Solution:
Notice that the question's condition mandates all blues to go to reds, but reds do not necessarily have to go to blue. Let us do casework on how many blues there are.
If there are no blues whatsoever, there is only one case. This case is valid, as all of the (zero) blues have gone to reds. (One could also view it as: the location of all the blues now were not previously red.) Thus, we have $1$.
If there is a single blue somewhere, there are $8$ cases - where can the blue be? Each of these is valid.
If there are two blues, again, every case is valid, and there are $\dbinom82=28$ cases.
If there are three blues, every case is again valid; there are $\dbinom83=56$ such cases.
The case with four blues is trickier. Let us look at all possible subcases.
If all four are adjacent (as in the diagram below), it is obvious: we can simply reverse the diagram (rotate it by $4$ units) to achieve the problem's condition. There are $8$ possible ways to have $4$ adjacent blues, so this subcase contributes $8$.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,0,0,1,1,1,1}; oct11(sus); [/asy]
If three are adjacent and one is one away (as shown in the diagram below), we can not rotate the diagram to satisfy the question. This subcase does not work.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,0,1,0,1,1,1}; oct11(sus); [/asy]
If three are adjacent and one is two away, obviously it is not possible as there is nowhere for the three adjacent blues to go.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,0,1,1,0,1,1}; oct11(sus); [/asy]
If there are two adjacent pairs that are $1$ apart, it is not possible since we do not have anywhere to put the two pairs.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,0,0,1,1,1}; oct11(sus); [/asy]
If there are two adjacent pairs that are $2$ apart, all of these cases are possible as we can rotate the diagram by $2$ vertices to work. There are $4$ of these cases.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,1,0,0,1,1}; oct11(sus); [/asy]
If there is one adjacent pair and there are two separate ones each a distance of $1$ from the other, this case does not work.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,0,1,0,1,1}; oct11(sus); [/asy]
If we have one adjacent pair and two separate ones that are $2$ away from each other, we can flip the diagram by $4$ vertices. There are $8$ of these cases.
[asy] import graph;  void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,0,1,1,0,1}; oct11(sus); [/asy]
Finally, if the red and blues alternate, we can simply shift the diagram by a single vertex to satisfy the question. Thus, all of these cases work, and we have $2$ subcases.
There can not be more than $4$ blues, so we are done.
Our total is $1+8+28+56+8+4+8+2=115$. There are $2^8=256$ possible colorings, so we have $\dfrac{115}{256}$ and our answer is $115+256=\boxed{371}$.
~Technodoggo
Let $r$ be the number of red vertices and $b$ be the number of blue vertices, where $r+b=8$. By the Pigeonhole Principle, $r\geq{b} \Longrightarrow b\leq4$ if a configuration is valid.
We claim that if $b\leq3$, then any configuration is valid. We attempt to prove by the following:
If there are \[b\in{0,1,2}\] vertices, then intuitively any configuration is valid. For $b=3$, we do cases:
If all the vertices in $b$ are non-adjacent, then simply rotating once in any direction suffices. If there are $2$ adjacent vertices, then WLOG let us create a set $\{b_1,b_2,r_1\cdots\}$ where the third $b_3$ is somewhere later in the set. If we assign the set as $\{1,2,3,4,5,6,7,8\}$ and $b_3\leq4$, then intuitively, rotating it $4$ will suffice. If $b_3=5$, then rotating it by 2 will suffice. Consider any other $b_3>5$ as simply a mirror to a configuration of the cases.
Therefore, if $b\leq3$, then there are $\sum_{i=0}^{3}{\binom{8}{i}}=93$ ways. We do count the [i]degenerate[/i] case.
Now if $b=4$, we do casework on the number of adjacent vertices.
0 adjacent: $\{b_1,r_1,b_2,r_2\cdots{r_4}\}$. There are 4 axes of symmetry so there are only $\frac{8}{4}=2$ rotations of this configuration.
1 adjacent: WLOG $\{b_1,b_2\cdots{b_3}\cdots{b_4}\}$ where $b_4\neq{8}$. Listing out the cases and trying, we get that $b_3=4$ and $b_4=7$ is the only configuration. There are $8$ ways to choose $b_1$ and $b_2$ and the rest is set, so there are $8$ ways.
2 adjacent: We can have WLOG $\{b_1,b_2\cdots{b_3},b_4\}$ or $\{b_1,b_2,b_3\cdots\}$ where $b_4\neq{8}$. The former yields the case $b_3=5$ and $b_4=6$ by simply rotating it 2 times. The latter yields none. There are 2 axes of symmetry so there are $\frac{8}{2}=4$ configurations.
3 adjacent: WLOG $\{b_1,b_2,b_3,b_4\cdots\}$ which intuitively works. There are $8$ configurations here as $b_1$ can is unique.
In total, $b=4$ yields $2+8+4+8=22$ configurations.
There are $22+93=115$ configurations in total. There are $2^8=256$ total cases, so the probability is $\frac{115}{256}$. Adding them up, we get $115+256=\boxed{371}$.

Answer: 371

<｜begin▁of▁sentence｜><｜User｜>Each vertex of a regular octagon is independently colored either red or blue with equal probability. The probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?<｜Assistant｜><think>

Okay, so I have this problem here about coloring the vertices of a regular octagon either red or blue, each with equal probability, and then figuring out the probability that there's some rotation of the octagon where all the blue vertices end up where there were originally red vertices. The answer needs to be in the form of a reduced fraction m/n and then m + n. Alright, let me try to unpack this step by step.

First, let me make sure I understand the problem correctly. We have a regular octagon, so 8 vertices. Each vertex is colored either red or blue, independently, with a 50-50 chance for each color. Then, we want the probability that there exists some rotation (so rotating the octagon by some multiple of 45 degrees, since it's regular) such that after that rotation, every blue vertex is now in a position that was originally red. Hmm, so essentially, after rotating, the set of blue vertices is a subset of the originally red vertices? Wait, not exactly. Let me parse that again.

"The probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices."

So, after rotating, each blue vertex must be in a position that was originally red. That is, the image of the blue vertices under the rotation must be a subset of the original red vertices. Alternatively, the rotation maps the blue vertices to positions that were red before any rotation. So, in other words, if we denote the original coloring as a set R of red vertices and B of blue vertices, then there exists a rotation r such that r(B) ⊆ R. Since the rotation is a bijection, this would mean that B is a subset of r^{-1}(R). Since R and B are complements (since every vertex is either red or blue), so R = complement of B. Wait, actually, R and B are complements, so R = V \ B where V is the set of all vertices. Therefore, r(B) ⊆ R = V \ B, so r(B) and B are disjoint? Wait, maybe I need to think more carefully.

Alternatively, perhaps another way: after rotating the octagon by some amount, all the blue vertices coincide with positions that were originally red. So, the rotated blue vertices don't overlap with any original blue vertices. So, the rotated blue vertices must be a subset of the original red vertices. Since the rotation is a symmetry of the octagon, it's a permutation of the vertices. Therefore, the condition is that there exists a rotation permutation π such that π(B) ⊆ R, where B is the set of originally blue vertices and R is the set of originally red vertices. Since R is the complement of B, this is equivalent to π(B) ∩ B = ∅. Because if π(B) is a subset of R, then their intersection with B must be empty.

Therefore, the condition is that there exists a rotation π such that π(B) and B are disjoint. So, the problem reduces to finding the probability that such a rotation exists. Got it.

So, the question is: given a random subset B of the 8 vertices (each vertex included with probability 1/2), what is the probability that there exists a rotation π (other than the identity? Wait, no, the identity is allowed too, but if you rotate by 0 degrees, then π(B) = B, so π(B) and B would overlap unless B is empty. Since B is random, maybe the identity is allowed, but in that case, the only way π(B) and B are disjoint is if B is empty. But since each vertex is colored independently, the probability that B is empty is 1/256, which is negligible. So probably, the non-identity rotations are the ones we need to consider. Hmm, but the problem says "can be rotated", so maybe including the identity. Let me check.

Wait, if we take the identity rotation, then π(B) = B, so the condition would require that all blue vertices are at positions where there were originally red vertices, but since they are the same positions, that would require that all blue vertices are in red positions, which would mean that B is a subset of R, but since R is the complement of B, that would require B is empty. So, only the empty set would satisfy that with the identity rotation. Since the problem says "can be rotated", so maybe we can choose any rotation, including identity, but the only case where identity works is when B is empty, which is possible but rare. So, perhaps the problem allows the identity rotation, but in most cases, we need a non-trivial rotation.

Alternatively, maybe the problem is intended to consider non-trivial rotations? The problem statement isn't entirely clear, but since it says "rotated", maybe excluding the identity? Hmm, but in math problems like this, usually rotations include the identity unless specified otherwise. Hmm. Well, perhaps we can just proceed and see.

So, the problem is to compute the probability over all colorings (so all possible subsets B of the 8 vertices) that there exists a rotation π (including identity?) such that π(B) and B are disjoint. Since each coloring is equally likely, the probability is equal to the number of subsets B for which such a π exists, divided by 2^8 = 256.

Therefore, we need to count the number of subsets B where there exists a rotation π such that π(B) ∩ B = ∅. Let me note that the rotations of the octagon form a group under composition, isomorphic to the cyclic group C8. The rotations are by multiples of 45 degrees, so rotation by k*45 degrees for k = 0,1,...,7. Each rotation corresponds to a shift by k positions around the octagon.

Therefore, the rotations are π_0 (identity), π_1 (shift by 1), π_2 (shift by 2), ..., π_7 (shift by 7). So, for each subset B, we need to check whether there exists some k in 0,...,7 such that π_k(B) ∩ B = ∅. So, the question is, how many subsets B have this property?

Alternatively, the complement event is that for all rotations π_k, π_k(B) intersects B. So, the subsets B where for every rotation π_k, π_k(B) and B share at least one common element. So, the problem is asking for 1 minus the probability that for all rotations π_k, π_k(B) intersects B. Wait, but actually, no, the problem is asking for the probability that there exists at least one rotation π_k such that π_k(B) and B are disjoint. So, the count is over all subsets B where this condition holds for at least one rotation.

Hmm, inclusion-exclusion might be needed here, but maybe there's a smarter way. Let me think.

Alternatively, perhaps Burnside's lemma could be helpful here? Burnside's lemma relates the number of orbits to the average number of fixed points. Wait, but I need to think if that applies here. Alternatively, maybe we can model this as a group action problem where the group is the rotation group acting on the subsets, but I need to see.

Alternatively, perhaps it's better to think in terms of the required condition. Let me consider that for a subset B, there exists a rotation π such that π(B) and B are disjoint. Let me note that π(B) is the set of vertices obtained by rotating B by π. So, if we can rotate B so that it doesn't overlap with itself, then such a rotation exists.

Alternatively, this is equivalent to saying that B is not a "fixed" subset under any rotation? Wait, not exactly. Wait, actually, the condition is that there exists a rotation that shifts B so that it doesn't intersect itself. Hmm.

Alternatively, perhaps the problem is similar to the concept of a "shift" in combinatorics where a set is shifted such that it doesn't overlap with its original position. For cyclic shifts, in this case.

Alternatively, maybe we can model the octagon as the cyclic group Z/8Z, with elements 0 through 7, and rotations correspond to adding a constant modulo 8. Then, the condition is that there exists some k in Z/8Z such that (B + k) ∩ B = ∅, where B + k is the set { (b + k) mod 8 | b ∈ B }.

Therefore, the problem reduces to counting subsets B of Z/8Z such that there exists a k where (B + k) ∩ B = ∅. So, how many subsets B have this property?

Alternatively, for each subset B, check if there exists a k such that B and B + k are disjoint. So, the question is, how many subsets B of Z/8Z have the property that they can be shifted by some k to be disjoint from themselves.

Alternatively, such subsets are called "shift-disjoint" or something? Maybe not a standard term, but perhaps there's a way to characterize them.

Alternatively, perhaps the complement is easier: compute the number of subsets B where for every k, B and B + k intersect. Then subtract that from the total to get the desired count. But inclusion-exclusion might be complicated here.

Alternatively, perhaps we can think in terms of the minimal period of the subset B. If the subset has a certain periodicity, then shifting by that period would map the subset onto itself, so in that case, shifting by half the period might disjoint it? Hmm, not sure.

Alternatively, let me think of specific cases. Let me consider small subsets first.

First, note that if B is empty, then any rotation would satisfy the condition, since the empty set disjoint from anything. So, the empty set counts. Similarly, if B is the entire set, then no rotation can make it disjoint from itself, except if you rotate by 0, but even then, it's the same set. So, the entire set does not satisfy the condition. Wait, but if B is the entire set, then π(B) is also the entire set, so their intersection is the entire set, so they can't be disjoint. So, the entire set is excluded.

Similarly, singleton sets: suppose B has one vertex. Then, rotating by 1 position would move that vertex to the next position, so the original B had one vertex, and the rotated B would have the next vertex. Since the original B had only one vertex, the rotated version would not overlap, so their intersection is empty. Therefore, any singleton set satisfies the condition, because you can rotate it by 1 to move it to a different position, hence disjoint from the original. Wait, actually, rotating a singleton set by any non-zero rotation would move it to a different position, so as long as the rotation is not 0, then the rotated singleton is disjoint from the original. Therefore, for any singleton set, there exists a rotation (in fact, 7 rotations) that would make them disjoint. Therefore, all singleton sets satisfy the condition.

Similarly, for pairs: suppose B has two vertices. Depending on their arrangement, maybe they can be rotated to be disjoint? Let's see. For example, if the two vertices are adjacent, then rotating by 1 would shift them to the next positions, so the original positions and the shifted positions would overlap by one vertex. Wait, so if the original pair is {0,1}, then shifting by 1 gives {1,2}, so the intersection is {1}, so not disjoint. Shifting by 2 would give {2,3}, which doesn't overlap with {0,1}. So, shifting by 2 would make them disjoint. Therefore, even adjacent pairs can be shifted to be disjoint. Wait, so perhaps any pair can be shifted to be disjoint?

Wait, let me check another configuration. Suppose the two vertices are opposite each other, like {0,4} in an octagon. Then, shifting by 4 would map {0,4} to {4,0}, which is the same set, so overlapping. But shifting by 1 would give {1,5}, which doesn't overlap with {0,4}. So, yes, shifting by 1 would make them disjoint. So, in that case, it's possible. Hmm, so maybe all pairs can be shifted to be disjoint?

Wait, let me think of another case. Suppose the two vertices are spaced by 2, like {0,2}. Then shifting by 1 would give {1,3}, which doesn't overlap with {0,2}, so that works. So, seems like for any pair, there exists a shift that moves them to not overlap with original. Therefore, all pairs satisfy the condition. Hmm, interesting.

Wait, so maybe for any subset B that is not the entire set, there exists a rotation that can shift it to be disjoint? Wait, but that can't be. Let me think of a subset that can't be shifted to be disjoint. For example, suppose B is the entire set except one vertex. Let's say B has 7 vertices. Then, any shift of B would also be 7 vertices. The intersection between B and its shift would be 7 + 7 - 8 = 6 vertices? Wait, maybe. Let me see. Suppose B is all vertices except vertex 0. Then shifting by 1 would give all vertices except vertex 7. The intersection would be all vertices except 0 and 7, so 6 vertices. So, the intersection is non-empty. Similarly, shifting by any k would result in a set missing a different vertex, so the intersection would be 6 vertices. Therefore, there is no shift that would make B and its shift disjoint. Therefore, such a subset (size 7) does not satisfy the condition. So, subsets of size 7 do not satisfy the condition. Similarly, subsets of size 8 (the entire set) also don't satisfy.

Similarly, subsets of size 6? Let me see. Suppose B has 6 vertices. Let's say it's missing two vertices. Depending on how the two missing vertices are arranged, maybe there's a shift that can move B such that the shifted B doesn't overlap with original? Wait, but if B has 6 vertices, then its complement has 2. To have π(B) disjoint from B, π(B) must be a subset of the complement of B, which has only 2 vertices. But π(B) has 6 vertices, so that's impossible. Therefore, any subset B with size greater than 4 cannot satisfy the condition, because its complement has less than 4 vertices, so you can't fit a shifted version of B (which has the same size) into the complement. Wait, actually, more precisely, if |B| = k, then to have π(B) disjoint from B, π(B) must be a subset of the complement of B, which has size 8 - k. Therefore, we must have that 8 - k ≥ k, so 8 ≥ 2k, so k ≤ 4. Therefore, only subsets with size at most 4 can possibly satisfy the condition. Because if the subset has size 5 or more, its complement is too small to contain a shifted copy of itself. Therefore, subsets of size 5,6,7,8 cannot satisfy the condition. So, only subsets of size 0 to 4 can possibly satisfy the condition. That's an important observation.

Therefore, the total number of subsets that can satisfy the condition is the sum over k=0 to 4 of the number of subsets of size k that can be shifted to be disjoint from themselves. So, we can restrict our attention to subsets of size 0 to 4.

Now, let's consider subsets of size 0: the empty set. As we said before, it trivially satisfies the condition because any shift would still be empty, so disjoint. So, that's 1 subset.

Subsets of size 1: as we saw earlier, all singleton subsets can be shifted to be disjoint. Since there are 8 subsets of size 1, all 8 satisfy the condition.

Subsets of size 2: as in the earlier examples, it seems like all subsets of size 2 can be shifted to be disjoint. Let me confirm with another example. Suppose the two vertices are opposite each other, like {0,4}. Shifting by 1 gives {1,5}, which doesn't intersect with {0,4}. Similarly, if they are two apart, like {0,2}, shifting by 1 gives {1,3}, which is disjoint. If they are adjacent, like {0,1}, shifting by 2 gives {2,3}, which is disjoint. So, indeed, any pair can be shifted to be disjoint. Therefore, all C(8,2)=28 subsets of size 2 satisfy the condition.

Subsets of size 3: Hmm, here it's possible that some subsets cannot be shifted to be disjoint. Let me see. Let's take an example. Suppose the three vertices are equally spaced, like {0, 3, 6}. Let's see if there's a shift that can move them to not overlap with the original. Let's try shifting by 1: {1,4,7}. The original set is {0,3,6}; the shifted set is {1,4,7}. These are disjoint? Yes, because none of 1,4,7 are in the original set. Wait, so actually, shifting by 1 would work here. Wait, so that set can be shifted to be disjoint. Hmm, interesting.

Wait another example: suppose three consecutive vertices, like {0,1,2}. Let's see if we can shift them so that the shifted set doesn't overlap. Let's try shifting by 3: {3,4,5}. The original set is {0,1,2}; the shifted set is {3,4,5}, which is disjoint. So that works. Alternatively, shifting by 4 would give {4,5,6}, which overlaps with the original at none. Wait, so that works too.

Hmm, maybe all triples can be shifted to be disjoint? Wait, let me think of a tricky case. Suppose the three vertices are spaced in such a way that every shift overlaps. Let me see. Suppose the subset is {0,1,4}. Let's see if there's a shift that can move this to be disjoint. Let's try shifting by 2: {2,3,6}. Original set is {0,1,4}; shifted is {2,3,6}. These are disjoint? Yes, no overlap. So that works. Hmm. Another example: {0,2,5}. Let's try shifting by 1: {1,3,6}. Original is {0,2,5}; shifted is {1,3,6}. Disjoint? Yes, no common elements. Hmm. Maybe all triples can be shifted to be disjoint?

Wait, perhaps so. Let me think of another configuration. Suppose the three vertices are {0,1,3}. Let's try shifting by 4: {4,5,7}. Original set {0,1,3} and shifted {4,5,7} are disjoint. So that works. Hmm. Maybe indeed all triples can be shifted to be disjoint. Wait, is there any subset of three vertices where no shift can make it disjoint?

Suppose the three vertices are arranged such that they are spaced 2 apart: {0,2,4,6} but that's four vertices. Wait, for three vertices, maybe {0,2,4}. Let's see. Shifting by 1: {1,3,5}. Original is {0,2,4}; shifted is {1,3,5}. Disjoint? Yes, so that works. Hmm. Alternatively, shifting by 3: {3,5,7}, which also disjoint. Hmm.

Alternatively, maybe a subset like {0,1,2}. Shifting by 3 gives {3,4,5}, which is disjoint. So, seems like all triples can be shifted to be disjoint. Wait, so maybe all subsets of size 3 satisfy the condition? Then, the number would be C(8,3)=56.

Wait, but let me think again. Suppose the subset is {0,1,4}. Let me check all possible shifts. Wait, but even if one shift works, that's enough. Since we only need existence. So, as long as there exists at least one shift that makes it disjoint, then it's okay. So, even if some shifts overlap, as long as there's one that doesn't, it's okay. So, perhaps all triples can be shifted to be disjoint. Hmm, maybe so. Let me see if there's a subset of three vertices where every shift overlaps with itself. Suppose such a subset exists. Then, for every shift k, the shifted subset shares at least one element with the original.

Suppose B is a 3-element subset such that for every k, B ∩ (B + k) ≠ ∅. Is that possible?

Let me suppose such a B exists. Let me try to construct it. Let me suppose B = {0,1, x}. Let me see if I can choose x such that for any shift k, shifting B by k will hit at least one element of B.

Wait, but this seems difficult. Let me try with B = {0,1,4}. Let's see for k=2: shifted B is {2,3,6}, which doesn't intersect with original B. So that's bad for the example. So that subset is okay. Let me try another B. Suppose B = {0, 2, 4}. Then shifting by 1 gives {1,3,5}, which is disjoint. So that's also okay. Hmm.

Alternatively, maybe B = {0, 1, 3}. Shifting by 2 gives {2,4,5}, which doesn't intersect with original. Hmm. Maybe it's impossible to have a 3-element subset where every shift overlaps? Maybe all 3-element subsets can be shifted to be disjoint. Therefore, all C(8,3)=56 subsets of size 3 satisfy the condition.

Moving on to subsets of size 4. Now, this is where it gets more interesting because subsets of size 4 can potentially have overlaps with their shifts. Let me see. Since the complement of a 4-element subset is also a 4-element subset, so to have a shift where π(B) is disjoint from B, π(B) must be exactly the complement of B. Because the complement has size 4, so π(B) must equal the complement. Therefore, for a subset B of size 4, the condition is equivalent to: there exists a rotation π such that π(B) = complement of B. Because π(B) has to be disjoint from B and of the same size, so it must be exactly the complement.

Therefore, for subsets of size 4, the condition is that B is rotated to its complement by some rotation. Therefore, such subsets must be "complemented" by a rotation. So, in other words, B is a subset such that its complement is a rotation of itself. Therefore, B is a rotation of its own complement. Therefore, the subset B must be such that there exists a rotation π with π(B) = complement of B. Therefore, applying π twice, we get π(π(B)) = π(complement of B) = complement of π^{-1}(B). Hmm, maybe getting too abstract.

Alternatively, let me note that if π(B) = complement of B, then applying π again, π^2(B) = π(complement of B) = complement of π^{-1}(B). Hmm, perhaps not helpful. Alternatively, since the rotation group is cyclic of order 8, so if such a rotation π exists, then the order of π must divide 8. Let me think of specific examples.

Suppose B is a subset of size 4 such that its complement is a rotation of B. Let me consider some examples. Let me take B as four consecutive vertices, say {0,1,2,3}. Then, the complement is {4,5,6,7}. Is the complement a rotation of B? Yes, rotating B by 4 positions gives {4,5,6,7}, which is exactly the complement. Therefore, this subset satisfies the condition. So, this works.

Another example: suppose B is every other vertex, like {0,2,4,6}. The complement is {1,3,5,7}, which is a rotation (shift by 1) of B? Wait, shifting B by 1 would give {1,3,5,7}, which is exactly the complement. Therefore, yes, so this also satisfies the condition.

Another example: suppose B is {0,1,4,5}. Let's see, the complement is {2,3,6,7}. Is the complement a rotation of B? Let's see. If we rotate B by 2, we get {2,3,6,7}, which is exactly the complement. So yes, that works.

Hmm, so maybe all subsets of size 4 have this property? Wait, no. Let me think of a subset of size 4 where the complement is not a rotation of B. Let me try to construct such a subset.

Suppose B = {0,1,2,4}. Let's see what its complement is: {3,5,6,7}. Is there a rotation that maps B to the complement? Let's check. Let me see if there's a shift k such that {0+k,1+k,2+k,4+k mod8} = {3,5,6,7}. Let's try k=3: shifting B by 3 gives {3,4,5,7}. The complement is {3,5,6,7}. So, {3,4,5,7} vs {3,5,6,7}. The difference is that instead of 6, we have 4. So, not equal. Try k=5: shifting B by 5 gives {5,6,7,1}. Which is {1,5,6,7}, which is not the complement. Shifting by 4: {4,5,6,0} = {0,4,5,6}, which is not the complement. Shifting by 6: {6,7,0,2} = {0,2,6,7}, not the complement. Hmm, shifting by 2: {2,3,4,6}, which is not the complement. Shifting by 1: {1,2,3,5}, nope. Shifting by 7: {7,0,1,3}, which is {0,1,3,7}, nope. So, seems like there is no shift that maps B to its complement. Therefore, this subset B={0,1,2,4} does NOT satisfy the condition. Therefore, not all subsets of size 4 satisfy the condition. So, only some of them do.

Therefore, for subsets of size 4, we need to count how many of them have the property that their complement is a rotation of themselves. So, how many 4-element subsets B of Z/8Z satisfy that complement(B) = π_k(B) for some k.

Alternatively, such subsets are called "rotationally complementary" or something? Maybe not standard. Let me think of how to count them.

First, note that if a subset B is such that its complement is a rotation of itself, then the rotation must be of order 2, because applying the rotation twice would map B to its complement's complement, which is B again. Wait, let me see:

Suppose π(B) = complement(B). Then, applying π again, π^2(B) = π(complement(B)). But complement(B) is π(B), so π(complement(B)) = complement(π^{-1}(B)). Hmm, this might not be straightforward. Alternatively, perhaps the rotation π must satisfy π^2(B) = B. Because π^2(B) = π(complement(B)) = complement(π^{-1}(B)). Hmm, maybe this is getting too convoluted.

Alternatively, perhaps the rotation that maps B to its complement must be such that applying it twice brings you back to the original set. Let me see. Suppose π(B) = complement(B). Then, π(complement(B)) = complement(complement(B)) = B. Therefore, π^2(B) = B. Therefore, the rotation π must be an involution (i.e., π^2 is the identity) or the order of π divides 2. But in the rotation group C8, the only elements of order dividing 2 are the identity and the rotation by 4 (since rotating by 4 twice gives rotation by 8, which is identity). Therefore, the only possible rotations π that satisfy π^2 = identity are π_0 (identity) and π_4 (rotation by 180 degrees).

Therefore, if there exists a rotation π such that π(B) = complement(B), then π must be either π_0 or π_4. But π_0(B) = B, so that would require B = complement(B), which is only possible if B is the entire set or empty, but since B has size 4, that's only possible if the entire set is partitioned into two equal parts, but in this case, B would have to be self-complementary, which would require that 4 = 8/2, so possible, but actually, B being self-complementary would mean B = complement(B), which would require that B is exactly half the vertices, but in such a way that it's equal to its complement, which is only possible if the set is its own complement, which would require that for every element in B, its complement is also in B, but since the complement is the entire set minus B, this is only possible if B is exactly half the vertices and symmetric in some way. Wait, actually, in a set with even size, a self-complementary set would need to be equal to its complement, which would require that the set is exactly half the universe, but since the complement is the other half, so the only way B = complement(B) is if the universe has even size and B is exactly half, but in that case, B can't be equal to its complement unless the universe is empty, which it's not. Wait, actually, no. Wait, if the universe has 8 elements, then the complement of B is the set of elements not in B. So, if B is equal to its complement, then B must be equal to the set of elements not in B, which implies that every element is in B if and only if it is not in B, which is only possible if the universe is empty. Therefore, there is no such non-empty set. Therefore, the only way π(B) = complement(B) with π being the identity is impossible unless B is empty, which it's not. Therefore, the only possible rotation is π_4 (rotation by 4 positions). So, the rotation by 180 degrees (4 steps). Therefore, for a subset B of size 4, the condition is equivalent to: rotating B by 4 positions gives its complement. Therefore, π_4(B) = complement(B). Therefore, such subsets B must satisfy that rotating them by 4 gives their complement. Therefore, B is a subset such that for every element x in B, x + 4 mod8 is in complement(B), and vice versa. Therefore, B and its complement are related by a 180-degree rotation. Therefore, such subsets are in bijection with subsets of 4 elements where each element is paired with its opposite element. Since rotating by 4 maps each element to its opposite. Therefore, for B to satisfy π_4(B) = complement(B), then for each element x in B, the opposite element x+4 must be in complement(B). Therefore, B must not contain both x and x+4 for any x. Wait, actually, more precisely, if x is in B, then x+4 must not be in B, because x+4 is in complement(B). Wait, but since complement(B) is exactly the set of elements not in B, so π_4(B) = complement(B) means that for any x, x is in π_4(B) iff x is in complement(B). Therefore, x is in π_4(B) means that x = y +4 mod8 for some y in B. Therefore, the set {y +4 | y ∈ B} = complement(B). Therefore, for each y in B, y +4 is not in B (since it's in complement(B)), and every element not in B is equal to y +4 for some y in B. Therefore, this implies that B is a set of four elements, none of which are opposite each other (since if y and y+4 were both in B, then y+4 would be in π_4(B), but also in complement(B), which would require y+4 not in B, contradiction). Moreover, the set B must consist of exactly one element from each of the four pairs of opposite vertices. Since the octagon has four pairs of opposite vertices: {0,4}, {1,5}, {2,6}, {3,7}. Therefore, to form such a subset B, we need to choose exactly one element from each pair. Because if you choose one from each pair, then rotating by 4 would take each chosen element to its opposite, which was not chosen, hence the rotated set would be exactly the complement. Therefore, the number of such subsets is 2^4 = 16. Because for each of the four pairs, you choose one of the two elements. Therefore, there are 16 subsets of size 4 that satisfy the condition that rotating by 4 gives the complement. Therefore, these are the only subsets of size 4 that satisfy the condition. Because only rotation by 4 can satisfy π(B) = complement(B), as we saw earlier. Therefore, the number of size 4 subsets that satisfy the condition is 16.

Wait, let me confirm this with an example. Take the pair {0,4}. If I choose 0, then I can't choose 4. Similarly for the other pairs. So, for each pair, choose one, so 2 choices per pair, 4 pairs, so 16 subsets. For example, choosing 0,1,2,3 would not be allowed because those are not choosing one from each pair. Wait, actually, {0,1,2,3} includes 0 (from pair {0,4}), 1 (from {1,5}), 2 (from {2,6}), 3 (from {3,7}), so that is a valid choice. Wait, but in this case, rotating by 4 would give {4,5,6,7}, which is exactly the complement. So that works. Similarly, choosing {0,1,2,7} would not be allowed because 7 is in the pair {3,7}, so if you choose 7, you can't choose 3. Wait, but in this case, you have 0,1,2,7. The pairs are {0,4}, {1,5}, {2,6}, {3,7}. So, for the last pair {3,7}, you chose 7 instead of 3. So that is allowed. Then rotating by 4 would give {4,5,6,3}, which is the complement of {0,1,2,7} (since the complement is {4,5,6,3}), so yes, that works. Therefore, indeed, each such subset where you pick one from each opposite pair will satisfy the condition. Therefore, there are 16 such subsets.

Therefore, subsets of size 4 contribute 16 to the total count.

Therefore, summing up all the valid subsets:

- Size 0: 1

- Size 1: 8

- Size 2: 28

- Size 3: 56

- Size 4: 16

Wait, hold on. Wait, earlier I thought that all subsets of size 3 satisfy the condition, but let me confirm again. Wait, in the case of subsets of size 3, the requirement is that there exists some rotation that shifts them to be disjoint from themselves. Since their complement is 5 elements, so shifting the subset (size 3) can be placed into the complement (size 5), so there is enough space. But does that always work?

Wait, actually, the logic earlier was that for a subset B of size k, to have a shift disjoint from itself, it suffices that the complement has size at least k, which for k=3, 8-3=5 ≥3, so possible. But does that guarantee existence?

Wait, but in the case of subsets of size 3, is there any subset where no shift can make it disjoint?

Suppose we have a subset B of size 3 such that for every shift k, B and B +k share at least one element. Let me try to construct such a subset.

Suppose B is {0,1,2}. Then shifting by 3 gives {3,4,5}, which is disjoint. So that works. How about a more spread out subset? Let's say {0, 2, 4}. Shifting by 1 gives {1,3,5}, which is disjoint. Hmm. Another example: {0,1,4}. Shifting by 2 gives {2,3,6}, which is disjoint. Hmm. Let me try a different configuration. Suppose B = {0, 3, 6}. Shifting by 1 gives {1,4,7}, which is disjoint. Hmm. Maybe it's impossible to have such a subset? Let me think of a subset where every shift overlaps.

Suppose B = {0, 1, 3}. Let's check all possible shifts:

Shift by 0: same as original, obviously overlaps.

Shift by 1: {1,2,4}. Original B is {0,1,3}. The intersection is {1}, so overlaps.

Shift by 2: {2,3,5}. Original has 3, so intersection {3}.

Shift by 3: {3,4,6}. Original has 3, so intersection {3}.

Shift by 4: {4,5,7}. No overlap with original {0,1,3}. So, shifting by 4 works. Therefore, even this set can be shifted to be disjoint.

Hmm, another try: Let me suppose B = {0, 2, 5}. Shifting by 1: {1,3,6}, which is disjoint from {0,2,5}. So that works. Hmm.

Wait, maybe it's actually impossible to have a 3-element subset that cannot be shifted to be disjoint. Let me think of a subset where every shift overlaps. Suppose such a subset exists. Let me suppose B = {a, b, c}. For every shift k, B +k shares at least one element with B.

This would mean that for every k, there exists some element in B such that (element - k) mod8 is also in B. Wait, that's equivalent to saying that for every k, there exists x in B such that x ≡ y -k mod8 for some y in B. Which is equivalent to saying that the difference between any two elements covers all possible shifts? Hmm, not sure.

Alternatively, perhaps such a subset would have to be such that the differences between elements cover all residues mod8? Not sure. Alternatively, maybe a subset forming an arithmetic progression?

Wait, let me try B = {0, 1, 2}. As before, shifting by 3 gives a disjoint set. So that doesn't work. How about B = {0, 1, 4}. Shifting by 2 gives disjoint. Hmm. Maybe it's impossible. Let me think of a different approach.

Suppose we have a 3-element subset B. The total number of possible shifts is 8. For each shift k from 1 to 7 (excluding 0), the shifted set B +k must overlap with B. To have no disjoint shift, all 7 shifts must overlap. Let me see how many overlaps are required.

Each shift k, the intersection B ∩ (B +k) must be non-empty. The total number of overlaps over all shifts is equal to the number of triples (x, y, k) such that x ∈ B, y ∈ B, and y = x +k mod8. For each pair of elements in B, their difference corresponds to a shift k where they overlap. So, for a subset B with 3 elements, there are 3*2 =6 differences (ordered pairs). Each difference corresponds to a shift k where the shift would cause an overlap. But since shifts can be in either direction, the difference y -x mod8 is a shift that would map x to y. Therefore, for each pair, two shifts (k and -k) would cause an overlap? Wait, maybe not exactly.

Wait, for each pair of distinct elements x and y in B, the shift k = y -x mod8 would cause x to map to y, hence overlapping. Similarly, the shift k = x - y mod8 would cause y to map to x, hence overlapping. Therefore, each pair contributes two shifts where the overlap occurs. However, for a subset of size 3, there are 3 pairs, so 6 shifts accounted for. But there are 7 non-zero shifts. Therefore, there is one shift remaining which is not covered by any pair's difference. Therefore, that shift would not cause an overlap, meaning that the shifted set would be disjoint. Therefore, such a shift must exist. Therefore, every 3-element subset has at least one shift that makes it disjoint from itself. Therefore, all subsets of size 3 satisfy the condition. Therefore, my initial thought was correct. Therefore, all C(8,3)=56 subsets of size 3 satisfy the condition.

Therefore, the total number of subsets that satisfy the condition is:

Size 0: 1

Size 1: 8

Size 2: 28

Size 3: 56

Size 4: 16

Total: 1 + 8 + 28 + 56 + 16 = let's compute:

1 + 8 = 9

9 +28 = 37

37 +56 = 93

93 +16 = 109

Therefore, 109 subsets satisfy the condition. Since the total number of subsets is 256, the probability is 109/256. But wait, hold on. Wait, 1+8+28+56+16 is indeed 109? Let me check again:

1 (size 0)

+8 (size1) =9

+28 (size2)= 37

+56 (size3)= 93

+16 (size4)= 109. Yes.

But wait, hold on. Wait, the problem states that each vertex is colored independently red or blue with equal probability. So, the total number of colorings is 2^8 =256, which is correct. The number of subsets B (since each coloring corresponds to a subset B of blue vertices) that satisfy the condition is 109. Therefore, the probability is 109/256. But the problem states "the probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices".

Wait, but in our analysis, we considered subsets B where there exists a rotation π such that π(B) is disjoint from B. However, the problem says "all of the blue vertices end up at positions where there were originally red vertices". The original red vertices are the complement of B. Therefore, π(B) must be a subset of the original red vertices, which is indeed the complement of B. Therefore, π(B) ⊆ complement(B), which is equivalent to π(B) and B being disjoint. So our analysis is correct. Therefore, the count is indeed 109 subsets. Therefore, the probability is 109/256.

But wait, 109 and 256: are they coprime? Let's check. 256 is 2^8. 109 is an odd number, so gcd(109,256)=1. Therefore, 109 and 256 are coprime. Therefore, m=109, n=256, so m +n= 109+256=365.

Wait, but before I conclude, let me double-check my count for subsets of size 4. Earlier, I concluded that there are 16 subsets of size 4 that satisfy the condition. Let me confirm again.

Each such subset must choose one from each of the four opposite pairs: {0,4}, {1,5}, {2,6}, {3,7}. For each pair, 2 choices, so 2^4=16. That seems correct. For example:

Choosing the first element of each pair: {0,1,2,3} – rotating by 4 gives {4,5,6,7}, which is the complement.

Choosing the second element of each pair: {4,5,6,7} – rotating by 4 gives {0,1,2,3}, which is the complement. Wait, but {4,5,6,7} is a valid subset, and it's counted in the 16. However, note that {4,5,6,7} is the complement of {0,1,2,3}, so when you rotate it by 4, you get the original set. Therefore, both subsets in each complementary pair are counted. Since each such subset is uniquely determined by choosing one from each pair, so 16 total. So that's correct.

Therefore, the total is indeed 109. Therefore, the answer should be 109 + 256 = 365? Wait, but 109 +256 is 365? Wait, 109 +256: 100+200=300, 9+56=65, so total 365. Hmm, but I have a feeling that maybe I made a mistake here because 365 is a familiar number, but let me check again.

Wait, but let me think again about subsets of size 4. The problem requires that there exists a rotation such that all blue vertices (the subset B) when rotated end up in originally red vertices (the complement of B). For subsets of size 4, this requires that there exists a rotation π where π(B) is exactly the complement of B, since they must be disjoint and of the same size. Therefore, only those subsets B where the complement is a rotation of B qualify. We counted those as 16. However, is there a possibility that some subsets of size 4 might be rotated by a different rotation (not 180 degrees) to a subset disjoint from themselves?

Wait, for example, suppose B is a subset of size 4 that is not of the form where its complement is a 180 rotation, but maybe there exists another rotation (not 180) such that π(B) is disjoint from B. Let me see.

Take for example the subset B = {0,1,2,3}. Its complement is {4,5,6,7}, which is a 180 rotation. So that's already counted. Now take another subset, say B = {0,1,2,4}. As before, its complement is {3,5,6,7}. Is there a rotation other than 180 that can map B to a subset disjoint from itself?

Let me try rotating B by 2: {2,3,4,6}. The original B is {0,1,2,4}. The intersection is {2,4}, so not disjoint. Rotating by 1: {1,2,3,5}. Intersection with B is {1,2}, not disjoint. Rotating by 3: {3,4,5,7}. Intersection with B is {4}, so not disjoint. Rotating by 5: {5,6,7,1}. Intersection with B is {1}, not disjoint. Rotating by 6: {6,7,0,2}. Intersection with B is {0,2}, not disjoint. Rotating by 7: {7,0,1,3}. Intersection with B is {0,1}, not disjoint. Rotating by 4: {4,5,6,7}, which is the complement. So, only rotation by 4 works, but since the complement is not a rotation of B (Wait, actually in this case, rotating by 4 does give the complement, but in this case, B was {0,1,2,4}, so rotating by 4 gives {4,5,6,7}, which is the complement. Wait, but in this case, does B satisfy the condition? Wait, according to our previous analysis, B must choose one from each opposite pair. However, in this case, B includes 0 and 4, which are opposites (0 and 4 are opposite?), Wait, in an octagon, vertex 0 and 4 are opposite (since 8/2=4). So, if B includes both 0 and 4, then it violates the condition of choosing only one from each pair. Therefore, this subset {0,1,2,4} is not among the 16 counted subsets, because it includes both 0 and 4. Therefore, in this case, rotating by 4 does give the complement, but since the subset includes both elements of a pair, then the complement would include their opposites? Wait, but in this case, rotating B by 4 gives the complement, but the subset B is not among the 16 because it contains both 0 and 4. Wait, so this is a contradiction?

Wait, let me check again. If B is {0,1,2,4}, then rotating by 4 gives {4,5,6,7}, which is indeed the complement of B. But according to our previous logic, such subsets must choose one from each opposite pair. However, in this case, the pair {0,4} has both 0 and 4 in B and its complement? Wait, no. B contains 0 and 4, so the complement does not contain them. Wait, but rotating B by 4 moves 0 to 4, which is also in B, so that would mean that 4 is in both B and the rotated B, so they would intersect. Wait, but in this case, rotating B by 4 gives {4,5,6,7}, which is the complement. The original B is {0,1,2,4}. The intersection between B and rotated B is {4}, so they are not disjoint. Wait, that's a problem. Wait, but the problem requires that the rotated B is disjoint from the original B. So, in this case, rotating B by 4 gives a set that is NOT disjoint from B, because 4 is in both. Therefore, this subset does NOT satisfy the condition, even though rotating by 4 maps it to its complement. Wait, but the complement of B is {3,5,6,7}, but rotating B by 4 gives {4,5,6,7}, which is not the complement. Wait, hold on, I made a mistake here.

Wait, B is {0,1,2,4}. The complement is all vertices not in B, which are 3,5,6,7. But rotating B by 4 would take each element x in B to x+4 mod8. So:

0+4=4,

1+4=5,

2+4=6,

4+4=0 mod8.

Wait, so rotating B by 4 gives {4,5,6,0}. Which is {0,4,5,6}. But the complement is {3,5,6,7}. So, {0,4,5,6} is not equal to the complement. Therefore, my mistake earlier: rotating B by 4 does not give the complement. Therefore, my previous analysis was correct. Therefore, subsets that are not choosing one from each pair do not satisfy the condition. Therefore, only the 16 subsets that choose one from each opposite pair satisfy that rotating by 4 gives the complement. Therefore, the count of 16 is correct.

Therefore, the total is indeed 109. Therefore, the probability is 109/256, so m + n = 109 + 256 = 365.

Wait, but I have a concern here. Let me check with an example of a subset of size 4 that is not in the 16, like {0,1,2,3}. Wait, {0,1,2,3} does choose one from each pair? Let's see:

The pairs are {0,4}, {1,5}, {2,6}, {3,7}. So, {0,1,2,3} chooses 0 (from first pair), 1 (second), 2 (third), 3 (fourth). None of the opposites (4,5,6,7) are included. Therefore, this is one of the 16 subsets. So rotating it by 4 gives {4,5,6,7}, which is the complement. So that works. So that is included in the 16.

Another example: {0,1,2,5}. Let's see the pairs:

0 is from first pair {0,4},

1 from {1,5}: but here we chose 1 instead of 5,

2 from {2,6},

5 is from {1,5} pair, but since we already chose 1, that's a problem. Wait, no, {0,1,2,5} includes 1 and 5? Wait, no, 5 is in the pair {1,5}. So if you choose 1, you can't choose 5. But in this subset, {0,1,2,5}, we have both 1 and 5? Wait, no, 5 is included. Wait, 5 is in the subset. Wait, but 1 and 5 are in the same pair {1,5}. Therefore, this subset includes both elements of a pair, so it's not among the 16. Therefore, rotating it by 4 would give {4,5,6,1}, which is {1,4,5,6}. The complement of the original subset {0,1,2,5} is {3,4,6,7}. So, the rotated set {1,4,5,6} is not the complement, so it doesn't satisfy the condition. Therefore, such subsets are excluded, so the count of 16 is correct.

Therefore, I think my calculation is correct. Therefore, the answer is 109 + 256 = 365. But wait, let me check once more for subsets of size 4. Let me compute how many subsets of size 4 are there: C(8,4)=70. Out of these, 16 satisfy the condition. So 16/70 is the probability for size 4. So total subsets satisfying the condition are 1+8+28+56+16=109. 109/256 is the probability. Since 109 is a prime number? Let me check. 109 divided by 2, no. 3? 1+0+9=10, not divisible by 3. 5, ends with 9. 7? 7*15=105, 109-105=4, not divisible. 11? 11*9=99, 10, no. So 109 is prime. Therefore, 109 and 256 are coprime. So m+n=365.

However, I recall that in some similar problems, the answer might be different, so I want to make sure I didn't miss anything.

Wait, let me think again about subsets of size 4. Suppose there is a subset B of size 4 which is fixed under some rotation, but not necessarily the 180 rotation. For example, suppose B is the four vertices of a square, like {0,2,4,6}. This subset is fixed under rotation by 2 (since rotating by 2 gives {2,4,6,0} which is the same set). Therefore, rotating by 2 would map B to itself, so the intersection is the entire set, so that's not disjoint. However, rotating by 1 would give {1,3,5,7}, which is the complement. Therefore, rotating by 1 would map B to its complement. Wait, so in this case, the complement is a rotation (by 1) of B. Therefore, this subset should be counted. But according to our previous analysis, {0,2,4,6} is choosing both elements from the pairs {0,4} and {2,6}, so it's not among the 16 subsets. Wait, but in this case, rotating by 1 gives the complement. Therefore, this subset does satisfy the condition, but it's not in the 16 we counted. Therefore, my previous analysis is wrong.

Hold on, this is a problem. Let me re-examine this example.

Take B = {0,2,4,6}. The complement is {1,3,5,7}. Rotating B by 1 gives {1,3,5,7}, which is exactly the complement. Therefore, this subset does satisfy the condition. However, according to my previous logic, such subsets must choose one from each opposite pair, but in this case, B contains both 0 and 4 (which are opposites), as well as 2 and 6 (also opposites). Therefore, it's selecting both elements from two of the opposite pairs. Therefore, it's not among the 16 subsets counted earlier. Therefore, my previous count was incorrect.

This means that there are more subsets of size 4 that satisfy the condition. So I need to revisit the counting for size 4 subsets.

Hmm, so my mistake was assuming that only subsets choosing one from each opposite pair would work, but in reality, there are other subsets where a different rotation (not 180 degrees) can map them to their complement.

Therefore, I need to find all subsets B of size 4 such that there exists some rotation π where π(B) = complement(B). This can happen for rotations other than 180 degrees.

Therefore, let's re-examine this.

Let me consider the subset B = {0,2,4,6}. As above, rotating by 1 maps it to the complement. So, in this case, the rotation by 1 is such that π_1(B) = complement(B). Therefore, this subset should be counted. Similarly, rotating by 3,5,7 would also map it to the complement? Let me see:

Rotating by 3: {3,5,7,1} which is the same as complement(B). So, yes, rotating by 1,3,5,7 would all map B to its complement. Therefore, this subset satisfies the condition.

Similarly, let's consider another subset: {0,1,4,5}. Let's see, complement is {2,3,6,7}. Rotating by 2: {2,3,6,7}, which is the complement. Therefore, rotating by 2 maps it to complement. Therefore, this subset also satisfies the condition. However, this subset is {0,1,4,5}. Looking at the opposite pairs:

- 0 and 4: both in B,

- 1 and 5: both in B,

- 2 and 6: neither in B,

- 3 and 7: neither in B.

Therefore, this subset includes both elements from two opposite pairs and none from the other two. Therefore, it's not among the 16 subsets that choose one from each pair. Therefore, my previous count missed these subsets.

Therefore, my earlier approach was incorrect. I need a better way to count all subsets of size 4 where there exists some rotation π with π(B) = complement(B).

Let me think of this as follows: For a subset B of size 4, the condition is that complement(B) is in the orbit of B under the rotation group. Since the rotation group acts on the subsets, the orbit of B under rotation includes all rotations of B. Therefore, complement(B) must be in the orbit of B. Therefore, the subset B must satisfy that its complement is a rotation of itself. Therefore, the subset B is such that its complement is in its rotation orbit.

Therefore, to count such subsets, we can consider the orbits of the rotation group on the subsets of size 4, and count those subsets whose orbit includes their complement.

Alternatively, perhaps it's better to think in terms of group actions and Burnside's lemma, but that might be complicated.

Alternatively, let me consider all possible subsets B of size 4 where complement(B) is a rotation of B. Let me think of the rotation that maps B to its complement. Let the rotation be by k steps. Then, π_k(B) = complement(B). Therefore, applying π_k twice, we get π_{2k}(B) = π_k(complement(B)) = complement(π_{-k}(B)). Hmm, not sure.

Alternatively, since π_k(B) = complement(B), then applying π_k again, π_{2k}(B) = complement(complement(B)) = B. Therefore, π_{2k}(B) = B. Therefore, the rotation by 2k must map B to itself. Therefore, the subset B is fixed under rotation by 2k. Therefore, the rotation by 2k is a symmetry of B.

Therefore, the subset B must be invariant under some rotation (specifically, rotation by 2k), and the rotation by k maps it to its complement.

Therefore, the possible rotations k must satisfy that 2k is an element of the stabilizer subgroup of B.

Alternatively, perhaps considering the possible rotation amounts k for which this can happen.

Let me consider possible rotation amounts k from 1 to 7 (since k=0 is identity which can't work unless B is self-complementary, which it can't be).

Let me consider the rotation by k=1:

If there exists a subset B where rotating by 1 gives the complement, then rotating by 2 would give rotating the complement by 1, which is π_1(complement(B)) = complement(π_{-1}(B)). Hmm, getting tangled.

Alternatively, let me consider specific cases where the rotation by k maps B to its complement.

Case 1: Rotation by 1.

Suppose π_1(B) = complement(B). Then, applying π_1 eight times brings us back to the original set, so π_8(B)=B. But π_1^8(B)=B. However, π_1^8 is the identity. So, that's consistent. But what does this imply about B?

If rotating by 1 each time cycles the elements, then for B to satisfy π_1(B) = complement(B), then applying π_1 again would give π_1(complement(B)) = complement(π_{-1}(B)) = complement(B rotated by -1). Hmm, not sure.

Alternatively, let me think of the elements of B as positions 0-7. If rotating B by 1 gives the complement, then for each element x in B, x+1 mod8 must be in complement(B). Therefore, x+1 is not in B. Similarly, for each x not in B, x+1 must be in B. This creates a sort of alternating pattern.

Wait, let's try to construct such a subset B.

Suppose B is such that if x is in B, then x+1 is not in B, and vice versa. Wait, but this would require that B and its complement are both unions of non-consecutive elements? Not sure.

Alternatively, let me try to build such a subset. Let me suppose that B = {0,2,4,6}. Then, rotating by 1 gives {1,3,5,7}, which is the complement. So this works. Similarly, rotating by 1 again would give {2,4,6,0}=B. So, in this case, rotating by 2 brings it back. Therefore, this subset works with k=1.

Similarly, another example: Let me try B = {0,1,4,5}. Rotating by 2 gives {2,3,6,7}, which is the complement. Therefore, this works with k=2.

Similarly, rotating by 3: Let me see if there's a subset B where rotating by 3 gives the complement. Let me try B = {0,1,2,3}. Rotating by 4 gives the complement, as before. Rotating by 3 would give {3,4,5,6}, which is not the complement. Hmm.

Alternatively, let me try B = {0,3,4,7}. Let's see, rotating by 1 gives {1,4,5,0}, which is {0,1,4,5}, not the complement. Not helpful.

Alternatively, perhaps another approach: For a subset B of size 4, the rotation that maps it to its complement must be such that the rotation amount k must satisfy that 2k is a multiple of 8/gcd(8, something). Hmm, perhaps considering the cycle structure.

Alternatively, note that in the example where B is {0,2,4,6}, rotating by 1 gives the complement. Here, the subset B is fixed under rotation by 2 (since rotating by 2 gives {2,4,6,0}=B). Therefore, the stabilizer subgroup of B includes rotation by 2. The rotation by 1 maps it to the complement, so the orbit of B under rotations includes both B and its complement. The size of the orbit is 8 divided by the stabilizer size. Since the stabilizer includes rotations by 0 and 2 (and possibly more?), let's see:

The stabilizer of B={0,2,4,6} is the set of rotations that map B to itself. Rotating by 2 maps it to itself, rotating by 4 maps it to itself as well? Rotating by 4 would map {0,2,4,6} to {4,6,0,2}=B. So, rotations by 0,2,4,6 stabilize B. Therefore, the stabilizer has size 4, so the orbit has size 8/4=2. The orbit consists of B and its complement. Therefore, in this case, the complement is in the orbit, so it works.

Similarly, for the subset {0,1,4,5}, let's see its stabilizer. Rotating by 2: {2,3,6,7}, which is the complement. Rotating by 4: {4,5,0,1}=B. So rotating by 4 stabilizes B. Rotating by 1: {1,2,5,6}, which is not B or its complement. Therefore, the stabilizer here is rotations by 0 and 4. So stabilizer size 2, so orbit size 4. The orbit includes B, rotated by 1, rotated by 2 (complement), rotated by 3, etc. Wait, but the complement is in the orbit, so that's okay.

Therefore, in general, for a subset B of size 4, if its orbit under rotation includes its complement, then it satisfies the condition. The number of such subsets is equal to the number of orbits where the orbit includes both B and its complement, multiplied by the size of the orbit, but perhaps it's better to count them by considering the possible rotation amounts.

Alternatively, perhaps the subsets B of size 4 that satisfy the condition are those that are "complementary" under some rotation, which can be of different rotation amounts. Let me consider the possible rotation amounts k for which there exists a subset B with π_k(B) = complement(B).

Let me consider possible k from 1 to 4 (since rotation by k and 8 -k are similar).

Case 1: k=1.

We have the example {0,2,4,6} works with k=1. Are there others?

Suppose B is such that rotating by 1 gives the complement. Let me try to construct another such subset.

Suppose B = {0,1,3,4}. Rotating by 1 gives {1,2,4,5}. The complement of B is {2,5,6,7}. Not equal. Doesn't work.

Alternatively, let me see if there's another subset besides the every-other vertex set.

Suppose B = {0,3,5,6}. Rotating by 1 gives {1,4,6,7}, complement of B is {1,2,4,7}. Not equal.

Hmm, maybe only the every-other set works for k=1. Let me see:

Suppose B is {0,2,4,6}. Rotating by 1 gives complement. Similarly, rotating the complement by 1 gives back B. So this is the only subset for k=1?

Alternatively, shifting the starting point: {1,3,5,7} is the complement, but rotating that by 1 gives {2,4,6,0}=B. So, the pair {B, complement(B)} forms an orbit of size 2 under rotation. Therefore, there is only one such orbit for k=1, containing two subsets. But since we are counting subsets, not orbits, this gives two subsets: {0,2,4,6} and {1,3,5,7}. Wait, but {1,3,5,7} is the complement of {0,2,4,6}, so they form a single orbit. The orbit has size 2 (since rotating by 2 steps takes B to itself, so the orbit size is 8 / stabilizer size. Wait, earlier we saw the stabilizer of B is size 4 (rotations by 0,2,4,6), so orbit size is 2. Therefore, the orbit contains B and its complement. Therefore, these two subsets form one orbit. Therefore, there are two subsets in this case.

Wait, but actually, the subset {1,3,5,7} is just the complement of {0,2,4,6}, so they are considered as two different subsets but in the same orbit. Therefore, for rotation k=1, there are two subsets that satisfy the condition (B and its complement), but since we are counting all subsets, both are included. So that's 2 subsets.

Case 2: k=2.

Let me consider subsets where rotating by 2 gives the complement.

Take the example B={0,1,4,5}. Rotating by 2 gives {2,3,6,7}, which is the complement. So this works. Let's see if there are others.

Another example: B={0,2,5,7}. Rotating by 2 gives {2,4,7,1}, which is {1,2,4,7}. The complement of B is {1,3,4,6}, which is different. Doesn't work.

Alternatively, B={0,1,3,6}. Rotating by 2 gives {2,4,5,0}, which is {0,2,4,5}. The complement is {1,3,6,7}, not matching.

Hmm, let's see if there's another subset. Let me try B={0,1,5,6}. Rotating by 2 gives {2,3,7,0}, which is {0,2,3,7}. The complement is {1,4,5,6}, not matching.

Alternatively, B={0,1,2,6}. Rotating by 2 gives {2,3,4,0}, which is {0,2,3,4}. The complement is {1,5,6,7}, nope.

Hmm, maybe only specific subsets work. Let me see the structure of B={0,1,4,5}. Rotating by 2 gives complement. Let me see the stabilizer of this set.

Rotating by 4: {4,5,0,1} which is B itself. So stabilizer includes rotation by 4. Rotating by 2 gives the complement. Rotating by 6 (which is -2 mod8) would give rotating complement by -2, which would be rotating B by 6? Let me see:

Rotating B by 4 gives B again. So stabilizer is rotations by 0 and 4. So the orbit size is 8/2=4. The orbit includes B, rotated by 1, rotated by 2 (complement), rotated by 3, etc. Therefore, the orbit has four elements. Therefore, the subset B and its complement are in the same orbit, but there are other subsets in the orbit as well. Therefore, how many subsets in this case?

The orbit has four subsets. Each of these subsets would satisfy that their complement is in the orbit, but not necessarily that the complement is exactly a rotation of themselves. Wait, but in this case, the complement is in the orbit, so there exists some rotation that maps it to the complement. Therefore, all subsets in the orbit satisfy the condition. Therefore, each such orbit contributes 4 subsets. Wait, but in this case, the orbit has four subsets, each of which when rotated appropriately can reach their complement. Therefore, all four subsets in the orbit satisfy the condition.

Therefore, to count all subsets of size 4 satisfying the condition, we need to consider all orbits where the orbit includes the complement of the subset, and count all subsets in such orbits.

This is getting complex. Maybe a better approach is required.

Alternative approach: For each possible rotation k from 1 to 4 (since rotations beyond 4 are inverses of earlier ones), count the number of subsets B of size 4 such that rotating by k gives the complement. Then use inclusion-exclusion to avoid overcounting subsets that work for multiple k.

Let me try this.

First, note that rotation by k and rotation by 8 -k are inverses, so they would give similar counts.

Let me consider rotation amounts k=1,2,3,4.

Case k=1:

We need subsets B where rotating by 1 gives the complement. Let's find how many such subsets exist.

For such a subset B, the condition is that for every x in B, x+1 mod8 is not in B (since it must be in the complement). Also, for every x not in B, x+1 must be in B.

This creates a system where the elements of B and its complement must alternate. Specifically, the subset B must consist of every other element, but shifted.

Wait, let's see:

Suppose B contains 0. Then 1 must be in the complement, so 1 is not in B. Then, since 1 is not in B, 2 must be in B. Then 3 must be in complement, so not in B. 4 must be in B, 5 not in B, 6 in B, 7 not in B. Then, since it's mod8, 0 must be in B (which it is). So this gives B={0,2,4,6}. Similarly, if we start with 1, then B would be {1,3,5,7}. These are the only two subsets that satisfy the condition for k=1. Therefore, there are 2 subsets for k=1.

Case k=2:

We need subsets B where rotating by 2 gives the complement.

Let me denote the rotation by 2 as π_2. So π_2(B) = complement(B).

This implies that π_2^2(B) = π_2(complement(B)) = complement(π_{-2}(B)). But π_2^2 is rotation by 4. So π_4(B) = complement(π_{-2}(B)). Hmm, not straightforward.

Alternatively, let's try to construct such subsets.

Let me suppose B is such that rotating by 2 gives the complement. Let me try the example B={0,1,4,5}.

Rotating by 2 gives {2,3,6,7}, which is the complement. So this works. Let's see the pattern here.

Looking at the elements:

0 is in B,

1 is in B,

2 is not,

3 is not,

4 is in B,

5 is in B,

6 is not,

7 is not.

So the pattern is two in, two out, two in, two out. Let me see if this is a requirement.

Suppose B has elements such that every pair of consecutive elements (with step 2) are included or excluded accordingly.

Alternatively, let me see if there's a general form.

Suppose B is composed of two pairs of consecutive elements separated by two elements. Like {0,1,4,5}. Similarly, {2,3,6,7} is the complement.

Another example: {1,2,5,6}. Rotating by 2 gives {3,4,7,0}, which would need to be the complement. The complement of {1,2,5,6} is {0,3,4,7}, which matches {0,3,4,7} only if rotated by 2 gives {3,4,7,0}, which is indeed the complement. Therefore, this works.

Similarly, {3,4,7,0} rotated by 2 gives {5,6,1,2}, which is the complement of {3,4,7,0} (since complement is {1,2,5,6}), so yes.

Therefore, these form orbits. Let me see how many such subsets exist.

The pattern is two consecutive elements, then two excluded, two included, two excluded. The starting point can be any of the 8 positions, but due to rotational symmetry, each distinct subset is counted multiple times.

For example, starting at 0: {0,1,4,5}

Starting at 1: {1,2,5,6}

Starting at 2: {2,3,6,7}

Starting at 3: {3,4,7,0}

Starting at 4: {4,5,0,1} (same as first one rotated by 4)

So, there are 4 distinct subsets in this pattern. Each is part of an orbit of size 4 under rotation. Therefore, there are 4 subsets for this case.

Wait, but let me count them explicitly:

1. {0,1,4,5}

2. {1,2,5,6}

3. {2,3,6,7}

4. {3,4,7,0}

These are four distinct subsets. Each of these, when rotated by 2, gives their complement. Therefore, there are 4 subsets for k=2.

Case k=3:

This is similar to k=1 since rotation by 3 is inverse of rotation by 5 (which is equivalent to -3 mod8). Let me see if there are subsets where rotating by 3 gives the complement.

Let me try constructing such a subset.

Suppose B is such that rotating by 3 gives the complement. Let me try the set {0,2,4,6} again. Rotating by 3 gives {3,5,7,1}, which is the complement. Wait, so rotating by 3 also works for this set. Indeed, rotating by 1,3,5,7 all give the complement. Therefore, the subsets {0,2,4,6} and {1,3,5,7} also satisfy the condition for k=3,5,7. Therefore, they are already counted in the k=1 case.

Are there other subsets for k=3?

Suppose we try another pattern. Let me attempt to create a subset where rotating by 3 gives the complement.

Let me suppose B = {0,1,3,6}. Rotating by 3 gives {3,4,6,1}. The complement of B is {2,4,5,7}, which is not equal to {1,3,4,6}. Doesn't work.

Alternatively, let me try B={0,1,2,7}. Rotating by 3 gives {3,4,5,2}. The complement is {3,4,5,6}, not matching.

Hmm, perhaps only the every-other set works for k=3 as well. Since rotating by 3 on {0,2,4,6} gives the complement, so those two subsets already counted in k=1 also satisfy for k=3. Therefore, no new subsets here.

Case k=4:

Rotation by 4 (180 degrees). This is its own inverse since rotating by 4 twice gives identity.

We need subsets B where rotating by 4 gives the complement. This was our initial case, where subsets choose one from each opposite pair. Let me recall that there are 16 such subsets.

Wait, let me confirm with an example. Take B={0,1,2,3}. Rotating by 4 gives {4,5,6,7}, the complement. Another example: {0,1,2,4} does not work, but {0,1,2,3} does. The 16 subsets are those choosing one from each of the four opposite pairs. Each pair contributes a binary choice, so 2^4=16.

Therefore, for k=4, there are 16 subsets.

Now, we need to consider overlaps between these cases. For instance, the subsets {0,2,4,6} and {1,3,5,7} are counted in both k=1 and k=4 (since rotating by 4 also maps them to their complements). Wait, rotating {0,2,4,6} by 4 gives {4,6,0,2}=B itself. Wait, no! Wait, rotating {0,2,4,6} by 4 gives {4,6,0,2}, which is the same as B. Therefore, rotating by 4 does NOT give the complement. Wait a second, this is a mistake.

Hold on, earlier I thought that rotating {0,2,4,6} by 4 would give the complement, but actually:

Original B = {0,2,4,6}

Rotating by 4: each element x becomes x+4 mod8.

0+4=4,

2+4=6,

4+4=0,

6+4=2.

Therefore, the rotated set is {4,6,0,2} which is exactly B itself. Therefore, rotating by 4 leaves B unchanged. Therefore, the complement is not obtained by rotating with k=4. Therefore, my earlier assertion was wrong. Therefore, these subsets are NOT included in the k=4 case.

Therefore, the subsets {0,2,4,6} and {1,3,5,7} are only counted in the k=1 and k=3 cases, not in k=4.

Therefore, the count for k=4 is indeed 16 subsets that choose one from each opposite pair, and these subsets are different from those in the k=1,2,3 cases.

Therefore, total subsets of size 4 satisfying the condition are:

From k=1: 2 subsets.

From k=2: 4 subsets.

From k=3: 0 new subsets (since they overlap with k=1).

From k=4: 16 subsets.

But wait, need to check for overlaps between these cases.

Are any subsets counted in both k=1 and k=2?

Take the subset from k=2: {0,1,4,5}. Does rotating by 1 give the complement?

Rotating by 1 gives {1,2,5,6}, which is not the complement (the complement is {2,3,6,7}), so no. Therefore, no overlap between k=1 and k=2 subsets.

Similarly, the subsets in k=2 are different from k=4's subsets.

The subsets in k=4 are those that choose one from each opposite pair, like {0,1,2,3}, which when rotated by 4 gives the complement. These are distinct from the k=1 and k=2 subsets.

Therefore, total subsets of size 4 satisfying the condition are 2 + 4 +16 =22?

Wait, but wait, let me recount:

Wait, for k=1, there are 2 subsets.

For k=2, there are 4 subsets.

For k=4, there are 16 subsets.

Total: 2+4+16=22.

But let me verify with an example:

Take the subset {0,1,2,3} (from k=4 case). Rotating by 4 gives the complement {4,5,6,7}, so it's valid.

The subset {0,1,4,5} (from k=2 case). Rotating by 2 gives the complement.

The subset {0,2,4,6} (from k=1 case). Rotating by 1 gives the complement.

These are all distinct. So 2+4+16=22.

Wait, but let me see if there are other rotation amounts like k=3 or k=5 or k=6 or k=7.

For k=5, which is equivalent to -3 mod8. Let me see if any subsets not already counted satisfy rotating by 5 gives complement.

Take the subset from k=1 case: {0,2,4,6}. Rotating by 5 gives {5,7,1,3}, which is the complement. So rotating by 5 also works for this subset, but it was already counted in k=1.

Similarly, rotating the k=2 subsets by 6 (which is -2 mod8) would also map to their complements, but they were already counted in k=2.

Therefore, the total unique subsets are indeed 2+4+16=22.

Wait, but let me check another example for k=4. Take the subset {0,1,2,4}. Wait, no, that wasn't a valid subset. Wait, the subsets for k=4 must choose one from each opposite pair. For example, {0,1,2,3} chooses 0 (not 4), 1 (not5), 2 (not6), 3 (not7). So that's valid. Another example: {0,1,6,7} would not be valid because for the pair {2,6}, it chooses 6, and for {3,7} chooses7, but then rotating by4 would give {4,5,2,3}, which is the complement? Wait, original B is {0,1,6,7}. The complement is {2,3,4,5}. Rotating B by4 gives {4,5,2,3}, which is indeed the complement. So that works. So that subset is counted in the 16.

Therefore, the total subsets of size4 are 22. Therefore, the total number of valid subsets is:

Size0:1

Size1:8

Size2:28

Size3:56

Size4:22

Total:1+8=9; 9+28=37; 37+56=93; 93+22=115.

Wait, so now the total is 115? But this contradicts my previous count. Hmm, so now I'm confused again.

Wait, let me recast:

Earlier, I thought size4 had 16, but now with the corrected analysis, it's 22. So total becomes 1+8+28+56+22=115.

But then the probability would be 115/256. Are 115 and 256 coprime? 115 factors into 5*23. 256 is 2^8. So yes, gcd is 1. So m+n=115+256=371.

But now I'm getting conflicting results based on different analyses. This suggests that my approach is flawed.

Perhaps I need a more systematic way.

Alternative approach: Instead of trying to count directly, use Burnside's lemma to count the number of subsets B where B and its complement are in the same orbit under rotation.

Wait, the problem requires that there exists a rotation π such that π(B) = complement(B). This is equivalent to B and complement(B) being in the same orbit under the rotation group.

Therefore, the number of such subsets is equal to the number of orbits where the orbit of B includes its complement, multiplied by the size of the orbit, but actually, each such subset is in an orbit that includes its complement. So the total number is the sum over all orbits O where O contains B and complement(B) for some B in O, of the size of O.

Alternatively, using Burnside's lemma, the number of orbits is equal to the average number of fixed points. But I need the number of subsets B where B and complement(B) are in the same orbit.

Alternatively, perhaps it's easier to compute the total number as follows:

For each rotation π (excluding identity?), count the number of subsets B such that π(B) = complement(B), then use inclusion-exclusion.

Wait, let me consider all rotations π (including identity). For each rotation π, the number of subsets B with π(B)=complement(B) is 2^{8 / gcd(8,k)} where k is the rotation amount? Not sure.

Alternatively, for a given rotation π_k (rotation by k steps), the number of subsets B where π_k(B) = complement(B) is equal to the number of subsets B such that B is a fixed point of the involution π_k composed with complementation.

Wait, this is getting too abstract. Let me try for each rotation π_k, find the number of subsets B where π_k(B) = complement(B).

For a given k, the equation π_k(B) = complement(B) can be rewritten as B ∪ π_k(B) = entire set, and B ∩ π_k(B) = empty set. Therefore, B and π_k(B) partition the vertex set.

Therefore, such subsets B must be a section of the rotation such that applying π_k twice brings you back to the original set. Since π_k^2(B) = π_k(complement(B)) = complement(π_{-k}(B)). For this to equal B, we need complement(π_{-k}(B)) = B, so π_{-k}(B) = complement(B). But since π_k(B)=complement(B), then π_{-k}(B)=complement(B)=π_k(B), so π_{-k}=π_k inverse, so π_{-k}(B)=π_k^{-1}(B)=complement(B). Hmm, not sure.

Alternatively, for the equation π_k(B) = complement(B) to hold, applying π_k twice gives π_{2k}(B) = π_k(complement(B)) = complement(π_{-k}(B)). But since π_k(B)=complement(B), then π_{-k}(B)=complement(π_k(B))=complement(complement(B))=B. Therefore, π_{-k}(B)=B. Therefore, B must be fixed under rotation by -k. Therefore, B is fixed under rotation by d = gcd(k,8). Let me see.

Let me consider rotation by k steps. The condition that π_{-k}(B)=B implies that B is fixed under rotation by k steps in the negative direction, i.e., rotation by -k. The set of rotations that fix B must include the subgroup generated by -k. The number of such subsets B is equal to the number of subsets fixed under rotation by d, where d = gcd(k,8). But I need to think carefully.

Alternatively, the condition π_k(B) = complement(B) and π_{-k}(B)=B implies that applying π_{2k} to B gives B back. Therefore, B must be fixed under rotation by 2k. The minimal rotation that fixes B divides 2k. This is getting too involved.

Perhaps it's better to compute for each rotation k from 0 to 7, the number of subsets B where π_k(B) = complement(B).

Let me start with k=0 (identity rotation). Then π_0(B)=B. So the equation becomes B = complement(B), which requires B is self-complementary. For a set of size 4, this is possible. The number of self-complementary subsets of size 4 is equal to the number of subsets B where B = complement(B). But this requires that for every element x, x is in B iff it is not in B, which is impossible unless the set is empty, which it's not. Therefore, there are 0 subsets for k=0.

k=1:

We need subsets B where rotating by 1 gives the complement. As before, the only such subsets are the two every-other-vertex sets: {0,2,4,6} and {1,3,5,7}. So 2 subsets.

k=2:

Need subsets where rotating by 2 gives complement. Earlier found 4 subsets: {0,1,4,5}, {1,2,5,6}, {2,3,6,7}, {3,4,7,0}. So 4 subsets.

k=3:

Similarly to k=1, since rotation by 3 is inverse of rotation by 5. The subsets here would be the same as for k=1, because rotating by 3 is equivalent to rotating by -5, but the condition π_3(B)=complement(B) would require similar alternating patterns. Let me check:

Suppose B is {0,2,4,6}. Rotating by 3 gives {3,5,7,1}, which is the complement. So this works. Similarly, {1,3,5,7} rotated by 3 gives {4,6,0,2}, which is the complement. Therefore, the same two subsets as in k=1 also satisfy k=3. Therefore, for k=3, there are 2 subsets.

k=4:

Rotation by 4. We need subsets where rotating by 4 gives the complement. These are the 16 subsets choosing one from each opposite pair.

k=5:

This is equivalent to rotation by -3. The subsets here would be the same as for k=3, so the two every-other-vertex subsets. So 2 subsets.

k=6:

Equivalent to rotation by -2. The subsets here would be the same as for k=2. So 4 subsets.

k=7:

Equivalent to rotation by -1. The subsets here would be the same as for k=1. So 2 subsets.

Therefore, summing over all rotations:

For each rotation k=0 to7:

k=0:0

k=1:2

k=2:4

k=3:2

k=4:16

k=5:2

k=6:4

k=7:2

Total: 0+2+4+2+16+2+4+2 = 32.

However, this counts for each rotation the number of subsets B where rotating by k gives the complement. But each subset B that satisfies the condition for some rotation is counted once for each rotation k that works for it.

For example, the every-other subsets {0,2,4,6} and {1,3,5,7} are counted for k=1,3,5,7 (four rotations each). Each of these subsets works for four different rotations (rotating by 1,3,5,7 all give the complement). Therefore, each is counted 4 times.

The subsets from k=2 and k=6 (the 4 subsets like {0,1,4,5}) are counted for k=2 and k=6 (two rotations each). Each such subset works for two rotations (k=2 and k=6), so counted twice.

The subsets from k=4 (the 16 subsets choosing one from each pair) are counted only for k=4, so once each.

Therefore, to get the total number of distinct subsets, we need to compute:

Let me denote:

Let A be the number of subsets that work for k=1,3,5,7 (the every-other sets): there are 2 subsets, each counted 4 times. So total count from them: 2*4=8.

Let B be the number of subsets that work for k=2,6: 4 subsets, each counted 2 times. Total count:4*2=8.

Let C be the number of subsets that work for k=4:16 subsets, each counted once. Total count:16*1=16.

Total over all rotations:8+8+16=32, which matches the earlier total.

Therefore, the actual number of distinct subsets is:

A:2 subsets,

B:4 subsets,

C:16 subsets,

Total:2+4+16=22 subsets.

Therefore, the correct count for size4 subsets is 22.

Therefore, the total number of subsets satisfying the condition is:

Size0:1,

Size1:8,

Size2:28,

Size3:56,

Size4:22,

Total:1+8=9; 9+28=37; 37+56=93; 93+22=115.

Therefore, the probability is 115/256. Since 115 and 256 are coprime (115 factors into 5×23; 256 is 2^8), so m=115, n=256, so m+n=115+256=371.

But now I'm getting 371 as the answer, which differs from my first calculation. This suggests that my initial analysis for size4 subsets was incorrect, and the correct count is 22, leading to 115 total subsets.

However, I need to verify this with another approach.

Alternative verification:

The total number of subsets B where there exists a rotation π with π(B)=complement(B).

This is equivalent to the number of subsets B such that B and its complement are in the same orbit under the rotation group.

The rotation group of the octagon is cyclic of order 8. The number of orbits of subsets of size 4 under rotation can be computed using Burnside's lemma, but this might be complex. However, perhaps we can use the formula:

The number of such subsets is equal to the number of orbits where the orbit of B includes its complement. For each orbit O, if O contains the complement of some element in O, then all elements in O satisfy the condition. Therefore, the total number is the sum over all orbits O of the size of O if O contains its complement, else 0.

Alternatively, the number is equal to the number of orbits O such that O is equal to its own complement orbit. Since complementation is an involution on the set of subsets.

This is getting too abstract. Alternatively, let me consider that for each subset B, either it and its complement are in the same orbit, or not. The total number we need is the number of subsets in orbits where B and complement(B) are in the same orbit.

The total number of such subsets is equal to the sum over all orbits O of |O| if O is closed under complementation, otherwise 0.

To compute this, we can use Burnside's lemma with the group action being rotation combined with complementation, but this is complicated.

Alternatively, given that through the previous analysis we found 115 subsets, and the total number of subsets of size up to 4 is 1+8+28+56+70=163 (wait, no, total subsets of size ≤4 is sum_{k=0}^4 C(8,k)=1+8+28+56+70=163. But our count was 115, which is less than that.

Alternatively, perhaps my second analysis is correct with 115, leading to 371. However, I need to check with an example.

Take the subset {0,1,2,3} (size4). Its complement is {4,5,6,7}. Rotating by4 gives the complement, so it is counted in the 16 subsets for k=4.

The subset {0,1,4,5} is counted in the k=2 case.

The subset {0,2,4,6} is counted in the k=1 case.

These are all distinct, so 16+4+2=22 for size4.

Therefore, I think the correct total is 115, so the answer is 371.

But now I'm confused because my initial analysis was 109, then 115. Which is correct?

Wait, let me recount the size4 subsets:

From k=1 (rotations 1,3,5,7): 2 subsets.

From k=2 (rotations 2,6):4 subsets.

From k=4:16 subsets.

Total:2+4+16=22. Yes.

Total subsets satisfying the condition:

Size0:1,

Size1:8,

Size2:28,

Size3:56,

Size4:22,

Total:1+8=9; 9+28=37; 37+56=93; 93+22=115.

Therefore, 115/256. So m+n=371.

However, I need to check if there are any other subsets of size4 that I missed.

Consider the subset {0,1,3,4}. Let's see if it can be rotated to its complement.

The complement is {2,5,6,7}.

Trying rotations:

Rotate by1: {1,2,4,5}. Intersection with original: {1,4}. Not disjoint.

Rotate by2: {2,3,5,6}. Intersection with original: {3}. Not disjoint.

Rotate by3: {3,4,6,7}. Intersection with original: {3,4}. Not disjoint.

Rotate by4: {4,5,0,1}. Intersection with original: {0,1,4,5} ∩ {0,1,3,4} = {0,1,4}. Not disjoint.

Rotate by5: {5,6,1,2}. Intersection: {1}. Not disjoint.

Rotate by6: {6,7,2,3}. Intersection: {3}. Not disjoint.

Rotate by7: {7,0,3,4}. Intersection: {0,3,4}. Not disjoint.

Therefore, this subset does not satisfy the condition, so it's correctly not counted.

Another test: subset {0,1,2,4} (size4). Its complement is {3,5,6,7}.

Rotating by4 gives {4,5,6,0}, which intersects with original at 0 and4. Not disjoint.

Rotating by1: {1,2,3,5}. Intersection with original: {1,2}. Not disjoint.

Rotating by2: {2,3,4,6}. Intersection: {2,4}. Not disjoint.

Rotating by3: {3,4,5,7}. Intersection: {4}. Not disjoint.

Rotating by5: {5,6,7,1}. Intersection: {1}. Not disjoint.

Rotating by6: {6,7,0,2}. Intersection: {0,2}. Not disjoint.

Rotating by7: {7,0,1,3}. Intersection: {0,1}. Not disjoint.

Therefore, this subset does not satisfy the condition, so it's correctly excluded.

Therefore, my count of 22 for size4 seems correct.

Therefore, the correct answer should be 115/256, so m+n=371.

But now I need to check if I made a mistake in the problem's interpretation.

Wait, the problem says "the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices".

This means that after rotation, every blue vertex is in a position that was originally red. So, the rotated blue set is a subset of the original red set. Since the original red set is the complement of the original blue set, this requires that rotated(B) ⊆ complement(B). Which is equivalent to rotated(B) and B are disjoint. So our analysis is correct.

Therefore, I think the correct answer is 371. However, I recall that in similar problems, the answer might be 109, but perhaps my second analysis is correct.

Wait, let me check the total number of subsets counted:

Size0:1

Size1:8

Size2:28

Size3:56

Size4:22

Total:115.

Total subsets:256.

Therefore, 115/256 reduces to 115/256, so m+n=371.

Alternatively, perhaps I made a mistake in the k=2 case. Let me recount the subsets for k=2.

For rotation by 2, the subsets are those where rotating by2 gives the complement. The examples were:

{0,1,4,5}, {1,2,5,6}, {2,3,6,7}, {3,4,7,0}.

These are four distinct subsets. Each of these subsets, when rotated by2, gives their complement. So that's 4 subsets.

Similarly, rotating by6 (which is -2) would also map them to their complements. So these are the only four subsets for this case.

Therefore, the count is correct.

Therefore, I think the correct answer is 371. However, I need to see if there is an error in considering the subsets for k=4.

The subsets for k=4 are those where rotating by4 gives the complement. These are subsets choosing one from each of the four opposite pairs. There are four pairs, each contributing two choices, so 2^4=16 subsets. This is correct.

Therefore, the total is indeed 22 for size4.

Therefore, I think the correct answer is 371. However, I might have made a mistake in the problem's original constraints.

Wait, the problem says "Each vertex of a regular octagon is independently colored either red or blue with equal probability." So each coloring is equally likely, and we need the probability that there exists a rotation making all blue vertices land on originally red positions.

This is exactly the same as the number of subsets B (blue vertices) for which there exists a rotation π with π(B) disjoint from B, divided by 256.

Our calculation of 115 subsets gives 115/256.

But let me see if I can find another way to compute this.

Alternative approach using inclusion:

The total number of subsets B where there exists a rotation π such that π(B) ∩ B = ∅.

This is equivalent to the union over all rotations π of the sets S_π = { B | π(B) ∩ B = ∅ }.

Therefore, by inclusion-exclusion, the count is:

|⋃_{π} S_π| = Σ|S_π| - Σ|S_{π1} ∩ S_{π2}| + Σ|S_{π1} ∩ S_{π2} ∩ S_{π3}| - ...

But this seems complicated, but maybe feasible for the rotation group of order8.

First, note that there are 8 rotations (including identity).

First, compute Σ|S_π|:

For each rotation π_k, |S_{π_k}| is the number of subsets B where π_k(B) ∩ B = ∅.

This is equivalent to B ⊆ complement(π_k(B)).

Which is equivalent to B being a subset of the complement of its own rotation.

Alternatively, for each rotation π, the condition is that B and π(B) are disjoint. Therefore, the number of such subsets B is equal to the number of subsets B where B ⊆ complement(π(B)).

Alternatively, since π is a permutation, this is equivalent to B being a subset of the complement of its image under π.

This is similar to choosing a subset B such that B and its image under π are disjoint.

The number of such subsets is equal to the number of subsets B where for each element x in B, π^{-1}(x) is not in B.

Wait, more precisely, for each x in B, π(x) is not in B.

Wait, let me think of it as a graph. The rotation π can be considered as a permutation consisting of cycles. For a rotation by k steps, the cycle structure depends on gcd(k,8).

For example:

- Rotation by 1: cycle structure is a single 8-cycle.

- Rotation by 2: two 4-cycles.

- Rotation by 4: four 2-cycles.

- Rotation by 3: gcd(3,8)=1, so single 8-cycle.

- etc.

The number of subsets B such that B and π(B) are disjoint is equal to the number of subsets independent in the graph where each element is connected to its image under π.

This is equivalent to the number of independent sets of size... Wait, actually, it's the number of subsets B such that no two elements in B are in the same orbit under π and adjacent in the cycle.

Wait, more systematically:

For a permutation π decomposed into cycles, the subsets B must not contain any two elements in the same cycle that are mapped to each other.

For a cycle of length m, the number of subsets B such that B and π(B) are disjoint is equal to the number of subsets of the cycle's elements with no two elements consecutive in the cycle (considering the cycle as a circular arrangement).

Wait, for a single cycle of length m, the number of subsets B where B and π(B) are disjoint is equal to the number of subsets where no two elements are consecutive in the cycle. Because if you have an element x in B, then π(x) cannot be in B, and since it's a cycle, this creates a non-consecutive condition.

The number of such subsets for a single m-cycle is equal to the Fibonacci number Fib(m+1). Wait, actually, for a linear arrangement it's Fib(n+1), but for a cycle it's Fib(n-1) + Fib(n+1) or something else. Let me recall:

The number of independent sets on a cycle graph C_m is Fib(m) + Fib(m-2). Wait, perhaps better to compute directly.

For a cycle of length m, the number of subsets with no two adjacent elements is Fib(m) + Fib(m-2). Wait, actually, the number is Fib(m-1) + Fib(m+1). Hmm, maybe I need to think differently.

Alternatively, for a cycle of m elements, the number of subsets with no two adjacent is equal to Fib(m) + Fib(m-2). Let me check for small m:

m=3 (triangle):

Possible subsets: {}, {1}, {2}, {3}, so 4 subsets. Fib(3)=2, Fib(1)=1, so 2+1=3, which is less than 4. Hmm, not matching.

Alternatively, the formula is Fib(m+1). For m=3, Fib(4)=3, but actual is 4. Hmm.

Wait, perhaps the formula is different. Let me compute for m=3:

Subsets with no two adjacent: the empty set, {1}, {2}, {3}. Total 4.

For m=4 (square):

Subsets: {}, {1}, {2}, {3}, {4}, {1,3}, {2,4}. Total 7.

Fib(5)=5, Fib(3)=2, so 5+2=7. So that works.

So for m=4, it's Fib(5)+Fib(3)=5+2=7.

Similarly for m=3: Fib(4)+Fib(2)=3+1=4, which matches.

Therefore, the formula is Fib(m+1) + Fib(m-1).

Wait, Fib(m+1) + Fib(m-1) = Fib(m) + Fib(m) (since Fib(m+1)=Fib(m)+Fib(m-1)), so maybe not. Wait, let me see:

Wait, for m=3:

Fib(4) + Fib(2) = 3 +1=4.

Fib(m+1) + Fib(m-1) where m=3: Fib(4)+Fib(2)=3+1=4.

Similarly for m=4: Fib(5)+Fib(3)=5+2=7.

Therefore, the general formula for the number of subsets of a cycle of length m with no two adjacent elements is Fib(m+1) + Fib(m-1). Wait, actually, that's equivalent to Fib(m-1) + Fib(m+1).

Alternatively, the number is Fib(m) + Fib(m-2). For m=3: Fib(3)=2, Fib(1)=1, so 3, which is not. Hmm.

Alternatively, perhaps the formula is Fib(m) + Fib(m-2). For m=4: Fib(4)=3, Fib(2)=1, so 4, but actual is7. No.

Alternatively, perhaps the number is Fib(m+1). For m=3, Fib(4)=3, but actual is4. No.

Alternatively, the number is Fib(m) + Fib(m-2). Hmm, not sure.

Alternatively, let me think recursively. Let C(m) be the number for a cycle of m nodes. For a cycle of m nodes, the number is equal to the number for a line of m nodes minus the cases where first and last are both chosen. The number for a line of m nodes is Fib(m+2). Wait, the number of independent sets on a path graph is Fib(n+2). For a cycle, it's Fib(n) + Fib(n-2). Hmm.

Alternatively, let me not dwell on the formula and compute for m=8.

But let's proceed with the problem.

For a rotation π_k, the permutation decomposes into cycles whose length is 8/gcd(k,8).

For example:

- For rotation by 1 (gcd=1), single cycle of length8.

- Rotation by2 (gcd=2), two cycles of length4.

- Rotation by4 (gcd=4), four cycles of length2.

Therefore, for each rotation π_k, the number of subsets B where B and π(B) are disjoint is the product over each cycle of the number of subsets for that cycle.

For a cycle of length m, the number of subsets is Fib(m+1) + Fib(m-1). Wait, as per earlier.

Wait, for a single cycle of length m, the number of subsets with no two adjacent is equal to Fib(m) + Fib(m-2). Wait, let me compute for m=8.

Alternatively, let me compute for each rotation:

First, rotation by 0 (identity). Then π_0(B) = B. So the condition is B ∩ B = empty, which requires B is empty. So only 1 subset (the empty set). But in our problem, we are considering all rotations except identity? No, the problem allows any rotation, including identity. However, for the identity rotation, the only subset satisfying B ∩ π_0(B)=∅ is the empty set.

But in our problem, we are counting all subsets B where there exists any rotation (including identity) such that π(B) and B are disjoint. Therefore, the empty set is counted via the identity rotation.

But let's proceed with computing |S_{π_k}| for each rotation π_k:

1. Rotation by 0 (identity):

As above, only the empty set satisfies B ∩ B = ∅. So |S_{π_0}| =1.

2. Rotation by1 (cycle of8):

The number of subsets B where B and π(B) are disjoint is the number of subsets of the 8-cycle with no two adjacent vertices. For a cycle of8, this is Fib(8) + Fib(6). Wait, using the earlier formula for m=8:

Fib(8)=21, Fib(6)=8, so 21+8=29? Wait, let me compute manually.

Alternatively, the number of independent sets on a cycle of8 nodes is known to be Fib(8) + Fib(6) =21 +8=29.

Wait, but let me compute it step by step.

The number of independent sets on a cycle of n nodes is equal to Fib(n-1) + Fib(n+1). For n=8:

Fib(7) + Fib(9). Fib(7)=13, Fib(9)=34, so 13+34=47? Hmm, conflicting results.

Alternatively, perhaps my initial approach is wrong. Let me use a different method.

For a cycle of n nodes, the number of independent sets is equal to Fib(n) + Fib(n-2). For n=8:

Fib(8)=21, Fib(6)=8, so 21+8=29.

Alternatively, let me think recursively. Let C(n) be the number for a cycle of n nodes, and L(n) for a line of n nodes.

We know that L(n) = Fib(n+2).

For a cycle, C(n) = L(n) - L(n-2). Because in the line, the first and last can both be chosen, but in the cycle they cannot. So subtract the cases where both first and last are chosen, which is L(n-2).

Thus, C(n) = L(n) - L(n-2) = Fib(n+2) - Fib(n).

For n=8:

L(8)=Fib(10)=55,

L(6)=Fib(8)=21,

Thus C(8)=55-21=34.

Wait, but this contradicts previous.

Alternatively, another formula: C(n) = L(n) - C(n-2). Hmm, not sure.

Alternatively, let me compute manually for small n.

For n=3 (triangle):

C(3)=4 (as before).

Using formula C(n)=L(n)-L(n-2):

L(3)=Fib(5)=5,

L(1)=Fib(3)=2,

Thus C(3)=5-2=3, but actual is4. So formula is wrong.

Hmm.

Alternatively, another approach: for a cycle of n nodes, the number of independent sets is Fib(n-1) + Fib(n+1).

For n=3: Fib(2)+Fib(4)=1+3=4. Correct.

For n=4: Fib(3)+Fib(5)=2+5=7. Correct.

For n=8: Fib(7)+Fib(9)=13+34=47.

Therefore, the correct formula is Fib(n-1)+Fib(n+1).

Thus for n=8, it's 47.

Therefore, for rotation by1 (8-cycle), the number of subsets B is47.

Wait, but this counts all subsets where no two adjacent in the cycle. However, in our problem, the condition is that B and π(B) are disjoint. Since π is rotation by1, π(B) is the set of next elements. So, the condition is that B and its rotation by1 are disjoint, which is equivalent to no two consecutive elements in B (since if x is in B, then x+1 mod8 cannot be in B). Therefore, this is exactly the independent sets on the cycle graph, which is47.

Therefore, |S_{π_1}|=47.

Similarly for rotations by3,5,7 (all give single 8-cycles), so each has47 subsets.

Rotation by2 (two 4-cycles):

Each 4-cycle contributes independently. For each 4-cycle, the number of subsets is C(4)= Fib(3)+Fib(5)=2+5=7.

Therefore, total for two independent 4-cycles:7*7=49.

Therefore, |S_{π_2}|=49.

Similarly for rotations by6 (same as rotation by-2), also two 4-cycles, so49.

Rotation by4 (four 2-cycles):

Each 2-cycle is a pair of opposite vertices. For a 2-cycle, the number of subsets where B and π(B) are disjoint: since π swaps the two elements. So, for each pair, the subsets can be:

- Choose none: allowed.

- Choose one: allowed (since the other is not chosen).

- Choose both: not allowed, since they are mapped to each other.

Thus, for each 2-cycle, there are3 choices (empty, {a}, {b}).

Therefore, for four independent 2-cycles, the total is3^4=81.

Therefore, |S_{π_4}|=81.

Rotation by other amounts:

Rotation by k=0:1 subset.

Rotations by1,3,5,7: each have47 subsets.

Rotations by2,6: each have49 subsets.

Rotation by4:81 subsets.

Total Σ|S_π| =1 +4*47 +2*49 +81.

Calculating:

4*47=188,

2*49=98,

So total:1+188=189; 189+98=287; 287+81=368.

Now, moving to the next term in inclusion-exclusion: subtract Σ|S_{π1} ∩ S_{π2}| for all pairs π1, π2.

This is going to be very complicated, but perhaps there's symmetry to exploit.

However, given the time constraints, perhaps the first term approximation is too rough, but given that the problem likely has a nice answer, and my previous detailed count gave 115, which is less than 368, but inclusion-exclusion would reduce the count.

Alternatively, perhaps the problem is designed such that the only subsets that satisfy the condition for some rotation are those counted in my second analysis (115), leading to 371.

Alternatively, perhaps my detailed count is correct, and the answer is 371.

However, given that in the initial problem statement, the answer is requested as m+n where the probability is m/n reduced. If the correct count is 115/256, then 115+256=371.

Alternatively, perhaps I made a mistake in the size4 count.

Wait, let me see the total number of subsets counted via the detailed analysis:

Size0:1,

Size1:8,

Size2:28,

Size3:56,

Size4:22,

Total:115.

Now, let me see what the first term of inclusion-exclusion (Σ|S_π|) gives 368, which counts all subsets that are "good" for at least one rotation, but with overlaps counted multiple times. The actual count is 115, which is much less than 368, so inclusion-exclusion is necessary but complex.

Given the time I've spent and the detailed analysis, I think my second count of 115 is correct, leading to 371. However, I recall that in similar problems, the answer might be 109, but perhaps I made an error in considering the subsets for rotation by2.

Wait, let me recount the subsets for rotation by2 (k=2):

The four subsets are:

1. {0,1,4,5}

2. {1,2,5,6}

3. {2,3,6,7}

4. {3,4,7,0}

These are four distinct subsets. Each of these subsets has the property that rotating by2 gives the complement. So when you rotate them by2, you get the complement. So they are valid.

Additionally, their complements are also in the list. For example, the complement of {0,1,4,5} is {2,3,6,7}, which is another subset in the list. So these form two pairs of complementary subsets. So total four subsets.

Therefore, the count is correct.

Similarly, the subsets for rotation by1 are two subsets and their complements.

Therefore, the total is indeed 22 for size4.

Therefore, I think the correct answer is 371.

However, I just found a resource that might be similar. Wait, perhaps the correct answer is 109, but I might have overcounted.

Wait, let me think of the total number of subsets that can be rotated to be disjoint from themselves:

For size0:1,

size1:8,

size2:28,

size3:56,

size4:16 (the ones choosing one from each pair).

Total:1+8+28+56+16=109.

This would give 109/256, leading to 365.

But this assumes that only the subsets choosing one from each pair for size4 work, which is incorrect because we have examples like {0,1,4,5} which also work but are not of that form.

Therefore, the error is in assuming that only the k=4 case contributes to size4 subsets, but in reality, there are more subsets from other rotations.

Therefore, the correct count must be 115, leading to 371.

Alternatively, perhaps the problem counts the identity rotation, so the empty set is counted, but in the problem statement, "all of the blue vertices end up at positions where there were originally red vertices". If B is empty, then after any rotation, the blue vertices (none) satisfy the condition trivially. So the empty set is valid.

Similarly, if B is the full set, then no rotation can move it to be disjoint, so it's excluded.

Therefore, my detailed count of 115 seems correct.

Therefore, I will go with 371 as the answer.

**Final Answer**
\boxed{371}
</think>

To solve the problem, we need to determine the probability that a randomly colored regular octagon can be rotated such that all blue vertices end up at positions where there were originally red vertices. This requires counting the number of subsets \( B \) of the vertices (blue vertices) such that there exists a rotation \( \pi \) where \( \pi(B) \) is disjoint from \( B \).

### Key Steps:

1. **Understanding the Problem**:
   - Each vertex is independently colored red or blue with equal probability.
   - We need the probability that there exists a rotation such that all blue vertices are mapped to originally red vertices.

2. **Subset Analysis**:
   - Subsets \( B \) of size 0, 1, 2, 3, or 4 can potentially satisfy the condition because larger subsets cannot be shifted to fit into their complements.
   - For subsets of size 4, the complement must be a rotation of the subset.

3. **Counting Valid Subsets**:
   - **Size 0**: 1 subset (empty set).
   - **Size 1**: All 8 subsets (any singleton can be rotated to be disjoint).
   - **Size 2**: All 28 subsets (any pair can be rotated to be disjoint).
   - **Size 3**: All 56 subsets (any triple can be rotated to be disjoint).
   - **Size 4**: Subsets must be such that their complement is a rotation. These include:
     - 2 subsets from rotation by 1 (every-other-vertex sets).
     - 4 subsets from rotation by 2 (specific patterns).
     - 16 subsets from rotation by 4 (choosing one from each pair of opposite vertices).

4. **Total Valid Subsets**:
   - Summing these, we get \( 1 + 8 + 28 + 56 + 22 = 115 \).

5. **Probability Calculation**:
   - The total number of subsets is \( 2^8 = 256 \).
   - The probability is \( \frac{115}{256} \), leading to \( m = 115 \) and \( n = 256 \).

6. **Final Answer**:
   - The sum \( m + n = 115 + 256 = 371 \).

\boxed{371}