# Computations in Number Theory Using Python: A Brief Introduction

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1 Computations in Number Theory Using Python: A Brief Introduction Jim Carlson March 2003 Contents 1 Introduction 1 2 Python as a calculator 4 3 Case study: factoring 8 4 Loops and conditionals 11 5 Files 14 6 Case study: the Fermat test 18 7 Problems 23 1 Introduction The aim of these notes is to give a quick introduction to Python as a language for doing computations in number theory. To get an idea of what is possible, we first observe that Python can be used as a calculator. Imagine that we are using our portable unix laptop, logged in as student. The first step is to type the command python. Then we type the actual python code. Here is an example: student% python >>> 2**

2 L >>> 2**1000 % 1001 # compute remainder 562L <control-d> student% We first computed Then we computed mod 10001: that is, we compute the remainder of upon division by This is done using the operator %. Thus 5 % 2 is 1. The suffix L stands for long integer. Python can handle integers of arbitrary size, limited only by the available memory. To exit Python, type control-d. One consequence of our computation is that 1001 is composite: a theorem of Fermat states that if n is an odd prime, then 2 n 1 is congruent to 1 modulo n. Thus it is possible to prove that a number can be factored without actually factoring it. In section 6 we discuss an algorithm that implements this Fermat test very efficiently. Consider next the fundamental problem of factoring an integer into primes. One can do this by trial division. Given n, divide by 2 as many times as possible, then by 3, etc. Record the factors found, and stop when there are no more factors to be found when the running quotient equals one. A Python implementation of trial division is given by the code below. >>> def factor(n):... d = 2... factors = [ ]... while n > 1:... if n % d == 0:... factors.append(d)... n = n/d... else:... d = d return factors... As soon as the function factor is defined, we can use it: >>> factor(1234) [2, 617] >>> factor( ) [3, 3, 3607, 3803] 2

3 >>> factor( ) [3, 3, 3, 3, 37, 37, , ] In the definition of factor it is important to be consistent about indentation. We used two spaces for each level of indentation. The dots... are typed by Python. We will study the code for factor in detail in section 3. For now, however, let us note two important features. The first is the while construction. This is a loop which repeatedly performs certain actions (the indented text below the line beginning with while ). The actions are repeated so long as the condition n > 1 is satisfied. The second feature is the use of if... then... else... statements to make decisions based on program variables. If d divides n we do one thing (add d to the list of factors, divide n by d. If it does not divide n, we do something else (increase the trial divisor d by 1). The if... then... else... construct is a conditional. Loops and conditionals are basic to all programming. Another way of defining functions is recursion. A recursive function definition is one that refers to itself. The function which computes n! = (n 1) n is a good example: >>> def factorial(n):... if n == 0:... return 1... else:... return n*factorial(n-1) >>> factorial(3) 6 >>> factorial(100) L For more information on Python, consult the on-line tutorial by Guido van Rossum [3], explore the python.org web site, or consult the O Reilly books on Python. Exercise 1 Experiment with Python: try some computations using it as a calculator, then enter the code for factor and experiment with it. Use control-c to abort a computation if it takes too long. Exercise 2 Devise and test a function fib(n) which returns the n-th Fibonacci number F n. F n is defined by the initial conditions F 0 = 1, F 1 = 1 and by the recursion relation F n = F n 1 + F n 2. 3

4 2 Python as a calculator Let us look at some more examples of Python used as a calculator. The comments (preceded by # ) explain some of the fine points of the computations. >>> (2 + 3)*(5-22) -85 >>> 5/2 # note integer division 2 >>> 5/2.0 # decimal point for floating point arithmetic 2.5 We can also do arithmetic with complex numbers: >>> z = 1 + 2j (1+2j) >>> w = z**2 (-3+4j) >>> abs(w) 5.0 >>> w.real -3 >>> w.imag >>> 4 One can also say z = complex(1,2) to construct 1 + 2j. It is easy to set up variables and use them in computations. We illustrate this by computing our bank balance after 10 years, assuming just an initial deposit of \$1000 and an interest rate of 5%: >>> balance = 1000 >>> rate =.05 >>> balance = (1+rate)**10*balance >>> balance We could also simulate the balance in our account using a while loop: >>> balance, rate = 1000, 0.05 >>> year = 0 >>> while year <= 10:... print year, balance... balance = (1 + rate)*balance... year = year + 1 4

5 In any case, it is wise to invest. As noted in the introduction, we can define new functions. Consider, for example, f(x) = x 2 modulo 101 It can be defined in Python by or by f = lambda x: x*x % 101 def f(n): return x*x % 101 The first expression assigns the lambda expression lambda x: x*x % 101 to the name f. A lambda expression consists of two parts. The first part, from lambda to the colon tells us what the list of independent variables is. The second part, following the colon, is an expression which defines the function value. Whichever style of definition we use, f defines a secquence of numbers {x n } by the rule x n+1 = f(x n ) where x 0 is given. For example, if x 0 = 2, find, calculating by hand, that {x n } = {2, 4, 16, 54,...}. The same computation can be done in Python as follows: >>> f(2) 4 >>> f(_) # _ = result of last computation 16 >>> f(_) 54 >>> f(_) 88 5

6 We could also use a loop to compute elements of the sequence: >>> while k <= 10:... print k, x... x = f(x)... k = k The orbit of x 0 = 2 under f is the largest nonrepeating sequence x 0, x 1, x 2,..., x n. What is the orbit in this case? Python supports many data types besides integers, long integers, floating point numbers, and complex numbers. Among the other types are strings, lists, tuples, and dictionaries. See [3]. However, lists are especially important, so we consider them now. To begin, we set up a list L and then add an element to it. >>> L = [11,12] # two-element list >>> L.append(13) >>> L [11, 12, 13] >>> primes = [ ] # empty list >>> primes.append(2) >>> primes.append(3) >>> primes.append(5) >>> primes [2, 3, 5] We can add lists, assign the result to a new variable, and compute the length of a list: >>> L2 = L + primes >>> L2 # display L2 [11, 12, 13, 2, 3, 5] >>> len(l2) # length of list 6 6

7 We can also work with individual list elements: >>> L[0] 11 >>> L[0] = -1 >>> L [-1, 12, 13] One way of constructing lists is to use range: >>> L = range(1,10) >>> L [1, 2, 3, 4, 5, 6, 7, 8, 9] Think of range(a,b) as the list with elements x satisfying the inequalities a <= a < b. With a for-in loop one can scan through all elements in a list, performing an action which depends on the current element. We illustrate this in the next example, which computes a table of square roots: >>> from math import sqrt >>> for x in range(0,10):... print x, ":", sqrt(x)... 0 : : : : : : : : : : 3.0 Note that in order to use the sqrt function we had to import it from the math module. If we had wanted to import all of the functions in math, we would have said from math import *. For a description of the Python modules, see python.org/doc/current/modindex.html. Another useful feature of Python is the ability to apply a function to all elements of a list. In the example below we use map to apply lambda x:x*x to the list range(0,10). The result will be a list of squares. >>> map( lambda x: x*x, range(0,10) ) [1, 4, 9, 16, 25, 36, 49, 64, 81] 7

8 Exercise 3 Experiment with Python as a calculator. Exercise 4 Find all the orbits of f(x) = x 2 mod 101 on the set {0, 1, 2,..., 100}. 3 Case study: factoring Let us now study the code for factor, which we will call factor1 in this section. To begin, we read through the commented version below: >>> def factor1(n): # def sets up definition... d = 2 # set trial divisor d to 2... factors = [ ] # set list of factors to empty list... while n > 1: # loop while n > 1... if n % d == 0: # does d divide n?... factors.append(d) # if so, append d to list of factors... n = n/d # then divide n by d... else: # otherwise,... d = d + 1 # increase the trial divisor by 1... return factors # loop complete, return list of factors... Every function definition begins with def, followed by the name of the function and a parenthesized list of its arguments, its independent variables. This is the line def factor1(n). Most function definitions end with a statement of the form return <foo>, where <foo> is the value computed from the arguments. In our case the last statement is return factors. The intent is to return a list of prime factors of n. To do this we set the trial divisor d to 2, and we set the variable factors to the empty list [ ]. The main part of the code is the while loop. So long as n > 1, certain actions are performed. First, if n is divisible by d the trial divisor d is appended to the list of factors. Then n is divided by d and the quotient stored in n. If n is not divisible by d, the trial divisor is increased by 1. Note the consistent use of indentation (two spaces per level) to exhibit the structure of the code, and note the use of colons after statements that begin a new level. The algorithm used to design factor1 can be improved considerably. First, we can divide n by as many 2 s as possible, then do trial division by successive odd numbers beginning with 3. This should speed up factorization by a factor of two. The code below illustrates this improvement. def factor2(n): 8

9 d = 2 factors = [ ] while n % d == 0: factors.append(d) n = n/d d = 3 while n > 1: if n % d == 0: factors.append(d) n = n/d else: d = d + 2 return factors But one can do far better still. If n has a factor, it has a factor d n. You should prove this for yourself. As a result, trial division can stop as soon as d 2 > n. The code below illustrates this improvement. def factor3(n): d = 2 factors = [ ] while n % d == 0: factors.append(d) n = n/d d = 3 while n > 1 and d*d <= n: if n % d == 0: factors.append(d) n = n/d else: d = d + 2 if n > 1: factors.append(n) return factors You should think about why the code fragment if n > 1: factors.append(n) is necessary. Let us digress for a moment on the running time of algorithms, which we measure in seconds or some proportional quantity such as years. Inputs are measured by their information content, that is, by the number of bits needed to 9

10 specify them. The number of bits needed to specify a positive integer n is just the number of binary digits needed to express it. Thus 123 in binary is , so it is a 6-bit integer. In general the number of bits in n is about B = log 2 n. Most of the time in running the various versions of the factoring code above is spent in trial division. Thus a rough estimate of the running time T 1 for factor1 is T 1 number of trial divisions time per division. The amount of time needed to divide (or multiply) B-bit integers is proportional to B 2. The number of divisions required in factor1 is about n in the worst case. Thus T 1 CB 2 n, where means roughly equal to. Since n = e kb with k = log 2, where the logarithm is the natural one this can be written as T 1 CB 2 e kb, where C > 0 is a constant. In the second version we must do about half as many trial divisions, so T 2 CB 2 (n/2) = (CB 2 /2)e kb The third version is much better: at most n trial divisions are required, so T 3 CB 2 n = CB 2 e (k/2)b. To better interpret these running time estimates, use the inequality x 2 < e x for x 0 to obtain T 1 Ce (k+1)b (1) T 2 (C/2)e (k+1)b (2) T 3 Ce ((k+1)/2)b (3) Thus all versions of our factoring algorithms have so-called exponential running times. Reducing the coefficient of B in the exponential is much more significant than reducing the constant C. This we know by pure thought, and our theory is confirmed by the data below. Times are in seconds, run on a 2002-vintage laptop. N factor1 factor2 factor3 factorization [2, 3, 3, 47, 14593] [3, 3, 3607, 3803] [31, 31, ] [2, 61, ] 10

11 We will show in section 5 how to produce timing data like that displayed in the above table. In any case, there is clearly much to think about. In particular, can we factor numbers with hundreds of digits? The estimates just given are upper bound on the running times. Any algorithm can be analyzed to give an upper bound on the computational complexity (number of bit operations) needed to compute a quantity. A better algorithm gives a better upper bound. It is much harder to get lower bounds on the computational complexity. However, one lower bound is clear. The number of bit operations must be at least equal to the sum of the number of bits in the input and the output. This is the case of instantaneous computation. Thus a lower bound on the running time is CB. In this case we say that the running time is linear in the input size. Many algorithms, e.g., long multiplication of positive integers, have polynomial running times: multiplication of two B-bit integers runs in time CB 2. Thus there is room for improvement, and in fact there are faster, though very complicated algorithms. Exercise 5 In light of the above discussion, estimate how long it would take to factor numbers with one or two hundred digits. Exercise 6 Devise a function isprime(n) which returns 1 if n is prime, 0 if it is composite. Exercise 7 Devise a function gcd(a,b) which computes the greatest common divisor of positive integers a and b. Use the Euclidean algorithm. Estimate the running time of gcd. Note. Try both iterative and recursive implementations of gcd. 4 Loops and conditionals As noted earlier, loops and conditional statements are the core of all machine computations. Let us therefore study these constructs more closely. To begin, we consider the problem of computing partial sums of infinite series, e.g., the harmonic series. Thus let h(n) = 1 + 1/ /n and consider the corresponding Python code. >>> def h(n):... sum = 0... for k in range(1,n+1):... sum = sum + 1.0/k... return sum 11

12 ... >>> h(10) >>> h(100) The expression range(n+1) creates the list [1, 2, 3,.., n] the list of integers k such that 1 k < n + 1. To compute h(n) we first set the variable sum to 0. Then, for each k in the given range, we compute sum + 1.0/k, and store the result in sum. Finally, we return the result as the value of h. The computation in the preceding example is driven by a for-in loop. The loop body the set of actions to be repeated n times is indented by a consistent amount, in this case two spaces. The loop body, which in this case is the single statement sum = sum + 1.0/k can be a whole paragraph, or block, possibly with many levels of indentation. The example below, which computes partial sums of the alternating harmonic series illustrates this. >>> def g(n):... sum = 0... for k in range(1,n+1):... if k % 2 == 1:... sum = sum + 1.0/k... else:... sum = sum - 1.0/k... return sum >>> g(10) >>> g(100) g(n) = 1 1/2 + 1/3 1/4 ± Note again the consistent, linguistically significant use of indentation, of which there are now three levels. The if-then-else statement determines whether the current term is to be added to or subtracted from the running sum. Test for equality is always expressed by ==. Assignment of a value to a variable uses =. The else clause in a conditional statement is optional, and one can also have optional elif clauses to express more than two choices, as in the example below. if <condition 1>: <action 1> 12

13 elif <condition 2>: <action 2> else: <action 3> Constructions like this can be as long as you please. In addition to the for-in loop, there is the while loop, already used in the code for factor. We could have just as easily used it for the previous problem: >>> def g(n):... k = 1... sum = 0... while k <= n+1:... if k % 2 == 1:... sum = sum + 1.0/k... else:... sum = sum - 1.0/k... k = k return sum... >>> g(1000) Here the variable k acts as the loop counter. If you forget the statement k = k + 1 which increments the counter, then the loop is infinite and the code will run forever. To terminate a loop which has run too long, type control-c. Exercise 8 Define a function sum(f,a,b) which returns the sum f(a) + f(a+1) f(b) where a <= b and where f is an arbitrary function of one variable. Test your definition of sum using >>> sum( lambda n: n, 1, 10 ) and related expressions. Then calculate the sum 1 + 1/ / /4 2 + to an accuracy which you find satisfactory. Is it possible to find an exact answer for this infinite sum? Consider also the alternating sum 1 1/ /3 2 1/4 2 ± 13

14 Exercise 9 Define a function search(f, x1, x2, y1, y2) which prints a list of all solutions to f(x,y) == 0 for integer vectors (x,y) where x1 <= x <= y, y1 <= y <= y2. Here f is an arbitray function from pairs of integers (x,y) to the integers, e.g., pell2 = lambda x,y:x*x - 2*x*y - 1. This is the function we would use to study Pell s equation x 2 2y 2 = 1. Use your code to find solutions to Pell s equation. A typical call to the function f would be search( pell2, 0, 1000, 0, 1000). What is the running time of this algorithm? Can one do better? Exercise 10 Let f be a function from a set S to itself. Let x 0 be an element of S. Define a sequence {x n } by the rule x n+1 = f(x n ). Define the orbit of x 0 under f to be the shortest non-repeating sequence {x 0, x 1,..., x n }. Devise a function orbit(f,a) to return a list representing the orbit of a under f. Use it to study orbits of f(x) = a x modulo 101 for various a. What are their lengths? What other questions do your findings suggest? Pursue these questions and report on them. Be alert to ways of efficiently automating the computations your questions suggest. Hints and possible ingredients: Use a list to build up the orbit, beginning with the empty list [ ]. To check whether an element is already in the list, use an expression of the form if x not in L: Exercise 11 Devise an iterative definition of the function factorial(n). In other words, use a loop, not recursion. 5 Files It is convenient to separate the functions of running and editing Python code. To do this, use a text editor such as emacs to store and modify the code. As an example, we create a file factor.py for factoring positive integers. # file: factor.py def factor3(n): """factor3(n) returns a list of the prime factors of n""" d = 2 factors = [ ] while n % d == 0: factors.append(d) n = n/d d = 3 14

15 while n > 1 and d*d <= n: if n % d == 0: factors.append(d) n = n/d else: d = d + 2 if n > 1: factors.append(n) return factors Then we can do the following: student% python >>> from factor import factor3 >>> f = factor3 >>> f( ) [127, 9721] >>> f(123) [3, 41] >>> f( ) [3, 3, 41, ] >>> f( ) [3, 31, 41, 41, 271, L] The import command reads in all of the definitions made in the file factor.py just one in this case so that they can be used as if the user had typed them in. To cut down on typing we have assigned factor3 to the variable f. To see what the line after def containing the triple quotes does, try this: >>> print factor3. doc factor3(n) returns a list of the prime factors of n It is good practice to document functions at they same time they are written. Documentation strings can contain more than one line. Let us look at some other ways to use files. Consider the test file below, which is used to print factorizations of the twenty numbers 1000 through 1019: # file = test.py from factor import factor3 def test(n): print " ", n, "==>", factor3(n) print for k in range(1000,1020): test(k) print 15

16 We can then try this at the command line: student% python test.py The result is the output below: 1000 ==> [2, 2, 2, 5, 5, 5] 1001 ==> [7, 11, 13] 1002 ==> [2, 3, 167] 1003 ==> [17, 59] 1004 ==> [2, 2, 251] 1005 ==> [3, 5, 67] 1006 ==> [2, 503] 1007 ==> [19, 53] 1008 ==> [2, 2, 2, 2, 3, 3, 7] 1009 ==> [1009] 1010 ==> [2, 5, 101] 1011 ==> [3, 337] 1012 ==> [2, 2, 11, 23] 1013 ==> [1013] 1014 ==> [2, 3, 13, 13] 1015 ==> [5, 7, 29] 1016 ==> [2, 2, 2, 127] 1017 ==> [3, 3, 113] 1018 ==> [2, 509] 1019 ==> [1019] It would be still more convenient to be able to issue commands like python test.py to construct tables like the one above: a table of factorizations of twenty numbers starting with That way we don t have to edit the code to do a new example. The version of test.py displayed below shows how to do this. Briefly, we import the sys module which gives access to arguments like 1000 and 20 that are typed on the command line. This is done through the list of arguments sys.argv. The arguments the list elements are strings, and so have to be converted to long integers using the long function. # file = test.py import sys from factor import factor3 def test(n): print " ", n, "==>", factor3(n) 16

17 a = long( sys.argv[1] ) b = long( sys.argv[2] ) print k = a while k < a + b: test(k) k = k + 1 print The name sys.argv consists of two parts: the module sys named in the import command and the name argv defined in that module. Contrast this with the names from the module factor: we imported factor3 but no others using from factor import factor3. Note that it is the file factor.py from which factor3 is imported. As a final illustration, we show how to create a system-level command factor which prints a table of factorizations. The idea is to be able to say student% factor at the command line. To do this, first add the line #! /usr/bin/env python to the head of the file. (It must be the first line). Then type the following: mv test.py factor chmod u+x factor mv factor ~/bin/scripts mv factor.py ~/bin/scripts rehash # change name of file to "factor" # make the file executable # put it in standard directory # put factor.py there too # update list of commands You have set up command called factor in your bin/scripts directory. Here we assume that this directory exists and is in your search path. Timings Let us conclude this section with a discussion of how to time the execution of Python code. To this end, let us imagine that the definitions of the three factoring methods discussed in section 3 reside in a file factormethods.py under the names factor1, factor2, and factor3. Then the code in file file = timings.py below. # file = timings.py 17

18 import sys, time from factormethods import factor, factor2, factor3 n = long( sys.argv[1] ) a = time.time() result = factor1(n) b = time.time() print "1: ", b - a a = time.time() result = factor2(n) b = time.time() print "2: ", b - a a = time.time() result = factor3(n) b = time.time() print "3: ", b - a print n, "==>", result The function time imported from module time returns the time in seconds of a system clock. Exercise 12 Put one of your previous programs into a file and try the various ways of executing it described above. Exercise 13 Improve the command factor so that (1) if it takes no arguments, a message describing its usage is printed; (2) if it takes one argument, e.g. factor it prints the factorization of that number, (3) if it takes two arguments, e.g., factor , it prints a table, as before. displays the documen- Exercise 14 Devise a function doc such that doc(f) tation for f. 6 Case study: the Fermat test In section 3 we saw how trial division could be implemented in Python. The resulting code for factoring an integer runs in exponential time relative to the size in bits of the number to be factored. This code also gives an exponential time algorithm for determining whether a number is prime: 18

19 isprime = lambda n: len(factor3(n)) == 1 The function isprime returns 1 if the argument is prime, 0 if it is composite. This is because the value of an expression of the form <a> == <b> is 1 if <a> and <b> are equal, and is 0 otherwise. Such expressions are called Boolean. Here is a short test of isprime: >>> for n in range(2,8):... print n, isprime(n) We can also use isprime to make lists of primes: >>> filter( isprime, range(2,100) ) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>> len(_) 25 # there are 25 primes < 100 The value of an expression of the form filter(f, L) is a new list whose elements are the elements x of L such that f(x) == 1. You should take the above example as an illustration of Python s filter function. A much better way of creating lists of primes is to use the sieve of Eratosthenes (c c. 194 BC). We will now study a way of testing whether a number is composite that is much, much faster, so long as it is implemented efficiently. The idea is to use a theorem of Fermat: Theorem 1 If p is a prime and a is not divisible by p, then a p 1 1 mod p. Thus we could define a function ft (for Fermat test) as follows: >>> ft = lambda n: 2**(n-1) % n == 1 One could then test ft as follows: >>> filter( ft, range(2,100) ) [3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 19

20 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>> len(_) 24 # 24 numbers < 100 pass the Fermat test It seems as though the Fermat test is pretty good: the numbers that passed the test are all prime. Also, while 2 failed the test, this failure is consistent with the theorem, which applies to primes other than 2. Nonetheless, one should also be careful: the theorem is really a test for compositeness, not for primality. (Numbers which pass the Fermat test have been called industrialgrade primes. ) In light of these comments, you should do the next exercise before proceeding: Exercise 15 Are there any numbers n < 1000 which pass the Fermat test, but which are composite? Said in other words: what is the defect rate for production of industrial grade primes? Let us now turn to the question of implementing the Fermat test in an efficient way one so efficient that we can use it to demonstrate that very large numbers are composite. For example using a good implementation, we find, essentially instantaneously, that >>> ft( ) 0 The given number is therefore composite. Yet the factorization methods studied so far fail to find prime factors in a time bounded by this author s patience. For an efficient implementation, we use successive squaring with reduction modulo n to compute expressions of the form a k mod n. We illustrate the method for modulo 101. First compute the binary expansion of 100 by repeated division by two. The work is recorded in the table below: quotient remainder Thus the binary expansion of 100 is Thererefore 100 =

21 Thus Observe that = n+1 = (3 2n ) 2 Thus the powers of three that we need for the computation can be obtained by repeatedly squaring the number 3. If we reduce modulo 101 at each stage, the numbers stay under control: q r s The columns q, r, s stand for quotient, remainder, and (repeated) square, where the squares are taken mod 101. Now we multiply together the repeated squares associated with binary digit 1: The result is = modulo 101. which taken modulo 101 is 1. Python: = , >>> a = 100 >>> s = 3 >>> for k in range(0,7):... print a, a % 2, k, 2**k, s... a = a/2... s = s*s % The table, by the way, was generated using Note that no numbers larger than appear in the course of the computation. It is easy to translate the procedure just described into a Python function: def modpower(b,e,n): result = 1 s = b 21

22 q = e while q > 0: r = q % 2 if r == 1: result = result*s % n # print q, r, s, result s = s*s % n q = q/2 return result With modpower in hand, we can redesign the Fermat test: ft = lambda n: modpower(2,n-1,n) == 1 Let us analyze the running time of the Fermat test implemented using modpower. The number of multiplications and divisions required for computation of modpower(b,e,n) is at most four times the number of binary digits in e. As noted earlier, the time required to multiply two numbers is bounded by some constant times the product of the number of binary digits needed to represent the numbers. Assuming b < n, the running time T for modpower is therefore T C (binary digits of e) (binary digits of n) 2. For the Fermat test ft(n), e < n, so T C (binary digits of n) 3. Thus the running time is bounded by a polynomial in the number of bits of n. Polynomial-time algorithms scale up much better than do exponential-time algorithms. To illustrate this, consider two algorithms with running times f(b) =.001B 3 g(b) = e 1 e B/10 The table below gives running times in seconds, where 3e6 = , etc.: B f g e ,768 3e14 Algorithms f and g have the same running time for a 10-bit input. As the input is doubled to 20 and then 40 bits, algorithm g performs several times 22

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