3 C4
we have So as (P0,C0) and (P4,C0)
assuming that the one with the smaller index has lesser weight PO is selected.In the program I have used random weights.from set S1 we select the pair PO,CO
S1 =(P0,C1),(P1,C1) and (P4,C1)
since P0 and P4 are already used in previous set we dont use them for checking in S1 so we have (P1,C1) as best.
S2=(P0,C2),(P1,C2) and (P4,C2) so we dont use P0,C2 and P1 and C2 because PO and P1 are already used in S1.
So we choose P4,C2
in S3 and S4 ae have (P0,C3),(P1,C3),(P4,C3) so we dont choose anything
and same in S4 also.
So answer is
(P0,C0),(P1,C1) and (P4,C2).
My program is trying to assign weights and I am trying to print the weights along with the sets.It doesnt work fine.I need help to write this program to do this.
Thanks.
Regards.
NP
What I have tried till now.
I have one more question could you help me with this.
I have an array list of this form.
package mypackage1;
import java.util.*;
public class DD
private int P;
private int C;
private int weight;
public void set_p(int P1)
P=P1;
public void set_c(int C1)
C=C1;
public void set_weight(int W1)
weight=W1;
public int get_p()
return P;
public int get_c()
return C;
public int get_x()
return weight;
public static void main(String args[])
ArrayList a=new ArrayList();
ArrayList min_weights_int=new ArrayList();
ArrayList rows=new ArrayList();
ArrayList temp=new ArrayList();
Hashtable h=new Hashtable();
String v;
int o=0;
DD[] d=new DD[5];
for(int i=0;i<4;i++)
d=new DD();
for(int i=0;i<4;i++)
d[i].set_p(((int)(StrictMath.random()*10 + 1)));
d[i].set_c((int)(StrictMath.random()*10 + 1));
d[i].set_weight(0);
System.out.println("Producers");
for(int i=0;i<4;i++)
System.out.println(d[i].get_p());
System.out.println("Consumers");
for(int i=0;i<4;i++)
System.out.println(d[i].get_c());
System.out.println("Weights");
for(int i=0;i<4;i++)
System.out.println(d[i].get_x());
for(int i=0;i<4;i++ )
int bi =d[i].get_c();
ArrayList row=new ArrayList();
for(int j=0;j<4;j++)
if( d[j].get_p() >=bi)
d[j].set_weight((int)(StrictMath.random()*10 + 1));
row.add("(" + bi + "," + d[j].get_p() + "," +d[j].get_x() + ")");
else
d[j].set_weight(0);
row.add("null");
rows.add(row);
System.out.println(rows);
int f=0;
for(Iterator p=rows.iterator();p.hasNext();)
temp=(ArrayList)p.next();
String S="S" +f;
h.put(S,temp);
String tt=new String();
for(int j=0;j<4;j++)
if(temp.get(j).toString() !="null")
// System.out.println("In if loop");
//System.out.println(temp.get(j).toString());
String l=temp.get(j).toString();
System.out.println(l);
//System.out.println("Comma matches" + l.lastIndexOf(","));
//System.out.println(min_weights);
f++;
for(Enumeration e=h.keys();e.hasMoreElements();)
//System.out.println("I am here");
int ii=0;
int smallest=0;
String key=(String)e.nextElement();
System.out.println("key=" + key);
temp=(ArrayList)h.get(key);
System.out.println("Array List" + temp);
for( int j=0;j<4;j++)
String l=(temp.get(j).toString());
if(l!="null")
System.out.println("l=" +l);
[\code]

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The most obvious approach is to systematically try
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In Java, you can implement the algorithm in much the same
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