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Week 3 - Graded Assignment 3

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Week 3 - Graded Assignment 3

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Course: Jan 2026 - Python

Topic: While Loop Iteration Counting | Marks: 3

Question 1

How many asterisks does the following code print


python
i = -3
while (i <= 13):
    print('**')
    print('*')
    i = i + 1

Status: Yes, the answer is correct. Score: Score: 3

Feedback: In every iteration 3 asterisks will be printed, 17 iterations happens, hence 17 * 3 = 51 asterisks will be printed.

Accepted Answers:

Topic: Logarithmic logic using Floor Division | Marks: 3

Question 2

Select the correct implementation of a program that accepts a positive integer

x

as input and prints the maximum value of the integer

y

such that

2^y \leq x

Sample Test Cases

```python
x = int(input())
y = 0
while x > 1:
x = x // 2
y = y + 1
print(y)
```python
x = int(input())
y = 0
while x >= 1:
x = x // 2
y = y + 1
print(y)
```python
x = int(input())
y = 0
while x > 1:
x = x / 2
y = y + 1
print(y)
```python
x = input()
y = 0
while x > 1:
x = x // 2
y = y + 1
print(y)

Status: Yes, the answer is correct. Score: Score: 3

Feedback: The basic mathematical idea is as follows. For any positive integer

x

, we can always find a

y

such that:

\displaystyle 2^y \leq x < 2^{y + 1}

That is, every number should lie between two consecutive powers of

2

. From this it follows that:

\displaystyle 2^{y - 1} \leq \left \lfloor \cfrac{x}{2} \right \rfloor < 2^y

The exercise for you is to connect this mathematical insight with the option (a). We have used floor function in the inequality. This corresponds to the // operator in Python. Why is it important to do floor division? Take the case of a number that is not a power of

2

, say

10

, and run the code given in option-(c). What happens?

Now, for the wrong options:

Option-(b) is quite close to option-(a), but has the >= sign in the while condition. This will make the loop go through one more iteration than what is needed.
Option-(c) does normal division and not floor division.

Why is option-(d) wrong? This is left as an exercise for you.

Accepted Answers:


python
x = int(input())
y = 0
while x > 1:
    x = x // 2
    y = y + 1
print(y)

Topic: State-based String Processing | Marks: 3

Question 3

Consider the following snippet of code.


python
name = input()
nick = ''    # there is no space between the quotes
space = ' ' # there is one space between the quotes
first_char = True
for char in name:
    if first_char == True:
        nick = nick + char
        first_char = False
    if char == space:
        first_char = True
print(nick)
^nick-generation-logic

What is the output for the following input?


plain
Albus Percival Brian Wulfric Dumbledore

```plain-text
Albus
```plain-text
Dumbledore
```plain-text
AP
```plain-text
APBWD

Status: Yes, the answer is correct. Score: Score: 3

Feedback: When does first_char become True? When does it become False? If you can answer these two questions, then you are halfway on the path to understanding the code. The names of variables can often help you figure out their purpose in the code. The two variables names — first_char and nick — have a clear meaning. Can you make an informed guess as to what they stand for?

Note This method of using a Boolean variable such as first_char to determine the state of the code is explained in detail in approach-2 of PPA-4 in the PPA-hints document. Refer to it for more details. Once you read that section, get back to this problem and identify what the state stands for here.

Accepted Answers:


plain
APBWD

Topic: Fibonacci Sequence Implementations | Marks: 3

Question 4

The first five terms of the Fibonacci sequence is given below.

\displaystyle \begin{gathered} F_1 = 1& F_2 = 1& F_3 = 2\ F_4 = 3& F_5 = 5 \end{gathered}

We wish to write a program that accepts a positive integer

n

as input and prints

F_n

as the output. Select all correct implementations of this program. (MSQ)

```python
n = int(input())
F_prev = 1
F_curr = 1
count = 2
while count < n:
temp = F_prev + F_curr
F_prev = F_curr
F_curr = temp
count += 1
print(F_curr)

```python
n = int(input())
if n <= 2:
print(1)
else:
F_prev = 1
F_curr = 1
count = 2
while count < n:
temp = F_prev + F_curr
F_prev = F_curr
F_curr = temp
count += 1
print(F_curr)
```python
n = int(input())
F_prev = 1
F_curr = 1
for i in range(n):
temp = F_prev + F_curr
F_prev = F_curr
F_curr = temp
print(F_curr)
```python
n = int(input())
F_prev = 1
F_curr = 1
for i in range(n - 2):
temp = F_prev + F_curr
F_prev = F_curr
F_curr = temp
print(F_curr)

Status: Yes, the answer is correct. Score: Score: 3

Feedback: F_prev and F_curr correspond to two consecutive terms in the sequence. F_prev + F_curr is the next term in the sequence. So, we need to perform the following updates:

(1) F_prev + F_curr will be the new F_curr and

(2) F_prev will be assigned the current value of F_curr

If this is not clear, consider the following table:

<table style="box-sizing: border-box;font-size: 16px;margin: 1em 0;border-collapse: collapse;width: fit-content;max-width: 100%;overflow-x: auto;display: block;font-variant-numeric: lining-nums tabular-nums;" border="1"><thead style="box-sizing: border-box;font-size: 16px;"><tr style="box-sizing: border-box;font-size: 16px;"><th style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"></th><th style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;">Current</th><th style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;">Next</th></tr></thead><tbody style="box-sizing: border-box;font-size: 16px;margin-top: 0.5em;border-top: thin solid #000;border-bottom: thin solid #000;"><tr style="box-sizing: border-box;font-size: 16px;"><td style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"><span style="box-sizing: border-box;font-size: 85%;font-family: monospace;padding: 1px 5.5px 3px 5.5px;background-color: rgba(0, 0, 0, 0.07);border-radius: 5px">F_prev</span></td><td style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"><span style="box-sizing: border-box;font-size: 85%;font-family: monospace;padding: 1px 5.5px 3px 5.5px;background-color: rgba(0, 0, 0, 0.07);border-radius: 5px">x</span></td><td style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"><span style="box-sizing: border-box;font-size: 85%;font-family: monospace;padding: 1px 5.5px 3px 5.5px;background-color: rgba(0, 0, 0, 0.07);border-radius: 5px">y</span></td></tr><tr style="box-sizing: border-box;font-size: 16px;"><td style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"><span style="box-sizing: border-box;font-size: 85%;font-family: monospace;padding: 1px 5.5px 3px 5.5px;background-color: rgba(0, 0, 0, 0.07);border-radius: 5px">F_curr</span></td><td style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"><span style="box-sizing: border-box;font-size: 85%;font-family: monospace;padding: 1px 5.5px 3px 5.5px;background-color: rgba(0, 0, 0, 0.07);border-radius: 5px">y</span></td><td style="box-sizing: border-box;font-size: 16px;border: thin solid #000;padding: 0.5em 0.5em 0.25em 0.5em;"><span style="box-sizing: border-box;font-size: 85%;font-family: monospace;padding: 1px 5.5px 3px 5.5px;background-color: rgba(0, 0, 0, 0.07);border-radius: 5px">x + y</span></td></tr></tbody></table>

If these operations are done sequentially, say (1) followed by (2), update (2) would fail as we no longer have access to the old value of F_curr. On the other hand, if (2) is performed first followed by (1), the first operation would fail as we no longer have access to the old value of F_prev.

In some sense, both these updates have to be done simultaneously. This is where a temp variable comes in handy. Convince yourself that the use of temp variable circumvents the problems posed by sequential updates.

As a final point, the interesting thing is that you don't need the if-else condition for

n = 1

and

n = 2

. Options (a) and (d) show the way to do it. Option-(b) is also correct, but makes the code less elegant. Why is option-(c) wrong? This is left as an exercise for you.

Accepted Answers:


python
n = int(input())
F_prev = 1
F_curr = 1
count = 2
while count < n:
    temp = F_prev + F_curr
    F_prev = F_curr
    F_curr = temp
    count += 1
print(F_curr)


python
n = int(input())
if n <= 2:
    print(1)
else:
    F_prev = 1
    F_curr = 1
    count = 2
    while count < n:
        temp = F_prev + F_curr
        F_prev = F_curr
        F_curr = temp
        count += 1
    print(F_curr)


python
n = int(input())
F_prev = 1
F_curr = 1
for i in range(n - 2):
    temp = F_prev + F_curr
    F_prev = F_curr
    F_curr = temp
print(F_curr)

Topic: While Loop Termination Logic | Marks: 2

Question 5

If the while loop of below code executes, when does it terminate?


python
x = int(input("Enter any number: "))
while(x % 5 != 0 and x % 10 != 0):
    x = int(input("Enter any number: "))
print("outside loop, the value of x is ", x)

Never terminates it is a infinite loop.
when we input a number which is multiple of 10.
when we input a number which is multiple of 5.
when we input a number which is not a multiple of both 5 and 10.

Feedback: while loop terminates if the condition x % 5 != 0 and x % 10 != 0 becomes False it happens only when we input a number which is multiple of 5. (multiple of 10 is always a multiple of 5).

Accepted Answers:

when we input a number which is multiple of 10.

when we input a number which is multiple of 5.

Topic: Loop Control (continue) and Vowel Filtering | Marks: 3

Question 6

What does the following code block print?


python
for i in 'We are in question one':
    if i == 'a' or i == 'e' or i == 'i' or i == 'o' or i == 'u':
        continue
    print(i, end = '')

W r n qstn n
Wrnqstnn
We are in question one
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Feedback: The for loop checks whether each character is a vowel or not. If the character is a vowel (a', 'e', 'i', 'o', 'u' ), the character is skipped and not printed on the console due to continue statement.

Accepted Answers:

W r n qstn n

Topic: String Multiplication for Repetition | Marks: 3

Question 7

Is it possible to get the below output without using any loop and without repeating the string Hello Python! in Python?


plain
Hello Python!
Hello Python!
Hello Python!
Hello Python!
Hello Python!
Hello Python!
Hello Python!
Hello Python!
Hello Python!
Hello Python!

True
False

Status: Yes, the answer is correct. Score: Score: 3

Feedback: Multiplying a string with a positive number n is equivalent to the string repeated n times one after another.

print(10*'Hello Python!\n')

Accepted Answers:

True

Topic: Range function Step Size Compatibility | Marks: 3

Question 8

A programmer wants to print a decreasing sequence. How many times does the print statement get executed? And why?


python
for i in range(10, 0, 1):
    print(i)

One time because i takes only the value 10 and thereafter it will be decremented
One time because i takes only the value 9 and thereafter it will be decremented
print statement will not be executed due to invalid end points
print statement will not be executed due to incompatible step size

Status: Yes, the answer is correct. Score: Score: 3

Feedback: The start, end and step parameters in range() are 10, 0 and 1. The variable i starts from 10. It should be incremented at each iteration by 1 and should end at 0, which is not possible. The range(10, 0, 1) returns no values, so the loop does not run and therefore print statement will not be executed.

Accepted Answers:

print statement will not be executed due to incompatible step size

Topic: C-style String Formatting (Padding) | Marks: 3

Question 9

Which is the correct option that can be used in order to add ‘n-1’ blank spaces after a given string T?

Note: Variable T is a single character string.

print("%ns"%T)
print("-ns"%T)
print("%-ns"%T)
print("-ns"%T)

Status: Yes, the answer is correct. Score: Score: 3

Feedback: For any string T and positive integer n ,print("%-ns"%T) format method adds (n-len(T)) blank space after string T and print("%ns"%T) adds n-len(T) blank space before string T . n-len(T) should be greater than 0.

According to the problem statement length of T is 1 so, option (c) is correct to add n-1 blank spaces after a given string T.

Accepted Answers:

print("%-ns"%T)

Topic: Nested Loop Execution Flow | Marks: 3

Question 10

What is the output of the following code


python
total = 0
for i in range(1, 5):
    for j in range(i):
        total = total + i
print(total)

Status: Yes, the answer is correct. Score: Score: 3

Feedback:

In the first iteration of outer loop i = 1, the inner loop iterates 1 time and the total is 1
In the second iteration of outer loop i = 2, the inner loop iterates 2 times and the total is 5
In the third iteration of outer loop i = 3, the inner loop iterates 3 times and the total is 14
In the fourth iteration of outer loop i = 4, the inner loop iterates 4 times and the total is 30

Accepted Answers:

Topic: Nested Range Space Analysis | Marks: 2

Question 11

Common data for the next 2 questions

Consider the following snippet of code.


python
for x in range(100):
    for y in range(100):
        if x != y:
            print(f'{x},{y}')

When the code given above is executed, how many lines will the output have? (NAT)

Your Answer: 9900

Status: Yes, the answer is correct. Score: Score: 2

Accepted Answers:

(Type: String) 9900

Topic: While Loop Equivalence to Nested Range | Marks: 2

Question 12

Among the code blocks given below, select the correct equivalent implementation of the code given in the common data.

```python
x, y = 0, 0
while x < 100:
while y < 100:
if x != y:
print(f'{x},{y}')
```python
x, y = 0, 0
while x < 100:
while y < 100:
if x != y:
print(x, y)
x += 1
y += 1
```python
x, y = 0, 0
while x < 100:
while y < 100:
if x != y:
print(x, y)
y += 1
x += 1
```python
x, y = 0, 0
while x < 100:
while y < 100:
if x != y:
print(x, y)
y += 1
x += 1
```python
x, y = 0, 0
while x < 100:
y = 0
while y < 100:
if x != y:
print(x, y)
y += 1
x += 1

Status: Yes, the answer is correct. Score: Score: 2

Feedback: The last option is the correct one. Now start from option (a) and go all the way up to option (d). Compare each of these options with option (e). This will reveal all the errors to you. A brief demonstration is given for your reference:

In some ways, option-(a) has some very serious errors. x and y are never incremented!
There is an attempt to correct this mistake in option-(b). Unfortunately, x and y are being incremented in the wrong place.
The same problem persists in option-(c).
Option-(d) corrects the position of increment operations. However, it still misses one crucial step, that of resetting y to zero within the inner loop.

Accepted Answers:


python
x, y = 0, 0
while x < 100:
    y = 0
    while y < 100:
        if x != y:
            print(x, y)
        y += 1
    x += 1

Topic: Sum of first N Odd Numbers (While) | Marks: 3

Question 13

Select all the snippets that prints the sum of the first n odd numbers starting from 1(including). Assume n is a positive integer and is already defined.

Example n = 5 , output: 1+3+5+7+9 = 25

```python
result = 0
i = 0
while i<n:
result+=2*i+1
i+=1
print(result)
```python
result = 0
i = 1
while i<n:
result+=2*i+1
i+=1
print(result)
```python
result = 0
i = 1
while i<n:
result+=2*i+1
i+=1
result+=1
print(result)
```python
result = 0
i = 0
while i<n:
if i%2 != 0:
result+=2*i+1
i+=1
print(result)
```python
result = 0
i = 0
while i<2*n+1:
if i%2 != 0:
result+=i
i+=1
print(result)
```python
result = 0
i = 0
while i<2*n+1:
if i%2 != 0:
result+=i
i+=1
print(result)
```python
result = 0
i = 0
while i<2*n+1:
if i%2 != 0:
result+=i
i+=1
print(result)

Accepted Answers:


python
result = 0
i = 0
while i<n:
    result+=2*i+1
    i+=1
print(result)


python
result = 0
i = 1
while i<n:
    result+=2*i+1
    i+=1
result+=1
print(result)


python
result = 0
i = 0
while i<2*n+1:
    if i%2 != 0:
      result+=i
    i+=1
print(result)

Topic: Sum of first N Odd Numbers (For) | Marks: 3

Question 14

Select all the snippets that prints the sum of the first n odd numbers starting from 1(including). Assume n is a positive integer and is already defined.

Example n = 5 , output: 1+3+5+7+9 = 25

```python
result = 0
for i in range(n):
result+=2*i+1
print(result)
```python
result = 0
for i in range(1,n+1):
result+=2*i-1
print(result)
```python
result = 0
for i in range(1,n):
result+=2*i+1
result+=1
print(result)
```python
result = 0
for i in range(1,n):
result+=2*i+1
print(result)
```python
result = 0
for i in range(-1,-n-1,-1):
result-=2*i+1
print(result)
```python
result = 0
for i in range(-1,-n):
result-=2*i+1
print(result)

Accepted Answers:


python
result = 0
for i in range(n):
    result+=2*i+1
print(result)


python
result = 0
for i in range(1,n+1):
    result+=2*i-1
print(result)


python
result = 0
for i in range(1,n):
    result+=2*i+1
result+=1
print(result)


python
result = 0
for i in range(-1,-n-1,-1):
    result-=2*i+1
print(result)

Topic: Data Processing Patterns (Aggregation/Filtering) | Marks: 3

Question 15

Consider the below code.


python
n = int(input())
value = 0
for i in range(n):
  num = int(input())
  if len(str(num))<2:
      value+= num
print(value)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Feedback: Output is the sum of single digit numbers

Accepted Answers:

Aggregation

Filtering

Question 16

Consider the below code.


python
sentence = input()
for word in sentence.split():
    if 'a' in word:
        print(word)
        break

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Accepted Answers:

Aggregation

Question 17

Consider the below code.


python
sentence = input()
for word in sentence.split():
    if 'a' in word:
        print(word*2)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Accepted Answers:

Filtering

Mapping

Question 18

Consider the below code.


python
sentence = input()
for word in sentence.split():
    if 'a' in word:
        print(word*2)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Accepted Answers:

Filtering

Mapping

Question 19

Consider the below code.


python
sentence = input()
value = ""
for char in sentence:
    if char in 'aeiou':
        value+=chr(ord(char)+1)
print(value)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Accepted Answers:

Aggregation

Filtering

Mapping

Question 20

Consider the below code.


python
n = int(input())
for i in range(n):
    print(int(input())*2)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Accepted Answers:

Mapping

Question 21

Consider the below code.


python
n = int(input())
for i in range(n):
  num = int(input())
  if len(str(num))<2:
      print(num)
      break

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Accepted Answers:

Aggregation

Question 22

Consider the below code.


python
n = int(input())
value = None
for i in range(n):
  num = int(input())
  if len(str(num))<2:
      value = num
print(value)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Accepted Answers:

Aggregation

Question 23

Consider the below code.


python
n = int(input())
value = None
for i in range(n):
  num = int(input())
  if value is None:
      value = num
  elif num < value:
      value = num
print(value)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Status: Yes, the answer is correct. Score: Score: 3

Accepted Answers:

Aggregation

Question 24

Consider the below code.


python
n = int(input())
value = None
for i in range(n):
  num = len(input())
  if value is None:
      value = num
  elif num/2 < value:
      value = num
print(value)

Select all the data processing pattern(s) found in the given code.

Aggregation
Filtering
Mapping
None of the above

Accepted Answers:

Aggregation

Question 25

Consider the below python code.


python
num = int(input())
i = num
result = 0
while True:
    i -= 4
    if i>10:
        continue
    if i<-10:
        break
    result += i
print(result)

Select all the possible outputs of the above code for any possible valid integer as input.

Hint: To solve this question you might want to try out with different values of num or automate that with another loop with different values for num. What range of values do you use for this automation?

5
4
0
-4
-6
-9
-10
-11
-12
-13

Status: Yes, the answer is correct. Score: Score: 4

Accepted Answers:

-4

-9

-10

-12

Question 26

Consider the below python code.


python
num = int(input())
i = num
result = 0
while True:
    i -= 5
    if i>15:
        continue
    if i<-15:
        break
    result -= i
print(result)

Select all the possible outputs of the above code for any possible valid integer as input.

-9
-7
-3
-2
1
3
4
5
8
9

Status: Yes, the answer is correct. Score: Score: 4

Accepted Answers:

-9

-3

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