Question

Python sequential and binary search help needed

Asked by: dave_the_bear

You should augment both functions to count the number of comparisons each makes.

To test your work, create a sorted list containing the 1000 elements 1..1000. Write code that searches the list for each value between 1 and 1000. Find the average number of comparisons for each of the techniques.

Sequential search:
def seqsearch(list , value):
    look = 0
    while look < len(list) and list[look] != value:
        look = look + 1
    if look == len(list):
        return "item not found"
    else:
        return look
 
Binary search:
def binarysearch(list ,value):
    left=0
    right=len(list)-1
    middle=(left + right)/2
    while value !=list[middle] and left <= right:
        if value < list[middle]:
                right=middle-1
        else:
                left=middle+1
        middle=(left + right)/2
    if value==list[middle]:
        return middle
    else:
        return "Item is not in list"

                                  
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Asked On
2009-10-22 at 08:14:00ID24834845
Tags

Python Programming Bubble Sort Selection Sort Sorting

Topics

Python Scripting Language

,

Miscellaneous Programming

Participating Experts
1
Points
250
Comments
7

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Answers

 

by: peprPosted on 2009-10-23 at 00:12:48ID: 25641916

Homework? And what is your question? ;)

 

by: dave_the_bearPosted on 2009-10-23 at 20:24:41ID: 25650598

Yup, Ive stated what needs to be done to each function provided.

 

by: peprPosted on 2009-10-24 at 11:19:07ID: 25653504

Well, I can lead you towards the solution. However, YOU have to do the steps. My questions:

  1.  Was the code written by your teacher as it is shown? If yes, I guess that the intention is to use Python as if it was a more general language, i.e. no special features specific to the Python. For example, the while loop is rarely used in Python as it has rather nice, generalized for loop (not of the C-language kind; the Python one is more abstract and more natural to be used).
     
  2. Should the generation of the list be done using the same approach? I.e. to make a list by appending elements in the counted while loop... 
  3. Do you know how to write and run a simple Python program? (I.e. text file with .py extension, launching the program...)
     
Your first step will be to implement generator of the list with values 1..1000. I suggest to write a function that returns the filled list. Actually, there already is such function in Python, but you should write it by hand anyway to understand better what is going on.

You should know that empty list can be constructed as:
    lst = []         # square brackets

Then there is the .append(element) method to add the element to the end of the list. This way you just initialize the empty list and put from 1 to 1000 to the end of the list. You can visualise the content of your lst via simply

   print lst

Try it, and ask your further (related) questions. (I mean ask it here -- as "author comments").

 

by: dave_the_bearPosted on 2009-10-29 at 09:07:18ID: 25695168

The binary search doesnt seem to be working? Have i put the counter in the right place?
It returns the following results:

No. of comparisons: 1
Item is not in list

For the sequential search im getting an invalid syntax error? at the else: after the print "No. of comparisons", count

def binarysearch(list ,value):
    left=0
    right=len(list)-1
    middle=(left + right)/2
    count=0 #Sets a counter
    while value !=list[middle] and left <= right:
        if value < list[middle]:
                count=count+1 #Counts amount of comparisons
                right=middle-1
        else:
                left=middle+1
        middle=(left + right)/2
    print "No. of comparisons:", count #Prints no. of comparisons made
    if value==list[middle]:
        return middle
    else:
        return "Item is not in list"
 
lista=[1,5,3,2,4,6,3]
print binarysearch(lista ,5)
 
--------------------------------------------------------------------
def seqsearch(list , value):
    look = 0
    count = 0 #Sets a counter
    while look < len(list) and list[look] != value:
        look = look + 1
    if look == len(list):
        count=count+1 #Counts amount of comparisons
        return "item not found"
    print "No. of comparisons:", count #Prints no. of comparisons made
    else:
        return look
    
lista=[1,5,3,2,4,6,3]
print seqsearch(lista ,3)

                                              
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by: dave_the_bearPosted on 2009-10-29 at 09:41:16ID: 25695538

OK i figured it out and fixed it, but now i dont know how to search the list range(1, 1001) for each value between 1 and 1000 or how to find the average number of comparisons for each of the techniques?

 

by: peprPosted on 2009-10-29 at 15:06:24ID: 25698731

Firstly, some comment. The identifier 'list' is used for a built-in data type. Nor you and nor your teacher should use 'list' as identifier for a variable. The lst is fine in such situations.

I guess you have found that the binary search requires the sorted list.

Now, you can generate the list simply by range(). For example

lst = range(1, 1001)     # is equal to:   lst = [1, 2, 3, ..., 999, 1000]

The question says in other words "do search for each element in the list". Now it is clear that you must do it in the loop. I suggest to use the Python for loop here like this. I will write it but I expect you to study it, understand it and write the similar loop without the for loop (using the while). See the snippet below.

The reporting could also be enhanced. The print could be moved from inside the functions. The count value can be passed also in return command.

lst = range(1, 1001)
 
for element in lst:
    print seqsearch(lst, element)

                                              
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by: dave_the_bearPosted on 2009-11-01 at 11:59:15ID: 31644575

Thanks, but seeing as it was an assignment i had to hand it in before this. Comments went over my head slightly.

20120131-EE-VQP-002

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