Major iTunes problem! Need Help!!!

I have completely cleaned out my iTunes library and all music files on my iPod, but it still says I am using half of my iPod storage. (8.79 out of 19 GIGs) I have uninstalled, re-installed, rebooted, shut down and still every time I plug in my iPod, it reads half full. Can anyone tell me how to get my meter back to zero?

Just run the iPod updater. One of the choices is to restore it back to factory defaults. That should wipe the drive. Then you can run the updater again and install the latest version of software for your particular iPod.
I had the same problem once before, and that took care of it, lickity split!
Good luck,
Mike

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One day I was casually syncing my iPod video with my computer to get a few new songs uploaded onto it.
Suddenly, the iPod freezes up, shuts itself off, restarts, and magically it has forgotten everything and pretends it is a brand new iPod.
No problem, I think, believing it will just need to load the songs again and everything will be fine.
However, this is not the case.
The iPod would then continue to load all the songs, shut down after loading them, restart and again pretend it is brand new and lad all the songs again.
This continues to happen again and again.
I try to update my iPod, figuring it may just be a version error or somesuch nonsense. The updater refuses to work and actually causes my iPod to crash, saying it has a diskwrite error.
It refuses to "System Restore" on my computer, so I run the "System Restore" on another window computer that I own. And it runs just fine, making the iPod load songs once more.
And I think, problem solved!
However, again, this is not the case.
I load it once, everything goes fine.
The nxt day, after buying more songs, I try to upload and after the upload, the iPod shut off, restarts, and pretends to be new once more.
I go through the "System Restore" process again, but afterwards, the iPod all out refuses to upload my music fully, shutting off about halfway through the process each time.
And now whenever it is plugged into my computer, windows says it is unreadable and must be reformatted, so I run the updater and it says that there is an "iPod manager internal error" which could basically be another language considering my only basic knowledge of computers.
Is there any way to repair my iPod at home? Or must it be sent in to Apple?

GREAT NEWS!
Last night, through some very thorough digging in the
"Search Discussions", I found the answer and now my
iPod is running perfect once more!
Thank you very much for your help b noir. Without
you, and many others on this board who had their
problems solved by equally helpful people, I would
still be digging through Apple's support section to
no avail.
PS. I found the answer in http://discussions.apple.com/thread.jspa?message
ID=1539429

Major iTunes problems- please help

Okay people, here's a good one for you. On my desktop I have the latest iTunes version, software, and all that jazz. When I hook up my 60GB iPod or my 40GB iPod iTunes locks up and refuses to sync my iPod. I have had this problem for quite a while now. Anyone have any suggestions? Now I have another problem. Since I cannot use my desktop I backed up my 11GB library to 3 DVDs so I can transfer my library to my Sony Vaio laptop. On that laptop I uninstalled iTunes from the C drive and installed the latest 7.3 iTunes on the D drive so I would have enough space. I finished restoring the first disc and then put in the second. A message popped up saying there was not enough room on the C drive for all of the music. What in the ****? I installed on the D drive! So then I looked at the shortcut icon's properties it said it was located on the C drive. I then did the 'explore' thing and it said iTunes was on the D drive. So according to the laptop iTunes is in two different places. Does anybody know what the **** is going on and can you please help me?
intel core 2 duo Windows Vista
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Malwarebytes Anti-Malware (Trial) 1.62.0.1300
www.malwarebytes.org
Database version: v2012.07.25.04
Windows 7 Service Pack 1 x64 NTFS
Internet Explorer 9.0.8112.16421
[administrator
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7/25/2012 8:34:03 PM
mbam-log-2012-07-25 (20-42-53).txt
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Scan options disabled: P2P
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Time elapsed: 7 minute(s), 52 second(s)
Memory Processes Detected: 0
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Memory Modules Detected: 0
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Registry Values Detected: 0
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Registry Data Items Detected: 0
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Folders Detected: 0
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johnnnyp wrote:
hi, i have a problem with my creative vision m. i wanted to watch a movie and it froze. i tried to shut it down but it wouldnt respond, and now is frozen. please help me if you can!
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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]

In your example you selected the pair with the greatest
distance from the first set, and the pair with the least
distance from the second. I don't see how the distance
function was used.
Also it's not clear to me that there is always a solution,
and, if there is, whether consistently choosing the
furthest or the closest pairs will always work.
The most obvious approach is to systematically try
all possibilities until the answer is reached, or there
are no possibilities left. This means backtracking whenever
a point is reached where you cannot continue. In this case
backtrack one step and try another possibility at this
step. After all possible choices of the previous step,
backtrack one more step and so on.
This seems rather involved, and it probably is.
Interestingly, if you know Prolog, it is ridiculously
easy because Prolog does all the backtracking for you.
In Java, you can implement the algorithm in much the same
way as Prolog implementations do it--keep a list of all the
choice points and work through them until success or there
are none left.
If you do know Prolog, you could generate lots of random
problems and see if there is always a solution.

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GB

Major iTunes problem! Need Help!!!

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