chronograph math problem

[gwnn]

I have been thinking about the following (pointless) math problem about my chronograph watch and I finally got it.

 

How can you set a standard 2-button chronograph to 0 in darkness (i.e. you don't see which state it is in)?

 

sorry this is the definition of 'standard 2-button chronograph' (i will try to keep it short):

 

the chronograph has a face that shows time and two buttons: "start/stop" (button a) and "reset/split" (button b). if you have a lot of time on your hand you will count 6 8 different states the chron. can be in:

 

wrong description hidden

 

0. initial

1. simple run

2. simple stop

3. background run

4. background run stopped

5. background run stopped and recalled

 

anyway this is what each button does:

 

state 0 1 2 3 4 5

but a 1 2 1 4 4 5

but b 0 3 0 1 5 0

 

 

I have this also in matrix form but I really don't want to put them in. Is there anyone still with me? The question is whether there is a string of a's and b's that gets you to 0 regardless of the initial state.

 

I guess the only remotely, potentially interested people in this problem are watch geeks but yea who knows?

 

edit: the revised states are:

 

0. initial

1. simple run

2. stopped in additive mode

3. running in additive mode (can no longer go to split mode)

4. background running in split mode

5. background stopped in split mode

6. running in split mode

7. stopped in split mode (can only be reset)

 

(split mode: you can freeze one point in time while it is running in the background

additive mode: you stop at one point in time and then can restart it to add more than one interval)

 

effects of buttons:

 

state 0 1 2 3 4 5 6 7

but a 1 2 3 2 5 5 7 7

but b 0 4 0 3 6 7 4 0

Edited January 6, 2012 by gwnn

[helene_t]

The sequence (a,b,b) takes all states to either 0 or 1, while (a,b,b,b) takes them to 0 or 3, so the question is if there is a sequence that takes 0 as well as 1 to 0, or, alternatively, takes 0 as well as 3 to 0.

 

It seems to me to be impossible. I am a rubbish mathematician but I am sure Han can prove this to be the case. (That it is impossible to solve the problem, not that I am a rubbish mathematician, I mean).

 

This R sniplet can be used to experiment:

 

a=0

b=1

f=function(s) {

res = 0:5

for(i in s) {

if(i==a) res = c(1,2,1,4,4,5)[res+1]

if(i==b) res = c(0,3,0,1,5,0)[res+1]

}

return(res)

}

> f(c(b,b,b))

[1] 0 3 0 1 0 0

[WellSpyder]

Yes, sounds like a completely pointless problem, but isn't that what maths is all about? :)

 

Anyway, I found it tough to follow your initial exposition of the problem, but the final description of what state results from applying each button to each initial state is entirely clear. Maybe someone can find a shorter solution, but I've got a 5-step one which I will add as soon as I've worked out how to produce hidden text.....

[gwnn]

BTW I found out a few other states, lol. Turns out a, a, a does not yield 1 and neither does a, b, b.

 

state 0 1 2 3 4 5 6 7

but a 1 2 3 2 5 5 7 7

but b 0 4 0 3 6 7 4 0

 

I had a 7 solution maybe your 5 still works?

[WellSpyder]

Yes, sounds like a completely pointless problem, but isn't that what maths is all about? :)

 

Anyway, I found it tough to follow your initial exposition of the problem, but the final description of what state results from applying each button to each initial state is entirely clear. Maybe someone can find a shorter solution, but I've got a 5-step one which I will add as soon as I've worked out how to produce hidden text.....

 

Sorry, still can't see how to do this - I thought there would be a simple "spoiler" button. Anyway, in trying to set out my solution more clearly I see it doesn't actually work - oops! So I'll have another look. I'd better do some work first, though....

[gwnn]

It is actually rubbish - if I start, stop, restart why can't I use the split button any more and if I press start, split, resume, why can't I use stop, restart? Does anyone have an expensive chrono watch and try?

[gwnn]

(spoiler) solution(/spoiler) with [ ] stuff around

[WellSpyder]

It is actually rubbish - if I start, stop, restart why can't I use the split button any more and if I press start, split, resume, why can't I use stop, restart? Does anyone have an expensive chrono watch and try?

How about getting a watch with a built-in light???

[gwnn]

:-) yes but that's not the point. Suppose I start the clock for my running but then I stop to tie my shoelaces. However, when I restart my watch I can't use the split time any more (aaab does nothing) But I know I am not making myself clear.

[Gerben42]

Take out the batteries and put them in again.

[Trinidad]

Take out the batteries and put them in again.

This separates the physicists from the mathematicians.

 

Rik

[kenberg]

I am having trouble following the current status of this problem. I came late but there are solutions and/or modifications?

 

With

state 0 1 2 3 4 5 6 7

but a 1 2 3 2 5 5 7 7

but b 0 4 0 3 6 7 4 0

 

I think aabbabaaabab (applied in order starting at the left) does it. It would not amaze me if something shorter does it. The aabb takes everything to either 0 or 3, after which

the remaining sequence

 

abaaabab

 

applied to 0 gives

14555770

 

and applied to 3 gives

20123320

 

I have checked it, but errors creep in for such things.

 

But there is a solution of length 7? or 5? As I say, I am not clear on the status.

[gwnn]

My solution is 7 long and works. How did you find it? Mine was mostly based on trial and error. I wonder if this is a known problem in matrix algebra. Basically you have two matrixes A and B and you have to find a way to combine them in a long row of matrix multiplication so that the result is a matrix with exactly one row of all ones (and the rest is zero).

[kenberg]

I found it while expecting to find a proof that it was not possible.

 

The general set up is as follows: Suppose A is some set of numbers from 0 through 7. Push one of the buttons. If you know that before the push the state is one of the numbers in A, then you know that after the push it will be one of the numbers obtained from A. We want a sequence of pushes that ultimately takes S={0,1,2,3,4,5,6,7} to {0}.

 

It was only a little playing around to see that we can quickly get from S to {0,3}. I expected to be able to create a collection C of subsets, each with two elements exactly, such that every application of either button would take each of these pairs into another one of the pairs. For example, b applied to {0,3} gives {0,3}, while a applied to {0,3} gives {1,2}. So then apply a and b to {1,2}. It only took a few minutes to find, contrary to my expectations, a sequence ending in {0,0}={0}. I stopped then. Presumably I could have found other sequences.

 

Definitely the Biblical Method, seek and ye shall find. No insight. If you vary the button functions I have to start from scratch.

[gwnn]

Mine was

babaabb

. Actually I found this for my initial conditions but it also works for the new ones.