From: ben.usenet@bsb.me.uk
Mike Terry writes:
> On 24/12/2022 14:21, Tim Rentsch wrote:
>> Ben Bacarisse writes:
>>
>>> Tim Rentsch writes:
>>>
>>>> ram@zedat.fu-berlin.de (Stefan Ram) writes:
>>>>
>>>>> ram@zedat.fu-berlin.de (Stefan Ram) writes:
>>>>>
>>>>>> Given n times of the 24-hour day, print their average.
>>>>>> For example, the average of "eight o'clock" and
>>>>>> "ten o'clock" (n=2) would be "nine o'clock".
>>>>>> (You can choose any representation, for example "HH:MM"
>>>>>> or "seconds since midnight".)
>>>>>
>>>>> Thanks for all replies!
>>>>>
>>>>> I waited a few days before answering to allow
>>>>> sufficient time to think about the problem.
>>>>>
>>>>> There were not enough tests written and run. As a result,
>>>>> the puzzle has not yet been solved (unless I have overlooked
>>>>> a contribution or misworded expectations).
>>>>>
>>>>> So, here are two possible test cases.
>>>>>
>>>>> average( 23.5, 1.5 )== 0.5
>>>>> average( 11.5, 13.5 )== 12.5
>>>>>
>>>>> (I use hours as units, so "0.5" means, "half past midnight".)
>>>>>
>>>>> I hope that these test cases encode sensible expectations
>>>>> for an average of two times on a 24-hour clock in the spirit
>>>>> of the example given in the OP, which was, "the average of
>>>>> eight o'clock and ten o'clock would be nine o'clock", since
>>>>> these test cases just have rotated that example by 3.5 and
>>>>> 15.5 hours.
>>>>>
>>>>> I believe that I have not seen an algorithm so far in this
>>>>> thread that would pass these tests.
>>>>
>>>> As before, the problem is underspecified.
>>>
>>> Some remarks not specifically in reply to you, Tim...
>> I didn't really say very much. Your comments are a lot more
>> interesting.
>>
>>> The input is a collection, t(n), of n > 1 numbers in [0, 24). The
>>> average should be a number, A, in [0, 24) that minimises
>>>
>>> Sum_{i=1,n} distance(A, t(i))
>>>
>>> (or Sum_{i=1,n} difference(A, t(i))^2 if you prefer to think in terms
>>> of variance). So far, this is just what an average is. The key point
>>> is what is the distance (or difference) whose sum (or sum of squares)
>>> we want to minimise? For times, I would say it is the length of the
>>> shorter arc round an imaginary 24-hour clock face.
>> Minimizing the square of the distance (aka length of the shorter arc)
>> is one metric, and I think a reasonable one. (Minimizing the absolute
>> value of the distance gives a result more like the median than the
>> mean, IIANM.)
>
> Yes - perhaps Ben was thinking of aking
>
> Sum_{i=1,n} signed_distance(A, t(i)) = 0
>
> rather than minimising the (absolute) distance. That would match up
> with minimising the sum of squares of the distance (mean). Or he just
> meant the median as you say, which is different from the mean. Both
> have plausible "average" properties,
Well, I was trying to be deliberately vague (hence my talking about "an
average" rather than mean, median or whatever) but I was no vague enough
with the formula.
I wasn't thinking of a zero net distance because some metrics (though
probably not applicable in this situation) don't always allow for a zero
sum. I should have referred to minimising some kind distance sum, be it
the absolute value of a signed sum or, as in the solution I was
proposing, the sum of directional differences (i.e. full vectors).
> and meet my "minimal" criteria
> for what I think anyone would require of an "average" applying to a
> modular arithmetic:
>
> - if all x_i are within 5 minutes of each other [ALLOWING FOR WRAPPING
> AROUND THE END OF THE MODULO LOOP] then the "average" will be within
> those same 5 minutes. NOT e.g. 12 hours or 4 hours or whatever away.
And when the x_i are widely spaced the average might be considered
nearly meaningless. If you have {0, 8, 16} hours then what is the
"average"? Are there cases where every set of circular values has a
meaningful "average"? If not, you really need more than a single number
answer.
(These are not questions to you, specifically, I'm interested to know of
interpretations that go beyond the way I see things.)
> - if all x_i are translated by a fixed offset like 1 hour, the average
> will translate by the same offset, i.e. it maintains the same "modulo"
> relation with the translated x_i.
>
> hmm, thinking now I'll add the similar:
>
> - if all x_i are "reflected" about a chosen time, then the new average
> will be the reflected time of the original average.
>
> (those aren't intended to be a definition of a "modular average", just
> to exclude suggestions that blatently lack the right symmetry for such
> an average.)
>
>>> The problem has a natural interpretation in terms of angles. Whatever
>>> the circular quantity is, convert the values to unit vectors round a
>>> circle. For times of day, just scale [0, 24) to [0, 2*pi). The
>>> average is then just the direction of the average vector, converted
>>> back to a time of day.
>> Averaging the "time unit vectors" is another plausible metric.
>
> Yes I think this one would work best in most situations, given that
> the OP didn't have a precise mathematical requirement for his solution
> - probably he wants something that just works reasonably as expected.
> (It's my favourite answer.)
>
> Another way of thinking of this approach is "balancing" the 24-hour
> modular circle, where we've put unit weights at each of the times x_i.
> E.g. we look for a balancing line through the centre of the circle.
Is that the same? For {0, 8, 16} any of the three lines through 0, 8 or
16 balances, but the vector sum interpretation is undefined. And for
{0, 6, 12, 18} there are two balance lines, neither of which seems to be
a natural answer for the average.
(Also, every balance line gives two answers, so how do you pick?)
> [Note times on the circle go from 1-24, not 1-12 like a normal clock.]
> Also, thinking like this, it's equivalent to (conventional) averaging
> of the sin of the angular differences from the average (balancing)
> point, since the sin will give the perpendicular distance of the
> weight from the balancing line. So it still fits with the general
> scheme of minimising distance(x_i, v)^2, but with a slightly different
> metric (distance function) in use.
>
> As sin(x) is close to x when x is small, the vector average solution
> would match closely with the mean (by arc length) solution when the
> times are close together, but diverge more noticeably when the times
> are spread out...
--
Ben.
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