Sorting

How fast can we sort?

• All the sorting algorithms we have seen so far are comparison sorts: only use comparisons to determine the relative order of elements.

• The best worst-case running time that we’ve seen for comparison sorting is O(nlgn).

– Is O(nlgn)the best we can do?

• Decision treescan help us answer this question. •E.g., insertion sort, merge sort, quicksort, heapsort.

Decision-tree example• Sort ⟨a1, a2, …, an⟩

Each internal node is labeled i:j for i, j {1, 2,…, n}.∈The left subtree shows subsequent comparisons if ai≤aj.The right subtree shows subsequent comparisons if ai≥aj.

• Sort ⟨a1, a2, a3 = ⟩• ⟨9, 4, 6 :⟩

• Sort ⟨a1, a2, a3 = ⟩• ⟨9, 4, 6 :⟩

• Sort ⟨a1, a2, a3 = ⟩• ⟨9, 4, 6 :⟩

Each leaf contains a permutation π(1), π(2),…, π(⟨ n) to indicate ⟩that the ordering aπ(1)≤aπ(2)≤...≤aπ(n)has been established.

• Sort ⟨a1, a2, a3 = ⟩• ⟨9, 4, 6 :⟩

Decision-tree model• A decision tree can model the execution of any

comparison sort:

• One tree for each input size n. (there is specific tree for different n values tree is not generic)

• View the algorithm as splitting whenever it compares two elements.

• The tree contains the comparisons along all possible instruction traces.

• The running time of the algorithm =the length of the path taken.

• Worst-case running time =height of tree.

Lower bound for decision-tree sorting

• Theorem. Any decision tree that can sort n elements must have height Ω(nlgn).

Proof. The tree must contain ≥n! leaves, since there are n! possible permutations.

• A height-h binary tree has ≤2h leaves. – Thus, n! ≤2h.

• h≥lg(n!) (lg is mono. İncreasing )• ≥lg ((n/e)n)(Stirling’s formula)• = nl(g n–lg e) lg e is constant • = Ω(nlg n)

monotonically increasing,increasing if for all x and y such that x ≤ y one has f(x) ≤ f(y), so f preserves the order

Lower bound for comparison sortingCorollary• Heapsort and merge sort are asymptotically

optimal comparison sorting algorithms.

Counting Sort

1 CountingSort(A, B, k)

2 for i=1 to k

3 C[i]= 0;

4 for j=1 to n

5 C[A[j]] += 1;

6 for i=2 to k

7 C[i] = C[i] + C[i-1];

8 for j=n downto 1

9 B[C[A[j]]] = A[j];

10 C[A[j]] -= 1;

What will be the running time?

Takes time O(k)

Takes time O(n)

Analysis

Running time

• If k= O(n), then counting sort takes Θ(n)time.• But, sorting takes Ω(nlgn) time!

Answer:• Comparison sorting takes Ω(nlgn)time.• Counting sort is not a comparison sort.• In fact, not a single comparison between

elements occurs!

• Why don’t we always use counting sort?• Because it depends on range k of elements• Could we use counting sort to sort 32 bit

integers? Why or why not?• Answer: no, k too large (232 = 4,294,967,296)

Stable sorting

• Counting sort is a stablesort: it preserves the input order among equal elements.

Dynamic setsDynamic sets

Sets that can grow, shrink, or otherwise change over time.

• Two types of operations:– queries return information about the set– modifying operations change the set

Common queriesSearch, Minimum, Maximum, Successor, Predecessor

Common modifying operationsInsert, Delete

Dictionary

DictionaryStores a set S of elements, each with an associated key (integer value).

Operations

Search(S, k): return a pointer to an element x in S with key[x] = k, or NIL if such an element does not exist.

Insert(S, x): inserts element x into S, that is, S ← S ⋃ {x}

Delete(S, x): remove element x from S

Search Insert Delete

linked list

sorted array

hash table

Θ(1) Θ(1)

Θ(n)

Θ(n)

Θ(n)Θ(log n)

Implementing a dictionary

Hash table Running times are average times and assume (simple) uniform hashing and a large enough table (for example, of size 2n)

Today Binary search trees

Θ(1) Θ(1) Θ(1)

Binary search trees

• Binary search trees are an important data structure for dynamic sets.

• They can be used as both a dictionary and a priority queue.

• They accomplish many dynamic-set operations in O(h) time, where h = height of tree.

5

Binary search trees• root of T denoted by root[T]• internal nodes have four fields:

key (and possible other satellite data)left: points to left childright: points to right childp: points to parent. p[root[T]] = NIL

1

73

84

Binary tree T

root[T]

= NIL

= NIL

Binary search trees

1

5

73

84

Binary tree T

root[T]

Keys are stored only in internal nodes!There are binary search trees which store keys only in the leaves …

Binary search trees

• a binary tree is– a leaf or– a root node x with a binary tree as its left and/or right child

Binary-search-tree property– if y is in the left subtree of x, then key[y] ≤ key[x]– if y is in the right subtree of x, then key[y] ≥ key[x]

Keys don’t have to be unique …

1

5

73

84

Binary search trees

1

5

73

84

1

3

7

85

4

height h = 2

height h = 4

Binary-search-tree property– if y is in the left subtree of x, then key[y] ≤ key[x]– if y is in the right subtree of x, then key[y] ≥ key[x]

Inorder tree walk• binary search trees are recursive structures

➨ recursive algorithms often work well

Example: print all keys in order using an inorder tree walk

InorderTreeWalk(x) 1. if x ≠ NIL2. then InorderTreeWalk(left[x])3. print key[x]4. InorderTreeWalk(right[x])

Correctness: follows by induction from the binary search tree property

Running time?

Intuitively, O(n) time for a tree with n nodes, since we visit and print each node once.

1

5

73

84

Inorder tree walkInorderTreeWalk(x) 1. if x ≠ NIL2. then InorderTreeWalk(left[x])3. print key[x]4. InorderTreeWalk(right[x])

TheoremIf x is the root of an n-node subtree, then the call InorderTreeWalk(x) takes Θ(n) time.

Proof:– T(n) takes small, constant amount of time on empty subtree

➨ T(0) = c for some positive constant c– for n > 0 assume that left subtree has k nodes, right subtree n-k-1

➨ T(n) = T(k) + T(n-k-1) + d for some positive constant d

use substitution method … to show: T(n) = (c+d)n + c ■

Querying a binary search treeTreeSearch(x, k) 1. if x = NIL or k = key[x]2. then return x3. if k < key[x]4. then return TreeSearch(left[x], k)5. else return TreeSearch(right[x], k)

Initial call: TreeSearch(root[T], k)

– TreeSearch(root[T], 4)– TreeSearch (root[T], 2)

Running time: Θ(length of search path)worst case Θ(h)

k

≤ k ≥ k

Binary-search-tree property

1

5

73

84

Querying a binary search tree – iteratively

TreeSearch(x, k) 1. if x = NIL or k = key[x]2. then return x3. if k < key[x]4. then return TreeSearch(left[x], k)5. else return TreeSearch(right[x], k)

IterativeTreeSearch(x, k) 1. while x ≠ NIL and k ≠ key[x]2. do if k < key[x]3. then x ← left[x]4. else x ← right[x]5. return x

• iterative version more efficient on most computers

k

≤ k ≥ k

Binary-search-tree property

Minimum and maximum• binary-search-tree property guarantees that

– the minimum key is located in the leftmost node

– the maximum key is located in the rightmost node

TreeMinimum(x) 1. while left[x] ≠ NIL2. do x ← left[x]3. return x

TreeMaximum(x) 1. while right[x] ≠ NIL2. do x ← right[x]3. return x

k

≤ k ≥ k

Binary-search-tree property

1

5

73

84

Minimum and maximum• binary-search-tree property guarantees that

– the minimum key is located in the leftmost node

– the maximum key is located in the rightmost node

TreeMinimum(x) 1. while left[x] ≠ NIL2. do x ← left[x]3. return x

TreeMaximum(x) 1. while right[x] ≠ NIL2. do x ← right[x]3. return x

k

≤ k ≥ k

Binary-search-tree property

Running time?Both procedures visit nodes on a downward path from the root ➨ O(h) time

Successor and predecessor• Assume that all keys are distinct

Successor of a node xnode y such that key[y] is the smallest key > key[x]

We can find y based entirely on the tree structure, no key comparisons are necessary …

if x has the largest key, then we say x’s successor is NIL

Successor and predecessorSuccessor of a node x

node y such that key[y] is the smallest key > key[x]

Two cases:

1. x has a non-empty right subtree ➨ x’s successor is the minimum in x’s right subtree

2. x has an empty right subtree ➨ x’s successor y is the node that x is the predecessor of

(x is the maximum in y’s left subtree)as long as we move to the left up the tree (move up through right children), we’re visiting smaller keys …

Successor and predecessorTreeSucessor(x) 1. if right[x] ≠ NIL2. then return TreeMinimum(right[x])3. y ← p[x]4. while y ≠ NIL and x = right[y]5. do x ← y6. y ← p[x]7. return y

– Successor of 15?– Successor of 6?– Successor of 4?

3

15

186

20177

42 13

9

Successor and predecessorTreeSucessor(x) 1. if right[x] ≠ NIL2. then return TreeMinimum(right[x])3. y ← p[x]4. while y ≠ NIL and x = right[y]5. do x ← y6. y ← p[x]7. return y

Running time?

3

15

186

20177

42 13

9

O(h)

InsertionTreeInsert(T, z) 1. y ← NIL2. x ← root[T]3. while x ≠ NIL4. do y ← x5. if key[z] < key[x]6. then x ← left[x]7. else x ← right[x]8. p[z] ← y9. if y = NIL10. then root[T] ← x 11. else if key[z] < key[y]12. then left[y] ← z13. else right[y] ← z

• to insert value v, insert node z with key[z] = v, left[z] = NIL, and right[z] = NIL

• traverse tree down to find correct position for z

13

7

6

15

InsertionTreeInsert(T, z) 1. y ← NIL2. x ← root[T]3. while x ≠ NIL4. do y ← x5. if key[z] < key[x]6. then x ← left[x]7. else x ← right[x]8. p[z] ← y9. if y = NIL10. then root[T] ← x 11. else if key[z] < key[y]12. then left[y] ← z13. else right[y] ← z

• to insert value v, insert node z with key[z] = v, left[z] = NIL, and right[z] = NIL

• traverse tree down to find correct position for z

3

18

2017

42

9

14zx

y = NILy

x y

x y

x y

= NILx

InsertionTreeInsert(T, z) 1. y ← NIL2. x ← root[T]3. while x ≠ NIL4. do y ← x5. if key[z] < key[x]6. then x ← left[x]7. else x ← right[x]8. p[z] ← y9. if y = NIL10. then root[T] ← z 11. else if key[z] < key[y]12. then left[y] ← z13. else right[y] ← z

Running time?

• to insert value v, insert node z with key[z] = v, left[z] = NIL, and right[z] = NIL

• traverse tree down to find correct position for z

3

18

2017

42

9 14 z

y

15

6

7

13

O(h)

7

Deletion• we want to delete node z

TreeDelete has three cases:

• z has no children

➨ delete z by having z’s parent point to NIL, instead of to z

3

15

186

209

18

Deletion• we want to delete node z

TreeDelete has three cases:

• z has no children

➨ delete z by having z’s parent point to NIL, instead of to z

• z has one child

➨ delete z by having z’s parent point to z’s child, instead of to z

3

15

6

209

7

• we want to delete node z

TreeDelete has three cases:

• z has no children

➨ delete z by having z’s parent point to NIL, instead of to z

• z has one child

➨ delete z by having z’s parent point to z’s child, instead of to z

• z has two children

➨ z’s successor y has either no or one child ➨ delete y from the tree and replace z’s key and satellite data with y’s

6

7

Deletion

3

15

209

18

• we want to delete node zTreeDelete has three cases:• z has no children

➨ delete z by having z’s parent point to NIL, instead of to z

• z has one child

➨ delete z by having z’s parent point to z’s child, instead of to z

• z has two children➨ z’s successor y has either no or one child

➨ delete y from the tree and replace z’s key and satellite data with y’s

Running time?

7

Deletion

3

15

209

18

O(h)

Minimizing the running time• all operations can be executed in time proportional to the height h of the

tree (instead of proportional to the number n of nodes in the tree)

Worst case:

Solution: guarantee small height (balance the tree) ➨ h = Θ(log n)

Next week: several schemes to balance binary search trees

Θ(n)

Balanced binary search treesAdvantages of balanced binary search trees

over linked lists

efficient search in Θ(log n) time

over sorted arrays

efficient insertion and deletion in Θ(log n) time

over hash tables

can find successor and predecessor efficiently in Θ(log n) time

Search Insert Delete

linked list

sorted array

hash table

binary search tree

Θ(1) Θ(1)

Θ(n)

Θ(n)

Θ(n)Θ(log n)

Implementing a dictionary

balanced binary search trees ➨ height is Θ(log n)

Θ(1) Θ(1) Θ(1)

Θ(height)Θ(height)Θ(height)