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Sahil
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Add recap Day 2 notes
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‎day3/recap.py

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"""
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program to check if the number (say 'n') given as input is a prime
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a prime no. is the one which has only 1 and the numbers itself as its divisors
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1 is not a prime no.
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eg. n = 2, 3, 5, 7, ...
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idea: start a loop from 2 to n/2, if n is divisble by the divisor => not prime
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because a no. cannot be divided by a no. > n/2 except n
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"""
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n = int(input())
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checked = 0
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for divisor in range(2, n//2 + 1):
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if (n % divisor == 0):
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print("No is not prime!")
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checked = 1
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# no need checking further, can break the loop here itself
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break
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# the above code doesn't say that 1 is not a prime because loop won't run even once
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# to handle that case, we can add the extra condition:
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if (n == 1):
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print("No is not prime!")
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checked = 1
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# in cases where checked is false or equal to 0, can say it is prime!
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if (not checked):
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print("No is prime!")
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"""
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program to print the following pattern:
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*
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* *
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* * *
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* * * *
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...
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idea: As you can observer, we are looping in two directions, row wise and column wise
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1st row has 1 star
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2nd row has 2 stars
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3rd row has 3 stars and so on...
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the outer loop would represent looping over the rows
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and in the inner loop, we will loop over the elements in the rows, i.e. columns
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Always better to use loop variable 'r' or 'row' for looping on rows
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and the loop variable 'c' or 'col' to loop on columns
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for easy understanding and debugging of your code !
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"""
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num_rows = n
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for row in range(1, num_rows + 1):
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# r_th row will have r stars
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for col in range(1, row + 1):
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print('*', end=' ')
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# endline
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print('')
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# is it possible to do the same thing using one loop, think?
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# idea is the same: print 1 char in 1st row, 2 chars in 2nd row, ...
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line_num = 1
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printed_on_cur_row = 0
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while (line_num <= num_rows):
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if (printed_on_cur_row < line_num):
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print('*', end=' ')
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printed_on_cur_row += 1
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else:
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print('')
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# reset count of stars on cur_row to be zero
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printed_on_cur_row = 0
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# increment the line_num, i.e. move to next row
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line_num += 1
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"""
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function to return maximum of three numbers
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idea: let numbers be a, b, c
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when can a be the max? when a >= b and a >= c
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similarly for b and c
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so, write three conditions, and return the answer if that condition is true
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also, you can use if elif else as well!
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"""
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def maximum_of_three(num_one, num_two, num_three):
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if (num_one >= num_two and num_one >= num_three):
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return num_one
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if (num_two >= num_one and num_two >= num_three):
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return num_two
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if (num_three >= num_one and num_three >= num_two):
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return num_three
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return -1
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n1 = 3
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n2 = 7
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n3 = 1
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res = maximum_of_three(n1, n2, n3)
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print(res)
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# idea: assign max = a, now check if b >= max, max = b, again if c >= max, max = c
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def maximum_of_three_simple(n1, n2, n3):
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maxi = n1
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if (n2 >= maxi):
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maxi = n2
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if (n3 >= maxi):
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maxi = n3
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return maxi
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res = maximum_of_three(n1, n2, n3)
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print(res)

‎lecture3.md

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## Day 2 Recap
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[Control Flow: Conditionals, Loops, Jump statements, functions](day3/recap.py)
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### Check if given number is a prime
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<details>
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<summary>
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Code
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</summary>
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```py
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"""
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program to check if the number (say 'n') given as input is a prime
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a prime no. is the one which has only 1 and the numbers itself as its divisors
17+
1 is not a prime no.
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eg. n = 2, 3, 5, 7, ...
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idea: start a loop from 2 to n/2, if n is divisble by the divisor => not prime
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because a no. cannot be divided by a no. > n/2 except n
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"""
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n = int(input())
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checked = 0
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for divisor in range(2, n//2 + 1):
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if (n % divisor == 0):
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print("No is not prime!")
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checked = 1
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# no need checking further, can break the loop here itself
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break
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# the above code doesn't say that 1 is not a prime because loop won't run even once
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# to handle that case, we can add the extra condition:
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if (n == 1):
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print("No is not prime!")
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checked = 1
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# in cases where checked is false or equal to 0, can say it is prime!
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if (not checked):
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print("No is prime!")
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```
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</details>
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### Printing star pattern
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<details>
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<summary>
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Code
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</summary>
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```py
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"""
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program to print the following pattern:
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*
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* *
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* * *
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* * * *
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...
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idea: As you can observer, we are looping in two directions, row wise and column wise
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1st row has 1 star
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2nd row has 2 stars
66+
3rd row has 3 stars and so on...
67+
68+
the outer loop would represent looping over the rows
69+
and in the inner loop, we will loop over the elements in the rows, i.e. columns
70+
71+
Always better to use loop variable 'r' or 'row' for looping on rows
72+
and the loop variable 'c' or 'col' to loop on columns
73+
for easy understanding and debugging of your code !
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"""
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num_rows = n
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for row in range(1, num_rows + 1):
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# r_th row will have r stars
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for col in range(1, row + 1):
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print('*', end=' ')
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# endline
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print('')
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# is it possible to do the same thing using one loop, think?
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# idea is the same: print 1 char in 1st row, 2 chars in 2nd row, ...
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line_num = 1
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printed_on_cur_row = 0
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while (line_num <= num_rows):
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if (printed_on_cur_row < line_num):
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print('*', end=' ')
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printed_on_cur_row += 1
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else:
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print('')
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# reset count of stars on cur_row to be zero
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printed_on_cur_row = 0
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# increment the line_num, i.e. move to next row
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line_num += 1
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```
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</details>
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### Find max of three numbers
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<details>
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<summary>
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Function 1
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</summary>
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```py
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"""
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function to return maximum of three numbers
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idea: let numbers be a, b, c
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when can a be the max? when a >= b and a >= c
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similarly for b and c
117+
so, write three conditions, and return the answer if that condition is true
118+
also, you can use if elif else as well!
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"""
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def maximum_of_three(num_one, num_two, num_three):
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if (num_one >= num_two and num_one >= num_three):
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return num_one
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if (num_two >= num_one and num_two >= num_three):
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return num_two
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if (num_three >= num_one and num_three >= num_two):
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return num_three
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return -1
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n1 = 3
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n2 = 7
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n3 = 1
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res = maximum_of_three(n1, n2, n3)
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print(res)
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```
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</details>
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<details>
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<summary>
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Function 2
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</summary>
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```py
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# idea: assign max = a, now check if b >= max, max = b, again if c >= max, max = c
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def maximum_of_three_simple(n1, n2, n3):
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maxi = n1
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if (n2 >= maxi):
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maxi = n2
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if (n3 >= maxi):
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maxi = n3
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return maxi
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res = maximum_of_three(n1, n2, n3)
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print(res)
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```
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</details>
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## More on Functions
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[Code](day3/functions.py)
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