Location Cube

Raster access speeds… and size limits

2011-09-08T08:06:00.001-07:00

I need to be able to handle large raster images, and to be able to find values from cells very quickly.

There are many ways to do this…. here’s a look at a few:

Method	Handles GeoCoords	Max Size	Read Access Speed
DOTSPATIAL Raster	Yes	Medium (9k x 9k)	2.6 million per second
C# dictionary	No	Small (5k x 5k)	50 million per second
EMGU IMGs	No	Large (30k x 30k)	3.2 million per second
C# BMP (.net)	No	Medium (9k x 9k)	1 million per second
C# FastBMP	No	Medium (9k x 9k)	33 million per second
Rasterlite	Yes	Large (internal tiling)	(to do)
GDAL	Yes	Large	(to do)
R (grids)	(to do)	(to do)	(to do)
XNA texture	No	Small (4k x 4k)	(to do)

DotSpatial

2011-09-06T09:43:00.001-07:00

So I’m checking out DotSpatial for use in a project.. and the first thing on my list is to check it can convert coordinate systems easily from WGS84 <> OSGB36.

Seems the current version’s inbuilt definition of OSGB36 isn’t correct.. and when using it with the Reproject method gives out some strange results. It’s also listed as failing the unit test currently.

However you can specify the details yourself.. and then it seems to work just fine… so here’s the code for taking a coordinate as OSGB36 and transforming it to WGS84…

PointD pd = new PointD();
pd.X = 326097.07;
pd.Y = 673630.932;

double [] pts_xy = new double[2];
pts_xy[0] = pd.X;
pts_xy[1] = pd.Y;

double [] pts_z = new double[1];
pts_z[0] = 0;

var pStart = new ProjectionInfo("+proj=tmerc +lat_0=49 +lon_0=-2 +k=0.9996012717 +x_0=400000 +y_0=-100000 +ellps=airy +towgs84=446.448,-125.157,542.060,0.1502,0.2470,0.8421,-20.4894 +units=m +no_defs");
var pEnd = KnownCoordinateSystems.Geographic.World.WGS1984;

Reproject.ReprojectPoints(pts_xy, pts_z, pStart, pEnd, 0, 1);
System.Console.WriteLine(pts_xy[0] + " " + pts_xy[1]);

WGS84 to OSGB36

2011-08-18T10:07:00.001-07:00

A transformation between WGS84 and OSBB36 requires a datum transformation. A 7 parameter transformation should give results accurate to about +/- 5m…

ArcGIS supports a number of 3 para, and 7 para transforms. CS2CS using Proj may also be used as follows :

cs2cs +proj=latlong +ellps=WGS84 +towgs84=0,0,0 +nodefs +to +proj=tmerc
+lat_0=49 +lon_0=-2 +k=0.9996012717 +x_0=400000 +y_0=-100000 +ellps=airy
+towgs84=446.448,-125.157,542.060,0.1502,0.2470,0.8421,-20.4894 +units=m
+no_defs

(thanks to Paul from http://osgeo-org.1803224.n2.nabble.com/OSGB36-td2062542.html for settings)

However for a centimetre accurate transformation an NTv2 grid is required. The OS provide an online coordinate transformation service here (a site for bulk conversions is also available)

http://gps.ordnancesurvey.co.uk/etrs89geo_natgrid.asp

It’s also possible to download the grid in NTv2 format from here:

http://www.ordnancesurvey.co.uk/oswebsite/gps/osnetfreeservices/furtherinfo/ostn02_ntv2.html

or directly from this link http://www.ordnancesurvey.co.uk/oswebsite/gps/docs/OSTN02_NTv2_DataFiles.zip

Test for a point in Edinburgh (Scotland)

WGS84 –3.185, 55.95
OS Online NTv2 Conversion > 326097.07,673630.932 [taken to be correct conversion]
ESRI (3 par option_1) > 326097.86,673625.357
ESRI (3 par option_5) > 326095.042,673619.452
ESRI(petroleum option)> 326096.974,673630.405
CS2CS (not using NTv2)> 326096.98,673630.4

So it would appear that ESRI’s petroleum option and CS2CS (as above without NTv2) and the OS Online conversion tool all agree to within 1 metre, while the 3 parameter version are out by at least 5m. The options relate to which parts of the country the re-projection has been set up for… this test would suggest option_1 works better for Scotland.

SpaceBook Project

2011-08-12T07:53:00.001-07:00

So we’ve arrived in Edinburgh and I’ve started working with the SB team… William’s away and kindly allowing me to use his office for the moment.

Useful Data sources (thanks to Guy):

MasterMap – building outlines from OS (downloaded thru EDINA)
LiDAR and 3D building (from DSM-DTM) polygons from http://landmap.ac.uk/

(The 3D buildings from Landmap SHP source)

GRASS – new viewshed function (ie not r.los)

r.viewshed (addon) – using faster algo (sweep method)

Install method for Windows summary – download GRASS source (all).. add r.viewshed source folders… change the MAKEFILE to incl r.viewshed – compile GRASS as per this page:

http://trac.osgeo.org/grass/wiki/CompileOnWindows

Using msys

/mswindows/osgeo4w/package.sh
the package error log goes into this folder:
\\OSGeo4W\usr\src\grass641\mswindows\osgeo4w as package.log

Manually signing an Android App

2011-06-23T20:02:00.001-07:00

1) Open CMD and then go to c:\Program Files\Java\jdk1.6.0_25\bin>
2) Setting up the keystore
keytool -genkey -v -keystore c:\android\manualkeystoredem
o -alias demo123 -keyalg RSA –keysize 2048 -validity 10000
enter a password (and again)
enter your name details (optional)
enter y to keep and progress…
re-renter your password

3) Export unsigned APK from Eclipse then sign it…
Export by Right Click > Android Tools >

jarsigner -verbose -keystore c:\android\manualkeystoregeosensorapp d:\geosensor.apk geosensor

4) check it’s signed OK by using jarsigner verify
jarsigner –verify –verbose d:\geosensor.apk

5) Zip align your code…
Go to Andriod-sdk-windows-tools

(It always uses an int value of 4 for byte ordering - this isn't API level etc)
zipalign -v 4 your_project_name-unaligned.apk your_project_name.apk

eg zipalign -v 4 d:\GeoSensorLogger.apk d:\GeoSensorLogger_signed.apk

Put the apk file on your phone.. or marketplace…

FB API Tips and Tricks

2011-05-24T18:16:00.001-07:00

Play around with GRAPH API and FQL API

https://www.simoncross.com/fb/graph

Also some useful videos here…

http://developers.facebook.com/videos/

FQL overview here:

http://developers.facebook.com/docs/reference/fql/

Eclipse IDE tips for Android Development

2011-05-17T16:27:00.001-07:00

Perspectives (top right of screen)

JAVA
DEBUG
DDMS – device info, processes, fake GPS (GPX,KML etc), fake tel info

- file explorer, memory allocation, threads,
logcat (logging output)

_______________________

Window> Andriod SDK AVD manager – setup emulators

_______________________

CTRL+SHIFT+T = open type (eg MyClass)
CTRL + SHIFT + R = open resource (eg main.xml)
CTRL+O – overview of methods in class.. can search with wildcards * too

SHIFT+ALT+R – rename a type and all xrefs to it
CTRL+SHIFT+O = organise the imports (eg add / remove android.os.Bundle)

Also if you add:

Button b;

then hit CTRL+SHIFT+O it’ll add import android.widget.Button;

Setting up Android Development tools on Windows

2011-05-17T02:21:00.001-07:00

Ok so here’s the quick how to:

1) Download and install Eclipse from here

http://www.eclipse.org/downloads/

Choose the Eclipse IDE for Java Developers 32 bit version
(64 bit version has issues with Android plugins at time of writing)

Just download.. unzip and put into a new folder on your PC.. say Program Files..

2) Download and install the Java JDK from here:

http://www.oracle.com/technetwork/java/javase/downloads/index.html

(again 32 bit option - x86)

Go with default install location is fine here..

3) Install the Android SDK from here:

http://developer.android.com/sdk/index.html

Choose the Windows Installer version as it is a lot easier in this case, as it pulls down various plug in options later in the steps.

If you get an error saying the JDK couldn’t be found then press the BACK button on the install dialogue box.. then push NEXT again.. this time it finds it.. weird install bug.

Change the install path!!! Take out the spaces otherwise you’ll get problems later on.. so make it start with c:\android\

Accept all the options and wait for download/install…

4) Next we need to configure the Eclipse ADT plug in…

Load Eclipse then go to the Help menu.. Choose Install new software

In the Work With box enter>>

https://dl-ssl.google.com/android/eclipse

Give it the Name: Android
Click OK and wait…

(If you have problems check this link which has details on a ZIP download alternative >> http://developer.android.com/sdk/eclipse-adt.html )

Select all, accept licences.. set it going an wait for it to complete.

Restart Eclipse…

Now go to Window > Preference
Choose Andriod > Specify your Android install location (eg C:\Android\AndroidSDK…)

Euler problem 41

2010-12-29T23:34:00.001-08:00

Last one for today… this isn’t fast code but works…

A bit of a speed up implemented by only checking those which are of length equal to their max character… so there’s not point checking 17 as it’s 2 in length.. so max permitted would be 12… however 132 is fine as it’s 3 in length and has a 3 in it!

_______________________

static void Main(string[] args)
{
long max=0;

            for (long i = 1; i < 987654321; i++)
            {
                int len = i.ToString().Length;

               //find max char
                int maxchar = 0;
                foreach (char c in i.ToString())
                {
                    if (c - 48 > maxchar)
                    {
                        maxchar = c - 48;
                    }
                }

//only check if this could be pandigital - by checking max char and len
//eg1 17 can't be pandigital as only 2 chars long (eg2 12 =OK, 13=not)

                string s = i.ToString();
                if (maxchar == len && isprime(i))
                {
                 if (s.Split('1').Length==2 && s.Split('2').Length==2 )
                 {
                    if (i.ToString().Length==2 && i > max) {max=i;}

                     if (s.Split('3').Length==2)
                     {
                         if (i.ToString().Length==3 && i > max) {max=i;}
                         if (s.Split('4').Length==2)
                         {
                             if (i.ToString().Length==4 && i > max) {max=i;}
                              if (s.Split('5').Length==2)
                              {
                                  if (i.ToString().Length==5 && i > max) {max=i;}
                                  if (s.Split('6').Length==2)
                                  {
                                      if (i.ToString().Length==6 && i > max) {max=i;}
                                       if (s.Split('7').Length==2)
                                       {
                                           if (i.ToString().Length==7 && i > max) {max=i;}
                                           if (s.Split('8').Length == 2)
                                           {
                                               if (i.ToString().Length == 8 && i > max) { max = i; }
                                               if (s.Split('9').Length == 2)
                                               {
                                                   if (i.ToString().Length == 9 && i > max) { max = i; }
                                                   break;
                                               }
                                           }
                                       }
                                  }
                              }
                         }
                     }
                 }
                }
            }
            Console.WriteLine(max);
            Console.ReadLine();

}

static bool isprime(long n)
{
if (n == 2 || n == 3 || n == 5) { return true; }

          if (n%2==0) {return false;}
          if (n%3 == 0) { return false;}
          if (n%5 == 0) { return false;}

          for (int t = 7; t < Math.Sqrt(n); t++)
          {
              if (n % t== 0)
              {
                  return false;
              }
          }

return true;

}

Euler Problem 40

2010-12-29T21:11:00.001-08:00

Brute force attack!!!

Keep making the string longer until you hit 1 million chars long.. then do the calc they want.. rem first place is s[0]…. and you’re sorted!

____

static void Main(string[] args)
{
    string s = "";
    int i=0;
    while (s.Length<1000001)
   {
       i++;
        s += i.ToString();
    }

    int sum = (Convert.ToInt32(s[0])-48) * (Convert.ToInt32(s[9])-48) * (Convert.ToInt32(s[99])-48) * (Convert.ToInt32(s[999])-48) * (Convert.ToInt32(s[9999])-48) * (Convert.ToInt32(s[99999])-48) * (Convert.ToInt32(s[999999])-48);


    Console.WriteLine("Answer>>"+sum);
    Console.ReadLine();

}

Euler Problem 39

2010-12-29T20:41:00.001-08:00

An easier one… some geometry and Pythog theroem..

I just made a big array, used that as a counter for each solution.. then scanned it to find the answer.

Just remember that the counter is not for c(the hyp) but for the perimeter length (p=a+b+c)…. !

____________

static void Main(string[] args)
{
int[] cc = new int[1001]; //somewhere to store the counts

           for (int a = 1; a < 999; a++)
           {
               for (int b = 1; b < 999; b++)
               {
                   double c = Math.Sqrt(a * a + b * b);

                   if (c%1==0 && (a+b+c <= 1000)) //must be int length sides and per<=1000
                   {

                       cc[Convert.ToInt32(a+b+c)]++;
                   }
               }
           }
           //now find largest value
           int maxp = 0;
           int maxn = 0;
           for (int i = 0; i < 1000; i++)
           {
               if (cc[i] > maxn)
               {
                   maxn = cc[i];
                   maxp = i;
               }
           }

Console.WriteLine("Answer>>"+maxp + " count of "+maxn);
Console.ReadLine();

}

Euler Problem 38

2010-12-29T19:54:00.001-08:00

So here you need to do some pandigital checks again like problem 32…

We know the answer must be bigger than that given in the example.. as I tried putting their example answer in to see if it was the answer they were after.. but no…! Anyway this means the final value is somewhere between 91873654 and 987654321… so that’s a start.

I was thinking the number of parts to split up the 9 digit number into may be 2,3,4.. as in 123,456789 would be 2 parts.. but to be a correct value then part 1 x 2 should equal part 2… (not here obviously as 123 * 2 = 246)…

My plan was to try this with 2 parts.. and if it didn’t work then look at 3 parts.. and so on.. but I ran it with 2 parts and got the answer so didn’t need to go any further than this!!

Not the smartest, or most elegant solution… but it did the job..

____________________

static void Main(string[] args)
{
//must be bigger than 918273654 as it's mentioned in question
//and I tested it as the answer and it wasn't..
for (int i = 987654321; i > 918273645; i--)
{
string s = i.ToString();

//contains 1-9 once - pandigital 9
if (
s.Split('1').Length == 2 &&
s.Split('2').Length == 2 &&
s.Split('3').Length == 2 &&
s.Split('4').Length == 2 &&
s.Split('5').Length == 2 &&
s.Split('6').Length == 2 &&
s.Split('7').Length == 2 &&
s.Split('8').Length == 2 &&
s.Split('9').Length == 2)
{

//check case of 2 parts
for (int n=1;n<9;n++)
{
int part1 = Convert.ToInt32(i.ToString().Substring(0,n));
int part2= Convert.ToInt32(i.ToString().Substring(n,i.ToString().Length-n));

if (part2==part1*2)
{
Console.WriteLine(i);
goto foo;
}

}

}//end of pandigital check
}

foo:
Console.WriteLine("Done!");
Console.ReadLine();

}

Euler Problem 37

2010-12-29T16:26:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=37

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

______

so the first thing I did wrong was to consider 1 to be a prime.. once I’d sorted that out I got this list.. which is 11 primes… so those must be what they’re after!

23
37
53
73
313
317
373
797
3137
3797
739397

_________________

static void Main(string[] args)
       {
           long sum = 0;
           int found = 0;
           long i = 11;
           while (found<11)
           {
              int c=0;
              i++;
              int l = i.ToString().Length;

for (int x = 0; x < l; x++)
{

long t1 = Convert.ToInt64(i.ToString().Substring(x, l - x));
long t2 = Convert.ToInt64(i.ToString().Substring(0, x+1));

                  if (isprime(t1) && isprime(t2) && t1 != 1 && t2 != 1 )
                  {
                      c++;
                  }
                  else
                  {
                      c = 0;
                  }

              }

               if (c == i.ToString().Length)
                   {
                       sum += i;
                       found++;
                       Console.WriteLine(i);
                   }

}

Console.WriteLine("Answer>>>"+sum);
Console.ReadLine();

}

static bool isprime(long n)
{
if (n == 2 || n == 3 || n == 5) { return true; }

          if (n%2==0) {return false;}
          if (n%3 == 0) { return false;}
          if (n%5 == 0) { return false;}

          for (int t = 7; t < Math.Sqrt(n); t++)
          {
              if (n % t== 0)
              {
                  return false;
              }
          }

return true;

}

Euler Problem 36

2010-12-29T02:29:00.001-08:00

Just count from 1 to n, find equivalent binary, check both base 10 and base 2 values are palindromic.. job done!

Interestingly in .NET to turn binary into base 10 then just do this!

Convert.ToInt32("1001",2)

1001 ==>> 9

________________

static void Main(string[] args)
       {
           int counter = 0;
           int sum = 0;

           for (int i = 1; i < 1000000; i++)
           {
               string binary = IntToString(i, new char[] { '0', '1' });

               if (palindromic(i.ToString()) && palindromic(binary))
               {
                   sum += i;
               }

}

Console.WriteLine(sum);
Console.ReadLine();

}

       static bool palindromic(string s)
       {
           string rev="";
           for (int i = s.Length;i>0;i--)
           {
               rev += s.Substring(i-1,1);
           }

           if (s == rev)
           {
               return true;
           }

return false;

}

Euler Problem 35

2010-12-28T22:57:00.001-08:00

Fairly straight forward this one…

________________

static void Main(string[] args)
{
int counter = 0;

            for (long i = 2; i < 1000000; i++)
            {
                string s = i.ToString();
                int l= s.ToString().Length;

int c = 0;

                for (int x = 0; x <l; x++)
                {
                    string sr = s.Substring(x, l - x) + s.Substring(0,x);
                    if (!isprime(Convert.ToInt32(sr)))
                    {
                        break;
                    }
                    else
                    {
                        c++;
                    }

}

if (c==l) {
counter++;}

}

Console.WriteLine(counter);
Console.ReadLine();

}

static bool isprime(int n)
{
if (n == 2 || n == 3 || n == 5) { return true; }

          if (n%2==0) {return false;}
          if (n%3 == 0) { return false;}
          if (n%5 == 0) { return false;}

          for (int t = 7; t < Math.Sqrt(n); t++)
          {
              if (n % t== 0)
              {
                  return false;
              }
          }

return true;

}

Euler Problem 34

2010-12-28T19:51:00.001-08:00

This was an easy one.. at last!

http://projecteuler.net/index.php?section=problems&id=34

______________________

static void Main(string[] args)
{

long bigsum = 0;

           for (long i = 3; i < 1000000; i++)
           {
               string s = i.ToString();
               long sum = 0;
               foreach (char c in s) //split up to chars
               {
                  sum+= factorial(c-48);   //convert from ASCII to numerical value of char (c)
                  if (sum > i) { break; }
               }

if (sum == i) {bigsum += sum;
}

}

Console.WriteLine(bigsum);
Console.ReadLine();

}

static long factorial(long n)
        {
            long tot = 1;
            for (long i = 1; i < n+1; i++)
            {
                tot *= i;

            }
            return tot;
        }

Euler Problem 33

2010-12-28T19:40:00.001-08:00

I found this puzzle quite tricky to understand… perhaps the question could be written more clearly…

Essentially you just need to find all the fractions which have no more than 2 digits in top/bottom (ie <99/99), which are equal to <1, and which have the same number in the top and bottom which when that number is removed from the fraction the result is the same as the full fraction… anything using factors of 10 is ignored.

So 10/20 = 1/2 (remove 0 from top and base) – but this simple example is ignored.. they only want more complex ones like…

16/64 = 0.25 … remove those common 6s and you’d have 1/4 which is also 0.25 so this what they’re after…! In fact you now only need find the other 3!

I did it by using MOD10 to get the last number, and (Convert.ToInt16(t/10)) to get the first… then just compare these to find when numbers match… could be done using string conversions otherwise..

_______________________

static void Main(string[] args)
{
double sum = 1;

for (int b = 1; b < 100; b++) //t/b has to be <1
{
for (int t = 1; t < b; t++) //only check up to 99/98
{
double f= Convert.ToDouble(t) / Convert.ToSingle(b);
double i= Convert.ToDouble(Convert.ToInt16(t/10));
double j= Convert.ToDouble(Convert.ToInt16(b/10));
double x = 0;

//skip easy ones that divide by 10 top and bottom
if ( (t%10!=0 && b%10!=0))
{
//(i == j || i == b % 10 || t % 10 == j || t % 10 == b % 10) )

if (i == j) { x = (t % 10) / (b % 10); }
if (i == b%10) { x = (t % 10) / j; }
if (t%10 == j) { x = i / (b % 10); }
if (t%10 == b%10) { x = i / j; }

if (x == f)
{
sum *= f;
Console.WriteLine(t + "/" + b);
}

}
}
}

Console.WriteLine("Answer>>" + 1/sum);
Console.ReadLine();

}

Euler Problem 32

2010-12-27T14:17:00.001-08:00

Been on hols… so mini break from this.. but had the chance to do one more…

Here we need to find the sum of all products which use numbers 1-9 ONCE in being formed from 2 multipliers and the product itself…

e.g. 4 x 1738 = 6952 => uses 1,2,3,4,5,6,7,8,9

Note we should only include each product (eg 6952) once in the final sum… so I used a list for this.

Other tips – the product and multipliers can be formed into a string.. whose length must be 9.. anything else would mean a double or missing value.

Also we only need test up to 9876 (we could do 9999 but it’d be pointless as there are repeated values there already!)

_________________________

static void Main(string[] args)
{

BigInteger sum = 0;
List<long> found = new List<long>();

           for (long a = 1; a < 9876; a++)
           {
               for (long b = 1; b < 9876; b++)
               {
                   long c = a * b;
                   if (contains1to9once(a, b, c))
                   {
                       if (!found.Contains(c))
                       {
                           sum += c;
                           found.Add(c);
                       }
                   }
               }
           }
           Console.WriteLine(sum);
           Console.ReadLine();

}

       static bool contains1to9once(long a, long b, long c)
       {
           string s = a.ToString() + b.ToString() + c.ToString();

           //has to be 9 chars long only with one of each
           if(s.Length==9 &&
           s.Split('1').Length==2 &&
           s.Split('2').Length==2 &&
           s.Split('3').Length==2 &&
           s.Split('4').Length==2 &&
           s.Split('5').Length==2 &&
           s.Split('6').Length==2 &&
           s.Split('7').Length==2 &&
           s.Split('8').Length==2 &&
           s.Split('9').Length==2)
           {

               return true;
           }
           return false;

}

Euler Problem 31

2010-12-19T02:40:00.001-08:00

How many ways using UK coins (1,2,5,10,20,50,100,200) can you make 200p?

I just used loops for each coin type.. not very fast but does the trick..

______

static void Main(string[] args)
{
int c = 1; //2 dollar coin = one way

            for (int p1 = 0; p1 < 201; p1++)
            {
                for (int p2 = 0; p2 < 201; p2 += 2)
                {
                    for (int p5 = 0; p5 < 201; p5 += 5)
                    {
                        for (int p10 = 0; p10 < 201; p10 += 10)
                        {
                            for (int p20 = 0; p20 < 201; p20 += 20)
                            {
                                for (int p50 = 0; p50 < 201; p50 += 50)
                                {
                                    for (int p100 = 0; p100 < 201; p100 += 100)
                                    {
                                        if (p1 + +p2 + p5 + p10 + p20 + p50 + p100 == 200)
                                        {
                                            c++;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
            Console.WriteLine(c);

Console.ReadLine();

}

Euler Problem 30

2010-12-19T00:57:00.001-08:00

Iterate.. split up to sep characters… power of 5 to each.. sum… is it equal to the current iteration variable.. if so.. add to the big sum.. repeat…

http://projecteuler.net/index.php?section=problems&id=30

____________

static void Main(string[] args)
{
int bigsum = 0;

for (int i = 2; i < 10000000; i++)
{

string s = i.ToString();
int sum = 0;

                for (int x=0;x<s.Length;x++)
                {
                    int c = Convert.ToInt32( s.Substring(x, 1));

sum += Convert.ToInt32(Math.Pow(c,5));
}

                if (sum == i)
                {
                    Console.WriteLine(i);
                    bigsum += sum;
                }

}

Console.WriteLine(">> "+bigsum);
Console.ReadLine();

}

Euler Problem 29

2010-12-18T16:07:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=29

This is easy … IF… you can handle big numbers…. which is possible in .Net4.. or.. using a BigInteger class..

I used the CodeProject class from Chew Keong TAN

http://www.codeproject.com/KB/cs/biginteger.aspx

then ran the simple iteration as follows…

__________________

static void Main(string[] args)
       {

           List<BigInteger> ans = new List<BigInteger>();

           for (int a = 2; a < 101; a++)
           {
               for (int b = 2; b < 101; b++)
               {

BigInteger bg = pow(a, b);

                   if (!ans.Contains(bg))
                   {
                       ans.Add(bg);
                   }

}

Console.WriteLine(ans.Count);

Console.ReadLine();

}

       static BigInteger pow(int a, int b)
       {
           BigInteger res = 1;

           for (int t = 1; t < b+1; t++)
           {
               res *= a;
           }

return res;
}

Euler Problem 28

2010-12-18T15:40:00.001-08:00

Hmmm.. so a spiral… going out from the centre…

Well this isn’t the smoothest solution but I noticed the following from the example given:

21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13

the numbers from the 4 diagonals are:

1,9,25 1,3,13 1,5,17 1,7,21

So I added another lap around the square and found the intervals between numbers were:

1,9,25,49     1,3,13,31      1,5,17,37    1,7,21,43
8 16   24          2 10 18        4 12 20      6 14 22
   8   8               8   8              8   8            8   8

Hmmmm.. so.. the first diagonal (1,9,25) just goes up in multiples of 8… and so do the others but from different starting points.

So I just made an app which cycled thru building the numbers for each diagonal and adding to a list.. then pull them in order from the list and calc the sum! The only thing to remember is that in the sum you should only count the 1 in the centre a single time.. so I just did a –3… found this out from testing it against their example of a 5x5 square.

Oh and the other thing is when to stop adding… so I just determined this by going until the max value in the first diagonal reached the square of the size…

Like I said.. a bit messy.. but got me the answer first time!

________

static void Main(string[] args)
       {
           int size= 1001;
           long d1 = 1;
           long d2 = 3;
           long d3 = 5;
           long d4 = 7;
           int i = 0;
           long offsetd2 = 2;
           long offsetd3 = 4;
           long offsetd4 = 6;

           List<long> l1 = new List<long>();
           List<long> l2 = new List<long>();
           List<long> l3 = new List<long>();
           List<long> l4 = new List<long>();

           l2.Add(1);
           l2.Add(3);
           l3.Add(1);
           l3.Add(5);
           l4.Add(1);
           l4.Add(7);

           while (d1<((size*size)+1))
           {
               offsetd2 += 8;
               offsetd3 += 8;
               offsetd4 += 8;

               d1 += 8 * i;
               d2 += offsetd2;
               d3 += offsetd3;
               d4 += offsetd4;

               l1.Add(d1);
               l2.Add(d2);
               l3.Add(d3);
               l4.Add(d4);

           i++;
           }

//sort out starting positions

long sum = 0;

           for (int w = 0; w < (i-1); w++)
           {
               sum += l1[w] + l2[w] + l3[w] + l4[w];
           }

Console.WriteLine(sum-3); //1 is only counted on one of the diagonals so -3 from total

Console.ReadLine();

}

Euler Problem 27

2010-12-18T14:10:00.001-08:00

Requires… a prime number solver… done that earlier.. so.. actually quite an easy problem..

static void Main(string[] args)
{

     // n² + an + b,
     int max = 0;
     int maxx = 0;

for (int a = -1000; a < 1000; a++)
{

         for (int b = -1000; b < 1000; b++)
         {
             max = 0;

             for (int n = 0; n < 500; n++) //number of primes in a row
             {
                int p= (n * n) + (a * n) + b;

                 if (isprime(p) && p> 1)
                 {
                     max++;

                     if (max > maxx)
                     {
                         maxx = max;

                         Console.WriteLine(a + "," + b + "," + maxx);
                     }
                 }
                 else
                 {
                     max = 0;
                     break;

                 }
             }
         }

}

Console.ReadLine();

}

Euler Problem 9

2010-12-18T13:59:00.001-08:00

Did this a while ago but didn’t post… so here it is…

static void Main(string[] args)
{

           for (int a = 1; a < 100000; a++)
           {
               for (int b = a; b < 100000; b++)
               {

                   double c = Math.Sqrt((a*a)+(b*b));
                   if (a + b + c == 1000)
                   {
                       Console.WriteLine(a + "," + b + "," + c);
                       Console.WriteLine(a * b * c);
                   }

}

Euler Problem 26

2010-12-17T13:35:00.001-08:00

Difficult!!

http://projecteuler.net/index.php?section=problems&id=26

Needed help with this one.. from here … thanks!!

http://universequeen.org/archives/133

____________________

static void Main(string[] args)
        {

            int max=0;
            int max_d=0;

            for (int d=2;d<1000;d++)
            {
                 double s = 1f/Convert.ToDouble(d);

                int l = reclencycle(d);

                if (l> max)
                {
                    max=l;
                    max_d=d;
                }

}

Console.WriteLine(max_d);
Console.ReadLine();

}

        static int reclencycle(int n)
        {

        int[] q = new int[1000];
        int[] r = new int[1000];

        r[0] = 1;
        q[0] = 0;

        for( int i=1; i<1000; i++ )
        {
            q[i] = r[i-1]*10/n;
            r[i] = r[i-1]*10 - q[i]*n;

            for( int j=1; j<i; j++ )
            {
                if( q[j] == q[i] && r[j] == r[i] )
                {
                    return (i-j);
                }
            }

        }

        return 0;
    }

Euler Problem 25

2010-12-16T17:23:00.001-08:00

Ahh.. back to good old problem 16 again.. and adapt the code…

http://projecteuler.net/index.php?section=problems&id=25

_______________

           int[] c = new int[1000];
            int[] d = new int[1000];
            int[] e = new int[1000];

            c[0] = 1; //starting condition
            d[0] = 1;
            e[0] = 0;

int counter = 2;// starts with 1,1,.. needs to be offset in count so F12=144

while (c[999]==0)
{

                counter++;
                //do mult in place
                for (int n = 0; n < 1000; n++)
                {
                    e[n] = d[n];
                    d[n] = c[n];
                    c[n] = c[n]+e[n];

}

                //sort out carries across to right
                for (int n = 0; n < 999; n++)
                {
                    while (c[n] > 9)
                    {
                        if (c[n] >= 10)
                        {
                            c[n + 1] += 1;
                            c[n] -= 10;
                        }
                    }

                    while (d[n] > 9)
                    {
                        if (d[n] >= 10)
                        {
                            d[n + 1] += 1;
                            d[n] -= 10;
                        }
                    }

                    while (e[n] > 9)
                    {
                        if (e[n] >= 10)
                        {
                            e[n + 1] += 1;
                            e[n] -= 10;
                        }
                    }

}

Euler Problem 24

2010-12-16T15:49:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=24

A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012 021 102 120 201 210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

_______________

Found this one kinda tricky in C#.. but in Python using the powerful itertools module it’s just 2 lines!

import itertools
print [d for d in itertools.permutations(range(10))][999999]

_________________

Euler Problem 23

2010-12-16T13:56:00.001-08:00

Abundant numbers… those whose factors summed are greater than the number.

The challenge here is to find all abundant numbers up to 28123, then to find all those numbers which can’t be written as the sum of two abundant numbers.

So 12 is the first abundant number, whose factors add up to 16… making the smallest number with a sum of abundant numbers = 24.

http://projecteuler.net/index.php?section=problems&id=23

This works.. but fairly dumb code…. still .. enough to get me to the next challenge!

_______________

static void Main(string[] args)
{
List<int> abundantnumbers = new List<int>();

int sum=0; //non abundant numbers factors

for (int i=1;i<28123;i++)
{

              if (i < sumofdivisors(divisors(i)))
              {
                  abundantnumbers.Add(i);
              }

          //can i be found using all abundant numbers found so far
              if (!issumofabundantnumners(abundantnumbers, i))
              {
                  sum += i;
              }

          }

Console.WriteLine(sum);
Console.ReadLine();

}

      static List<int> divisors(int n)
      {
          List<int> div = new List<int>();
          for (int d = 1; d < n; d++)
          {
              if (n % d == 0) { div.Add(d); }
          }

return div;

}

      static int sumofdivisors(List<int> div)
      {
          int sum = 0;

          foreach (int i in div)
          {
              sum += i;
          }

return sum;

}

      static bool issumofabundantnumners(List<int> ab, int i)
      {
          foreach (int i1 in ab)
          {

foreach (int i2 in ab)
{

                  if (i == (i1 + i2))
                  {
                      return true;
                  }

}
}

return false;

}

Euler Problem 22

2010-12-16T13:07:00.001-08:00

So you download a list of names.. and have to order them.. then calculate a value based on the characters (a-z) and position in the list… EASY! or is it????

Yes it is… C# has all the tools for this… build a list..then sort it.. pretty much job done… the only tip is to remember that the first place in the list (pos=0) is the 1st place in the list!!!

http://projecteuler.net/index.php?section=problems&id=22

______________

static void Main(string[] args)
        {
        List<string> names = new List<string>();

        //all names are in 1 line!
        System.IO.StreamReader sr = new System.IO.StreamReader("o:\\names.txt");

        foreach (string s in sr.ReadLine().Split(','))
        {
            names.Add(s);
        }

names.Sort();

long wordssum=0;

        for (int i = 0; i < names.Count; i++)
        {
            wordssum += wordvalue(names[i]) * (i+1);
        }

Console.WriteLine(wordssum);
Console.ReadLine();

}

    static int wordvalue(string s)
      {
        // a=1 b=2 etc
          int val=0;

          foreach (char c in s.ToLower())
          {
              if (c > 96 && c < 123)
              {
                  val += Convert.ToInt16(c) - 96;
              }
          }

return val;
}

Euler Problem 21

2010-12-15T19:56:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=21

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

___________________

static void Main(string[] args)
{
long sumofallpairs = 0;

        for (int n = 1; n < 10000; n++)
        {
            if (isamicablepair(n))
            {
                sumofallpairs +=n;
            }

}

Console.WriteLine(sumofallpairs);

Console.ReadLine();

}

    static bool isamicablepair (int n)
{
    int sum1 = 0;
    int sum2 = 0;

    foreach (int i in divisors(n))
    {
        sum1 += i;
    }

    foreach (int i in divisors(sum1))
    {
        sum2 += i;
    }

    if (n == sum2 && n != sum1)
    {
        return true;
    }
    else
    {
        return false;
    }

}

    static List<int> divisors(int n)
    {
        List<int> div = new List<int>();
        for (int d = 1; d < n; d++)
        {
            if (n % d == 0) { div.Add(d); }
        }

return div;

}

Euler Problem 20

2010-12-15T01:02:00.001-08:00

Ahhh now this one uses something we built before.. around problem 16 I think… just a slight change needed… job done!! Nice to have an easy one for a change!

Problem = sum the numbers from the answer of 100!

http://projecteuler.net/index.php?section=problems&id=20

_________

static void Main(string[] args)
{
int[] c = new int[999];

c[0] = 1; //starting condition

            for (int i = 1; i < 100; i++)   //go to the power of 1000
            {
                //do mult in place
                for (int n = 0; n < 999; n++)
                {
                    c[n] *= i;
                }

                //sort out carries across to right
                for (int n = 0; n < 999; n++)
                {
                    while (c[n]>9)
                    {
                    if (c[n] >= 10)
                    {
                        c[n + 1] += 1;
                        c[n] -= 10;
                    }
                    }
                }

}

//now add up all the columns
long sum = 0;

            for (int n = 0; n < 999; n++)
            {
                sum += c[n];
            }

Console.WriteLine(sum);
Console.ReadLine();

}

Euler Problem 19

2010-12-15T00:47:00.001-08:00

How many Sundays on the first of the month in the 20th Century from 1 Jan 1901 to 31 Dec 2000?

http://projecteuler.net/index.php?section=problems&id=19

Luckily C# can do this really easily!!

static void Main(string[] args)
{

            int counter=0;
            DateTime d = new DateTime(1901,1,1);

            while (d.Year<2001)
            {
                if (d.DayOfWeek == DayOfWeek.Sunday && d.Day == 1)
                {
                    counter++;
                }

d= d.AddDays(1);
}

Console.WriteLine(counter);
Console.ReadLine();

}

Euler Problem 18

2010-12-15T00:38:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=18

OK so a triangle.. of costs.. and you need to find the most expensive way across it… so the best way is to work backwards from all possible endings and calculate the most expensive way to get there.

A simple example

2 5

7 3 5

Ending positions are 7,3,5,6…

To get to 7 we can have arrived from 2 or 5.. the most expensive route would be 5+7=13

To get to 3 we can have either come from 2 or 5.. the most expensive route would be 3+5 = 8..

etc to get a new triangle of max routes which looks like this..

9 10

7 3 5

giving us the answer of 11… for the most expensive path thru the triangle….. so repeat this formula for any size of triangle.. using code as follows:

______________________

static void Main(string[] args)
{

int[,] a= new int[15,15];

a[0,0]=75;
a[0,1]=95; a[1,1]=64;
a[0,2]=17; a[1,2]=47; a[2,2]=82;
a[0,3]=18; a[1,3]=35; a[2,3]=87; a[3,3]=10;
a[0,4]=20; a[1,4]=04; a[2,4]=82; a[3,4]=47; a[4,4]=65;
a[0,5]=19; a[1,5]=01; a[2,5]=23; a[3,5]=75; a[4,5]=03; a[5,5]=34;
a[0,6]=88; a[1,6]=02; a[2,6]=77; a[3,6]=73; a[4,6]=07; a[5,6]=63; a[6,6]=67;
a[0,7]=99; a[1,7]=65; a[2,7]=04; a[3,7]=28; a[4,7]=06; a[5,7]=16; a[6,7]=70; a[7,7]=92;
a[0,8]=41; a[1,8]=41; a[2,8]=26; a[3,8]=56; a[4,8]=83; a[5,8]=40; a[6,8]=80; a[7,8]=70; a[8,8]=33;
a[0,9]=41; a[1,9]=48; a[2,9]=72; a[3,9]=33; a[4,9]=47; a[5,9]=32; a[6,9]=37; a[7,9]=16; a[8,9]=94; a[9,9]=29;
a[0,10]=53; a[1,10]=71; a[2,10]=44; a[3,10]=65; a[4,10]=25; a[5,10]=43; a[6,10]=91; a[7,10]=52; a[8,10]=97; a[9,10]=51; a[10,10]=14;
a[0,11]=70; a[1,11]=11; a[2,11]=33; a[3,11]=28; a[4,11]=77; a[5,11]=73; a[6,11]=17; a[7,11]=78; a[8,11]=39; a[9,11]=68; a[10,11]=17; a[11,11]=57;
a[0,12]=91; a[1,12]=71; a[2,12]=52; a[3,12]=38; a[4,12]=17; a[5,12]=14; a[6,12]=91; a[7,12]=43; a[8,12]=58; a[9,12]=50; a[10,12]=27; a[11,12]=29; a[12,12]=48;
a[0,13]=63; a[1,13]=66; a[2,13]=04; a[3,13]=68; a[4,13]=89; a[5,13]=53; a[6,13]=67; a[7,13]=30; a[8,13]=73; a[9,13]=16; a[10,13]=69; a[11,13]=87; a[12,13]=40; a[13,13]=31;
a[0,14]=04; a[1,14] = 62; a[2,14] = 98; a[3,14] = 27; a[4,14] = 23; a[5,14] = 09; a[6, 14] = 70; a[7, 14] = 98; a[8, 14] = 73; a[9,14] = 93; a[10, 14] = 38; a[11, 14] = 53; a[12, 14] = 60; a[13, 14] = 04; a[14, 14] = 23;

//start at the last row - work back to top of triangle from all end possibilities
for (int row = 13; row > -1; row--)
{
    //now work thru all the columns
    for (int col = 0; col < 14; col++)
    {
        //find in the row below the max val possible from option1 or option 2 (l or r linked cell)
        if (a[col,row+1] > a[col + 1,row+1])
        {
            a[col,row] = a[col,row+1] + a[col,row];
        }
        else
        {
            a[col,row] = a[col+1, row+1] + a[col,row];
        }

}

Console.WriteLine(a[0,0]);
Console.ReadLine();

}

Euler Problem 17

2010-12-14T13:51:00.001-08:00

How many letters are used in counting from 1 to 1000 inclusive?

Inc the ands, but not spaces or hyphens!

Full details here: http://projecteuler.net/index.php?section=problems&id=17

the approach here is to work out how many times each sequence occurs.. so there are 900 occurrences of the word “hundred” for example…

In the end the answer is 21,124 characters!

_____

static void Main(string[] args)
{

            string a1 = "onetwothreefourfivesixseveneightnine";
            string b1="teneleventwelvethirteenfourteenfifteensixteenseventeeneighteenninetine";
            string b2 = "twentythirtyfortyfiftysixtyseventyeightyninety";
            string c1 = "hundred";
            string c2 ="and";
            string c3 = "onethousand";

int ans = (90*a1.Length) + (10*b1.Length) + (100*b2.Length)
+(c1.Length*900) + (a1.Length*100) + (c3.Length) + (c2.Length*99*9);

Console.WriteLine(ans);
Console.ReadLine();

}

Euler Problem 16

2010-12-14T02:48:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=16

What is the sum of all the numbers in the answer from 2^1000?

This is tricky because it’s a big number!

C# .NET3.5 does this…1.0715086071862673E+301
…which means I don’t have all the digits to add up…

As an aside.. here’s what python does when you ask it for 2 to the power 1000…

>>> print pow(2,1000)
10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376

No problem! Just need to add those up……

Anyway back to C#… how to cope? We need to make an array.. and handle it ourselves like this..:

static void Main(string[] args)
{
int[] c = new int[999];

c[0] = 2; //starting condition

           for (int i = 1; i < 1000; i++)   //go to the power of 1000
           {
               //do mult in place
                for (int n=0;n<999;n++)
               {
                   c[n] *= 2;
               }


               //sort out carries across to right
               for (int n=0;n<999;n++)
               {

                   if (c[n] >= 10)
                   {
                       c[n+1] += 1;
                       c[n] -= 10;
                   }
                   }

}

//now add up all the columns
long sum = 0;

           for (int n = 0; n < 999; n++)
           {
              sum+= c[n];
           }

Console.WriteLine(sum);

Console.ReadLine();

}

Project Euler–Problem 15

2010-12-14T00:15:00.001-08:00

This was a tough one! So the problem is to find how many ways to get from one side of a 20x20 chessboard without backtracking.

As a simple example.. starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner.

So the answer is to think about the problem rather differently… in fact the question is to use 20 Rights and 20 Downs to get across the chessboard… so how many combinations are there to arrange 20R and 20D?? So now it’s a combination problem which means we can use some stats..

The formula is n! / r! (n-r)!

Where n is 40 and r is 20… so 40!/(20! * 20!)

Which gives us: 137846528820 ways to get there!

In fact Google’s online calculator can do this for you….

Just type 40 choose 20 into Google!!

Euler problem 14

2010-12-12T21:17:00.001-08:00

Easy as long as you remember to use LONG and not INT!

http://projecteuler.net/index.php?section=problems&id=14

The following iterative sequence is defined for the set of positive integers:

n n/2 (n is even)
n 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

________________

static void Main(string[] args)
      {
          long maxl = 0;
          long maxn = 0;
          for (long s = 6; s < 1000000; s++)
          {
             long x = s;
              long l = 1;
              while (x > 1)
              {
                  l++;
                  if (x % 2 == 0)
                  {
                      x = x / 2;
                  }
                  else
                  {
                      x = 3 * x + 1;
                  }

              }
              if (l > maxl)
              {
                  maxn = s; //starting number
                  maxl = l;
              }
          }
      Console.WriteLine(maxl + ","+ maxn);
      Console.ReadLine();

}

Euler–problem 13

2010-12-12T14:26:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=13

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

_______________

Another great one!

This time all the numbers are 100 digits long.. so don’t fit into an int, or even a long…

C# can handle is using BigInt when using .NET 4.. but trying to do it another way I thought about using doubles, and just adding a decimal point to each number… we only need to first ten digits from the sum… not pretty… but it works and only took 2 mins to code up using TextPad’s macros to do the formatting for me!!

______

static void Main(string[] args)
{

     double a1 = 3710728753.3902102798797998220837590246510135740250;
     double a2 = 4637693767.7490009712648124896970078050417018260538;
     double a3 = 7432498619.9524741059474233309513058123726617309629;
     double a4 = 9194221336.3574161572522430563301811072406154908250;
     double a5 = 2306758820.7539346171171980310421047513778063246676;
     double a6 = 8926167069.6623633820136378418383684178734361726757;
     double a7 = 2811287981.2849979408065481931592621691275889832738;
     double a8 = 4427422891.7432520321923589422876796487670272189318;
     double a9 = 4745144573.6001306439091167216856844588711603153276;
     double a10 = 7038648610.5843025439939619828917593665686757934951;
     double a11 = 6217645714.1856560629502157223196586755079324193331;
     double a12 = 6490635246.2741904929101432445813822663347944758178;
     double a13 = 9257586771.8337217661963751590579239728245598838407;
     double a14 = 5820356532.5359399008402633568948830189458628227828;
     double a15 = 8018119938.4826282014278194139940567587151170094390;
     double a16 = 3539866437.2827112653829987240784473053190104293586;
     double a17 = 8651550600.6295864861532075273371959191420517255829;
     double a18 = 7169388870.7715466499115593487603532921714970056938;
     double a19 = 5437007057.6826684624621495650076471787294438377604;
     double a20 = 5328265410.8756828443191190634694037855217779295145;
     double a21 = 3612327252.5000296071075082563815656710885258350721;
     double a22 = 4587657617.2410976447339110607218265236877223636045;
     double a23 = 1742370690.5851860660448207621209813287860733969412;
     double a24 = 8114266041.8086830619328460811191061556940512689692;
     double a25 = 5193432545.1728388641918047049293215058642563049483;
     double a26 = 6246722164.8435076201727918039944693004732956340691;
     double a27 = 1573244438.6908125794514089057706229429197107928209;
     double a28 = 5503768752.5678773091862540744969844508330393682126;
     double a29 = 1833638482.5330154686196124348767681297534375946515;
     double a30 = 8038628759.2878490201521685554828717201219257766954;
     double a31 = 7818283375.7993103614740356856449095527097864797581;
     double a32 = 1672632010.0436897842553539920931837441497806860984;
     double a33 = 4840309812.9077791799088218795327364475675590848030;
     double a34 = 8708698755.1392711854517078544161852424320693150332;
     double a35 = 5995940689.5756536782107074926966537676326235447210;
     double a36 = 6979395067.9652694742597709739166693763042633987085;
     double a100 = 4105268470.8299085211399427365734116182760315001271;
     double a37 = 6537860736.1501080857009149939512557028198746004375;
     double a38 = 3582903531.7434717326932123578154982629742552737307;
     double a39 = 9495375976.5105305946966067683156574377167401875275;
     double a40 = 8890280257.1733229619176668713819931811048770190271;
     double a41 = 2526768027.6078003013678680992525463401061632866526;
     double a42 = 3627021854.0497705585629946580636237993140746255962;
     double a43 = 2407448690.8231174977792365466257246923322810917141;
     double a44 = 9143028819.7103288597806669760892938638285025333403;
     double a45 = 3441306557.8016127815921815005561868836468420090470;
     double a46 = 2305308117.2816430487623791969842487255036638784583;
     double a47 = 1148769693.2154902810424020138335124462181441773470;
     double a48 = 6378329949.0636259666498587618221225225512486764533;
     double a49 = 6772018697.1698544312419572409913959008952310058822;
     double a50 = 9554825530.0263520781532296796249481641953868218774;
     double a51 = 7608532713.2285723110424803456124867697064507995236;
     double a52 = 3777424253.5411291684276865538926205024910326572967;
     double a53 = 2370191327.5725675285653248258265463092207058596522;
     double a54 = 2979886027.2258331913126375147341994889534765745501;
     double a55 = 1849570145.4879288984856827726077713721403798879715;
     double a56 = 3829820378.3031473527721580348144513491373226651381;
     double a57 = 3482954382.9199918180278916522431027392251122869539;
     double a58 = 4095795306.6405232632538044100059654939159879593635;
     double a59 = 2974615218.5502371307642255121183693803580388584903;
     double a60 = 4169811622.2072977186158236678424689157993532961922;
     double a61 = 6246795719.4401269043877107275048102390895523597457;
     double a62 = 2318970677.2547915061505504953922979530901129967519;
     double a63 = 8618808822.5875314529584099251203829009407770775672;
     double a64 = 1130673970.8304724483816533873502340845647058077308;
     double a65 = 8295917476.7140363198008187129011875491310547126581;
     double a66 = 9762333104.4818386269515456334926366572897563400500;
     double a67 = 4284628018.3517070527831839425882145521227251250327;
     double a68 = 5512160354.6981200581762165212827652751691296897789;
     double a69 = 3223819573.4329339946437501907836945765883352399886;
     double a70 = 7550616496.5184775180738168837861091527357929701337;
     double a71 = 6217784275.2192623401942399639168044983993173312731;
     double a72 = 3292418570.7147349566916674687634660915035914677504;
     double a73 = 9951867143.0235219628894890102423325116913619626622;
     double a74 = 7326746080.0591547471830798392868535206946944540724;
     double a75 = 7684182252.4674417161514036427982273348055556214818;
     double a76 = 9714261791.0342598647204516893989422179826088076852;
     double a77 = 8778364618.2799346313767754307809363333018982642090;
     double a78 = 1084880252.1674670883215120185883543223812876952786;
     double a79 = 7132961247.4782464538636993009049310363619763878039;
     double a80 = 6218407357.2399794223406235393808339651327408011116;
     double a81 = 6662789198.1488087797941876876144230030984490851411;
     double a82 = 6066182629.3682836764744779239180335110989069790714;
     double a83 = 8578694408.9552990653640447425576083659976645795096;
     double a84 = 6602439640.9905389607120198219976047599490197230297;
     double a85 = 6491398268.0032973156037120041377903785566085089252;
     double a86 = 1673093931.9872750275468906903707539413042652315011;
     double a87 = 9480937724.5048795150954100921645863754710598436791;
     double a88 = 7863916702.1187492431995700641917969777599028300699;
     double a89 = 1536871371.1936614952811305876380278410754449733078;
     double a90 = 4078992311.5535562561142322423255033685442488917353;
     double a91 = 4488991150.1440648020369068063960672322193204149535;
     double a92 = 4150312888.0339536053299340368006977710650566631954;
     double a93 = 8123488067.3210146739058568557934581403627822703280;
     double a94 = 8261657077.3948327592232845941706525094512325230608;
     double a95 = 2291880205.8777319719839450180888072429661980811197;
     double a96 = 7715854250.2016545090413245809786882778948721859617;
     double a97 = 7210783843.5069186155435662884062257473692284509516;
     double a98 = 2084960398.0134001723930671666823555245252804609722;
     double a99 = 5350353422.6472524250874054075591789781264330331690;

     double sum = a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 + a10
          + a11 + a12 + a13 + a14
          + a15
     + a16 + a17       + a18 + a19        + a20 + a21       + a22 + a23
     + a24 + a25       + a26 + a27       + a28 + a29       + a30 + a31
     + a32 + a33       + a34 + a35       + a36 + a100       + a37 + a38
     + a39 + a40       + a41 + a42       + a43 + a44       + a45 + a46
     + a47 + a48       + a49 + a50       + a51 + a52       + a53 + a54
     + a55 + a56       + a57 + a58       + a59 + a60       + a61 + a62
     + a63 + a64       + a65 + a66       + a67 + a68       + a69 + a70
     + a71 + a72       + a73 + a74       + a75 + a76       + a77 + a78
     + a79 + a80       + a81 + a82       + a83 + a84       + a85 + a86
     + a87 + a88       + a89 + a90       + a91 + a92       + a93 + a94
     + a95 + a96       + a97 + a98       + a99;

Console.WriteLine(sum);
Console.ReadLine();

}

Euler–Project Number 12

2010-12-12T14:02:00.001-08:00

http://projecteuler.net/index.php?section=problems&id=12

The sequence of triangle numbers is generated by adding the natural numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

____________________________

Ok so this was a bit tricky… the dumb approach to just add up numbers and keep trying for factors takes WAY too long.. so you need to use the summation formula:

Sum of numbers up to n = n(n+1)/2

Then remember you only need to find half of the factors… so when factorising 12 you’d get:

1,2,3,4,6,12 – 1 and 12 are obvious so start with 2 factors…

then work your way up to sqrt(12) to get 2,3 as factors…
their counter parts will be 6,4 – but you don’t need to know them.. only that they exist.. so in other words just double up the number of factors found up to sqrt(12) = 4

So total number of factors for 12 is 2 + 4 = 6

________________

CODE:

static void Main(string[] args)
      {
          int t;
          int f = 0;

          for (int q = 2; q < 1000000000; q++)
          {
               //sum of all numbers up to t using summation formula
              t = q * (q + 1) / 2;
          //factors

              f = 2; //1 and the number t to be added as factors
              for (int z = 2; z < (Math.Sqrt(t)+1); z++)
              {
                  if (t % z == 0) { f+=2; }

              }
         if (f >= 500)
          {
              Console.WriteLine(t +","+ f);
          }

}

Console.WriteLine(">>");

Console.ReadLine();

}

MTB Skills - phundamentals

2010-12-12T02:39:00.001-08:00

great course today... at bottle lake...

Heals down, elbow out, find the the light hands balance point!

Euler 11

2010-12-12T02:34:00.001-08:00

There's a cool website full of maths and programming challenges.. which I've started working through.. Dave told me about it.. and suggested documenting the solutions.. so here it goes.. I've done the previous 11 but not saved the code.. but from now on I'll document it here.

This is the solution to problem 11...

http://projecteuler.net/index.php?section=problems&id=11

In the 20x20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 63 78 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?

___________

static void Main(string[] args)
        {
            string[] a = new string[20];
           string[,] aa = new string[20,20];

a[0]="08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08";
a[1]="49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00";
a[2]="81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65";
a[3]="52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91";
a[4]="22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80";
a[5]="24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50";
a[6]="32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70";
a[7]="67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21";
a[8]="24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72";
a[9]="21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95";
a[10]="78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92";
a[11]="16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57";
a[12]="86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58";
a[13]="19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40";
a[14]="04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66";
a[15]="88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69";
a[16] ="04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36";
a[17]="20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16";
a[18]="20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54";
a[19]="01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48";

int max=0;
int r = 0;
//set up - split up
foreach (string c in a)
{
     for (int n=0;n<20;n++)
     {
        aa[r,n]=c.Split(' ')[n];
     }
     r++;
}
//horiz
for (int x=0;x<16;x++)
{
    for (int y = 0; y < 20; y++)
    {
        int tot = Convert.ToInt16(aa[x,y]) *Convert.ToInt16(aa[x + 1,y])*Convert.ToInt16( aa[x + 2,y]) *Convert.ToInt16( aa[x + 3,y]);
        if (tot>max) {max=tot;}

}
}

//vert
for (int x = 0; x < 20; x++)
{
    for (int y = 0; y < 16; y++)
    {
        int tot = Convert.ToInt16(aa[x, y]) * Convert.ToInt16(aa[x, y + 1]) * Convert.ToInt16(aa[x, y + 2]) *Convert.ToInt16( aa[x, y + 3]);
        if (tot > max) { max = tot; }

}
}

//diag tl - br
//vert
for (int x = 0; x < 16; x++)
{
    for (int y = 0; y < 16; y++)
    {
        int tot = Convert.ToInt16(aa[x, y]) * Convert.ToInt16(aa[x + 1, y + 1]) * Convert.ToInt16(aa[x + 2, y + 2]) * Convert.ToInt16(aa[x + 3, y + 3]);
        if (tot > max) { max = tot; }

}
}

//diag tl - br
//vert
for (int x = 0; x < 16; x++)
{
    for (int y = 0; y < 16; y++)
    {
        int tot = Convert.ToInt16(aa[x + 3, y]) * Convert.ToInt16(aa[x + 2, y + 1]) * Convert.ToInt16(aa[x + 1, y + 2]) * Convert.ToInt16(aa[x, y + 3]);
        if (tot > max) { max = tot; }

}
}

Console.WriteLine(max);

// Console.WriteLine(max);

Console.ReadLine();

}

Nokia N82 Python DataLogger Code (audio, image, GPS)

2009-10-13T18:58:00.000-07:00

Python code for Nokia N82 phone to record the GPS location, Cell ID, and an image from front camera every 3 seconds. It also records a sound file continuously and records the playhead position into the log file to ensure a tightly coupled link between audio and location data.

You will to ceate a "datalogger" folder on your phone's micro-SD card before using the application for the first occasion. This can be done while the phone is connected to a PC.

Code used in FOSS4G academic paper => http://2009.foss4g.org/researchpapers/#researchpaper_3

(Code is availale for use in your project under MIT Licence)

- - - - - - - -

import e32, appuifw, graphics
import key_codes
import positioning
import sensor
import location
import audio
import sysinfo
import camera
import time
import audio

counter =0

def handle_redraw(rect):
canvas.clear()
canvas.blit(img)

def startrecording():
global S
global counter
t=int(time.time())
filename="e:\\datalogger\\snd_"+str(t)+".wav"
S=audio.Sound.open(filename)
S.record()
img.point ((10,20),0xff0000,width=20)
handle_redraw(())

def stoprecording():
global S
S.stop()
S.close()

def fncSavedata (data):
io=open("e:\\datalogger\\gps_travel.log",'a')
io.write(data +'\n')
io.close
return

def cb(event):

global counter
counter = counter + 1

t=int(time.time())
sat=event["satellites"]
tm=sat["time"]
tot=sat["satellites"]
used=sat["used_satellites"]
pos = event["position"]
lat=pos["latitude"]
lng=pos["longitude"]
alt=pos["altitude"]
vdop=pos["vertical_accuracy"]
hdop=pos["horizontal_accuracy"]
crse=event["course"]
hdg=crse["heading"]
hdgacc=crse["heading_accuracy"]
spd=crse["speed"]

sdb=sysinfo.signal_dbm()
loc = location.gsm_location()
sloc = str(loc)
sloc = sloc.replace(',','/')

global S
sndtime=int(S.current_position() /1000)

s = "gps:"+str(t)+","+str(sndtime)+','+str(tm)+","+str(tot)+"," +str(used)+","+str(lat)+","+str(lng)+ "," + str(alt)
s = s +"," + str(vdop) + "," + str(hdop) + "," + str(hdg) + "," + str(hdgacc) + "," + str(spd) + "gsm:"+str(sloc)+","+str(sdb)
fncSavedata(s)

pic=camera.take_photo('RGB',(320,240),0,'none','auto','auto',1)
sp="e:\\datalogger\\pic" + str(t) + '.jpg'
pic.save(sp)

img.clear()
img.text((40,34),u'Log ID: ' + str(counter),0xff0000,font='normal')
img.text((40,64),u'Sats: ' + str(used) + "/" + str(tot) ,0xff0000,font='normal')
img.text((40,94),u'Spd: ' + str(int(spd)),0xff0000,font='normal')
img.text((40,124),u'Snd: ' + str(sndtime),0xff0000,font='normal')
handle_redraw(())

canvas = appuifw.Canvas(event_callback = None)
appuifw.app.body = canvas
appuifw.app.title = u"Data logger"
w,h = canvas.size
img=graphics.Image.new((w,h))

startrecording()

positioning.set_requestors([{"type":"service","format":"application","data":"test_app"}])
positioning.select_module(270526860)
positioning.position(course=1,satellites=1,callback=cb,interval=1000000,partial=1)

app_lock = e32.Ao_lock()
app_lock.wait()

stoprecording()

__________________________

Copyright (c) 2009 GRCNZ,University of Canterbury Geog Dept
Permission is hereby granted, free of charge, to any personobtaining a copy of this software and associated documentationfiles (the "Software"), to deal in the Software withoutrestriction, including without limitation the rights to use,copy, modify, merge, publish, distribute, sublicense, and/or sellcopies of the Software, and to permit persons to whom theSoftware is furnished to do so, subject to the followingconditions:
The above copyright notice and this permission notice shall beincluded in all copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIESOF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE ANDNONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHTHOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY,WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISINGFROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OROTHER DEALINGS IN THE SOFTWARE.

2009-01-22T17:21:00.000-08:00

To turn a S60 Python .py file into something installable (.SIS).. you can used Ensemble.

You need to install OpenSSL too, and put include it in your Path.

Then run ensemble and turn your .py into a .sis.
Include any permissions your SIS will need... like access to GPS and networks etc.
Then once you have a SIS file sign it using Symbian online open signing.. again with all permissions required.

Then install to phone.

NOTE... if Python is installed to C on your phone (Phone memory) then your SIS should be installed there too otherwise you may just get the appl flash up a blank white screen for a second then close again.

eg of syntax for all permissions you can get on Nokia N82 GPS camera phone.

D:\iArchive\mobiles>ensymble_python2.4-0.26.py py2sis --caps=LocalServices+Location+NetworkServices+PowerMgmt+ProtServ+R
eadUserData+SurroundingsDD+SWEvent+UserEnvironment+WriteUSerData+ReadDeviceData+TrustedUI+WriteDeviceData o:\GeogDatalog
ger.py o:\GeogDatalogger.sis

ensymble_python2.4-0.26.py: warning: no application version given, using 1.0.0
ensymble_python2.4-0.26.py: warning: no UID given, using auto-generated test UID 0xe72d1e27
ensymble_python2.4-0.26.py: warning: no certificate given, using insecure built-in one

Then sign the SIS using Symbian Open Signing online...

Cell Positions

2008-10-15T16:24:00.000-07:00

http://spectrumonline.med.govt.nz/licence-search.html

Enter into LICENCE either:

telecom
vodafone

The main license types for cell phone base stations are SZ, SY, and SX (relates to transmit power, SZ is most common).

From here you can drill down to find LAT/LNG for each transmitter.

Google Maps - Quadtree tiling system

2008-09-18T19:59:00.000-07:00

So Google and other online mapsite using a tiling system to transfer large images.

These images are cut into smaller ones, based on a nested hierachy.... known as Quadtree. Good explaination here

Google use q,r,t,s (0-3) to denote one of 4 corners of a tile.

So to find your location run this Python code....

Run it like this:

python googletiles.py 17 -43.525 172.616

Spherical Mercator (ESPG:900913) coordinates for lat/lon:
(19215525.222771712, -5392226.689463486)
17/128383/47899 ( TileMapService: z / x / y )
Google: 128383 83172
Quadkey: 31311030123311311 ( 14851095925 )

Substitute the QUADKEY values as follows
0=q
1=r
2=t
3=s

Then type that list of letters into this page...

Here's an example for Christchurch, NZ
http://pallit.lhi.is/bigice/makeme.php?startPic=tsrsrrqsqrtssrrsrr&hTiles=14&vTiles=14

___________
Other useful info on tiling

cut your own tiles - x,y info
http://makemeamap.com/cgi-bin/mercator4.pl

view tiles on one page
http://pallit.lhi.is/bigice/bigpic.html

how it works
http://www.maptiler.org/google-maps-coordinate-system-projection-epsg-900913-3785/

overlay on web your own image and create the tiles for it
http://open.atlas.free.fr/GMapsTransparenciesImgOver.php

use python to create a local cache of WMS tiles
http://www.tilecache.org/

Some Mapping links - most with source code included!

2008-09-17T04:12:00.000-07:00

Display OpenStreetMap on a DEM!

http://wiki.openstreetmap.org/index.php/Kosmos#Download

- - - -
Adding MS VE to a .NET application!

http://blogs.msdn.com/virtualearth3d/archive/2008/04/19/creating-the-most-basic-windows-form.aspx

- - - -
3D and graphics tools

http://www.grime.net/gistools/

------

http://www.alpix.com/3d/TerrainViewer/index.html

3d mini terrain viewer

- - - -
http://blogs.msdn.com/coding4fun/archive/2007/10/18/5506286.aspx

connecting Wii to MS VE

The power of Folksonomies - Tags from FLickR shown on a map

2008-09-11T20:04:00.001-07:00

Google Map Hacking

2008-06-23T22:53:00.000-07:00

http://econym.googlepages.com/index.htm

Great set of examples and explanations for anyone working on Google Map hacking!

Open mapping sources

2008-06-17T20:45:00.000-07:00

Two open map sources

http://www.openstreetmap.org/

http://openaerialmap.org/

Earth Bridge (GPS to Google Earth)

2008-06-17T20:17:00.000-07:00

Earth Bridge... link GPS to Google Earth...
Takes standard NMEA output....

http://mboffin.com/earthbridge/

Coastal GIS tools

2008-06-15T18:27:00.001-07:00

Just stumbled across these... some coastal tools for GIS people.

http://ecoastal.usace.army.mil/tools.asp

Particularly interested in the 3D viewer and comparing volume changes between LiDAR dataset from different times.

GPS Art

2008-05-25T17:01:00.001-07:00

A briefcase + GPS + many batteries + DHL + the World = self portrait

http://biggestdrawingintheworld.com/d rawing.aspx

Population movement tracking in malls

2008-05-20T15:21:00.000-07:00

Using people's phones as a way to track movements.

http://technology.timesonline.co.uk/tol/news/tech_and_web/article3945496.ece

Here the IMEI numbers unique to each phone are determined from shoppers so that the number of shoppers, country the phone is registered, and length of time spent in the mall can be stored. The trends can be used to extend shopping hours, or change the language of signs if it's found one nation is well represented in a shopping mall.

TCPIP sockets in c#

2008-04-28T14:08:00.000-07:00

Many great working examples on TCPIP sockets... in C#

http://www.java2s.com/Code/CSharp/Network/AsyncTcpServer.htm

- - -
/*C# Network Programming by Richard BlumPublisher: Sybex ISBN: 0782141765*/
using System;
using System.Net;
using System.Net.Sockets;
using System.Text;
public class FixedTcpSrvr
{
private static int SendData(Socket s, byte[] data)
{
int total = 0;
int size = data.Length;
int dataleft = size;
int sent;
while (total < size)
{
sent = s.Send(data, total, dataleft, SocketFlags.None);
total += sent;
dataleft -= sent;
}
return total;
}
private static byte[] ReceiveData(Socket s, int size)
{
int total = 0;
int dataleft = size;
byte[] data = new byte[size];
int recv;
while (total < size)
{
recv = s.Receive(data, total, dataleft, 0);
if (recv == 0)
{
data = Encoding.ASCII.GetBytes("exit ");
break;
}
total += recv;
dataleft -= recv;
}
return data;
}
public static void Main()
{
byte[] data = new byte[1024];
IPEndPoint ipep = new IPEndPoint(IPAddress.Any, 12008);
Socket newsock = new Socket(AddressFamily.InterNetwork,
SocketType.Stream, ProtocolType.Tcp);
newsock.Bind(ipep);
newsock.Listen(10);
Console.WriteLine("Waiting for a client...");
Socket client = newsock.Accept();
IPEndPoint newclient = (IPEndPoint)client.RemoteEndPoint;
Console.WriteLine("Connected with {0} at port {1}",
newclient.Address, newclient.Port);
string welcome = "Welcome to my test server";
data = Encoding.ASCII.GetBytes(welcome);
int sent = SendData(client, data);
for (int i = 0; i < 5; i++)
{
data = ReceiveData(client, 9);
Console.WriteLine(Encoding.ASCII.GetString(data));
}
Console.WriteLine("Disconnected from {0}", newclient.Address);
client.Close();
newsock.Close();
}
}

My Location - Google Mobile Maps

2008-04-01T00:54:00.000-07:00

So you don't have GPS but your phone can run JAVA apps... or Python apps..

Google Maps for mobiles can use the GSM tower (cell) id to look up roughly where you are... it plots you on Google Map (for mobile) to within a few km (at worst).

So if Google have access to some database of Cell Tower ID - Lat/Long can we access that data too?

Well.. you can buy it.. or try to use Google's dataset.

In the background the Google maps application uses this website to post your Cell ID

http://www.google.com/glm/mmap

Neil Young written a demo C# app to make use of this to get a lat/long for a given Cell id.
(copy of source at end of this article)
http://maps.alphadex.de/datafiles/fct0e1b11782832f02.cs

Excellent article on his webstie: http://maps.alphadex.de/index.php?section=mylocation

- - - - - -

Yahoo have something similar but no where near as populated or dense cell data.
http://developer.yahoo.com/yrb/zonetag/locatecell.html

- - -

using System;
using System.Collections.Generic;
using System.Text;
using System.Net;
using System.IO;
using System.Diagnostics;
/*
* Sample code to obtain geo codes from a cell info
* "GSM/UMTS" setting revealed by smuraro, thanks!
*/
/* (c) "Neil Young" (neil.young@freenet.de)
*
* This script/program is provided "as is".
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; either version 3 of the License, or
* (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* GNU General Public License, see .
*/
namespace GMM {
class Program {
static byte[] PostData(int MCC, int MNC, int LAC, int CID, bool shortCID) {
/* The shortCID parameter follows heuristic experiences:
* Sometimes UMTS CIDs are build up from the original GSM CID (lower 4 hex digits)
* and the RNC-ID left shifted into the upper 4 digits.
*/
byte[] pd = new byte[] {
0x00, 0x0e,
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
0x00, 0x00,
0x00, 0x00,
0x00, 0x00,
0x1b,
0x00, 0x00, 0x00, 0x00, // Offset 0x11
0x00, 0x00, 0x00, 0x00, // Offset 0x15
0x00, 0x00, 0x00, 0x00, // Offset 0x19
0x00, 0x00,
0x00, 0x00, 0x00, 0x00, // Offset 0x1f
0x00, 0x00, 0x00, 0x00, // Offset 0x23
0x00, 0x00, 0x00, 0x00, // Offset 0x27
0x00, 0x00, 0x00, 0x00, // Offset 0x2b
0xff, 0xff, 0xff, 0xff,
0x00, 0x00, 0x00, 0x00
};
bool isUMTSCell = ((Int64)CID > 65535);
if (isUMTSCell)
Console.WriteLine("UMTS CID. {0}", shortCID ? "Using short CID to resolve." : "");
else
Console.WriteLine("GSM CID given.");
if (shortCID)
CID &= 0xFFFF; /* Attempt to resolve the cell using the GSM CID part */
if ((Int64)CID > 65536) /* GSM: 4 hex digits, UTMS: 6 hex digits */
pd[0x1c] = 5;
else
pd[0x1c] = 3;
pd[0x11] = (byte)((MNC >> 24) & 0xFF);
pd[0x12] = (byte)((MNC >> 16) & 0xFF);
pd[0x13] = (byte)((MNC >> 8) & 0xFF);
pd[0x14] = (byte)((MNC >> 0) & 0xFF);
pd[0x15] = (byte)((MCC >> 24) & 0xFF);
pd[0x16] = (byte)((MCC >> 16) & 0xFF);
pd[0x17] = (byte)((MCC >> 8) & 0xFF);
pd[0x18] = (byte)((MCC >> 0) & 0xFF);
pd[0x27] = (byte)((MNC >> 24) & 0xFF);
pd[0x28] = (byte)((MNC >> 16) & 0xFF);
pd[0x29] = (byte)((MNC >> 8) & 0xFF);
pd[0x2a] = (byte)((MNC >> 0) & 0xFF);
pd[0x2b] = (byte)((MCC >> 24) & 0xFF);
pd[0x2c] = (byte)((MCC >> 16) & 0xFF);
pd[0x2d] = (byte)((MCC >> 8) & 0xFF);
pd[0x2e] = (byte)((MCC >> 0) & 0xFF);
pd[0x1f] = (byte)((CID >> 24) & 0xFF);
pd[0x20] = (byte)((CID >> 16) & 0xFF);
pd[0x21] = (byte)((CID >> 8) & 0xFF);
pd[0x22] = (byte)((CID >> 0) & 0xFF);
pd[0x23] = (byte)((LAC >> 24) & 0xFF);
pd[0x24] = (byte)((LAC >> 16) & 0xFF);
pd[0x25] = (byte)((LAC >> 8) & 0xFF);
pd[0x26] = (byte)((LAC >> 0) & 0xFF);
return pd;
}
static void Main(string[] args) {
if (args.Length < 4) {
Console.WriteLine("Usage: gmm MCC MNC LAC CID [\"shortcid\"]");
return;
}
string shortCID = ""; /* Default, no change at all */
if (args.Length == 5)
shortCID = args[4].ToLower();
try {
String url = "http://www.google.com/glm/mmap";
HttpWebRequest req = (HttpWebRequest)WebRequest.Create(new Uri(url));
req.Method = "POST";

int MCC = Convert.ToInt32(args[0]);
int MNC = Convert.ToInt32(args[1]);
int LAC = Convert.ToInt32(args[2]);
int CID = Convert.ToInt32(args[3]);
byte[] pd = PostData(MCC, MNC, LAC, CID, shortCID == "shortcid");
req.ContentLength = pd.Length;
req.ContentType = "application/binary";
Stream outputStream = req.GetRequestStream();
outputStream.Write(pd, 0, pd.Length);
outputStream.Close();
HttpWebResponse res = (HttpWebResponse)req.GetResponse();
byte[] ps = new byte[res.ContentLength];
int totalBytesRead = 0;
while (totalBytesRead < ps.Length) {
totalBytesRead += res.GetResponseStream().Read(ps, totalBytesRead, ps.Length - totalBytesRead);
}
if (res.StatusCode == HttpStatusCode.OK) {
short opcode1 = (short)(ps[0] << 8 ps[1]);
byte opcode2 = ps[2];
System.Diagnostics.Debug.Assert(opcode1 == 0x0e);
System.Diagnostics.Debug.Assert(opcode2 == 0x1b);
int ret_code = (int)((ps[3] << 24) (ps[4] << 16) (ps[5] << 8) (ps[6]));

if (ret_code == 0) {
double lat = ((double)((ps[7] << 24) (ps[8] << 16) (ps[9] << 8) (ps[10]))) / 1000000;
double lon = ((double)((ps[11] << 24) (ps[12] << 16) (ps[13] << 8) (ps[14]))) / 1000000;
Console.WriteLine("Latitude: {0}, Longitude: {1}", lat, lon);

Process p = new Process();
p.StartInfo.FileName = "iexplore";
Console.WriteLine("\nClose map window to exit\n");
p.StartInfo.Arguments = String.Format(
"http://maps.google.de/maps?f=q&hl=de&q={0},{1}&ie=UTF8&z=15",
lat.ToString().Replace(',','.'), lon.ToString().Replace(',','.'));
p.Start();
p.WaitForExit();
}
else
Console.WriteLine("Error {0}", ret_code);
}
else
Console.WriteLine("HTTP Status {0} {1}", res.StatusCode, res.StatusDescription);
}
catch (Exception) {
throw;
}
}
}
}

Urban wiki

2008-03-10T15:08:00.001-07:00

http://wiki.urban.cens.ucla.edu/index.php?title=Main_Page

location based.. sensor nets

Mobile City

2008-03-09T14:48:00.000-07:00

http://www.spatialmetro.org/

http://liftlab.com/think/nova/2008/03/04/notes-from-the-mobile-city-conference/

City information layers available on mobiles.. and using those patterns...

NZMG - WGS84 Projections

2008-02-25T12:23:00.000-08:00

NZMG was the official New Zealand mapping projection until 2001.

From the LINZ website (NZ mapping body):

"NZMG is not based on a geometric projection (transverse Mercator is based on a cylinder). Instead it uses a complex-number polynomial expansion. This has the advantage of exhibiting minimal scale distortion over New Zealand; however it is a projection unique to New Zealand and so can be difficult to use or program into computer software or positioning devices. "

In packages such as ESRI ArcGIS you can reproject data easily...... however if you're writting your own software you need to find some libraries to do this.

PROJ4 is one such library. It can be downloaded with GDAL / OGR as part of FW TOOLS.
http://fwtools.maptools.org/

Here there are 2 tools to assist in transformations... PROJ and CS2CS.
CS2CS caters with changes in DATUM.

To convert from NZMG to WGS84 requires a change in datum.

There are 3 methods to do this...

a) 3 parameter shift
b) 7 parameter shift
c) grid (NTv2)

a) is only good to the nearest 100m or so...
b) is only good to 5m or so..
c) uses a look up grid and should be good to nearest few cm

The look up grid can be downloaded from here: ftp://ftp.remotesensing.org/proj

Syntax to convert a user entered co-ordinate from WGS84 lat/long to NZMG x,y (m):

C:\Program Files\FWTools2.1.0\bin>cs2cs +proj=latlong +datum=WGS84 +to +proj=nzmg +datum=nzgd49 +nadgrids=o:/nzgd2kgrid0005.gsb

170 -43 [enter]
2265438.60 5796423.71 0.00
[ctrl+z]

This should give you metre accuracte conversions.

_____

To carry out from C# using FW TOOLS... you will need to check the C# example from Tamas Szekeres. This can be found in the FWTOOLS / csharp /apps folder.. and is entitled OSRTransform.cs.

A few modifications are required to work with NZMG and WGS84... as follows:

OSGeo.OSR.SpatialReference src = new OSGeo.OSR.SpatialReference("");

src.ImportFromProj4("+proj=latlong +datum=WGS84");

OSGeo.OSR.SpatialReference dst = new OSGeo.OSR.SpatialReference("");

dst.ImportFromProj4("+proj=nzmg +x_0=25100000.0000000000 +y_0=6023150.0000000000 +datum=nzgd49 +nadgrids='o:\nzgd2kgrid0005.gsb'");

CoordinateTransformation ct = new OSGeo.OSR.CoordinateTransformation(src, dst);

double[] p = new double[3];

p[0] = 172.585303;

p[1] = -43.510183;

p[2]=0; //needed even if 0 value

ct.TransformPoint(p); Console.WriteLine("x:" + p[0] + " y:" + p[1]) ;

___
This should give you results accurate to a few metres.

List of great web sites relating to GIS and GPS

2008-02-19T17:40:00.000-08:00

http://www.tec.army.mil/gis/

Awesome listing of GIS tools, blogs, web pages, development examples...
A great starting point in any search for GIS or GPS tools...

Open Source GIS tools - starting page

2008-02-19T12:38:00.001-08:00

Intro to GRASS, GDAL, PROJ, and other useful open source GIS tools.

http://casoilresource.lawr.ucdavis.edu/drupal/node/105

Spoken word

2008-02-12T01:18:00.000-08:00

Books read by volunteers....

Can download to Nokia's Audio books tool for S60 phones!

http://librivox.org/

Garmin get talking

2008-02-12T00:38:00.001-08:00

So.. Nokia are getting mapping...

Why shouldn't the mapping people get talking...

Check this out! New from Garmin in the near future!

The Nuviphone!

http://www.electronista.com/articles/08/01/30/garmin.nuviphone/

Nokia gets mapping big time!

2008-02-12T00:29:00.000-08:00

OK.. so Nokia (phones) are getting into mapping...

They brought out the 6610 with GPS (as used by Ewan McGreggor on long way down) :-http://www.mobilewhack.com/nokia-6110-navigator-takes-the-long-way-down-with-ewan-mcgregor-and-charley-boorman/

Then the multimedia computer that was the N95.. with all it troubles and slow lock on to GPS signals.

The N82 which improved on the N95 GPS performance...
An Updated N95 8GB which sorted out the original N95 GPS troubles....
An update to the N95 to get it back on track....

They've purchased NavTec for a few billion http://www.pcworld.com/article/id,137887-c,mapping/article.html
These guys make the maps! Next time you're using Google Maps.. check the bottom line to see if NavTec are mentioned! If they're not chances are that TeleAtlas are.. oh yer. they've just been bought by TomTom a GPS giant!

So what's next for Nokia? Well it seems the next mobile mapping software will be aimed at pedestrians... http://www.nokia.com/betalabs/maps

Mobile Phones as Environmental Sensors

2008-02-04T12:47:00.000-08:00

Mobile phones are a form of ubiqutous computing.
The connected crowd (Rheingold).

Sensors on phones include:
sound
vision
location (GPS, cell-id, wifi)
movement (accelerometers)
ad hoc bluetooth devices

Now projects using mobiles to collect environmental data are becoming common place:
Cellphones to monitor highway traffic

Cellphones to detect dirty bombs?

Cellphones to track air pollution

Capturing ambient intelligence