# Digital clock problem - not homework- any way to solve other than brute counting?

I was responding to a math problem posted in Games when the thread got deleted, because (probably) it looked like a homework problem to the mod that deleted it, and in all fairness that’s kind of what it smelled like.

Anyway, as I tied to formulate a way to approach the OP’s question it appeared to me that the only way you could approach the problem was by some sort of brute force counting approach. I then wondered if any of the math mavens here could show how there was some more elegant advanced mathematical procedure to solve the problem that would take away all that toil.

The OPs question is gone but here’s what I remember.

A digital clock display varies in brightness over a 24 hour day. Assuming there are 7 illuminated line elements to each LED segment (IE maximum brightness if it was displaying an “8” with all 7 segments lit up) and there are 4 changing segments in the 4 digit display mantissa.

Question 1: What is the single point in the minute to minute display during a 12 hour day that the display will experience the greatest change in brightness going from one hour-minute display to the other?

Question 2: What is the single point in the minute to minute display during a 24 hour day (assuming clock is displaying military time in this scenario) that the display will experience the greatest change in brightness going from one hour-minute display other other ?

My brute force take is to assign 1-7 power values to each number

blank is lowest (no bar ) - 1 is next (1 bar) 8 is strongest (7 bars)

Assign total power values to 720 discrete (12 hours x 60 minutes) 4 digit mantissa states for a 12 hour clock and 1440 discrete 4 digit mantissa states for 24 hour clock.

Then count the power levels for each discrete state and calculate the change in LED illumination power as the clock changes states to find the greatest change.
I don’t see any way to do this other than brute force counting.

Can any of you math mavens do better with your mad math skillz?

Analyze the number of leds lit for each number. This assumes the leds all have the same output, which isn’t a given. This is the first step, I leave the rest for you.

I already stated that determination in the OP.

While the geometry of LED number displays can vary, let’s assume for the sake of this problem argument that all the LED segments have same light output.

And to be clear I’m not looking for the specific answer as much as a determination that it is solvable by other than by means of brute force calculation.

2 bars.

I think it’s possible to significantly reduce the amount of counting necessary. Consider that most of the minute-to-minute display changes are going to be the change of only the last digit, with a smaller number of two-digit changes, followed by an even smaller number of 3- or 4-digit changes. Since 9 and 0 use the same number of elements, each 1-digit change is also represented among the 2, 3 and 4 digit changes (e.g., the change from 1 to 2 in 12:01 -> 12:02 = 12:19 -> 12:20), you can disregard all the 1-digit changes. This means you only have to find the biggest from the following pool of changes:

-five 2-digit changes per hour
-one 3-digit change between each hour (for a 12-hour clock, there would be 11 or 12 of these depending on whether you stop it at 11:59 or let it hit 12:00 again, and 23 or 24 for a 24-hour clock)
-one (two) four-digit changes (9:59 -> 10:00 and 19:59 -> 20:00),

For a total of 17 or 18 changes for a 12-hour clock, and 30 or 31 changes for a 24-hour clock.

:o …umm yeah 2 bars

I believe the biggest increase in illumination is from 7:59 to 8:00.

First three assumptions:
Times like 1:00 are written as 1:00 and not 01:00
The digit 6 has the top bar illuminated so it has six of the seven bars lit.
The digit 9 does not have the bottom bar lit so it has five bars in it.

The first thing I did was look at all possible changes that could take place in a single spot. We have 0->1, 1->2, … 9->0, plus the 5->0 for :59 changing to :00. We’ll handle the 2-> 1 change below.

These changes are

0 to 1 -4
1 to 2 3
2 to 3 0
3 to 4 -1
4 to 5 1
5 to 6 or 0 1
6 to 7 -3
7 to 8 4
8 to 9 -2
9 to 0 1

So any time the units digit changes from 7 to 8 we get a 4 increase. Any time we get a change from x,9 to x+1,0, we get 1 plus the change from x to x+1. But for minutes this only goes through 49 to 50 so the biggest change just in the minute displays is 19 to 20 (+4) or any x7 to x8 (+4).

This leaves us only the twelve hour changes to count, and the biggest one is 7:59 to 8:00 which has a change of 4+1+1 = 6.

I leave the 24 hour clock and biggest decrease to others.

I am not attempting to solve the OP (math is not my strong suite), but I suspect that the appropriate and elegant mathematical solution will involve graph theory and constructing a network of some sort. Or maybe using a big matrix, then reducing that, somehow.

I would go for a computational simulation, personally. CPU time is cheap

Si

I’m still calculating the odds of someone asking a clock question on this particular weekend that has nothing at all to do with Daylight Savings Time.

You can use a light meter.

The biggest change in brightness is going from 7:59 (13 segments) to 8:00 (19 segments), an increase of 6 segments. (The biggest decrease in brightness is 4 segments, of which there are many.)

I have posted an Excel workbook (339K) that shows the derivation.

Figuring the answer for 24-hour time is left as an exercise.

Sorry, I reread the OP and my method is brute force counting. But I don’t think there’s an elegant way to do this.

When I bought a new clock-radio, I noticed that something was different. When it displays “7”, it uses 4 segments, not 3.

XXXXX
XxxxX
XxxxX
xxxxX
xxxxX
xxxxX
xxxxX

If you’re really high class, does your digital clock cross its sevens?

Thanks to all for thier assistance!

It seems like you overlooked the fact that sometimes a 2 can change into a 1 (going from 12:59 to 1:00) and a 1 can turn into a blank (same).

Also, most digital clocks I have seen illuminate the bottom segment for 9.