Hi and thanks for looking!
Background
I have a computing task that requires either a lot of time, or parallel computing.
Specifically, I need to loop through a list of about 50 images, Base64 encode them, and then calculate the Levenshtein distance between each newly encoded item and values in an XML file containing about 2000 Base64 string-encoded images in order to find the string in the XML file that has the smallest Lev. Distance from the benchmark string.
A regular foreach
loop works, but is too slow so I have chosen to use PLINQ to take advantage of my Core i7 multi-core processor:
Parallel.ForEach(candidates, item => findImage(total,currentWinner,benchmark,item));
The task starts brilliantly, racing along at high speed, but then I get an "Out of Memory" exception.
I am using C#, .NET 4, Forms App.
Question
How do I tweak my PLINQ code so that I don't run out of available memory?
Update/Sample Code
Here is the method that is called to iniate the PLINQ foreach
:
private void btnGo_Click(object sender, EventArgs e)
{
XDocument doc = XDocument.Load(@"C:\Foo.xml");
var imagesNode = doc.Element("images").Elements("image"); //Each "image" node contains a Base64 encoded string.
string benchmark = tbData.Text; //A Base64 encoded string.
IEnumerable<XElement> candidates = imagesNode;
currentWinner = 1000000; //Set the "Current" low score to a million and bubble lower scores into it's place iteratively.
Parallel.ForEach(candidates, i => {
dist = Levenshtein(benchmark, i.Element("score").Value);
if (dist < currentWinner)
{
currentWinner = dist;
path = i.Element("path").Value;
}
});
}
. . .and here is the Levenshtein Distance Method:
public static int Levenshtein(string s, string t) {
int n = s.Length;
int m = t.Length;
var d = new int[n + 1, m + 1];
// Step 1
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
// Step 2
for (int i = 0; i <= n; d[i, 0] = i++)
{
}
for (int j = 0; j <= m; d[0, j] = j++)
{
}
// Step 3
for (int i = 1; i <= n; i++)
{
//Step 4
for (int j = 1; j <= m; j++)
{
// Step 5
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
// Step 6
d[i, j] = Math.Min(
Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
d[i - 1, j - 1] + cost);
}
}
// Step 7
return d[n, m];
}
Thanks in advance!