Mixing
A question about lonely runner conjecture
$begingroup$
After wikipedia:
"Consider $k$ runners on a circular track of unit length. At $t = 0$, all runners are at the same position and start to run; the runners' speeds are pairwise distinct. A runner is said to be lonely at time $t$ if he is at a distance of at least $1/k$ from every other runner at time $t$. The lonely runner conjecture states that each runner is lonely at some time."
I gave some thoughts to it by myself. Could you please tell me if my thinking, presented below, is correct?
Let's simplify the problem making following assumption:
$m_{i}$ - number of full circles which runner $i$ runs within some period of time (let's call it $delta_{t}$), $m_{i}$ is a natural number such that $gcd({m_i})=1$, $i = 1,..,k$.
After time:
$T_{j} = mathrm{lcm}({m_i, i=1,..,ktext{ and }i neq j})$ all of the runners, apart from the runner $j$, cross starting point. Runner $j$ is apart from the rest (i.e. from the starting point) by some distance $d_{j}$.
It is ease to notice that $forall j quad exists n in N$: after time $nT_{j}$ we get: $d_{j} > 1/k$.
Is my thinking correct? That would be a very short proof (assuming of course boundaries put on $m_{i}$).
combinatorics number-theory open-problem
edited Apr 9 '16 at 16:36
user26857
39.5k124283
asked Apr 7 '16 at 9:41
PawelBPawelB
113
$endgroup$
add a comment |
$begingroup$
After wikipedia:
"Consider $k$ runners on a circular track of unit length. At $t = 0$, all runners are at the same position and start to run; the runners' speeds are pairwise distinct. A runner is said to be lonely at time $t$ if he is at a distance of at least $1/k$ from every other runner at time $t$. The lonely runner conjecture states that each runner is lonely at some time."
I gave some thoughts to it by myself. Could you please tell me if my thinking, presented below, is correct?
Let's simplify the problem making following assumption:
$m_{i}$ - number of full circles which runner $i$ runs within some period of time (let's call it $delta_{t}$), $m_{i}$ is a natural number such that $gcd({m_i})=1$, $i = 1,..,k$.
After time:
$T_{j} = mathrm{lcm}({m_i, i=1,..,ktext{ and }i neq j})$ all of the runners, apart from the runner $j$, cross starting point. Runner $j$ is apart from the rest (i.e. from the starting point) by some distance $d_{j}$.
It is ease to notice that $forall j quad exists n in N$: after time $nT_{j}$ we get: $d_{j} > 1/k$.
Is my thinking correct? That would be a very short proof (assuming of course boundaries put on $m_{i}$).
combinatorics number-theory open-problem
edited Apr 9 '16 at 16:36
user26857
39.5k124283
asked Apr 7 '16 at 9:41
PawelBPawelB
113
$endgroup$
1
$begingroup$
Compare also with this question.
$endgroup$
– Dietrich Burde
Apr 7 '16 at 12:06
add a comment |
$begingroup$
After wikipedia:
"Consider $k$ runners on a circular track of unit length. At $t = 0$, all runners are at the same position and start to run; the runners' speeds are pairwise distinct. A runner is said to be lonely at time $t$ if he is at a distance of at least $1/k$ from every other runner at time $t$. The lonely runner conjecture states that each runner is lonely at some time."
I gave some thoughts to it by myself. Could you please tell me if my thinking, presented below, is correct?
Let's simplify the problem making following assumption:
$m_{i}$ - number of full circles which runner $i$ runs within some period of time (let's call it $delta_{t}$), $m_{i}$ is a natural number such that $gcd({m_i})=1$, $i = 1,..,k$.
After time:
$T_{j} = mathrm{lcm}({m_i, i=1,..,ktext{ and }i neq j})$ all of the runners, apart from the runner $j$, cross starting point. Runner $j$ is apart from the rest (i.e. from the starting point) by some distance $d_{j}$.
It is ease to notice that $forall j quad exists n in N$: after time $nT_{j}$ we get: $d_{j} > 1/k$.
Is my thinking correct? That would be a very short proof (assuming of course boundaries put on $m_{i}$).
combinatorics number-theory open-problem
edited Apr 9 '16 at 16:36
user26857
39.5k124283
asked Apr 7 '16 at 9:41
PawelBPawelB
113
$endgroup$
After wikipedia:
"Consider $k$ runners on a circular track of unit length. At $t = 0$, all runners are at the same position and start to run; the runners' speeds are pairwise distinct. A runner is said to be lonely at time $t$ if he is at a distance of at least $1/k$ from every other runner at time $t$. The lonely runner conjecture states that each runner is lonely at some time."
I gave some thoughts to it by myself. Could you please tell me if my thinking, presented below, is correct?
Let's simplify the problem making following assumption:
$m_{i}$ - number of full circles which runner $i$ runs within some period of time (let's call it $delta_{t}$), $m_{i}$ is a natural number such that $gcd({m_i})=1$, $i = 1,..,k$.
After time:
$T_{j} = mathrm{lcm}({m_i, i=1,..,ktext{ and }i neq j})$ all of the runners, apart from the runner $j$, cross starting point. Runner $j$ is apart from the rest (i.e. from the starting point) by some distance $d_{j}$.
It is ease to notice that $forall j quad exists n in N$: after time $nT_{j}$ we get: $d_{j} > 1/k$.
Is my thinking correct? That would be a very short proof (assuming of course boundaries put on $m_{i}$).
combinatorics number-theory open-problem
combinatorics number-theory open-problem
edited Apr 9 '16 at 16:36
user26857
39.5k124283
asked Apr 7 '16 at 9:41
PawelBPawelB
113
edited Apr 9 '16 at 16:36
user26857
39.5k124283
asked Apr 7 '16 at 9:41
PawelBPawelB
113
edited Apr 9 '16 at 16:36
user26857
39.5k124283
edited Apr 9 '16 at 16:36
user26857
39.5k124283
edited Apr 9 '16 at 16:36
user26857
39.5k124283
39.5k124283
asked Apr 7 '16 at 9:41
PawelBPawelB
113
asked Apr 7 '16 at 9:41
PawelBPawelB
113
asked Apr 7 '16 at 9:41
PawelBPawelB
113
113
1
$begingroup$
Compare also with this question.
$endgroup$
– Dietrich Burde
Apr 7 '16 at 12:06
add a comment |
1
$begingroup$
Compare also with this question.
$endgroup$
– Dietrich Burde
Apr 7 '16 at 12:06
1
1
$begingroup$
Compare also with this question.
$endgroup$
– Dietrich Burde
Apr 7 '16 at 12:06
$begingroup$
Compare also with this question.
$endgroup$
– Dietrich Burde
Apr 7 '16 at 12:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As far as I can see, your proof is entirely correct; however, it is not a proof of the full lonely runner conjecture, but rather a proof of a particular easy special case.
In the original statement of the problems, the speeds of the runners are allowed to be arbitrary real numbers. It is a theorem that it is in fact enough to take the speeds of the runners to be rational.
What you have proved is that the conjecture is true if the speeds of the runners are taken to be $1/m_i$, where $m_1,dots,m_k$ are pairwise coprime integers. This is not enough to prove the full conjecture.
As an example of why your approach can't prove the full conjecture, let $v_i=i$ for $i=0,dots,k$. Suppose the runners have been running for time $alpha$. Then, by the Dirichlet approximation theorem, there exist integers $p,q$ with $1le qle k$ such that
$$
|qalpha - p|<frac1k
$$
$qalpha$ measures the distance along the track that runner $q$ has run. Subtracting $p$ means subtracting off a whole number multiple of full laps, so $|qalpha-p|$ can measure the distance between runner $q$ and runner $0$ (who is stuck at point $0$ forever). So in this case there is always some runner within $1/k$ of runner $0$ (where there are $k+1$ runners).
In your case, however, no such bound holds. Indeed, you have situations where all but one of the runners are clustered together at the same point but one of them is far away from the others. So you are considering a special case that can't hope to prove other cases like the one outlined above.
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
$endgroup$
1
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
1
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
1
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
1
$begingroup$
@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
|
show 7 more comments
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$begingroup$
As far as I can see, your proof is entirely correct; however, it is not a proof of the full lonely runner conjecture, but rather a proof of a particular easy special case.
In the original statement of the problems, the speeds of the runners are allowed to be arbitrary real numbers. It is a theorem that it is in fact enough to take the speeds of the runners to be rational.
What you have proved is that the conjecture is true if the speeds of the runners are taken to be $1/m_i$, where $m_1,dots,m_k$ are pairwise coprime integers. This is not enough to prove the full conjecture.
As an example of why your approach can't prove the full conjecture, let $v_i=i$ for $i=0,dots,k$. Suppose the runners have been running for time $alpha$. Then, by the Dirichlet approximation theorem, there exist integers $p,q$ with $1le qle k$ such that
$$
|qalpha - p|<frac1k
$$
$qalpha$ measures the distance along the track that runner $q$ has run. Subtracting $p$ means subtracting off a whole number multiple of full laps, so $|qalpha-p|$ can measure the distance between runner $q$ and runner $0$ (who is stuck at point $0$ forever). So in this case there is always some runner within $1/k$ of runner $0$ (where there are $k+1$ runners).
In your case, however, no such bound holds. Indeed, you have situations where all but one of the runners are clustered together at the same point but one of them is far away from the others. So you are considering a special case that can't hope to prove other cases like the one outlined above.
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
$endgroup$
1
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
1
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
1
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
1
$begingroup$
@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
|
show 7 more comments
$begingroup$
As far as I can see, your proof is entirely correct; however, it is not a proof of the full lonely runner conjecture, but rather a proof of a particular easy special case.
In the original statement of the problems, the speeds of the runners are allowed to be arbitrary real numbers. It is a theorem that it is in fact enough to take the speeds of the runners to be rational.
What you have proved is that the conjecture is true if the speeds of the runners are taken to be $1/m_i$, where $m_1,dots,m_k$ are pairwise coprime integers. This is not enough to prove the full conjecture.
As an example of why your approach can't prove the full conjecture, let $v_i=i$ for $i=0,dots,k$. Suppose the runners have been running for time $alpha$. Then, by the Dirichlet approximation theorem, there exist integers $p,q$ with $1le qle k$ such that
$$
|qalpha - p|<frac1k
$$
$qalpha$ measures the distance along the track that runner $q$ has run. Subtracting $p$ means subtracting off a whole number multiple of full laps, so $|qalpha-p|$ can measure the distance between runner $q$ and runner $0$ (who is stuck at point $0$ forever). So in this case there is always some runner within $1/k$ of runner $0$ (where there are $k+1$ runners).
In your case, however, no such bound holds. Indeed, you have situations where all but one of the runners are clustered together at the same point but one of them is far away from the others. So you are considering a special case that can't hope to prove other cases like the one outlined above.
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
$endgroup$
1
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
1
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
1
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
1
$begingroup$
@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
|
show 7 more comments
$begingroup$
As far as I can see, your proof is entirely correct; however, it is not a proof of the full lonely runner conjecture, but rather a proof of a particular easy special case.
In the original statement of the problems, the speeds of the runners are allowed to be arbitrary real numbers. It is a theorem that it is in fact enough to take the speeds of the runners to be rational.
What you have proved is that the conjecture is true if the speeds of the runners are taken to be $1/m_i$, where $m_1,dots,m_k$ are pairwise coprime integers. This is not enough to prove the full conjecture.
As an example of why your approach can't prove the full conjecture, let $v_i=i$ for $i=0,dots,k$. Suppose the runners have been running for time $alpha$. Then, by the Dirichlet approximation theorem, there exist integers $p,q$ with $1le qle k$ such that
$$
|qalpha - p|<frac1k
$$
$qalpha$ measures the distance along the track that runner $q$ has run. Subtracting $p$ means subtracting off a whole number multiple of full laps, so $|qalpha-p|$ can measure the distance between runner $q$ and runner $0$ (who is stuck at point $0$ forever). So in this case there is always some runner within $1/k$ of runner $0$ (where there are $k+1$ runners).
In your case, however, no such bound holds. Indeed, you have situations where all but one of the runners are clustered together at the same point but one of them is far away from the others. So you are considering a special case that can't hope to prove other cases like the one outlined above.
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
$endgroup$
As far as I can see, your proof is entirely correct; however, it is not a proof of the full lonely runner conjecture, but rather a proof of a particular easy special case.
In the original statement of the problems, the speeds of the runners are allowed to be arbitrary real numbers. It is a theorem that it is in fact enough to take the speeds of the runners to be rational.
What you have proved is that the conjecture is true if the speeds of the runners are taken to be $1/m_i$, where $m_1,dots,m_k$ are pairwise coprime integers. This is not enough to prove the full conjecture.
As an example of why your approach can't prove the full conjecture, let $v_i=i$ for $i=0,dots,k$. Suppose the runners have been running for time $alpha$. Then, by the Dirichlet approximation theorem, there exist integers $p,q$ with $1le qle k$ such that
$$
|qalpha - p|<frac1k
$$
$qalpha$ measures the distance along the track that runner $q$ has run. Subtracting $p$ means subtracting off a whole number multiple of full laps, so $|qalpha-p|$ can measure the distance between runner $q$ and runner $0$ (who is stuck at point $0$ forever). So in this case there is always some runner within $1/k$ of runner $0$ (where there are $k+1$ runners).
In your case, however, no such bound holds. Indeed, you have situations where all but one of the runners are clustered together at the same point but one of them is far away from the others. So you are considering a special case that can't hope to prove other cases like the one outlined above.
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
edited Jan 28 at 13:27
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
answered Apr 7 '16 at 10:02
John GowersJohn Gowers
18.3k44472
18.3k44472
1
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
1
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
1
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
1
$begingroup$
@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
|
show 7 more comments
1
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
1
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
1
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
1
$begingroup$
@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
1
1
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
$begingroup$
Thanks! I am aware of course, as you stated, it is only a special case. This simple thought crossed my mind as I was reading about lonely runner conjecture and I wanted to make sure I am correct/share it.
$endgroup$
– PawelB
Apr 7 '16 at 10:13
1
1
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
OK. You're correct.
$endgroup$
– John Gowers
Apr 7 '16 at 10:16
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
$begingroup$
@John Gowers, Sorry because this may be a stupid question, but may you explain why $|qalpha-p|<frac{1}{k}$ could tell us the method of PawelB's does not make sense in general? I think this is not relevant to the situation: $v_i=i$ for $i=0,1,...,n$ . I think this situation arrive the minimum $frac{1}{n+1}$ is just come from a explicit calculate. I need more details...
$endgroup$
– Hu xiyu
Oct 28 '17 at 17:54
1
1
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
$begingroup$
@Huxiyu Are you asking me for a strategy to prove the conjecture? If I knew the answer, then it wouldn't be a conjecture!
$endgroup$
– John Gowers
Oct 28 '17 at 18:24
1
1
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@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
$begingroup$
@quantum That's a misquotation of what I said, and not what I was getting at at all. Note that the time $alpha$ can be any real number, not just a rational one. There is no contradiction with the conjecture, because the number of runners is not $k$ but $k+1$ (OP's proof was for $k$ runners, but we could easily add another runner with speed $0$ and their bound -- which this example shows does not hold in the general case -- would still hold.)
$endgroup$
– John Gowers
Jan 28 at 12:25
|
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Compare also with this question.
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– Dietrich Burde
Apr 7 '16 at 12:06