已知a、b为实数，且满足a=√(b-3)+√(3-b)+2,求√(ab)*√(ab+1/a+b)的值。_百度知道
已知a、b为实数，且满足a=√(b-3)+√(3-b)+2,求√(ab)*√(ab+1/a+b)的值。
括号里的是根号下的部分
提问者采纳
特殊值b=3,a=2根(ab)=根6，ab=6,1/a=1/2,b=3根6*根(19/2)=根57
其他类似问题
其他2条回答
√(b-3)和√(3-b)同时有意义，所以b=3因此a=2代入上式可得√57
要使函数有意义，则{b-3≥0，,3-b≥0}→b=3，a=2，可直接算出√(ab)*√(ab+1/a+b)=√6×√6+1/2+3=√57
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出门在外也不愁Structure and Interpretation
of Computer Programs
We have now considered the elements of programming: We have used
primitive arithmetic operations, we have combined these operations, and
we have abstracted these composite operations by defining them as compound
procedures.
But that is not enough to enable us to say that we know
how to program.
Our situation is analogous to that of someone who has
learned the rules for how the pieces move in chess but knows nothing
of typical openings, tactics, or strategy.
Like the novice chess
player, we don't yet know the common patterns of usage in the domain.
We lack the knowledge of which moves are worth making (which
procedures are worth defining).
We lack the experience to predict the
consequences of making a move (executing a procedure).
The ability to visualize the consequences of the actions under
consideration is crucial to becoming an expert programmer, just as it
is in any synthetic, creative activity.
In becoming an expert
photographer, for example, one must learn how to look at a scene and
know how dark each region will appear on a print for each possible
choice of exposure and development conditions.
Only then can one
reason backward, planning framing, lighting, exposure, and development
to obtain the desired effects.
So it is with programming, where we
are planning the course of action to be taken by a process and where
we control the process by means of a program.
To become experts, we
must learn to visualize the processes generated by various types of
procedures.
Only after we have developed such a skill can we learn
to reliably construct programs that exhibit the desired behavior.
A procedure is a pattern for the local evolution of a
computational process.
It specifies how each stage of the process is
built upon the previous stage.
We would like to be able to make
statements about the overall, or global, behavior of a
process whose local evolution has been specified by a procedure.
is very difficult to do in general, but we can at least try to
describe some typical patterns of process evolution.
In this section we will examine some common ``shapes'' for processes
generated by simple procedures.
We will also investigate the
rates at which these processes consume the important computational
resources of time and space.
The procedures we will consider
are very simple.
Their role is like that played by test patterns in
photography: as oversimplified prototypical patterns, rather than
practical examples in their own right.
Figure 1.3:&&A linear recursive process for computing 6!.
We begin by considering the factorial function, defined by
There are many ways to compute factorials.
One way is to make use of
the observation that n! is equal to n times (n - 1)! for
any positive integer&n:
Thus, we can compute n! by computing (n - 1)! and multiplying the
result by n.
If we add the stipulation that 1! is equal to 1,
this observation translates directly into a procedure:
(define&(factorial&n)
&&(if&(=&n&1)
&&&&&&(*&n&(factorial&(-&n&1)))))
We can use the substitution model of
section& to watch this procedure in action
computing 6!, as shown in figure&.
Now let's take a different perspective on computing factorials.
could describe a rule for computing n! by specifying that we
first multiply 1 by 2, then multiply the result by 3, then by 4,
and so on until we reach n.
More formally, we maintain a running product, together with a counter
that counts from 1 up to n.
We can describe the computation by
saying that the counter and the product simultaneously change from one
step to the next according to the rule
and stipulating that n! is the value of the product when
the counter exceeds n.
Figure 1.4:&&A linear iterative process for computing 6!.
Once again, we can recast our description as a procedure for computing
factorials:
(define&(factorial&n)
&&(fact-iter&1&1&n))
(define&(fact-iter&product&counter&max-count)
&&(if&(&&counter&max-count)
&&&&&&product
&&&&&&(fact-iter&(*&counter&product)
&&&&&&&&&&&&&&&&&(+&counter&1)
&&&&&&&&&&&&&&&&&max-count)))
As before, we can use the substitution model to visualize the process
of computing 6!, as shown in figure&.
Compare the two processes.
From one point of view, they seem hardly
different at all.
Both compute the same mathematical function on the
same domain, and each requires a number of steps proportional to n
to compute n!.
Indeed, both processes even carry out the same
sequence of multiplications, obtaining the same sequence of partial
On the other hand, when we consider the ``shapes'' of the
two processes, we find that they evolve quite differently.
Consider the first process.
The substitution model reveals a shape of
expansion followed by contraction, indicated by the arrow in
The expansion occurs as the
process builds up a chain of deferred operations (in this case,
a chain of multiplications).
The contraction occurs as
the operations are
actually performed.
This type of process, characterized by a chain of
deferred operations, is called a recursive process.
out this process requires that the interpreter keep track of the
operations to be performed later on.
In the computation of n!,
the length of the chain of deferred multiplications, and hence the amount
of information needed to keep track of it, grows linearly with n
(is proportional to n), just like the number of steps.
Such a process is called a linear recursive process.
By contrast, the second process does not grow and shrink.
step, all we need to keep track of, for any n, are the current
values of the variables product, counter, and max-count.
We call this an iterative process.
In general, an
iterative process is one whose state can be summarized by a fixed
number of state variables, together with a fixed rule that
describes how the state variables should be updated as the process
moves from state to state and an (optional) end test that specifies
conditions under which the process should terminate.
In computing
n!, the number of steps required grows linearly with n.
Such a process is
called a linear iterative process.
The contrast between the two processes can be seen in another way.
the iterative case, the program variables provide a complete
description of the state of the process at any point.
If we stopped
the computation between steps, all we would need to do to resume the
computation is to supply the interpreter with the values of the three
program variables.
Not so with the recursive process.
In this case
there is some additional ``hidden'' information, maintained by the
interpreter and not contained in the program variables, which
indicates ``where the process is'' in negotiating the chain of
deferred operations.
The longer the chain, the more information must
be maintained.
In contrasting iteration and recursion, we must be careful not to
confuse the notion of a recursive process with the notion of a
recursive procedure.
When we describe a procedure as recursive,
we are referring to the syntactic fact that the procedure definition
refers (either directly or indirectly) to the procedure itself.
when we describe a process as following a pattern that is, say,
linearly recursive, we are speaking about how the process evolves, not
about the syntax of how a procedure is written.
It may seem
disturbing that we refer to a recursive procedure such as fact-iter as generating an iterative process.
However, the process
really is iterative: Its state is captured completely by its three
state variables, and an interpreter need keep track of only three
variables in order to execute the process.
One reason that the distinction between process and procedure may be
confusing is that most implementations of common languages (including
Ada, Pascal, and C) are designed in such a way that the
interpretation of any recursive procedure consumes an amount of memory
that grows with the number of procedure calls, even when the process
described is, in principle, iterative.
As a consequence, these
languages can describe iterative processes only by resorting to
special-purpose ``looping constructs'' such as do, repeat,
until, for, and while.
The implementation of Scheme
we shall consider in chapter&5 does not share this defect.
execute an iterative process in constant space, even if the iterative
process is described by a recursive procedure.
An implementation with
this property is called tail-recursive.
With a tail-recursive
implementation, iteration can be expressed using the ordinary
procedure call mechanism, so that special iteration constructs are
useful only as syntactic sugar.
Exercise 1.9.&&Each of the following two procedures defines a method for adding two
positive integers in terms of the procedures inc,
which increments its argument by 1, and dec, which decrements
its argument by 1.
(define&(+&a&b)
&&(if&(=&a&0)
&&&&&&(inc&(+&(dec&a)&b))))
(define&(+&a&b)
&&(if&(=&a&0)
&&&&&&(+&(dec&a)&(inc&b))))
Using the substitution model, illustrate the process generated by each
procedure in evaluating (+ 4 5).
Are these processes
iterative or recursive?
Exercise 1.10.&&The following procedure computes a mathematical function called
Ackermann's function.
(define&(A&x&y)
&&(cond&((=&y&0)&0)
&&&&&&&&((=&x&0)&(*&2&y))
&&&&&&&&((=&y&1)&2)
&&&&&&&&(else&(A&(-&x&1)
&&&&&&&&&&&&&&&&&(A&x&(-&y&1))))))
What are the values of the following expressions?
Consider the following procedures, where A is the procedure
defined above:
(define&(f&n)&(A&0&n))
(define&(g&n)&(A&1&n))
(define&(h&n)&(A&2&n))
(define&(k&n)&(*&5&n&n))
Give concise mathematical definitions for the functions computed by
the procedures f, g, and h for positive integer
values of n.
For example, (k n) computes 5n2.
Another common pattern of computation is called tree recursion.
As an example, consider computing the sequence of Fibonacci numbers,
in which each number is the sum of the preceding two:
In general, the Fibonacci numbers can be defined by the rule
We can immediately translate this definition into a recursive
procedure for computing Fibonacci numbers:
(define&(fib&n)
&&(cond&((=&n&0)&0)
&&&&&&&&((=&n&1)&1)
&&&&&&&&(else&(+&(fib&(-&n&1))
&&&&&&&&&&&&&&&&&(fib&(-&n&2))))))
Figure 1.5:&&The tree-recursive process generated in computing (fib 5).
Consider the pattern of this computation.
To compute (fib 5),
we compute (fib 4) and (fib 3).
To compute (fib 4),
we compute (fib 3) and (fib 2).
In general, the evolved
process looks like a tree, as shown in figure&.
Notice that the branches split into two at each level (except at the
bottom); this reflects the fact that the fib procedure calls
itself twice each time it is invoked.
This procedure is instructive as a prototypical tree recursion, but it
is a terrible way to compute Fibonacci numbers because it does so much
redundant computation.
Notice in figure& that
the entire computation of (fib 3) -- almost half the work -- is
duplicated.
In fact, it is not hard to show that the number of times
the procedure will compute (fib 1) or (fib 0) (the number
of leaves in the above tree, in general) is precisely
Fib(n + 1).
To get an idea of how bad this is, one can show that the
Fib(n) grows exponentially with n.
More precisely
(see exercise&),
Fib(n) is the closest
integer to n /5, where
is the golden ratio, which satisfies the equation
Thus, the process uses a number of steps that grows exponentially
with the input.
On the other hand, the space required grows only
linearly with the input, because we need keep track only of which
nodes are above us in the tree at any point in the computation.
general, the number of steps required by a tree-recursive process will be
proportional to the number of nodes in the tree, while the space
required will be proportional to the maximum depth of the tree.
We can also formulate an iterative process for computing the
Fibonacci numbers.
The idea is to use a pair of integers a and
b, initialized to
Fib(1) = 1 and
Fib(0) = 0,
and to repeatedly apply the simultaneous
transformations
It is not hard to show that, after applying this transformation n
times, a and b will be equal, respectively, to
Fib(n + 1) and
Thus, we can compute Fibonacci numbers iteratively using
the procedure
(define&(fib&n)
&&(fib-iter&1&0&n))
(define&(fib-iter&a&b&count)
&&(if&(=&count&0)
&&&&&&(fib-iter&(+&a&b)&a&(-&count&1))))
This second method for computing
Fib(n) is a linear iteration.
difference in number of steps required by the two methods -- one linear in n,
one growing as fast as
Fib(n) itself -- is enormous, even for
small inputs.
One should not conclude from this that tree-recursive processes are
When we consider processes that operate on hierarchically
structured data rather than numbers, we will find that tree recursion
is a natural and powerful tool. But even in numerical operations,
tree-recursive processes can be useful in helping us to understand and
design programs.
For instance, although the first fib procedure
is much less efficient than the second one, it is more
straightforward, being little more than a translation into Lisp of the
definition of the Fibonacci sequence.
To formulate the iterative
algorithm required noticing that the computation could be recast as an
iteration with three state variables.
It takes only a bit of cleverness to come up with the iterative
Fibonacci algorithm.
In contrast, consider the
following problem: How many different ways can we make change of $ 1.00,
given half-dollars, quarters, dimes, nickels, and pennies?
generally, can we write a procedure to compute the number of ways to
change any given amount of money?
This problem has a simple solution as a recursive procedure.
we think of the types of coins available as arranged in some order.
Then the following relation holds:
The number of ways to change amount a using n kinds of coins equals
the number of ways to change amount a using all but the first
kind of coin, plus
the number of ways to change amount a - d using all n kinds of
coins, where d is the denomination of the first kind of coin.
To see why this is true, observe that the ways to make change can be
divided into two groups: those that do not use any of the first kind
of coin, and those that do.
Therefore, the total number of ways to
make change for some amount is equal to the number of ways to make
change for the amount without using any of the first kind of coin,
plus the number of ways to make change assuming that we do use the
first kind of coin.
But the latter number is equal to the number of
ways to make change for the amount that remains after using a coin of
the first kind.
Thus, we can recursively reduce the problem of changing a given amount
to the problem of changing smaller amounts using fewer kinds of coins.
Consider this reduction rule carefully, and convince yourself that we
can use it to describe an algorithm if we specify the following
degenerate cases:
If a is exactly 0, we should count that as 1 way to make change.
If a is less than 0, we should count that as 0 ways to make change.
If n is 0, we should count that as 0 ways to make change.
We can easily translate this description into a recursive
procedure:
(define&(count-change&amount)
&&(cc&amount&5))
(define&(cc&amount&kinds-of-coins)
&&(cond&((=&amount&0)&1)
&&&&&&&&((or&(&&amount&0)&(=&kinds-of-coins&0))&0)
&&&&&&&&(else&(+&(cc&amount
&&&&&&&&&&&&&&&&&&&&&(-&kinds-of-coins&1))
&&&&&&&&&&&&&&&&&(cc&(-&amount
&&&&&&&&&&&&&&&&&&&&&&&&(first-denomination&kinds-of-coins))
&&&&&&&&&&&&&&&&&&&&&kinds-of-coins)))))
(define&(first-denomination&kinds-of-coins)
&&(cond&((=&kinds-of-coins&1)&1)
&&&&&&&&((=&kinds-of-coins&2)&5)
&&&&&&&&((=&kinds-of-coins&3)&10)
&&&&&&&&((=&kinds-of-coins&4)&25)
&&&&&&&&((=&kinds-of-coins&5)&50)))
(The first-denomination procedure takes as input the number of
kinds of coins available and returns the denomination of the first
Here we are thinking of the coins as arranged in order from
largest to smallest, but any order would do as well.)
We can now
answer our original question about changing a dollar:
(count-change&100)
Count-change generates a tree-recursive process with
redundancies similar to those in our first implementation of fib.
(It will take quite a while for that 292 to be computed.)
the other hand, it is not obvious how to design a better algorithm
for computing the result, and we leave this problem as a challenge.
The observation that a tree-recursive process may be highly
inefficient but often easy to specify and understand has led people to
propose that one could get the best of both worlds by designing a
``smart compiler'' that could transform tree-recursive procedures into
more efficient procedures that compute the same result.
Exercise 1.11.&&A function f is defined by the rule that f(n) = n if n&3 and
f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) if n& 3.
Write a procedure that
computes f by means of a recursive process.
Write a procedure that
computes f by means of an iterative process.
Exercise 1.12.&&The following pattern of numbers is called
Pascal's triangle.
The numbers at the edge of the triangle are all 1, and
each number inside the triangle is the sum of the two numbers above it.
Write a procedure that computes elements of Pascal's triangle by means
of a recursive process.
Exercise 1.13.&&Prove that
Fib(n) is the closest integer to n/5,
(1 + 5)/2.
(1 - 5)/2.
induction and the definition of the Fibonacci numbers (see
section&) to prove that
Fib(n) = (n
The previous examples illustrate that processes can differ
considerably in the rates at which they consume computational
resources.
One convenient way to describe this difference is to use
the notion of order of growth to obtain a gross measure of the
resources required by a process as the inputs become larger.
Let n be a parameter that measures the size of the problem, and let
R(n) be the amount of resources the process requires for a problem
of size n.
In our previous examples we took n to be the number
for which a given function is to be computed, but there are other
possibilities.
For instance, if our goal is to compute an
approximation to the square root of a number, we might take n to be
the number of digits accuracy required.
For matrix multiplication we
might take n to be the number of rows in the matrices.
In general
there are a number of properties of the problem with respect to which
it will be desirable to analyze a given process.
Similarly, R(n)
might measure the number of internal storage registers used, the
number of elementary machine operations performed, and so on.
computers that do only a fixed number of operations at a time, the
time required will be proportional to the number of elementary machine
operations performed.
We say that R(n) has order of growth (f(n)), written
R(n) = (f(n)) (pronounced ``theta of f(n)''), if there are
positive constants k1 and k2 independent of n such that
for any sufficiently large value of n.
words, for large n, the value R(n) is sandwiched between k1f(n)
and k2f(n).)
For instance, with the linear recursive process for computing
factorial described in section& the
number of steps grows proportionally to the input n.
steps required for this process grows as (n).
We also saw
that the space required grows as (n).
For the iterative
factorial, the number of steps is still (n) but the space is
(1) -- that is, constant. The tree-recursive Fibonacci computation requires
(n) steps and space (n), where
golden ratio described in section&.
Orders of growth provide only a crude description of the behavior of a
For example, a process requiring n2 steps and a process
requiring 1000n2 steps and a process requiring 3n2 + 10n + 17 steps
all have (n2) order of growth.
On the other hand, order of
growth provides a useful indication of how we may expect the behavior
of the process to change as we change the size of the problem.
(n) (linear) process, doubling the size will roughly double the amount
of resources used.
For an exponential process, each increment in
problem size will multiply the resource utilization by a constant
In the remainder of section&
we will examine two
algorithms whose order of growth is logarithmic, so that doubling the
problem size increases the resource requirement by a constant amount.
Exercise 1.14.&&Draw the tree illustrating the process generated by the count-change procedure of section& in making
change for 11 cents.
What are the orders of growth of the space and
number of steps used by this process as the amount to be changed
increases?
Exercise 1.15.&&The sine of an angle (specified in
radians) can be computed by making use of the approximation
sufficiently small, and the trigonometric identity
to reduce the size of the argument of sin.
purposes of this exercise an angle is considered ``sufficiently
small'' if its magnitude is not greater than 0.1 radians.) These
ideas are incorporated in the following procedures:
(define&(cube&x)&(*&x&x&x))
(define&(p&x)&(-&(*&3&x)&(*&4&(cube&x))))
(define&(sine&angle)
&&&(if&(not&(&&(abs&angle)&0.1))
&&&&&&&angle
&&&&&&&(p&(sine&(/&angle&3.0)))))
a.&&How many times is the procedure p
applied when (sine 12.15) is evaluated?
b.&&What is the order of growth in space and number of steps (as a
function of&a) used by the process generated by the sine
procedure when (sine a) is evaluated?
Consider the problem of computing the exponential of a given number.
We would like a procedure that takes as arguments a base b and a
positive integer exponent n and computes bn.
One way to do this
is via the recursive definition
which translates readily into the procedure
(define&(expt&b&n)
&&(if&(=&n&0)
&&&&&&(*&b&(expt&b&(-&n&1)))))
This is a linear recursive process, which requires (n) steps
and (n) space.
Just as with factorial, we can readily
formulate an equivalent linear iteration:
(define&(expt&b&n)
&&(expt-iter&b&n&1))
(define&(expt-iter&b&counter&product)
&&(if&(=&counter&0)
&&&&&&product
&&&&&&(expt-iter&b
&&&&&&&&&&&&&&&&(-&counter&1)
&&&&&&&&&&&&&&&&(*&b&product))))&
This version requires (n) steps and (1) space.
We can compute exponentials in fewer steps by using successive
For instance, rather than computing b8 as
we can compute it using three multiplications:
This method works fine for exponents that are powers of 2.
also take advantage of successive squaring in computing exponentials
in general if we use the rule
We can express this method as a procedure:
(define&(fast-expt&b&n)
&&(cond&((=&n&0)&1)
&&&&&&&&((even?&n)&(square&(fast-expt&b&(/&n&2))))
&&&&&&&&(else&(*&b&(fast-expt&b&(-&n&1))))))
where the predicate to test whether an integer is even is defined in terms of
the primitive procedure remainder by
(define&(even?&n)
&&(=&(remainder&n&2)&0))
The process evolved by fast-expt grows logarithmically with n
in both space and number of steps.
To see this, observe that
computing b2n using fast-expt requires only one more
multiplication than computing bn.
The size of the exponent we can
compute therefore doubles (approximately) with every new
multiplication we are allowed.
Thus, the number of multiplications
required for an exponent of n grows about as fast as the logarithm
of n to the base 2.
The process has (log n)
The difference between (log n) growth and (n) growth
becomes striking as n becomes large.
For example, fast-expt
for n = 1000 requires only 14 multiplications. It is also possible to use the idea of
successive squaring to devise an iterative algorithm that computes
exponentials with a logarithmic number of steps
(see exercise&), although, as is often
the case with iterative algorithms, this is not written down so
straightforwardly as the recursive algorithm.
Exercise 1.16.&&Design a procedure that evolves an iterative exponentiation process
that uses successive squaring and uses a logarithmic number of steps,
as does fast-expt.
(Hint: Using the observation that
= (b2)n/2, keep, along with the exponent n and the
base b, an additional state variable a, and define the state
transformation in such a way that the product a bn is unchanged
from state to state.
At the beginning of the process a is taken to
be 1, and the answer is given by the value of a at the end of the
In general, the technique of defining an invariant
quantity that remains unchanged from state to state is a powerful way
to think about the design of iterative algorithms.)
Exercise 1.17.&&The exponentiation algorithms in this section are based on performing
exponentiation by means of repeated multiplication.
In a similar way,
one can perform integer multiplication by means of repeated addition.
The following multiplication procedure (in which it is assumed that
our language can only add, not multiply) is analogous to the expt procedure:
(define&(*&a&b)
&&(if&(=&b&0)
&&&&&&(+&a&(*&a&(-&b&1)))))
This algorithm takes a number of steps that is linear in b.
Now suppose we include, together with addition, operations double,
which doubles an integer, and halve, which divides an (even)
integer by 2.
Using these, design a multiplication procedure analogous
to fast-expt that uses a logarithmic number of steps.
Exercise 1.18.&&Using the results of exercises&
and&, devise a procedure that generates an iterative
process for multiplying two integers in terms of adding, doubling, and
halving and uses a logarithmic number of steps.
Exercise 1.19.&&
There is a clever algorithm for computing the Fibonacci numbers in
a logarithmic number of steps.
Recall the transformation of the state variables
a and b in the fib-iter process of
section&: a
a + b and b
Call this transformation T, and observe that applying T over
and over again n times, starting with 1 and 0, produces the pair
Fib(n + 1) and
In other words, the Fibonacci
numbers are produced by applying Tn, the nth power of the
transformation T, starting with the pair (1,0).
Now consider T
to be the special case of p = 0 and q = 1 in a family of
transformations Tpq, where Tpq transforms the pair (a,b)
according to a
bq + aq + ap and b
that if we apply such a transformation Tpq twice, the effect is
the same as using a single transformation Tp'q' of the same form,
and compute p' and q' in terms of p and&q.
This gives us an
explicit way to square these transformations, and thus we can compute
Tn using successive squaring, as in the fast-expt
procedure.
Put this all together to complete the following procedure,
which runs in a logarithmic number of steps:
(define&(fib&n)
&&(fib-iter&1&0&0&1&n))
(define&(fib-iter&a&b&p&q&count)
&&(cond&((=&count&0)&b)
&&&&&&&&((even?&count)
&&&&&&&&&(fib-iter&a
&&&&&&&&&&&&&&&&&&&b
&&&&&&&&&&&&&&&&&&&&??&&&&&&&;&compute&p'
&&&&&&&&&&&&&&&&&&&&??&&&&&&&;&compute&q'
&&&&&&&&&&&&&&&&&&&(/&count&2)))
&&&&&&&&(else&(fib-iter&(+&(*&b&q)&(*&a&q)&(*&a&p))
&&&&&&&&&&&&&&&&&&&&&&&&(+&(*&b&p)&(*&a&q))
&&&&&&&&&&&&&&&&&&&&&&&&p
&&&&&&&&&&&&&&&&&&&&&&&&q
&&&&&&&&&&&&&&&&&&&&&&&&(-&count&1)))))
The greatest common divisor (GCD) of two integers a and b is
defined to be the largest integer that divides both a and
b with no remainder.
For example, the GCD of 16 and 28 is 4.
In chapter&2,
when we investigate how to implement rational-number arithmetic, we
will need to be able to compute GCDs in order to reduce
rational numbers to lowest terms.
(To reduce a rational number to
lowest terms, we must divide both the numerator and the denominator by their
For example, 16/28 reduces to 4/7.)
One way to find the
GCD of two integers is to factor them and search for common
factors, but there is a famous algorithm that is much more efficient.
The idea of the algorithm is based on the observation that, if r is
the remainder when a is divided by b, then the common divisors of
a and b are precisely the same as the common divisors of b and
Thus, we can use the equation
to successively reduce the problem of computing a GCD to the
problem of computing the GCD of smaller and smaller pairs of
For example,
reduces GCD(206,40) to GCD(2,0), which is 2.
possible to show that starting with any two positive integers and
performing repeated reductions will always eventually produce a pair
where the second number is 0.
Then the GCD is the other
number in the pair.
This method for computing the GCD is
known as Euclid's Algorithm.
It is easy to express Euclid's Algorithm as a procedure:
(define&(gcd&a&b)
&&(if&(=&b&0)
&&&&&&(gcd&b&(remainder&a&b))))
This generates an iterative process, whose number of steps grows as
the logarithm of the numbers involved.
The fact that the number of steps required by Euclid's Algorithm has
logarithmic growth bears an interesting relation to the Fibonacci
Lam&'s Theorem: If Euclid's Algorithm requires k steps to
compute the GCD of some pair, then the smaller number in the pair
must be greater than or equal to the kth Fibonacci
We can use this theorem to get an order-of-growth estimate for Euclid's
Algorithm.
Let n be the smaller of the two inputs to the
procedure.
If the process takes k steps, then we must have
the number of steps k grows as the logarithm (to the base
Hence, the order of growth is (log n).
Exercise 1.20.&&The process that a procedure generates is of course dependent on the
rules used by the interpreter.
As an example, consider the iterative
gcd procedure given above.
Suppose we were to interpret this procedure using normal-order
evaluation, as discussed in section&.
(The normal-order-evaluation rule for if is described in
exercise&.)
substitution method (for normal order), illustrate the process
generated in evaluating (gcd 206 40) and indicate the
remainder operations that are actually performed.
How many remainder operations are actually performed
in the normal-order evaluation of (gcd 206&40)?
In the applicative-order evaluation?
This section describes two methods for checking the primality of an
integer n, one with order of growth (n), and a
``probabilistic'' algorithm with order of growth (log n).
exercises at the end of this section suggest programming
projects based on these algorithms.
Since ancient times, mathematicians have been fascinated by problems
concerning prime numbers, and many people have worked on the problem
of determining ways to test if numbers are prime.
to test if a number is prime is to find the number's divisors.
following program finds the smallest integral divisor (greater than 1)
of a given number&n.
It does this in a straightforward way, by
testing n for divisibility by successive integers starting with 2.
(define&(smallest-divisor&n)
&&(find-divisor&n&2))
(define&(find-divisor&n&test-divisor)
&&(cond&((&&(square&test-divisor)&n)&n)
&&&&&&&&((divides?&test-divisor&n)&test-divisor)
&&&&&&&&(else&(find-divisor&n&(+&test-divisor&1)))))
(define&(divides?&a&b)
&&(=&(remainder&b&a)&0))
We can test whether a number is prime as follows: n is prime if
and only if n is its own smallest divisor.
(define&(prime?&n)
&&(=&n&(smallest-divisor&n)))
The end test for find-divisor is based on the fact that if n
is not prime it must have a divisor less than or equal to
means that the algorithm need only test divisors between 1 and
Consequently, the number of steps required to identify
n as prime will have order of growth (n).
The (log n) primality test is based on a result from number
theory known as Fermat's Little Theorem.
Fermat's Little Theorem: If n is a prime number and
a is any positive integer less than n, then a raised to the
nth power is congruent to a modulo n.
(Two numbers are said to be congruent modulo n if
they both have the same remainder when divided by n.
remainder of a number a when divided by n is also referred to as
the remainder of a modulo n, or simply as a
modulo n.)
If n is not prime, then, in general, most of the numbers a& n will not
satisfy the above relation.
This leads to the following algorithm for
testing primality: Given a number n, pick a random number a & n and
compute the remainder of an modulo n.
If the result is not equal to
a, then n is certainly not prime.
If it is a, then chances are good
that n is prime.
Now pick another random number a and test it with the
same method.
If it also satisfies the equation, then we can be even more
confident that n is prime.
By trying more and more values of a, we can
increase our confidence in the result.
This algorithm is known as the
Fermat test.
To implement the Fermat test, we need a procedure that computes the
exponential of a number modulo another number:
(define&(expmod&base&exp&m)
&&(cond&((=&exp&0)&1)
&&&&&&&&((even?&exp)
&&&&&&&&&(remainder&(square&(expmod&base&(/&exp&2)&m))
&&&&&&&&&&&&&&&&&&&&m))
&&&&&&&&(else
&&&&&&&&&(remainder&(*&base&(expmod&base&(-&exp&1)&m))
&&&&&&&&&&&&&&&&&&&&m))))&&&&&&&&
This is very similar to the fast-expt procedure of
It uses successive squaring, so
that the number of steps grows logarithmically with the
The Fermat test is performed by choosing at random a number a
between 1 and n - 1 inclusive and checking whether the remainder
modulo n of the nth power of a is equal to a.
The random
number a is chosen using the procedure random, which we assume is
included as a primitive in Scheme. Random returns a
nonnegative integer less than its integer input.
Hence, to obtain a random
number between 1 and n - 1, we call random with an input of
n - 1 and add 1 to the result:
(define&(fermat-test&n)
&&(define&(try-it&a)
&&&&(=&(expmod&a&n&n)&a))
&&(try-it&(+&1&(random&(-&n&1)))))
The following procedure runs the test a given number of times, as
specified by a parameter.
Its value is true if the test succeeds
every time, and false otherwise.
(define&(fast-prime?&n&times)
&&(cond&((=&times&0)&true)
&&&&&&&&((fermat-test&n)&(fast-prime?&n&(-&times&1)))
&&&&&&&&(else&false)))
The Fermat test differs in character from most familiar algorithms, in
which one computes an answer that is guaranteed to be correct.
the answer obtained is only probably correct.
More precisely, if n
ever fails the Fermat test, we can be certain that n is not prime.
But the fact that n passes the test, while an extremely strong
indication, is still not a guarantee that n is prime.
What we would
like to say is that for any number n, if we perform the test enough
times and find that n always passes the test, then the probability
of error in our primality test can be made as small as we like.
Unfortunately, this assertion is not quite correct.
There do exist
numbers that fool the Fermat test: numbers n that are not prime and
yet have the property that an is congruent to a modulo n for
all integers a & n.
Such numbers are extremely rare, so the Fermat
test is quite reliable in practice.
There are variations of the Fermat test that cannot be fooled.
these tests, as with the Fermat method, one tests the primality of an
integer n by choosing a random integer a&n and checking some
condition that depends upon n and a.
exercise& for an example of such a test.)
other hand, in contrast to the Fermat test, one can prove that, for
any n, the condition does not hold for most of the integers a&n
unless n is prime.
Thus, if n passes the test for some random
choice of&a, the chances are better than even that n is prime.
n passes the test for two random choices of a, the chances are better
than 3 out of 4 that n is prime. By running the test with more and
more randomly chosen values of a we can make the probability of
error as small as we like.
The existence of tests for which one can prove that the chance of
error becomes arbitrarily small has sparked interest in algorithms of
this type, which have come to be known as probabilistic
algorithms.
There is a great deal of research activity in this area,
and probabilistic algorithms have been fruitfully applied to many
Exercise 1.21.&&Use the smallest-divisor procedure to find the smallest divisor
of each of the following numbers: 199, .
Exercise 1.22.&&Most Lisp implementations include a primitive called runtime
that returns an integer that specifies the amount of time the system
has been running (measured, for example, in microseconds).
following timed-prime-test procedure, when called with an
integer n, prints n and checks to see if n is prime.
prime, the procedure prints three asterisks followed by the amount of time
used in performing the test.
(define&(timed-prime-test&n)
&&(newline)
&&(display&n)
&&(start-prime-test&n&(runtime)))
(define&(start-prime-test&n&start-time)
&&(if&(prime?&n)
&&&&&&(report-prime&(-&(runtime)&start-time))))
(define&(report-prime&elapsed-time)
&&(display&&&***&&)
&&(display&elapsed-time))
Using this procedure, write a procedure search-for-primes that
checks the primality of consecutive odd integers in a specified range.
Use your procedure to find the three smallest primes larger than 1000;
larger than 10,000; larger than 100,000; larger than 1,000,000.
the time needed to test each prime.
Since the testing algorithm has
order of growth of (n), you should expect that testing
for primes around 10,000 should take about 10 times as long
as testing for primes around 1000.
Do your timing data bear this out?
How well do the data for 100,000 and 1,000,000 support the n
prediction?
Is your result compatible with the notion that programs
on your machine run in time proportional to the number of steps
required for the computation?
Exercise 1.23.&&The smallest-divisor procedure shown at the start of this section
does lots of needless testing: After it checks to see if the
number is divisible by 2 there is no point in checking to see if
it is divisible by any larger even numbers.
This suggests that the
values used for test-divisor should not be 2, 3, 4, 5, 6,
..., but rather 2, 3, 5, 7, 9, ....
To implement this
change, define a procedure next that returns 3 if its input is
equal to 2 and otherwise returns its input plus&2.
Modify the smallest-divisor procedure to use (next test-divisor) instead
of (+ test-divisor 1).
With timed-prime-test
incorporating this modified version of smallest-divisor, run the
test for each of the 12 primes found in
exercise&.
Since this modification halves the
number of test steps, you should expect it to run about twice as fast.
Is this expectation confirmed?
If not, what is the observed ratio of
the speeds of the two algorithms, and how do you explain the fact that
it is different from 2?
Exercise 1.24.&&Modify the timed-prime-test procedure of
exercise& to use fast-prime? (the
Fermat method), and test each of the 12 primes you found in that
Since the Fermat test has (log n) growth, how
would you expect the time to test primes near 1,000,000 to compare
with the time needed to test primes near 1000?
Do your data bear this
Can you explain any discrepancy you find?
Exercise 1.25.&&Alyssa P. Hacker complains that we went to a lot of extra work in
writing expmod.
After all, she says, since we already know how
to compute exponentials, we could have simply written
(define&(expmod&base&exp&m)
&&(remainder&(fast-expt&base&exp)&m))
Is she correct?
Would this procedure serve as well for our fast prime
Exercise 1.26.&&Louis Reasoner is having great difficulty doing
exercise&.
His fast-prime? test
seems to run more slowly than his prime? test.
Louis calls his
friend Eva Lu Ator over to help.
When they examine Louis's code, they
find that he has rewritten the expmod procedure to use an
explicit multiplication, rather than calling square:
(define&(expmod&base&exp&m)
&&(cond&((=&exp&0)&1)
&&&&&&&&((even?&exp)
&&&&&&&&&(remainder&(*&(expmod&base&(/&exp&2)&m)
&&&&&&&&&&&&&&&&&&&&&&&(expmod&base&(/&exp&2)&m))
&&&&&&&&&&&&&&&&&&&&m))
&&&&&&&&(else
&&&&&&&&&(remainder&(*&base&(expmod&base&(-&exp&1)&m))
&&&&&&&&&&&&&&&&&&&&m))))
``I don't see what difference that could make,'' says Louis.
``By writing the procedure like that, you have
transformed the (log n) process into a (n) process.''
Exercise 1.27.&&Demonstrate that the Carmichael numbers listed in
footnote& really do fool
the Fermat test.
That is, write a procedure that takes an integer n
and tests whether an is congruent to a modulo n for every
a&n, and try your procedure on the given Carmichael numbers.
Exercise 1.28.&&One variant of the Fermat test that cannot be fooled is called the
Miller-Rabin test (Miller 1976; Rabin 1980).
This starts from
an alternate form of Fermat's Little Theorem, which states that if n
is a prime number and a is any positive integer less than n, then
a raised to the (n - 1)st power is congruent to 1 modulo&n.
the primality of a number n by the Miller-Rabin test, we pick a
random number a&n and raise a to the (n - 1)st power modulo&n
using the expmod procedure.
However, whenever we perform the
squaring step in expmod, we check to see if we have discovered a
``nontrivial square root of 1 modulo&n,'' that is, a number not
equal to 1 or n - 1 whose square is equal to 1 modulo&n.
possible to prove that if such a nontrivial square root of 1 exists,
then n is not prime.
It is also possible to prove that if n is an
odd number that is not prime, then, for at least half the numbers
a&n, computing an-1 in this way will reveal a nontrivial
square root of 1 modulo&n.
(This is why the Miller-Rabin test
cannot be fooled.)
Modify the expmod procedure to signal if it
discovers a nontrivial square root of 1, and use this to implement
the Miller-Rabin test with a procedure analogous to fermat-test.
Check your procedure by testing various known primes and non-primes.
Hint: One convenient way to make expmod signal is to have it