I need help with this please. Essentially I need to complete each of the functions listed within the directions. The main functions are just for testing the listed functions. A sample output would be greatly appreciated. Thank you.

Abstract

In this programming assignment, you will implement a Fibonacci function that avoids repet- itive computation by computing the sequence linearly from the bottom up: F(0) through F(n). The calculation of the F(n) will provide a key to encrypt a simple text message. You will also overcome the limitations of C's 32-bit integers by storing very large integers in arrays of indi- vidual digits.

By completing this assignment, you will gain experience crafting algorithms of moderate com- plexity, develop a deeper understanding of integer type limitations, become acquainted with unsigned hexadecimal integers, and reinforce your understanding of dynamic memory manage- ment in C. In the end, you will have a very fast program for computing huge 40 hexadecimal digit sequences of Fibonacci numbers that also encrypts short text messages.

Interestingly, this problem will be limited to 40 hexadecimal digit numbers, from the outset, thru the whole program. This will mimic the performance constraints of some old cryptographic equipment (KW-26) that generated key strings based on a 2 number input to start a continuous chain of some type of calculations to generate long apparently random number sequences. This 40 digit number will be used as an encryption key for a text message, read from a user specified file.

Attachments

kw26.h, kw26-m{1-6}.c, kw26-m{1-6}.log, kw26-m4.err

Deliverables

kw26.c (Note: Pay attention - as the filename matters!)

1 Overview 1.1 Computational Considerations for Recursive Fibonacci

We've seen in class that calculating Fibonacci numbers with the most straightforward recursive implementation of the function is prohibitively slow, as there is a lot of repetitive computation:

int fib(int n) { 

 //base cases: F(0) = 0, F(1) = 1 

 if (n < 2) return n; 

 //definition of Fibonacci: F(n) = F(n - 1) + F(n - 2) 

 return fib(n - 1) + fib(n - 2); } 

This recursive function has an exponential runtime. It is possible to achieve linear runtime by building from the base cases, F(0) = 0 and F(1) = 1, toward the desired result, F(n). This avoids the expensive and exponentially explosive recursive function calls.

This assignment will emulate some aspects of hardware encryption from the 1960s, specifically the KW-26, while the KW-26 used 45 digit round-robin counters and a bit of other hardware, this as- signment will use 40 hexadecimal digit counters, with two initialization vectors, the cryptoVariable and the hwConfigVariable. Each of these vectors will be 40 hexadecimal digits. All subsequent products will be 40 hexadecimal digits. In the event of overflow the overflow product will be ig- nored.

Once the cryptoVariable and the hwConfigVariable have been read and created, respectively, they will be decimally added to produce a Fibonacci sum. All subsequent 40 hexadecimal digit integers will be the sum of the two previous 40 hexadecimal digit integers. (This ensures that the digits after F(2) will be unique and the full 40 digits.) The math is shown below.

f0 = hwConfigV ariable f1 = cryptoV ariable f2 = f1 + f0

. fn = fn1 + fn2

Note that 40 hexadecimal digits does not fit into any standard C variable data type. (See Section 7, Representing huge integers in C for a detailed explanation on how to add large integers using a created data type.)

Careful review shows that by placing the 40 hexadecimal digit integers into an array, with the least significant digit in the leading digit it will be possible to add the two 40 digit numbers together, if added one digit at a time from the first element in the array to the last element in the array. Arithmetically speaking the most significant digit will be in the most significant slot in the array.

For example, the decimal number 12,567 would be parsed one digit at a time into an array named x containing 7 in x[0], 6 in x[1], 5 in x[2], 2 in x[3], and 1 in x[4]. All 40 hexadecimal digits will be stored in an array using the following data structure to hold the pointer to the malloced buffer of 40 digits.

typedef struct Int40 { 

 // a dynamically allocated array to hold a 40 // digit integer, stored in reverse order int *digits; 

} Int40;

2 Attachments 2.1 Header File (kw26.h)

This assignment includes a header file, kw26.h, which contains the definition for the Int40 struct, as well as functional prototypes for all the required functions in this assignment. You should #include this header file from your kw26.c source file, like so:

 #include "kw26.h" 

2.2 Test Cases

This assignment comes with multiple sample main files (kw26-m1-6.c), which you can compile with your kw26.c source file. For more information about compiling projects with multiple source files, see Section 5, Compilation and Testing (Linux/Mac Command Line).

2.3 Sample Output Files

Also included are a number of sample output files that show the expected results of executing your program (kw26-main1-6.log & kw26-m4.err).

2.4 Disclaimer

The test cases included with this assignment are by no means comprehensive. Please be sure to develop your own test cases, and spend some time thinking of edge cases that might break each of the required functions.

3 Function Requirements

In the source file you submit, kw26.c, you must implement the following functions. You may im- plement any auxiliary functions you need to make these work, as well. Notice that none of your functions should print anything to the screen or STDOUT.

Int40 *kw26Add(Int40 *p, Int40 *q); 

Description: Return a pointer to a new, dynamically allocated Int40 struct that contains the result of adding the 40 digit integers represented by p and q.

Special Notes: If a NULL pointer is passed to this function, simply return NULL. If any dy- namic memory allocation functions fail within this function, also return NULL, but be care- ful to avoid memory leaks when you do so.

Hint: Before adding two huge integers, you will want to create an array to store the result. Re- member that all integers in this problem are 40 digits long. In the event that the most sig- nificant digits (MSD) result in a carry, the carry will be ignored. For example, if the MSD of the two inputs are 9 and 7, the resultant MSD will be 6 with a carry of 1 for the MSD + 1 digit. (916 + 716 = 1016, therefore 6 is the MSD and the 1 is ignored.)1

Returns: A pointer to the newly allocated Int40 struct, or NULL in the special cases mentioned above.

Int40 *kw26Destroyer(Int40 *p); 

Description: Destroy any and all dynamically allocated memory associated with p. Avoid seg- mentation faults and memory leaks.

Returns: NULL

Int40 *parseString(char *str); 

Description: Convert a number from string format to Int40 format. (For example function calls, see kw26-m1.c.)

Special Notes: If the empty string () is passed to this function, treat it as a zero (0). If any dynamic memory allocation functions fail within this function, or if str is NULL, return NULL, be careful to avoid memory leaks when you do so. You may assume the string will only contain ASCII digits 0 through 9 and the letters A thru F in either upper or lower case, for a minimum of 40 digits. In the event that 40 digits are not in the input string, print an error message to STDERR and fill with leading zeroes. Also, if there are more than 40 digits in the input string use the first 40 digits in the string.

Returns: A pointer to the newly allocated Int40 struct, or NULL if dynamic memory allocation fails or if the input str is NULL.

1The subscript of 16 indicate base 16. However the two expressions 1016 and 10x are equivalent and used equally often.

Int40 *fibKw26(int n, Int40 *first, Int40 *second); 

Description: This is your Fibonacci function. Pay careful attention the F(0) is initialized with the hwConfigVariable and F(1) is initialized with the cryptoVariable. Implement an iterative solution that runs in O(n) time and returns a pointer to a Int40 struct that contains F(n). Be sure to prevent memory leaks before returning from this function.

Space Consideration: When computing F(n) for large n, its important to keep as few Fibonacci numbers in memory as necessary at any given time. For example, in building up to F(10000), you wont want to hold Fibonacci numbers F(0) through F(9999) in memory all at once. Find a way to have only a few Fibonacci numbers in memory at any given time over the course of a single call to fibKw26(...).

Special Notes: Remember that n is the second parameter passed as an input argument to the program. You may assume that n is a non-negative integer. If any dynamic memory alloca- tion functions fail within this function, return NULL, but be careful to avoid memory leaks when you do so.

Returns: A pointer to an Int40 representing F(n), or NULL if dynamic memory allocation fails.

Int40 *encrypt(Int40 *key, Int40 *inputText); 

Description: This is the encryption function. It will take the inputText pointer and XOR the in- putText data with the key that had been generated using the fibKw26 function. This process will produce an encrypted message.

Special Notes: It is interesting to note that encrypted data can produce the original plain text by simply XORing the encrypted data with the key, which this assignment demon- strates can be reproduced on demand.

Returns: A pointer to an Int40 representing the encrypted plain text.

void kw26Rating(); 

STDERR output: Outputs the following items to STDERR, delimited by a semicolon ;: 1. NID

2. A difficulty rating of how difficult you found this assignment on a scale of 1.0 (ridicu- lously easy) through 5.0 (insanely difficult).

3. Duration, in hours, of the time you spent on this assignment.

The first argument to this function is the pointer to the kw26RatingStruct which is defined in the kw26.h include file. Make sure to output those items to STDERR. Each element should be terminated or delimited by a ;.

Int40* loadHwConfigVariable(int seed); 

Returns: A pointer to an Int40 array of 40 digits. If the input variable seed is not zero, the ran- dom number generator will be seeded with the time of the epoch which is expressed as an integer value of the number of seconds since 01-01-1970, otherwise the random number gen- erator will not be used. In the event the seed is zero, the function should return 40 digits, each with the value of 1. In the event the seed is greater than 0, the 40 digits will be initial- ized in 8 groups of the same 5 random digits. For example, if the five unique random digits were 16597, the result would be those 5 digits concatenated 8 times to produce:

1659716597165971659716597165971659716597 Returns NULL if there is an error during creation or initialization of the hwConfigVariable.

Note: It is acceptable to use decimal values for the 5 random digits, that is, there is no require- ment to use 5 random hexadecimal digits.

Int40* loadCryptoVariable(char *cryptoVariableFilename); 

Returns: A pointer to an Int40 array of 40 random hexadecimal digits read in from the crypto- VariableFilename. Returns NULL if there is an error during initialization of the cryptoVari- able or in the file I/O.

Int40* loadPlainText(char *plainTextFilename); 

Note: Reading the input text will require casting the character or text to an integer compatible format. Therefore it is likely to have an internal function roughly equivalent to the previ- ously define funtion, parseString.

Returns: A pointer to an Int40 array of 40 characters read in from the plainTextFilename. Re- turns NULL if there is an error during conversion of the characters of the plain text or in the file I/O.

Special Cases: There are two special cases. In the event there are less than 40 characters in the plainTextFilename this function

should pad with the equivalent of spaces, that is a zero. In the event there are more than 40 characters in the plainTextFilename, the trailing

characters should be deleted.

4 Compilation and Testing (Linux/Mac Command Line)

To compile multiple source files (.c files) at the command line:

gcc kw26.c kw26-m1.c By default, this will produce an executable file called a.out that you can run by typing, e.g.:

./a.out

If you want to name the executable something else, use:

 gcc kw26.c kw26-m1.c -o kw26-01.exe 

...and then run the program using:

 ./kw26-01.exe 

Running your program could potentially dump a lot of output to the screen. If you want to redi- rect your output to a text file in Linux, its easy. Just run the program using the following:

./kw26-01.exe > whatever.txt This will create a file called whatever.txt that contains the output from your program.

Linux has a helpful command called diff for comparing the contents of two files, which is really helpful here since weve provided sample output files. You can see whether your output matches ours exactly by typing, e.g.:

diff whatever.txt kw26-output01.txt If the contents of whatever.txt and kw26-output01.txt are exactly the same, diff wont have any output. It will just look like this:

 mcalpin@eustis:~$ diff whatever.txt kw26-output01.txt mcalpin@eustis:~$ _ 

If the files differ, it will spit out some information about the lines that arent the same. For exam- ple:

 mcalpin@eustis:-$ diff kw26-m2.log kw26-m2.bogus.log 2c2 < 19427194271942719427194271942719427194271942719427 --- 

 > 49427194271942719427194271942719427194271942719427 mcalpin@eustis:~$ _ 

4.1 Deliverables

Submit a single source file, named kw26.c, via Webcourses. The source file should contain defi- nitions for all the required functions (listed above), as well as any auxiliary functions you need to make them work.

Your source file must not contain a main() function. Do not submit additional source files, and do not submit a modified kw26.h header file. Your program must compile and run on Eustis to receive credit. Programs that do not compile will receive an automatic zero. Specifically, your program must compile without any special flags, as in:

 gcc kw26.c kw26-m1.c 

Be sure to include your name and NID as a comment at the top of your source file.

5 Grading

Scoring will be based on the following rubric: Table 1: Grading Rubric

Percentage	Description
-100	Cannot compile on Eustis
- 10	Late
- 30	Cannot convert a string to the correct Int40 (both numeric and input text)
-5	Cannot create the correct Int40 from loadCrypto- Variable
- 10	Cannot create an Int40 from loadHwConfigVari- able(w/o seed)
-5	Creates an Int40 from loadHwConfigVariable(w/o seed), but it is not a repeating pattern
- 10	Cannot create an Int40 from loadHwConfigVari- able(w/ seed)
-5	Creates the same Int40 from subsequent calls to loadHwConfigVariable(w/ seed)
- 10	Cannot create a valid array of hexadecimal repre- sentations of plain text data
- 10	Does not correctly encrypt plain text with the asso- ciated F[N] keystream
- 10	Cannot manage memory for Int40s - (no partial credit)
- 20	Crashes when adding Int40 numbers
- 10	Adds Int40 numbers, but gets the wrong answer
- 10	Adds Int40y numbers correctly, but cannot calcu- late fibKw26
- 10	Calculates fibKw26, but uses the wrong base cases
- 10	The large n case must be forced to stop
-5	The large n case finishes, but takes longer than 5 seconds
- 10	Does not output kw26Rating data or outputs data without the ; delimiter
-5	Outputs kw26Rating data, but not to STDERR

Your grade will be based primarily on your programs ability to compile and produce the exact output expected. Even minor deviations (such as capitalization or punctuation errors) in your output will cause your programs output to be marked as incorrect, resulting in severe point de- ductions. The same is true of how you name your functions and their parameters. Please be sure to follow all requirements carefully and test your program thoroughly.

Please note that you will not receive credit for test cases that call your Fibonacci function if that functions runtime is worse than O(n), or if your program has memory leaks that slow down exe-

cution. In grading, programs that take longer than a fraction of a second per test case (or perhaps a whole second or two for very large test cases) will be terminated.

Your kw26.c must not include a main() function. If it does, your code will fail to compile during testing, and you will receive zero credit for the assignment.

Special Restrictions: As always, you must avoid the use of global variables, mid-function variable declarations, and system calls (such as system(pause)).

6 Submission Instructions

The assignment shall be submitted via WebCourses.

7 Representing huge integers in C

Any linear Fibonacci function has a big problem, though, which is perhaps less obvious than the original runtime issue: when computing the sequence, we quickly exceed the limits of C's 32-bit integer representation. On most modern systems, the maximum int value in C is 232 1, or 2,147,483,647. 2 The first Fibonacci number to exceed that limit is F(47) = 2,971,215,073. Ob- viously, this will not support a 40 hexadecimal digit number calculation.

This problem is exacerbated by the fact that all the numbers used in this problem will be 40 hex- adecimal digits long. The maximum value of 1640 is 4bitsPerHexdigit 40 digits or 160 bits. The decimal equivalent is:

1640 = 1.4615 1048

Even C's 64-bit unsigned long long int type is only guaranteed to represent non-negative in- tegers up to and including 18,446,744,073,709,551,615 which is 264 1.3 The Fibonacci number F(93) is 12,200,160,415,121,876,738, which can be stored as an unsigned long long int. How- ever, F(94) is 19,740,274,219,868,223,167, which is too big to store in any of C's extended integer data types.

To overcome this limitation, we will represent integers in this program using arrays, where each index holds a single digit of an integer.4 For reasons that will soon become obvious, we will store integers in reverse order in these arrays. So, for example, the numbers 2,147,483,648 and 10,0087 would be represented as:

Storing these integers in reverse order makes it really easy to add two of them together. The ones digits for both integers are stored at index [0] in their respective arrays, the tens digits are at in- dex [1], the hundreds digits are at index [2], and so on. How convenient!

2To see the upper limit of the int data type on your system, #include , then printf(%d , INT_MAX);

3To see the upper limit of the unsigned long long int data type on your system, #include , then printf(%llu , ULLONG_MAX);

4Yes, there is a lot of wasted space with this approach. We only need 4 bits to represent all the hexadecimal digits in the range 0 through F, yet the int type on most modern systems is 32 bits.

Figure 1: Two numbers stored in array - LSD first

So, to add these two numbers together, we add the values at index [0] (8 + 7 = 15), throw down the 5 at index [0] in some new array where we want to store the sum, carry the 1, add it to the values at index [1] in our arrays (1 + 4 + 8 = 13), and so on:

Figure 2: Calculating the sum of two numbers (LSD first)

Note that the examples shown are for small sequences of digits. For all numbers in this program, we will use this array representation for integers containing 40 hexadecimal digits. The arrays will be allocated dynamically.

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%x", p->digits[i]);

printf(" ");

}

int main(void)

{

Int40 *p;

Int40 *q;

Int40 *r;

//40 digits

kw26Print(p = parseString("0123456789abcdef0123456789abcdef01234567"));

kw26Destroyer(p);

//<40 digits

kw26Print(p = parseString("354913546879519843519843548943513179"));

kw26Destroyer(p);

//>40 digits

kw26Print(p = parseString("012345678901234567890123456789012345678901234567899999"));

kw26Destroyer(p);

//Null digits

kw26Print(p = parseString(NULL));

kw26Destroyer(p);

//40 digit add

p = parseString("ffffffffffffffffffffffffffffffffffffffff"); //40 fs

q = parseString("1"); //just a single 1

r = kw26Add( p, q);

kw26Print(p);

kw26Print(q);

kw26Print(r);

//40 digit add

p = parseString("7777777777777777777777777777777777777777"); //40 7s

q = parseString("5555555555555555555555555555555555555555"); //40 5s

r = kw26Add( p, q);

kw26Print(p);

kw26Print(q);

kw26Print(r);

return 0;

}

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%x", p->digits[i]);

printf(" ");

}

int main(void)

{

Int40 *p;

// Load crypto variable from the file named cryptoVarFile

kw26Print(p = loadCryptoVariable("cryptoVarFile"));

kw26Destroyer(p);

// Load the unseeded HW config variable

// NB The seeded variable should have a repeating

// pattern of 5 random digits - 8 times.

// NB The unseeded variable should have the digit "1"

// 40 times. The 0 parameter below is unseeded.

kw26Print(p = loadHWConfigVariable(0));

kw26Destroyer(p);

// Load the SEEDED HW config variable

// SEEDED random number generator should generate

// a different sequence of numbers in every run

// Remember the number should be 5 random digits that

// repeat 8 times for 40 digits.

kw26Print(p = loadHWConfigVariable(1));

kw26Destroyer(p);

return 0;

}

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%x", p->digits[i]);

printf(" ");

}

int main(void)

{

Int40 *p;

Int40 *q;

Int40 *r;

// Load crypto variable from the file named cryptoVarFile

// Load the unseeded HW config variable

// Unseeded random number generator should generate

// the same sequence of numbers in every run

// NB The unseeded variable should have a repeating

// pattern of 5 "random" digits - 8 times.

p = loadCryptoVariable("cryptoVarFile");

q = loadHWConfigVariable(0);

/*

p = parseString("7656362109723093382194813785549032874959");

q = parseString("1942719427194271942719427194271942719427");

*/

r = kw26Add( p, q);

kw26Print(p);

kw26Print(q);

kw26Print(r);

return 0;

}

#include

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%x", p->digits[i]);

printf(" ");

}

int main(int argc, char *argv[])

{

Int40 *p;

Int40 *q;

Int40 *r;

int seed = 0;

int fNum = 0;

char *cryptoFilename;

// Check if command line parms entered

// IF none ->EXIT but leave a message

if (argc == 1)

{

printf(" \t usage: ");

printf("\t kw26-m4 fibonacciNumber cryptoFilename seedType ");

exit(1);

}

fNum = atoi(argv[1]);

cryptoFilename = argv[2];

seed = atoi(argv[3]);

// Load crypto variable from the file named cryptoFilename

// Load the HW config variable F[0]

p = loadHWConfigVariable(seed);

// Load the Crypto Variable F[1]

q = loadCryptoVariable(cryptoFilename);

r = fibKw26(fNum, p, q);

kw26Print(p); //print HW config variable F[0]

kw26Print(q); //print crypto variable F[1]

kw26Print(r); //print F[n]

kw26Rating(); //print the rating (NID,difficulty, & hours)

return 0;

}

#include

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%x", p->digits[i]);

printf(" ");

}

int main(int argc, char *argv[])

{

Int40 *p;

Int40 *q;

Int40 *r;

int seed = 0;

int fNum = 0;

char *cryptoFilename;

// Check if command line parms entered

// IF none ->EXIT but leave a message

if (argc == 1)

{

printf(" \t usage: ");

printf("\t kw26-m5 cryptoFilename seedType ");

exit(1);

}

cryptoFilename = argv[1];

seed = atoi(argv[2]);

// Load crypto variable from the file named cryptoFilename

// Load the HW config variable F[0]

p = loadHWConfigVariable(seed);

kw26Print(p); //print HW config variable F[0]

// Load the Crypto Variable F[1]

q = loadCryptoVariable(cryptoFilename);

kw26Print(q); //print crypto variable F[1]

printf("F(5) ");

r = fibKw26(5, p, q);

kw26Print(r); //print F[n]

printf("F(10) ");

r = fibKw26(10, p, q);

kw26Print(r); //print F[n]

printf("F(20) ");

r = fibKw26(20, p, q);

kw26Print(r); //print F[n]

return 0;

}

#include

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%x", p->digits[i]);

printf(" ");

}

// print an Int40 with TWO hex chars for all 40 ints

void kw26PrintBy2X(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%02X", p->digits[i]);

printf(" ");

}

int main(int argc, char *argv[])

{

Int40 *p; //F[0]

Int40 *q; //F[1]

Int40 *r; //F[x]

Int40 *t; // used for plain text

Int40 *encryptedText; // crypto data

Int40 *decryptedText; // un-crypto data

int seed = 0;

int fNum = 0;

char *cryptoFilename;

char *inputPlainTextFilename;

// Check if command line parms entered

// IF none ->EXIT but leave a message

if (argc == 1)

{

printf(" \t usage: ");

printf("\t kw26-m6 cryptoFilename inputPlainTextFilename F[N] seedType ");

exit(1);

}

cryptoFilename = argv[1];

inputPlainTextFilename = argv[2];

fNum = atoi(argv[3]);

seed = atoi(argv[4]);

// Load crypto variable from the file named cryptoFilename

// Load the HW config variable F[0]

p = loadHWConfigVariable(seed);

kw26Print(p); //print HW config variable F[0]

// Load the Crypto Variable F[1]

q = loadCryptoVariable(cryptoFilename);

kw26Print(q); //print crypto variable F[1]

printf("F(%d) ", fNum);

r = fibKw26(fNum, p, q);

kw26Print(r); //print F[fNum] one hex digit at a time

kw26PrintBy2X(r); //print F[fNum] two hex digits per each input digit

// load the plain text

t = loadPlainText(inputPlainTextFilename);

kw26PrintBy2X(t); //print two hex digits per each input character

//encrypt the plain text, t, with the key, r

encryptedText = encrypt(r, t);

kw26PrintBy2X(encryptedText); //print two hex digits per each input character

//decrypt the encrypted text, encryptedText, with the key, r

decryptedText = encrypt( encryptedText, r);

kw26PrintBy2X(decryptedText); //print two hex digits per each input character

return 0;

}

#ifndef __KW26_H

#define __KW26_H

#define MAX40 40

typedef struct Int40

{

// a dynamically allocated array to hold a 40

// digit integer, stored in reverse order

int *digits;

} Int40;

typedef struct kw26RatingStruct{

char *NID; //pointer to a malloc'ed buffer for the NID

float degreeOfDifficulty; //1.0 for super easy to 5.0 for insanity++

float duration; // Hours spent writing, reading,

// designing & building the code

} kw26RatingStruct;

char *cryptoVariableFilename; // for the filename

int seed;//to seed the RNG or not

int nFib; //control the number of Fibonacci numbers to calculate

// F[0] is loaded with the cryptovariable

Int40 *cryptoVariable; // 40 digits used to start the F[x]

// F[1] is loaded with the hwConfigvariable

Int40 *hwConfigVariable; // 40 digits of psuedo or real

// randomness to start the F[x]

// Functional Prototypes

Int40 *kw26Add(Int40 *p, Int40 *q);

Int40 *kw26Destroyer(Int40 *p);

Int40 *fibKw26(int n, Int40 *first, Int40 *second);

void kw26Rating(void);

Int40 *parseString(char *str);

Int40 *loadHWConfigVariable(int doSeed);

Int40 *loadCryptoVariable(char *inputFilename);

Int40 *loadPlainText(char *inputFilename);

Int40 *encrypt(Int40 *key, Int40 *inputText);

#endif

#include

#include

#include "kw26.h"

// print an Int40 (followed by a newline character)

void kw26Print(Int40 *p)

{

int i;

if (p == NULL)

{

printf("(null pointer) ");

return;

}

for (i = MAX40 - 1; i >= 0; i--)

printf("%d", p->digits[i]);

printf(" ");

}

int main(void)

{

Int40 *p;

Int40 *q;

// Infinite loop.

while(1)

{

p = parseString("1010101010101");

// Use this block of code if your parseString function is broken

/*

p = malloc(sizeof(Int40));

p->digits = malloc(40*sizeof(int));

for (int j=0; j<40; j++)

{

p->digits[j] = j%16;

}

*/

// If your add function is broken then comment out both

// q = kw26Add(p,p) & kw26Destroyer(q)

q = kw26Add(p,p);

kw26Destroyer(p);

kw26Destroyer(q);

}

return 0;

}