Two digital clocks

[MatthewV]

So I someone had two digital clocks, what would be the fewest possible lighted areas which could still give all the correct times at a glance?

The number on the clocks look like this:


Both clocks keep perfect time. They are 12-hour clocks.

This problem came to me while sitting in a recliner looking at the microwave and oven clocks. I do not have the solution yet (and I really must do my homework first!)

My starting idea: The time on the clocks have a known offset

[Changabooniggiwan*]

Not wishing to sound rude, MatthewW, but I'm completely confused as to what your puzzle actually requires. Could you rephrase it?

[Jack_Ian]

It just needs one lighted area. Just enough to see my watch in the dark.
I must admit I was confused by this puzzle too. I fail to see the significance of needing two clocks. Also do they still need to show the numbers or can the blocks be considered as 7-bit binary numbers?

[Zag]

I believe he is saying that each numeric position has the 7 lights that make up the number, but some of them are not working. What is the fewest number of lights, total, that you need working. You can set the clocks to be, say, 1:11 off from each other, which means, I think, that you only need one of the lights in each digit on each clock working, so 8 lights total.

[Lepton*]

I frustrated myself for about an hour last night trying to figure this out. I'm somewhat confident that it can be done with 12 lighted areas. The first digit only needs one, as it's either null or 1. The second digit varies from 0 to 9, so it needs 4 lighted areas. The third needs only 3, and the fourth needs 4 again.

The question, though, is which lighted areas? You need two clocks with a time offset, because there is no way to choose 4 lighted areas from the 7 of a single digit such that each of the 10 numbers is uniquely specified.

Zag: which lighted areas? If you choose the bottom left area, is that 0/1, 2/3, 6/7, or 8/9? In any case, you're "wasting" a lighted area on the first digit -- you only need a single lighted area to differentiate between null and 1.

An absolute minimum would be log_2(60*12), which gives 8. But to do it in less than 12, you'd need some sort of interference between the four different digits, and I don't see how to do that.

There's a very real chance I don't understand what's being asked.

[Zag]

Oh, I hadn't actually figured it out, yet. I was waiting for confirmation from MattV that I was at least trying to solve the right problem.

I was assuming a 24-hour clock, which is why you still need 2 for the first digit.

Of course you need more lights than I was saying -- considering the lights simply as bits:

The first digit has 3 states, so you'll need a minimum of 2 lights.
The second and fourth digits have 10 states, so you'll need at least 4 lights each.
The third digit has only 6 states, so you should be able to do it with only 3 lights.

so, 13 lights total. I'll leave the solution of which 13 to the grad student, but I'm sure it can be done. I sort of doubt you even need two clocks, so maybe this isn't the problem MattV intended.

[Jack_Ian]

[MatthewV]

[Zag]

[MatthewV]

The solution for the 12 hour clock and the solution for a 24 hour clock will be different for the first two digits. After that, they will be the same.

[Zag]

Ok. Some real analysis. First I'm just going to consider the second and fourth digits, which can go from 0 to 9. With 4 bits, we should be able to distinguish between 16 values, so 10 values should be easy.

Edit: HAH! I think I proved it to be impossible, but I rambled along on my journey to get there. You might want to skip to the summary.

(Considering only one digit) my first thought was that you could choose 3 lights on one of the clocks, and then set the other clock to a known offset so that you only need one light on the other clock to differentiate. However, I can show that won't work.

I'll use letters for the the 7 lights top to bottom and left to right (so the three horizontal ones are A D G.

Now, consider, for each light, what numbers activate it. We'd like to find lights that activate on exactly 5 of the numbers.

A. (top horizontal) 0 2 3 5 6 7 8 9 (or eight of the numbers)
B. (top left vert) 0 4 5 6 8 9 (six! more promising)
C. (top right vert) 0 1 2 3 4 7 8 9 (eight again)
D. (mid horizontal) 2 3 4 5 8 9 (six)
E. (bot left vert) 0 2 6 8 (four)
F. (bot right vert) 0 1 3 4 5 6 7 8 9 (yuck, everything but 2)
G. (bot horizontal) 0 2 3 5 6 8 9

So, we want to pick three of these lights that leave us with no more than two numbers in any one identical set. Then we can differentiate with one light on the other clock.

The most obvious choice is the pick the lights B, D, and E, but 4, 5, and 9 are identical with that arrangement.

BDG: 5, 6, 8, 9 are identical

BEG: 0, 6, 8 are identical

DEG: 268 and 359 are identical sets

I didn't try other combinations, since separating out only 2 of the 10 digits didn't seem significant enough.

-----------

So let's find a set of two lights that give us groups of no more than 4, and we'll try to differentiate with two lights on the other clock.


BD: leaves 4 5 6 8 9 together
BG: leaves 0 5 6 8 9 together
DG: leaves 2 3 5 6 8
BE: promising: {068} {137} {459} are identical groups. Suppose we use the same two lights on the other clock, and it were set to show the number x before the real number,
x=1 no, 06 are still the same (because 59 are the same)
x=2 no, 08 become 86
x=3 no, 06 become 73
x=4 no, 17 become 73
x=5 no, 68 become 13
(by modulo symmetry, the rest won't work.)

EG: also promising: {0268} {147} {359}
Same analysis, assuming EG on both clocks, and an offset of x
x=1: 06 -> 95
x=2: 08 -> 86
x=3: 26 -> 93
x=4: 06 -> 62
x=5: 08 -> 53

Ok. So, let's assume that the correct clock uses BE and the one that is set back by x uses EG

BE: {068} {137} {459} {2}
EG: {0268} {147} {359}

x=0: 68->68
x=1: 06->95
x=2: 08->86
x=3: 68->35
x=4: 06->62
x=5: 08->53
x=6: 68->02
x=7: 06->39
x=8: 08->20
x=9: 06->17

Ugh.

DE: {17} {3459} {268} {0} also might do it, either against itself or one of the other two

DE against itself
x=1: 45->34
x=2: 39->17
x=3: 59->26
x=4: 39->95
x=5: 49->94

sigh
BE: {068} {137} {459} {2} against
DE: {17} {3459} {268} {0}
x=0: 45->45
x=1: 45->34
x=2: 08->86
x=3: 68->35
x=4: 06->62
x=5: 08->53
x=6: 59->93 (rats! so close!)
x=7: 06->39
x=8: 06->28
x=9: 06->17

DE: {17} {3459} {268} {0}
EG: {0268} {147} {359} against
this can't possibly have any positive results, because we would need an offset that puts each member of the DE group {3459} into a different group of EG, but there are only three groups in EG.

So it must use of the other pairs of lights. At a quick glance (just looking for both-lights-on combo)
AB: {05689}
AC: {023789}
AD: {23589}
AE: {0268} {3579} {14} (Impossible unless another set has groupings with counts 3, 3, 2, 2)
AF: {0356789}
AG: {0235689}

BC: {0489} {1237} {56}
BD: (45689}
BE: (above) {068} {137} {459} {2}
BF: {045689}
BG: {05689}

CD: {23489}
CE: {13479} (C and ~E)
CF: {0134789}

DE: (above) {17} {3459} {268} {0}
DF: {34589}
DG: {235689}

EF: {134579} ~E and F
EG: (above) {0268} {147} {359}

FG: {35689}

Well, I'm done poking at this. Unless I've made a mistake (which seems highly likely), there's no way to differentiate a digit with only 4 lights total between two clocks. It feels wrong, somehow, but I believe it is exhaustive.

Summary:
Although 4 bits seems to be enough to represent 10 different states, with the lights as listed, you would have to be able to find two lights that split the numbers up into groups with no more than 4 values in each group (where all the numbers in a group appear identical with only those lights available). There are only 5 such pairs of lights. (If any group has more than 4 members, then there will be no way to differentiate amongst them with only 2 more bits of information.)

AE: {0268} {3579} {14}
BC: {0489} {1237} {56}
BE: {068} {137} {459} {2}
DE: {17} {3459} {268} {0}
EG: {0268} {147} {359}

We then need to take each of these groups and either find one additional light which splits them up into groups of no more than two (impossible) or else find a combination of lights on one clock and lights on another clock, plus a known offset, which gives a unique combination for every value.

AE and BC can't possibly be used, because they each have two groups of 4. There is no way to "dole out" all four values of both the groups of 4 into different groups in the other set, even using an offset. (This is the hardest concept to explain, but if you had done all the hand-work I did above, it would be clear. I only understood it about halfway through it all.) Consider when I was comparing DE to BE, and the {3459} group of DE. There would have to be an offset that puts the 3 into one of the BE groups, the 4 into another, the 5 into another, and the 9 into another. Otherwise you'll have two digits that look identical on one clock also look identical on the other clock. It could be possible to do this, as long as the offset is converting one of 3459 to be 2, the single-member group. However, consider AE's two groups {0268} and {3579}. Whatever offset correctly splits up {0268} into the 4 separate groups can not possibly also split up {3579}.

So, just considering the light combos BE, DE, and EG, I considered that one clock has one set and the other clock has the other set, and there is some offset x in the time between them. I tried all 6 combinations with all possible offset and was able to identify one pair of numbers that are identical on both clocks. Unless I made a mistake (which is highly likely) I think it is impossible with only 4 lights on 2 clocks, to distinguish all numbers uniquely.

I suspect that if you draw the 9 without the bottom line, it will be possible. (That is, a 9 uses the lights ABCDF instead of ABCDFG.) That approach divides up the groups much more favorably.

[Zag]

[Jack_Ian]

[Jake2*]

Nice analysis, Zag.

For the third digit, which only ranges between 0 and 5, I think I can confirm that it can be done with the minimum three bits. For example, use lights A and B on one clock and light E on the other, no time offset required. Don't think you can do it with just one clock.

[Jake2*]

Oh, and I found that so quickly that I suspect that whatever time offset you end up using, there will be a 3-light solution for the third digit.

[Jake2*]

*sigh*
Scratch that, you can do it on one clock, a couple of ways, e.g. with lights A, B and E.

[Lepton*]

Zag, I've come to a similar conclusion. I guess that, at this point, we have the hypothesis that the minimum of lighted areas required to tell time on a 12-hour clock is 1+5+3+5 = 14. This could be done, for example, with A / ABCDE / ABE / ABCDE on a single clock. Therefore, the clock offset approach offers no clear benefit. I wonder if this is a general rule, or a specific consequence of the patterns we use for numbers.

[Lepton*]

btw, one cannot "bleed" information from the 3rd digit into the 4th by using an offset of 5 (or whatever) and looking to see if the 3rd digit changes. This is the case because it adds a completely new "bit" of information to the 3rd digit, and so you no longer have uniqueness for each of the (now 12) possibilities.

[lostdummy]

Just checked this, and it seems to me that there is version of this problem that could benefit from 'time difference' idea.

If we say that there is just ONE clock, but has option to show two world times (for example by pressing a button - this actually exists on some digital clocks), and we can set independently those two world times.

I think that such clock could do with less active segments than just regular digital clock as before. In previous posts I understood that best case if we look just at minutes (0..59) is 3 segments active for first digit (0..5) and 5 segments needed for second digit (0..9), for total of 8 segments.

With this clock, I *think* that I found solution (was trying again just for minutes, so two digits) that require only 5 active segments:

1) set 2 minutes time difference between clock times
2) first digit need BEG segments (same as before for 0..5)
3) second digit need just B&G segments!

I tried to test most cases and it seemed to work.

For example, most duplicated case for 2nd digit are 0,5,6,8,9 which all light both segments. But their +2 versions are 2,7,8,0,1 respectively, and only 0-8 and 1-7 result again in same segments, but in both pairs only one case also increased first digit, so it is possible to differentiate.

It may be strange why first digit require more segments when it only show numbers 0..5, and main reason for that is previous case when there is carry over. When I tried case with both digits using only segments B&G, and +22 instead of +2, problem was for example to differ between 06 and 08.

In case with 3+2 segments and +2, 06+2=08, 08+2=10 ... they result in same right digit (8 and 0 light both segments) but differ in left digit (1 light no segment, 0 light both).

In case with 2+2 segments and +22, 06+22=28, 08+22=30 ... again same right digit, but this time left digits 2 and 3 are NOT different in segments they light if only 2 segments B&G are on left digit (both 2 and 3 light G only).

What could possibly make only 2+2 segments possible for minutes is some carry over to hours, but I still did not check that one.