Dynamic registration

[Carty]

Hello,
Im actually a new kid on programming and developed a small app for pocket pc and now trying to implement a dynamic registration method.. So please help me how to calculate the registration key using a rpn string.. More details are given below.. Its gonna be from Handango.com
Im able to get the C/DUI i.e:The name of the owner, and so give me an example project on VB.net to calculate the following..

Here is an example of how Handango uses the RPN string you give us to compute the correct dynamic registration code for a user based on the C/DUI on that user's handheld device.
For our example:C/DUI = "Will P"
RPN string = "i 0 == 111 * key + c 2 * +"

1. We convert each character of the C/DUI to ASCII value.
   W = 87 decimal
   i = 105 decimal
   l = 108 decimal
   l = 108 decimal
   [space] = 32 decimal
   P = 80 decimal
2. We apply the RPN string to the ASCII value of each character in the C/DUI. The RPN string is actually a mathematical formula written in reverse Polish notation. Where 'i' is the position of the character in the C/DUI starting at 0, and 'c' is the ASCII value of the character in the C/DUI. (We will discuss 'key' later in this example.)
3. We start out by setting the variable ?key? = 0.
4. Then we apply the RPN string to the first character, 'W'. For 'W' i=0 since 'W' is the first character of the C/DUI and we start counting at 0, and c=87 since 'W' has an ASCII value of 87.
5. Now apply the RPN string. The RPN string is evaluated by placing each operand in it on a stack, when an operator is reached, it is applied to the last two operands on the stack. Those operands are popped from the stack and the result is pushed back onto the stack.
6. We process the first operand 'i' in the RPN string by pushing it onto the stack, now stack = i.
7. Then we process the next operand in the RPN string, '0' by pushing it onto the stack, now stack = i 0.
8. Next we come to the '==' which is an operator that works just like the C operator. We pop the last two operands from the stack which were i and 0 and we apply the operator '==' to them. Since i equals 0, the result is true. So, 1 is pushed back onto the stack, now stack = 1.
9. Then we process the next operand in the RPN string which is '111' by pushing it onto the stack, making stack = 1 111.
10. Next, we come to the '*' which is an operator representing multiplication. We pop the last two operands from the stack, which were 1 and 111, and we multiply them. The result is 111 which is pushed back onto the stack, making stack = 111.
11. Then we process the next operand in the RPN string, which is 'key', by pushing it onto the stack, making stack = 111 key.
12. Next we come to the '+' which is an operator representing addition. We pop the last two operands from the stack, which are 111 and ?key?, and we add them together. Since key=0, and 111 + 0 = 111, the result is 111 which is pushed back onto the stack, making stack = 111.
13. Then we process the next operand in the RPN string which is 'c' by pushing it onto the stack, making stack = 111 c.
14. Then we process the next operand in the RPN string, which is '2', by pushing it onto the stack, making stack = 111 c 2.
15. Next we come to the '*' which is an operator representing multiplication. We pop the last two operands from the stack, which are c and 2, and we multiply them. Since c=87 and 2*87=174, the result is 174 which is pushed back onto the stack, making stack = 111 174.
16. Next we come to the '+' which is an operator representing addition. We pop the last two operands from the stack, which were 111 and 174, and we add them together. 111+174=285. The result is 285, which is pushed back onto the stack, making stack = 285.
17. Now we have reached the end of the RPN string so the result of applying the RPN string to the first character 'W' of the C/DUI is 285.
18. We set key=285. This concludes applying the RPN string to 'W'.
19. Now we apply the RPN string to the next character 'i', which is the second character in the C/DUI, so i=1 since we start counting at 0. The ASCII value of 'i' is 105. So, c=105 and from processing 'W' key=285.
20. We process the first operand 'i' in the RPN string by pushing it onto the stack, now stack = i.
21. Then we process the next operand in the RPN string, '0' by pushing it onto the stack. Now stack = i 0.
22. Next we come to the '==' which is an operator that works just like the C operator. We pop the last two operands from the stack which were i and 0 and we apply the operator '==' to them. Since i equals 1, the result is false, so 0 is pushed back onto the stack. Now stack = 0
23. Then we process the next operand in the RPN string, which is '111', by pushing it onto the stack giving, stack = 0 111.
24. Next we come to the '*' which is an operator representing multiplication. We pop the last two operands from the stack, which were 0 and 111, and we multiply them. The result is 0 which is pushed back onto the stack, making stack = 0.
25. Then we process the next operand in the RPN string, which is 'key', by pushing it onto the stack, making stack = 0 key.
26. Next we come to the '+' which is an operator representing addition. We pop the last two operands from the stack, which were 0 and key, and we apply addition to them. Since key=285, and 285 + 0 = 285,the result is 285 which is pushed back onto the stack, making stack = 285.
27. Then we process the next operand in the RPN string, which is 'c', by pushing it onto the stack, making stack = 285 c.
28. Then we process the next operand in the RPN string, which is '2', by pushing it onto the stack, making stack = 285 c 2.
29. Next we come to the '*' which is an operator representing multiplication. We pop the last two operands from the stack, which were c and 2, and we multiply them. Since c=105 and 2*105=210, the result is 210, which is pushed back onto the stack, making stack = 285 210.
30. Next we come to the '+' which is an operator representing addition. We pop the last two operands from the stack, which were 285 and 210, and we add them together. 285+210=495. The result is 495, which is pushed back onto the stack, making stack = 495.
31. Now we have reached the end of the RPN string so the result of applying the RPN string to the second character 'i' of the C/DUI is 495.
32. We set key=495. This concludes applying the RPN string to 'i'
33. Now we apply the RPN string to the next character 'l', which is the third character in the C/DUI, so i=2 since we start counting at 0. The ASCII value of 'l' is 108. So, c=108 and from processing 'l' key=495.
34. Applying the RPN string as in the previous step, we get key=495 + 2*108 = 711.
35. We set key=711. This concludes applying the RPN string to the first 'l'.
36. Now we apply the RPN string to the next character 'l', which is the fourth character in the C/DUI, so i=3 since we start counting at 0. The ASCII value of 'l' is 108. So, c=108 and from processing 'l' key=711.
37. Applying the RPN string as in the previous step, we get key=711 + 2*108 = 927.
38. We set key=927. This concludes applying the RPN string to the second 'l'.
39. Now we apply the RPN string to the next character [space], which is the fifth character in the C/DUI, so i=4 since we start counting at 0. The ASCII value of [space] is 32. So, c=32 and from processing [space] key=927.
40. Applying the RPN string as in the previous step, we get key=927 + 2*32 = 991.
41. We set key=991. This concludes applying the RPN string to [space].
42. Now we apply the RPN string to the next character 'P', which is the sixth character in the C/DUI, so i=5 since we start counting at 0. The ASCII value of 'P' is 80. So, c=80 and from processing 'P' key=991.
43. Applying the RPN string as in the previous step, we get key=991 + 2*80 = 115.
44. Since we are at the end of the C/DUI, we get key = 1151. Next we add a 0 to the beginning to make it a five-digit number, thus our dynamic registration code for a C/DUI of 'Will P' is 01151.
    • So for C/DUI = "Will P' and RPN String = "i 0 == 111 * key + c 2 * +"
    • The dynamic registration code = "01151"
    Note: The RPN String "i 0 == 111 * key + c 2 * +" Adds up the ASCII values of all the characters in the C/DUI, multiplies that by 2 and adds 111 to that.
    • For "Will P": 87 + 105 + 108 + 108 + 32 + 80 = 520; 520 * 2 = 1040; 1040 + 111 = 1151.
    • An RPN string of "i 0 == 111 * key + c 3 * +" would add up the ASCII values of all the characters in the C/DUI. Multiply that by 3 and add 111 to that.
    • So, for "Will P": 520*3=1560; 1560 + 111 = 1671.
    • An RPN string of "i 0 == 987 * key + c +" would add up the ASCII values of all the characters in the C/DUI and add 987 to that.
    • So for "Will P": 520 + 987 = 1507.

Hope you would have got an idea of calculating the reg. key but i dont know how to do this in VB.net and so, can u pleassssse help me providing a sample to take an rpn and owner name and provide the calculated reg key.. Thanx a milllllion in advance..

Regards
Carty..