Time between Military Hours

I've been writing a program that calculates the hours and minutes between 2 military times.

For example the hours and minutes between 0900 and 1730; 1730 and 0900 the next day. Now I just converted all the times to seconds because I find that to be easier.

this is what I do..

hours for time 1 = time1 / 100
minutes for time 1 = time1 % 100
calculate total no of seconds = (hours * 3600) + (minutes * 60)

hours for time 2 = time2 / 100
minutes for time 2 = time2 % 100
calculate total no of seconds = (hours * 3600) + (minutes * 60)

Now this is where I have trouble.. I know I can do it with an if statement but we aren't allowed to use that yet...

remaining total seconds = seconds in 24 hours - abs((total num of secs for time 1 - total num of secs for time 2)

number of minutes = remaing total seconds / 60
number of hours = num of minutes / 60

minutes left over = number of minutes % 60

Then I output hours and minutes like 8 hours and 30 minutes

So if I take away the number of seconds in a day in this: remaining total seconds = seconds in 24 hours - abs((total num of secs for time 1 - total num of secs for time 2)

Then when I want to find the number of hours and minutes between 0900 -> 1730 I get 8 hours and 30 minutes which is correct.

But if I do 1730 -> 0900 the next morning I still get 8 hours 30 minutes.

And if I leaving the number of seconds in a day in I get the correct answer for 1730 -> 0900 but not for 0900 -> 1730.

I would appreciate any help you can give me this is a puzzle for me..

it is assumed that time 2 is greater than time 1
(actually you need to check that this holds but since you must not use ifs you cant do the check here)

.) convert time 1 to minutes
.) convert time 2 to minutes

tmp = time2- time1;

if i am not mistaken the difference in hours an minutes would be
hours = tmp / 60;
minutes = tmp % 60; // where this is the remainder of the division

Thats the problem it's suppose to work when time1 is greater and less than time2.

Perhaps you could do this (in pseudocode):

Code:

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time2 = (however you found time2);
time2 += (1 day);
timediff = (time2 - time1) % (1 day);

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I think that should work, though I can't really explain it. Something along the lines of... add a day to time2, so if it's smaller than time1 originally then you'll get the time difference from the next day same time, but if it's greater than time1 originally then the % (1 day) will have the effect of basically cutting off the extra 1 day you added to it. An if statement would be more intuitive to use though; the concept of this particular solution here is still a little hazy in my own mind :D Interesting problem.

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...when time1 is greater and less than time2.

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time1 is greater and less than time2?

And/or, as I understood it.

Again, pencil and paper (or word processor) should be of help here. Assume I give you two times in terms of just military time and don't give you a date. The difference is just the larger minus the smaller. But now assume there is a date attached. If the dates are the same, then the difference is still just the larger minus the smaller. But, if the dates are different, then you need to account for that difference. So first determine which date comes firt. Subtract the time from 2400 for that date and store as part1 and add the time from the latter date to part1 to get total. If there are any full days between the two dates, account for that, too.


DateTime date1 = Feb 28, 2005 1320
DateTime date2 = Feb 26, 2005 1910

date1.year == date2.year
date1.month == date2.month
but date1.day is after date2.day, therefore date1 is "larger"/"later"/whatever.
So 2400 - 1910 = 450 + 1320 = 1810 but date1.day is separated from date2 by 1 full day, so need to add 2400 for a result that the two times are 4210 min apart in military time lingo or 42 h and 10 min apart in routine speak.

Now if military time is entered as a single entity then will need to separate digits into separate hour and minute components, whereas if entries are separate to begin with, that won't be necessary.

Now put this all in the appropriate loops and if/else control statements and create an addition/subtraction mechanism to account for base 60 instead of base 10 when it comes time to add/subtract minutes and I think the process can be accomplished.

>>and if/else control statements
Aye, there's the catch.

Quote:

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Now this is where I have trouble.. I know I can do it with an if statement but we aren't allowed to use that yet...

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These 'no ifs' assignments seem to be quite the fad these days, although they do make life unnecessarily difficult. Perhaps, the code would execute slightly faster without having to deal with the branching, but code readability gets shot to hell (as you can see in my example above). Perhaps the idea is to test problem solving skills rather than to test good programming practices.

I've had similar situations where professors won't allow concepts that haven't been taught yet to be used in current assignments. Its usually because the teacher wants to reinforce the topic at hand. For example this question can be done easily in only a line or two with if statements, but I doubt that has anything do do with what this weeks topic is.

Are you only given the two military times with no dates attached where its assumed that there is never more than 24 hours between two times? If so, then another way to solve this is with a for loop that starts at the first time and ends at the second time and counts the minutes until the loop is done. For example, say you were counting the number of days between two calendar dates and you've converted the dates to day number (Jan 1=1, Dec 31=365) then the loop would be like this:

Code:

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int numdays=0;
for(int x=day1; x!=day2; x++)
{
    numdays++;
    x%=365;
}
// numdays should now have the total numbers of days between day1 and day2

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Note, I haven't tested this code so it might not work perfectly, but you should get the general idea.

I don't believe this offers any improvement. Rather, you'll get a semi-infinite loop when day2 < day1 (eventually the loop will stop, when the counter wraps around and comes back). Anyway, this introduces a test condition. If you could use for, you could fake an if statement like so:

Code:

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for(; day2 < day1; )
{
  day2 += (1 day);
  break;
}

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I still see no solution other than the use of modulus.

Well, didn't say it was an improvement, just another idea ;)

It won't be infinite because x will revert to zero as soon as it hits 365 due to the x%=365 line.

Have to admit that I didn't read the OP close enough to pick up the "no ifs" qualification. Sorry about that.
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>>It won't be infinite because x will revert to zero as soon as it hits 365 due to the x%=365 line.

True, it won't be an infinite loop, but when day2 is earlier than day1 then numdays will always == 365 - day2 at the end of the loop, not the difference between day1 and day2.
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I don't see a way to do this if you can't use ifs and you can't be sure that a > b and you have to hard code
result = a - b.

Not to beat a dead horse because Hunter2's solution is more efficient than mine, but my code should work. For example, say we are finding the number of days between May 1st to April 1st which is a difference of 335 days. Ok, day1 would equal 121 and day2 would equal 91. The variable x would start at 121 and increment, when it hits 365 it will revert to zero since 365%365==0. At this point the loop would have executed 244 times thus numdays will equal 244. Now the loop will keep executing 91 more times incrementing numdays 91 more times at which point x will equal day2. So numdays will have a grand total of 244+91=335 which is the correct answer.

>>It won't be infinite because x will revert to zero as soon as it hits 365 due to the x%=365 line.
Oops, missed the mod line. Guess it does work after all :D

>>will always == 365 - day2
No, it will be the number of days from day2 to day1 the next year - the loop condition is !=, not <.

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I don't see a way to do this if you can't use ifs and you can't be sure that a > b and you have to hard code
result = a - b.

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:confused: Does my solution not work? It seemed to work for the 2 test cases I worked out on paper.

As far as I can tell your solution works and is quick and efficient. I doubt there's a better solution than that. Only problem is if he's rarely worked with modulus and he has to turn around and explain his reasoning to his teacher :D

Hehehe, after some thinking I came up with an even more twisted solution:

Code:

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int diff = ((abs(day2 - day1)/(day2 - day1) + 1) / 2 * 365 + day2) - day1

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In more bite-sized pieces (or 4-bite pieces, depending on platform):

Code:

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int diff = day2 - day1;
int sign = abs(diff) / diff;
int shouldWeAdd365OrNot = (sign + 1) / 2;  //1 for yes, 0 for no
int correctedDay2 = day2 + (shouldWeAdd365OrNot * 365);

int finalDiff = correctedDay2 - day1;

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Of course, abs() probably uses if internally, but since it's apparently allowed for the assignment.. heck, why not? :) And, as far as I can see, it strictly uses features that have already been demonstrated to be allowed (i.e. no mod involved, in case that's banned).

I admit my statement "always == 365 - day2" was wrong, and I admit if day1 is May 1st and day2 is April 1st of the preceding year, then the difference is 335 days. However, if day1 and day2 are in the same year, then the difference isn't 335, it's 30. That's the labeling problem identified earier, and if an if statement isn't possible then this approach won't work when in cases when dates are in same year and when dates are in different years.


Likewise, let's say that time1 is 6AM and time2 is 9AM on the same day. Ideally, you'd say 9 - 6 = 3 and you'd be correct. But, what if time1 is 9AM and time2 is 6AM on the same day and your code is hardcoded to use time2 - time1. Now 6 - 9 = -3, which isn't correct, but you could use absolute value to get the correct answer, though there are problems with this approach for times that aren't on the same day. How about using the following with time1 = 9AM and time2 = 6AM of the same day (so we know without a doubt the difference in the two times is 3 hours)?

time1 = 9
time2 = 6
time2 += 24 means time2 = 6 + 24 = 30;
timediff = (time2 - time1) % 24 = (30 - 9) % 24 = 21 % 24 = 21??????????

Which is why I said I can't see a solution that works for all cases of time1 and time2 without using an if statement somewhere.

Edit : I haven't looked at the latest post closely yet, and it's getting busy here at work, so I may not be able to post on that til later.

Edit : Just a brief run I guess. So, my analysis of Hunter2s latest post.

Given:
day1 = day 200 of yr a;
day2 = day 100 of yr a;

The actual difference is 100 days because number of days can't be negative.

Calculated distance per Hunter2 last post:
int diff = day2 - day1 = 100 - 200 = -100;
int sign = abs(diff) / diff = 100/-100 = -1;
int shouldWeAdd365OrNot = (sign + 1) / 2 = (-1 + 1)/2 = 0/2 = 0; //1 for yes, 0 for no
int correctedDay2 = day2 + (shouldWeAdd365OrNot * 365) = 100 + (0 * 365) = 100;

int finalDiff = correctedDay2 - day1 = 100 - 200 = -100????????

I think the confusion lies in what it means for time2<time1. As far as I can tell, if the numerical value of time2 is less than time1 it is assumed that it is the next day. So at no point would 9am to 6am be three hours but will always be 21 hours in which case both hunter's and my code will always work. If this is not the case than I'm not sure how to solve the problem because the OP didn't tell us how the program is to know if its the same day or not.

>>int finalDiff = correctedDay2 - day1 = 100 - 200 = -100????????
Whoops. The problem lies in:

Code:

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int shouldWeAdd365OrNot = (sign + 1) / 2 = (-1 + 1)/2 = 0/2 = 0;

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Since this should be 1 when time2 < time1, i.e. sign == -1, this should be:

Code:

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int shouldWeAdd365OrNot = ((-sign + 1) / 2);  //1 for yes, 0 for no

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Quote:

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but you could use absolute value to get the correct answer, though there are problems with this approach for times that aren't on the same day.

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Indeed, I made the assumption that negative time differences are always interpreted as time2 being the next day, since given the input format there would be no way to determine the intention if the other interpretation were possible.