As 0^2 is 0, I never even considered it to be a perfect square so I stand corrected.
My error; however my code can very easily be changed to cater for it.
As you say it is slow, but it was coded from an observation not from any mathematical set of algorithms, that is, every _line_no_ is the perfect square of the odd number series...
Yes, it was a long post. I hope a few people found the narrative informative.
My code rejected 0 since your code didn't process it correctly. If you'd like to modify my script to accept 0, here is a diff showing the changes needed:
Your code used the observation that the sum of all of the odd numbers from 0 to x is a perfect square. My code used the observation that the square root of any positive x digit number has a square root between 10^((x-1)/2) and 10^(((x-1)/2)+1)-1 inclusive (except for the number 0 [since 10^0 is 1; not 0]).
Your code starts with the lowest perfect square and calculates successive perfect squares until it has a perfect square that is greater than or equal to the number being checked. My code performs a binary search of the possible square roots that could match the target number until it determines that there is a square root in range that is the square root of the target or determines that one does not exist.
For a 63-bit integer value, your code may need to loop through millions of possible perfect squares. A binary search like my code uses needs to loop through no more than 32 possible square roots. My code needs less than that since it restricts the high and low possible square roots based on the number of digits in the target number.
OK, here is another take at finding the square root: it does not use any operation other than subtraction/addition and it is no approximation method.
The math is simple: we know that
or, in the special case of b = 1:
which we can more conveniently write as:
This means that
In less mathematical terms: we have two counters, x and y. We start with x=0, y=1 and we add them both to a sum, incrementing the counters at each pass. Here is a little table:
As you can see the "y"-column always holds the root to the square in the "sum"-column.
To find the root i simply reversed that and subtracted (instead of added) from the value originally given down to 0. The code is very simple:
I have to find last delimiter in each line of a file and store the value after the last '/' in a variable in ksh script...Pls Pls help me:(The file is as shown below:
/opt/apps/cobqa/apps/abadv/bind/advc0007.bnd
/opt/apps/cobqa/apps/abbrio/bind/naac6115.bnd... (5 Replies)
I want to print only the lines that meet the criteria : "worde:" and "wordo;"
I got this far:
sed -n '/\(*\)\1e:\1o;/p;'
But it doesn't quite work.
Can someone please perfect it and tell me exactly how its a fixed version/what was wrong with mine?
Thanks heaps, (1 Reply)
Hi All,
I have a text file which looks like this:
computer programming
systems engineering
I want to get rid of these square brackets and also the text that is inside these brackets. So that my final text file looks like this:
computer programming
systems engineering
I am using... (3 Replies)
I've got an aix-box somewhere on the network and a PC on my desk. Nothing fancy so far.
The PC is made dual-boot:
- windowsXP with putty & winSCP
or
- slackware 13 with xfce4 installed.
The aix-box runs DB2 v8.2 and I've installed db2top to monitor the database.
db2top is a character... (0 Replies)
Here's my work of testing whether a number input is perfect or not..
echo Enter a number
read no
i=1
ans=0
while
do
if
then
ans='expr $ans + $i'
fi
i='expr $i + 1'
done
if
then
echo $no is perfect
else
echo $no is NOT perfect
fi (12 Replies)
Hi All,
I have a file of the following format.
<?xml version='1.0' encoding='utf-8'?>
<tomcat-users>
<role rolename="tomcat"/>
<role rolename="role1"/>
<role rolename="manager"/>
<role rolename="admin"/>
<user username="tomcat" password="tomcat" roles="tomcat"/>
<user... (5 Replies)