CS 2073 Engineering Programming
Topics and Review Questions for
Final Exam, Fall 2005
Selected Answers

Date and length: Thursday, 8 December 2005, 1:30 pm - 4:15 pm.

Sample Questions:

basics:

    Write C code that will start with a certain number of seconds, and will convert to hours, minutes, and seconds, where "hours" and "minuters" are as large as possible:

    Two versions
    /* Hours, minutes and seconds */
#include <stdio.h>

int main() {
   int time = 74747;  /* or some other value in seconds */
   int hours = time/3600;
   int minutes = (time - hours*3600)/60;
   int seconds = (time - hours*3600 - minutes*60);
   printf("%i seconds is\n%5i hours,\n%5i minutes,\n%5i seconds\n",
      time, hours, minutes, seconds);
}
-------------------------------------------------------------------
/* Hours, minutes and seconds */
#include <stdio.h>

int main() {
   int time = 74747;  /* or some other value in seconds */
   int hours = time/3600;
   int minutes = time%3600/60;
   int seconds = (time%3600)%60;
   printf("%i seconds is\n%5i hours,\n%5i minutes,\n%5i seconds\n",
      time, hours, minutes, seconds);
}

    74747 seconds is
   20 hours,
   45 minutes,
   47 seconds

if-else:

  1. Write a short if statement (no else) that will check if an int size is inside the range from 2 to 10 (inclusive) and will print "In range" if this is so.

  2. Write another short if statement (no else) that will check if an int size is not inside the range from 2 to 10 (inclusive) and will print "Out of range" if this is so.

    Is size between 2 and 10?
    / range.c: is size between 2 and 10 (inclusive)?
#include <stdio.h>

int main() {
   while(1) {
      int size;
      scanf("%i", &size);
      if (size == 0) break;
      
      /* Question 2 above */
      if (  2 <= size  &&   size <= 10)  printf("%i is in range\n", size);

      /* Question 3 above (3 separate answers -- study them all!) */
      if (!(2 <= size  &&   size <= 10)) printf("%i out of range (version 1)\n", size);
      if (!(2 <= size) || !(size <= 10)) printf("%i out of range (version 2)\n", size);
      if (  2 > size   ||   size > 10)   printf("%i out of range (version 3)\n", size);
   }
}
    A run (input in bold)
    1
1 out of range (version 1)
1 out of range (version 2)
1 out of range (version 3)
2
2 is in range
6
6 is in range
10
10 is in range
11
11 out of range (version 1)
11 out of range (version 2)
11 out of range (version 3)
0

functions:

  1. Write a function that will take double x and double y as input formal parameters. Then the function should return the quadrand number that the point(x, y) lies in (1-4). In case the point is on an axis, return 0. Try this out with inputs: (1, 2), (-1, 3), (3, -2), (-1, -4), (0, 5), (-3, 0), (0, 0).

    Note below that there are many possible ways to write the function quadrant. I gave two versions that are sort of at the extremes.

    Quadrant: first version Second version plus run
    /* find quadrant of a pair of points */
#include <stdio.h>
int quadrant(double x, double y);

int main() {
   double x;
   double y;
   int quad;
   while (1) {
      scanf("%lf%lf", &x, &y);
      if (x == 0 && y == 0) break;
      quad = quadrant(x, y);
      /* quad = quadrant2(x, y); */
      printf("(%5.2f, %5.2f)", x, y);
      if (quad != 0)
         printf(" is in quadrant: %i\n", quad);
      else
         printf(" is on an axis\n");
   }
}

int quadrant(double x, double y) {
   if (x == 0 || y == 0) return 0;
   if (x >  0 && y >  0) return 1;
   if (x <  0 && y >  0) return 2;
   if (x <  0 && y <  0) return 3;
   if (x >  0 && y <  0) return 4;
}

    int quadrant2(double x, double y) {
   /* default: if x == 0 or y == 0*/
   int result = 0;
   if (x > 0) {
      if (y > 0) result = 1;
      else if (y < 0) result = 4;
   }
   else if (x < 0) {
      if (y > 0) result = 2;
      else if (y < 0) result = 3;
   }
   return result;
}
Results of a run: input in bold:
1 2
( 1.00,  2.00) is in quadrant: 1
-1 3
(-1.00,  3.00) is in quadrant: 2
3 -2
( 3.00, -2.00) is in quadrant: 4
-1 -4
(-1.00, -4.00) is in quadrant: 3
0 5
( 0.00,  5.00) is on an axis
-3 0
(-3.00,  0.00) is on an axis
0 0

Loops:

  1. Translate the following while loop to a for loop that does the same thing:

  2. Translate the following for loop to a while loop that does the same thing:

  3. Use a for loop to calculate 100000 terms of the following sum. Compare the result with pi*pi/6.0.

    Approximate pi Multiple runs
    // pi_calc.c: series for pi
#include <stdio.h>
#include <math.h>

int main() {
   double sum = 0.0;
   int n = 1000000;
   double pi = atan(1)*4.0;
   int i;
   for (i = 1; i < n; i++)
      sum = sum + 1.0/((double)i*i);
   printf("# terms:   %18i\n", n);
   printf("Final sum: %18.15f\n", sum);
   printf("Exact ans: %18.15f\n", pi*pi/6.0);
   printf("Pi value:  %18.15f\n", pi);
}

    # terms:               100000
Final sum:  1.644924066798243
Exact ans:  1.644934066848226
Pi value:   3.141592653589793

# terms:              1000000
Final sum:  1.644933066847771
Exact ans:  1.644934066848226
Pi value:   3.141592653589793

# terms:             10000000
Final sum:  1.644933966847308
Exact ans:  1.644934066848226
Pi value:   3.141592653589793

    Two slight modifications to the above formulas give much more accurate approximations to Pi: Calculate Pi.

  4. Write a code segment in C that will start with an amount of money (a double amount), and a number of years (an int years), and an interest rate (a double rate). Use a loop to get the amount of money at the end of this time, applying the interest each year. Compare this amount with the amount you would have if you applied the interest 4 times a year. (You apply 0.25*rate a number times equal to 4*years.)

    Compounding Interest Run of the program
    /* compound interest yearly & quarterly */
#include <stdio.h>

int main() {
   double amount; /* initial amount of money */
   double amount_save; /* saved amount  */
   int years;     /* years to apply interest */
   double rate; /* interest rate in percent */
   int i;              /* loop variable */
   scanf("%lf%i%lf", &amount, &years, &rate);
   amount_save = amount;

   /* apply interest each year */
   for(i = 1; i <= years; i++) {
      amount = amount + amount*rate/100.0;
      printf("End of year %2i:", i);
      printf(" amount: %10.6f\n", amount);

   }
   printf("\n");

   /* apply 0.25*interest 4 times per year */
   amount = amount_save;
   for(i = 1; i <= years*4; i++) {
      amount = amount + amount*rate/4/100.0;
      if (i%4 != 0) { /* all but end of year */
         printf("  End of quarter %2i:", i);
         printf(" amount: %10.6f\n", amount);
      }
      else { /* end of year */
         printf("End of year %2i:", i/4);
         printf(" amount: %10.6f\n", amount);
      }
   }
   printf("\n");
}


    1000 4 10
End of year  1: amount: 1100.000000
End of year  2: amount: 1210.000000
End of year  3: amount: 1331.000000
End of year  4: amount: 1464.100000

  End of quarter  1: amount: 1025.000000
  End of quarter  2: amount: 1050.625000
  End of quarter  3: amount: 1076.890625
End of year  1: amount: 1103.812891
  End of quarter  5: amount: 1131.408213
  End of quarter  6: amount: 1159.693418
  End of quarter  7: amount: 1188.685754
End of year  2: amount: 1218.402898
  End of quarter  9: amount: 1248.862970
  End of quarter 10: amount: 1280.084544
  End of quarter 11: amount: 1312.086658
End of year  3: amount: 1344.888824
  End of quarter 13: amount: 1378.511045
  End of quarter 14: amount: 1412.973821
  End of quarter 15: amount: 1448.298166
End of year  4: amount: 1484.505621
(Only print at end of year)
1000 10 30
End of year  1: amount: 1300.000000
End of year  2: amount: 1690.000000
End of year  3: amount: 2197.000000
End of year  4: amount: 2856.100000
End of year  5: amount: 3712.930000
End of year  6: amount: 4826.809000
End of year  7: amount: 6274.851700
End of year  8: amount: 8157.307210
End of year  9: amount: 10604.499373
End of year 10: amount: 13785.849185

End of year  1: amount: 1335.469141
End of year  2: amount: 1783.477826
End of year  3: amount: 2381.779599
End of year  4: amount: 3180.793154
End of year  5: amount: 4247.851100
End of year  6: amount: 5672.874058
End of year  7: amount: 7575.948244
End of year  8: amount: 10117.445090
End of year  9: amount: 13511.535700
End of year 10: amount: 18044.238970

    See Hi-tech version of compounding interest.

  5. Write a code segment that will use the following equations to approximate pi to 15 digits:

    Then replace a1 and b1 by a0 and b0, and repeat the last two equations. Keep repeating until a1 and b1 are within 1.0e-15 of one another. The final answer should be the average of a1 and b1.

    Calculate PI Execution of the program
    #include <stdlib.h>
#include <math.h>

int main() {
   double a0 = 2*sqrt(3);
   double b0 = 3;
   double a1, b1;
   int count = 0;
   printf(" iter       a1               b1\n\n");
   while (1) {
         a1 = 2*a0*b0/(a0 + b0);
         b1 = sqrt(a1*b0);
         printf("# %2i: %18.15f  ", ++count, a1);
         printf("%18.15f\n", b1);
         if (fabs(a1 - b1) < 1.0e-14) break;
         a0 = a1;
         b0 = b1;
   }
   printf("Final Answer: %18.15f\n",
   (a1 + b1)/2);
}
     iter       a1                  b1

#  1:  3.215390309173472   3.105828541230249
#  2:  3.159659942097500   3.132628613281238
#  3:  3.146086215131435   3.139350203046867
#  4:  3.142714599645368   3.141031950890509
#  5:  3.141873049979824   3.141452472285462
#  6:  3.141662747056848   3.141557607911857
#  7:  3.141610176604689   3.141583892148318
#  8:  3.141597034321526   3.141590463228050
#  9:  3.141593748771351   3.141592105999271
# 10:  3.141592927385096   3.141592516692157
# 11:  3.141592722038614   3.141592619365384
# 12:  3.141592670701998   3.141592645033691
# 13:  3.141592657867844   3.141592651450767
# 14:  3.141592654659306   3.141592653055036
# 15:  3.141592653857171   3.141592653456104
# 16:  3.141592653656637   3.141592653556371
# 17:  3.141592653606504   3.141592653581437
# 18:  3.141592653593970   3.141592653587704
# 19:  3.141592653590837   3.141592653589270
# 20:  3.141592653590053   3.141592653589662
# 21:  3.141592653589858   3.141592653589759
# 22:  3.141592653589808   3.141592653589784
# 23:  3.141592653589796   3.141592653589790
Final Answer:  3.141592653589793

Arrays:

  1. Write a C function that will take two 1-dimensional int arrays and their size as formal parameters, and will compute and return their inner product. If int a[] = {1, 2, 3}; and int b[] = {6, 7, 8};, then the inner product (a*b) is 1*6 + 2*7 + 3*8. Execute your function with the above example and with another longer example.

    Inner product Rest of code and run
    #include <stdio.h> /* for input/output */
void print_array(char name, int x[], int n);
int inner_product(int x[], int y[], int n);

int main() {
   /* initialize vectors using {} initializer */
   /* first pair of test vectors */
   int a[] = {1, 2, 3};
   int b[] = {6, 7, 8};
   /* second pair of test vectors */
   int c[] = {5, 4, 3, 2, 1};
   int d[] = {6, 7, 8, 9, 10};
   int inn_prod;

   print_array('a', a, 3);
   print_array('b', b, 3);
   inn_prod = inner_product(a, b, 3);
   printf("Inner product is: %i\n\n", inn_prod);

   print_array('c', c, 5);
   print_array('d', d, 5);
   inn_prod = inner_product(c, d, 5);
   printf("Inner product is: %i\n\n", inn_prod);

}

    /* calcuate inner product of two vectors */
int inner_product(int x[], int y[], int n) {
   int i;
   int result = 0;
   for (i = 0; i < n; i++) {
      result = result + x[i]*y[i];
   }
   return result;
}

/* print_array: print a 1-dimensional array */
void print_array(char name, int x[], int n) {
   int i;
   printf("Array %c: ", name);
   for (i = 0; i < n; i++) {
      printf("%3i", x[i]);
      if (i < n - 1) printf(", ");
      else printf("\n");
   }
}
Results of a run:
Array a:   1,   2,   3
Array b:   6,   7,   8
Inner product is: 44

Array c:   5,   4,   3,   2,   1
Array d:   6,   7,   8,   9,  10
Inner product is: 110

  2. Start with a 2-dimensional array int m[N][N], where there is a #define N 16, say. For each location m[i][j], put the number i*j into it. Then use a separate function (and pass the array as a parameter) to print the array in the form of a multiplication table, leaving off row 0 and column 0. [Oops. This was in the Review for Exam 2. The answer is there also.]

  3. Consider the declaration:

    In this case, months[1], months[2], ... are each strings that can be printed with a %s format. First use a for loop to print them.

    Now create a array of random numbers of the same size as months:

    Finally, sort the array ran into order, carrying the array months along with it. The result is to randomize the order of the months.

    Randomize the order of an array Execution of the program
    #include <stdlib.h>
#include <time.h>

int main() {
   // array of strings, each string an array of char
   char *months[] = {"January",
      "February", "March", "April", "May",
      "June", "July", "August", "September",
      "October", "November", "December"};

   // array of random doubles
   double ran[12];
   int i, dum;
   srand((long)time(NULL));
   // print the strings, using %s for string
   printf("Months:\n");
   for(i = 0; i < 12; i++)
      printf("  %s\n", months[i]);

   // load and print the doubles, random order
   printf("\nRandom doubles (unsorted):\n");
   for (i = 0; i < 12; i++)
      ran[i] = rand()/(double)RAND_MAX;
   for(i = 0; i < 12; i++)
      printf("%20.15f\n", ran[i]);

   // sort doubles into decreasing order
   // use bubblesort, that interchanges elements
   // interchange months array exactly the same way
   for (dum = 0; dum < 11; dum++)
      for (i = 0; i < 11; i++)
         if (ran[i] < ran[i+1]) {
            char *tempm;
            // interchange ran array elts
            double temp = ran[i];
            ran[i] = ran[i+1];
            ran[i+1] = temp;

            // interchange months array elts
            tempm = months[i];
            months[i] = months[i+1];
            months[i+1] = tempm;
         }

   // print the doubles, now decreasing order
   printf("\nRandom doubles (sorted):\n");
   for(i = 0; i < 12; i++)
      printf("%20.15f\n", ran[i]);

   // print the months, now randomized order
   printf("\nMonths (randomized)\n");
   for(i = 0; i < 12; i++)
      printf("  %s\n", months[i]);

   printf("\n");
}
    Months:
  January
  February
  March
  April
  May
  June
  July
  August
  September
  October
  November
  December

Random doubles (unsorted):
   0.141427402916098
   0.607038361303060
   0.389743974613838
   0.405499067811993
   0.256928145539448
   0.091640621932056
   0.825441365514622
   0.986272125498518
   0.176089033566457
   0.313722355437336
   0.632441553116050
   0.371727083051450

Random doubles (sorted):
   0.986272125498518
   0.825441365514622
   0.632441553116050
   0.607038361303060
   0.405499067811993
   0.389743974613838
   0.371727083051450
   0.313722355437336
   0.256928145539448
   0.176089033566457
   0.141427402916098
   0.091640621932056

Months (randomized)
  August
  July
  November
  February
  April
  March
  December
  October
  May
  September
  January
  June

Pointers:

  1. No questions, just referring you to: Pointer Page
structs:

  1. No questions, just referring you to: Structs and Fractions.
  2. In class we covered an example that used a struct and also an array of structs: Letter Frequency Example.