This is the third post on a small series about transferring large files with WCF using streaming :
I am closing this series of posts by sharing a skeleton for the client side handling of the uploaded and downloaded files. Please note that the following are excerpt from a larger context and so will not compile in your environment as is: they are only intended to give a rough idea of how the files can be handled.
Download
private void IssueDownloadRequest(string localFile, string serviceUrl, FileDownloadMessage request)
{
this.service = this.proxy.OpenProxy(serviceUrl);
try
{
using (FileDownloadReturnMessage response = this.service.DownloadFile(request))
{
if (response != null && response.FileByteStream != null)
{
SaveFile(response.FileByteStream, localFile);
}
}
this.proxy.CloseProxy(this.service);
}
catch (Exception e)
{
throw new FileTransferProxyException("Error while downloading the file", e);
}
finally
{
// we expect the stream returned from the server to be closed by the
// server itself so nothing to be done with it here. Just abort
// the proxy if needed.
this.proxy.AbortProxyIfNeeded(this.service);
}
}
private static void SaveFile(Stream saveFile, string localFilePath)
{
const int bufferSize = 65536; // 64K
using (FileStream outfile = new FileStream(localFilePath, FileMode.Create))
{
byte[] buffer = new byte[bufferSize];
int bytesRead = saveFile.Read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
outfile.Write(buffer, 0, bytesRead);
bytesRead = saveFile.Read(buffer, 0, bufferSize);
}
}
}
Upload
public void UploadFile(string localFileName, string serviceUrl, FileTypeEnum fileType)
{
this.service = this.proxy.OpenProxy(serviceUrl);
try
{
using (Stream fileStream = new FileStream(localFileName, FileMode.Open, FileAccess.Read))
{
var request = new FileUploadMessage();
string remoteFileName = null;
if (fileType == FileTypeEnum.Generic)
{
// we are using the service as a "FTP on WCF"
// give the remote file the same name as the local one
remoteFileName = Path.GetFileName(localFileName);
}
var fileMetadata = new FileMetaData(localFileName, remoteFileName, fileType);
request.MetaData = fileMetadata;
request.FileByteStream = fileStream;
this.service.UploadFile(request);
this.proxy.CloseProxy(this.service);
}
}
catch (IOException)
{
throw new FileTransferProxyException("Unable to open the file to upload");
}
catch (Exception e)
{
throw new FileTransferProxyException(e.Message);
}
finally
{
this.proxy.AbortProxyIfNeeded(this.service);
}
}
Project Euler Problem 9 introduces the interesting mathematical concept of Pythagorean Triple. Let’s have a look at the question:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a² + b² = c²For example, 3² + 4² = 9 + 16 = 25 = 5².
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
As usual, I first show my solution and then comment on the approach:
#light
// generate triplets using Euclid's Formula
let pythagorean_triplets top =
[ for m in 1 .. top do
for n in 1 .. m-1 do
let a = m*m-n*n
let b = 2*m*n
let c = m*m+n*n
yield [a;b;c] ]
// multiply all the values of a list
let multiply_list list =
List.fold (fun acc elem -> acc*elem) 1 list
// find a triplet where the sum of values
// is equal to a given number
let find_triplet_with_sum sum =
pythagorean_triplets sum
|> List.find (fun [a;b;c] -> a+b+c=sum)
let problem_9 () =
find_triplet_with_sum 1000
|> multiply_list
Let’s go through the solution step by step:
- The first function,
pythagorean_tripletsuses Euclid’s Formula to enumerate all possible Pytagorean Triplets up to a given threshold. The formula can be summarized like this:
a = m² – n², b = 2mn, c = m² + n², where m and n are positive integers with m > n.It’s interesting to note that each item of the list is itself a list of three numbers. I could have generated a tuple, but since in the end I need to multiply all the 3 values together, the list was more straightfoward to use.
multiply_listis just a convenience function to multiply all the elements of a list togetherfind_triplet_with_sumdoes the heavy work of generating the actual triplets picking the first one where the sum is equal to a given value.- At the end I just put the pieces together and solve the problem.
One caveat: Euclid’s Formula does not generate all Pytagorean Triplets(there are other formulas that do). I can say that I have been lucky that it generated the one requested by the Project’s Euler problem. The code above might not work if the goal is changed to find a triplet with a different product.
See all Project Euler problems solutions.
After two easy problems, it’s now time for Problem 8.
At first, this problem also seemed to me more complex than it really is. Let’s have a look at the question:
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
So, basically what we need to do is:
- consider all chunks of this list of numbers that can be created by taking 5 consecutive digits from the beginning till the end (there will be n-5 chunks if n is the number of digits in the list)
- for each chunk, multiply all the digits together (that’s n-1 products per chunk)
- pick the value corresponding to the greatest product.
At first I have considered resolving an easier problem first, like: find the chunk whose sum is bigger and then calculate that product. Assuming that sum is cheaper to calculate than product (and it is), we would have had (n-5) * (n-1) sums and just 1 product. However you can easily find examples where the sum of the digits is the same while the product is different: e.g. 4 + 5 = 7 + 2 = 9 but 4 * 5 = 20 > 7 * 2 = 14. Therefore that was a dead end.
A more mathematically sound approach to find the chunk giving the highest product might have been to first locate the chunk with the highest sum of logarithms of each digit and then calculate the corresponding product. However, a quick test revealed that logarithm is way slower than product.
Finally, a couple of micro-optimizations that occurred to me.
- You can skip all chunks containing at least one 0 (trivially, the product will be zero)
- You don’t really have to multiply all the digits for each chunk (that is (n-5) * (n-1) products): once you have computed the product of the first chunk, the product of the next chunk would be the product of the previous chunk times the sixth digit divided by the first digit, and so on for subsequent chunks.
Eventually, I did not try to implement these two things, since brute force, which I am about to show you, is so quick that it’s wasn’t really worth it:
#light
let digits = [| 7; 3; 1; 6; 7; 1; 7; 6; 5; 3; 1; 3; 3; 0; 6; 2; 4; 9;
1; 9; 2; 2; 5; 1; 1; 9; 6; 7; 4; 4; 2; 6; 5; 7; 4; 7; 4; 2; 3; 5;
5; 3; 4; 9; 1; 9; 4; 9; 3; 4; 9; 6; 9; 8; 3; 5; 2; 0; 3; 1; 2; 7;
7; 4; 5; 0; 6; 3; 2; 6; 2; 3; 9; 5; 7; 8; 3; 1; 8; 0; 1; 6; 9; 8;
4; 8; 0; 1; 8; 6; 9; 4; 7; 8; 8; 5; 1; 8; 4; 3; 8; 5; 8; 6; 1; 5;
6; 0; 7; 8; 9; 1; 1; 2; 9; 4; 9; 4; 9; 5; 4; 5; 9; 5; 0; 1; 7; 3;
7; 9; 5; 8; 3; 3; 1; 9; 5; 2; 8; 5; 3; 2; 0; 8; 8; 0; 5; 5; 1; 1;
1; 2; 5; 4; 0; 6; 9; 8; 7; 4; 7; 1; 5; 8; 5; 2; 3; 8; 6; 3; 0; 5;
0; 7; 1; 5; 6; 9; 3; 2; 9; 0; 9; 6; 3; 2; 9; 5; 2; 2; 7; 4; 4; 3;
0; 4; 3; 5; 5; 7; 6; 6; 8; 9; 6; 6; 4; 8; 9; 5; 0; 4; 4; 5; 2; 4;
4; 5; 2; 3; 1; 6; 1; 7; 3; 1; 8; 5; 6; 4; 0; 3; 0; 9; 8; 7; 1; 1;
1; 2; 1; 7; 2; 2; 3; 8; 3; 1; 1; 3; 6; 2; 2; 2; 9; 8; 9; 3; 4; 2;
3; 3; 8; 0; 3; 0; 8; 1; 3; 5; 3; 3; 6; 2; 7; 6; 6; 1; 4; 2; 8; 2;
8; 0; 6; 4; 4; 4; 4; 8; 6; 6; 4; 5; 2; 3; 8; 7; 4; 9; 3; 0; 3; 5;
8; 9; 0; 7; 2; 9; 6; 2; 9; 0; 4; 9; 1; 5; 6; 0; 4; 4; 0; 7; 7; 2;
3; 9; 0; 7; 1; 3; 8; 1; 0; 5; 1; 5; 8; 5; 9; 3; 0; 7; 9; 6; 0; 8;
6; 6; 7; 0; 1; 7; 2; 4; 2; 7; 1; 2; 1; 8; 8; 3; 9; 9; 8; 7; 9; 7;
9; 0; 8; 7; 9; 2; 2; 7; 4; 9; 2; 1; 9; 0; 1; 6; 9; 9; 7; 2; 0; 8;
8; 8; 0; 9; 3; 7; 7; 6; 6; 5; 7; 2; 7; 3; 3; 3; 0; 0; 1; 0; 5; 3;
3; 6; 7; 8; 8; 1; 2; 2; 0; 2; 3; 5; 4; 2; 1; 8; 0; 9; 7; 5; 1; 2;
5; 4; 5; 4; 0; 5; 9; 4; 7; 5; 2; 2; 4; 3; 5; 2; 5; 8; 4; 9; 0; 7;
7; 1; 1; 6; 7; 0; 5; 5; 6; 0; 1; 3; 6; 0; 4; 8; 3; 9; 5; 8; 6; 4;
4; 6; 7; 0; 6; 3; 2; 4; 4; 1; 5; 7; 2; 2; 1; 5; 5; 3; 9; 7; 5; 3;
6; 9; 7; 8; 1; 7; 9; 7; 7; 8; 4; 6; 1; 7; 4; 0; 6; 4; 9; 5; 5; 1;
4; 9; 2; 9; 0; 8; 6; 2; 5; 6; 9; 3; 2; 1; 9; 7; 8; 4; 6; 8; 6; 2;
2; 4; 8; 2; 8; 3; 9; 7; 2; 2; 4; 1; 3; 7; 5; 6; 5; 7; 0; 5; 6; 0;
5; 7; 4; 9; 0; 2; 6; 1; 4; 0; 7; 9; 7; 2; 9; 6; 8; 6; 5; 2; 4; 1;
4; 5; 3; 5; 1; 0; 0; 4; 7; 4; 8; 2; 1; 6; 6; 3; 7; 0; 4; 8; 4; 4;
0; 3; 1; 9; 9; 8; 9; 0; 0; 0; 8; 8; 9; 5; 2; 4; 3; 4; 5; 0; 6; 5;
8; 5; 4; 1; 2; 2; 7; 5; 8; 8; 6; 6; 6; 8; 8; 1; 1; 6; 4; 2; 7; 1;
7; 1; 4; 7; 9; 9; 2; 4; 4; 4; 2; 9; 2; 8; 2; 3; 0; 8; 6; 3; 4; 6;
5; 6; 7; 4; 8; 1; 3; 9; 1; 9; 1; 2; 3; 1; 6; 2; 8; 2; 4; 5; 8; 6;
1; 7; 8; 6; 6; 4; 5; 8; 3; 5; 9; 1; 2; 4; 5; 6; 6; 5; 2; 9; 4; 7;
6; 5; 4; 5; 6; 8; 2; 8; 4; 8; 9; 1; 2; 8; 8; 3; 1; 4; 2; 6; 0; 7;
6; 9; 0; 0; 4; 2; 2; 4; 2; 1; 9; 0; 2; 2; 6; 7; 1; 0; 5; 5; 6; 2;
6; 3; 2; 1; 1; 1; 1; 1; 0; 9; 3; 7; 0; 5; 4; 4; 2; 1; 7; 5; 0; 6;
9; 4; 1; 6; 5; 8; 9; 6; 0; 4; 0; 8; 0; 7; 1; 9; 8; 4; 0; 3; 8; 5;
0; 9; 6; 2; 4; 5; 5; 4; 4; 4; 3; 6; 2; 9; 8; 1; 2; 3; 0; 9; 8; 7;
8; 7; 9; 9; 2; 7; 2; 4; 4; 2; 8; 4; 9; 0; 9; 1; 8; 8; 8; 4; 5; 8;
0; 1; 5; 6; 1; 6; 6; 0; 9; 7; 9; 1; 9; 1; 3; 3; 8; 7; 5; 4; 9; 9;
2; 0; 0; 5; 2; 4; 0; 6; 3; 6; 8; 9; 9; 1; 2; 5; 6; 0; 7; 1; 7; 6;
0; 6; 0; 5; 8; 8; 6; 1; 1; 6; 4; 6; 7; 1; 0; 9; 4; 0; 5; 0; 7; 7;
5; 4; 1; 0; 0; 2; 2; 5; 6; 9; 8; 3; 1; 5; 5; 2; 0; 0; 0; 5; 5; 9;
3; 5; 7; 2; 9; 7; 2; 5; 7; 1; 6; 3; 6; 2; 6; 9; 5; 6; 1; 8; 8; 2;
6; 7; 0; 4; 2; 8; 2; 5; 2; 4; 8; 3; 6; 0; 0; 8; 2; 3; 2; 5; 7; 5;
3; 0; 4; 2; 0; 7; 5; 2; 9; 6; 3; 4; 5; 0|]
let calculate_max_chunk_product chunk_size =
seq {
for i in 0 .. (Array.length digits - chunk_size) do
let chunk = Array.sub digits i chunk_size
yield Array.fold (fun acc elem -> acc * elem) 1 chunk
}
|> Seq.max
let problem_8 () =
calculate_max_chunk_product 5
As you can see, we are generating a sequence of products and then pick the maximum value. Each product is created by first taking a chunk of the array (using the Array.sub) and then using Array.fold to compute the product of all elements in the chunk.
Again, I have tried to keep the imperative part to a minimum and I was able to get by without using any mutable variable.
See all Project Euler problems solutions.
