Question 1 was from the first midterm in Winter 2005
Question 2 was from an exam in Fall 2004 (first time those students saw Game of Life)
Question 3 was based on a question from the textbook (page 101)
Question 4 was part from the textbook (page 69), part from a Winter 2005 quiz
Question 5 was new


1.  20 pts, 5 pts per fill-in

	// if nterms = 1 this should compute 1
	// if nterms = 2 this should compute 1 + 1/4
	// if nterms = 3 this should compute 1 + 1/4 + 1/9
	// if nterms = 4 this should compute 1 + 1/4 + 1/9 + 1/16
	// each new term adds an inverse square

	double sum = 0;  		// need to initialize sum

	int i= 1;			// 1/(i*i) is bad if i=0

	while (i <= nterms)		// <= to get correct number of terms
	{
		term = 1.0 / (i*i);	// 1/(i*i) is integer division
		sum  = sum + term;
		i    = i + 1;
	}


2.  20 pts, 4 pts per fill-in

					// sum contains the sum of "live"
					// cells in the 3x3 grid around ic,jc
	return sum - table[ic][jc];  	// must exclude middle cell from total

// There are various ways to implement the rules.  One possibility:

	if (count == 3)			// when there are 3 neighbors
	{
		updated[i][j] = 1;	// next generation is always alive
	}
	else if (count == 2 and table[i][j] == 1 ) // 2 neighbors & alive
	{
		updated[i][j] = 1;	// means it stays alive next generation
	}


3.   20 pts, roughly 2 pts per error, multiple solutions possible

	#include<iostream>
	using namespace std;

	int main()
	{
		double x;
		cout << "Input x ";
		cin >> x;		// was cin << x; 	(syntax)

		double p=1;		// was double p=0; 	(logic)
		double sum=1;		// was double sum; 	(logic)

		int i=2;		// was not present 	(syntax/logic)

		while (i<=N)		// removed ; 		(syntax)
		{
			p = p * x * x;	// was just p=p*x; 	(logic)
			sum = sum + p;
			i = i + 2;	// was not present 	(logic)
		}

		cout << "Sum = " << sum << endl;	// added ; (syntax)

		return 0;
	}


4.   20 pts divided 2/2/3/3/2/2/3/3

	22 base 10 = 10110 base 2. 	// 16 + 4 + 2

	44 base 10 = 101100 base 2.  	// Note 44 = 2*22, so add a zero!

	AB base 16 = 10101011 base 2.  	// AB is a 2-digit number, not A * B
					// A = 1010, B = 1011

	FFE3 + 4FA9 = 14F8C.		// partial credit given for 14F9C
					// but 3 + 9 = C without carry
					// E + A     = 8 carry a 1 (base 16)
					// F + F + 1 = F carry a 1 
					// F + 4 + 1 = 4 carry a 1

	double p = 1 + 2 + 3/4 + 5;	// 3/4 is integer division = 0
					// proper answer is 8.0 (p is a double)
					// 8 (written as an integer) accepted

	int k = 10 % 3;			// % is remainder after division
					// answer = 1

	int j = 13/5 + 13%5		// 13/5 = 2 (drop fraction)
					// 13%5 = 3 (remainder)
					// answer = 5

	int n = 1-(static_cast<int>(3*5/4.0)/2);
		// start with inner most parenthesis just like in math
		// 3 * 5 is 15
		// 15 / 4.0 is 3.75 (not integer division, keeps fraction)
		// integer part of 3.75 is 3 (static_cast<int> drops fraction)
		// integer division 3/2 is 1
		// 1 - 1 gives answer = 0


5.  20 pts, various ways to write the if-condition are possible

	#include<iostream>
	using namespace std;

	int main()
	{
		int age;
		cout << " Enter your age ";
		cin >> age;

		cout << "Hello, you are old enough to purchase ";

		if (age >= 21)
		{
			cout << "booze and cigarettes." << endl;
		}
		else if (age < 18)
		{
			cout << "neither booze nor cigarettes." << endl;
		}
		else  // case of age >= 18 and age < 21
		{
			cout << "cigarettes, but not booze." << endl;
		}
	}