CMU 15-112: Fundamentals of Programming and Computer Science
Class Notes: Loops

• Use for loops to repeat actions a specified number of times
• Use while loops to repeat actions until a condition is met

1. Predict
2. for loops and range
3. Check 2.1
4. Check 2.2
5. nested for loops
6. Check 2.3
7. while loops
8. Check 2.4
9. break and continue
10. isPrime
11. nthPrime

1. Predict
   Predict the output of the following code snippet and write your guess in the editor.
   for i in range(5): print(i)
   import sys def set_certificate(certificate_div_id, certificate): document[certificate_div_id].textContent = certificate def get_student_code(student_code_div_id): raw_student_code = document[student_code_div_id].textContent return window.patchCodeToCheckTimeout(raw_student_code, 'check_timeout();'); def make_certificate(student_code_div_id, certificate_div_id): student_code = get_student_code(student_code_div_id) certificate = [] output = "failure" if student_code != "": output = "success" certificate.append((output, type(output))) set_certificate(certificate_div_id, str(certificate))
   
   Compare
   Now run the code and see if you were right!
   for i in range(5): print(i)


2. for loops and range
   # A for loop repeats an action a specific number of times # based on the provided range def sumFromMToN(m, n): total = 0 # note that range(x, y) includes x but excludes y for x in range(m, n+1): total += x return total print(sumFromMToN(5, 10) == 5+6+7+8+9+10)
   
   Actually, we don't need a loop here...
   def sumFromMToN(m, n): return sum(range(m, n+1)) print(sumFromMToN(5, 10) == 5+6+7+8+9+10) # And we can even do this with a closed-form formula, # which is the fastest way, but which doesn't really # help us demonstrate loops. :-) def sumToN(n): # helper function return n*(n+1)//2 def sumFromMToN_byFormula(m, n): return (sumToN(n) - sumToN(m-1)) print(sumFromMToN_byFormula(5, 10) == 5+6+7+8+9+10)
   
   What if we omit the first parameter?
   def sumToN(n): total = 0 # range defaults the starting number to 0 for x in range(n+1): total += x return total print(sumToN(5) == 0+1+2+3+4+5)
   
   What if we add a third parameter?
   def sumEveryKthFromMToN(m, n, k): total = 0 # the third parameter becomes a step for x in range(m, n+1, k): total += x return total print(sumEveryKthFromMToN(5, 20, 7) == (5 + 12 + 19))
   
   Sum just odd numbers from m to n:
   # We can also change the step by changing the inside of the loop def sumOfOddsFromMToN(m, n): total = 0 for x in range(m, n+1): if (x % 2 == 1): total += x return total print(sumOfOddsFromMToN(4, 10) == sumOfOddsFromMToN(5,9) == (5+7+9))
   
   Now do it backwards!
   # Here we will range in reverse # (not wise in this case, but instructional) def sumOfOddsFromMToN(m, n): total = 0 for x in range(n, m-1, -1): if (x % 2 == 1): total += x return total print(sumOfOddsFromMToN(4, 10) == sumOfOddsFromMToN(5,9) == (5+7+9))


3. Check 2.1


4. Check 2.2


5. Nested for loops
   # We can put loops inside of loops to repeat actions at multiple levels # This prints the coordinates def printCoordinates(xMax, yMax): for x in range(xMax+1): for y in range(yMax+1): print("(", x, ",", y, ") ", end="") print() printCoordinates(4, 5)
   
   How about some stars?
   def printStarRectangle(n): # print an nxn rectangle of asterisks for row in range(n): for col in range(n): print("*", end="") print() printStarRectangle(5)
   
   And again:
   # What would this do? Be careful and be precise! def printMysteryStarShape(n): for row in range(n): print(row, end=" ") for col in range(row): print("*", end=" ") print() printMysteryStarShape(5)


6. Check 2.3


7. while loops
   # use while loops when there is an indeterminate number of iterations def leftmostDigit(n): n = abs(n) while (n >= 10): n = n//10 return n print(leftmostDigit(72658489290098) == 7)
   
   Example: nth positive integer with some property
   # eg: find the nth number that is a multiple of either 4 or 7 def isMultipleOf4or7(x): return ((x % 4) == 0) or ((x % 7) == 0) def nthMultipleOf4or7(n): found = 0 guess = -1 while (found <= n): guess += 1 if (isMultipleOf4or7(guess)): found += 1 return guess print("Multiples of 4 or 7: ", end="") for n in range(15): print(nthMultipleOf4or7(n), end=" ") print()
   
   Misuse: While loop over a fixed range
   # sum numbers from 1 to 10 # note: this works, but you should not use "while" here. # instead, do this with "for" (once you know how) def sumToN(n): # note: even though this works, it is bad style. # You should do this with a "for" loop, not a "while" loop. total = 0 counter = 1 while (counter <= n): total += counter counter += 1 return total print(sumToN(5) == 1+2+3+4+5)


8. break and continue
   # continue, break, and pass are three keywords used in loops # in order to change the program flow for n in range(200): if (n % 3 == 0): continue # skips rest of this pass elif (n == 8): break # skips rest of entire loop else: pass # does nothing! pass is a placeholder, not needed here print(n, end=" ") print()
   
   Infinite "while" loop with break:
   # Note- this is advanced content, as it uses strings. It's okay # to not fully understand the content below. def readUntilDone(): linesEntered = 0 while (True): response = input("Enter a string (or 'done' to quit): ") if (response == "done"): break print(" You entered: ", response) linesEntered += 1 print("Bye!") return linesEntered linesEntered = readUntilDone() print("You entered", linesEntered, "lines (not counting 'done').")


9. Check 2.4


10. isPrime
    # Note: there are faster/better ways. We're just going for clarity and simplicity here. def isPrime(n): if (n < 2): return False for factor in range(2,n): if (n % factor == 0): return False return True # And take it for a spin for n in range(100): if isPrime(n): print(n, end=" ") print()
    
    fasterIsPrime:
    # Note: this is still not the fastest way, but it's a nice improvement. def fasterIsPrime(n): if (n < 2): return False if (n == 2): return True if (n % 2 == 0): return False maxFactor = round(n**0.5) for factor in range(3,maxFactor+1,2): if (n % factor == 0): return False return True # And try out this version: for n in range(100): if fasterIsPrime(n): print(n, end=" ") print()
    
    Verify fasterIsPrime is faster:
    def isPrime(n): if (n < 2): return False for factor in range(2,n): if (n % factor == 0): return False return True def fasterIsPrime(n): if (n < 2): return False if (n == 2): return True if (n % 2 == 0): return False maxFactor = round(n**0.5) for factor in range(3,maxFactor+1,2): if (n % factor == 0): return False return True # Verify these are the same for n in range(100): assert(isPrime(n) == fasterIsPrime(n)) print("They seem to work the same!") # Now let's see if we really sped things up import time bigPrime = 499 # Try 1010809, or 10101023, or 102030407 print("Timing isPrime(",bigPrime,")", end=" ") time0 = time.time() print(", returns ", isPrime(bigPrime), end=" ") time1 = time.time() print(", time = ",(time1-time0)*1000,"ms") print("Timing fasterIsPrime(",bigPrime,")", end=" ") time0 = time.time() print(", returns ", fasterIsPrime(bigPrime), end=" ") time1 = time.time() print(", time = ",(time1-time0)*1000,"ms")


11. nthPrime
    def isPrime(n): if (n < 2): return False if (n == 2): return True if (n % 2 == 0): return False maxFactor = round(n**0.5) for factor in range(3,maxFactor+1,2): if (n % factor == 0): return False return True # Adapt the "nth" pattern we used above in nthMultipleOf4or7() def nthPrime(n): found = 0 guess = 0 while (found <= n): guess += 1 if (isPrime(guess)): found += 1 return guess # and let's see a list of the primes for n in range(10): print(n, nthPrime(n)) print("Done!")