Unit 4 - HW Quiz

Unit 4 - Iteration:

• This is the homework quiz for unit 4, iterations
• 4 multiple choice questions
• 2 programming hacks
• 1 bonus programming hack (required to get above 0.9)

Question 1:

What does the following code print?

A. 5 6 7 8 9

B. 4 5 6 7 8 9 10 11 12

C. 3 5 7 9 11

D. 3 4 5 6 7 8 9 10 11 12

Explain your answer. (explanation is graded not answer)

for (int i = 3; i <= 12; i++) {
   System.out.print(i + " ");
}

Answer: D The code iterates through all the numbers from 3 to 12 and prints out the index. It index starts at three, and therefore prints out three, then iterates by one and ends at 12. D is the only option that does this.

Bonus:

• Explain the difference between using a variable like i inside a for loop, vs. using a variable that exists in the code itself for a while loop

Using a new variable ensures that the loop functions properly, because when a variable that exists in the code itself is used, things can change, as changing one part of the code results in alteration of another part that can have unexpected results.

Question 2:

How many times does the following method print a “*” ?

A. 9

B. 7

C. 8

D. 6

Explain your answer. (explanation is graded not answer)

for (int i = 3; i < 11; i++) {
   System.out.print("*");
}

C. The code iterates beginning at 3 and ends at 11, index incrementing by one and printing out a * each time. It prints out the * 11-3 = 8 times.

Question 3:

What does the following code print?

A. -4 -3 -2 -1 0

B. -5 -4 -3 -2 -1

C. 5 4 3 2 1

Explain your answer. (explanation is graded not answer)

int x = -5;
while (x < 0)
{
   x++;
   System.out.print(x + " ");
}

A. It prints out the index starting at -5 and is incremented by one each time. The increment is done before printing so it starts at -4.

Question 4:

What does the following code print?

A. 20

B. 21

C. 25

D. 30

Explain your answer. (explanation is graded not answer)

int sum = 0;

for (int i = 1; i <= 5; i++) {
    if (i % 2 == 0) {
        sum += i * 2;
    } else {
        sum += i;
    }
}

System.out.println(sum);

This sums up the numbers from one two five but doubles the even numbers. 1 + 4 + 3 + 8 + 5 = 21.

Loops HW Hack

Easy Hack

• Use a while loop to find the numbers from 1-50 that are divisible by 3 or 5, then store them into a list (make sure to print it out at the end)
• Use a for loop to do the same thing detailed above

for (int i = 1; i < 51; i++) {
    if (i % 3 == 0) {
        System.out.println(i + " is divisible by 3");
    }
}

int j = 1;
while (j < 51) {
    if (j % 5 == 0) {
        System.out.println(j + " is divisible by 5");
    }
    j += 1;
}

3 is divisible by 3
6 is divisible by 3
9 is divisible by 3
12 is divisible by 3
15 is divisible by 3
18 is divisible by 3
21 is divisible by 3
24 is divisible by 3
27 is divisible by 3
30 is divisible by 3
33 is divisible by 3
36 is divisible by 3
39 is divisible by 3
42 is divisible by 3
45 is divisible by 3
48 is divisible by 3


5 is divisible by 5
10 is divisible by 5
15 is divisible by 5
20 is divisible by 5
25 is divisible by 5
30 is divisible by 5
35 is divisible by 5
40 is divisible by 5
45 is divisible by 5
50 is divisible by 5

Harder Hack

Palindromes are numbers that have the same value when reversed (ex: “123321” or “323”). Create a program that uses a while loop that outputs all palindromes in any given list.

Sample Input: test_list = [5672, 235, 5537, 6032, 317, 8460, 1672, 8104, 7770, 4442, 913, 2508, 1116, 9969, 9091, 522, 8756, 9527, 7968, 1520, 4444, 515, 2882, 6556, 595]

Sample Output: 4444, 515, 2882, 6556, 595

public static boolean isPalindrome(int number) {
    String str = Integer.toString(number);
    String reversedStr = new StringBuilder(str).reverse().toString();
    return str.equals(reversedStr);
}


int[] test_list = {5672, 235, 5537, 6032, 317, 8460, 1672, 8104, 7770, 4442, 913, 2508, 1116, 9969, 9091, 522, 8756, 9527, 7968, 1520, 4444, 515, 2882, 6556, 595};

for (int number : test_list) {
    if (isPalindrome(number)) {
        System.out.println(number + " is a palindrome.");
    }
}

4444 is a palindrome.
515 is a palindrome.
2882 is a palindrome.
6556 is a palindrome.
595 is a palindrome.

Bonus Hack (for above 0.9)

Use a for loop to output a spiral matrix with size n

Example:

Sample Input: n = 3

Output: [[1, 2, 3], [8, 9, 4], [7, 6, 5]]

static int n = 15;
static int[][] matrix = new int[n][n];
static int nextInt = 1;

int top = 0, bottom = n - 1, left = 0, right = n - 1;

while (nextInt <= n * n) {
    for (int col = left; col <= right && nextInt <= n * n; col++) {
        matrix[top][col] = nextInt++;
    }
    top++;

    for (int row = top; row <= bottom && nextInt <= n * n; row++) {
        matrix[row][right] = nextInt++;
    }
    right--;

    for (int col = right; col >= left && nextInt <= n * n; col--) {
        matrix[bottom][col] = nextInt++;
    }
    bottom--;

    for (int row = bottom; row >= top && nextInt <= n * n; row--) {
        matrix[row][left] = nextInt++;
    }
    left++;
}

for (int[] row : matrix) {
    for (int value : row) {
        System.out.print(value + "\t");
    }
    System.out.println();
}

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	
56	57	58	59	60	61	62	63	64	65	66	67	68	69	16	
55	104	105	106	107	108	109	110	111	112	113	114	115	70	17	
54	103	144	145	146	147	148	149	150	151	152	153	116	71	18	
53	102	143	176	177	178	179	180	181	182	183	154	117	72	19	
52	101	142	175	200	201	202	203	204	205	184	155	118	73	20	
51	100	141	174	199	216	217	218	219	206	185	156	119	74	21	
50	99	140	173	198	215	224	225	220	207	186	157	120	75	22	
49	98	139	172	197	214	223	222	221	208	187	158	121	76	23	
48	97	138	171	196	213	212	211	210	209	188	159	122	77	24	
47	96	137	170	195	194	193	192	191	190	189	160	123	78	25	
46	95	136	169	168	167	166	165	164	163	162	161	124	79	26	
45	94	135	134	133	132	131	130	129	128	127	126	125	80	27	
44	93	92	91	90	89	88	87	86	85	84	83	82	81	28	
43	42	41	40	39	38	37	36	35	34	33	32	31	30	29