Turning a Biased Coin into an Unbiased one Deterministically











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Non-deterministic Exact Algorithm



There is a simple algorithm to turn a biased coin into a fair one:




  1. Flip the coin twice.

  2. Identify HT with H and TH with T.

  3. Discard cases HH and TT.


This algorithm produces a perfectly fair coin, but it is non-deterministic.



Deterministic Approximation



I also know it is possible to approximate a fair coin with a deterministic algorithm:



Let $C_0$ be the biased coin and define $C_1$ by flipping $C_0$ twice. $C_1$ is H if $C_0$ was HH or TT, and $C_1$ is T if $C_0$ was HT or TH.



We can see that if the probability that $C_0$ was heads is $p$, then the probability that $C_1$ is heads is $p_1 = 1 - 2p(1 - p)$. This is a parabola connecting $(0,1), (.5,.5)$, and $(1,1)$ and we can see that if we assume $0<p<1$ then the function has a fixed point at $0.5$. Since $0.5 < p_1 < p$ if $p>0.5$ and $0.5<p_1<1$ if $p<0.5$, then we can see that a fixed point iteration with $0<p<1$ will always converge to $0.5$. Therefore, we can find a deterministic $C_i$ that is arbitrarily fair (defined by flipping $C_{i-1}$ twice).



My Problem



I am trying to find out if, given some biased coin with rational probability of heads $p$, we can construct an algorithm to solve this problem that is both deterministic and exact. Does anyone have any insights?



(Note that the algorithm only has to work for a fixed probability $p$, since as pointed out in the comments and answer, there are some $p$, e.g., $p = 1/3$ for which there is no such algorithm.)










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  • 1




    Do we know $p$ beforehand? Can the algorithm depend on $p$? There is no algorithm that works for all $p$, but for some values of $p$, such as $frac{1}{4}$ it is possible to have a deterministic algorithm.
    – Todor Markov
    Nov 16 at 11:04








  • 1




    Yes, we know $p$ beforehand, so we can take as much time as we need to pre-compute, say, a decision tree. I'll update this in the question.
    – helper
    Nov 16 at 14:44















up vote
4
down vote

favorite
3












Non-deterministic Exact Algorithm



There is a simple algorithm to turn a biased coin into a fair one:




  1. Flip the coin twice.

  2. Identify HT with H and TH with T.

  3. Discard cases HH and TT.


This algorithm produces a perfectly fair coin, but it is non-deterministic.



Deterministic Approximation



I also know it is possible to approximate a fair coin with a deterministic algorithm:



Let $C_0$ be the biased coin and define $C_1$ by flipping $C_0$ twice. $C_1$ is H if $C_0$ was HH or TT, and $C_1$ is T if $C_0$ was HT or TH.



We can see that if the probability that $C_0$ was heads is $p$, then the probability that $C_1$ is heads is $p_1 = 1 - 2p(1 - p)$. This is a parabola connecting $(0,1), (.5,.5)$, and $(1,1)$ and we can see that if we assume $0<p<1$ then the function has a fixed point at $0.5$. Since $0.5 < p_1 < p$ if $p>0.5$ and $0.5<p_1<1$ if $p<0.5$, then we can see that a fixed point iteration with $0<p<1$ will always converge to $0.5$. Therefore, we can find a deterministic $C_i$ that is arbitrarily fair (defined by flipping $C_{i-1}$ twice).



My Problem



I am trying to find out if, given some biased coin with rational probability of heads $p$, we can construct an algorithm to solve this problem that is both deterministic and exact. Does anyone have any insights?



(Note that the algorithm only has to work for a fixed probability $p$, since as pointed out in the comments and answer, there are some $p$, e.g., $p = 1/3$ for which there is no such algorithm.)










share|cite|improve this question




















  • 1




    Do we know $p$ beforehand? Can the algorithm depend on $p$? There is no algorithm that works for all $p$, but for some values of $p$, such as $frac{1}{4}$ it is possible to have a deterministic algorithm.
    – Todor Markov
    Nov 16 at 11:04








  • 1




    Yes, we know $p$ beforehand, so we can take as much time as we need to pre-compute, say, a decision tree. I'll update this in the question.
    – helper
    Nov 16 at 14:44













up vote
4
down vote

favorite
3









up vote
4
down vote

favorite
3






3





Non-deterministic Exact Algorithm



There is a simple algorithm to turn a biased coin into a fair one:




  1. Flip the coin twice.

  2. Identify HT with H and TH with T.

  3. Discard cases HH and TT.


This algorithm produces a perfectly fair coin, but it is non-deterministic.



Deterministic Approximation



I also know it is possible to approximate a fair coin with a deterministic algorithm:



Let $C_0$ be the biased coin and define $C_1$ by flipping $C_0$ twice. $C_1$ is H if $C_0$ was HH or TT, and $C_1$ is T if $C_0$ was HT or TH.



We can see that if the probability that $C_0$ was heads is $p$, then the probability that $C_1$ is heads is $p_1 = 1 - 2p(1 - p)$. This is a parabola connecting $(0,1), (.5,.5)$, and $(1,1)$ and we can see that if we assume $0<p<1$ then the function has a fixed point at $0.5$. Since $0.5 < p_1 < p$ if $p>0.5$ and $0.5<p_1<1$ if $p<0.5$, then we can see that a fixed point iteration with $0<p<1$ will always converge to $0.5$. Therefore, we can find a deterministic $C_i$ that is arbitrarily fair (defined by flipping $C_{i-1}$ twice).



My Problem



I am trying to find out if, given some biased coin with rational probability of heads $p$, we can construct an algorithm to solve this problem that is both deterministic and exact. Does anyone have any insights?



(Note that the algorithm only has to work for a fixed probability $p$, since as pointed out in the comments and answer, there are some $p$, e.g., $p = 1/3$ for which there is no such algorithm.)










share|cite|improve this question















Non-deterministic Exact Algorithm



There is a simple algorithm to turn a biased coin into a fair one:




  1. Flip the coin twice.

  2. Identify HT with H and TH with T.

  3. Discard cases HH and TT.


This algorithm produces a perfectly fair coin, but it is non-deterministic.



Deterministic Approximation



I also know it is possible to approximate a fair coin with a deterministic algorithm:



Let $C_0$ be the biased coin and define $C_1$ by flipping $C_0$ twice. $C_1$ is H if $C_0$ was HH or TT, and $C_1$ is T if $C_0$ was HT or TH.



We can see that if the probability that $C_0$ was heads is $p$, then the probability that $C_1$ is heads is $p_1 = 1 - 2p(1 - p)$. This is a parabola connecting $(0,1), (.5,.5)$, and $(1,1)$ and we can see that if we assume $0<p<1$ then the function has a fixed point at $0.5$. Since $0.5 < p_1 < p$ if $p>0.5$ and $0.5<p_1<1$ if $p<0.5$, then we can see that a fixed point iteration with $0<p<1$ will always converge to $0.5$. Therefore, we can find a deterministic $C_i$ that is arbitrarily fair (defined by flipping $C_{i-1}$ twice).



My Problem



I am trying to find out if, given some biased coin with rational probability of heads $p$, we can construct an algorithm to solve this problem that is both deterministic and exact. Does anyone have any insights?



(Note that the algorithm only has to work for a fixed probability $p$, since as pointed out in the comments and answer, there are some $p$, e.g., $p = 1/3$ for which there is no such algorithm.)







probability combinatorics algorithms






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edited Nov 16 at 14:49

























asked Nov 16 at 7:04









helper

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  • 1




    Do we know $p$ beforehand? Can the algorithm depend on $p$? There is no algorithm that works for all $p$, but for some values of $p$, such as $frac{1}{4}$ it is possible to have a deterministic algorithm.
    – Todor Markov
    Nov 16 at 11:04








  • 1




    Yes, we know $p$ beforehand, so we can take as much time as we need to pre-compute, say, a decision tree. I'll update this in the question.
    – helper
    Nov 16 at 14:44














  • 1




    Do we know $p$ beforehand? Can the algorithm depend on $p$? There is no algorithm that works for all $p$, but for some values of $p$, such as $frac{1}{4}$ it is possible to have a deterministic algorithm.
    – Todor Markov
    Nov 16 at 11:04








  • 1




    Yes, we know $p$ beforehand, so we can take as much time as we need to pre-compute, say, a decision tree. I'll update this in the question.
    – helper
    Nov 16 at 14:44








1




1




Do we know $p$ beforehand? Can the algorithm depend on $p$? There is no algorithm that works for all $p$, but for some values of $p$, such as $frac{1}{4}$ it is possible to have a deterministic algorithm.
– Todor Markov
Nov 16 at 11:04






Do we know $p$ beforehand? Can the algorithm depend on $p$? There is no algorithm that works for all $p$, but for some values of $p$, such as $frac{1}{4}$ it is possible to have a deterministic algorithm.
– Todor Markov
Nov 16 at 11:04






1




1




Yes, we know $p$ beforehand, so we can take as much time as we need to pre-compute, say, a decision tree. I'll update this in the question.
– helper
Nov 16 at 14:44




Yes, we know $p$ beforehand, so we can take as much time as we need to pre-compute, say, a decision tree. I'll update this in the question.
– helper
Nov 16 at 14:44










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










An algorithm that works for any $p$ doesn't exist. It is possible to solve the problem for some $p$.



Let $p$ be the probability of heads.



Consider a deterministic algorithm $F$ which uses $n$ tosses. Denote $A$ the set of all possible sequences of $n$ coin tosses ($|A| = 2^n$). The $F$ is essentially a function $F: A to {0, 1}$, which returns 0 (heads) for some sequences of $n$ tosses, and 1 (tails) for the rest.



Our deterministic algorithm defines two sets, $H = {v in A | F(v) = 0}$ - the set of $n$-toss sequences where out algorithm decided heads, and $T = A setminus H$, such that $mathbb{P}[H] = mathbb{P}[T] = 0.5$.



On the other hand, for any sequence $v in A$ denote $h(v)$ be the number of tails in $v$.
$$mathbb{P}[H] = sum_{v in H}mathbb{P}[v] = sum_{v in H} p^{h(v)}(1-p)^{n - h(v)} {hspace{2cm}} [1]$$



We need to make this $frac{1}{2}$.



Let $p = frac{r}{q}$ when fully reduced. Then
$$p^{h(v)}(1-p)^{n - h(v)} = frac{r^{h(v)}(q-r)^{n-h(v)}}{q^n}$$



Let's count how many times each possible value of $h(v)$ appears in the sum [1]:



Denote $c_k = |{v | v in H text{ and } h(v)=k }|$. Then [1] becomes



$$mathbb{P}[H] = sum_{k=0}^n c_k frac{r^{k}(q-r)^{n-k}}{q^n} =
frac{sum_{k=0}^n c_k r^{k}(q-r)^{n-k}}{q^n} = frac{1}{2}$$



An algorithm that works for all $p$ doesn't exist, because if $q$ is odd, there is no way to introduce a factor of $2$ in the denominator of this expression. The best we can do is, given a $p$, try to find an algorithm that works for that specific $p$, or determine that such doesn't exist.



Since there are $n choose k$ sequences of coin tosses with length $n$ and $k$ heads, we also need $c_k leq {n choose k}$.



If $q$ is even, and we have a solution to the following equation, satisfying the bounds on $c_k$
$$sum_{k=0}^n c_k r^{k}(q-r)^{n-k} = frac{q^n}{2}$$



Then we can simply include $c_k$ sequences containing $k$ heads in $H$ for each $k$ to get our algorithm.



Note that this equation has finitely many potential solutions, so we can simply try them all.






share|cite|improve this answer










New contributor




Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.














  • 1




    The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
    – Qidi
    Nov 16 at 9:44










  • My bad I missed that. Thanks for the tip.
    – Todor Markov
    Nov 16 at 10:17










  • You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
    – Michal Adamaszek
    Nov 16 at 10:57










  • @Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
    – Todor Markov
    Nov 16 at 11:02












  • Updated answer to include the case when we want an algorithm for a specific $p$.
    – Todor Markov
    Nov 16 at 11:44













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote



accepted










An algorithm that works for any $p$ doesn't exist. It is possible to solve the problem for some $p$.



Let $p$ be the probability of heads.



Consider a deterministic algorithm $F$ which uses $n$ tosses. Denote $A$ the set of all possible sequences of $n$ coin tosses ($|A| = 2^n$). The $F$ is essentially a function $F: A to {0, 1}$, which returns 0 (heads) for some sequences of $n$ tosses, and 1 (tails) for the rest.



Our deterministic algorithm defines two sets, $H = {v in A | F(v) = 0}$ - the set of $n$-toss sequences where out algorithm decided heads, and $T = A setminus H$, such that $mathbb{P}[H] = mathbb{P}[T] = 0.5$.



On the other hand, for any sequence $v in A$ denote $h(v)$ be the number of tails in $v$.
$$mathbb{P}[H] = sum_{v in H}mathbb{P}[v] = sum_{v in H} p^{h(v)}(1-p)^{n - h(v)} {hspace{2cm}} [1]$$



We need to make this $frac{1}{2}$.



Let $p = frac{r}{q}$ when fully reduced. Then
$$p^{h(v)}(1-p)^{n - h(v)} = frac{r^{h(v)}(q-r)^{n-h(v)}}{q^n}$$



Let's count how many times each possible value of $h(v)$ appears in the sum [1]:



Denote $c_k = |{v | v in H text{ and } h(v)=k }|$. Then [1] becomes



$$mathbb{P}[H] = sum_{k=0}^n c_k frac{r^{k}(q-r)^{n-k}}{q^n} =
frac{sum_{k=0}^n c_k r^{k}(q-r)^{n-k}}{q^n} = frac{1}{2}$$



An algorithm that works for all $p$ doesn't exist, because if $q$ is odd, there is no way to introduce a factor of $2$ in the denominator of this expression. The best we can do is, given a $p$, try to find an algorithm that works for that specific $p$, or determine that such doesn't exist.



Since there are $n choose k$ sequences of coin tosses with length $n$ and $k$ heads, we also need $c_k leq {n choose k}$.



If $q$ is even, and we have a solution to the following equation, satisfying the bounds on $c_k$
$$sum_{k=0}^n c_k r^{k}(q-r)^{n-k} = frac{q^n}{2}$$



Then we can simply include $c_k$ sequences containing $k$ heads in $H$ for each $k$ to get our algorithm.



Note that this equation has finitely many potential solutions, so we can simply try them all.






share|cite|improve this answer










New contributor




Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.














  • 1




    The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
    – Qidi
    Nov 16 at 9:44










  • My bad I missed that. Thanks for the tip.
    – Todor Markov
    Nov 16 at 10:17










  • You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
    – Michal Adamaszek
    Nov 16 at 10:57










  • @Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
    – Todor Markov
    Nov 16 at 11:02












  • Updated answer to include the case when we want an algorithm for a specific $p$.
    – Todor Markov
    Nov 16 at 11:44

















up vote
1
down vote



accepted










An algorithm that works for any $p$ doesn't exist. It is possible to solve the problem for some $p$.



Let $p$ be the probability of heads.



Consider a deterministic algorithm $F$ which uses $n$ tosses. Denote $A$ the set of all possible sequences of $n$ coin tosses ($|A| = 2^n$). The $F$ is essentially a function $F: A to {0, 1}$, which returns 0 (heads) for some sequences of $n$ tosses, and 1 (tails) for the rest.



Our deterministic algorithm defines two sets, $H = {v in A | F(v) = 0}$ - the set of $n$-toss sequences where out algorithm decided heads, and $T = A setminus H$, such that $mathbb{P}[H] = mathbb{P}[T] = 0.5$.



On the other hand, for any sequence $v in A$ denote $h(v)$ be the number of tails in $v$.
$$mathbb{P}[H] = sum_{v in H}mathbb{P}[v] = sum_{v in H} p^{h(v)}(1-p)^{n - h(v)} {hspace{2cm}} [1]$$



We need to make this $frac{1}{2}$.



Let $p = frac{r}{q}$ when fully reduced. Then
$$p^{h(v)}(1-p)^{n - h(v)} = frac{r^{h(v)}(q-r)^{n-h(v)}}{q^n}$$



Let's count how many times each possible value of $h(v)$ appears in the sum [1]:



Denote $c_k = |{v | v in H text{ and } h(v)=k }|$. Then [1] becomes



$$mathbb{P}[H] = sum_{k=0}^n c_k frac{r^{k}(q-r)^{n-k}}{q^n} =
frac{sum_{k=0}^n c_k r^{k}(q-r)^{n-k}}{q^n} = frac{1}{2}$$



An algorithm that works for all $p$ doesn't exist, because if $q$ is odd, there is no way to introduce a factor of $2$ in the denominator of this expression. The best we can do is, given a $p$, try to find an algorithm that works for that specific $p$, or determine that such doesn't exist.



Since there are $n choose k$ sequences of coin tosses with length $n$ and $k$ heads, we also need $c_k leq {n choose k}$.



If $q$ is even, and we have a solution to the following equation, satisfying the bounds on $c_k$
$$sum_{k=0}^n c_k r^{k}(q-r)^{n-k} = frac{q^n}{2}$$



Then we can simply include $c_k$ sequences containing $k$ heads in $H$ for each $k$ to get our algorithm.



Note that this equation has finitely many potential solutions, so we can simply try them all.






share|cite|improve this answer










New contributor




Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.














  • 1




    The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
    – Qidi
    Nov 16 at 9:44










  • My bad I missed that. Thanks for the tip.
    – Todor Markov
    Nov 16 at 10:17










  • You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
    – Michal Adamaszek
    Nov 16 at 10:57










  • @Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
    – Todor Markov
    Nov 16 at 11:02












  • Updated answer to include the case when we want an algorithm for a specific $p$.
    – Todor Markov
    Nov 16 at 11:44















up vote
1
down vote



accepted







up vote
1
down vote



accepted






An algorithm that works for any $p$ doesn't exist. It is possible to solve the problem for some $p$.



Let $p$ be the probability of heads.



Consider a deterministic algorithm $F$ which uses $n$ tosses. Denote $A$ the set of all possible sequences of $n$ coin tosses ($|A| = 2^n$). The $F$ is essentially a function $F: A to {0, 1}$, which returns 0 (heads) for some sequences of $n$ tosses, and 1 (tails) for the rest.



Our deterministic algorithm defines two sets, $H = {v in A | F(v) = 0}$ - the set of $n$-toss sequences where out algorithm decided heads, and $T = A setminus H$, such that $mathbb{P}[H] = mathbb{P}[T] = 0.5$.



On the other hand, for any sequence $v in A$ denote $h(v)$ be the number of tails in $v$.
$$mathbb{P}[H] = sum_{v in H}mathbb{P}[v] = sum_{v in H} p^{h(v)}(1-p)^{n - h(v)} {hspace{2cm}} [1]$$



We need to make this $frac{1}{2}$.



Let $p = frac{r}{q}$ when fully reduced. Then
$$p^{h(v)}(1-p)^{n - h(v)} = frac{r^{h(v)}(q-r)^{n-h(v)}}{q^n}$$



Let's count how many times each possible value of $h(v)$ appears in the sum [1]:



Denote $c_k = |{v | v in H text{ and } h(v)=k }|$. Then [1] becomes



$$mathbb{P}[H] = sum_{k=0}^n c_k frac{r^{k}(q-r)^{n-k}}{q^n} =
frac{sum_{k=0}^n c_k r^{k}(q-r)^{n-k}}{q^n} = frac{1}{2}$$



An algorithm that works for all $p$ doesn't exist, because if $q$ is odd, there is no way to introduce a factor of $2$ in the denominator of this expression. The best we can do is, given a $p$, try to find an algorithm that works for that specific $p$, or determine that such doesn't exist.



Since there are $n choose k$ sequences of coin tosses with length $n$ and $k$ heads, we also need $c_k leq {n choose k}$.



If $q$ is even, and we have a solution to the following equation, satisfying the bounds on $c_k$
$$sum_{k=0}^n c_k r^{k}(q-r)^{n-k} = frac{q^n}{2}$$



Then we can simply include $c_k$ sequences containing $k$ heads in $H$ for each $k$ to get our algorithm.



Note that this equation has finitely many potential solutions, so we can simply try them all.






share|cite|improve this answer










New contributor




Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









An algorithm that works for any $p$ doesn't exist. It is possible to solve the problem for some $p$.



Let $p$ be the probability of heads.



Consider a deterministic algorithm $F$ which uses $n$ tosses. Denote $A$ the set of all possible sequences of $n$ coin tosses ($|A| = 2^n$). The $F$ is essentially a function $F: A to {0, 1}$, which returns 0 (heads) for some sequences of $n$ tosses, and 1 (tails) for the rest.



Our deterministic algorithm defines two sets, $H = {v in A | F(v) = 0}$ - the set of $n$-toss sequences where out algorithm decided heads, and $T = A setminus H$, such that $mathbb{P}[H] = mathbb{P}[T] = 0.5$.



On the other hand, for any sequence $v in A$ denote $h(v)$ be the number of tails in $v$.
$$mathbb{P}[H] = sum_{v in H}mathbb{P}[v] = sum_{v in H} p^{h(v)}(1-p)^{n - h(v)} {hspace{2cm}} [1]$$



We need to make this $frac{1}{2}$.



Let $p = frac{r}{q}$ when fully reduced. Then
$$p^{h(v)}(1-p)^{n - h(v)} = frac{r^{h(v)}(q-r)^{n-h(v)}}{q^n}$$



Let's count how many times each possible value of $h(v)$ appears in the sum [1]:



Denote $c_k = |{v | v in H text{ and } h(v)=k }|$. Then [1] becomes



$$mathbb{P}[H] = sum_{k=0}^n c_k frac{r^{k}(q-r)^{n-k}}{q^n} =
frac{sum_{k=0}^n c_k r^{k}(q-r)^{n-k}}{q^n} = frac{1}{2}$$



An algorithm that works for all $p$ doesn't exist, because if $q$ is odd, there is no way to introduce a factor of $2$ in the denominator of this expression. The best we can do is, given a $p$, try to find an algorithm that works for that specific $p$, or determine that such doesn't exist.



Since there are $n choose k$ sequences of coin tosses with length $n$ and $k$ heads, we also need $c_k leq {n choose k}$.



If $q$ is even, and we have a solution to the following equation, satisfying the bounds on $c_k$
$$sum_{k=0}^n c_k r^{k}(q-r)^{n-k} = frac{q^n}{2}$$



Then we can simply include $c_k$ sequences containing $k$ heads in $H$ for each $k$ to get our algorithm.



Note that this equation has finitely many potential solutions, so we can simply try them all.







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edited Nov 16 at 14:55





















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answered Nov 16 at 9:36









Todor Markov

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  • 1




    The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
    – Qidi
    Nov 16 at 9:44










  • My bad I missed that. Thanks for the tip.
    – Todor Markov
    Nov 16 at 10:17










  • You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
    – Michal Adamaszek
    Nov 16 at 10:57










  • @Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
    – Todor Markov
    Nov 16 at 11:02












  • Updated answer to include the case when we want an algorithm for a specific $p$.
    – Todor Markov
    Nov 16 at 11:44
















  • 1




    The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
    – Qidi
    Nov 16 at 9:44










  • My bad I missed that. Thanks for the tip.
    – Todor Markov
    Nov 16 at 10:17










  • You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
    – Michal Adamaszek
    Nov 16 at 10:57










  • @Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
    – Todor Markov
    Nov 16 at 11:02












  • Updated answer to include the case when we want an algorithm for a specific $p$.
    – Todor Markov
    Nov 16 at 11:44










1




1




The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
– Qidi
Nov 16 at 9:44




The problem says $p$ is rational. I think you can modify your arguments to $mathbb{P}[H]$ has denominator of the form $q^m$, which is an odd number, where $p = frac{q'}{q}$, so there is a solution only when $p=frac{1}{2}$.
– Qidi
Nov 16 at 9:44












My bad I missed that. Thanks for the tip.
– Todor Markov
Nov 16 at 10:17




My bad I missed that. Thanks for the tip.
– Todor Markov
Nov 16 at 10:17












You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
– Michal Adamaszek
Nov 16 at 10:57




You mean there are some $p$ for which there is no algorithm. For some $p$ it is possible, for instance $p=1/4$.
– Michal Adamaszek
Nov 16 at 10:57












@Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
– Todor Markov
Nov 16 at 11:02






@Michael Adamaszek Yes. The way I understand the problem, you don't actually know $p$ beforehand to be able to tune your algorithm to it. Indeed, both example algorithms shown in the first post work for any $p$ unaltered. But it is worth clarifying that.
– Todor Markov
Nov 16 at 11:02














Updated answer to include the case when we want an algorithm for a specific $p$.
– Todor Markov
Nov 16 at 11:44






Updated answer to include the case when we want an algorithm for a specific $p$.
– Todor Markov
Nov 16 at 11:44




















 

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