Chapter 1. Problem solving

What is an Algorithm?

Explaining how an algorithm works is like telling a story. Each algorithm introduces some novel concept or innovation that improves over ordinary solutions. In this chapter we explore several solutions to a simple problem to explain the factors that affect an algorithm’s performance. Along the way I will introduce techniques used to analyze an algorithm’s performance, independent of its implementation, though I will always provide empirical evidence from actual implementations.

Tip

An algorithm is a step-by-step problem solving method implemented as a computer program that returns a correct answer in a predictable amount of time. The study of algorithms is concerned with both correctness (will this algorithm work for all input?) and performance (is this the most efficient way to solve this problem?).

Let’s walk through an example of a problem solving method to see what this looks like in practice. What if you wanted to find the largest value in an unordered list? Each Python list in Figure 1-1 is a problem instance, that is, the input processed by an algorithm (shown as a cylinder); the correct answer appears on the right. How is this algorithm implemented? How would it perform on different problem instances? Can you predict the time needed to find the largest value in a list of one million elements?

Algorithm Processing
Figure 1-1. Three different problem instances processed by an algorithm

An algorithm is more than just a problem solving method; the program also needs to complete in a predictable amount of time. The built-in Python function max() already solves this problem. However, it can be hard to predict an algorithm’s performance on problem instances containing random data, so it’s worth identifying problem instances whose structure is regular. Table 1-1 shows the results of timing max() on two kinds of problem instances of size N, one where the list contains ascending integers and one where the list contains descending integers. You can execute the ch01.timing program to generate this table. While you will get different values in the table, based on the configuration of your computing system, you can verify the following two statements:

  • The timing for max() on ascending values is always slower than on descending values once N is large enough; and

  • As N increases ten-fold in subsequent rows, the corresponding time for max() also appears to increase ten-fold, with some deviation to be expected from live performance trials.

For this problem, the maximum value is returned and the input is unchanged. In some cases, the algorithm updates the problem instance directly instead of computing a new value — for example, sorting a list of values as you will see in chapter five. Throughout this book, we use N to represent the size of a problem instance.

Table 1-1. Executing max() on two kinds of problem instances of size N (time in ms)
N Ascending values Descending values

100

0.001822

0.001760

1,000

0.015898

0.015302

10,000

0.161746

0.157824

100,000

1.688498

1.673066

1,000,000

22.442488

21.952842

When it comes to timing, the key points are that:

  • You can’t predict in advance the value of T(1,000), that is, the time required by the algorithm to solve a problem instance of size 1,000, because computing platforms vary and there may be different programming languages used; however,

  • Once you empirically determine T(1,000), you can predict T(100,000), that is, the time to solve a problem instance one hundred times larger.

Now that we’ve walked through a sample algorithm, it’s important to understand the concept of correctness, whether this algorithm will work for all inputs. We will spend more time in chapter two explaining how to compare the behavior of different algorithms that solve the exact same problem. While the mathematics of algorithms can get quite tricky, I will provide specific examples to always connect the abstract concepts with real-world problems.

The Almighty Array

The standard way to judge the efficiency of an algorithm is to count how many computing operations it requires. But this is exceptionally hard to do! Computers have a central processing unit (CPU) that executes machine instructions that perform mathematical computations (like add and multiply), assign values to CPU registers, and compare two values with each other. Modern programming languages are either compiled into machine instructions (like C or C++) or an intermediate byte code representation (like Python and Java). The Python interpreter (which is itself a C program) executes the byte code and built-in functions, such as min() and max(), are implemented in C and ultimately compiled into machine instructions for execution. Whew! That is a lot to take in!

Complications when counting statements
Figure 1-2. Counting operations or instructions is complicated

In Figure 1-2, the Python function, f(), is compiled into 19 byte code instructions. This byte code is executed by a Python interpreter, written in C, which includes the implementation of the built-in max() function. This interpreter is itself compiled into machine instructions, and the jle (“jump less than or equal”) machine instruction corresponds to the C program statement that compares whether cmp > 0.

It is nearly impossible to count the total number of executed machine instructions for an algorithm, not to mention that modern day CPUs can execute billions of instructions per second! Instead, we will count the number of times a key operation is invoked for each algorithm, which could be “the number of times two elements in a list are compared with each other” or “how many times a function is called.” In this discussion of max(), the key operation is “how many times the less-than (<) operator is invoked.” We will expand on this counting principle in chapter two.

At the end of every chapter in the book, I pose challenge problems that give you a chance to put your new knowledge to the test. I encourage you to try these on your own before you review my sample solutions.

Now is a good time to lift up the hood on the max algorithm to see why it behaves the way it does. All algorithms in this chapter will use Python.

Finding the largest value in an arbitrary list

Consider the following (flawed) Python implementation that attempts to find the largest value in an arbitrary list by comparing each value in A against my_max, updating my_max as needed when larger values are found.

def flawed(A):
  my_max = 0        1
  for v in A:       2
    if my_max < v:
      my_max = v    3
  return my_max
1

my_max is a variable that holds the maximum value; here my_max is initialized to 0.

2

The for loop defines a variable v that iterates over each element in A. The if statement executes once for each value, v.

3

Update my_max if v is larger.

Central to this solution is the less-than operator (<) that compares two numbers to determine whether a value is smaller than another.

>>> p = [1, 5, 2, 9, 3, 4]
>>> flawed(p)
9

flawed correctly determines the largest value for p, invoking the less-than operator six times, once for each of its values. flawed invokes the less-than operator N times on a problem instance of size N. In Figure 1-3, as v takes on successive values from A, you can see that my_max is updated three times to determine the largest value in A.

LargestVisualization
Figure 1-3. Visualizing the execution of flawed

This implementation is flawed because it assumes that at least one value in A is greater than 0. Computing flawed([-5,-3,-11]) returns 0, which is incorrect. One common fix is to initialize my_max to the smallest possible value, such as my_max = float('-inf'). This approach is still flawed since it would return this value on an empty problem instance. Let’s fix this defect now.

Counting key operations

Since the largest value must be an element in A, largest selects the first element of A as my_max, checking other elements to see if any is larger.

def largest(A):
  my_max = A[0]                 1
  for idx in range(1, len(A)):  2
    if my_max < A[idx]:
      my_max = A[idx]           3
   return my_max
1

Set my_max to the first element in A, found at index position 0.

2

idx takes on integer values from 1 up to, but not including, len(A).

3

Update my_max if the element in A at position idx is larger.

When given a non-empty list, largest returns the largest element in the list. Now that we have a correct Python implementation of our algorithm, can you determine how many times less-than is invoked in this new algorithm? Right! N–1 times. We have fixed the flaw in the algorithm and improved its performance (admittedly, by just a tiny bit).

Why is it important to count the uses of less-than? This is the key operation used when comparing two elements. All other program statements (such as for or while loops) are arbitrary choices during implementation, based on the program language used. We will expand on this idea in the next chapter, but for now counting key operations is sufficient.

Models can predict algorithm performance

What if someone shows you a different algorithm for this same problem? How would you determine which one to select? Consider the following alternate algorithm that repeatedly checks each element in the list to see if it is larger than all other elements in the same list. Will this algorithm return the correct result? How many times does it invoke less-than on a problem of size N?

def alternate(A):
  for v in A:
    none_greater_than_v = True        1
    for x in A:
      if v < x:
        none_greater_than_v = False   2
        break
    if none_greater_than_v:
      return v	                      3
  return None                         4
1

When iterating over each v, assume it could be the largest.

2

If v is smaller than another element, x, stop and record that v is not greatest.

3

If none_greater_than_v is true, return v since it is maximum.

4

This final statement is here in case A is an empty list.

alternate attempts to find a value, v, in A such that no other value, x, in A is greater. The implementation uses two nested for loops. This time it’s not so simple to compute how many times less-than is invoked, because the inner for loop stops as soon as an x is found that is greater than v. Also, the outer for loop stops once the maximum value is found. Figure 1-4 visualizes executing alternate on our list example.

Alternate Visualization
Figure 1-4. Visualizing the execution of alternate

For this problem-instance, less-than is invoked 14 times. But you can see that this total count depends on the specific values in the list A. What if the values were in a different order? Can you think of an arrangement of values that requires the least number of less-than invocations? Such a problem instance would be considered a best case for alternate. For example, if the first value in A is the largest of the N values, then the total number of calls to less-than is always N.

Tip

Best Case — a problem instance of size N that requires the least amount of work performed by the algorithm.

Worst Case — a problem instance of size N that demands the most amount of work.

Let’s try to identify a worst case problem instance for alternate that requires the most number of calls to less-than. More than just ensuring that the largest value is the last value in A, in a worst case problem instance for alternate, the values in A must appear in ascending order.

Figure 1-5 shows a best case on the left (p = [9,5,2,1,3,4]) and a worst case on the right (p = [1,2,3,4,5,9]).

Best and Worst Case
Figure 1-5. Visualizing the execution of alternate on best and worst cases

In the best case, there are six calls to less-than; if there were N values in a best case, then the total number of invocations to less-than would be N. It’s a bit more complicated for the worst case. In Figure 1-5 you can see there are a total of 26 calls to less-than when the list of N values is in ascending sorted order. With a little bit of mathematics, I can show that for N elements, this will always be (N2 + 3*N - 2)/2. The following table presents empirical evidence on largest and alternate on worst case problem instances of size N.

Table 1-2. Comparison of largest with alternate
N Largest (# less-than) Alternate (# less-than) Largest (time in ms) Alternate (time in ms)

8

7

43

0.00

0.00

16

15

151

0.00

0.00

32

31

559

0.00

0.01

64

63

2143

0.00

0.04

128

127

8383

0.01

0.15

256

255

33151

0.01

0.59

512

511

131839

0.02

2.31

1024

1023

525823

0.05

9.17

2048

2047

2100223

0.10

36.60

For small problem instances, it doesn’t seem all that bad, but as we double the size of the problem instances, you can see that the number of less-than invocations for alternate nearly quadruples, far surpassing the count for largest. The next two columns in Table 1-2 show the performance of these two implementations on 100 random trials of problem instances of size N. The completion time for alternate quadruples as well.

Note

There are many different approaches to evaluate runtime performance of a target algorithm. In this book, I execute repeated random problem instances of size N and measure the time (in milliseconds) required by an algorithm to process each one. From this set of runs, I select the quickest completion time (i.e., the smallest). This is preferable to simply averaging the total running time over all runs, since that might skew the results.

Counting the number of less-than invocations helps explain the behaviors of largest and alternate. As N doubles, the number of calls to less-than doubles for largest but quadruples for alternate. This behavior is consistent as N increases and you can use this information to predict the run-time performance of these two algorithms on larger-sized problems.

Congratulations! You’ve just performed a key step in algorithmic analysis – judging the relative performance of two algorithms by counting the number of times a key operation is performed. You can certainly go and implement both variations and measure their respective elapsed times on problems instances that double in size; but it actually isn’t necessary since the model predicts this behavior, and confirms that largest is the more efficient algorithm of the two.

Table 1-3 shows that largest is always significantly slower than max(), typically four times slower. Ultimately the reason is that Python is an interpreted language, and built-in functions, such as max(), will always outperform an identical Python implementation. What you should observe is that in all cases, as N doubles, the corresponding performance of largest and max - for both best case and worst case - also doubles.

Table 1-3. Performance of largest and max() on best and worst cases
N Largest Worst Case Max Worst Case Largest Best Case Max Best Case

4,096

0.20

0.05

0.14

0.05

8,192

0.40

0.11

0.29

0.10

16,384

0.80

0.21

0.57

0.19

32,768

1.60

0.41

1.14

0.39

65,536

3.21

0.85

2.28

0.78

131,072

6.46

1.73

4.59

1.59

262,144

13.06

3.50

9.32

3.24

524,288

26.17

7.00

18.74

6.50

Table 1-3 shows it is possible to predict the time required to solve problems instances of increasing size. Now let’s change the problem slightly to make it more interesting.

Find two largest values in arbitrary list

Devise an algorithm that finds the two largest values in an arbitrary list. Perhaps you can modify the existing largest algorithm with just a few tweaks.

Why don’t you take a stab at solving this modified problem and come back here with your solution.

def two_largest(A):
  my_max,second = A[:2]                1
  if my_max < second:
    my_max,second = second,my_max

  for idx in range(2, len(A)):
    if my_max < A[idx]:                2
      my_max,second = A[idx],my_max
    elif second < A[idx]:              3
      second = A[idx]
  return (my_max, second)
1

Determine my_max and second of first two elements, swapping if out of order.

2

If A[idx] is a new max value, then my_max becomes that max value, and second becomes the old my_max.

3

If a value is larger than second (but smaller than my_max), only update second.

two_largest feels similar to largest: compute my_max and second to be the first two elements in A, properly ordered. Then for each of the remaining elements in A (how many? N–2, right!) if you find an A[idx] larger than my_max, adjust both (my_max, second) otherwise check to see whether second alone needs updating.

Counting the number of times less-than is invoked is more complicated because, once again, it depends on the values in the problem instance.

In the best case, two_largest performs the fewest less-than invocations when if statement 2 is true. When A is in ascending sorted order, the less-than at 2 is always true, so it is invoked N–2 times; don’t forget to add 1 because of the use of less-than at 1. In the best case, therefore, you only need N–1 invocations of less-than to determine the top two values. The less-than at 3 is never used in the best case.

For two_largest, can you construct a worst case problem instance? Try it yourself: it happens whenever less-than at 2 is False.

I bet you can see that whenever the list A is in descending order, two_largest requires the most uses of less-than. In particular, for the worst case, less-than is used twice each time through the for loop, or 1 + 2*(N–2) = 2*N–3 times. Somehow this feels right, doesn’t it? If you need N–1 uses of less-than to find the largest value in A, perhaps you truly do need an additional N–2 comparisons (leaving out the largest value, of course) to also find the second largest value.

To summarize:

  • For best case, two_largest finds both values in N–1 uses of less-than.

  • In worst case, two_largest finds both values in 2N–3 uses of less-than.

Note

If you determine the worst case for a particular algorithm on a given problem, perhaps a different algorithm solving the same problem could do better.

Are we done? is this the “best” algorithm to solve the problem of finding the two largest values in an arbitrary list? We can choose to prefer one algorithm over another based on a number of factors:

  • Required extra storage - Does an algorithm need to make a copy of the original problem instance?

  • Programming effort - How few lines of code must the programmer write?

  • Mutable input - Does the algorithm modify the input provided by the problem instance in place, or does it leave it alone?

  • Speed - Does an algorithm outperform the competition?

Let’s investigate three different algorithms to solve this exact same problem, starting with two brief algorithms with no explicit less-than usage (making it hard to know in advance how well they perform relative to largest).

two_largest_sorting creates a new list containing the values in A in increasing order, grabs the final two values and reverses them. two_largest_extra_storage uses max() to find the maximum value in A, removes it from a copy of A and then finds the max of that reduced list to be second. Both of these algorithms require extra storage and only work on lists with more than one value.

We can compare these algorithms against largest by their run-time performance; but first, I want to introduce one final algorithm that uses far fewer less-than invocations than largest, and basketball fans will find its logic familiar.

def two_largest_sorting(A):
  return tuple(reversed(sorted(A)[-2:]))     1

def two_largest_extra_storage(A):
  my_max = max(A)                            2
  copy = list(A)
  copy.remove(my_max)                        3
  return (my_max, max(copy))                 4
1

Sort A to create a new list whose values are in ascending order, select the last two values, reverse their order and return as a tuple.

2

Uses built-in max() function to find largest.

3

Create a copy of the original A and remove my_max.

4

Return a tuple containing my_max and the largest value in copy.

Tournament algorithm

A single-elimination tournament consists of a number of teams competing to see who is the champion. Ideally, the number of teams is a power of 2, like 16 or 64. The tournament is built from a sequence of rounds where all remaining teams in the tournament are paired up to play a match; match losers are eliminated, while winners advance to the next round. The final team remaining is the tournament champion.

Consider the problem instance p=[3,1,4,1,5,9,2,6] with N=8 values. Figure 1-6 shows the single-elimination tournament whose initial round consists of comparing neighboring values using less-than, and the larger values advance in the tournament. In the Elite Eight round, four teams are eliminated, leaving values [3, 4, 9, 6]. In the Final Four round, the values [4, 9] advance and in the Championship round, 9 is declared the victor.

Tournament
Figure 1-6. A tournament with eight initial values

In this tournament, there are seven less-than invocations, which should be reassuring, since this means the largest value was found with N–1 uses of less-than, as we had discussed earlier. If you store each of these N-1 matches, then you can quickly locate the second largest value, as I now show.

Where can the second largest value be “hiding” once 9 is declared the champion? Select 4 as the candidate second-largest value, since this was the value that lost in the Championship round. But the largest value, 9, had two earlier matches, so you must check the other two losing values – value 6 in the Final Four round and value 5 in the Elite Eight round. Thus the second largest value is 6.

For eight initial values, you need just 2 additional less-than invocations – (is 4 < 6) and (is 6 < 5) – to determine that 6 is the second largest value. Because 8=23, you need 3-1=2 comparisons. In general, for N = 2K, you need an additional K–1 comparisons, where K is the number of rounds.

When there are 8=23 initial values, the algorithm constructs a tournament with 3 rounds. To double the number of values in the tournament, you only need one additional round of matches; in other words, with each new round K, you can add 2K more values. Want to find the largest of 64 values? you only need 6 rounds.

Hierarchy
Figure 1-7. A tournament with thirty-two values

To determine the number of rounds given any N, we turn to the logarithm function, which is the opposite of the exponent operation. With N=8 initial values, there are 3 rounds required for the tournament, since 23=8 and log2(8) = 3. In this book, and traditionally in computer science, the log() operator is in base 2, unless otherwise specified.

When N is a power of 2 – like 64 or 65,536 – the number of rounds in a tournament is log(N), which means the number of extra less-than usages is log(N)–1.

I have just sketched an algorithm to compute the largest and second largest value in A using just N–1+log(N)–1 = N+log(N)-2 less-than invocations for any N that is a power of 2.

Is this tournament_two algorithm practical? Will it outperform two_largest? If you only count the number of times less-than is invoked, tournament_two is noticeably better. two_largest requires 131,069 less-than operations on problems of size N while tournament_two only requires 65,536 + 162 = 65,550, just about half. But there is more to the story.

def tournament_two(A):
  N = len(A)
  winner = [None] * (N-1)          1
  loser = [None] * (N-1)
  prior = [-1] * (N-1)

  idx = 0
  for i in range(0, N, 2):         2
    if A[i] < A[i+1]:
      winner[idx] = A[i+1]
      loser[idx] = A[i]
    else:
      winner[idx] = A[i]
      loser[idx] = A[i+1]
    idx += 1

  m = 0                            3
  while idx < N-1:
    if winner[m] < winner[m+1]:    4
      winner[idx] = winner[m+1]
      loser[idx]  = winner[m]
      prior[idx]  = m+1
    else:
      winner[idx] = winner[m]
      loser[idx]  = winner[m+1]
      prior[idx]  = m
    m += 2                         5
    idx += 1

  largest = winner[idx-1]
  second = loser[idx-1]            6
  m = prior[idx-1]
  while m >= 0:
    if second < loser[m]:          7
      second = loser[m]
    m = prior[m]

  return (largest, second)
1

Create arrays for winner and loser values, and index of prior matches.

2

Populate first N/2 winner/loser pairs using N/2 invocations of less-than.

3

Pair up winners to find new winner and record prior match index.

4

An additional N/2-1 invocations of less-than.

5

Advance m by two to find next pair of winners. When idx reaches N-1, winner[idx-1] is largest.

6

Initial candidate for second largest, but must check all others that lost to largest to find actual second largest.

7

log(N)-1 additional invocations of less-than.

Table 1-4 reveals that tournament_two is almost ten times slower than any of its competitors! Let’s record the total time it takes to solve 100 random problem instances (of size N growing from 1,024 to 16,384). While we are at it, I will include the performance results of two_largest_sorting and two_largest_extra_storage as well.

Table 1-4. Table comparing run-time performance (in milliseconds) of all four algorithms
N extra_storage two_largest sorting tournament_two

1,024

0.00

0.01

0.01

0.06

2,048

0.01

0.01

0.02

0.11

4,096

0.01

0.02

0.03

0.22

8,192

0.02

0.05

0.08

0.47

16,384

0.05

0.10

0.17

1.17

32,768

0.13

0.20

0.38

3.22

65,536

0.23

0.40

0.85

8.80

131,072

0.50

0.84

1.84

42.33

262,144

1.28

1.85

4.25

*

524,288

4.39

5.03

10.61

*

Ok, that’s a whole lot of numbers to throw at you! Table 1-4 is organized so the fastest algorithm (extra_storage) is on the left while the slowest algorithm (tournament_two) is on the right; in fact it is so slow, I do not even run it on the larger problem instances. Figure 1-8 visualizes the run-time trends as the problem size instance grows ever larger.

Hierachy
Figure 1-8. Runtime-performance comparison

While tournament_two is clever, it is significantly slower than the other three alternatives. I can also claim that on large enough problem size instances (in this case, 262,144 and larger) tournament_two is not a reasonable algorithm to use, because there are algorithms that continue to have acceptable performance even on large problem instances. Because the graphed behavior of two_largest and extra_storage are so similar, they appear to have more in common with each other than they do with sorting, although I won’t be able to completely explain why until you read chapter two. Now it’s time to explain the two useful fundamental concepts that explain the inherent complexity of an algorithm.

Time Complexity and Space Complexity

It can be hard to count the number of elementary operations (such as addition, variable assignment, control logic), because of the difference in programming languages, plus the fact that some languages are executed by an interpreter, such as Java and Python. But if you could count the total number of operations executed by an algorithm, the goal of time complexity is to come up with a function C(N) that counts the number of elementary operations performed by an algorithm as a function of N, the size of a problem instance.

Each elementary operation takes a fixed amount of time, t, based on the CPU executing the operation. We model the time to perform the algorithm as T(N) = t * C(N).

The following small example provides an insight that the structure of a program is critical. For functions f0, f1, f2, and f3, you can exactly compute how many times each one executes the operation ct = ct + 1 based on the input size, N. Table 1-5 contains the counts for a few values of N.

Four implementations
Figure 1-9. Four different functions with different performance profiles

The count for f0 is always the same, independent of N. The count for f2 is always seven times greater than f1, and both of them double in size as N doubles. In contrast, the count for f3 increases far more rapidly; as you have seen before, as N doubles, the value of f3(N) quadruples. Here, f1 and f2 are more similar to each other than they are to f3. In the next chapter, we will explain the importance of for loops, and nested for loops in evaluating an algorithm’s performance.

Table 1-5. Counting operations in four different functions
N f0 f1 f2 f3

512

2

512

3584

262144

1024

2

1024

7168

1048576

2048

2

2048

14336

4194304

When evaluating an algorithm, we also have to consider space complexity, which accounts for extra memory required by an algorithm based on the size, N, of a problem instance. Memory is a generic term for data stored in the file system or the Random Access Memory (RAM) of a computer. two_largest has minimal space requirements – three variables my_max, second, and the iterator variable idx. No matter the size of the problem instance, its extra space never changes. We say that this means the space complexity is independent of the size of the problem instance, or constant. In contrast, tournament_two constructs a tournament starting initially with N/2 Match objects, ultimately constructing a total of N–1 of them. As N increases, the total extra storage increases in a manner that is directly proportional to the size of the problem instance. Building the tournament structure is going to slow tournament_two down, when compared against two_largest.

Still, there is one observation that holds. If you note in Table 1-4, you can see that the timing results for columns extra_storage and two_largest more-or-less double in subsequent rows. This means that in both cases, the total time appears to be directly proportional to the size of the problem instance which is doubling in subsequent rows. This is an important observation, since these are both more efficient than sorting which appears to follow a different, less efficient trajectory. tournament_two is still the least efficient, with a runtime performance that more than doubles, growing so rapidly that I don’t bother executing it for large problem instances. In the next chapter, I will present the mathematical tools necessary to analyze algorithm behavior properly.

Note

The Tournament algorithm proposed here only works for an even number of values, N. How would you modify the algorithm to handle an odd number of N values?

Conclusion

This chapter provides an introduction to the rich and diverse field of algorithms. I showed how to model the performance of an algorithm on a problem instance of size N by counting the number of key operations it performs. You can also empirically evaluate the runtime performance of the implementation of an algorithm. In both cases, you can determine the order of growth of the algorithm as the problem instance size N doubles.

I introduced several key concepts, including:

  • Time Complexity as estimated by counting the number of key operations executed by an algorithm on a problem instance of size N

  • Space Complexity as estimated by the amount of memory required when an algorithm executes on a problem instance of size N

In the next chapter, I will introduce the mathematical tools of Asymptotic Analysis that will complete my explanation of the techniques needed to properly analyze algorithms.

Challenge Questions

  1. Palindrome Word Detector: A palindrome word reads the same backward as forward, such as “madam”. Devise an algorithm that validates whether a word of N characters is a palindrome. Confirm empirically that it outperforms the following two alternatives:

    def is_palindrome1(w):
      """Create slice with negative step and confirm equality with w."""
      return w[::-1] == w
    
    def is_palindrome2(w):
      """Strip outermost characters if same, return false when mismatch."""
      while len(w) > 1:
        if w[0] != w[-1]:     # if mismatch, return False
          return False
        w = w[1:-1]           # strip characters on either end; repeat
    
      return True             # must have been palindrome

    Once you have this problem working, modify it to detect palindrome strings with spaces, punctuation, and mixed capitalization. For example, the following string should classify as a palindrome: A man, a plan, a canal. Panama!

  2. Linear Time Median: A wonderful algorithm exists that efficiently locates the median value in an arbitrary list (for simplicity, assume list is odd size). Review the code (TBA) and count the number of times less-than is invoked. This implementation rearranges the arbitrary List as it processes.

    A different implementation requires extra storage. Compare its performance with the one above.

  3. Counting Sort: If you know that an arbitrary list, A, only contains non-negative positive integers from 0 to M, then the following algorithm will properly sort A using just an extra storage of size M.

    def countingSort(A, M):
      counts = [0] * M
      for v in A:
        counts[v] += 1
    
      pos = 0
      v = 0
      while pos < len(A):
        for idx in range(counts[v]):
          A[pos+idx] = v
        pos += counts[v]
        v += 1

    This code has nested loops - a for loop within a while loop. However, you can demonstrate that A[pos+idx] = v only executes N times.

    Conduct a performance analysis to demonstrate that the time to sort N integers in the range from 0 to M doubles as the size of N doubles.

    You can eliminate the inner for loop, and improve the performance of this operation, using the ability in Python to replace a sublist using sublist[left:right] = [2, 3, 4]. Make the change and empirically validate that it too doubles as N doubles, with over a 30% improvement in speed.

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