Transcript Class 20: Quick Sorting Queen’s University, Belfast, Northern Ireland CS200: Computer Science University of Virginia Computer Science David Evans http://www.cs.virginia.edu/evans.

Class 20:
Quick
Sorting
Queen’s University, Belfast, Northern Ireland
CS200: Computer Science
University of Virginia
Computer Science
David Evans
http://www.cs.virginia.edu/evans
TuttleSort
(define (TuttleSort cf lst)
(if (null? lst) lst
(let ((most (find-most cf lst)))
(cons
most
(TuttleSort cf
(delete lst most))))))
(define (find-most cf lst)
(insertl
(lambda (c1 c2)
(if (cf c1 c2) c1 c2))
lst
(car lst)))
TuttleSort is (n2)
If we double the length of the list, we
amount of work sort does approximately
quadruples.
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Insert Sort
(define (insertel cf el lst)
(define (insertsort cf lst)
(if (null? lst)
(if (null? lst)
(list el)
null
(if (cf el (car lst))
(insertel cf
(cons el lst)
(car lst)
(cons (car lst)
(insertsort cf
(insertel cf el
(cdr lst)))))
(cdr lst))))))
insertsort is (n2)
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Divide and Conquer
• Both TuttleSort and InsertSort divide the
problem of sorting a list of length n into:
– Sorting a list of length n-1
– Doing the right thing with one element
• Hence, there are always n steps
– And since each step is  (n), they are  (n2)
• To sort more efficiently, we need to divide
the problem more evenly each step
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Can we do better?
(insertel < 88
(list 1 2 3 5 6 23 63 77 89 90))
Suppose we had procedures
(first-half lst)
(second-half lst)
that quickly divided the list in two halves?
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Insert Halves
(define (insertelh cf el lst)
(if (null? lst)
(list el)
(let ((fh (first-half lst))
(sh (second-half lst)))
(if (cf el (car fh))
(append (cons el fh) sh)
(if (null? sh)
(append fh (list el))
(if (cf el (car sh))
(append (insertelh cf el fh) sh)
(append fh (insertelh cf el sh))))))))
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Evaluating insertelh
> (insertelh < 3 (list 1 2 4 5 7))
(define (insertelh cf el lst)
|(insertelh #<procedure:traced-<> 3 (1 2 4 5 7))
(if (null? lst)
| (< 3 1)
(list el)
| #f
(let ((fh (first-half lst))
| (< 3 5)
| #t
(sh (second-half lst)))
| (insertelh #<procedure:traced-<> 3 (1 2 4))
(if (cf el (car fh))
| |(< 3 1)
(append (cons el fh) sh)
| |#f
(if (null? sh)
| |(< 3 4)
(append fh (list el))
| |#t
(if (cf el (car sh))
| |(insertelh #<procedure:traced-<> 3 (1 2))
(append (insertelh cf el fh) sh)
| | (< 3 1)
| | #f
(append fh (insertelh cf el sh))))))))
| | (< 3 2)
| | #f
| | (insertelh #<procedure:traced-<> 3 (2))
| | |(< 3 2)
| | |#f
| | (2 3)
Every time we call
, the size
| |(1 2 3)
of the list is approximately halved!
| (1 2 3 4)
|(1 2 3 4 5 7)
(1 2 3 4 5 7)
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insertelh
How much work is insertelh?
Suppose first-half and second-half are  (1)
Each time we call
insertelh, the size of
lst halves. So, doubling
the size of the list only
increases the number of
calls by 1.
List Size
1
2
4
8
16
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(define (insertelh cf el lst)
(if (null? lst)
(list el)
(let ((fh (first-half lst))
(sh (second-half lst)))
(if (cf el (car fh))
(append (cons el fh) sh)
(if (null? sh)
Number of insertelh applications
(append fh (list el))
1
(if (cf el (car sh))
2
(append (insertelh cf el fh) sh)
3
(append fh
4
(insertelh cf el sh))))))))
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How much work is insertelh?
Suppose first-half and second-half are  (1)
Each time we call
insertelh, the size of
lst halves. So, doubling
the size of the list only
increases the number of
calls by 1.
List Size
1
2
4
8
16
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insertelh is  (log2 n)
Number of insertelh applications
1
2
3
4
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log2 a = b
means
b
2 =a
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insertsorth
Same as insertsort, except uses insertelh
(define (insertsorth cf lst)
(if (null? lst)
null
(insertelh cf
(car lst)
(insertsorth
cf
(cdr lst)))))
(define (insertelh cf el lst)
(if (null? lst)
(list el)
(let ((fh (first-half lst))
(sh (second-half lst)))
(if (cf el (car fh))
(append (cons el fh) sh)
(if (null? sh)
(append fh (list el))
(if (cf el (car sh))
(append (insertelh cf el fh) sh)
(append fh
(insertelh cf el sh))))))))
insertsorth would be (n log2 n)
if we have fast first-half/second-half
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Is there a fast first-half procedure?
• No!
• To produce the first half of a list length n,
we need to cdr down the first n/2 elements
• So:
– first-half is  (n)
– insertelh calls first-half every time…so
– insertelh is  (n) *  (log2 n) =  (n log2 n)
– insertsorth is  (n) *  (n log2 n) =  (n2 log2 n)
Yikes! We’ve done all this work, and its still worse than our simple TuttleSort!
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The Great
Lambda
Tree
of Ultimate
Knowledge
and Infinite
Power
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Sorted Binary Trees
e
l
left
A tree containing
all elements x such
that (cf x el) is true
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right
A tree containing
all elements x such
that (cf x el) is false
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Tree Example
3
cf: <
5
2
4
1
null
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8
7
null
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Representing Trees
(define (make-tree left el right)
(list left el right))
left and right are trees
(null is a tree)
(define (get-left tree)
(first tree))
tree must be a non-null tree
(define (get-element tree)
(second tree))
tree must be a non-null tree
(define (get-right tree)
(third tree))
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tree must be a non-null tree
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Trees as Lists
(define (make-tree left el right)
(list left el right))
5
2
1
(define (get-left tree)
(first tree))
(define (get-element tree)
(second tree))
(define (get-right tree)
(third tree))
8
(make-tree (make-tree (make-tree null 1 null)
2
null)
5
(make-tree null 8 null))
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insertel-tree
(define (insertel-tree cf el tree)
If the tree is null, make a new tree
(if (null? tree)
with el as its element and no left or
(make-tree null el null)
right trees.
(if (cf el (get-element tree))
(make-tree
(insertel-tree cf el (get-left tree)) Otherwise, decide
if el should be in
(get-element tree)
the left or right subtree.
(get-right tree))
insert it into that
subtree, but leave the
(make-tree
other subtree unchanged.
(get-left tree)
(get-element tree)
(insertel-tree cf el (get-right tree))))))
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How much work is insertel-tree?
Each time we call
(define (insertel-tree cf el tree)
insertel-tree, the size of
(if (null? tree)
the tree. So, doubling
(make-tree null el null)
(if (cf el (get-element tree))
the size of the tree only
(make-tree
increases the number of
(insertel-tree cf el (get-left tree))
calls by 1!
(get-element tree)
(get-right tree))
insertel-tree is
(make-tree
(get-left tree)
(log2 n)
(get-element tree)
(insertel-tree cf el (get-right tree))))))

log2 a = b
means 2b = a
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insertsort-tree
(define (insertsort cf lst)
(if (null? lst) null
(insertel cf (car lst)
(insertsort cf (cdr lst)))))
(define (insertsort-worker cf lst)
(if (null? lst) null
(insertel-tree cf (car lst)
(insertsort-worker cf (cdr lst)))))
No change…but insertsort-worker evaluates to a tree not a list!
(((() 1 ()) 2 ()) 5 (() 8 ()))
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extract-elements
We need to make a list of all the tree
elements, from left to right.
(define (extract-elements tree)
(if (null? tree)
null
(append (extract-elements (get-left tree))
(cons (get-element tree)
(extract-elements (get-right tree))))))
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How much work is
insertsort-tree?
(define (insertsort-tree cf lst)
(define (insertsort-worker cf lst)
(if (null? lst)
null
(insertel-tree cf
(car lst)
(insertsort-worker cf (cdr lst)))))
(extract-elements (insertsort-worker cf lst)))
(n) applications
of insertel-tree
each is (log n)
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(n log2 n)
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Growth of time to sort random list
12000
n2
TuttleSort
10000
8000
6000
4000
2000
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90
82
74
66
58
50
42
34
26
18
10
2
0
n log2 n
insertsort-tree
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Comparing sorts
> (testgrowth tuttlesort)
n = 250, time = 110
n = 500, time = 371
n = 1000, time = 2363
n = 2000, time = 8162
n = 4000, time = 31757
(3.37 6.37 3.45 3.89)
> (testgrowth insertsort)
n = 250, time = 40
n = 500, time = 180
n = 1000, time = 571
n = 2000, time = 2644
n = 4000, time = 11537
(4.5 3.17 4.63 4.36)
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> (testgrowth insertsorth)
n = 250, time = 251
n = 500, time = 1262
n = 1000, time = 4025
n = 2000, time = 16454
n = 4000, time = 66137
(5.03 3.19 4.09 4.02)
> (testgrowth insertsort-tree)
n = 250, time = 30
n = 500, time = 250
n = 1000, time = 150
n = 2000, time = 301
n = 4000, time = 1001
(8.3 0.6 2.0 3.3)
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Can we do better?
• Making all those trees is a lot of work
• Can we divide the problem in two halves,
without making trees?
Continues in Lecture 21…
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