Clock Confusion puzzle: I came across this today.
Clock confusion
Clock
We all know that homemade presents are supposed to be the best, but the clock your Aunty Mabel made you is a little hard to get used to — the hour and minute hands are exactly the same! You can muddle through most of the time but sometimes, say 26 minutes past 2 or just after 12 minutes past 5, you can't tell which of the two times it is. In the 12-hour period between noon and midnight, how many moments are there when it is not possible to tell the time on this clock?
Originally posted by sonhouseHow precise are we getting here? At 12:05, technically you could tell because the hour hand should be 2.5 degrees off of 12, but practically speaking it would be impossible to tell between that and 1:00
Clock Confusion puzzle: I came across this today.
Clock confusion
Clock
We all know that homemade presents are supposed to be the best, but the clock your Aunty Mabel made you is a little hard to get used to — the hour and minute hands are exactly the same! You can muddle through most of the time but sometimes, say 26 minutes past 2 or just after 12 nd midnight, how many moments are there when it is not possible to tell the time on this clock?
*edit* My math was slightly wrong, but my point stands.
Originally posted by forkedknightJust to the minute, no need to count between minutes, there is no second hand.
How precise are we getting here? At 12:05, technically you could tell because the hour hand should be 2.5 degrees off of 12, but practically speaking it would be impossible to tell between that and 1:00
*edit* My math was slightly wrong, but my point stands.
So how many minutes out of that 720 minute set cannot be resolved to a real time hack?
Originally posted by sonhouse12
Clock Confusion puzzle: I came across this today.
Clock confusion
Clock
We all know that homemade presents are supposed to be the best, but the clock your Aunty Mabel made you is a little hard to get used to — the hour and minute hands are exactly the same! You can muddle through most of the time but sometimes, say 26 minutes past 2 or just after 12 ...[text shortened]... nd midnight, how many moments are there when it is not possible to tell the time on this clock?
Originally posted by uzlessNot true; the position of the hour hand between two numbers depends on the position of the minute hand. You would not confuse 9:30 with 6:45 because the minute hand points directly at the 6 in the first case and in the second the hour hand would almost be at 7.
all
Originally posted by AThousandYoungCan we assume we cannot tell the difference between morning and afternoon without a clock?
Not true; the position of the hour hand between two numbers depends on the position of the minute hand. You would not confuse 9:30 with 6:45 because the minute hand points directly at the 6 in the first case and in the second the hour hand would almost be at 7.
Originally posted by TheMaster37You are in an isolated cell, with just a light and no windows. Your clock is atomic regulated but you just don't know which is hour hand and which is minute hand. Also, 12 is definitely wrong.
Can we assume we cannot tell the difference between morning and afternoon without a clock?
Originally posted by greenpawn34This is a special LCD clock, it's a big screen so it's just an image of a clock, they are on the same plane.
The minute hand is always on top of the hour hand.
(Go and check with the clocks in your house.)
So by looking closer at the clock you an determine which is which
and tell the correct time.
Originally posted by sonhouseA far more interesting problem than I first thought!
Clock Confusion puzzle: I came across this today.
Clock confusion
Clock
We all know that homemade presents are supposed to be the best, but the clock your Aunty Mabel made you is a little hard to get used to — the hour and minute hands are exactly the same! You can muddle through most of the time but sometimes, say 26 minutes past 2 or just after 12 ...[text shortened]... nd midnight, how many moments are there when it is not possible to tell the time on this clock?
The first thing to note is that we're looking for clock positions that are symmetrical. The trick is, symmetrical with respect to what? In this case, we need the hour and minute hands to be interchangeable. The simplest example of this is 12:00, where it doesn't matter which hand is which since they both point straight up at the 12. Are there any more positions like this? It turns out that there are lots, but the number is not intuitive.
One major wrinkle in this problem is the fact that the minute hand completes twelve full circles before the hour hand completes its first. This is of course obvious, but it complicates the equations somewhat. However, thinking about symmetry lead me to a graphical solution, which I think is the simplest approach here.
If we draw a graph of the rotation of the minute hand with respect to the rotation of the hour hand (hour hand rotation on the x-axis), we get a shark-tooth pattern that looks something like this:
|. / . . . /
| / . . ./
|/____/____
(Sorry about the dots, they're just spacers...)
Basically, the minute hand will rotate from 0 to 2*pi radians by the time the hour hand rotates (1/12)*2*pi = pi/6 radians, and then it goes back to 0 and starts again. As stated above, we need the clock positions to be symmetrical such that we can interchange the hour and minute hands and the clock will read the same. To find these points of symmetry, we simply draw a new graph with the hour hand rotation on the y-axis and the minute hand rotation on the x-axis and superimpose it on the original (if your left hand were the graph, this is this is equivalent to starting with your hand outstretched with your palm facing out, rotating your hand 90 degrees to the right, and then flipping your hand so your palm faces inward - note the position of your thumb in both cases). What you end up with is diamond-shaped grid with numerous points of intersection. Some of these intersections are important and some aren't, but I'll get to all that after lunch.
More to come shortly! 🙂
- Joined
- 26 Apr 03
- Moves
- 26771
Interesting approach PBE6, I got 66 positions, with the following proof:
It simplifies things to work with the position of a hand represented by a number between 0 and 12, where 0 and 12 are midnight.
The following equation will produce such a number:
Pos_1 = H + X/12 + F/12
where H and X are Integers between 0 and 11 (inclusive), and 0<=F<1
In such a position, H will correspond to hours, and x/12 + f/12 to "fractions of an hour", the definition ensures that the fractional part cannot exceed 1.
So let us assume Hand_A of the clock is at such a position, and first of all let us assume that it is the Hour Hand.
We know that Hour = H, and, working in fractional minutes we can state:
Minutes = 5(X + F)
(note that Minutes cannot exceed 60)
If Hand_A is the Hour Hand, then Hand_B must be the minute hand, and in that case, to show 5(X + F) minutes, it must be at:
Pos_2 = X + F
But, assuming this is an ambiguous time, Hand B could alternatively be an Hour Hand
Noting that X ranges from 0 to 11 and F from O to 1, so from the mapping of position to time, if this is the case the time will be X hours, and F*60 minutes.
So if Hand B showed the hour, the Hand_A would be the minute hand, to show F*60 minutes, it would point to:
Pos_3 = 12F
Because they refer to the same hand, Pos_1 and Pos_3 must be equivalent, so:
12F = H + X/12 + F/12
which rearranges to:
F = (H + X/12)*144/143
We can therefore cancel F out of the first equation and rearrange to give:
Pos_1 = (12H + X)*144/143
So, every combination of H and X is an ambiguous time, giving 12^2 = 144 combinations.
Of these 144, for some Pos_1 and Pos_2 will be identical, meaning that while we don't know which hand is the hour and which is the minute we do know the time. Noting that in Pos 1, H is the hour, and in Pos 2 X is the hour, and for the times to be identical the hours must be equivalent, therefore for these times X=H
So there are 12 combinations where the two hands are in the same place
This leaves 144 - 12 = 132 combinations
But, because there are two ways round the hands can go for each position, there are only half this number of unique combinations so:
Unique, ambiguous hand combinations = 132/2 = 66
- Joined
- 26 Apr 03
- Moves
- 26771
sorry folks, there were many errors in the first versionand I ran out of time to edit it.
Interesting approach PBE6, I got 132 moments, and 66 possible hand configurations, with the following proof:
It simplifies things to work with the position of a hand represented by a number between 0 and 12, where 0 and 12 are midnight.
The following equation will produce such a number:
Pos_1 = H + X/12 + F/12
where H and X are Integers between 0 and 11 (inclusive), and 0<=F<1
In such a position, H will correspond to hours, and x/12 + f/12 to "fractions of an hour", the definition ensures that the fractional part cannot exceed 1.
So let us assume Hand_A of the clock is at such a position, and first of all let us assume that it is the Hour Hand.
We know that Hour = H, and, working in fractional minutes we can state:
Minutes = 5(X + F)
(note that Minutes cannot exceed 60)
If Hand_A is the Hour Hand, then Hand_B must be the minute hand, and in that case, to show 5(X + F) minutes, it must be at:
Pos_2 = X + F
But, assuming this is an ambiguous time, Hand B could alternatively be an Hour Hand
Noting that X ranges from 0 to 11 and F from O to 1, so from the mapping of position to time, if this is the case the time will be X hours, and F*60 minutes.
So if Hand B showed the hour, the Hand_A would be the minute hand, to show F*60 minutes, it would point to:
Pos_3 = 12F
Because they refer to the same hand, Pos_1 and Pos_3 must be equivalent, so:
12F = H + X/12 + F/12
which rearranges to:
F = (H + X/12)*12/143
We can therefore cancel F out of the first equation and rearrange to give:
Pos_1 = (144H + 12X)/143
Similarly we can rearrange the second equation to give the pleasingly symmetric:
Pos_2 = (12H + 144X)/143
So, every combination of H and X is an ambiguous time, giving 12^2 = 144 ambiguous moments.
Of these 144, for some Pos_1 and Pos_2 will be identical, meaning that while we don't know which hand is the hour and which is the minute we do know the time. Noting that in Pos 1, H is the hour, and in Pos 2 X is the hour, and for the times to be identical the hours must be equivalent, therefore for these times X=H
So there are 12 moments where the two hands are in the same place
This leaves 144 - 12 = 132 moments
But, if we assume the hands are identical these, there are only half this number of unique hand patterns so:
Unique, ambiguous hand patterns= 132/2 = 66
ambiguous moments = 132