Transcript lecture4-2001ppt

Lecture 4
Material in the textbook on
Pages 44-46, 50-53 of 2nd Edition
Sections 1.2.4 and 1.2.6
+ Hanoy towers
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Review: Recursive Process
(define (exp-R a b) ; computes ab
(if (= b 0)
1
(* a (exp-R a (- b 1))))) ; ab = a * ab-1
(exp-R 3 4)
(* 3 (exp-R 3 3))
(* 3 (* 3 (exp-R 3 2)))
(* 3 (* 3 (* 3 (exp-R 3 1))))
(* 3 (* 3 (* 3 (* 3 (exp-R 3 0)))))
(* 3 (* 3 (* 3 (* 3 1))))
(* 3 (* 3 (* 3 3)))
(* 3 (* 3 9))
(* 3 27)
• Space b <= R(b) <= b which is Q(b)
81
• Time b <= R(b) <= 2b which is Q(b)
Linear Recursive Process
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Review – Iterative process
(define (exp-iter a b product)
(if (= b 0)
product
(exp-iter a (- b 1) (* a product))))
(define (exp-I a b) (exp-iter a b 1))
(exp-I 3 4)
(exp-iter 3
(exp-iter 3
(exp-iter 3
(exp-iter 3
(exp-iter 3
81
4
3
2
1
0
1)
3)
9)
27)
81)
• Space Q(1)
• Time Q(b)
Linear Iterative Process
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Another algorithm for computing ab
• If b is even, then ab = (a2)(b/2)
• If b is odd, then ab = a*a(b-1)
Note that here, we reduce the problem in half in one step.
(define (exp-fast a b) ; computes ab
(cond
((= b 0) 1)
((even? b) (exp-fast (* a a) (/ b 2)))
(else (* a (exp-fast a (- b 1)))))))
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The conditional form
(cond (<test-1> <consequent-1>)
(<test-2> <consequent-2>)
….
(<test-n> <consequent-n>)
(else
<consequent-else>))
(define (abs x)
(cond ((> x 0) x)
((= x 0) 0)
(else
((< x (0) x))))
(- x))))
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(exp-fast 3 56)
(define (exp-fast a b)
(cond ((= b 0) 1)
((even? b) (exp-fast (* a a) (/ b 2)))
(else (* a (exp-fast a (- b 1)))))))
(exp-fast 3 56) ; compute 3^56
(exp-fast 9 28)
(exp-fast 81 14)
(exp-fast 6561 7)
6561 * (exp-fast 6561 6)
6561 * (exp-fast 43046721 3)
6561 * 43046721 * (exp-fast 43046721 2)
6561 * 43046721 * (exp-fast 1853020188851841 1)
6561 * 43046721 * 1853020188851841 * (exp-fast .. 0)
6561 * 43046721 * 1853020188851841
523347633027360537213511521
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How much time does exp-fast take?
Denote T(b) the number of arithmetic operations it takes
to compute (exp-fast a b).
If b is even:
and if b is odd then:
T(b) = T(b/2)+2
T(b) = T((b-1)/2)+3
T(b) <= T(b/2)+O(1)
T(1) = O(1)
Conclusion: T(b)=O(log b)
The analysis is tight.
The order of growth in time and space is Q(log b) -logarithmic.
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Comparing the three exponentiation procedures
Assume a,b are integers, written in binary with 400 digits.
a = 100101010101010111110100110101….
b = 101001010101011000101001010101….
2400 <= a,b <= 2401
Time
Space
exp-R (recursive)
Q(b)
Q(b)
exp-I (iterative)
Q(b)
Q(1)
Q(log b) Q(log b)
exp-fast
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Is exp-R feasible?
exp-R takes Q(b) space.
We need at least 2400 storage bits.
That’s about 2370 giga bits.
Each gigabit costs a dollar…
Never mind. Let’s go to the dealer
Sorry, that’s more the number of particles
in the universe…..
Absolutely infeasible !!!!
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Is exp-I feasible?
exp-I takes at least 2400 operations.
We can run 1 billion (109 ) operations a second.
We need about 2370 seconds.
That’s about 2343 years.
That’s about 2340 millenniums.
Might be longer then the universe age….
Might be longer than the time our plant will last….
Infeasible !!!!
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Let’s buy a faster computer
and make exp-I feasible.
Our new computer can run giga billion (1018 ) operations a
second. Absolutely the last word in the field of computing.
We need about 2340 seconds.
That’s about 2313 years.
That’s about 2310 millenniums.
Does not help much.
Infeasible !!!!
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Exp-fast is feasible.
We use a first generation pc, manufactured at 1977 and
executing one operation a second.
We need about 1200 operations.
That’s about 20 minutes.
We need 1200 storage bits.
Feasible !!!!
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Let’s buy a faster computer..
We use a second generation pc, manufactured at 1987 and
executing one million operations a second.
We need about 1200 operations.
That’s so much less than a second that we do not bother
counting it.
We still need 1200 storage bits.
Very feasible !!!!
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Towers of Hanoi
• Three posts, and a set of different size disks
• A disk can be only on a larger size disk.
• At the beginning all the disks are on the left post.
The goal is to move the disks one at a time, while preserving
these conditions, until the entire stack has moved from one
post to another
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Use our paradigm
• Wishful thinking:
• Smaller problem: A problem with one disk less
• How do we use it ?
Move n-1 disks from peg A to peg B
Move the largest from peg A to peg C
Move n-1 disks from peg B to peg C
We solve 2 smaller problems !
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Towers of Hanoi
(define (move-tower size from to aux)
(cond ((= size 1) (one-move from to))
(else (move-tower (- size 1) from aux to)
(one-move from to)
(move-tower (- size 1) aux to from))))
(define (one-move from to)
(display "Move top disk from ")
(display from)
(display " To ")
(display to)
(newline))
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Towers of Hanoi -- trace
(move-tower 3
Move top disk
Move top disk
Move top disk
Move top disk
Move top disk
Move top disk
Move top disk
2 1 3)
from 2
from 2
from 1
from 2
from 3
from 3
from 2
to
to
to
to
to
to
to
1
3
3
1
2
1
1
(move-tower 2 2 3 1)
(move-tower 2 3 1 2)
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3
333
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Tree Recursion
(mt 3 2 1 3)
(mt 2 2 3 1) (move-one 2 1) (mt 2 3 1 2)
(move-one 2 3)
(mt 1 2 1 3)
(mt 1 1 3 2)
(mt311)2 1 3)
(mt 1 3(move-one
2 1)
(move-one 2 1)(move-one 1 3)(move-one 3 2) (move-one 2 1)
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Orders of growth for towers of Hanoi
Denote by T(n) be the number of steps that we need to take to solve the
case for n disks.
T(n) = 2T(n-1) + 1
T(1) = 1
This solves to:
T(n) = 2n - 1
exponential
For the space complexity we have
S(n) = S(n-1) + O(1)
S(n) = O(n)
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Moral
If you wish to see the result
Of your programming efforts
Better think ahead
What your algorithm
Is
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Primality Testing
n is a prime iff its only divisors are 1 and n
(define (divides? a b)
(= (remainder b a) 0))
(define (smallest-divisor n)
(define (find-divisor n i)
(cond ((divides? i n) i)
(else (find-divisor n (+ i 1)))))
(find-divisor n 2))
(define (prime? n)
(= n (smallest-divisor n)))
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(Prime? 7)
(= 7 (smallest-divisor 7))
(= 7 (find-divisor 7 2))
(= 7 (cond (divides? 2 7) 2)
(else (find-divisor 7 3))))
(= 7 (find-divisor 7 3))
(= 7 (cond (divides? 3 7) 3)
(else (find-divisor 7 4))))
(= 7 (find-divisor 7 4))
(= 7 (find-divisor 7 5))
(= 7 (find-divisor 7 6))
(= 7 (find-divisor 7 7))
(= 7 7)
#t
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Analysis
Time complexity:
T(n) = T(n-1) + O(1)
T(n)= Q(n) – linear in n
In fact, for every prime n, the running time is n.
If n is a 400 digit number, that’s very bad.
Absolutely infeasible.
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Primality Testing - II
n is a prime iff its only divisors are 1 and n
Iff it has no divisors between 2 and (sqrt n)
(define (divides? a b)
(= (remainder b a) 0))
(define (smallest-divisor n)
(define (find-divisor n i)
(cond ((> i (sqrt n)) n)
((divides? i n) i)
(else (find-divisor n (+ i 1)))))
(find-divisor n 2))
(define (prime? n)
(= n (smallest-divisor n)))
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Analysis
• Correctness:
If n is not a prime, then n=a * b for a,b>1.
Then at least one of them is n.
So n must have a divisor smaller then n.
• Time complexity:
Q (n) . For a number n, we test at most n
numbers to see if they divide n.
If n is a 800 digit number, that’s very bad.
Absolutely infeasible.
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The Fermat Primality Test
Fermat’s little theorem:
If n is a prime number then:
an = a (mod n) for every 0 < a < n, integer
The Fermat Test:
Do 1000 times:
Pick a random a < n and compute an (mod n)
If  a then for sure n is not a prime.
If all 1000 tests passed, declare that n is a prime.
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Computing ab (mod m) fast.
(define (expmod a b m) ; computes ab (mod m)
(cond
((= b 0) 1)
((even? b)
(remainder (expmod
(remainder (* a a) m)
(/ b 2)
m) m))
(else
(remainder (* a (expmod a (- b 1) m))
m))))
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Implementing Fermat test
(define (one-test n)
(define (test a)(= (expmod a n n) a))
(test (+ 1 (random (- n 1)))))
(define (many-tests n t); calls one-test t times
(cond ((= t 0) true)
((one-test n) (many-test n (- t 1)))
(else false)))
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Time complexity
To test if n is a prime. We run 100 tests.
Each takes about log(n) multiplcations.
T(n) = O(log n)
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Some mathematical facts
Fermat’s theorem: Every prime will always pass
the test.
Definition: A Carmichael number, is a number
such that
•n is Composite, and
•n always passes the test.
For every a, an = a (mod n)
A fact: If n is not a Carmichael number then
for at least half of the choices of a, an <>
a (mod n).
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Correctness
Suppose we do the test t=100 times.
• If n is a prime we are never wrong.
• If n is a composite and not a Carmichael number
we are wrong with probability at most 2-100 .
Error probability smaller than the chance the hardware
is faulty.
• If n is a Carmichael number, we are always wrong
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A probabilistic algorithm
An algorithm that uses random coins,
and for every input gives the right answer
with a good probability.
Even though Carmichael numbers are very rare
Fermat test is not good enough.
There are inputs on which it is wrong.
There are modifications of Fermat’s test,
that for every input give the right answer,
with a high probability.
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