6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
6.001 Notes,Section 4.1
Slide 4.1.1
In this lecture,we are going to take a careful look at the kinds
of procedures we can build,We will first go back to look very
carefully at the substitution model,to see how the rules for
evaluation help us determine the evolution of a process
controlled by a procedure,We will then see how different kinds
of process evolution can occur,with different demands in terms
of computational resources,And we will see how to
characterize those differences formally in terms of orders of
growth of a process,Finally,we will explore examples of
procedures from different classes of growth,helping you to
begin to relate how choices in procedure design may affect
performance of the actual use of the procedure,
Slide 4.1.2
Now that we have seen two different implementations of the
same idea,with different behaviors,we need a way of more
formally characterizing why that difference occurs,To do that,
we are going to go back to our substitution model,
Now we are going to think of our model as a set of rewrite
rules,that is,as a set of rules that formally specify,given an
expression of a particular form,rewrite it in the following form,
Slide 4.1.3
So elementary expressions in this viewpoint are just "left
alone",that is,numbers,names of built-in procedures,and
lambda expressions are left as is,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.1.4
If our expression is a name that we created ourselves,we will
rewrite in place of it the value that was associated with that
name during the definition,
Slide 4.1.5
For the special form if we use the rule we just saw,We
evaluate the predicate clause,and based on its value,we either
rewrite the if expression by its consequent or its alternative,
Slide 4.1.6
And finally,combinations,We use the rule that we first
evaluate the operator expression to get the procedure,and we
evaluate the operands to get the set of arguments,If we have a
primitive procedure,we are just going to,do the right thing”,
Otherwise we are going to replace this entire expression with
the body of the compound expression,with the arguments
substituted for their associated parameters,
Slide 4.1.7
Given that model,we can more formally capture the evolution
of a process,We can talk about the order of growth of a process
as these rewrite rules evolve,This measures how much of a
particular resource a process is going to take as these rules
evolve,where typically we measure either space or time as the
resource,
More formally,let n be a parameter that measures the size of
the problem of interest,In the case of factorial,this would be
the size of the input argument,We let R(n) denote the amount
of resources we will need to solve a problem of this size,where
as noted,the resources are usually space and time,We are
interested in characterizing R(n) for large values of N,that is,in
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
the asymptotic limit,We say that R(n) has order of growth,Theta of f of n if we can find a function f(n) that is
roughly of the same order as the needed resource,where the more formal definition is shown in the
Slide 4.1.8
In more common terms,this says that when measuring the
amount of space we need,we want to find a function that
measures the number of deferred operations,as a function of the
size of the problem,up to a constant factor,For measuring the
amount of time,we want to find a function that measures the
number of basic or primitive steps that the rewrite rules go
through,again as a function of the size of the problem,
Slide 4.1.9
So let's use our rewrite rule idea,together with this notion of
measuring the amount of resource we need as an order of
growth in the size of the problem,Here is our recursive factorial
procedure,.,
Slide 4.1.10
..and here is a partial trace of fact using those rewrite rules,
We start with (fact 4),That reduces to evaluating an if
expression,Since the predicate is not true,this reduces to
evaluating the alternative statement,which is a combination of
a multiplication and another evaluation of fact,
Notice how the orange colored lines capture the evolution of the
rewrite rules,Note how (fact 4) reduces to a deferred operation
and an evaluation of (fact 3) and this further reduces to another
deferred operation and a subsequent call to fact,Notice the
shape of this process,it grows out with a set of deferred
operations until it reaches a base case,and then it contracts back
in,completing each deferred operation,Using this,we can see that the amount of space grows linearly with the
size of the argument,That is,with each increase in the argument size,I add a constant amount of space
requirement,In terms of the number of operations,we can see that we basically need twice as many steps as the
size of the problem,one to expand out,and one to reduce back down,This is also a linear growth,since the
constant 2 does not change with problem size,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.1.11
And let's compare that our iterative factorial,Here is our code
again,and here is a partial trace of the rewrite evolution of that
process,
Here we see the different shape of the process,and our order of
growth analysis captures that difference,In particular,there are
no deferred operations,and you can see that the maximum
amount of space I need at any stage,that is,the maximum
amount of space used by any rewrite,is independent of the size
of the problem,We say that it is constant,as it does not grow
with n,In terms of time,there is basically one operation for
each increment in the argument,so this is again a linear order
of growth,
Slide 4.1.12
So we can formally capture this,as shown,Fact has linear
growth in space,written as shown,because the maximum
amount of space needed by any rewrite stage is a linear multiple
of the size of the argument,It also has linear growth in time,
because as we saw it takes a number of basic steps that is also a
linear multiple of the size of the argument,
Slide 4.1.13
On the other hand,iterative-fact has no deferred
operations,and is constant in space,written as shown,while as
we argued,the time order of growth is linear,Notice that the
fact that the recursive version took 2n steps and the iterative
version took n steps doesn't matter in terms of order of growth,
Both are said to be linear,
Thus we see that we can formally characterize the difference
between these two processes,which have different shapes of
evolution,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.1.14
So why have we done this? The primary goal is to allow you to
start to recognize different kinds of behaviors,and the
associated procedure forms that give rise to those behaviors,
This will start to allow you to work backwards,in designing a
procedure,by enabling you to visualize the consequences of
performing a computation in different ways,
6.001 Notes,Section 4.2
Slide 4.2.1
Having seen two different kinds of processes,one linear,and
one constant,we want to fill out our repertoire of processes,by
looking at other kinds of processes,The next one is an example
of an exponential process,and deals with a classic function
called Fibonacci,Its definition is that if its argument is 0,its
value is 0,if its argument is 1,its value is 1,and for all other
positive integer arguments,its value is the sum of its values for
the two preceding arguments,
We would like to write a procedure to compute Fibonacci,and
in particular see how it gives rise to a different kind of
behavior,
Slide 4.2.2
To solve this problem,let's use our tool of wishful thinking,
Here,that wishful thinking says,let's assume that given an
argument n,we know how to solve Fibonacci for any smaller
sized argument,Using that idea,we can then work out a
solution to the problem,With this in hand,it is clear that the
solution to the general problem is just to solve two smaller
sized problems,then just add the results together,Note that in
this case we are using wishful thinking twice,not once,as in
our previous examples,
Here is a procedure that captures this idea,
First,we introduce a new expression,called a cond
expression,Cond uses the following rules of evaluation,The cond consists of a set of clauses,each of which
has within it,a predicate clause,and one or more subsequent expressions,Cond proceeds by first evaluating the
predicate of the first clause,in this case (= n 0),If it is true,then we evaluate in turn each of the other
expressions in this clause,returning the value of the last one as the value of the whole cond,In the case of the
first clause of this cond that is the expression 0,If the predicate of the first clause is false,we move to the next
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
clause of the cond and repeat the process,This continues for however many clauses are contained in the cond
until either a true predicate is reached,or we reach a clause with the special keyword else,In this latter case,
that predicate is treated as true,and the subsequent expressions are evaluated,
Given that form for cond,we can see how our fibonacci procedure captures the idea we expressed,We
keep recursively solving smaller sized problems until we get to a base case,Notice that here we have two base
cases,not one,
Also notice that while this is a recursive procedure,it is different from our earlier ones,Here,there are two
recursive calls to the procedure in the body,rather than just one,Our question is whether this leads to a different
kind of behavior,or a different order of growth,
Slide 4.2.3
This does give rise to a different kind of behavior,which we
can easily see with the illustrated diagram,To solve a problem
of size 4,notice that we have to solve two problems,one of size
3 and one of size 2,and each of these requires solving two
smaller problems,and so on,
This gives rise to a kind of tree of things we have to do,and
each recursive call gives rise to two subproblems of smaller
size,This leads to a different order of growth,
Slide 4.2.4
To measure the order of growth,let's let t of n denote the
number of time steps we need to solve a problem of size n,
From our tree,we see that to do this,we need to solve a
problem of size n-1,and a problem of size n-2,We could
actually work out a detailed analysis of this relationship,but for
our purposes,we can approximate this as roughly the same as
solving two problems of size n-2,Expanding a step further,this
is roughly the same as solving 4 problems of size n-4 and
roughly the same as solving 8 problems of size n-6,A little
math shows that in general this reduces to 2 to the power of n/2
steps,
This is an example of an exponential order of growth,and this
is very different from what we saw earlier,To convince yourself of this,assume that each step takes one second,
and see how much time it would take for an exponential process as compared to a linear one,as n gets large,
In terms of space,our tree shows us that we have basically one deferred operation for step,or in other words,the
maximum depth of that tree is linear in the size of the problem,and the maximum depth exactly captures the
maximum number of deferred operations,
6.001 Notes,Section 4.3
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.1
Let's take another quick look at how we can create procedures
with different orders of growth to compute the same function,
As a specific example,suppose we want to compute
exponentials,such as a raised to the b power,but to do so only
using the simpler operations of multiplication and addition,
How might we use the tools we have been developing to
accomplish this?
Slide 4.3.2
So,recall the stages we used to solve problems like this,we
will use some wishful thinking to assume that solutions to
simpler versions of the problem exist; we will then decompose
the problem into a simpler version of the same problem,plus
some other simple operations,and we will use this to construct
a solution to the more general problem; and finally,we will
determine the smallest sized subproblem into which we want to
do decomposition,Let's look at using these tools on the
problem of exponentiation,
Slide 4.3.3
Wishful thinking is our tool for using induction,It says,let's
assume that some procedure my-expt exists,so that we can
use it,but that it only solves smaller versions of the same
problem,
Slide 4.3.4
Given that assumption,we can then turn to the tricky part,
which is determining how to decompose the problem into a
simpler version of the same problem,
Here is one method,Notice that a to the power b
mathematically is just b products of a,But this we can also
group as a times b-1 products of a,and that latter we
recognize as a smaller version of the same exponentiation
problem,Thus,we can reduce a to the bth power as a times a
to the b-1st power,Thus we can reduce the problem to a
simpler version of the same problem,together with some simple
operations,in this case,an additional multiplication,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.5
Given that idea,here is the start of our procedure for my
expt,Notice how it recursively uses my-expt to solve
the subproblem,then multiplies the result by a to get the full
solution,
Slide 4.3.6
Of course,as stated,that procedure will fail,as it will recurse
indefinitely,To stop unwinding into simpler versions of the
same problem,we need to find a smallest size subproblem,
Here,that is easy,Anything to the 0th power is just 1!
Slide 4.3.7
And thus we can add our base case to our procedure,creating
the same kind of procedure we saw earlier for factorial,
Note how the base case will stop unwinding the computation
into simpler cases,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.8
You could run the substitution model on this procedure to both
verify that it works,and to determine its growth pattern,
However,the form is very familiar,and we can thus deduce by
comparison to factorial that this procedure has linear
growth in both space and time,The time is easy to see,there is
one additional step for each increment in the size of the
argument,And for space,we see that there is one deferred
operation for each recursive call of the procedure,so we will
stack up a linear number of such deferred operations until we
get down to the base case,at which point we can start
completing the deferred multiplications and gathering up the
answer,
Slide 4.3.9
As we saw earlier,we expect there are other ways of creating
procedures to solve the same problem,With factorial,we used
the idea of a table of state variables,with update rules for
changing the values of those variables,We should be able to do
the same thing here,
Slide 4.3.10
So recall the idea of our table,We set up one column for each
piece of information that we will need to complete a step in the
computation,and we use one row for each such step,The idea
here is pretty straightforward,We know that a to the bth
power is just b successive multiplications by a,So we can just
do the multiplication,keep track of the product accumulated so
far,and how many multiplications we have left,Thus after one
step,we will have a product of a and b-1 things left to do,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.11
After the next step,we will multiply our current product by a
and keep track of that new result,plus the fact that we have one
less thing to do,
Slide 4.3.12
And this process we can continue until,.,
Slide 4.3.13
..,we reach a point where we have no further multiplies to do,
Slide 4.3.14
Now,what are the stages of this computation? To get the next
value for the product,we take the current value of the product
and the value of a,multiply together,and keep,That updates
one of the state variables,To get the next value of counter,we
simply subtract 1,since we have done one more of the
multiplications,That updates the other state variable,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.15
We know that when the counter gets down to zero,there are no
more multiplications to do,so we are done,And when we get
there,the answer is in the product column,
Slide 4.3.16
And finally,we see that the starting point is just that anything to
the zeroth power is 1,
Slide 4.3.17
So now we can capture this in a procedure,As with
factorial,we are going to use a helper procedure,And
as in that case,we can see that this procedure checks for a base
case,and if there,just returns the value of prod,Otherwise,it
reduces the computation to a simpler version of the same
computation,with a new value for prod and a new value for
count,both obtained by using the update rules we just saw,
Slide 4.3.18
And what is the order of growth here? In time,this is still
linear,as there is one subcomputation to perform for each
increment in the size of b,In space,however,we again see that
there are no deferred operations here,so this is constant,
Thus,just as with factorial,we see that we can create
different procedures to compute exponentiation,with different
classes of behaviors,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
6.001 Notes,Section 4.4
Slide 4.4.1
So now we have seen three very different kinds of procedures,
ones that give rise to constant behavior,ones that give rise to
linear growth,and ones that give rise to exponential growth,
Let's finish up by looking at a fourth kind of procedure,with yet
a different kind of behavior and thus a different cost in terms of
time and space requirements,
Slide 4.4.2
Here the problem is that we want to compute exponentials,that
is a to the power of b,where b is an integer,but where our
basic primitive operations are just multiplication,addition,and
simple tests,
Slide 4.4.3
As with previous algorithms,the key is to find a way to reduce
this problem to a combination of simpler problems,either
simpler versions of the same problem or more basic operations,
For this problem,here is the trick I can play,If b is an even
integer,then I can recognize that raising a to the power of b is
the same as first square a,then raising that result to the power
of b/2,Notice what I have done,Squaring a is just a single
multiplication,and reducing b by 2 is a simple operation,But
by doing this,I have reduce the size of the problem by a half,I
have only b/2 things left to consider,Of course,notice that I
am relying on b as an even integer here,since in that case b/2 is also an integer,and I can use the same
machinery to solve this smaller problem,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.4.4
But what if b is odd? In that case,the same trick won't work; as
I would reduce the exponentiation to a problem I don't know
how to solve,So instead I can use a different variant on wishful
thinking,similar to our factorial case,Here,I can reduce the
problem to multiplying a by the result of raising a to the
power b-1,that is to a simpler,or smaller,version of the same
problem,
Slide 4.4.5
Notice the effect of doing this,If b is odd,then in one step I
reduce the problem to the case where b is even,which means in
the next step,I reduce the problem size by half,Thus,no matter
what value b has,after at most two steps,the problem size is
halved,and after at most another two steps,it is halved again,
and so on,
Slide 4.4.6
With those ideas,we can build the procedure shown,The cases
clearly reflect the reasoning I just used,using one recursive
application of the procedure for even cases,and a different one
for odd cases,Of course,there is a base case to terminate the
reduction of the problem to simpler versions,
The form of this procedure is a bit different,In one case,it
looks like an iterative call (with just a change in the parameters
of the procedure),in the other; it looks like a recursive call
(with a deferred operation and a reduction in parameters),Thus
I expect that there will be some deferred operations in the
application of this procedure,but perhaps not as many as in
previous examples,
But what happens in terms of time? Since I am reducing the problem in half,I would hope that this leads to better
performance than just reducing the problem by 1,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.4.7
So let's measure the order of growth of this procedure,Here,n
measures the size of the problem,which means the size of the
argument b,We know that if the argument is even,in one step
the problem size is reduced by 2,If the argument is odd,in one
step it is reduced by 1,making it even so that in a second step,
it is reduced by 2,Thus in at most 2 steps,the problem size is
cut in half,After another 2 steps,it is cut in half again,and thus
after 2k steps,the problem is reduced by a factor of 2^k,or cut
in half k times,
How do we find the number of times we need to do this? We
are done when the problem size is just 1,and that happens when
k = log n,So this procedure has a different behavior,it is
logarithmic,This is in fact a very efficient procedure,and to convince yourself of this,try the same trick of seeing
how long it would take to solve a problem of different sizes,using one step per second,comparing it to a linear and
an exponential process,
The same kind of reasoning will show you that the space requirements also grow logarithmically with the size of
the problem,
Slide 4.4.8
To summarize,we have now seen how our substitution model
helps to explain the evolution of a process that occurs when a
procedure is applied,Using this model,we have seen that very
different kinds of behavior can occur,even for procedures
computing the same abstract function,and we saw how to
characterize those differences in terms of orders of growth in
space and time,Already,you can see that there are different
classes of algorithms,constant,linear,exponential,and
logarithmic,Part of your task is to learn how to recognize the
kinds of procedures that are associated with these different
behaviors,and to learn how to use that knowledge in designing
efficient procedures for particular problems,
6.001 Notes,Section 4.5
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.1
Let's take one more look at this idea of creating procedures to
solve a problem,that have different behaviors in terms of how
their underlying process evolves,For a different example,let’s
look at a mathematical problem known as Pascal’s triangle,
The elements of Pascal’s triangle are shown here,The first row
has just a single element,the second row has two elements,the
third row has three elements,and so on,Clearly there is a
structure to these elements,which we need to understand,
Slide 4.5.2
So,let’s structure this problem a bit,Let’s order the rows,and
enumerate them,labeling the first row n=0,the second row n=1,
and so on (we will see that this choice of labeling leads to a
cleaner description of the problem),Using this labeling,we
also see that the n’th row has n+1 elements in it,Let’s use the
notation P(j,n) to denote the j’th element of the n’th row,Our
goal is to determine how to compute all elements of each row,
Slide 4.5.3
Traditionally,Pascal’s triangle is constructed by noting that the
first and last element of each row (except the first) is a 1,and
by noting that to get any other element in a row,we add the
corresponding element of the previous row and its predecessor
together,If you use this rule of thumb you can verify that this
works by generating the first few rows of the triangle,So we
have an informal specification of how to generate the triangle,
Slide 4.5.4
So we can proceed in a straightforward manner by capturing
that idea in a procedure,Note the form,we have two base
cases,one for when we are generating the first element of a
row,and one for the last element of the row,
Otherwise,we simply rely on smaller versions of the same
computation to do the job,we compute the same element of the
previous row,and its predecessor,and then add them together,
You can verify that this is correct by trying some examples,
though clearly the
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.5
Slide 4.5.6
So what kind of process does this procedure engender? Well,it
looks a lot like our computation of Fibonacci,and that is a pretty
good clue,Just like Fibonacci,there are two recursive calls to
smaller versions of the same process at each stage,In fact,a
similar analysis will show that this is process that is exponential
in time,and linear in space,We have already suggested that
exponential algorithms are costly,In the case of Fibonacci,we
didn’t look for any other way of structuring the problem,but let’s
try to do better for Pascal,
Slide 4.5.7
To do better,we have to go back to the original problem,A
little information from combinatorics tells us that in actuality,
Pascal’s triangle is capturing the number of different ways of
choosing a set of j objects from a set of n objects (this isn’t
obvious,by the way,so just accept this as a fact),Thus,the
first element of a row is the number of ways of picking no
objects,which is by definition 1,The last element of a row is
the number of ways of picking a set of n objects from a set of n
objects,Since the order in which we pick the objects doesn’t
matter,there is only one size n subset of a set of size n,hence
this element is 1,
The general case is the number of different subsets of size j that
can be created out of a set of size n,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.8
So what is that number?
Well,we can pick the first element by selecting any of the n
elements of the set,
The second element has n-1 possibilities,and so on,until we
have picked out j objects,That product we can represent as a
fraction of two factorial products,as you can see,
Now,we are not quite done,because we said the order in which
we pick the objects doesn’t matter,Using the method we just
described,we could pick the same set of j objects in j! different
orderings,so to get the number of distinct subsets of size j,we
factor this out,as shown,
Slide 4.5.9
Now this leads to a straightforward way to compute Pascal,We
can just rely on the factorial code that we created earlier,Here
we use factorial three times,once for the numerator,and twice
in the denominator,Notice how our procedural abstraction
nicely isolates the details of fact from its use in this case,and
the code here cleanly expresses the idea of computing Pascal
based on factorial,
Slide 4.5.10
So do we do any better with this version of Pascal?
Sure! We know that this version of fact is linear,Our Pascal
implementation uses fact three different evaluations of fact,but
this is still linear in time,and a similar analysis lets us conclude
that this is linear in space as well,While it might take three
times as long to run than just running fact,what we are
concerned with is the general order of growth; or how these
processes’ computational needs change with changing problem
size,and both are linear in that change,
Slide 4.5.11
Well,earlier we saw that there were different ways of
computing factorial,Suppose we use the iterative version,
instead of the recursive one?
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.12
The same effect takes place,Now we have a version of Pascal
that is linear in time and constant in space,relying on the fact
that factorial under this version is also linear in time and
constant in space,
Slide 4.5.13
Finally,if the use of an iterative factorial procedure gives us
better performance,why not just do this computation directly as
well? By that,we mean just compute the two products,one for
the numerator and one for the denominator,and save those
extra multiplications that we are doing in the extra call to
factorial,After all,we are just computing a product of terms
that we know we are simply going to factor out,
Slide 4.5.14
We can do the same analysis for this version,Our help
procedure is clearly constant in space,and linear in time,Our
version of Pascal uses it twice,so while it does take less actual
time than the previous version,it still has the same general
behavior,linear in time and constant in space,Thus,in
practical terms,this may be the best version,but theoretically it
has the same class of behavior as our previous version,
Slide 4.5.15
And that leads to our concluding point,First,we stress that the
same problem may have many different solutions,and that
these solutions may have very different computational
behaviors,Some problems are inherently exponential in
nature,Others may have straightforward solutions that have
exponential behavior,but often some additional thought can
lead to more efficient solutions,Part of our goal is to get you to
recognize different classes of behavior,and to learn how to use
the properties of standard algorithms to help you design
efficient solutions to new problems,
6.001 Notes,Section 4.1
Slide 4.1.1
In this lecture,we are going to take a careful look at the kinds
of procedures we can build,We will first go back to look very
carefully at the substitution model,to see how the rules for
evaluation help us determine the evolution of a process
controlled by a procedure,We will then see how different kinds
of process evolution can occur,with different demands in terms
of computational resources,And we will see how to
characterize those differences formally in terms of orders of
growth of a process,Finally,we will explore examples of
procedures from different classes of growth,helping you to
begin to relate how choices in procedure design may affect
performance of the actual use of the procedure,
Slide 4.1.2
Now that we have seen two different implementations of the
same idea,with different behaviors,we need a way of more
formally characterizing why that difference occurs,To do that,
we are going to go back to our substitution model,
Now we are going to think of our model as a set of rewrite
rules,that is,as a set of rules that formally specify,given an
expression of a particular form,rewrite it in the following form,
Slide 4.1.3
So elementary expressions in this viewpoint are just "left
alone",that is,numbers,names of built-in procedures,and
lambda expressions are left as is,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.1.4
If our expression is a name that we created ourselves,we will
rewrite in place of it the value that was associated with that
name during the definition,
Slide 4.1.5
For the special form if we use the rule we just saw,We
evaluate the predicate clause,and based on its value,we either
rewrite the if expression by its consequent or its alternative,
Slide 4.1.6
And finally,combinations,We use the rule that we first
evaluate the operator expression to get the procedure,and we
evaluate the operands to get the set of arguments,If we have a
primitive procedure,we are just going to,do the right thing”,
Otherwise we are going to replace this entire expression with
the body of the compound expression,with the arguments
substituted for their associated parameters,
Slide 4.1.7
Given that model,we can more formally capture the evolution
of a process,We can talk about the order of growth of a process
as these rewrite rules evolve,This measures how much of a
particular resource a process is going to take as these rules
evolve,where typically we measure either space or time as the
resource,
More formally,let n be a parameter that measures the size of
the problem of interest,In the case of factorial,this would be
the size of the input argument,We let R(n) denote the amount
of resources we will need to solve a problem of this size,where
as noted,the resources are usually space and time,We are
interested in characterizing R(n) for large values of N,that is,in
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
the asymptotic limit,We say that R(n) has order of growth,Theta of f of n if we can find a function f(n) that is
roughly of the same order as the needed resource,where the more formal definition is shown in the
Slide 4.1.8
In more common terms,this says that when measuring the
amount of space we need,we want to find a function that
measures the number of deferred operations,as a function of the
size of the problem,up to a constant factor,For measuring the
amount of time,we want to find a function that measures the
number of basic or primitive steps that the rewrite rules go
through,again as a function of the size of the problem,
Slide 4.1.9
So let's use our rewrite rule idea,together with this notion of
measuring the amount of resource we need as an order of
growth in the size of the problem,Here is our recursive factorial
procedure,.,
Slide 4.1.10
..and here is a partial trace of fact using those rewrite rules,
We start with (fact 4),That reduces to evaluating an if
expression,Since the predicate is not true,this reduces to
evaluating the alternative statement,which is a combination of
a multiplication and another evaluation of fact,
Notice how the orange colored lines capture the evolution of the
rewrite rules,Note how (fact 4) reduces to a deferred operation
and an evaluation of (fact 3) and this further reduces to another
deferred operation and a subsequent call to fact,Notice the
shape of this process,it grows out with a set of deferred
operations until it reaches a base case,and then it contracts back
in,completing each deferred operation,Using this,we can see that the amount of space grows linearly with the
size of the argument,That is,with each increase in the argument size,I add a constant amount of space
requirement,In terms of the number of operations,we can see that we basically need twice as many steps as the
size of the problem,one to expand out,and one to reduce back down,This is also a linear growth,since the
constant 2 does not change with problem size,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.1.11
And let's compare that our iterative factorial,Here is our code
again,and here is a partial trace of the rewrite evolution of that
process,
Here we see the different shape of the process,and our order of
growth analysis captures that difference,In particular,there are
no deferred operations,and you can see that the maximum
amount of space I need at any stage,that is,the maximum
amount of space used by any rewrite,is independent of the size
of the problem,We say that it is constant,as it does not grow
with n,In terms of time,there is basically one operation for
each increment in the argument,so this is again a linear order
of growth,
Slide 4.1.12
So we can formally capture this,as shown,Fact has linear
growth in space,written as shown,because the maximum
amount of space needed by any rewrite stage is a linear multiple
of the size of the argument,It also has linear growth in time,
because as we saw it takes a number of basic steps that is also a
linear multiple of the size of the argument,
Slide 4.1.13
On the other hand,iterative-fact has no deferred
operations,and is constant in space,written as shown,while as
we argued,the time order of growth is linear,Notice that the
fact that the recursive version took 2n steps and the iterative
version took n steps doesn't matter in terms of order of growth,
Both are said to be linear,
Thus we see that we can formally characterize the difference
between these two processes,which have different shapes of
evolution,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.1.14
So why have we done this? The primary goal is to allow you to
start to recognize different kinds of behaviors,and the
associated procedure forms that give rise to those behaviors,
This will start to allow you to work backwards,in designing a
procedure,by enabling you to visualize the consequences of
performing a computation in different ways,
6.001 Notes,Section 4.2
Slide 4.2.1
Having seen two different kinds of processes,one linear,and
one constant,we want to fill out our repertoire of processes,by
looking at other kinds of processes,The next one is an example
of an exponential process,and deals with a classic function
called Fibonacci,Its definition is that if its argument is 0,its
value is 0,if its argument is 1,its value is 1,and for all other
positive integer arguments,its value is the sum of its values for
the two preceding arguments,
We would like to write a procedure to compute Fibonacci,and
in particular see how it gives rise to a different kind of
behavior,
Slide 4.2.2
To solve this problem,let's use our tool of wishful thinking,
Here,that wishful thinking says,let's assume that given an
argument n,we know how to solve Fibonacci for any smaller
sized argument,Using that idea,we can then work out a
solution to the problem,With this in hand,it is clear that the
solution to the general problem is just to solve two smaller
sized problems,then just add the results together,Note that in
this case we are using wishful thinking twice,not once,as in
our previous examples,
Here is a procedure that captures this idea,
First,we introduce a new expression,called a cond
expression,Cond uses the following rules of evaluation,The cond consists of a set of clauses,each of which
has within it,a predicate clause,and one or more subsequent expressions,Cond proceeds by first evaluating the
predicate of the first clause,in this case (= n 0),If it is true,then we evaluate in turn each of the other
expressions in this clause,returning the value of the last one as the value of the whole cond,In the case of the
first clause of this cond that is the expression 0,If the predicate of the first clause is false,we move to the next
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
clause of the cond and repeat the process,This continues for however many clauses are contained in the cond
until either a true predicate is reached,or we reach a clause with the special keyword else,In this latter case,
that predicate is treated as true,and the subsequent expressions are evaluated,
Given that form for cond,we can see how our fibonacci procedure captures the idea we expressed,We
keep recursively solving smaller sized problems until we get to a base case,Notice that here we have two base
cases,not one,
Also notice that while this is a recursive procedure,it is different from our earlier ones,Here,there are two
recursive calls to the procedure in the body,rather than just one,Our question is whether this leads to a different
kind of behavior,or a different order of growth,
Slide 4.2.3
This does give rise to a different kind of behavior,which we
can easily see with the illustrated diagram,To solve a problem
of size 4,notice that we have to solve two problems,one of size
3 and one of size 2,and each of these requires solving two
smaller problems,and so on,
This gives rise to a kind of tree of things we have to do,and
each recursive call gives rise to two subproblems of smaller
size,This leads to a different order of growth,
Slide 4.2.4
To measure the order of growth,let's let t of n denote the
number of time steps we need to solve a problem of size n,
From our tree,we see that to do this,we need to solve a
problem of size n-1,and a problem of size n-2,We could
actually work out a detailed analysis of this relationship,but for
our purposes,we can approximate this as roughly the same as
solving two problems of size n-2,Expanding a step further,this
is roughly the same as solving 4 problems of size n-4 and
roughly the same as solving 8 problems of size n-6,A little
math shows that in general this reduces to 2 to the power of n/2
steps,
This is an example of an exponential order of growth,and this
is very different from what we saw earlier,To convince yourself of this,assume that each step takes one second,
and see how much time it would take for an exponential process as compared to a linear one,as n gets large,
In terms of space,our tree shows us that we have basically one deferred operation for step,or in other words,the
maximum depth of that tree is linear in the size of the problem,and the maximum depth exactly captures the
maximum number of deferred operations,
6.001 Notes,Section 4.3
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.1
Let's take another quick look at how we can create procedures
with different orders of growth to compute the same function,
As a specific example,suppose we want to compute
exponentials,such as a raised to the b power,but to do so only
using the simpler operations of multiplication and addition,
How might we use the tools we have been developing to
accomplish this?
Slide 4.3.2
So,recall the stages we used to solve problems like this,we
will use some wishful thinking to assume that solutions to
simpler versions of the problem exist; we will then decompose
the problem into a simpler version of the same problem,plus
some other simple operations,and we will use this to construct
a solution to the more general problem; and finally,we will
determine the smallest sized subproblem into which we want to
do decomposition,Let's look at using these tools on the
problem of exponentiation,
Slide 4.3.3
Wishful thinking is our tool for using induction,It says,let's
assume that some procedure my-expt exists,so that we can
use it,but that it only solves smaller versions of the same
problem,
Slide 4.3.4
Given that assumption,we can then turn to the tricky part,
which is determining how to decompose the problem into a
simpler version of the same problem,
Here is one method,Notice that a to the power b
mathematically is just b products of a,But this we can also
group as a times b-1 products of a,and that latter we
recognize as a smaller version of the same exponentiation
problem,Thus,we can reduce a to the bth power as a times a
to the b-1st power,Thus we can reduce the problem to a
simpler version of the same problem,together with some simple
operations,in this case,an additional multiplication,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.5
Given that idea,here is the start of our procedure for my
expt,Notice how it recursively uses my-expt to solve
the subproblem,then multiplies the result by a to get the full
solution,
Slide 4.3.6
Of course,as stated,that procedure will fail,as it will recurse
indefinitely,To stop unwinding into simpler versions of the
same problem,we need to find a smallest size subproblem,
Here,that is easy,Anything to the 0th power is just 1!
Slide 4.3.7
And thus we can add our base case to our procedure,creating
the same kind of procedure we saw earlier for factorial,
Note how the base case will stop unwinding the computation
into simpler cases,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.8
You could run the substitution model on this procedure to both
verify that it works,and to determine its growth pattern,
However,the form is very familiar,and we can thus deduce by
comparison to factorial that this procedure has linear
growth in both space and time,The time is easy to see,there is
one additional step for each increment in the size of the
argument,And for space,we see that there is one deferred
operation for each recursive call of the procedure,so we will
stack up a linear number of such deferred operations until we
get down to the base case,at which point we can start
completing the deferred multiplications and gathering up the
answer,
Slide 4.3.9
As we saw earlier,we expect there are other ways of creating
procedures to solve the same problem,With factorial,we used
the idea of a table of state variables,with update rules for
changing the values of those variables,We should be able to do
the same thing here,
Slide 4.3.10
So recall the idea of our table,We set up one column for each
piece of information that we will need to complete a step in the
computation,and we use one row for each such step,The idea
here is pretty straightforward,We know that a to the bth
power is just b successive multiplications by a,So we can just
do the multiplication,keep track of the product accumulated so
far,and how many multiplications we have left,Thus after one
step,we will have a product of a and b-1 things left to do,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.11
After the next step,we will multiply our current product by a
and keep track of that new result,plus the fact that we have one
less thing to do,
Slide 4.3.12
And this process we can continue until,.,
Slide 4.3.13
..,we reach a point where we have no further multiplies to do,
Slide 4.3.14
Now,what are the stages of this computation? To get the next
value for the product,we take the current value of the product
and the value of a,multiply together,and keep,That updates
one of the state variables,To get the next value of counter,we
simply subtract 1,since we have done one more of the
multiplications,That updates the other state variable,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.3.15
We know that when the counter gets down to zero,there are no
more multiplications to do,so we are done,And when we get
there,the answer is in the product column,
Slide 4.3.16
And finally,we see that the starting point is just that anything to
the zeroth power is 1,
Slide 4.3.17
So now we can capture this in a procedure,As with
factorial,we are going to use a helper procedure,And
as in that case,we can see that this procedure checks for a base
case,and if there,just returns the value of prod,Otherwise,it
reduces the computation to a simpler version of the same
computation,with a new value for prod and a new value for
count,both obtained by using the update rules we just saw,
Slide 4.3.18
And what is the order of growth here? In time,this is still
linear,as there is one subcomputation to perform for each
increment in the size of b,In space,however,we again see that
there are no deferred operations here,so this is constant,
Thus,just as with factorial,we see that we can create
different procedures to compute exponentiation,with different
classes of behaviors,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
6.001 Notes,Section 4.4
Slide 4.4.1
So now we have seen three very different kinds of procedures,
ones that give rise to constant behavior,ones that give rise to
linear growth,and ones that give rise to exponential growth,
Let's finish up by looking at a fourth kind of procedure,with yet
a different kind of behavior and thus a different cost in terms of
time and space requirements,
Slide 4.4.2
Here the problem is that we want to compute exponentials,that
is a to the power of b,where b is an integer,but where our
basic primitive operations are just multiplication,addition,and
simple tests,
Slide 4.4.3
As with previous algorithms,the key is to find a way to reduce
this problem to a combination of simpler problems,either
simpler versions of the same problem or more basic operations,
For this problem,here is the trick I can play,If b is an even
integer,then I can recognize that raising a to the power of b is
the same as first square a,then raising that result to the power
of b/2,Notice what I have done,Squaring a is just a single
multiplication,and reducing b by 2 is a simple operation,But
by doing this,I have reduce the size of the problem by a half,I
have only b/2 things left to consider,Of course,notice that I
am relying on b as an even integer here,since in that case b/2 is also an integer,and I can use the same
machinery to solve this smaller problem,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.4.4
But what if b is odd? In that case,the same trick won't work; as
I would reduce the exponentiation to a problem I don't know
how to solve,So instead I can use a different variant on wishful
thinking,similar to our factorial case,Here,I can reduce the
problem to multiplying a by the result of raising a to the
power b-1,that is to a simpler,or smaller,version of the same
problem,
Slide 4.4.5
Notice the effect of doing this,If b is odd,then in one step I
reduce the problem to the case where b is even,which means in
the next step,I reduce the problem size by half,Thus,no matter
what value b has,after at most two steps,the problem size is
halved,and after at most another two steps,it is halved again,
and so on,
Slide 4.4.6
With those ideas,we can build the procedure shown,The cases
clearly reflect the reasoning I just used,using one recursive
application of the procedure for even cases,and a different one
for odd cases,Of course,there is a base case to terminate the
reduction of the problem to simpler versions,
The form of this procedure is a bit different,In one case,it
looks like an iterative call (with just a change in the parameters
of the procedure),in the other; it looks like a recursive call
(with a deferred operation and a reduction in parameters),Thus
I expect that there will be some deferred operations in the
application of this procedure,but perhaps not as many as in
previous examples,
But what happens in terms of time? Since I am reducing the problem in half,I would hope that this leads to better
performance than just reducing the problem by 1,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.4.7
So let's measure the order of growth of this procedure,Here,n
measures the size of the problem,which means the size of the
argument b,We know that if the argument is even,in one step
the problem size is reduced by 2,If the argument is odd,in one
step it is reduced by 1,making it even so that in a second step,
it is reduced by 2,Thus in at most 2 steps,the problem size is
cut in half,After another 2 steps,it is cut in half again,and thus
after 2k steps,the problem is reduced by a factor of 2^k,or cut
in half k times,
How do we find the number of times we need to do this? We
are done when the problem size is just 1,and that happens when
k = log n,So this procedure has a different behavior,it is
logarithmic,This is in fact a very efficient procedure,and to convince yourself of this,try the same trick of seeing
how long it would take to solve a problem of different sizes,using one step per second,comparing it to a linear and
an exponential process,
The same kind of reasoning will show you that the space requirements also grow logarithmically with the size of
the problem,
Slide 4.4.8
To summarize,we have now seen how our substitution model
helps to explain the evolution of a process that occurs when a
procedure is applied,Using this model,we have seen that very
different kinds of behavior can occur,even for procedures
computing the same abstract function,and we saw how to
characterize those differences in terms of orders of growth in
space and time,Already,you can see that there are different
classes of algorithms,constant,linear,exponential,and
logarithmic,Part of your task is to learn how to recognize the
kinds of procedures that are associated with these different
behaviors,and to learn how to use that knowledge in designing
efficient procedures for particular problems,
6.001 Notes,Section 4.5
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.1
Let's take one more look at this idea of creating procedures to
solve a problem,that have different behaviors in terms of how
their underlying process evolves,For a different example,let’s
look at a mathematical problem known as Pascal’s triangle,
The elements of Pascal’s triangle are shown here,The first row
has just a single element,the second row has two elements,the
third row has three elements,and so on,Clearly there is a
structure to these elements,which we need to understand,
Slide 4.5.2
So,let’s structure this problem a bit,Let’s order the rows,and
enumerate them,labeling the first row n=0,the second row n=1,
and so on (we will see that this choice of labeling leads to a
cleaner description of the problem),Using this labeling,we
also see that the n’th row has n+1 elements in it,Let’s use the
notation P(j,n) to denote the j’th element of the n’th row,Our
goal is to determine how to compute all elements of each row,
Slide 4.5.3
Traditionally,Pascal’s triangle is constructed by noting that the
first and last element of each row (except the first) is a 1,and
by noting that to get any other element in a row,we add the
corresponding element of the previous row and its predecessor
together,If you use this rule of thumb you can verify that this
works by generating the first few rows of the triangle,So we
have an informal specification of how to generate the triangle,
Slide 4.5.4
So we can proceed in a straightforward manner by capturing
that idea in a procedure,Note the form,we have two base
cases,one for when we are generating the first element of a
row,and one for the last element of the row,
Otherwise,we simply rely on smaller versions of the same
computation to do the job,we compute the same element of the
previous row,and its predecessor,and then add them together,
You can verify that this is correct by trying some examples,
though clearly the
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.5
Slide 4.5.6
So what kind of process does this procedure engender? Well,it
looks a lot like our computation of Fibonacci,and that is a pretty
good clue,Just like Fibonacci,there are two recursive calls to
smaller versions of the same process at each stage,In fact,a
similar analysis will show that this is process that is exponential
in time,and linear in space,We have already suggested that
exponential algorithms are costly,In the case of Fibonacci,we
didn’t look for any other way of structuring the problem,but let’s
try to do better for Pascal,
Slide 4.5.7
To do better,we have to go back to the original problem,A
little information from combinatorics tells us that in actuality,
Pascal’s triangle is capturing the number of different ways of
choosing a set of j objects from a set of n objects (this isn’t
obvious,by the way,so just accept this as a fact),Thus,the
first element of a row is the number of ways of picking no
objects,which is by definition 1,The last element of a row is
the number of ways of picking a set of n objects from a set of n
objects,Since the order in which we pick the objects doesn’t
matter,there is only one size n subset of a set of size n,hence
this element is 1,
The general case is the number of different subsets of size j that
can be created out of a set of size n,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.8
So what is that number?
Well,we can pick the first element by selecting any of the n
elements of the set,
The second element has n-1 possibilities,and so on,until we
have picked out j objects,That product we can represent as a
fraction of two factorial products,as you can see,
Now,we are not quite done,because we said the order in which
we pick the objects doesn’t matter,Using the method we just
described,we could pick the same set of j objects in j! different
orderings,so to get the number of distinct subsets of size j,we
factor this out,as shown,
Slide 4.5.9
Now this leads to a straightforward way to compute Pascal,We
can just rely on the factorial code that we created earlier,Here
we use factorial three times,once for the numerator,and twice
in the denominator,Notice how our procedural abstraction
nicely isolates the details of fact from its use in this case,and
the code here cleanly expresses the idea of computing Pascal
based on factorial,
Slide 4.5.10
So do we do any better with this version of Pascal?
Sure! We know that this version of fact is linear,Our Pascal
implementation uses fact three different evaluations of fact,but
this is still linear in time,and a similar analysis lets us conclude
that this is linear in space as well,While it might take three
times as long to run than just running fact,what we are
concerned with is the general order of growth; or how these
processes’ computational needs change with changing problem
size,and both are linear in that change,
Slide 4.5.11
Well,earlier we saw that there were different ways of
computing factorial,Suppose we use the iterative version,
instead of the recursive one?
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 4.5.12
The same effect takes place,Now we have a version of Pascal
that is linear in time and constant in space,relying on the fact
that factorial under this version is also linear in time and
constant in space,
Slide 4.5.13
Finally,if the use of an iterative factorial procedure gives us
better performance,why not just do this computation directly as
well? By that,we mean just compute the two products,one for
the numerator and one for the denominator,and save those
extra multiplications that we are doing in the extra call to
factorial,After all,we are just computing a product of terms
that we know we are simply going to factor out,
Slide 4.5.14
We can do the same analysis for this version,Our help
procedure is clearly constant in space,and linear in time,Our
version of Pascal uses it twice,so while it does take less actual
time than the previous version,it still has the same general
behavior,linear in time and constant in space,Thus,in
practical terms,this may be the best version,but theoretically it
has the same class of behavior as our previous version,
Slide 4.5.15
And that leads to our concluding point,First,we stress that the
same problem may have many different solutions,and that
these solutions may have very different computational
behaviors,Some problems are inherently exponential in
nature,Others may have straightforward solutions that have
exponential behavior,but often some additional thought can
lead to more efficient solutions,Part of our goal is to get you to
recognize different classes of behavior,and to learn how to use
the properties of standard algorithms to help you design
efficient solutions to new problems,
