I just attempted to type "cal 2015" at the command line prompt to view this year's calendar. But I mistyped it as "cal 201". However, it gave exactly the same result. So I tried taking it a step further, with "cal 20". Not the same. Oh well. For which years would you get the same result if you left off any number of digits from the right? Assume the "cal" command allowed years of any length, rather than being limited to four-digit years. What about leaving digits off the left? What about deleting any subset of digits whatsoever? (You must leave at least one digit, since "cal" with no argument shows just the current month. And the last remaining digit can't be 0 since there was no year 0.) (Note that "cal" switches to the Julian calendar before 1752. Feel free to disregard this to make the problem more elegant, but please say so if you do so. Thanks.)
Well, I get errors when I type that in. We must be on different operating systems. On Sun, Sep 13, 2015 at 11:56 AM, Keith F. Lynch <kfl@keithlynch.net> wrote:
I just attempted to type "cal 2015" at the command line prompt to view this year's calendar. But I mistyped it as "cal 201". However, it gave exactly the same result.
So I tried taking it a step further, with "cal 20". Not the same. Oh well.
For which years would you get the same result if you left off any number of digits from the right? Assume the "cal" command allowed years of any length, rather than being limited to four-digit years.
What about leaving digits off the left? What about deleting any subset of digits whatsoever?
(You must leave at least one digit, since "cal" with no argument shows just the current month. And the last remaining digit can't be 0 since there was no year 0.)
(Note that "cal" switches to the Julian calendar before 1752. Feel free to disregard this to make the problem more elegant, but please say so if you do so. Thanks.)
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The Gregorian calendar has a period of 400 years, so cal is a function of (year mod 400). Similarly, 'mistyped cal' is a function of (year mod 4000), so the solution set to your puzzle also depends only on the residue class of the year (mod 4000), ignoring the initial finite segment corresponding to the Julian calendar. This reduces the puzzle to a finite check. Sincerely, Adam P. Goucher
Sent: Sunday, September 13, 2015 at 5:56 PM From: "Keith F. Lynch" <kfl@KeithLynch.net> To: math-fun@mailman.xmission.com Subject: [math-fun] Calendar puzzle
I just attempted to type "cal 2015" at the command line prompt to view this year's calendar. But I mistyped it as "cal 201". However, it gave exactly the same result.
So I tried taking it a step further, with "cal 20". Not the same. Oh well.
For which years would you get the same result if you left off any number of digits from the right? Assume the "cal" command allowed years of any length, rather than being limited to four-digit years.
What about leaving digits off the left? What about deleting any subset of digits whatsoever?
(You must leave at least one digit, since "cal" with no argument shows just the current month. And the last remaining digit can't be 0 since there was no year 0.)
(Note that "cal" switches to the Julian calendar before 1752. Feel free to disregard this to make the problem more elegant, but please say so if you do so. Thanks.)
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On Sun, 13 Sep 2015, Keith F. Lynch wrote:
Assume the "cal" command allowed years of any length, rather than being limited to four-digit years.
Sometime around 4140 AD, the Gregorian rule will lose synchronization with the equinoxes and an extra Feb 29 will be needed. So the result depends on what fix humanity adopts. -- Tom Duff. Those that can, do. Those that can't, distribute.
Assuming that homo sapiens is still around ... WFL On 9/14/15, Tom Duff <td@pixar.com> wrote:
On Sun, 13 Sep 2015, Keith F. Lynch wrote:
Assume the "cal" command allowed years of any length, rather than being limited to four-digit years.
Sometime around 4140 AD, the Gregorian rule will lose synchronization with the equinoxes and an extra Feb 29 will be needed. So the result depends on what fix humanity adopts.
-- Tom Duff. Those that can, do. Those that can't, distribute.
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The doomsday argument suggests considering oneself as a random sample from the total number of humans to be born. Estimating the number of humans born so far at 100 billion and the number of annual births at ~0.13 billion per year (assuming stabilization around 10 billion, if current trends continue) a 95% confidence interval suggests that humans will be around at least 20 more years but not more than 30,000 years. Adjust assumptions to suit. Charles Greathouse Analyst/Programmer Case Western Reserve University On Mon, Sep 14, 2015 at 4:54 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Assuming that homo sapiens is still around ... WFL
On 9/14/15, Tom Duff <td@pixar.com> wrote:
On Sun, 13 Sep 2015, Keith F. Lynch wrote:
Assume the "cal" command allowed years of any length, rather than being limited to four-digit years.
Sometime around 4140 AD, the Gregorian rule will lose synchronization with the equinoxes and an extra Feb 29 will be needed. So the result depends on what fix humanity adopts.
-- Tom Duff. Those that can, do. Those that can't, distribute.
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I don't have any better method of estimating these things. But the best method may not be that good. I strongly suspect any such methods will have very little reliability, just because conditions for permitting or excluding global annihilation seem to change very fast, plus be accelerating nowadays. —Dan
On Sep 14, 2015, at 7:52 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
The doomsday argument suggests considering oneself as a random sample from the total number of humans to be born. Estimating the number of humans born so far at 100 billion and the number of annual births at ~0.13 billion per year (assuming stabilization around 10 billion, if current trends continue) a 95% confidence interval suggests that humans will be around at least 20 more years but not more than 30,000 years. Adjust assumptions to suit.]
Let (G, *) be a finite group. 1. Then (say by left multiplication) G acts as a group LG of permutations on itself: LG := {L_g | g in G} is a subgroup of S(G), where L_g(x) := gx for all x in G and all g in G, and S(G) is the group of all permutations of the set G. 2. Definition: ----------- A left-invariant metric D on G is one such that for every g in G, the permutation L_g: G -> G is an isometry. 3. There is always a left-invariant metric D on G. For, pick any metric d on G, and average it over the action of LG: D(x,y) := (1/|G|) Sum_{g in G} d(L_g(x), L_g(y)) Then it readily follows that D(L_g(x), L_g(y)) = D(x,y) for every g in G. 4. Question: --------- Which finite groups G have some left-invariant metric D whose full group of isometries is precisely the left multiplications Isom((G,D)) = {L_g | g in G} ??? 5. Example: -------- Let G be the alternating group A_3. Every left-invariant metric D makes G into an equilateral triangle. But every equilateral triangle has a 6-element isometry group, S_3. —Dan
participants (7)
-
Adam P. Goucher -
Charles Greathouse -
Dan Asimov -
Fred Lunnon -
James Buddenhagen -
Keith F. Lynch -
Tom Duff