# CMU 15-112: Fundamentals of Programming and Computer Science Check2 Practice (check2 in class on Tue 6-Sep)

• To start:
1. Create a folder named 'week2'
3. Edit check2.py using pyzo
• Do not use strings, lists, or recursion this week.
• Do not hardcode the test cases in your solutions.

1. Week1 time report and brief reflection
First, if you have not done so already, take a brief moment to fill out this week1 time report and brief reflection. Thanks!

2. Loops notes and videos
Next, carefully watch the videos and review the notes on Loops. Be sure to thoroughly understand them before proceeding to the next step!

3. Code Tracing
What will this code print? Figure it out by hand, then run the code to confirm. Then slightly edit the code and try again.
• Trace #1 of 3:
def ct1(m, n): total = 0 for x in range(m, n+1, 3): print('x =', x) total += x for y in range(m, m+2): print('y = ', y) total += y return total print(ct1(1,9))

• Trace #2 of 3:
def ct2(n): k = 0 total = 0 while (n >= k): print('k =', k) for i in range(k): total += n%10 n //= 10 print(i, n%10, total) k += 1 print('total =', total) print(ct2(1234))

• Trace #3 of 3:
def ct3(z): total = 0 for y in range(z,1,-1): if (y % 2 == 0): print('skip y =', y) continue total += y if (total > 20): print('break at y =', y) break return total print(ct3(10))

4. Reasoning Over Code
Find parameter(s) to the following functions so that they return True. Figure it out by hand, then run the code to confirm. There may be more than one correct answer for each function, and you can provide any one of them.
• RC #1 of 2:
def rc1(n): if ((not isinstance(n, int)) or (n > 100)): return False total = 0 while (n > 0): total = 10*total + n%10 n //= 10 return (total == 42)

• RC #2 of 2:
def f(n): if (n == 0): return 1 n = abs(n) count = 0 while (n > 0): count += 1 n //= 10 return count def rc2(m): if (not(isinstance(m, int)) or (m < 0)): return False start = 0 while True: count = 0 for n in range(start, start+3): count += f(n) if (count > 9): break start += 1 return (m == start)

5. digitCount(n)
Write the function digitCount(n) that takes a possibly-negative int and returns the number of digits in it. So, digitCount(12323) returns 5, digitCount(0) returns 1, and digitCount(-111) returns 3. One way you could do this would be to return len(str(abs(n))), but you cannot do that, since you may not use strings here! This can be solved with logarithms, but seeing as this is "loops week", you should instead simply repeatedly remove the ones digit until you cannot.

6. hasConsecutiveDigits(n)
Write the function hasConsecutiveDigits(n) that takes a possibly- negative int value n and returns True if that number contains two consecutive digits that are the same, and False otherwise.

7. gcd(m, n)
[Note: to receive any credit, you must solve this problem using Euclid's algorithm, and by no other means. In particular, do not just loop through all integers less than min(m,n) and find the common factors that way -- it is much too slow!]
According to Euclid, the greatest common divisor, or gcd, can be found like so:
gcd(x,y) == gcd(y, x%y)
We can use that to quickly find gcd's. For example:
gcd(270,250) == gcd(250, 20) # 270 % 250 == 20
== gcd(20, 10) # 250 % 20 == 10
== gcd(10, 0) # 20 % 10 == 0

When we get to gcd(x,0), the answer is x. So gcd(270, 250) is 10. With this in mind, write the function gcd(x,y) that takes two positive integers x and y and returns their gcd using Euclid's gcd algorithm.

8. nthPrime(n) and
Write the function nthPrime(n) which takes a non-negative int n and returns the nth prime number. We start counting at 0, so nthPrime(0) returns 2, nthPrime(1) returns 3, and so forth. You should use the faster isPrime method from the course notes. Of course, do not just copy-paste that code, but be certain to thoroughly understand it and be able to easily and quickly and correctly reproduce it.