Chapter 4: Sorting

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# Chapter 4: Sorting - PowerPoint PPT Presentation

Chapter 4: Sorting Sorting is an important application to study Sorting is used in numerous applications As time as gone on, technology has permitted easier sorting

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## Chapter 4: Sorting

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Chapter 4: Sorting
• Sorting is an important application to study
• Sorting is used in numerous applications
• As time as gone on, technology has permitted easier sorting
• For instance, it is no longer necessary (usually) to worry about how to keep portions of a list on disk because memory was too small in size
• From an algorithmic perspective, we find many different strategies for sorting
• Each has its own advantages and disadvantages
• We will study several sorting algorithms here and determine their average and worst-case complexities
• We will also note the amount of temporary storage required for each so that we can compare their space complexity as well as their time complexity
• Many of the algorithms we will view will be limited to sorting lists implemented as arrays, but some also accommodate linked lists
Insertion Sort
• int shiftVacRec(Element[ ] E,
• int vacant, Key x)

int xLoc;

if (vacant == 0)

xLoc = vacant;

else if (E[vacant-1].key <= x)

xLoc = vacant;

else

E[vacant] = E[vacant-1];

xLoc = shiftVacRec(E,

vacant-1, x);

return xLoc

• Basic idea:
• given a new item, insert it into an already ordered list
• repeat this process for each item in the list until the list is sorted
• for instance, insert the second element with respect to the first, then insert the third element with respect to the first and second (which are already sorted), etc
• The insertion portion can be done either iteratively or recursively (see example to the right)
• Notice that the given code has two comparisons (in bold), one being the base case
Complexity
• The recurrence equation for shiftVacRec is easy to derive:
• T(n) = T(n-1) + 1 with the base case being T(0) = 1
• Again remember that we are only counting comparisons
• T(n) = T(n-1) + 1 = T(n-2) + 2 = T(n-3) + 3 = … = T(0) + n = 1 + n   (n)
• Therefore, to insert 1 item into an already sorted list is in  (n)
• How do we perform the insertion sort? We use shiftVacRec n-1 times, once for each item in the array
• However, notice that n, from T(n) grows (or changes)
• The number of items in the array for shiftVacRec to work with increases from 1 for the first iteration to n-1 for the last
• So in fact, we have, in the worst case, a sequence of 1 comparison the first time, 2 the second, 3 the third, etc… up to n-1 the last time
• We saw previously that the summation from 1 to n-1 = n * (n – 1) / 2
• So, insertion sort takes ½*(n2-n) which is   (n2)
If n is large, the number of recursive calls makes our shiftVacRec inefficient because of the overhead of a procedure call

Instead we might choose to implement insertion sort iteratively by replacing procedure calls with a loop

we will find other sorting algorithms where we must use recursion, but here we do not HAVE to

This code is given to the right

It should be easy to see that the while loop will execute at most 1 time for the first iteration of xindex, 2 times for the second, etc… and n-1 times for the last, so iterative insertion sort does the same amount of work,   (n2)

void insertionSort(Element[ ] E, int n)

int xindex, xLoc; Element current

for(xindex=1;xindex<n;xindex++)

current = E[xindex];

xLoc = shiftVac(E, xindex, x);

E[xLoc] = current;

return;

int shiftVac(Element[ ] E, int xindex, x)

int vacant, xLoc;

vacant = xLoc;

while (vacant > 0)

if(E[vacant-1] <= x)

xLoc = vacant; break;

E[vacant]=E[vacant-1];

vacant--;

return xLoc;

Iterative Insertion Sort
Average Case Complexity
• The average case complexity is not as easy to determine
• First, assume that no values of the array are equal
• Next, assume that a value to be inserted has an equal chance of being inserted in any location
• There are i+1 locations to insert the ith item (before the first, between first and second, etc, after last) – see figure below
• In order to determine the average case, we compute the average of inserting i into each of the i+1 possibilities
• For each of the i+1 possibilities, there is 1 comparison (the if-statement) and there is a 1 / (i +1) chance so we have:
• 1/(i+1) * i=1j(i + 1) =
• 1/(i+1) * i=1j (i) + j * (1 / (i+1) ) =
• 1 / (i + 1) * (i+1)*i/2 + i) =
• i / 2 + i / (i + 1) to insert the ith item
• (remember i ranges from 1 to n-1)
Average Case Continued
• So, all of insertion sort takes
• A(n) = i=1n-1 [i / 2 + i / (i + 1) ] =
• i=1n-1 [i / 2 + 1 - 1 / (i + 1) ] =
• [(n-1)*n / 2] / 2 + n – 1 - i=1n-1 1 / (i + 1)
• The first term above is (n2-n)/4, the second term is n-1, and the last term is roughly equal to ln n
• see Example 1.7 on pages 26-27 for an explanation
• Therefore, insertion sort, on average, takes 
• (n2 - n)/4 + n – 1 + ln n = (n2 – 5*n – 4) / n + ln n 
• (n2)/4 so insertion sort’s average case is   (n2)
• Insertion sort has a space complexity of (n) because it is an “in-place” sort, that is, it does not copy the array and only needs 1 temporary variable, for x (the item being inserted)
Lower Bound on “in-place” Sorts
• The idea behind insertion sort is to sort “in-place” by taking one element and finding its proper position
• Other sorting algorithms are similar
• The common aspect is that these sorting algorithms place 1 item in its proper position for each iteration
• What is the lower bound on such an algorithm?
• Consider that in such an algorithm, we are comparing some portion of an already sorted array to a new item
• Such a comparison must encompass all items in the already sorted portion of the array to an item to be placed
• If the comparison is to the entire array (as is the case with Selection Sort) then we have n comparisons performed n-1 times
• If the comparison is to a growing portion of the array, we have a sequence of comparisons of 1 + 2 + 3 + … + n-1 = (n*n-1)/2
• Therefore, the minimum amount of work in which exactly 1 item is placed per iteration is   (n2)
• Does this mean insertion sort is an optimal sort?
Let us apply recursion to sorting as follows:

The sorting algorithm is solved by dividing the problem into

Dividing the array into smaller arrays

Sorting the smaller arrays

Combining the sorted smaller arrays into a larger array

See the algorithm to the right

Our algorithm’s complexity is described by the following recurrence relation

T(n) = D(n) +  S(size part i) + C(n)

We must determine how to divide the arrays (D), how to sort each one (S), and how to combine them when done (C)

We might find that S is simplified by recursion

Solve(I)

n = size(I)

if (n <= smallsize)

solution = directlySolve(I);

else

divide I into I1, …, Ik.

for each i in {1, …, k}

Si = solve(Ii);

solution = combine(S1, …, Sk);

return solution;

Divide and Conquer Sorts
Quicksort
• In Quicksort, we use the exact strategy described previously as follows:
• Before dividing, move elements so that, given some element x, all elements left of x are less than E[x] and all elements right of x are greater than E[x]
• Now we divide by simply repeating this on the left hand side and right hand side of x
• Combining is not necessary (it is simply a return statement)
• This brings about some questions:
• What element should be x? Is the choice of x important?
• How do we move elements so that x is positioned correctly?
• What kind of complexity will this algorithm yield?
Quicksort Continued
• We divide Quicksort into two procedures, the main Quicksort procedure finds x and recursively calls itself with the left-hand and right-hand sides
• Partition will be used to find move the elements of the array around so that x falls in its proper place with respect to all elements < x and all elements > x
• That is, E[x] < E[y] where x < y and E[z] < E[x] where z < x
• Partition does most of the work of the Quicksort algorithm
• How might Partition work?
• Move from right-to-left until we find an item less than E[x], we move that item into x
• Move from left-to-right (after x) until we find an item greater than E[x] and move that to the newly freed position
• Repeat until we meet in the middle somewhere and place E[x] there
• Partition will take n-1 comparisons
Quicksort Algorithm

void quicksort(Element[ ] E, int first, int last)

int splitPoint, Element pivot

if(first < last)

pivot = E[first];

splitPoint = partition(E, pivot, first, last);

E[splitPoint] = pivot;

quicksort(E, first, splitPoint – 1);

quicksort(E, splitPoint+1, last);

return;

int partition(Element[ ] E,

Element pivot, int first, int last)

int low = first, high = last;

int lowVac, highVac;

lowVac = low; highVac = high;

while (low < high)

while(E[highVac]>pivot)

highVac--;

E[lowVac] = E[high];

while(E[lowVac]<pivot)

lowVac++;

E[highVac]=E[low];

low=lowVac; high=highVac-1;

return low;

Quicksort Analysis
• Quicksort has a recurrence equation of
• T(n) = T(n-r-1) + T(r) + n - 1
• Where r is the number of elements to Pivot’s right
• notice the “- 1” in “n – r – 1” because pivot is already in its proper place
• How do we solve T(n) when r changes each time?
• The worst case occurs when r=1 or r=n–1, so let’s use r = 1
• This gives us T(n) = T(n-1) + T(1) + n – 1 with T(1) = 0
• So T(n) = T(n-1) + n - 1 = T(n-2) + 2*n - 2 = T(n-3) + 3*n - 3 = … = T(1) + (n-1)*n - n = n*(n-1) – n = n2 – 2*n
• So, Quicksort’s worst case is in   (n2)
Quicksort Analysis continued
• What about Quicksort’s average case?
• This will happen when r > 1 and r < n-1 but what will r be?
• On average, r will be between these two extremes and the average value of r is then
• 1/n * {i=1 to n-1} i = n*(n-1)/2*n = (n-1) / 2
• So, the average case complexity has the following recurrence equation:
• T(n) = T((n-1)/2) + T((n-1)/2) + n – 1 = 2 * T((n-1)/2) + n – 1
• Using the Master Theorem from chapter 3, we have f(n) = n – 1, b = 2 and c = 2 so that E = 1 and that means that T(n) is in   (n log n)
• A more formal analysis is given on pages 167-168 if you want to see the math!
• What is Quicksort’s space usage? Partition is done in place and so the only space required is the array plus a few temporary variables, or Quicksort’s space usage is   (n)
Improving Quicksort
• Even though Quicksort has a poor worst-case complexity (as opposed to some other sorting algorithms), the partition strategy is faster than most other sorting mechanisms, so Quicksort is a desired sort as long as we can prevent the worst case from arising
• How?
• There are numerous improvements:
• Make sure the pivot is a good one, possibly by selecting 3 values (at random, or the first 3, or the first, middle and last in the array) and finding the median. This causes partition to be slightly more complicated, but not any more computationally complex
• Remove the subroutines from partition (the version presented in these notes is like that, the version in the textbook uses the subroutines)
• For small arrays, use a different sort
• Optimize the stack space so that the recursive calls do not slow down the process
Merging Sorted Sequences
• Recall from our earlier idea to use divide and conquer to sort, we need a way to combine the sorted subarrays
• We did not need this in Quicksort because Quicksort didn’t really need a combining step
• How can we do this?
• Let’s assume we have two sorted subarrays, A and B
• We want to sort them into a new array C which is equal in size to A + B
• Which item goes into C[0]? It will either be A[0] or B[0]
• Which item goes into C[1]? It will either be A[1] or B[0] if A[0] has already been placed, or A[0] or B[1] if B[1] has already been placed
• etc…
• Until we have placed all of A or B into C, then the remainder of the merge requires copying the rest of whichever array still remains
Recursive Merge

void merge(Element[ ] a, b, c, int

sizeA, sizeB, currentA, currentB,

currentC)

if(currentA = = sizeA)

for(int k=currentB; k<sizeB; k++)

c[currentC++] = b[k];

else if(currentB = = sizeB)

for(int k=currentA; k<sizeA, k++)

c[currentC++] = a[k];

else if(a[currentA] < b[currentB]

c[currentC++] = a[currentA++];

merge(a, b, c, sizeA, sizeB,

currentA, currentB, currentC);

else

c[currentC++] = b[currentB++];

merge(a, b, c, sizeA, sizeB,

currentA, currentB, currentC);

return

• Merge algorithm (to the right)
• moves the rest of A or B into C if the other array is done, or finds the smaller of a[currentA], b[currentB] and moves it into c, and then recursively calls itself with the rest of a and b
• The recurrence equation for merge is
• T(n) = T(n -1) + 1
• The base case = 0, however what n is the base case? It depends on when we run out of one of the two arrays, but this will be no greater than n, so T(n)   (n) and in the worst case, T(n) = n - 1
Iterative Merge
• The iterative version of merge works similar to the recursive one, move one element into array c based on whether the current element of a or b is smaller, and repeat until one array is “emptied out” and then move the remainder of the other array into c
• see algorithm 4.4, page 172
• This requires n total copies (number of elements in a and b = n) but the number of comparisons varies depending on when the first array is “emptied”
• We have the same situation as with the recursive version T(n) = n – 1 in the worst case and T(n)   (n) in any case
Optimality of Merge
• Is there any better way to merge two arrays?
• Consider the following two arrays:
• A = [1, 2, 3, 4, 5, 6, 7, 8];
• B = [9, 10, 11, 12, 13, 14, 15, 16];
• We can merge these two arrays with 1 comparison (how?)
• In the worst case, our merge algorithm does the least amount of work required, consider these arrays:
• A = [1, 3, 5, 7, 9, 11, 13, 15];
• B = [2, 4, 6, 8, 10, 12, 14, 16];
• We could not improve over merge in this case because every pair of A[i] and B[j] will have to be compared where i and j are equal or off by one
• Thus, there are n-1 comparisons in the worst case (n-2 is possible in the worst case if A and B differ in size by 1 element)
Merge’s Space Usage
• While Merge gives us an optimal worst-case merger, it does so at a cost of space usage
• As seen in the previous examples, we have two arrays of size n combined to merge into array C
• Array C then must be of size n
• So Merge takes 2 * n space as opposed to n for Quicksort and Insertion sort
• Could we improve?
• In a non-worst case situation, yes, by not copying all of the second array into C, but instead, copying what we have in C back into the original array
• There is a discussion of this on pages 173-174 if you are interested
Now that we have a merge algorithm, we can define the rest of the sorting algorithm

Recall that we need a mechanism to

Divide arrays into subarrays

Sort each subarray

Combine the subarrays

Merge will combine the subarrays

Sorting each subarray is done when two sorted subarrays are merged, so we won’t need a sort

Dividing arrays into two subarrays will be done by finding the midpoint of the two arrays and calling the divide/sort/combine procedure recursively with the two halves

Mergesort
The Mergesort algorithm is given to the right

The midpoint will either be at n/2 or (n-1)/2 creating two subarrays that will be (n-1)/2 in size, or n/2 in size

To simplify, we will consider these to be floor(n/2) and ceiling(n/2)

Each recursive call on a subarray of size k will require at most k-1 comparisons to merge them

Since all subarrays at any recursive level will sum up to a total of n array elements, the number of comparisons at that level is n-1

The recurrence relation is then

T(n) = T(floor(n/2)) + T(ceil(n/2)) + n-1

To simplify, T(n) = 2 * T(n/2) + n - 1

The base case T(1) = 0

With f(n) = n – 1, b = 2, c = 2, we have E = 1 and the Master Theorem tells us that T(n)   (n log n)

Mergesort Continued
• void mergeSort(Element[] E,
• int first, int last)

if (first < last)

int mid = (first+last)/2;

mergeSort(E, first, mid);

mergeSort(E, mid+1, last);

merge(E, first, mid, last);

return;

Recursion Tree for Mergesort
• We can also visualize Mergesort’s complexity through the recursion tree to obtain a more precise complexity
• Notice that each level doubles the number of recursive calls but at each level, the amount of work needed is x fewer comparisons where x doubles per level (1, 2, 4, 8, etc)
• Thus, Mergesort will require the following amount of work:  (i = 0 to ceiling (log n)) (n – 2i) = (n – 1) + (n – 2) + (n – 4) + (n – 8) + … + (n – (n – 1)) + (n – n)
•  (i = 0 to ceiling (log n)) (n) =
• n log n
•  (i = 0 to ceiling (log n)) 2i =
• 2log n+1 – 1 =
• (n + 1) – 1 = n
• So, we get a worst case
• complexity of n log n – n
• In fact, mergesort’s worst case
• complexity is between
• ceiling(n log n – n + 1) &
• ceiling(n log n – .914 n)
Lower Bound for Sorting with Comparisons
• Consider our worst-case complexities for sorting:
•  (n2) for in-place sorts that move 1 item at a time
•  (n log n) for divide and conquer based sorts
• Can we do better for any sorting algorithm that compares pairs of values to find their positions?
• The reason we ask the question as above is that we will next see a sort that doesn’t compare values against themselves
• The answer to our question is unfortunately no, we cannot do better
• Why not?
• Consider for n items
• there is n! possible permutations of those n items
• for n = 3, we would have 6 combinations:
• {x1, x2, x3}, {x1, x3, x2}, {x2, x1, x3}, {x2, x3, x1}, {x3, x1, x2}, {x3, x2, x1}
• let’s arrange the possibilities in a tree where we traverse the tree to make the fewest comparisons – this is known as a decision tree
Our Decision Tree (for n = 3)
• First, compare x1 and x2
• if x1 < x2 take left branch else take right branch
• On left, compare x2 and x3, on right, compare x1 and x3
• On left, if x2 < x3 take left branch, sequence is {x1, x2, x3}
• On right, if x1 < x3 take left branch, sequence is {x2, x1, x3}
• We might need another comparison
• How many comparisons might we have to make and why?

The height of this tree is log n!

because there are n! leaf nodes

So, the maximum number of

comparisons is log n! since we

might not know the sequence

until reaching the leaves of the tree

How much is log n! ?

Lower Bound for Worst Case
• We see from the previous slide that the least number of comparisons needed to sort by comparing individual array elements is log n!
• How does log n! compare to n log n, our current lower bound for worst case?
• n! = n * (n – 1) * (n – 2) * … * 3 * 2 * 1
• log n! = log (n * (n – 1) * (n – 2) * … * 2 * 1) =

log n + log n – 1 + log n – 2 + … + log 2 + log 1 =

 (i = 1 to n) log i >= 1n log x dx = log e * 1n ln x dx =

log e * (x ln x – x) |1n = log e * (n ln n – n + 1) = n log n – n log e + log e

• log e < 1.443, so we have log n! >= n log n – 1.443 * n + 1.443
• for our worst case complexity, we round off to the nearest integer and so log n! >= ceiling (n log n – 1.443 * n) – we can omit the final + 1.443 as it is merely a constant
• So our lower bound for a worst case complexity is ceiling (n log n – 1.443 * n)
Lower Bound for Average Case
• Can we improve over n log n for an average case complexity?
• The proof for this is given on pages 180-181
• However, we can simplify this proof by reconsidering the decision tree
• In any decision tree, the leaves will all occur at either level log n! or (log n!) – 1
• So, our average case complexity will be between log n! and log n! – 1 >= n log n – 1.443 * n so, like our worst case complexity, the lower bound for average case complexity is n log n – 1.443 * n
• Notice in this case that we do not need to take the ceiling since average case complexity does not have to be an integer value
Mergesort vs. Quicksort
• We have two sorting algorithms (so far) that can give us n log n average case complexity, but:
• Mergesort requires twice the amount of storage space
• Quicksort cannot guarantee n log n complexity
• Mergesort’s merge operation is more time consuming than Quicksort’s partition operation even though they are both in  (n)
• In practice, Mergesort does 30% fewer comparisons in the worst case than Quicksort does in the average case, but because Quicksort does far fewer element movement, Quicksort turns out to often be faster
• But what if we want a guaranteed better performance than Mergesort? Quicksort cannot give us that.
• So we turn to a third n log n sort, Heapsort
• Heapsort is interesting for two reasons
• It guarantees n log n performance in average and worst case, like Mergesort, but is faster than Mergesort
• It does no recursion, thus save stack space and the overhead of a stack
Heapsort and Heaps
• You might recall from 364 a Heap:
• A binary tree stored in an array with the Heap property
• A value stored in a heap will be greater than values stored in the value’s node’s subtrees
• The tree must be a left-complete tree, which means that all leaf nodes are on the bottom two levels such that the nodes on the lowest level have no open nodes to their left
• In an array, a tree has the following pattern:
• Node i has children in position i*2 and i*2+1
• To satisfy the second attribute of a heap above, it means that the tree will be stored in array locations 0..n-1 for n nodes (or 1..n if we are using a language other than Java/C/C++)
• Because of the first attribute, we know the largest value will be at the root of the heap, so Heapsort iteratively removes the root of the heap and restructures the heap until the heap is empty
• Thus, Heapsort will sort in descending order
• NOTE: we can change the heap property so that a node is less than any value in its subtree to sort in ascending order
Heapsort
• The Heapsort algorithm itself is simple once you have the Heap ADT implemented:
• Given an array A of elements
• for(i=0;i<a.length;i++)
• for(i=a.length-1;i>=0;i--)
• a[i]=heap.delete( );
• That is, take the original array and build the heap by adding 1 element at a time
• Each time a new value is added to the heap, the heap is restructured to ensure the heap structure such that it is left-complete and each node is greater than all nodes in the subtree
• Now, refill array a by removing the largest item from the heap, restructure the heap, and repeat until the heap is empty

22 19 14 10 6 4 1

22 19 14 10 6 4 1 16

22 19 14 16 6 4 1 10

Adding to the Heap
• Consider a heap as stored in the array given below
• We now want to add a new value, 16
• Start by placing 16 at the end of the array and then walk the value up in position until it reaches its proper place
• If the array currently has n items, then insert 16 at location n and set temp = n
• Now, compare 16 to its parent (which will be at temp / 2)
• If (heap[temp] > heap[temp / 2]) then this value is greater than the parent, swap the two
• Continue until either heap[temp] < heap[temp / 2] or temp = = 0 (we have reached the root of the tree)

Original Heap

16 inserted

16 walked into place

22 19 14 16 6 4 1 10

19 16 14 10 6 4 1

Deleting from the Heap
• We only want to remove the largest element from the heap, which by definition must be at the root (index 0)
• Store heap[0] in a temporary variable to be returned
• Now, restructure the heap by moving the item at heap[n-1] (the last element in the array) to the root and walking it down into its proper position. How?
• Let temp = 0
• Compare heap[temp] with heap[temp*2] and heap[temp*2+1] – that is, with its two children
• If heap[temp] is not the largest of the three, swap heap[temp] with the larger of the two children and repeat until either the value is at a leaf, or is greater than its two children and thus in its proper place
• Return the old root (temporary value) and subtract 1 from n

Return 22

Walk down 10

After walking 10 into place

Heapsort Code and Analysis
• The student is invited to examine the algorithm as given on page 184 and 186 and the code on page 190 and 191
• It is not reproduced here for brevity (also, the code in the book is not the best version!)
• What is the complexity of this algorithm?
• Since the heap is always a balanced tree, it should be obvious that the heap’s height will be log n or log n – 1
• To add an element requires walking it up from a leaf node to its proper position, which at most, will be the root, or log n operations
• To remove an element requires deleting the root, moving the last item to the root and walking it down to its proper position, which at most will be a leaf, or log n operations
Analysis Continued
• How many times do we perform walkup and walkdown during a sort? Once per add and once per delete
• How many times do we add? n times
• How many times do we delete? n times
• Since walkup and walkdown are both log n operations, it takes  (n log n) to build the heap and  (n log n) to remove all items from the heap
• So, Heapsort is in   (n log n) in the worst case
• A more formal analysis is given on pages 190-191 where we see that the actual number of comparisons is roughly 2 * (n log n – 1.443 * n)
• What about the average case?
• We must determine the average amount of work performed by walkup and walkdown
• Let’s assume that all elements will differ in the array to be sorted
• Then, to insert element i, there is a 1 / (i + 1) chance that the element will fall between any two other elements (as we saw with insertion sort)
• However, unlike insertion sort, the amount of work does not range from 1 to i comparisons but instead from 1 to log i comparisons
Analysis Continued
• The average amount for any given walk up or walk down is 1 / (j + 1) *  (i = 1 to j) log i =

(j log j – 1.443 * j) / (j + 1)  log j – 1.443

• We must now sum this for all j from 1 to n twice (once for each walk up and for each walk down)
• So, the average case complexity for Heapsort =
• 2 *  [(i = 1 to n) log i – 1.443] =
• 2 * ( (i = 1 to n) log i – 1.443 * n) =
• 2 * (n log n – 1.443 * n – 1.443 * n) = 2 * (n log n – 2 * 1.443 * n)
• Thus, the only change in complexity between the worst and average cases is a doubling of 1.443 in the latter term
• So, the average case of Heapsort is in   (n log n)
Improving Heapsort
• Walkup requires fewer comparisons than walkdown
• walkup compares a given value against its parent, walkdown compares a given value against both children
• When a heap is large,
• the amount of walking down a value might be improved by adding some guessing in terms of how far down the value might be walked
• rather than walking it down the whole tree, we might walk it down some distance, and then bubble up a larger value at a leaf level
• a value walked down from root to leaf takes  2 * log n comparisons
• if we can walk it halfway down and bubble up a value from the leaf, we only need 2 * (log n) / 2 + 1 * (log n) / 2 comparisons = 3 * log n / 2 comparisons = log (n3) / 2
• How much of an improvement is this over log n?
• If n = 1000, the normal walkdown takes 20 comparisons, the bubble up walkdown takes 15
• But this improvement is risky, what if we don’t need to do any bubbling up? See pages 192-196 for a more detailed analysis
The Shell Sort
• Shell Sort algorithm is somewhat like the Insertion Sort
• It is an in-place sort
• Keys are compared so that smaller keys are moved before larger keys
• The main difference is that in Shell Sort, the values being compared are not necessarily next to each other
• Instead, we start by comparing values in intervals and lower the interval distance
• For instance, compare A[0], A[5], A[10], A[15] and compare A[1], A[6], A[11], A[16] and compare A[2], A[7], A[12], A[17] and compare A[3], A[8], A[13], A[18] and compare A[4], A[9], A[14], A[19]
• Next, lower the interval to be 3
• Next, lower the interval to be 2
• Finally, lower the interval to be 1
• We already know from earlier that an in-place sort has a worst-case complexity in  (n2) so is this an improvement?
• It turns out that it is because we are not moving a single value into its proper place with each pass through the list, but instead moving several values to their proper place within the given intervals
• We then repeat
• We do not have to repeat the process n times either, but we have to pick a proper interval to make it work
• We won’t go over the algorithm in detail (see section 4.10 if you are interested) but we will note its analysis next…
Shell Sort Analysis
• The exact worst case performance of shell sort has yet to be proven because it is not known what set of intervals is best
• It has been shown that if only two intervals are used, first when 1.72 * (n 1/3) and 1, then the performance is roughly n5/3
• It is also known that for intervals of 2k – 1 for 1 <= k <= log n, the performance is in O(n3/2)
• Finally, there is a sequence of intervals that gives Shell Sort a performance in O(n (log n)2)
• To determine how good this is, we know the following
• n log n < n (log n)2 < n3/2 < n5/3 < n2
• So, Shell Sort improves over some of the algorithms we have seen in this chapter without the overhead of additional memory space or  (n) operations per iteration (as with Mergesort or Heapsort)
Bucket Sorting
• Recall earlier that we proved the worst case lower bound for a sort that compares values is n log n
• But not all sorting algorithms must compare values against other values to be sorted
• How can we get around this?
• Let’s sort a list of values by analyzing the key of each value and placing it in a bucket (or pile)
• We can now sort only those keys in a given pile
• Then “scoop” up all of the piles keeping them in their proper order
• If we can create a pile using some n operations for n keys, and scoop them up in n operations, two thirds of our work is in   (n)
• Can we also sort each pile in   (n)? Only if we can do so without comparing the values in that pile
• Or if we can keep each pile small enough so that the k log k operations on the given pile (where k is the size of the pile) is small enough to keep the entire solution in   (n)
• Radix sort is an example of a bucket sort where the “sort a pile” step is omitted
• However, the “distribute keys to a pile” and “scoop up pile” steps must be repeated
• The good news: because the “distribute” and “scoop” steps are   (n), and because the number of repetitions is a constant based not on n, but on the size of the keys, then the Radix sort is   (n)
• The bad news: in many cases, the algorithm may be difficult to implement and the amount of work required to “distribute” and “scoop” is in   (n) with a large constant multiplier
• The result is interesting: an algorithm with a worst case complexity in   (n) but with a run-time substantially longer than sorts in   (n log n)
The Radix Sort Algorithm
• For the purpose of describing the algorithm
• we will assume that we are dealing with int keys that are no longer than k digits
• create 10 FIFO queues, queue[0]..queue[9]
• for i = k downto 1 do
• for j = 0 to n-1 do
• temp = jth value in the list
• peel off digit i from temp and place temp into queue[i]
• for j=0 to 9
• remove all items from queue[j] and return them to the original list in the order they were removed
Believe it or not…
• The algorithm really works! We show by induction:
• After one pass, the values are sorted by their final digit
• Assuming that after pass k-1, the values are sorted from their second to last digits, then
• If we remove all items from queue 0, then they are already sorted from second to last digit. Since they all start with 0, they are sorted correctly. Next we remove all items from queue 1 (already sorted), etc… and so we will have a completely sorted list
• Example 
• The complexity is easy to figure out from the pseudocode earlier:
• In sorting int values
• We create 10 queues (0 comparisons)
• We iterate for each digit
• for int values, assume 10 digits since int values are stored in 32 bits, giving a range of roughly –2 billion to +2 billion
• The inner loop requires taking each key (n of them), peeling off the current digit, and determining which queue it is placed into – n comparisons
• this assumes a (1) enqueue operation and a (1) “peel digit off” operation
• Now we have to remove every item from every queue, this requires n dequeues, again assuming (1) for dequeue
• Thus, the algorithm has 2 * n + 2 *  (1) operations per iteration and there are 10 iterations or roughly 20 * n comparisons which is  (n) !
• What if we were dealing with floats or doubles instead of int values?
• What if we were dealing with strings instead of int values?