Question

i have a problem with the code i have written to find the squareroot of a number
here is the code http://pastebin.com/0Dr5yAkk
the problem is that say if we enter 4 the computer should leave the loop for ans=2 but it takes one more guess...(please just make one change in code increase the ans by 0.001 instead of 0.01)

Answer 1

there should be in the reading materials for the course an explanation of dealing with floating point errors. i think the 2008 course had this mentioned in one of the lectures, i would bet that one of the early 2011 lectures does also. I can't recall but there is probably an explanation in the Tutorial in the Python documentation which should be installed on your computer; also there is probably something in "How to Think Like a Computer Scientist...."

your problem is at line 9 and it is due to floating point inaccuracies.

Answer 2

even if the problem is in line 9 , then say for x=4 when the ans reaches value 2 it must exit the while loop but rather it makes one more guess 2.001 and then exits the loop(i had checked by introducing a print command in the while loop portion)

Answer 3

when i run it, it stops at ans = 2.0
http://dpaste.com/803065/

Answer 4

thanks anyways but i forgot to mention just increase the ans by 0.001 and then see the difference. it`ll give 2.001 as the ans for x=4
and by the way what did you do in line 7 could you explain
can tell me a more competent (perhaps a better one) code that will also take into consideration the floating point error.

Answer 5

line 7 is prints a formatted string
7.1.3, 7.1.3.1 and 7.3.1.2 in the python standard library documentation shows how to use the string format() method. \t is a tab character
http://docs.python.org/library/string.html#format-string-syntax
http://docs.python.org/release/1.6/ref/strings.html

Answer 6

again, the problem is because of floating point errors. this shows why ans has to go to 2.001 - http://dpaste.com/803795/

1.007 cannot be accurately stored as a binary number - the value that gets stored is a little less than 1.007. after that, ans never catches up, it is always a bit less than it should be so it has to go one more iteration to satisfy the the while condition:
http://dpaste.com/803800/

Answer 7

this is why it is sometimes necessary to define an accuracy tolerance for testing floating point results. - http://dpaste.com/803804/

Answer 8

thank you !
finally i got the concept of floating point error