Maximum number of umbrellas that can be added in a one kilometer beach?
$begingroup$
Suppose we have a beach of length $1-$km. Suppose one Day $0$, the beach is empty.
One day $1$, a family comes and puts their umbrella at some point in the beach. This point is fixed forever and cannot be moved.
On day $2$, a new family comes and puts their umbrella in a new coordinate. This umbrella is now fixed forever. However, we are guaranteed that the intervals $[0,frac{1}{2}], [frac{1}{2},1]$ each have $1$ umbrella; the old and the new one.
In general, on day $i$, a new family comes in and fixes a new umbrella. We're guranteed that the interval $[0,frac{1}{i}], ..., [1-frac{1}{i}, 1]$ all have one umbrella each.
The question is, what is the maximum day $t$ such that we cannot add a new umbrella where the $t$ sub intervals would have $1$ umbrella each?
The question answer say $18$, but I don't see how this is true?! Shouldn't we be able to keep going forever?
puzzle combinatorial-geometry
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
$endgroup$
|
show 5 more comments
$begingroup$
Suppose we have a beach of length $1-$km. Suppose one Day $0$, the beach is empty.
One day $1$, a family comes and puts their umbrella at some point in the beach. This point is fixed forever and cannot be moved.
On day $2$, a new family comes and puts their umbrella in a new coordinate. This umbrella is now fixed forever. However, we are guaranteed that the intervals $[0,frac{1}{2}], [frac{1}{2},1]$ each have $1$ umbrella; the old and the new one.
In general, on day $i$, a new family comes in and fixes a new umbrella. We're guranteed that the interval $[0,frac{1}{i}], ..., [1-frac{1}{i}, 1]$ all have one umbrella each.
The question is, what is the maximum day $t$ such that we cannot add a new umbrella where the $t$ sub intervals would have $1$ umbrella each?
The question answer say $18$, but I don't see how this is true?! Shouldn't we be able to keep going forever?
puzzle combinatorial-geometry
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
$endgroup$
1
$begingroup$
Can you present a way to place umbrellas with which we can keep going for ever while satisfying the interval condition?
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:41
$begingroup$
Are the families trying to fit this pattern or is the problem claiming it always works? For example, suppose after day $2$ there is an umbrella just slightly to the left of $frac{1}{2}$ and another just slightly to the right of $frac{1}{2}$ so that both are contained in $(frac{1}{3},frac{2}{3})$. How can we place a third umbrella so that the condition is met?
$endgroup$
– John Douma
Dec 20 '18 at 16:58
$begingroup$
The families are trying to fit this pattern for as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 16:59
$begingroup$
So they can agree on where to put their umbrellas from the start to keep this going as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 17:00
1
$begingroup$
I have seen this problem before, and the proof is by distinguishing many many cases and brute force/exhaustion. There is nothing deep, just track which are the possible intervals for the 1st, 2nd, 3rd, etc. umbrellas after the 1st, 2nd, 3rd, etc. round.
$endgroup$
– Servaes
Dec 20 '18 at 17:59
|
show 5 more comments
$begingroup$
Suppose we have a beach of length $1-$km. Suppose one Day $0$, the beach is empty.
One day $1$, a family comes and puts their umbrella at some point in the beach. This point is fixed forever and cannot be moved.
On day $2$, a new family comes and puts their umbrella in a new coordinate. This umbrella is now fixed forever. However, we are guaranteed that the intervals $[0,frac{1}{2}], [frac{1}{2},1]$ each have $1$ umbrella; the old and the new one.
In general, on day $i$, a new family comes in and fixes a new umbrella. We're guranteed that the interval $[0,frac{1}{i}], ..., [1-frac{1}{i}, 1]$ all have one umbrella each.
The question is, what is the maximum day $t$ such that we cannot add a new umbrella where the $t$ sub intervals would have $1$ umbrella each?
The question answer say $18$, but I don't see how this is true?! Shouldn't we be able to keep going forever?
puzzle combinatorial-geometry
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
$endgroup$
Suppose we have a beach of length $1-$km. Suppose one Day $0$, the beach is empty.
One day $1$, a family comes and puts their umbrella at some point in the beach. This point is fixed forever and cannot be moved.
On day $2$, a new family comes and puts their umbrella in a new coordinate. This umbrella is now fixed forever. However, we are guaranteed that the intervals $[0,frac{1}{2}], [frac{1}{2},1]$ each have $1$ umbrella; the old and the new one.
In general, on day $i$, a new family comes in and fixes a new umbrella. We're guranteed that the interval $[0,frac{1}{i}], ..., [1-frac{1}{i}, 1]$ all have one umbrella each.
The question is, what is the maximum day $t$ such that we cannot add a new umbrella where the $t$ sub intervals would have $1$ umbrella each?
The question answer say $18$, but I don't see how this is true?! Shouldn't we be able to keep going forever?
puzzle combinatorial-geometry
puzzle combinatorial-geometry
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
asked Dec 20 '18 at 16:36
AspiringMatAspiringMat
540518
540518
1
$begingroup$
Can you present a way to place umbrellas with which we can keep going for ever while satisfying the interval condition?
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:41
$begingroup$
Are the families trying to fit this pattern or is the problem claiming it always works? For example, suppose after day $2$ there is an umbrella just slightly to the left of $frac{1}{2}$ and another just slightly to the right of $frac{1}{2}$ so that both are contained in $(frac{1}{3},frac{2}{3})$. How can we place a third umbrella so that the condition is met?
$endgroup$
– John Douma
Dec 20 '18 at 16:58
$begingroup$
The families are trying to fit this pattern for as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 16:59
$begingroup$
So they can agree on where to put their umbrellas from the start to keep this going as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 17:00
1
$begingroup$
I have seen this problem before, and the proof is by distinguishing many many cases and brute force/exhaustion. There is nothing deep, just track which are the possible intervals for the 1st, 2nd, 3rd, etc. umbrellas after the 1st, 2nd, 3rd, etc. round.
$endgroup$
– Servaes
Dec 20 '18 at 17:59
|
show 5 more comments
1
$begingroup$
Can you present a way to place umbrellas with which we can keep going for ever while satisfying the interval condition?
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:41
$begingroup$
Are the families trying to fit this pattern or is the problem claiming it always works? For example, suppose after day $2$ there is an umbrella just slightly to the left of $frac{1}{2}$ and another just slightly to the right of $frac{1}{2}$ so that both are contained in $(frac{1}{3},frac{2}{3})$. How can we place a third umbrella so that the condition is met?
$endgroup$
– John Douma
Dec 20 '18 at 16:58
$begingroup$
The families are trying to fit this pattern for as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 16:59
$begingroup$
So they can agree on where to put their umbrellas from the start to keep this going as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 17:00
1
$begingroup$
I have seen this problem before, and the proof is by distinguishing many many cases and brute force/exhaustion. There is nothing deep, just track which are the possible intervals for the 1st, 2nd, 3rd, etc. umbrellas after the 1st, 2nd, 3rd, etc. round.
$endgroup$
– Servaes
Dec 20 '18 at 17:59
1
1
$begingroup$
Can you present a way to place umbrellas with which we can keep going for ever while satisfying the interval condition?
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:41
$begingroup$
Can you present a way to place umbrellas with which we can keep going for ever while satisfying the interval condition?
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:41
$begingroup$
Are the families trying to fit this pattern or is the problem claiming it always works? For example, suppose after day $2$ there is an umbrella just slightly to the left of $frac{1}{2}$ and another just slightly to the right of $frac{1}{2}$ so that both are contained in $(frac{1}{3},frac{2}{3})$. How can we place a third umbrella so that the condition is met?
$endgroup$
– John Douma
Dec 20 '18 at 16:58
$begingroup$
Are the families trying to fit this pattern or is the problem claiming it always works? For example, suppose after day $2$ there is an umbrella just slightly to the left of $frac{1}{2}$ and another just slightly to the right of $frac{1}{2}$ so that both are contained in $(frac{1}{3},frac{2}{3})$. How can we place a third umbrella so that the condition is met?
$endgroup$
– John Douma
Dec 20 '18 at 16:58
$begingroup$
The families are trying to fit this pattern for as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 16:59
$begingroup$
The families are trying to fit this pattern for as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 16:59
$begingroup$
So they can agree on where to put their umbrellas from the start to keep this going as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 17:00
$begingroup$
So they can agree on where to put their umbrellas from the start to keep this going as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 17:00
1
1
$begingroup$
I have seen this problem before, and the proof is by distinguishing many many cases and brute force/exhaustion. There is nothing deep, just track which are the possible intervals for the 1st, 2nd, 3rd, etc. umbrellas after the 1st, 2nd, 3rd, etc. round.
$endgroup$
– Servaes
Dec 20 '18 at 17:59
$begingroup$
I have seen this problem before, and the proof is by distinguishing many many cases and brute force/exhaustion. There is nothing deep, just track which are the possible intervals for the 1st, 2nd, 3rd, etc. umbrellas after the 1st, 2nd, 3rd, etc. round.
$endgroup$
– Servaes
Dec 20 '18 at 17:59
|
show 5 more comments
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$begingroup$
Can you present a way to place umbrellas with which we can keep going for ever while satisfying the interval condition?
$endgroup$
– Benedict Randall Shaw
Dec 20 '18 at 16:41
$begingroup$
Are the families trying to fit this pattern or is the problem claiming it always works? For example, suppose after day $2$ there is an umbrella just slightly to the left of $frac{1}{2}$ and another just slightly to the right of $frac{1}{2}$ so that both are contained in $(frac{1}{3},frac{2}{3})$. How can we place a third umbrella so that the condition is met?
$endgroup$
– John Douma
Dec 20 '18 at 16:58
$begingroup$
The families are trying to fit this pattern for as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 16:59
$begingroup$
So they can agree on where to put their umbrellas from the start to keep this going as long as possible
$endgroup$
– AspiringMat
Dec 20 '18 at 17:00
1
$begingroup$
I have seen this problem before, and the proof is by distinguishing many many cases and brute force/exhaustion. There is nothing deep, just track which are the possible intervals for the 1st, 2nd, 3rd, etc. umbrellas after the 1st, 2nd, 3rd, etc. round.
$endgroup$
– Servaes
Dec 20 '18 at 17:59