SICP - Solution: Exercise 1.7

# SICP - Solution: Exercise 1.7

##### September 29, 2018

Exercise 1.7

The good-enough? test used in computing square roots will not be very effective for finding the square roots of very small numbers. Also, in real computers, arithmetic operations are almost always performed with limited precision. This makes our test inadequate for very large numbers. Explain these statements, with examples showing how the test fails for small and large numbers. An alternative strategy for implementing good-enough? is to watch how guess changes from one iteration to the next and to stop when the change is a very small fraction of the guess. Design a square-root procedure that uses this kind of end test. Does this work better for small and large numbers?

Solution

This is a very interesting exercise because it forces to look at “behind the scenes” at how computer handle certain kinds of numbers. It also shows that you need to have this kind of knowledge in order to develop correct programs.

First, let’s experiment with a few cases in the interpreter:

(sqrt 1234567890123)
> 1111111.1061109055

(sqrt 12345678901234)
does not finish

(sqrt 12345678901230)
> 3513641.828819494

(sqrt 0.00000000123)
> 0.031250013107186406

(square (sqrt 0.00000000123))
> 0.000976563319199322


From this we can see two problems:

• With large numbers, most of the time the computation doesn’t finish
• With small numbers, the result can be very inaccurate, by multiple orders of magnitude

In order to understand what is happening, we need to look at how real numbers are encoded in computers, more specifically “floating point” encoding in this case.

Key facts about floating point numbers:

• because each number is encoded on a finite number of bits, the number of floating point numbers that can be represented in a computer is finite.
• Most of the time, a floating point number is an approximation of a real number. This causes rounding issues.
• As the size of the number represented increases, the size of the “gap” between two consecutive numbers will increase step by step.

Squeezing infinitely many real numbers into a finite number of bits requires an approximate representation. Although there are infinitely many integers, in most programs the result of integer computations can be stored in 32 bits. In contrast, given any fixed number of bits, most calculations with real numbers will produce quantities that cannot be exactly represented using that many bits. Therefore the result of a floating-point calculation must often be rounded in order to fit back into its finite representation. This rounding error is the characteristic feature of floating-point computation.

#### Large numbers #

For most numbers above a certain size of digits, the computation of the square root will never finish.

When tracing the program step by step, we can see that this condition happens for large numbers when the guess is getting very close to the actual result. Because of rounding errors, the function (improve guess x) can’t improve the guess anymore as the smallest possible difference between $guess^2$ and $x$ is larger than 0.001. This is because with numbers of this order magnitude, the distance between two consecutive floating point numbers is larger than 0.001.

For example, here is the trace for (sqrt 12345678901234):

iteration guess (improve guess x) (- (square guess) x)
0 1.0 6172839450617.5 -12345678901233.0
1 6172839450617.5 3086419725309.75 3.8103946883087414e+25
2 3086419725309.75 1543209862656.875 9.525986720768766e+24
3 1543209862656.875 771604931332.4375 2.3814966801891054e+24
4 771604931332.4375 385802465674.21875 5.953741700441899e+23
5 385802465674.21875 192901232853.10938 1.4884354250796105e+23
6 192901232853.10938 96450616458.55469 3.7210885623903846e+22
7 96450616458.55469 48225308293.27734 9.302721402889541e+21
8 48225308293.27734 24112654274.63867 2.3256803476359655e+21
9 24112654274.63867 12056327393.319334 5.8142008382257175e+20
10 12056327393.319334 6028164208.659653 1.4535501786922326e+20
11 6028164208.659653 3014083128.3297105 3.6338751380886356e+19
12 3014083128.3297105 1507043612.1639276 9.084684758802912e+18
13 1507043612.1639276 753525902.074542 2.2711681032851973e+18
14 753525902.074542 376771142.9778979 5.677889394183511e+17
15 376771142.9778979 188401955.01397642 1.4194414850197034e+17
16 188401955.01397642 94233741.7076047 3.548295097418716e+16
17 94233741.7076047 47182376.47141617 8867652397314325.0
18 47182376.47141617 23722017.581583798 2213830970589212.0
19 23722017.581583798 12121224.408177208 550388439239736.8
20 12121224.408177208 6569870.942541849 134578402252156.9
21 6569870.942541849 4224503.311279173 30817525300421.72
22 4224503.311279173 3573450.5222935397 5500749325774.699
23 3573450.5222935397 3514142.3366335463 423869734045.97266
24 3514142.3366335463 3513641.86446291 3517460886.28125
25 3513641.86446291 3513641.8288200637 250472.396484375
26 3513641.8288200637 3513641.8288200637 0.001953125
27 3513641.8288200637 3513641.8288200637 0.001953125
28 3513641.8288200637 3513641.8288200637 0.001953125

If we are lucky, the rounding errors give that (- (square guess) x) is evaluated to exactly 0.0 and the evaluation stops.

If we are not lucky, the gap between two consecutive numbers around (square guess) is more than 0.001 and the assertion good-enough? will never become true. improve has reached a fixed point due to the rounding error, and will always return the same number that is larger than 0.001. For example:

(improve 3513641.8288200637 12345678901234) -> 3513641.8288200637


Increasing the precision to 0.00000001 will even makes things worse, as it triggers issues with even smaller number for x.

#### Small numbers #

The problem is this case is different. We have hardcoded the number of digits of precision we want. It means that we can’t have an accurate answer if x is smaller than the precision of 0.001.

(sqrt 0.00000000123456) = 0.0312500131557789 - Error: 0.0009765620876763541


Looking at the trace, we see that it stops iterating quickly, because the test to check the accuracy of the result think that this is good enough:

iteration guess (improve guess x) (- (square guess) x)
0 1.0 0.50000000061728 0.99999999876544
1 0.50000000061728 0.2500000015432 0.24999999938272
2 0.2500000015432 0.12500000324072 0.06249999953704
3 0.12500000324072 0.06250000655859987 0.01562499957562001
4 0.06250000655859987 0.0312500131557789 0.0039062495852650275
5 0.0312500131557789 0.015625026330841132 0.0009765620876763541

It is like asking to measure the size of a coin, plus or minus one meter. The result can be technically correct, it is still not very useful.

#### Alternative strategy #

The first step is to redefine good-enough? based on the definition from the problem statement:

(define (good-enough? previous-guess guess)
(< (abs (/ (- guess previous-guess) guess)) 0.00000000001))


The number is somewhat arbitrary 0.00000000001 is based on a few trial and error for the upcoming tests. Then we just need to adapt the function sqrt-iter to provide the correct argument:

(define (sqrt-iter guess x)
(if (good-enough? guess (improve guess x))
guess
(sqrt-iter (improve guess x) x)))


Although we could improve the code to avoid computing (improve guess x) twice, the method to do this is not learned yet learned at this time in the book.

The complete solution will look like:

(define (square x) (* x x))

(define (good-enough? previous-guess guess)
(< (abs (/ (- guess previous-guess) guess)) 0.00000000001))

(define (sqrt-iter guess x)
(if (good-enough? guess (improve guess x))
guess
(sqrt-iter (improve guess x) x)))

(define (improve guess x)
(average guess (/ x guess)))

(define (average x y)
(/ (+ x y) 2))

(define (sqrt x)
(sqrt-iter 1.0 x))


It is interesting to note that the new good-enough? does not depend on x anymore.

Now we can try large numbers:

(sqrt 123456789012345) = 11111111.061111081 - Error: 0.015625


and small numbers:

(sqrt 0.00000000123456) = 3.51363060095964e-05 - Error: 4.1359030627651384e-25


In both cases, the error is small relative to the size of the number computed.