Spring 2006CISC101 - Prof. McLeod1 Announcements Assn 4 is posted. Note that due date is the 12 th (Monday) at 7pm. (Last assignment!) Final Exam on June

Spring 2006 CISC101 - Prof. McLeod 1

Announcements• Assn 4 is posted. Note that due date is the 12th

(Monday) at 7pm. (Last assignment!)

• Final Exam on June 15, 9am. Room GOO247.

• Assn 3 solution is posted.


Last Time• Variable Scope and Lifetime

• Sorting


Today• Searching

– Sequential search– Binary search

• Source and effects of round-off error.

• Start working on assignment?

• Exam prep (mostly Thursday).


Searching

• Sequential search pseudocode:

• Loop through array starting at the first element until the value of target matches one of the array elements.

• If a match is not found, return –1.


Sequential Search

• As code:

public int seqSearch(int[] A, int target) {

for (int i = 0; i < A.length; i++)if (A[i] == target) return i;

return –1;}


Sequential Search, Cont.• Works on any kind of dataset.• Finds the first matching element – does not say

anything about how many matches there are!

• Best case – one comparison.• Worst case (target not found) – n comparisons (n

is the same as A.length.

• On average - n/2 comparisons.


Searching in an Ordered Dataset• How do you find a name in a telephone book?


Binary Search

• Binary search pseudocode. Only works on ordered sets:

a) Locate midpoint of array to search.b) Determine if target is in lower half or upper half of

array.• If in lower half, make this half the array to search.• If in upper half, make this half the array to search.• Loop back to step a), unless the size of the array to

search is one, and this element does not match, in which case return –1.


Binary Search – Cont.public int binSearch (int[] A, int key) {

int lo = 0;int hi = A.length - 1;int mid = (lo + hi) / 2;

while (lo <= hi) {if (key < A[mid])hi = mid - 1;else if (A[mid] < key)lo = mid + 1;else return mid;mid = (lo + hi) / 2;}return -1;

}


Binary Search – Cont.• For the best case, the element matches right at

the middle of the array, and the loop only executes once.

• For the worst case, key will not be found and the maximum number of loops will occur.

• Note that the loop will execute until there is only one element left that does not match.

• Each time through the loop the number of elements left is halved, giving the progression below for the number of elements:


Binary Search – Cont.• Number of elements to be searched -

progression:

• The last comparison is for n/2m, when the number of elements is one (worst case).

• So, n/2m = 1, or n = 2m.• Or, m = log2(n).• So, the algorithm loops log(n) times for the worst

case.

m

nnnnn2

...,,2

,2

,2

, 32


Binary Search - Cont.

• Binary search with log(n) iterations for the worst case is much better than n iterations for sequential search!

• Major reason to sort datasets!


From Before:• A float occupies 4 bytes• A double occupies 8 bytes of memory• (A byte is 8 bits, where a bit is zero or one.)


Storage of Real Numbers• So, a real number can only occupy a finite

amount of storage in memory.• This effect is very important for two kinds of

numbers:– Numbers like 0.1 that can be written exactly in base

10, but cannot be stored exactly in base 2.– Real numbers (like or e) that have an infinite number

of digits in their “real” representation can only be stored in a finite number of digits in memory.

• And, we will see that it has an effect on the accuracy of mathematical operations.


Storage of “Real” or “Floating-Point” Numbers - Cont.

• Consider 0.1:

(0.1)10 = (0.0 0011 0011 0011 0011 0011…)2

• What happens to the part of a real number that cannot be stored?

• It is lost - the number is truncated.• The “lost part” is called the “roundoff error”.



• Back to the storage of 0.1:• Compute:

• And, compare to 1000.

float sum = 0;for (int i = 0; i < 10000; i++) sum += 0.1F;System.out.println(sum);

10000

1

1.0i



• Prints a value of 999.9029 to the screen.• If sum is declared to be a double then the value:

1000.0000000001588 is printed to the screen.• So, the individual roundoff errors have piled up to

contribute to a cumulative error in this calculation.

• As expected, the roundoff error is smaller for a double than for a float.

• On the following charts, the x-axis is the counter, i, from the code shown above.


float AccumulationProgress of float Roundoff Error

-1.20E-01

-1.00E-01

-8.00E-02

-6.00E-02

-4.00E-02

-2.00E-02

0.00E+00

2.00E-02

4.00E-02

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


double AccumulationProgress of double Roundoff Error

-2.00E-11

0.00E+00

2.00E-11

4.00E-11

6.00E-11

8.00E-11

1.00E-10

1.20E-10

1.40E-10

1.60E-10

1.80E-10

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


Roundoff Error

• This error is referred to in two different ways:

• The absolute error:

absolute error = |x - xapprox|

• The relative error:

relative error = (absolute error) |x|


Roundoff Error - Cont.

• So for the calculation of 1000 as shown above, the errors are:

• The relative error on the storage of 0.1 is the absolute error divided by 1000.

Type Absolute Relativefloat 0.0971 9.71E-5double 1.588E-10 1.588E-13


The Effects of Roundoff Error• Roundoff error can have an effect on any

arithmetic operation carried out involving real numbers.

• For example, consider subtracting two numbers that are very close together:

• Use the function

for example. As x approaches zero, cos(x) approaches 1.

)cos(1)( xxf


The Effects of Roundoff Error• Using double variables, and a value of x of

1.0E-12, f(x) evaluates to 0.0.• But, it can be shown that the function f(x) can also

be represented by f’(x):

• For x = 1.0E-12, f’(x) evaluates to 5.0E-25.• The f’(x) function is less susceptible to roundoff

error.

)cos(1)(sin)()('

2

xxxfxf


The Effects of Roundoff Error - Cont.• Another example. Consider the smallest root of

the polynomial: ax2+bx+c=0:

• What happens when ac is small, compared to b?• It is known that for the two roots, x1 and x2:

aacbbx

242

1

acxx 21


The Effects of Roundoff Error - Cont.• Which leads to an equation for the root which is

not as susceptible to roundoff error:

• This equation approaches –c/b instead of zero when ac << b2.

acbbcx

4221


The Effects of Roundoff Error - Cont.• The examples above show what can happen

when two numbers that are very close are subtracted.

• Remember that this effect is a direct result of these numbers being stored with finite accuracy in memory.


The Effects of Roundoff Error - Cont.• A similar effect occurs when an attempt is made to add a

comparatively small number to a large number:

boolean aVal = ((1.0E10 + 1.0E-20)==1.0E10);System.out.println(aVal);

• Prints out true to the screen• Since 1.0E-20 is just too small to affect any of the bit

values used to store 1.0E10. The small number would have to be about 1.0E-5 or larger to affect the large number.

• So, keep this behaviour in mind when designing expressions!


The Effect on Summations• Taylor Series are used to approximate many

functions. For example:

• For ln(2):

1

1)1()1ln(i

ii

ixx

...41

31

211)1()2ln(

1

1

i

i

i


The Effect on Summations• Since we cannot loop to infinity, how many terms

would be sufficient?• Since the sum is stored in a finite memory space,

at some point the terms to be added will be much smaller than the sum itself.

• If the sum is stored in a float, which has about 7 significant digits, a term of about 1x10-8 would not be significant. So, i would be about 108 - that’s a lot of iterations!


The Effect on Summations - Cont.• On testing using a float, it took 33554433

iterations and 25540 msec to compute! (sum no longer changing, value = 0.6931375)

• Math.log(2) = 0.6931471805599453• So, roundoff error had a significant effect and the

summation did not even provide the correct value. A float could only provide about 5 correct significant digits, tops.

• For double, about 1015 iterations would be required! (I didn’t try this one…)

• So, this series does not converge quickly!


The Effect on Summations - Cont.

• Here is another way to compute natural logs:

• Using x = 1/3 will provide ln(2).

0

12

1212

11ln

i

ixix

x


The Effect on Summations - Cont.

• For float, this took 8 iterations and <1msec (value = 0.6931472).

• Math.log(2) = 0.6931471805599453• For double, it took 17 iterations, <1 msec to give

the value = 0.6931471805599451• Using the Windows calculator ln(2) =

0.693147180559945309417232121458177 (!!)• So, the use of the 17 iterations still introduced a

slight roundoff error.• The second version is much faster!!


Numeric Calculations• Error is introduced into a calculation through two

sources (assuming the formulae are correct!):– The inherent error in the numbers used in the

calculation.– Error resulting from roundoff error.

• Often the inherent error dominates the roundoff error.

• But, watch for conditions of slow convergence or ill-conditioned matrices, where roundoff error will accumulate and end up swamping out the inherent error.


Numeric Calculations - Cont.• Once a number is calculated, it is very important

to be able to estimate the error using both sources, if necessary.

• The error must be known in order that the number produced by your program can be reported in a valid manner.

• This is a non-trivial topic in numeric calculation that we will not discuss in this course.

Documents

Spring 2006CISC101 - Prof. McLeod1 Announcements Assn 4 is posted. Note that due date is the 12 th (Monday) at 7pm. (Last assignment!) Final Exam on June