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"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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hai all,
i need some algorithm to find the day code(1 to 366) based on the date.
for example, the code for feb 2nd 2007 will be
31+2=33.
for march 2nd 2007 will be 31+28+2=61
and for march 2nd 2008 will be 31+29+2=62 since 2008 is a leap year.
iam sure that, this is not a very big task. but, how simple our logic is, matters.
any ideas please?
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chandu004 wrote: hai all,
i need some algorithm to find the day code(1 to 366) based on the date.
for example, the code for feb 2nd 2007 will be
31+2=33.
for march 2nd 2007 will be 31+28+2=61
and for march 2nd 2008 will be 31+29+2=62 since 2008 is a leap year.
iam sure that, this is not a very big task. but, how simple our logic is, matters.
any ideas please?
There probably is a way, in the language that you use, to subtract two dates and return the result in days. Just subtract 31 Dec of the previous year from the date that you want...
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oh cool,
nice technique.
how ever, i strained my brain for half an hour and wrote a three lined logic to achieve this.
thank you.
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century years rules (of course if you need to deal with):
(1) Century years aren't leap years (though they're divisible by 4).
(2) Century years divisible by 400 (as 2000, for instance) are leap years.
Maybe you have to further strain your brain.
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.
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yes pallini,
i certainly had to deal with it and iam aware of it.
thank you.
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the main important thing how to prove the correctness of ur function
researche in mobile databases
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just supply some valid values to it and see.
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I also recomend testing invalid values and testing for proper failures as well.
Quite often we just test for success and don't test for failure. Failing in a predictable manner of often times more important than being correct. Correct is always predictable. Failing often times is not.
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this validation logic is implemented before the algorithm in my application.
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chandu004 wrote: just supply some valid values to it and see.
Don't let Prof. Knuth know your secret. 
"Normal is getting dressed in clothes that you buy for work and driving through traffic in a car that you are still paying for, in order to get to the job you need to pay for the clothes and the car and the house you leave vacant all day so you can afford to live in it." - Ellen Goodman
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
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The Gregorian calender doesn't have any leapyear exception rules for above 400. IF they did, IIRC the next toggle points would have been 2000 and 10,000. Instead we're adding single leap seconds as needed at new years to keep everything in sync.
--
If you view money as inherently evil, I view it as my duty to assist in making you more virtuous.
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dan neely wrote: IIRC the next toggle points would have been 2000 and 10,000
I doubt that; Wikipedia[^] gives some arguments for a leapyear correction every 4000
or 8000 years.
dan neely wrote: Instead we're adding single leap seconds as needed at new years to keep everything in sync
That can not be correct: with a one second correction every year
you would need 60 * 60 * 24 = 86400 years to shift time by one day, not quite sufficient.
[added] Again Wikipedia[^] explains
the purpose of leap seconds.[/added]
-- modified at 18:06 Thursday 1st November, 2007
Luc Pattyn [Forum Guidelines] [My Articles]
this months tips:
- use PRE tags to preserve formatting when showing multi-line code snippets
- before you ask a question here, search CodeProject, then Google
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Hai friends,
Here is my logic to find the day code.
if any body can please suggest a better logic, or if any body can validate the logic and find any bugs there, i would apprieciate it.
int offset[12]={0,31,59,90,120,151,181,212,243,273,304,334},l=0;<br />
if((((curyear%4==0) && (curyear%100!=0)) || (curyear%400==0)) && (curmonth>1))<br />
l=1;<br />
daycode=offset[curmonth]+curdate+l;
opinions please.
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Why not just use DateTime.DayOfYear?
Luc Pattyn [Forum Guidelines] [My Articles]
this months tips:
- use PRE tags to preserve formatting when showing multi-line code snippets
- before you ask a question here, search CodeProject, then Google
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of course,
but the user will supply date month and year separately.
if i had to use CTime class, then, i had to copy these values to that CTime right?
but i think even that can be possible.
thanks for your suggestion.
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int d=new DateTime(2007, 10, 26).DayOfYear;
Luc Pattyn [Forum Guidelines] [My Articles]
this months tips:
- use PRE tags to preserve formatting when showing multi-line code snippets
- before you ask a question here, search CodeProject, then Google
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thats simply wonderful,
thank you.
--------------------------------------------
Suggestion to the members:
Please prefix your main thread subject with [SOLVED] if it is solved.
thanks.
chandu.
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Here are the x and y values:
y = 25, x = 0
y = 100, x = 50
y = 400, x = 100
What I want is a direct formula that gives continuous values
x = f(y)
that satisfies the condition.
I tried everything including logarithms but could not get a formula yet.
Here is what I am trying to achieve:
If you look at the slider control in Microsoft Office 2007 the slider ranges from 25% to 400% with 100% being in the middle. You can smoothly zoom the UI by moving the slider.
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There are an infinite number of answers to your question, here is one:
x=50(y-25)(y-400)/(-75*300)+100(y-25)(y-100)/(375*300)
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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Yes, I realize that so I added more information to my original post. It has been a while soince I did maths so I have forgot all the terminology and I am not able to express my problem properly. 
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From reading your modified posting, it appears that what you want to preserve is equal ratios for equal distance from the centre (the limits are 1/4 x and 4x relative to the centre), so probably 1/2 x and 2x (i.e. 50 and 200) should also be equal distances from the centre. In this case exponentials are your friend and you should use Luc's solution.
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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Hi,
one exponential relation that satisfies your numbers is:
y = 25 * 2 ^ (x/25)
so the inverse formula would be:
x=25*log(y/25)/log(2)
BTW: it does not matter what the base of the log function is (2, e, 10, whatever).
Luc Pattyn [Forum Guidelines] [My Articles]
this months tips:
- use PRE tags to preserve formatting when showing multi-line code snippets
- before you ask a question here, search CodeProject, then Google
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Awesome, that does the job. Thank you!
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Hi,
for this solultion you could check the very good book numerical recipies in c. You can use the fitting algorithm decribed inside.
http://www.nr.com/
Also you can check Math.Net (for .Net)
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