Directions: Fill in answers on the pages below. Don't spend too much time on any one problem.
Points for each problem: 1-15, 2-20, 3-25, 4-15, 5-25.

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1. (15) Simple Arrays: Fill in the rest of the C program below, so that it will use a loop to add the square of each number in the array and then print the sum. (The program should be complete with all necessary declarations.)

   #include <stdio.h>
   
   int main() {
      int p[] = {2, 3, 5, 7, 11, 13, 17};
      /* Use a loop to add the square of each number in this array. */
      /* Then print the sum. */
      int sum = 0;
      int i;
      (for i = 0; i < 7; i++)
         sum = sum + p[i]*p[i];
      printf("Sum: %i\n", sum);
   }
   

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2. (20) Arrays as parameters: Fill in the rest of the C program below by writing a function printp that will take the array p as an actual parameter and will print the elements of p, on one line, with a comma but no space after each number except for the last one, and with a newline at the end. (No comma at the end.) Your program should print the following when it runs:
   2,3,5,7,11,13,17

   #include <stdio.h>
   /* give a prototype of the function printp on the next line */
   void printp(int p[]);
   
   int main() {
      int p[] = {2, 3, 5, 7, 11, 13, 17};
      /* call the function printp on the next line */
      printp(p);
   }
   
   /* give the definition of the function printp below */
   void printp(int p[]) {
      int i;
      for (i = 0; i < 7; i++) {
         printf("%i", p[i];
         if (i != 6) printf(",");
         else printf("\n");
      }
   }
   

3. (25) Gambling at Roulette: In the American version of Roulette, if you bet $1 on a single number, then if your number doesn't come up, they take your money. If your number does come up, then they add $35 to your $1, and you have $36 left. Your number comes up 1/38 of the time, that is, with probability 1/38 = 0.0263158. (In this game there are the 36 numbers 1 through 36 to bet on, plus 0 and 00. A rotating ball will fall into a slot labeled with one of these numbers. The game would be exactly fair if the probability of winning was 1/36 = 0.0277778. The difference is the house edge.)

   You are to write a program that simulates betting repeatedly on a single number. You start with $1000 (and we'll drop the $-sign). Just initialize a variable such as amount to 1000. You bet 20 at a time, inside a loop. If you lose, your amount of money drops by 20. If you win, your amount of money goes up by 20*36 = 720. I made a mistake in the problem statement here: This program is keeping track of the gains and loses, so in case of a win the amount of money goes up by 20*35 = 700, and NOT by 720. The loop continues until one of two things happens: either your money drops below 20 (actually to 0, since everything here occurs in increments of 20), so you can't bet any more, or your money goes up to 2000 or above, so that you have doubled your stake, and you quit happy.

   It's easy to simulate something randomly happening with probability 1/38: just produce a random double uniformly distributed between 0 and 1. If this random double is less than 1/38, you win, and otherwise you lose.

   Write your program into the framework below. I have inserted the random number generator that we have often used, along with its initialization.

   You should count the number of bets needed until you either at least double your money or lose all your money, and print this number. Because of the way this is set up, when the program ends, you will either have 0 amount, or if you double your money, you may have considerably more than 2000 when you quit. Thus you should also print the final value of amount.

   #include <stdio.h>  /* for input/output */
   #include <stdlib.h> /* for rand() and srand() */
   #include <time.h>   /* for time() in srand */
   
   double rand_double();
   
   int main() {
      int amount = 1000;
      int bets = 0;
      srand((long)time(NULL));
      while (amount >= 20 && amount < 2000) {
         bets++;
         if(rand_double() < 1.0/38.0)
            amount = amount + 700;
         else
            amount = amount - 20;
      }
      printf("# bets: %i, final amount: %i\n", bets, amount);
   }
      
   double rand_double() {
      return rand()/(double)RAND_MAX;
   }
   
   

   Results of 42 runs: # bets: 50, final amount: 0 # bets: 165, final amount: 2020      # bets: 158, final amount: 0 # bets: 302, final amount: 0 # bets: 50, final amount: 0 # bets: 90, final amount: 2080 # bets: 50, final amount: 0 # bets: 86, final amount: 0 # bets: 177, final amount: 2500 # bets: 86, final amount: 0 # bets: 40, final amount: 2360 # bets: 265, final amount: 2180 # bets: 374, final amount: 0 # bets: 223, final amount: 2300 # bets: 122, final amount: 0 # bets: 50, final amount: 0 # bets: 86, final amount: 0 # bets: 194, final amount: 0 # bets: 50, final amount: 0 # bets: 50, final amount: 0 # bets: 86, final amount: 0

   # bets: 194, final amount: 0 # bets: 265, final amount: 2180 # bets: 50, final amount: 0 # bets: 12, final amount: 2200 # bets: 11, final amount: 2220 # bets: 12, final amount: 2200 # bets: 147, final amount: 2380 # bets: 71, final amount: 2460 # bets: 48, final amount: 2200 # bets: 50, final amount: 0 # bets: 86, final amount: 0 # bets: 50, final amount: 0 # bets: 18, final amount: 2080 # bets: 89, final amount: 2100 # bets: 50, final amount: 0 # bets: 266, final amount: 0 # bets: 45, final amount: 2260 # bets: 40, final amount: 2360 # bets: 50, final amount: 0 # bets: 22, final amount: 2000 # bets: 57, final amount: 2020

   Other versions of the loop in this problem:

   Instead of:
   
        while (amount >= 20 && amount < 2000) {
           . . .
        }
   
   You could use some of these variations:
   
        for (count = 0; amount >= 20 && amount < 2000; count++) {
           . . .
        }
   
        while(1) {
           . . .
           if (!(amount >= 20 && amount < 2000)) break;
        }
   
        while(1) {
           . . .
           if (!(amount >= 20) || !(amount < 2000)) break;
        }
   
        while(1) {
           . . .
           if (amount < 20) || amount >= 2000) break;
        }
   
        while(1) {
           . . .
           if (amount < 20) break; 
           if (amount >= 2000) break;
        }
   
   
   Can also use:
        for( ; ; ) in place of while(1) 

   Next I expanded the program to the following, so that I could do 100000000 experiments at a time:

   #include <stdio.h>  /* for input/output */
   #include <stdlib.h> /* for rand() and srand() */
   #include <time.h>   /* for time() in srand */
   #define N 100000000
   
   double rand_double();
   
   int main() {
      int amount = 1000;
      int bets = 0;
      int i;
      int wins = 0;
      int loses = 0;
      double ave_win = 0.0;
      rand((long)time(NULL));
      for(i = 0; i < N; i++) {
         amount = 1000;
         bets = 0;
         while (amount >= 20 && amount < 2000) {
            bets++;
            if(rand_double() < 1.0/38.0)
               amount = amount + 700;
            else
               amount = amount - 20;
         }
         if (amount == 0) loses++;
         else wins++;
         ave_win += (double)amount/N;
      }
      printf("wins: %i, loses: %i, ave win: %0.5f\n", wins, loses, ave_win);
   }
   
   double rand_double() {
      return rand()/(double)RAND_MAX;
   }
   
   Results of a run (100000000 experiments):
   wins: 40836946, loses: 59163054, ave win: 906.42298
   
   Results of runs in Monte Carlo (100000000 experiments, changing 1.0/38.0 to 1.0/37.0 above):
   wins: 42864608, loses: 57135392, ave win: 951.83830
   

   Further Discussion: The above experiment show that betting in this fashion, for every $1000 we bet, we win about $906 on the average. This means the house is getting about 9.4% of our money.

   Similarly, if we play in Monte Carlo, $1000 get us $952 back, so the house there is only getting about 4.8% of our money.

   For each single bet, the house percentage is easy to compute. In Las Vegas, if we bet $1 on a number, then with probability 1/38 we win $36 and the rest of the time we win nothing. Thus our expected value after betting $1 is to end up with 36/38 of a dollar, or $0.9473684. So in Las Vegas the house percentage on this game is 5.26316%. Our game above didn't achieve this percentage, since it was not optimal play.

   If you bet on an even number or on a color in Las Vegas, you double your money with probability 18/38 and you lose with probability 20/38, so here your expected value if you bet $1 is 2*(18/38) = 36/38, which is the same answer as before.

   In Monte Carlo, the expected value when you bet $1 is 36/37 = 0.97297297, so the house percentage is 2.7027%.

4. (15) Macros: Consider the program below that uses a preprocessor macro, along with the ? : version of an if-else

   #include <stdio.h>
   #define MAX(x,y)  (x > y) ? (x) : (y)
   
   int main() {
      int a = 6;
      int b = 4;
      int max = MAX(a+6, b+7);
      printf("The max of (a+6) and (b+7) is: %i\n", max);
   }
   

   1. Could the macro MAX be used to find the maximum of two doubles? (Give more than just a yes or no answer.) Yes. Just because!
      But seriously folks, the substitution puts in an if-else statement that works whether the variables are typed int or double.

   2. Give below a macro MIN that will find the minimum of two numbers.
      #define MIN(x,y) (x < y) ? (x) : (y)

   3. Give what will be substituted in place of MAX(a+6, b+7) in the program above.
      ( a + 6 > b + 7 ) ? ( a + 6 ) : ( b + 7 )

5. (25) Pointers, and 2-dimensional arrays: Consider the program below, which defines a 4-by-6 2-dimensional array, with various numbers in it. The function search searches the whole array for a specific number, in the case below the number 24. This number is found in row 2 and column 4, as is shown by the printout. The function returns a 1 if the value is found, and a 0 if the value is not found. For this problem, you are to add to this program: specifically, you are to add two additional pointer parameters to the function search. These parameters will allow you to get two numbers out of the search function and back into the main function, namely the row and column numbers of the where the value being searched for is to be found, if it is in the array. You will have to make changes to the prototype, to the main function, and to the search function. Show all these changes on the listing below.

   #include <stdio.h>
   int search(int x, int val[4][6], int *a, int *b);
   
   int main() {
      int val[4][6] = {{1, 8, 27, 64, 125, 216},
                       {1, 7, 19, 37,  61,  91},
                       {1, 6, 12, 18,  24,  30},
                       {1, 5,  6,  6,   6,   6}};
      int x = 24;
      int found;
      int a, b;
      found = search(x, val, &a, &b);
      if (found == 1) {
         printf("%i found\n", x);
         printf("In main: the location of %i is (%i, %i)\n",
                  x, a, b);
      }
      else printf("%i NOT found\n", x);
   }
   
   int search(int x, int val[4][6], int *a, int *b) {
      int i, j;
      for (i = 0; i < 4; i++)
         for (j = 0; j < 6; j++)
            if (x == val[i][j]) {
               *a = i; *b = j;  
               printf("In search: the location of %i is (%i, %i)\n",
                  x, i, j);
               return 1;
            }
      return 0;
   }
   
   Results of a run: 
   In search: the location of 24 is (2, 4)
   24 found
   In main: the location of 24 is (2, 4)