Class Notes: Loops

1. for loops and range [Pre-reading]
2. nested for loops [Pre-reading]
3. while loops [Pre-reading]
4. break and continue
5. isPrime
6. nthPrime

for loops and range [Pre-reading]

  video

# 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)

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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)

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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)

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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))

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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))

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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))

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Yet more ways to approach the problem:   video

Nested for loops [Pre-reading]

  video

# 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(f"( {x} , {y} )  ", end="")
        print()

printCoordinates(4, 5)

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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)

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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)

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while loops [Pre-reading]

  video

# 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)

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Example: nth non-negative 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()

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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)

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break and continue

  video

# 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()

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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').")

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isPrime

  video

# 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()

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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()

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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")

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nthPrime

  video

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!")

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