1935: "2018"
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 Ken_g6
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Re: 1935: "2018"
No, see, if 2018 were a leap year, we'd get a chance to elect a better President than our current one. But we don't, so it's not.
Re: 1935: "2018"
Also, it's the World Cup of Association Soccerball.
ETA: but, yeah, this stuff is really hard in odd decades.
ETA: but, yeah, this stuff is really hard in odd decades.
xtifr wrote:... and orthogon merely sounds undecided.
Re: 1935: "2018"
orthogon wrote:Also, it's the World Cup of Association Soccerball.
ETA: but, yeah, this stuff is really hard in odd decades.
And the current decade is certainly one of the oddest in ... well, decades.
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Re: 1935: "2018"
If 2018 is a large number to cryptographers, then we really should go back to sending messages by pigeon.
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 Soupspoon
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Re: 1935: "2018"
RFCs 1149, 2549 and 6214…moody7277 wrote:If 2018 is a large number to cryptographers, then we really should go back to sending messages by pigeon.
Re: 1935: "2018"
moody7277 wrote:If 2018 is a large number to cryptographers, then we really should go back to sending messages by pigeon.
It's as large as the font size you choose. (with apologies to The Phantom Tollbooth)
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Re: 1935: "2018"
2018 is not divisible by ANY of the first 49 Mersenne primes. But how about the larger ones? We will probably have to try all of them, just to make sure that 2018 is cryptologically secure.
Re: 1935: "2018"
orthogon wrote:ETA: but, yeah, this stuff is really hard in odd decades.
If it's odd, subtract 2 years and then do the same test as for the even decades. I wouldn't call that really hard –it just takes time to internalize.
Re: 1935: "2018"
Flumble wrote:orthogon wrote:ETA: but, yeah, this stuff is really hard in odd decades.
If it's odd, subtract 2 years and then do the same test as for the even decades. I wouldn't call that really hard –it just takes time to internalize.
I find working modulo 20 works just fine  the only time I might have trouble with it is in 2100, and I can handle the Y21C problems if they arise...
Re: 1935: "2018"
Some of us are old enough to have the full association "leap year = presidential election = Olympics" permanently etched on a portion of our brains.
Re: 1935: "2018"
Presidential elections are held in leap years to allow the maximum amount of time to decide who is the best candidate.
 Soupspoon
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Re: 1935: "2018"
Yup. That works!qvxb wrote:Presidential elections are held in leap years to allow the maximum amount of time to decide who is the best candidate.
Re: 1935: "2018"
Wait, so everyone can't do 2018/4 in their heads? I guess arithmetic isn't that big of a thing.
 Zamfir
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Re: 1935: "2018"
I am not even sure what "doing" a year would be, in my head or otherwise.
Is it something dirty, because I am not allowed those?
Is it something dirty, because I am not allowed those?
Re: 1935: "2018"
ericgrau wrote:Wait, so everyone can't do 2018/4 in their heads? I guess arithmetic isn't that big of a thing.
You don't need to do any arithmetic to work out leap years. You just have to look at the digits of the year and follow a checklist. Like this...
 If the final two digits of the year are zero, use only the first two digits of the year in the following steps,
treating them as if they were the last two digits of the year.  Is the last digit odd? If so, not a leap year.
 Is the last digit is a 2 or 6 and the secondlast digit even? If so, not a leap year.
 Is the last digit a 0, 4, or 8 and the secondlast digit odd? If so, not a leap year.
 Any remaining year is a leap year.
 Soupspoon
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Re: 1935: "2018"
Your process isn't strictly resilient to 5+ digit years. (Nor less than four, nor many of the years we have already seen with four digits, and just don't try applying it to a Roman coin with "<foo> BC" marked on it. )
Re: 1935: "2018"
Then add a rule 0: Pad or truncate the front of the year until it's four digits.
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Re: 1935: "2018"
BrianM wrote:ericgrau wrote:Wait, so everyone can't do 2018/4 in their heads? I guess arithmetic isn't that big of a thing.
You don't need to do any arithmetic to work out leap years. You just have to look at the digits of the year and follow a checklist. Like this...(Reference lists of odd and even digits sold separately.)
 If the final two digits of the year are zero, use only the first two digits of the year in the following steps,
treating them as if they were the last two digits of the year. Is the last digit odd? If so, not a leap year.
 Is the last digit is a 2 or 6 and the secondlast digit even? If so, not a leap year.
 Is the last digit a 0, 4, or 8 and the secondlast digit odd? If so, not a leap year.
 Any remaining year is a leap year.
I think... teaching basic math would be easier.
2000/4=500
18/4=4.5. Wait is this step too hard? I'm scared. This is all a joke right?
2018/4=504.5. Not divisible by 4.
Re: 1935: "2018"
ericgrau wrote:I think... teaching basic math would be easier.
I would say that is basic math. You know all multiples of 100 are divisible by 4, so to tell if a number is divisible by 4 you just need to check if the number formed by the last two digits is. Brian is only writing out what happens with those.
ericgrau wrote:Wait, so everyone can't do 2018/4 in their heads?
Keep in mind not everyone learns these things the same way. Sure, maybe you can tell that means April in your head, but to many Americans that would represent the 2018th month which is a lot harder to interpret.
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Re: 1935: "2018"
chenille wrote:but to many Americans that would represent the 2018th month which is a lot harder to interpret.
Stupid Americans. It's so obviously the 2018th day of April!
(Which is the 9th of October, 5 years hence, roughly three instances out of every four; and 8th Oct the other times.)
 gmalivuk
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Re: 1935: "2018"
1900 and 2100 are divisible by 4 but not leap years.ericgrau wrote:↶BrianM wrote:ericgrau wrote:Wait, so everyone can't do 2018/4 in their heads? I guess arithmetic isn't that big of a thing.
You don't need to do any arithmetic to work out leap years. You just have to look at the digits of the year and follow a checklist. Like this...(Reference lists of odd and even digits sold separately.)
 If the final two digits of the year are zero, use only the first two digits of the year in the following steps,
treating them as if they were the last two digits of the year. Is the last digit odd? If so, not a leap year.
 Is the last digit is a 2 or 6 and the secondlast digit even? If so, not a leap year.
 Is the last digit a 0, 4, or 8 and the secondlast digit odd? If so, not a leap year.
 Any remaining year is a leap year.
I think... teaching basic math would be easier.
2000/4=500
18/4=4.5. Wait is this step too hard? I'm scared. This is all a joke right?
2018/4=504.5. Not divisible by 4.
Re: 1935: "2018"
gmalivuk wrote:1900 and 2100 are divisible by 4 but not leap years.
(Googles) Ah of course 3/4 of every '00 year breaks the rule. Because being off by 0.75 days a century would have been horrible come the year 3000. "Winter came a week early, noooo"
Ok, I'll worry about it in 83 years.
chenille wrote:ericgrau wrote:I think... teaching basic math would be easier.
I would say that is basic math. You know all multiples of 100 are divisible by 4, so to tell if a number is divisible by 4 you just need to check if the number formed by the last two digits is. Brian is only writing out what happens with those.
So... I have to learn that rule to tell me that a number is divisible by 4. And the rule for the next situation, and the rule for the next situation after that... x 1,000. No thanks. Arithmetic is easier.
Last edited by ericgrau on Sat Dec 30, 2017 9:52 pm UTC, edited 1 time in total.
Re: 1935: "2018"
If it’s a multiple of 100: can you halve it twice and still have a multiple of 100? If yes then it’s a leap year, else it’s not.
Else if it’s not a multiple of 100, can you halve it twice and still have a whole number? If yes then it’s a leap year, else it’s not.
All the math ability you need is the ability to read numbers and the ability to halve them.
Else if it’s not a multiple of 100, can you halve it twice and still have a whole number? If yes then it’s a leap year, else it’s not.
All the math ability you need is the ability to read numbers and the ability to halve them.
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Re: 1935: "2018"
Um, "Except only every 4th '00 is a leap year" is easier to remember. 400,800,1200,1600,2000,2400 are still leap years.
So we really dodged a bullet in the year 2000. I might have had to waste time learning about the exception. Except I just did... dammit.
Something tells me this still isn't accurate enough. When's the next '00 year that we have to add/subtract a leap year to get it perfect with the earth's orbit around the sun? I leave it as an exercise for the reader.
So we really dodged a bullet in the year 2000. I might have had to waste time learning about the exception. Except I just did... dammit.
Something tells me this still isn't accurate enough. When's the next '00 year that we have to add/subtract a leap year to get it perfect with the earth's orbit around the sun? I leave it as an exercise for the reader.
Re: 1935: "2018"
Soupspoon wrote:Your process isn't strictly resilient to 5+ digit years. (Nor less than four, nor many of the years we have already seen with four digits, and just don't try applying it to a Roman coin with "<foo> BC" marked on it. )
I suppose we could change rule 1 to read:
 If the year is larger than 1582 and the final two digits of the year are zero, ignore the last two digits of the year in the following steps,and instead treat the thirdlast and fourthlast digits as if they were the last two digits of the year.
Re: 1935: "2018"
ericgrau wrote:Something tells me this still isn't accurate enough. When's the next '00 year that we have to add/subtract a leap year to get it perfect with the earth's orbit around the sun? I leave it as an exercise for the reader.
The commonly proposed rule is to make every year that's an exact multiple of 4000 a nonleap year, making the average length of the year 365.24225 days.
An alternative proposal is to ditch the current practice of skipping leap years at the end of each century and use a more accurate system of skipping a leapyear whenever the year is exactly divisible by 128, making the average calendar year 365.2421875 days.
My personal preference is to ditch the "divisiblebyfour" system and just make any year divisible by 5 or 19 a leapyear. This makes the average calendar year 365.2421056263 days long.
The mean tropical year is measured at 365.24219 days, but is gradually getting smaller because the slowing of the earth's rotation is making the days longer. This means that in the longrun my system would be more accurate than either of those proposals.
(Note, when I say "divisible by 5 or 19" I mean an inclusive or, so that any year divisible by both 5 and 19 would also be a leap year.)
Re: 1935: "2018"
BrianM wrote:(1582 is the year the Gregorian calendar replaced the Julian calendar. In October.)
In some places.
Re: 1935: "2018"
Flumble wrote:orthogon wrote:ETA: but, yeah, this stuff is really hard in odd decades.
If it's odd, subtract 2 years and then do the same test as for the even decades. I wouldn't call that really hard –it just takes time to internalize.
Yes, yes; when I say "really hard", I just mean that it feels way harder than it ought to be, and that's the point I interpreted the comic as making. I can't reliably identify multiples of 4 on sight  I have to go through mental steps of the kind you describe. And things like the World Cup are even harder, since you're looking for odd multiples of 2, and that thinking somehow pollutes my reasoning regarding the easy case (leap years /Olympics). (Working out whether it's the Rugby World Cup is third year work).
xtifr wrote:... and orthogon merely sounds undecided.
Re: 1935: "2018"
ericgrau wrote:Um, "Except only every 4th '00 is a leap year" is easier to remember. 400,800,1200,1600,2000,2400 are still leap years.
So we really dodged a bullet in the year 2000. I might have had to waste time learning about the exception. Except I just did... dammit.
Something tells me this still isn't accurate enough. When's the next '00 year that we have to add/subtract a leap year to get it perfect with the earth's orbit around the sun? I leave it as an exercise for the reader.
In the year 4916 (which would normally be a leap year) the calendar will be 1 day ahead of Nature. It takes 3333.3 years for the Gregorian calendar to accumulate 1 day of error, even using the "only century years divisible by 400 are leap years" rule. So the next "exception" to standard calendar rules will be in 4916 which will need to be skipped as a leap year.
See: http://skylights.org/2016/02/29/qaleapdayexplained/
Of course by that time, climate change will obviate any need to distinguish between calendar and season.
Re: 1935: "2018"
orthogon wrote:Yes, yes; when I say "really hard", I just mean that it feels way harder than it ought to be, and that's the point I interpreted the comic as making.
Yes, yes, we're on the same page regarding that. I mean that it won't feel hard anymore when you internalize "tens are odd, check if unit is 2 or 6/subtract 2 from units and check if multiple of 4", but that takes voluntary effort/repetition/time or a weird situation in which you have to perform that mental process lots of times. That said, doing calcudokus on a daily basis does help with keeping your arithmetic up to speed.
Re: 1935: "2018"
Halve it. Is the result odd?orthogon wrote:I can't reliably identify multiples of 4 on sight
Is that official? Seems to me that in addition to the every four except every four hundred, we need an additional leap day every (about) three thousand years. Maybe it would be easier to just slow down the earth's rotation, so we don't have to do math.Heimhenge wrote:So the next "exception" to standard calendar rules will be in 4916 which will need to be skipped as a leap year.
Sure, it would take a lot of math to do this, but only the engineers would have to do it, and they'd only have to do it once. Otherwise, all of humanity has to deal with dividing by 2889 (or by tomorrow, 2890, depending on from when the discrepancy is noted). So, we'd save a lot of math that way, math that could be put to better use helping robots take over the world.
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Re: 1935: "2018"
I find it funny that you can spend a couple thousand years hashing out the most accurate calendar in human history, and people still complain  some that it's not accurate enough and others that it's too accurate and makes them think too much.
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Re: 1935: "2018"
2018 bits of enthalpy. Nobody will ever bruteforce that.
 gmalivuk
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Re: 1935: "2018"
Wikipedia's value for the tropical year is 365.2421875 days, which differs from the Gregorian definition by 1 day every 3200 years.Heimhenge wrote:↶ericgrau wrote:Um, "Except only every 4th '00 is a leap year" is easier to remember. 400,800,1200,1600,2000,2400 are still leap years.
So we really dodged a bullet in the year 2000. I might have had to waste time learning about the exception. Except I just did... dammit.
Something tells me this still isn't accurate enough. When's the next '00 year that we have to add/subtract a leap year to get it perfect with the earth's orbit around the sun? I leave it as an exercise for the reader.
In the year 4916 (which would normally be a leap year) the calendar will be 1 day ahead of Nature. It takes 3333.3 years for the Gregorian calendar to accumulate 1 day of error, even using the "only century years divisible by 400 are leap years" rule. So the next "exception" to standard calendar rules will be in 4916 which will need to be skipped as a leap year.
If instead we switched to centuries divisible by five as the exceptions to the exception (instead of centuries divisible by 4), years would average 365.242 days and we'd get an additional 2133 years before being off by a day (in the other direction).
If we added another rule saying years divisible by 5000 got two leap days, the average calendar year would be 365.2422 days, and we'd get 80,000 years between additional adjustments.
Of course, by that time the day itself will be more than a full second longer than it is now, which is a much more significant effect than any of these smaller calendar adjustments.
 RAGBRAIvet
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Re: 1935: "2018"
Simple solution — take the last two digits of the calendar year (2018 would be 18, for example). Can you divide that twodigit number evenly by four? If the answer is 'yes', then it's a leap year.
 RAGBRAIvet
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Re: 1935: "2018"
qvxb wrote:Presidential elections are held in leap years to allow the maximum amount of time to decide who is the best candidate.
And even with that, we still blew it in 2016, didn't we?
 GlassHouses
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Re: 1935: "2018"
ericgrau wrote:gmalivuk wrote:1900 and 2100 are divisible by 4 but not leap years.
(Googles) Ah of course 3/4 of every '00 year breaks the rule. Because being off by 0.75 days a century would have been horrible come the year 3000. "Winter came a week early, noooo"
Why do you think the Gregorian calendar reform happened in the first place? Those 0.75 days per century add up, and by 1582 they had been adding up for fifteen centuries. Props to Pope Gregory and his advisors for coming up with a scheme that fixed the calendar, rather than just kicking the can down the road.
Re: 1935: "2018"
RAGBRAIvet wrote:Simple solution — take the last two digits of the calendar year (2018 would be 18, for example). Can you divide that twodigit number evenly by four? If the answer is 'yes', then it's a leap year.
... unless the last two digits are both zero. In which case you do this with the two digits preceding those.
For example; the year 2000 was a leap year, but the year 2100 won't be a leap year.
Re: 1935: "2018"
BrianM wrote:The mean tropical year is measured at 365.24219 days, but is gradually getting smaller because the slowing of the earth's rotation is making the days longer. This means that in the longrun my system would be more accurate than either of those proposals.
In discussions like this one the mean Gregorian year length is often compared to the mean tropical year. However, that comparison may not be appropriate. The Gregorian calendar was created to stabilise the date of Easter, which is tied to the date of the (northern hemisphere) spring equinox. So rather than using the mean tropical year we should be using the vernal equinox year, which is (to 6 decimal places) 365.242374 days.
Astronomer Duncan Steel has discussed this point in several publications, including his popular book Marking Time: The Epic Quest to Invent the Perfect Calendar. Here's a short PDF that summarises his argument: The proper length of a calendar year. There's also some interesting information in his story about the the nonimplemented 33year English Protestant Calendar; there are further details of this intriguing theory in Marking Time.
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