Description
Problem 1 This is a straightforward version of the classic game “Snake”. A snake moves only from left to right, growing itself whenever it eats the bait. The snake is on a line of n discrete spaces. At each time point, the snake moves one space. Write a C program snake.c that reads a natural number n (at least 5) and prints the line that the snake is on as it moves. Assume the snake has an initial body length of 2 and its head is at position 3 (indices start from 0). The line has n spaces, and there are baits on every odd space that the snake hasn’t visited yet. Print the subsequent states in new lines; the game ends when the snake’s head reaches the end of the line. Denote snake’s head by “H”, body by “X”, bait by “.”, empty space character by “_”. For n==5, the output will look like the following (each line ends with \n character): For n==15, the output will look like the following (each line ends with \n character): Page 3 of 7 Problem 2 a) Create the file functions.h, which contains the following declarations: I) void square(int w); II) void spiral(int w); III) void rotation(int w); Assume w>0 b) Implement all the functions above (explained below) in the file functions.c Note: You are to submit this file (along with functions.h file) containing only your implemented functions (that is, you must delete the test cases portion and the main function). However, you must keep the required included libraries. c) Submit functions.zip, which contains the files functions.c, and functions.h Here are the objectives of the three functions: Assume w>1. I) void square(int w); prints the numbers 1,2,….,w2in a square shape (of size wxw) on w lines of output Examples: void square(2)prints: 1 2 3 4 void square(5) prints: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Page 4 of 7 void square(6) prints: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 II) void spiral(int w); prints the numbers 1,2,….,w2in a “spiral” order (compared with the square shape from part I above) on one line. Examples: void spiral(2)prints: 1 2 4 3 Because void spiral(5) prints: 1 2 3 4 5 10 15 20 25 24 23 22 21 16 11 6 7 8 9 14 19 18 17 12 13 Because: Page 5 of 7 void spiral(6) prints: 1 2 3 4 5 6 12 18 24 30 36 35 34 33 32 31 25 19 13 7 8 9 10 11 17 23 29 28 27 26 20 14 15 16 22 21 Because: III) void rotation(int w); prints the numbers 1,2,….,w2in a “rotation” order on one line (compared with the square shape). Each rotation is 90 degrees for some incremental offsets. Examples: void rotation(2)prints: 1 2 4 3 void rotation(5) prints: 1 5 25 21 2 10 24 16 3 15 23 11 4 20 22 6 7 9 19 17 8 14 18 12 13 void rotation(6) prints: 1 6 36 31 2 12 35 25 3 18 34 19 4 24 33 13 5 30 32 7 8 11 29 26 9 17 28 20 10 23 27 14 15 16 22 21 Explore the pattern before you start coding Notes: Each number printed is followed by a space (including the last number in each line). Each line ends with \n character Page 6 of 7 Problem 3 a) Create the file fun.h, which contains the following declarations: I) int isSophieGermainPrime(int p); II) int base2nat(int bs, int num); III) int nat2base(int bs, int num); b) Implement all the functions above (explained below) in the file fun.c Note: You are to submit this file (along with fun.h file) containing only your implemented functions (that is, you must delete the test cases portion and the main function). However, you should keep the required included libraries. c) Submit fun.zip which contains the files fun.c, and fun.h I) Definition: A Sophie Germain1 prime is a [positive] prime number p such that 2p + 1 is also a prime number. For example, 2 is a Sophie Germain prime since both 2 and 2(2) + 1 = 5 are prime numbers. Task: Create the function int isSophieGermainPrime(int n) which determines if an integer n is a Sophie Germain prime. The function should return 1 if n is a Sophie Germain prime and 0 otherwise. Assumptions: You may assume that the values entered are valid integers such that the magnitude of 2p + 1 is at most 109 . 1 Page 7 of 7 II) When you see a number such as 734, it is generally assumed that you are using the base 10 number system (also known as the decimal system). That is: 734 = 7*102 + 3*101 + 4*100 It is, however, possible to use any number as a base. For example, assuming we are in a base 5 number system, the notation 2301 would generate the decimal number 326: 2*53 + 3*52 + 0*51 + 1*50 = 326 (this equation is in decimal) Note that when using base 10 we have precisely 10 unique digits for each position, that is 0,1,2,3,4,5,6,7,8,9. Similarly, base 5 only allows for 5 digits 0,1,2,3,4 (To represent the “normal” (i.e. decimal) value 5 in base 5 we would write 10; 6 would be represented as 11, 7 would be 12 etc. To see this, consider 12 (in base 5), which means we compute 1*51 + 2*50 to get 7). Task: Create the function // pre: 1<bs<10 and num>0 a valid integer in base bs int base2nat(int bs, int num) which returns a positive integer representing the decimal value of num (num is in base bs). Assumptions: You may assume that the values entered are valid integers such that the magnitude of num is at most 109 . III) // pre: 1<bs<10 and num>0 int nat2base(int bs, int num); It takes a base (bs) and a non-negative integer (num) in decimals and returns the value in base bs. The following code will help you with testing 1. #include 2. #include 3. #include “fun.h ” 4. 5. int main(void){ 6. assert(isSophieGermainPrime(11)); 7. assert(isSophieGermainPrime(41)); 8. assert(base2nat(5,23114)==1659); 9. assert(base2nat(7,1)==1); 10. assert(base2nat(3,1211012)==1328); 11. assert(base2nat(8,715)==461); 12. assert(nat2base(5,1659)==23114); 13. assert(nat2base(9,1331)==1738); 14. }
