Homework Q&A, Introduction, Exact numbers, Floating-point numbers, Overflow, Underflow, Round-off Adding numbers of widely different magnitude, Tradeoffs
Any general questions? We'll answer them.
Don't forget that you should now be using abstract functions, where applicable. As previously described, you may choose to develop your programs the "old-fashioned" way and then edit them to use any appropriate abstract functions. By the way, you might find the following built-in abstract function useful:
; build-list : natnum (natnum -> alpha) -> list-of-alpha
; (build-list n f) = (list (f 0) ... (f (sub1 n)))
Mathematical programs in Scheme can sometimes be unacceptably slow because they calculate exact answers. For example, consider finding the integral of a curve using the standard beginning calculus method, adding the areas of many small trapezoids under the curve. Scheme would accurately calculate these areas in a result like 3141592654/2718281828. Working with such fractions can be slow, so Scheme can instead work to return an approximate result, like #i1.155727350137, much more quickly. (The #i stands for inexact.) (A partial explanation: Consider multiplying two fractions. This really means multiplying twice, once each for the denominators and numerators, plus also simplifying the result.) The idea is that the answer might be inexact, but hopefully the error is small. Today's lab discusses this trade-off between efficiency and accuracy, and also demonstrates how sometimes computers give answers that are just flat-out wrong.
While there is no bound on how big or tiny a mathematical number can be, unfortunately every computer does have a bound on how big or tiny a number it can represent internally.
You have probably already noticed that Scheme uses "arbitrary precision arithmetic": It can represent very large integers accurately. For instance, ten to the eightieth power plus seventeen can be typed directly into Scheme, 100000000000000000000000000000000000000000000000000000000000000000000000000000017, and the value stored exactly. The only limit to such numbers in Scheme is the actual amount of memory of the computer. To do: Try it!
In order to represent these large, exact-precision integers, Scheme is essentially keeping a list of the digits. When it adds or multiplies two exact-precision numbers together, Scheme uses the grade-school algorithms.
There is an alternate way to represent numbers: one can assume that all integers have (say) 9 digits. Knowing the size of all numbers, we can use better algorithms for addition and multiplication, and implement these algorithms in hardware. This makes addition and multiplication of numbers simpler and faster, and might be sufficient for most programmer's needs. Of course the drawback is that only about 109 numbers can be expressed in this manner.
This fixed-size representation doesn't apply only to integers,
but can be applied to fractional numbers.
Most computers nowadays represent numbers by using
about 15 significant digits.
In order to express fractional amounts,
of numbers, scientific notation is used. For instance,
#i6.022e23 is 6.022 with the decimal place shifted
23 places to the right (i.e., 60.22 hexillion).
Just as the mantissa (the "6.022" part) is restricted to about 15 digits,
the exponent (the "23" part) is also restricted in size; we'll see how
big it can be in a moment.
This method of representing numbers is called
floating-point representation,
since the decimal point "floats" among the
15 significant digits, depending on the exponent.
Notice that inside the computer, each floating-point number needs
the same, fixed number of digits to be stored. This convenience
sometimes makes up for the fact that such numbers might only be
approximations to the numbers we're actually interested in.
Note: The actual size restrictions are on the internal binary number representation, not on the printed decimal number representation.
In Scheme, floating-point numbers are called "inexact" numbers because they aren't exact (duh!). In some languages, they are referred to as "reals", because they approximate the real numbers, or as "floats", short for floating-point.
In addition to arbitrary-precision rational numbers,
Scheme also uses floating-point numbers whenever a
number is typed in as a decimal point.
To do:
Evaluate (/ 2 37) and compare it to (/ #i2.0 #i37.0).
(If you really need them,
there are also functions exact->inexact and
inexact->exact to convert between the two formats.)
Using fixed-width numbers up to a billion
might be fine for most everyday calculations, but what
happens when you try to add a couple large 9-digit numbers and
get a 10-digit number?
The result can no longer be stored as an exact integer;
this is called overflow.
What happens next depends on the particular system.
The program might halt; it might procede with
the result tagged as a special, anomolous value
perhaps written +Infinity; or (most frightful yet)
it might procede with some totally spurious number without
even informing anybody that all subsequent calculations are garbage.
Just as with integers, it is possible to think of numbers whose exponent is larger than can be stored in a computer's floating point number; calculations whose answer exceeds the largest expressible number is again called overflow.
For any number x, if we double x and then
divide by two, hopefully we get x back.
But here's where overflow can get in the way:
If x is large enough that
doubling it causes an overflow, then
doubling x followed by halving the result will likely not
give us x back.
To do: Let's take a number, and keep doubling it until (using this approach) we find an overflow:
(define (find-huge x)
(cond
[(= x (/ (* x 2) 2)) (find-huge (* x 2))]
[else x]))
(define huge (find-huge #i1.0))
What is the value of huge?
What is the representation of (* 2 huge), which can't
be represented accurately?
Underflow occurs when a number is too tiny to be expressed as a floating-point number; i.e., when you get a non-zero number which is very close to zero. Q: What number do you get when you type in the number #i1e-400? Underflow is not the same as "overflow in the negative direction", that is obtaining a very large negative number than cannot be represented.
To do: Write a function find-tiny which finds a number which is within a factor of two of the smallest (positive) representable fixpoint number. This should be similar to find-huge.
There is an IEEE floating point standard which also incorporates gradual underflow, a scheme to try to go underflow "softly" from a non-zero value to zero (in some situations this difference is a qualitative change, as well as quantitative one).
Sometimes, when dividing one number with 10 decimal places by another with 10 decimal places, the the result requires 11 or more decimal places to store the answer. No problem, you round the last digit, and you have an answer that's pretty darn close to exact. The problem is, round-off error can accumulate.
To do:
The following function adds n copies of m:
; add-copies : natnum num -> num
; Returns the addition of n copies of m.
(define (add-copies n m)
(cond
[(zero? n) 0]
[else (+ (add-copies (sub1 n) m) m)]))
What is the difference between the following, for example?
(add-copies 185 1/185)
(add-copies 185 (/ #i1.0 #i185.0))
Write a similar function to subtract n copies of m from 1 and try similar examples. What are the results?
If we are using only 15 significant digits, then we run
into problems when adding numbers which vary by more than a
factor of 1016:
#i1.0e16 + 1 should be #i1.00000000000000001e16,
but if you only have 15 digits the closest answer is #i1.0e16,
which is what you started with!
You might think this error is reasonable; being wrong by one part
in 1016 (ten million billion)
is admittedly close enough for government work.
However, small errors can accumulate.
To do/Q:
Load
/home/comp210/public_html/Labs/lab11/sum.ss
into DrScheme.
Use the functions sum-r and sum-l
(defined in sum.ss) to
add some example lists of numbers:
list1 list2, and list3.
These are all different arrangements of the same numbers, so you
would want the same result for each. Is that what you get?
Use the function order (defined in sum.ss) to order the numbers in increasing magnitude. (E.g., (order list1). Since the lists contain the same numbers, it doesn't matter which you use.) What's the sum of this list?
Does ordering elements by magnitude always solve the problem?
Compute sums of the following list, named janus after
the two-faced Roman god:
(define janus (list 99
#i3.2e+270
-99
#i-2.4e+270
1234
4
#i-8e+269
-4))
; (sum-r janus) = ??
; (sum-l janus) = ??
; (sum-r (order janus)) = ??
; (sum-l (order janus)) = ??
; The correct sum = ??
What's going on?
Do you still trust your computer to give you reasonably correct answers
as much as you did 30 seconds ago?
For the curious... Can you modify the small-magnitudes-first rule of thumb to better handle all cases?
For the curious... There are some subtleties with rounding. For example, if the answer is halfway between the two choices, which way do you round? E.g., should -2.5 be rounded to -2 or -3? Methods such as "always round down" or "always around away from 0" suffer from the flaw that all round-off errors, though individually tiny, may all accumulate in the same direction, leading to results that drift away from the true answer. One solution is to round to the nearest even number. Hopefully (though not guaranteedly) this allows two round-off errors to cancel each other half the time, and therefore the total round-off error grows much more slowly. (Will it still accumulate at all, on average?)
So if Scheme has arbitrary precision integers and rationals, why do we ever bother with floating-point arithmetic? Like any interesting choice, there is a trade-off between the two alternatives (otherwise, it wouldn't be interesting). Using arbitrary precision arithmetic can sometimes slow us down, and often the added accuracy isn't enlighting.
To do:
Consider writing our own version of expt to raise
one number to another, where the second number is a nice, positive
integer. You should be able to whip this out in no time:
; my-expt : number natnum[>=1] -> number
; Computes n to the given power.
(define (my-expt n power)
(cond [(zero? power) 1]
[else (* n (my-expt n (sub1 power)))]))
Now, we try our function on two versions of the same number, one inexact and one exact:
(my-expt (+ 1 #i1e-12) 30) ; 1e-12 is inexact (floating-point)
(my-expt (+ 1 (expt 10 -12) 30) ; (expt 10 -12) is exact
Which answer is more useful to you?
(Check out the interesting symmetry of the numerator
in this second version.)
Try similar examples with a much larger power. Is the result familiar?
Computers must give consistent results for even bizarre math questions like 1/0.
To do:
Try examples like (/ 1 0), (/ #i1.0 #i0.0), (/ #i-1.0 #i0.0), and (/ #i1.0 #i-0.0). Note that we can tell the difference between #i0.0 and #i-0.0!
What about (/ 0 0) and (/ #i0.0 #i0.0)? The latter returns something read as "Not a Number".
To understand more about floating-point numbers, take Comp 320 or Elec 320. To understand more about the consequences of imprecise arithmetic, and how to still compute valid results, take Caam 353.