How many balls will be left at the end of this process?
$begingroup$
Consider having $N$ colored balls. Each color has at least $N/2k$ and at most $N/k$ balls in the beginning, for some parameter $kll N$.
At each iteration, we remove $k$ balls with different colors, choosing from the colors that have most balls (breaking ties arbitrarily).
The process ends when there are less than $k$ colors still present.
For example, consider $k=3$ and having $4$ balls from color $1$, $2$ balls from colors $2$, $3$ and $4$ and a single ball from color $5$.
Then the process may be (where each tuple is the frequency of each color):
$(4,2,2,2,1)to(3,2,1,1,1)to(2,1,1,1,0)to(1,1,0,0,0)$.
That is, we are left with just one ball in this example.
Can we guarantee that the process terminates with at most $k-1$ balls?
Clearly, this cannot hold without the bound on the number of balls from each color. E.g., if we had $6$ balls (which is more than $13/3=N/k$) from the first color instead of $3$ we could get:
$(6,2,2,2,1)to(5,2,1,1,1)to(4,1,1,1,0)to(3,1,0,0,0)$.
It seems that, for all examples I tested, we are always left with at most $k-1$ balls from different colors. Is that true for any distribution that satisfies the size constraints?
algorithms recursion integers recursive-algorithms
asked Dec 6 '18 at 14:46
R BR B
1,3111331
$endgroup$
add a comment |
$begingroup$
Consider having $N$ colored balls. Each color has at least $N/2k$ and at most $N/k$ balls in the beginning, for some parameter $kll N$.
At each iteration, we remove $k$ balls with different colors, choosing from the colors that have most balls (breaking ties arbitrarily).
The process ends when there are less than $k$ colors still present.
For example, consider $k=3$ and having $4$ balls from color $1$, $2$ balls from colors $2$, $3$ and $4$ and a single ball from color $5$.
Then the process may be (where each tuple is the frequency of each color):
$(4,2,2,2,1)to(3,2,1,1,1)to(2,1,1,1,0)to(1,1,0,0,0)$.
That is, we are left with just one ball in this example.
Can we guarantee that the process terminates with at most $k-1$ balls?
Clearly, this cannot hold without the bound on the number of balls from each color. E.g., if we had $6$ balls (which is more than $13/3=N/k$) from the first color instead of $3$ we could get:
$(6,2,2,2,1)to(5,2,1,1,1)to(4,1,1,1,0)to(3,1,0,0,0)$.
It seems that, for all examples I tested, we are always left with at most $k-1$ balls from different colors. Is that true for any distribution that satisfies the size constraints?
algorithms recursion integers recursive-algorithms
asked Dec 6 '18 at 14:46
R BR B
1,3111331
$endgroup$
add a comment |
$begingroup$
Consider having $N$ colored balls. Each color has at least $N/2k$ and at most $N/k$ balls in the beginning, for some parameter $kll N$.
At each iteration, we remove $k$ balls with different colors, choosing from the colors that have most balls (breaking ties arbitrarily).
The process ends when there are less than $k$ colors still present.
For example, consider $k=3$ and having $4$ balls from color $1$, $2$ balls from colors $2$, $3$ and $4$ and a single ball from color $5$.
Then the process may be (where each tuple is the frequency of each color):
$(4,2,2,2,1)to(3,2,1,1,1)to(2,1,1,1,0)to(1,1,0,0,0)$.
That is, we are left with just one ball in this example.
Can we guarantee that the process terminates with at most $k-1$ balls?
Clearly, this cannot hold without the bound on the number of balls from each color. E.g., if we had $6$ balls (which is more than $13/3=N/k$) from the first color instead of $3$ we could get:
$(6,2,2,2,1)to(5,2,1,1,1)to(4,1,1,1,0)to(3,1,0,0,0)$.
It seems that, for all examples I tested, we are always left with at most $k-1$ balls from different colors. Is that true for any distribution that satisfies the size constraints?
algorithms recursion integers recursive-algorithms
asked Dec 6 '18 at 14:46
R BR B
1,3111331
$endgroup$
Consider having $N$ colored balls. Each color has at least $N/2k$ and at most $N/k$ balls in the beginning, for some parameter $kll N$.
At each iteration, we remove $k$ balls with different colors, choosing from the colors that have most balls (breaking ties arbitrarily).
The process ends when there are less than $k$ colors still present.
For example, consider $k=3$ and having $4$ balls from color $1$, $2$ balls from colors $2$, $3$ and $4$ and a single ball from color $5$.
Then the process may be (where each tuple is the frequency of each color):
$(4,2,2,2,1)to(3,2,1,1,1)to(2,1,1,1,0)to(1,1,0,0,0)$.
That is, we are left with just one ball in this example.
Can we guarantee that the process terminates with at most $k-1$ balls?
Clearly, this cannot hold without the bound on the number of balls from each color. E.g., if we had $6$ balls (which is more than $13/3=N/k$) from the first color instead of $3$ we could get:
$(6,2,2,2,1)to(5,2,1,1,1)to(4,1,1,1,0)to(3,1,0,0,0)$.
It seems that, for all examples I tested, we are always left with at most $k-1$ balls from different colors. Is that true for any distribution that satisfies the size constraints?
algorithms recursion integers recursive-algorithms
algorithms recursion integers recursive-algorithms
asked Dec 6 '18 at 14:46
R BR B
1,3111331
asked Dec 6 '18 at 14:46
R BR B
1,3111331
asked Dec 6 '18 at 14:46
R BR B
1,3111331
asked Dec 6 '18 at 14:46
R BR B
1,3111331
asked Dec 6 '18 at 14:46
R BR B
1,3111331
1,3111331
add a comment |
add a comment |
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