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Comparing the larger number to the smaller, it is (2.36/1.36)*10^19651 larger. 10^19728 is just less than 2^64K. This can be expressed exactly by an extended arithmetic binary number 64K BITS long, or 8K BYTES long, or 2048 DWORDS long.
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Thank you for taking the time to answer my question. I understand it now. 
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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Ian Uy wrote: Value 1 = 1.36 x 10^58
Value 2 = 2.36 x 10^19,709
Is there any mathematical technique or approach to simplify this?
In that case Value2 - Value1 = Value2 accurate to 19,651 decimal places.
Of course, one would have to question whether percentage increase is a meaningful method of displaying the difference.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Ian Uy wrote: Is there any mathematical technique or approach to simplify this?
Sure!
Given V1 = 1.36 x 10^58 and V2 = 2.36 x 10^19709,
and %Increase = [(V2 - V1)/V1] x 100:
1. (V2 - V1)/V1 = V2/V1 - V1/V1 = V2/V1 - 1
2. V2/V1 = 10^(log V2 - log V1)
3. Subtract 1, then multiply by 100 to get %
so,
1. (2.36 x 10^19709 - 1.36 x 10^58)/1.36 x 10^58 =
(2.36 x 10^19709/1.36 x 10^58) - 1
2. V2/V1 = 10^(log( 2.36 x 10^19709 ) - log( 1.36 x 10^58)) =
10^ (19709.00864 - 58.13354) = 10^ 19650.87510
3. % = (10^19650.87510 -1) x 100.
You're still going to have a hell of a time getting a computer to calculate a number that large, but the use of logarithms reduces the size of the intermediate results to a manageable magnitude.
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
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Roger Wright wrote: You're still going to have a hell of a time getting a computer to calculate a number that larg
I beg to differ, it does not take that long. I made the calculations (I am only reporting the digit count of the various values, but I can report the actual values if desired):
14.984 msec to calculate (14,984 seconds for 1000 loops).
Chars in 10^56 = 57
Chars in 136*10^56 (=V1) = 63
Chars in 10^19707 = 19708
Chars in 236*10^19707 (=V2) = 19714
Chars in (V2-V1) = 19714
Chars in quotient of (V2-V1)/V1 = 19652
Chars in remainder of (V2-V1)/V1 = 63
Chars in quotient of V2/V1 = 19652
Chars in remainder of V2/V1 = 63
Dave Augustine.
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How to implement TREND function of MS Excel in c#.net
thanks
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See, for instance, [^], [^], [^].
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[Image resize DLL]
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Hi ALL
I request some one can help me in generating the different combinations with the given string.
I am having string "ABC" and it need to generate the combinations with the same length string
Example: I am having string like "ABC" and generates Combinations
"ABC"
"ACB"
"BAC"
"BCA"
"CAB"
"CBA"
Same way if I choose 4 length string "ABCD" it should generate like below.
"ABCD"
"ABDC"
"ACBD"
"ACDB"
"ADBC"
"ADCB"
"BACD"
"BADC"
"BCAD"
"BCDA"
"BDAC"
"BDCA"
......should generate sixteen combinations..
length of the string can be dinamic(can be 5 , 6 or more).
Please can some one help me to build algorithm to generate this type of combinations in c#.
Thanks in Advance
Nivas
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SatyaVas wrote: ......should generate sixteen combinations..
There are 24 combinations in this case. It's 4! (4 factorial).
The C++ standard library already contains functions to do this. This code will pull it off:
#include "stdafx.h"
#include <iostream>
#include <algorithm>
void main()
{
using namespace std;
char initial[] = "ABCD";
char *pEnd = initial + sizeof(initial)/sizeof(initial[0])-1;
do
{
cout << initial << endl;
}
while (next_permutation(initial, pEnd));
}
If you look at the source code that implements next_permutation you can easily write a C# version.
Steve
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does anyone have a good formula for calculating azimuth between two city lat and lons
thanks
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winburn wrote: does anyone have a good formula for calculating azimuth between two city lat and lons
Google is your friend. Try the following search terms (without quotes).
formula latitude longitude azimuth
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Good Day,
Given a pseudo code below:
<br />
main()<br />
{<br />
for(int i=0;i<n;i++)><br />
{<br />
x[i]=func1(func2(x[i]));<br />
}<br />
}<br />
<br />
func1(x[])<br />
{<br />
for(int i=0;i<length(x);i++)><br />
}<br />
<br />
func2(x[])<br />
{<br />
for(int i=0;i<length(x);i++)><br />
}<br />
How can I compute for the Asymptotic notation? I know that the loop in main() is O(n) and both func1 and func2 are O(n). But How can I combine the 3? Can I even combine the 3 to get O(N^3)? or it will be just O(3N)?
Thanks!
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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What would be your analysis of this?
main()
{
for(int i=0;i<n;i++)>
{
for(int j=0;j<n;j++)>
{
for(int k=0;k<n;k++)>
{
}
}
}
}
Edit: Ah, better.
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O(n). Right? 
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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How many times do we count to n?
Hint: it's not 3 
main()
{
int count = 0;
for(int i=0;i<n;i++)>
{
for(int j=0;j<n;j++)>
{
for(int k=0;k<n;k++)>
{
count++;
}
}
}
}
If this were a complexity of O(n) then count would be 3n. However it's not, what's the value of count at the end?
I'm not downing you, I'm just trying to help you come to the conclusion yourself.
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hmm. So its O(N^3).
Can you please give a code example that will yield 3N? So I can properly differentiate the two.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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main()
{
int count = 0;
for(int i=0;i<n;i++)>
{
count++;
}
for(int i=0;i<n;i++)>
{
count++;
}
for(int i=0;i<n;i++)>
{
count++;
}
}
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Ah I see.
Thank you!
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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So as you can see now.
When you nest loops you multiply the complexity.
When you have loops side by side, you add.
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When you say we add, that means N+N+N or simply 3N.
Correct me if I'm wrong but there is no such thing as O(3N) but only O(N)?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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That's exactly correct, on both counts.
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leppyr64 wrote: When you nest loops you multiply the complexity.
If the number of iterations on an inner loop is proportional to the size of the problem, and will be the same for all iterations of the outer loop, then the complexity of the outer loop is a factor of N greater than that of the inner loop. The same will be true if the number of iterations for the inner loop will be uniformly distributed over a range, the top of which is proportional to the size of the problem.
There will be occasions, however, where nested loops aren't nearly as bad as they appear because, while the worst-case time of the inner loops is proportional to the size of the problem, they will almost always execute much faster. In Shell Sort, for example, there are three nested loops. In the last iteration of the outer loop, the middle loop will execute N-1 times and the inner loop may (worst-case) execute O(N) times, but the worst-case time is far less than O(N^2).
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Hmmm, how about this nested loop?
main()
{
List<int[]> n = new List<int[]>();
for(int i=0;i<n.count;i++)>
{
n[i]=new int[100];
for(int j=0;j<n[i].length;j++)>
{
for(int k=0;k<n[i].length;k++)>
{
}
}
}
}
Will this still be O(N^3)? The two inner loops is not proportional to the size of the problem.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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This is when you can start talking about O(N*M^2), where M is the "usual" size of an array stuffed in the list. In reality you'd probably have some idea about how many elements are put in the arrays. Say you were storing four integers in each array, and had hundreds of these arrays in the list, you would probably refer to it as O(N), as the M is fairly insignificant. If the arrays contained n.Length items, then it would be ^3...
Big-O notation is a very rough analysis, you are generally interested in the highest power.
Depends on data too, some sort algorithms are worst case O(N^2), but the likely case is much gentler.
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