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 with size greater than 4 cannot satisfy the condition, because their complement is smaller than themselves, so you can't fit the shifted subset into the complement. Wait, actually, more precisely, if |B| + |π(B)| > 8, then their intersection must be non-empty. Since π(B) is just a rotation, so |π(B)| = |B|. Therefore, if |B| > 4, then |B| + |B| > 8, so by pigeonhole principle, B and π(B) must intersect. Therefore, for any subset with size greater than 4, there is no rotation π such that π(B) and B are disjoint. Therefore, such subsets cannot satisfy the condition. Therefore, only subsets of size 0,1,2,3,4 can possibly satisfy the condition. Wait, but even some subsets of size 4 might not satisfy the condition. Let me see.

Wait, for subsets of size 4, if they form a "diameter" or something? Let me think. Suppose B is a set of four vertices, spaced every other vertex, like {0,2,4,6}. Then, shifting by 1 would give {1,3,5,7}, which is disjoint from the original. So that works. Alternatively, if B is four consecutive vertices, like {0,1,2,3}, then shifting by 4 would map them to {4,5,6,7}, which is disjoint. So that works. Hmm, so maybe all subsets of size 4 can be shifted to be disjoint? Wait, let me see a different configuration. Suppose B is {0,1,3,4}. Let me see if there's a shift that can make it disjoint. Let's try shifting by 2: {2,3,5,6}. The original B has 0,1,3,4. The shifted set has 2,3,5,6. The intersection is {3}, so not disjoint. Shifting by 1: {1,2,4,5}. Intersection with original is {1,4}. Shifting by 3: {3,4,6,7}. Intersection is {3,4}. Shifting by 4: {4,5,6,7}. Intersection is {4}. Shifting by 5: {5,6,7,0}. Intersection is {0}. Shifting by 6: {6,7,0,1}. Intersection is {0,1}. Shifting by 7: {7,0,2,3}. Intersection is {0,3}. Hmm, so in this case, is there any shift that makes it disjoint? Let's see shifting by 2 gives intersection at 3, shifting by 5 gives intersection at 0, shifting by 6 gives 0 and 1, shifting by 7 gives 0 and 3. Shifting by 4 gives intersection at 4. Hmm, so actually, in this case, there is no shift that makes the shifted B disjoint from original B? Wait, so this subset of size 4 cannot be shifted to be disjoint? That's a problem. Wait, let me check again.

Original B: {0,1,3,4}

Shift by 2: each element +2 mod8: 2,3,5,6. So the intersection with original B is {3}, so not disjoint.

Shift by 5: each element +5 mod8: 5,6,8 mod8=0, 9 mod8=1. So {5,6,0,1}. The intersection with original B is {0,1}, so not disjoint.

Shift by 6: +6 mod8: 6,7,9=1,10=2 mod8. So {6,7,1,2}. Intersection with B is {1}.

Shift by 7: +7 mod8: 7,0,10=2,11=3 mod8. So {7,0,2,3}. Intersection with B is {0,3}.

Shift by 1: {1,2,4,5}. Intersection with B is {1,4}.

Shift by 3: {3,4,6,7}. Intersection {3,4}.

Shift by 4: {4,5,7,0}. Wait, 0+4=4, 1+4=5, 3+4=7, 4+4=8 mod8=0. So {4,5,7,0}. Intersection with B is {0,4}.

Hmm, so in all shifts, there is at least one overlapping element. Therefore, this subset B={0,1,3,4} cannot be shifted to be disjoint from itself. Therefore, some subsets of size 4 do not satisfy the condition. Therefore, my previous thought that all subsets of size ≤4 satisfy the condition is incorrect. So, need to be careful.

Therefore, the problem is more nuanced. So, perhaps the key is to characterize all subsets B for which there exists a rotation π such that π(B) and B are disjoint. Let me think of another approach.

Alternatively, note that the condition π(B) ∩ B = ∅ is equivalent to B being a subset of the complement of π(B). Since π is a rotation, which is a bijection, so π(B) has the same size as B. Therefore, the complement of π(B) has size 8 - |B|. Therefore, for B to be a subset of the complement of π(B), we must have |B| ≤ 8 - |B|, so |B| ≤4. Therefore, only subsets of size at most 4 can satisfy the condition. Which aligns with the previous observation. So, subsets of size 5,6,7,8 cannot satisfy the condition, so they can be excluded. So, the possible subsets that can satisfy the condition are those with |B| from 0 to 4. However, even within those, as we saw, some subsets of size 4 do not satisfy the condition. Therefore, we need to count, for each size from 0 to 4, the number of subsets of that size which can be shifted to be disjoint from themselves.

Alternatively, perhaps for a subset B of size k ≤4, the number of subsets where such a shift exists can be calculated by considering for each possible shift, the subsets that are disjoint from their shift, but then using inclusion-exclusion over shifts. Hmm, but this might get complicated.

Alternatively, perhaps we can model this as follows: For a subset B, the existence of a shift k such that B and B +k are disjoint. Let me think of the shifts as elements of Z/8Z. For each subset B, define the set of shifts k such that B and B +k are disjoint. We need the subsets B where this set is non-empty.

Alternatively, for each subset B, the set of k where B +k is disjoint from B is the set of k such that (B +k) ∩ B = ∅. Let me note that this is equivalent to k ∉ (B - B), where (B - B) is the set of differences between elements of B. Wait, maybe not exactly, but there's a connection here with additive combinatorics.

Alternatively, perhaps for each subset B, the set of k such that B and B +k are disjoint is equal to the set of k not in the difference set of B. Wait, the difference set of B is {b - b' | b, b' ∈ B}, but here we are looking at k such that for all b ∈ B, (b -k) ∉ B. Wait, that is, k ∉ (B - B). Hmm, actually, let me see:

If (B +k) ∩ B = ∅, then for all b ∈ B, (b +k) mod8 ∉ B. Which is equivalent to saying that for all b ∈ B, -k ∉ (B - b). Hmm, perhaps this is getting too convoluted.

Alternatively, perhaps it's better to think in terms of the complement. Let me denote the complement of B as R = V \ B. Then, the condition that there exists a rotation π such that π(B) ⊆ R is equivalent to π(B) ⊆ R, which is equivalent to B ⊆ π^{-1}(R) = π^{-1}(V \ B) = V \ π^{-1}(B). Therefore, B and π^{-1}(B) are disjoint. Wait, but π^{-1}(B) is just the rotation in the opposite direction. So, this might not help much.

Alternatively, perhaps we can think of the problem as follows: For a subset B, the minimal number of shifts needed to cover the entire octagon? Not sure.

Alternatively, perhaps the problem is similar to the concept of a "code" with certain distance properties. If we think of the subset B as a code, then the condition that a shift exists so that B and its shift are disjoint is equivalent to the code having a certain minimum distance? Hmm, maybe not exactly.

Alternatively, let me think of the problem in terms of graph theory. Consider the octagon as a cycle graph C8. Then, the problem is about subsets of vertices such that there exists a rotation (i.e., a graph automorphism) that maps the subset to another subset disjoint from it. Hmm, but I don't see an immediate graph theory angle here.

Alternatively, perhaps we can use generating functions or something. Let me think of the total number of subsets B where there exists a shift k such that B and B +k are disjoint. Let me denote the number of such subsets as N. Then, the probability is N / 256, and we need to compute N.

To compute N, perhaps we can use inclusion over all possible shifts. For each shift k from 1 to 7 (since shift 0 would only allow B empty set), count the subsets B such that B and B +k are disjoint. But since shifts can overlap in their conditions, inclusion-exclusion would be necessary. Hmm, this might be a way.

Let me formalize this. Let S be the set of all subsets of the octagon's vertices. For each non-zero shift k (i.e., k =1,2,...,7), let A_k be the family of subsets B such that B and B +k are disjoint. Then, the subsets we want are the union over all A_k for k=1 to 7, plus the empty set (since the empty set is in every A_k). Wait, actually, the empty set is in all A_k, since shifting the empty set gives the empty set, which is disjoint from itself (since empty set intersected with anything is empty). Wait, actually, the problem allows the identity rotation? Wait, if we include shift 0, then A_0 would be the family of subsets B where B and B +0 (i.e., B itself) are disjoint, which only occurs when B is empty. So, if we consider all shifts including 0, then the total desired subsets are the union over k=0 to 7 of A_k. However, since the problem says "can be rotated", which might include the identity rotation (shift 0). But in that case, the only subset in A_0 is the empty set. However, as we saw earlier, subsets like singletons are in A_1, A_2, etc. So, perhaps the total desired subsets are the union over all A_k for k=0,...,7. But since A_0 only contains the empty set, which is already included in all other A_k (since empty set shifted by any k is still empty, which is disjoint from original empty set). Wait, actually, the empty set is in every A_k. So, the union over all A_k is just the union over k=1 to 7 of A_k, since A_0 adds nothing new except the empty set which is already in all A_k. Hmm, actually, no. Wait, the empty set is in all A_k, but other subsets may be in multiple A_k. So, to compute the total number of subsets in the union, we can use inclusion-exclusion over the shifts.

Alternatively, perhaps it's better to model this as follows:

The total number N is equal to the number of subsets B such that there exists some k ∈ Z/8Z where B ∩ (B +k) = ∅. To compute this, we can use the principle of inclusion-exclusion over the shifts. However, inclusion-exclusion over 8 shifts (including shift 0) might be complicated, but perhaps manageable.

Alternatively, maybe we can compute for each subset B, the number of shifts k for which B and B +k are disjoint, and then use some generating function approach. Hmm, not sure.

Alternatively, perhaps we can note that for a subset B, the number of shifts k where B and B +k are disjoint is equal to the number of k such that k ∉ (B - B), where (B - B) is the difference set. Wait, let me think again.

Suppose that for a given subset B, the set of forbidden shifts k are those for which there exists b, b' ∈ B such that b +k ≡ b' mod8. That is, k ≡ b' - b mod8. Therefore, the forbidden shifts are exactly the differences between elements of B. Therefore, the allowed shifts (those k where B and B +k are disjoint) are precisely those k not in the difference set of B. Therefore, the set of allowed shifts is the complement of the difference set of B in Z/8Z.

Therefore, the condition that there exists at least one allowed shift k is equivalent to the difference set of B not covering the entire group Z/8Z. In other words, the difference set of B must not be the entire group. Therefore, the subsets B we are counting are exactly those subsets whose difference sets are not the entire Z/8Z. Wait, but actually, the allowed shifts are the complement of the difference set, so if the difference set is not the entire group, then there exists some k not in the difference set, so such a shift exists. Therefore, the number N is equal to the total number of subsets B (excluding those whose difference sets cover the entire group). Wait, but actually, the empty set has an empty difference set, so it's included. The entire set has difference set the entire group, so it's excluded. So, N = total subsets - subsets whose difference sets are the entire Z/8Z. Hmm, but this might not be straightforward, since calculating the number of subsets whose difference sets cover the entire group is non-trivial.

Alternatively, perhaps this is a better approach. Let me recall that in additive combinatorics, a subset B of Z/8Z is called a "complete difference set" if its difference set covers the entire group. So, we need to count the number of subsets which are not complete difference sets. However, I don't know the exact count of such subsets, but maybe for Z/8Z, it's manageable.

Alternatively, perhaps it's easier to compute the total number of subsets B where there exists a shift k such that B and B +k are disjoint, by considering for each possible shift k, the number of subsets B that are disjoint from their shift by k, and then using inclusion-exclusion over the shifts. Let me try this approach.

Let me denote for each shift k ∈ {1,2,...,7} (since shift 0 is trivial and only allows the empty set), let A_k be the set of subsets B such that B and B +k are disjoint. Then, the total number N we want is |A_1 ∪ A_2 ∪ ... ∪ A_7|. To compute this, inclusion-exclusion formula is:

|A_1 ∪ ... ∪ A_7| = Σ|A_k| - Σ|A_k ∩ A_j| + Σ|A_k ∩ A_j ∩ A_m| - ... + (-1)^{n+1}|A_1 ∩ ... ∩ A_7}|.

This seems complicated, but maybe manageable for Z/8Z.

First, let's compute |A_k| for a single shift k. For a fixed shift k, how many subsets B are there such that B and B +k are disjoint?

Note that the condition is that for every element b in B, the element (b +k) mod8 is not in B. So, this is equivalent to B being a subset of the graph where no two elements are k apart. Wait, more precisely, the subset B must not contain any pair of elements differing by k. Wait, actually, more strictly, since if you have any element b in B, then (b +k) must not be in B. So, this is similar to a independent set in a graph where edges connect each vertex to the one k steps ahead. Since the octagon is a cycle, shifting by k corresponds to a step of k edges. So, for shift k, the condition is that B is an independent set in the graph consisting of edges {i, i +k mod8} for each i.

However, since k and 8 might have a common divisor, the graph is a union of cycles. For example, if k=1, the graph is the octagon itself, so a cycle of length 8. If k=2, then the graph is two cycles of length 4 each (since stepping by 2 in an 8-cycle splits it into two 4-cycles). Similarly, for k=4, it's four edges (since stepping by 4 in 8 vertices gives 4 diameters). Wait, let me confirm:

For shift k, the graph is a collection of cycles, each of length 8 / gcd(k,8). So, for example:

- For k=1: gcd(1,8)=1, so one cycle of length 8.

- For k=2: gcd(2,8)=2, so two cycles of length 4.

- For k=3: gcd(3,8)=1, so one cycle of length 8.

- For k=4: gcd(4,8)=4, so four cycles of length 2.

- Similarly, k=5 is same as k=3 (since 5 mod8 is equivalent to -3), so gcd(5,8)=1, cycle length 8.

- k=6: gcd(6,8)=2, so two cycles of length 4.

- k=7: gcd(7,8)=1, so cycle length 8.

Therefore, for each shift k, the graph is a union of cycles as above, and the independent sets in such a graph can be counted using standard formulas.

The number of independent sets on a cycle of length n is known to be Fibonacci(n+2). Wait, actually, the number of independent sets on a cycle graph C_n is F_{n} + F_{n-2} or something? Let me recall:

For a path graph P_n, the number of independent sets is F_{n+2}. For a cycle C_n, it's F_{n} + F_{n-2} or similar? Wait, actually, the formula is F_{n} + F_{n-2}? Wait, let me check for small n.

Wait, for a cycle of 3 nodes (triangle), the independent sets are:

- All single vertices: 3

- The empty set: 1

- No two vertices can be chosen together, so total 4. Fibonacci(3 + 2) = Fibonacci(5)=5, which is not matching. Hmm, so maybe different formula.

Alternatively, the number of independent sets on a cycle C_n is F_{n} + F_{n-2}? For n=3, that would be F_3 + F_1 = 2 + 1 = 3, which is still less than 4. Hmm, perhaps another approach.

Alternatively, the number of independent sets on a cycle C_n is equal to the number for a path P_n minus the cases where the first and last nodes are both selected. Wait, maybe better to use recurrence.

Let me recall that for a cycle C_n, the number of independent sets is equal to F_{n-1} + F_{n+1}. Wait, let me see for n=3:

F_2 + F_4 = 1 + 3 = 4, which matches. For n=4: F_3 + F_5 = 2 + 5 =7. Let's check:

For C4, the independent sets:

- Empty set:1

- Single vertices:4

- Pairs of opposite vertices:2 (since in a square, two pairs of opposites)

Total:1+4+2=7. Yes, that works. So the formula is F_{n-1} + F_{n+1} for a cycle C_n.

Alternatively, another way, the number is F_{n} + F_{n-2}?

Wait, for n=3: F_3 + F_1 = 2 +1=3, which is not 4. Hmm, so perhaps the first formula is correct: F_{n-1} + F_{n+1}.

Alternatively, let me just use the standard formula. The number of independent sets on a cycle graph C_n is indeed F_{n} + F_{n-2}, but maybe my initial terms are off. Let me check the standard result.

Wait, according to some references, the number of independent sets on a cycle graph C_n is equal to F_{n-1} + F_{n+1}. For example, for n=3: F_2 + F_4 =1 +3=4, which is correct. For n=4: F_3 + F_5=2+5=7, which is correct. For n=5: F_4 + F_6=3+8=11. Let me check C5:

Independent sets on C5:

- Empty set:1

- Single vertices:5

- Pairs: each pair of non-adjacent vertices. In a pentagon, each vertex has two neighbors, so non-adjacent are two apart. The number of such pairs:5 (each vertex can pair with the one two steps away, but each pair counted twice? Wait, actually in C5, the number of edges is 5, and the number of independent pairs is 5 (each vertex has two non-adjacent vertices, but this counts each pair twice, so total 5). Wait, actually, for each vertex, two non-adjacent vertices, so total 5*2/2=5 pairs.

- Triples: cannot have three vertices all non-adjacent in C5. Because in a pentagon, any three vertices must have at least two adjacent. So no triples.

- So total is 1+5+5=11, which matches. So the formula holds.

Therefore, the formula is F_{n-1} + F_{n+1}, where F_n is the nth Fibonacci number with F_1=F_2=1, F_3=2, etc.

Therefore, for a cycle of length m, the number of independent sets is F_{m-1} + F_{m+1}.

Therefore, returning to our problem, for a shift k with gcd(k,8)=d, the graph is a union of d cycles each of length 8/d. Therefore, the number of independent sets is the product over each cycle's independent sets. Since the cycles are disjoint, the total independent sets are the product of the independent sets for each cycle.

Therefore, for shift k with gcd(k,8)=d, the number of subsets B such that B and B +k are disjoint is equal to [F_{(8/d)-1} + F_{(8/d)+1}]^d.

Wait, let me clarify:

Suppose that the shift k divides the octagon into d cycles each of length m=8/d. Then, for each cycle, the number of independent sets is F_{m-1} + F_{m+1}, and since the cycles are independent, the total is [F_{m-1} + F_{m+1}]^d.

Therefore, for each shift k:

- If gcd(k,8)=1 (so m=8, d=1), then the number is F_7 + F_9. Let's compute Fibonacci numbers:

F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34.

Thus, for gcd=1, the number is (F_7 + F_9) =13 +34=47? Wait, no, wait the formula was F_{m-1} + F_{m+1} where m=8. So F_{7} + F_{9}=13 +34=47. So yes, 47.

Wait, but hold on, the formula was for a single cycle of length m=8, so the number of independent sets is F_{7} + F_{9}=47. So that's for shift k with gcd(k,8)=1 (i.e., k=1,3,5,7).

- For gcd(k,8)=2 (so d=2, m=4). Then, each cycle has length 4, so the number per cycle is F_{3} + F_{5}=2 +5=7. Since there are two cycles, total is 7^2=49.

- For gcd(k,8)=4 (so d=4, m=2). Each cycle is a cycle of length 2 (i.e., an edge). The number of independent sets on a cycle of length 2: F_{1} + F_{3}=1 +2=3. Since there are four cycles, total is 3^4=81.

Wait, but hold on, a cycle of length 2 is just two nodes connected in a cycle, so each cycle of length 2 has two nodes connected to each other. The independent sets on a 2-node cycle are:

- Empty set,

- {first node},

- {second node}.

So 3 independent sets, which matches 3. So, yes, 3^4=81 for gcd=4.

Similarly, for gcd= other divisors? Well, 8's divisors are 1,2,4,8. Since k is between 1 and7, gcd can't be 8.

Therefore, summarizing:

For a shift k:

- If gcd(k,8)=1 (k=1,3,5,7): number of subsets B with B ∩ (B +k)=∅ is 47.

- If gcd(k,8)=2 (k=2,6): number is 49.

- If gcd(k,8)=4 (k=4): number is 81.

Wait, let me confirm for k=4. Since shifting by 4, the graph becomes four cycles of length 2 (each pair of opposite vertices). So each pair is a cycle of 2. The independent sets for each pair are 3 (as above). So total is 3^4=81. That's correct.

Similarly, for k=2, which splits into two cycles of 4 nodes each. Each 4-cycle contributes 7 independent sets, so 7^2=49. Correct.

Therefore, for each shift k:

- k=1,3,5,7: 47 subsets each.

- k=2,6: 49 subsets each.

- k=4:81 subsets.

Therefore, for each shift, the number of subsets in A_k is as above. However, note that the empty set is included in all of these counts, since the empty set is an independent set in every graph.

Now, moving on to inclusion-exclusion. Let me denote the shifts as k=1 to 7. Let me first compute the first term: Σ|A_k|.

There are four shifts with gcd=1 (each contributing 47), two shifts with gcd=2 (each 49), and one shift with gcd=4 (81). So total Σ|A_k| = 4*47 + 2*49 +1*81 = let's compute:

4*47 = 188

2*49 = 98

1*81 =81

Total: 188 +98 = 286; 286 +81= 367.

So first term is 367.

Now, the second term is Σ|A_k ∩ A_j| over all 1 ≤ i < j ≤7. This is going to be complicated, but perhaps we can find a pattern or formula.

The intersection A_k ∩ A_j corresponds to subsets B such that B is disjoint from both B +k and B +j. Equivalently, B must be an independent set for both shifts k and j. This is equivalent to B being an independent set in the graph which is the union of the edges from both shifts k and j.

Alternatively, the combined graph for shifts k and j would have edges corresponding to both shifts. The structure of this graph depends on the relationship between k and j.

This seems very involved. Maybe there's a smarter way. Alternatively, perhaps we can use the principle that for each subset B, the number of shifts k for which B is in A_k is equal to the number of k not in the difference set of B. Therefore, the total over all subsets B of the number of shifts k for which B ∈ A_k is equal to Σ_{k=1}^7 |A_k| = 367 (excluding shift 0? Wait, actually, in our previous calculation, we considered shifts from 1 to7, so 7 shifts. Wait, but actually, in the problem, the shift can be any rotation, including shift 0. Wait, but in our current inclusion-exclusion, we are considering shifts 1 to7, since shift0 only allows the empty set. Hmm, perhaps I need to be careful here.

Alternatively, perhaps instead of inclusion-exclusion over all shifts, we can use Möbius inversion over the rotation group. Wait, maybe that's overcomplicating.

Alternatively, perhaps we can use generating functions or consider the problem in terms of the cycle index of the group action. Wait, Burnside's lemma might be applicable here. Let me think.

Wait, Burnside's lemma says that the number of orbits is equal to the average number of fixed points. But how does that relate here?

Alternatively, perhaps we can model this problem as counting the number of subsets B such that there exists a rotation π where π(B) is disjoint from B. This is equivalent to the existence of some group element (rotation) such that π(B) ∩ B = ∅. So, in other words, the subset B is not "fixed" in the sense that it doesn't intersect all its rotations. Hmm, not sure.

Alternatively, perhaps consider that each subset B has a stabilizer subgroup under the rotation action, but I don't see the connection.

Alternatively, perhaps think of the problem as follows: For each non-empty subset B (since empty set is trivial), the minimal number of shifts needed to cover all elements? Not sure.

Alternatively, perhaps it's better to proceed with inclusion-exclusion, even though it's tedious. Let me try to compute the second term Σ|A_k ∩ A_j|.

To compute |A_k ∩ A_j|, we need the number of subsets B such that B is disjoint from both B +k and B +j. This requires that for all b ∈ B, b +k ∉ B and b +j ∉ B. Additionally, since the shifts can interact, we have to ensure that the combined constraints are satisfied.

Alternatively, this is equivalent to B being an independent set in the graph where edges connect each vertex to the vertices k and j steps ahead. The structure of this graph depends on the steps k and j.

This is getting very case-based. Perhaps we can consider different cases based on the values of k and j.

Alternatively, note that the problem is symmetric with respect to rotation, so the value of |A_k ∩ A_j| depends only on the difference between k and j, or their relationship in terms of divisors.

Alternatively, let me pick specific shifts and compute |A_k ∩ A_j| for some pairs, then see if a pattern emerges.

Let me first consider two shifts k and j that are related by a common divisor. For example, let's take k=1 and j=2.

To compute |A_1 ∩ A_2|, which is the number of subsets B such that B is disjoint from B+1 and B+2.

This means that for any b in B, neither b+1 nor b+2 can be in B. So, the subset B must not contain any elements within 1 or 2 steps of any of its elements. Wait, actually, more precisely, the conditions are:

- If b is in B, then b+1 mod8 and b+2 mod8 cannot be in B.

But also, since the shift by 1 requires that B and B+1 are disjoint, so B cannot contain any two elements differing by 1. Similarly, shift by 2 requires that B and B+2 are disjoint, so B cannot contain any two elements differing by 2. Therefore, the subset B must not have any two elements differing by 1 or 2. So, the elements of B must be at least 3 apart from each other.

Wait, actually, more precisely, the difference between any two elements must not be 1 or 2 mod8. Wait, but since differences can be in either direction, so the minimal distance between any two elements must be at least 3.

Therefore, such subsets are independent sets in the graph where edges connect vertices at distance 1 or 2. Let me think of this graph.

In an octagon, connecting each vertex to its 1st and 2nd neighbors. So each vertex is connected to the next two vertices. This creates a graph where each node has degree 4? Wait, no, each node is connected to two previous and two next? Wait, no, if we consider undirected edges for differences of 1 and 2, then each vertex is connected to the vertices ±1 and ±2 mod8. So each vertex has degree 4. The graph is the union of two cycles: the original octagon (edges of length 1) and the "square" connections (edges of length 2). Hmm, actually, the graph is the octagon with edges also connecting every other vertex, forming a octagon with "chords" of length 2. This graph is known as the Wagner graph, which is a 3-regular graph? Wait, no, each vertex has degree 4, so not Wagner. Wait, maybe it's the cube graph? No, the cube is 3-regular. Hmm.

Alternatively, perhaps it's a 4-regular graph on 8 nodes. The number of independent sets in such a graph might be tricky to compute, but maybe manageable for small n.

Alternatively, perhaps we can model this as a graph and compute the number of independent sets recursively or via inclusion-exclusion.

Alternatively, let me try to count manually for this specific case (shifts 1 and 2).

We need subsets B where no two elements are 1 or 2 apart. Let me think of the octagon as vertices 0-7.

Let me try to count the number of such subsets.

First, the empty set is allowed.

Single elements: Any single vertex is allowed. There are 8.

Pairs: Need to choose two vertices with at least 3 apart. Let's see, for vertex 0, possible pairs are with 3,4,5 (since 0+3=3, 0+4=4, 0+5=5; but 0 and 3 are 3 apart, which is allowed, 0 and 4 are 4 apart (diametrically opposite), which is allowed, 0 and 5 is 5 apart, which is equivalent to 3 apart the other way. Wait, actually, the minimal distance between 0 and 5 is 3 (since 8-5=3). So, the minimal distance must be at least 3. So, for two vertices, they must be at least 3 apart in either direction.

So, the number of pairs: For each vertex, the forbidden vertices are those at distance 1,2, so 4 vertices (left1, left2, right1, right2). So, for each vertex, there are 8 -1 -4 =3 vertices allowed. But since this is circular, need to adjust for overlap.

Alternatively, the number of pairs with distance at least 3:

In an octagon, the number of pairs at distance exactly d is 8 for each d=1,2,3,4 (since distance 4 is diameter). Wait, actually, for distance 1: 8 pairs (each edge).

Distance 2: 8 pairs (each two apart).

Distance 3: 8 pairs (three apart).

Distance 4:4 pairs (diameters).

Wait, actually, for distance 3, each pair is counted twice? Wait, no, in an octagon, for distance 3 from a vertex, there are two vertices at distance 3 (clockwise and counter-clockwise), but since distance is minimal, the distance between 0 and 3 is 3, and between 0 and 5 is also 3 (since 8-5=3). So, each vertex has two vertices at distance 3. So total pairs at distance 3: 8*2 /2 =8.

Similarly, distance 4 is diameter, so 4 pairs.

Therefore, the number of pairs with distance ≥3 is the pairs at distance 3 or 4: 8 +4=12. So 12 pairs.

Triplets: Need to choose three vertices, each pair at least distance 3 apart. This is more complex.

Let me see if such triplets exist. Let's try to place three vertices each at least 3 apart.

Suppose we pick vertex 0. Then the next available vertex can be at 3. Then from 3, the next must be at least 3 away from both 0 and 3. From 3, the next could be 6 (distance 3 from 3, distance 6 from 0 (which is 2 mod8, but minimal distance is 2? Wait, 6 is 6 away from 0, but minimal distance is min(6, 2)=2, which is less than 3. So that's not allowed. Hmm.

Alternatively, starting at 0, next at 4 (distance 4). Then from 4, next must be at least 3 away from 4 and 0. From 4, the next could be 7 (distance 3 from 4, distance 7 from 0 (which is 1 mod8, minimal distance 1, which is too close to 0). Not good.

Alternatively, 0, 3, and 6: distance between 0 and 3 is 3, 3 and 6 is 3, but 0 and 6 is 6 mod8, minimal distance 2, which is too close. So that's invalid.

Hmm, maybe it's impossible to have three vertices all at least distance 3 apart in an octagon. Let me see:

Suppose we have three vertices. The total "space" they occupy with their forbidden zones (distance 1 and 2) would be 3*(1+2)*2 = 18, but since the octagon only has 8 vertices, this is overlapping a lot. Maybe it's impossible.

Wait, let me try another arrangement. Let's pick vertices 0, 3, and 6. As before, 0 and 3 are okay, 3 and 6 are okay, but 0 and 6 are distance 6 (which is equivalent to 2 the other way), so too close. Not allowed.

Another try: 0, 4, and 1. Wait, 0 and 1 are too close. Not allowed.

Alternatively, 0, 3, and 7. Distance from 0 to 3 is 3, 0 to7 is 1 (too close). Nope.

Hmm, maybe it's impossible to have three vertices. Let me see:

Suppose we have three vertices. Each vertex forbids the next two vertices in each direction. So, placing the first vertex at 0 forbids 1,2,7,6. Then the next vertex must be at 3,4, or5. Suppose we place at 3, which forbids 4,5,2,1 (but 1 and 2 already forbidden). Then the next vertex must be at 6 or 5 or something, but 5 is forbidden by 3, 6 is forbidden by 0. Hmm, seems impossible.

Therefore, perhaps there are no triplets. Similarly, larger subsets are impossible. So the only subsets in A_1 ∩ A_2 are the empty set, single vertices (8), and pairs (12). So total is 1 +8 +12=21.

Wait, but let me confirm the pairs. Earlier, I thought there are 12 pairs, but let me recount.

Each pair must be at distance 3 or 4.

Distance 3: For each vertex, two vertices at distance 3, so total pairs: 8*2/2=8.

Distance 4: diameters, 4 pairs.

Total pairs: 8 +4=12. So that's correct.

Therefore, total subsets in A_1 ∩ A_2 is 1 +8 +12=21.

Hmm, so |A_1 ∩ A_2|=21.

Wait, but let me see if there's another way to compute this. Since the constraints are that B must be an independent set in the graph where edges are steps of 1 and 2. The graph is 4-regular, as each node connected to ±1, ±2.

Alternatively, perhaps the number of independent sets can be calculated via recursion or something, but for now, let's take 21 as the value.

Now, let's see another intersection, say A_1 ∩ A_3.

Shifts k=1 and j=3. So, the conditions are that B is disjoint from B+1 and B+3.

Therefore, for any b in B, b+1 and b+3 cannot be in B. Additionally, since the shift by 3 also imposes that b-3 mod8 cannot be in B (since if b is in B, then (b-3)+3 =b must not be in B if (b-3) is in B). Wait, actually, the conditions are:

- If b ∈ B, then b+1 ∉ B and b+3 ∉ B.

But also, since shift by 3: B and B+3 are disjoint, so if c ∈ B, then c -3 ∉ B.

Therefore, the forbidden differences are 1 and 3. So, the subset B must not contain any two elements differing by 1 or 3.

Therefore, the minimal distance between any two elements must be more than 1 and not equal to 3. Wait, but difference of 3 is forbidden, so differences must not be 1 or 3. So allowed differences are 2,4,5,6,7 mod8. But minimal distances would be 2,4, etc.

Wait, perhaps it's easier to model this as a graph where edges connect vertices at differences 1 and 3, then count independent sets.

Alternatively, let me try to count manually.

Again, starting with the empty set (1), single vertices (8).

Pairs: Need two vertices with difference not 1 or 3. Let's see:

Possible differences allowed are 2,4,5,6,7. But minimal differences would be 2,4, or 5 (since 5 mod8 is equivalent to -3, which is difference 5, but minimal distance is 3 the other way? Wait, minimal distance is the smallest of |a-b| and 8 - |a-b|. So difference of 5 has minimal distance 3 (since 8-5=3). So differences of 5 are equivalent to minimal distance 3, which is forbidden. Similarly, difference of 6 is minimal distance 2 (8-6=2), so difference 6 is equivalent to minimal distance 2, which is allowed? Wait, no, the forbidden differences are 1 and 3. The actual difference of 2 is allowed, but difference of 6 would have minimal distance 2, which is allowed, but the actual difference is 6, which is not 1 or 3, so allowed.

Wait, perhaps I need to think in terms of actual differences, not minimal distances. The condition is that the difference between elements cannot be exactly 1 or 3 mod8. So, for two elements a and b, |a - b| mod8 cannot be 1 or 3. So, allowed differences are 0,2,4,5,6,7 mod8. But since we are considering pairs, difference 0 is same element, so allowed differences for pairs are 2,4,5,6,7. But since difference 5 is equivalent to -3 mod8, which is forbidden? Wait no, the forbidden differences are exactly 1 and 3. So difference of 5 is allowed, since it's not 1 or 3. So, pairs can have differences of 2,4,5,6,7.

But minimal distance for difference 5 is 3 (since 8-5=3), but the actual difference is 5, which is allowed. So, the minimal distance is not the concern, just the actual difference.

Therefore, for pairs:

The number of pairs with difference not equal to 1 or 3 mod8.

Total pairs: C(8,2)=28.

Forbidden pairs: those with difference 1 or 3. There are 8 differences of 1 (the edges) and 8 differences of 3 (since for each vertex, the vertex three ahead). So total forbidden pairs: 8 +8=16. Thus allowed pairs:28-16=12. Wait, but this contradicts previous count where for shifts 1 and 2, we had 12 pairs. Wait, but in this case, the forbidden differences are 1 and 3, so 16 forbidden pairs, so 28-16=12 allowed pairs. So same as before. Wait, but in the previous case with shifts 1 and 2, the forbidden differences were 1 and 2, so forbidden pairs would be difference 1 and 2, which is 8+8=16, so allowed pairs also 12. So in both cases, the number of pairs is 12. Interesting.

But in the first case (shifts 1 and 2), the minimal distances had to be at least 3, but here, with shifts 1 and 3, the allowed pairs can have minimal distance 2 (difference of 6, which is minimal distance 2). So the actual count is same, but the structure is different.

However, for triplets, the counts might differ.

Continuing with the current example (shifts 1 and3):

Triplets: Need three vertices where no two differ by 1 or 3.

Let me try to construct such a triplet.

Start with 0. Then cannot have 1,3,7 (since 0-1 mod8 is 7, which is allowed? Wait, differences are absolute. Wait, differences of 1 or 3 in either direction are forbidden. So from 0, forbidden vertices are 1,7 (difference 1), and 3,5 (difference 3). So forbidden are 1,3,5,7. So available vertices are 2,4,6.

Pick 2 next. Now from 2, forbidden differences: 1 and3. So forbidden vertices are 2±1=1,3 and 2±3=5,7. But already forbidden from 0. Also, from 2, the forbidden vertices are 1,3,5,7. So remaining available are 4,6.

Pick 4 next. Check differences with 0 and 2:

Difference between 4 and0 is4 (allowed), between4 and2 is2 (allowed). So okay.

Now, check if adding 4 introduces new forbidden vertices. From 4, forbidden are 4±1=3,5 and 4±3=1,7. Already forbidden.

Now, can we add another vertex? The remaining available is 6.

Check 6 with 0: difference 6 (allowed), with 2: difference4 (allowed), with4: difference2 (allowed). So adding 6 gives a quartet {0,2,4,6}. Let's check all differences:

0-2:2, 0-4:4, 0-6:6; 2-4:2, 2-6:4; 4-6:2. None of these differences are 1 or3. So this quartet is allowed.

Wait, so there exists a quartet. Hmm, so maybe my previous assumption was wrong.

Wait, so {0,2,4,6} is a valid subset. Let's see:

Shifting by1: {1,3,5,7}, which is disjoint from original.

Shifting by3: {3,5,7,1}, also disjoint. So yes, this subset is in A_1 ∩ A_3.

Therefore, there are larger subsets. So my previous approach was flawed.

This shows that the count is more involved. Let me try again.

Perhaps it's better to model this as a graph where edges represent forbidden adjacencies (differences of 1 or3) and then count independent sets.

The graph for shifts 1 and3 has edges between vertices differing by1 or3. Let me see the structure.

Starting at 0 connected to1,7 (diff1) and3,5 (diff3).

1 connected to0,2 and4,6 (diff1 and3).

2 connected to1,3 and5,7.

3 connected to2,4 and0,6.

4 connected to3,5 and1,7.

5 connected to4,6 and0,2.

6 connected to5,7 and2,4.

7 connected to6,0 and3,5.

This graph is quite connected. Each node has degree 4.

Looking for independent sets in this graph.

The independent sets include:

- The empty set.

- All single vertices (8).

- Pairs: as calculated, 12.

- Triplets: Let's see if there are any.

Take vertices 0,2,4. Check edges:

0 and2: difference2 (allowed), 0 and4: difference4 (allowed), 2 and4: difference2 (allowed). So this is an independent set.

Similarly, 0,2,4,6 is also independent.

Wait, so {0,2,4,6} is an independent set of size4.

Similarly, {1,3,5,7} is another.

Are there triplets not in the quartet?

Like {0,2,6}: differences 2,4,6. All allowed. So yes, that's a triplet.

Wait, 0 to2:2, 0 to6:6 (allowed), 2 to6:4 (allowed). So yes.

Similarly, {0,2,6} is a triplet.

Similarly, {0,4,6}: differences4,2,2 (wait 4 to6 is2, which is allowed). So yes.

Hmm, so there are multiple triplets and even a quartet.

This is getting too time-consuming to count manually. Perhaps there's a better way.

Alternatively, notice that the graph for shifts 1 and3 is actually two disjoint squares. Wait, let me see:

Looking at the connections:

0 connected to1,7,3,5.

If I follow the connections, perhaps the graph is composed of two squares?

Wait, let me see the connections for even and odd vertices.

Vertices 0,2,4,6 are even; 1,3,5,7 are odd.

Looking at edges between even vertices:

0 connected to2? Difference2, which is not forbidden (since forbidden differences are1 and3). Wait, in the graph for shifts1 and3, edges are only for differences1 and3. So difference2 is not an edge. So even vertices are connected only if their difference is1 or3. The difference between 0 and2 is2, so no edge. Similarly, 0 and4 differ by4, no edge. 0 and6 differ by6 (equiv to -2), no edge. So even vertices are not connected among themselves via edges. Wait, but 0 and3 are connected because difference3. Wait, 0 is even, 3 is odd.

Wait, actually, the even and odd vertices form two separate components?

Wait, let me see:

Starting at0 (even):

Neighbors are1 (odd),7 (odd),3 (odd),5 (odd). So all neighbors of an even node are odd.

Similarly, neighbors of an odd node (say1) are0 (even),2 (even),4 (even),6 (even). So the graph is bipartite between even and odd vertices, with edges between evens and odds that are difference1 or3.

Therefore, the graph is bipartite with partitions Evens {0,2,4,6} and Odds {1,3,5,7}, and edges between even and odd nodes that are 1 or3 apart.

Therefore, the independent sets in a bipartite graph can be calculated as the sum over subsets of one partition with no edges between them. Since it's bipartite, the independent sets can be formed by choosing any subset of Evens and any subset of Odds such that no edges between them. Wait, but actually, in a bipartite graph, an independent set can include nodes from both partitions as long as no two are adjacent. So the maximum independent set could be one of the partitions, but there are also smaller ones.

Alternatively, the number of independent sets in a bipartite graph can be calculated via the product of the number of independent sets in each partition considering dependencies. Hmm, perhaps not straightforward.

Alternatively, since the graph is bipartite between Evens and Odds, with edges between certain pairs, the number of independent sets is equal to the number of subsets of Evens times the number of subsets of Odds that don't conflict, but this is still complex.

Alternatively, since the graph is bipartite, the number of independent sets is equal to the product of (1 + number of independent sets on each side). Wait, no, that's not correct.

Alternatively, let me think of it as two sets E and O, with edges between them. An independent set can choose any subset of E and any subset of O such that no node in the chosen E is connected to any node in the chosen O.

Therefore, the total number is the sum over all subsets S of E and T of O such that there are no edges between S and T.

This is equivalent to the number of pairs (S,T) where S ⊆ E, T ⊆ O, and for all s ∈ S, t ∈ T, there is no edge between s and t.

This is equal to the product over each node in E and O of (1 + x), but with dependencies. Hmm, perhaps not helpful.

Alternatively, let me note that the graph is actually a 4x4 bipartite graph with specific edges. Let me see the adjacency between E and O.

Each even node (0,2,4,6) is connected to four odd nodes:

- 0 is connected to1,3,5,7 (diff1 and3: 0+1=1, 0-1=7; 0+3=3, 0-3=5).

Wait, actually:

Wait, for even node 0:

Difference1: 0±1=1 and7.

Difference3:0±3=3 and5.

So connected to1,3,5,7.

Similarly, even node2:

Difference1:2±1=1,3.

Difference3:2±3=5,7.

Thus connected to1,3,5,7.

Wait, same as 0? Wait, 2+3=5, 2-3= -1 ≡7 mod8. So yes, connected to1,3,5,7.

Wait, similarly, even node4:

4±1=3,5; 4±3=1,7. So connected to1,3,5,7.

Even node6:

6±1=5,7; 6±3=3,9≡1 mod8. So connected to1,3,5,7.

Therefore, every even node is connected to all four odd nodes (1,3,5,7).

Similarly, each odd node is connected to all even nodes:

Take odd node1:

Difference1:0,2.

Difference3: 1±3=4, -2≡6 mod8. So connected to0,2,4,6.

Similarly for other odds.

Therefore, the bipartite graph between E and O is a complete bipartite graph K_{4,4}. Because every even node is connected to every odd node.

Wait, that can't be. Wait, if every even node is connected to all four odd nodes, and vice versa, then yes, it's a complete bipartite graph K_{4,4}.

Therefore, the graph is a complete bipartite graph between two sets of four nodes each.

In a complete bipartite graph K_{n,n}, the number of independent sets is (2^n +1)^2. Wait, let me think.

In a complete bipartite graph K_{n,n}, an independent set can be any subset of one partition or the other, or a mixture as long as no edges exist. But in a complete bipartite graph, any independent set can contain at most all of one partition. Because if you have any nodes from both partitions, since every node in one is connected to every node in the other, you can't have any nodes from both.

Wait, actually, in a complete bipartite graph K_{m,n}, the independent sets are exactly the subsets contained entirely within one partition or the other, plus the empty set. Wait, no, actually, you can have subsets that include some from one partition and some from the other, as long as they don't have edges between them. But in complete bipartite, any non-empty subset from both partitions would have edges between them. Therefore, the only independent sets are the subsets entirely within E, entirely within O, or the empty set.

Wait, that's correct. Because if you have any vertex from E and any from O, they are connected. Therefore, the independent sets are:

- All subsets of E,

- All subsets of O,

- Plus the empty set (already counted in both).

Wait, actually, the subsets of E and subsets of O already include the empty set. So total independent sets are (2^4 + 2^4 -1) = 16 +16 -1=31? Wait, no, the total is the number of subsets of E plus subsets of O minus the overlap (the empty set counted twice). So total is (2^4 + 2^4 -1)= 31. Alternatively, since independent sets are either subsets of E or subsets of O, so total is 2^4 +2^4 -1=31 (subtracting the duplicate empty set).

Wait, but actually, the empty set is included in both, so total is 16 +16 -1=31.

But in our case, the graph is K_{4,4}, so the number of independent sets is 2^4 +2^4 -1=31.

Wait, but let me confirm with small n. For K_{1,1} (an edge), the independent sets are {}, {a}, {b}, so 3=2+2-1=3. Correct. For K_{2,2}, the independent sets are subsets of first partition (4), subsets of second (4), minus 1 for overlap: total 7. Indeed, in K_{2,2}, the independent sets are:

From first partition: {}, {a}, {b}, {a,b}

From second partition: {}, {c}, {d}, {c,d}

Total without overlap: 4+4-1=7. Yes. So formula holds.

Therefore, for K_{4,4}, it's 16+16-1=31 independent sets.

Therefore, returning to our problem, the number of independent sets in the graph for shifts1 and3 is 31. Therefore, |A_1 ∩ A_3|=31.

Wait, but earlier I found the quartet {0,2,4,6} which is a subset of E, so that's allowed. Similarly, the subset {1,3,5,7} is a subset of O. So that's correct.

Therefore, the count is 31.

Hmm, so this is different from the previous intersection (A1 ∩ A2)=21. So intersections can vary depending on the shifts.

This shows that the inclusion-exclusion terms are going to be complicated because the intersections depend on the specific shifts chosen.

Given the time constraints, perhaps this problem requires a more advanced combinatorial approach or leveraging group actions and Burnside's lemma.

Alternatively, perhaps the total number of subsets B that can be shifted to be disjoint is equal to the number of subsets that are not "complete" in some sense. Wait, but I need to think differently.

Alternatively, let me think of the problem in terms of the complement. Let me denote the complement of B as R. The condition is that there exists a rotation π such that π(B) ⊆ R. Since R is the complement of B, this is equivalent to π(B) ⊆ R = V \ B, so π(B) and B are disjoint. Therefore, the condition is equivalent to B and π(B) being disjoint, which is the same as before.

Alternatively, the problem is equivalent to asking for the number of subsets B such that B is not a "blocking set" for the rotation group, meaning that there exists a rotation that avoids B.

Alternatively, perhaps using generating functions with the cycle index of the rotation group.

The cycle index of the cyclic group C8 acting on the vertices can be used to count the number of subsets fixed by each group element, which might help in applying Burnside's lemma. However, I need to see how this relates to our problem.

Wait, Burnside's lemma says that the number of orbits is equal to 1/|G| times the number of fixed points of each group element. But how does this relate to our problem?

Alternatively, perhaps we can model the problem as follows: We want the number of subsets B such that there exists a group element (rotation) π where π(B) is disjoint from B. This is equivalent to the existence of a π such that π(B) ⊆ complement of B.

Alternatively, for each subset B, define its "avoidance set" as the set of rotations π where π(B) ⊆ complement(B). We need the number of B where this avoidance set is non-empty.

Alternatively, perhaps using the principle from probability: the probability that a random subset B has a non-empty avoidance set is equal to the expectation over π of the probability that π(B) ⊆ complement(B), but adjusted for overlap.

Wait, linearity of expectation might help here. The expected number of rotations π that avoid B is equal to the sum over π of the probability that π(B) ⊆ complement(B). Then, the probability that there exists at least one such π is less than or equal to this expectation. But we need the exact probability.

Alternatively, the expectation is Σ_{π} Pr[π(B) ⊆ R], where R is the complement of B. Since each element of π(B) must be in R, which is equivalent to B being disjoint from π^{-1}(B). Wait, perhaps this is getting too tangled.

Alternatively, note that for a random subset B, the probability that a particular rotation π (non-identity) satisfies π(B) ∩ B = ∅ is equal to the probability that for all vertices v, if v is in B then π^{-1}(v) is not in B. Since π is a permutation, this is equivalent to B being a subset that doesn't intersect with its own π^{-1} image. Since π is a rotation, π^{-1} is also a rotation.

The probability that B and π(B) are disjoint is equal to the probability that for every vertex v, v and π^{-1}(v) are not both in B. Since each vertex v and π^{-1}(v) form an orbit under the rotation? Wait, depending on the rotation.

Wait, for a rotation by k steps, the permutation π_k consists of cycles whose length is the order of k in Z/8Z, which is 8/gcd(k,8). Therefore, the permutation decomposes into cycles of length m=8/gcd(k,8). For each cycle, the condition that B and π_k(B) are disjoint requires that in each cycle, the subset B does not contain any two elements that are in the same position in their cycle.

Wait, perhaps more precisely, for each cycle in the permutation π_k, the subset B must not contain any pair of elements that are in the same position in their cycle. Wait, maybe better to think in terms of the cycle structure.

Suppose that the rotation π_k decomposes into cycles of length m. For example, if k=1, the entire permutation is a single 8-cycle. For each cycle, the condition that B and π_k(B) are disjoint requires that in each cycle, the subset B does not contain any element and its successor in the cycle. Wait, actually, more precisely, since π_k shifts each element along the cycle, the condition that B and π_k(B) are disjoint requires that for any element v in B, the element π_k(v) is not in B. This is equivalent to B being an independent set in the directed cycle where edges go from v to π_k(v). Which is similar to the independent sets we considered earlier.

Therefore, the probability that a random subset B satisfies B ∩ π_k(B)=∅ is equal to the number of independent sets in the directed cycle graph induced by π_k divided by 2^8.

Wait, but since the subset is chosen independently, the probability would be the number of independent sets divided by 2^8. However, the number of independent sets for a cycle of length m is F_{m-1} + F_{m+1}, as before. Therefore, for a rotation with cycle length m, the probability is [F_{m-1} + F_{m+1}]/2^m, but since there are multiple cycles, it's [ (F_{m-1} + F_{m+1}) / 2^m ]^d where d is the number of cycles.

Wait, let me clarify:

If the rotation π_k decomposes into d cycles each of length m (where m =8/d), then the number of subsets B where B and π_k(B) are disjoint is [F_{m-1} + F_{m+1}]^d, as previously established. Therefore, the probability for that rotation is [F_{m-1} + F_{m+1}]^d / 2^8.

Therefore, the expected number of rotations π_k (including identity) for which B satisfies the condition is equal to the sum over all rotations π of [number of suitable B]/2^8, but actually, for each rotation π, the probability that B is suitable is as above.

Wait, actually, the expectation over all subsets B of the number of rotations π that satisfy the condition is equal to the sum over all rotations π of the probability that B is suitable for π.

Therefore, the expected number is Σ_{π} [ (number of B suitable for π) / 2^8 ].

But we need the probability that this number is at least 1, which is different.

However, perhaps using the principle of inclusion-exclusion, the probability that there exists at least one π is equal to the sum over π of Pr[π works] minus sum over π<π' of Pr[π and π' work] + ... etc.

But this is exactly the inclusion-exclusion formula, which is what I was trying to compute earlier, but it's very involved.

Alternatively, perhaps the total number of subsets B that are "shiftable" (i.e., can be rotated to be disjoint) is equal to the sum over all non-empty rotations π of the number of subsets suitable for π, divided by the number of rotations? No, that would be overcounting.

Alternatively, perhaps using Möbius inversion over the subgroup lattice, but this is getting too advanced.

Alternatively, perhaps the total number is equal to the sum_{d |8} something, but I need to think.

Alternatively, let me consider all possible subsets B and count for each B the number of shifts k where B can be shifted by k to be disjoint. Then, the total over all B is equal to the sum_{k=1}^7 |A_k|. Then, the number of B with at least one shift is equal to the inclusion-exclusion formula.

But since I need the exact count, perhaps it's better to use the formula:

Number of B with at least one shift = Σ|A_k| - Σ|A_k ∩ A_j| + Σ|A_k ∩ A_j ∩ A_m| - ... + (-1)^{n+1}|A_1 ∩ ... ∩ A_7}|.

Given the complexity, perhaps the problem is designed so that the answer can be computed using Burnside's lemma or another method.

Wait, let me think differently. Let me consider all possible subsets B and their rotations. For each subset B, the number of rotations π such that π(B) is disjoint from B is equal to the number of k where B and B +k are disjoint. Let me denote this count as N(B).

Then, the total number of pairs (B, π) where π(B) is disjoint from B is equal to Σ_{B} N(B) = Σ_{π} |A_π|, where A_π is the number of subsets suitable for π. As computed before, for each rotation π (excluding identity), the number |A_π| depends on its cycle type.

As calculated earlier, for each rotation with shift k:

- If gcd(k,8)=1 (4 shifts), each has |A_k|=47.

- gcd=2 (2 shifts), each has 49.

- gcd=4 (1 shift), has 81.

Including the identity shift π_0, which only allows B=∅, so |A_0|=1.

Therefore, the total over all rotations (including identity) is 1 +4*47 +2*49 +1*81 =1 +188 +98 +81= 1+188=189; 189+98=287; 287+81=368.

Therefore, Σ_{π} |A_π|=368.

But this counts the total number of pairs (B, π) where π(B) is disjoint from B. Each subset B contributes N(B) to this total, where N(B) is the number of rotations π that work for it.

We need the number of subsets B where N(B) ≥1. Let me denote this as X. Then,

Σ_{B} N(B) = Σ_{B} [number of π for which π works on B] = 368.

But we need X = number of B with N(B) ≥1.

We also know that the total number of subsets is 256. The subsets where N(B)=0 are those for which no rotation π (including identity) satisfies the condition. The only subset where identity works is B=∅, which is counted in the N(B) for π=0. But for non-empty B, N(B) counts rotations excluding identity? Wait, actually, the identity rotation π_0 requires that π_0(B)=B is disjoint from B, which only occurs when B is empty. So for non-empty B, N(B) counts the number of non-identity rotations π where π(B) is disjoint from B.

Therefore, the subsets with N(B)=0 are those non-empty subsets for which no non-identity rotation works, plus the empty set? Wait, no. The empty set has N(B)=8 (all rotations work, since shifting empty set is empty, which is disjoint from itself). Wait, actually, for the empty set, every rotation π satisfies π(B)=∅ ⊆ R=V\∅=V. So N(∅)=8 (all rotations). For the full set B=V, N(B)=0, since no rotation can make π(B)=V disjoint from B=V.

Therefore, the total Σ N(B) = 368 includes all subsets:

- For B=∅: contributes 8.

- For other subsets: their count of working rotations.

We need X = number of B with N(B) ≥1. This includes all subsets except those with N(B)=0. The subsets with N(B)=0 are those for which no rotation (including identity) works except possibly the identity. But for non-empty B, the identity only works if B is empty. So non-empty subsets with N(B)=0 are those for which no non-identity rotation works. So total subsets with N(B)=0 are:

- The full set B=V, which has N(B)=0.

- Any other subset B where no rotation (except possibly identity) works. Since identity only works for B=∅, these are subsets where no rotation (including non-identity) works except possibly identity, but since they are non-empty, identity doesn't work. So they have N(B)=0.

Therefore, the number of subsets with N(B)=0 is equal to the number of subsets B (including V) for which no rotation π (including non-identity) satisfies π(B) disjoint from B.

Therefore, X = total subsets - subsets with N(B)=0 =256 - Y, where Y is the number of subsets with N(B)=0.

But how to compute Y?

Alternatively, note that:

Σ_{B} N(B) = Σ_{B} [number of π where π works on B] = 368.

But also,

Σ_{B} N(B) = Σ_{π} |A_π| =368.

We can also write:

Σ_{B} N(B) = Σ_{B≠∅} N(B) + N(∅) = Σ_{B≠∅} N(B) +8 =368.

Thus,

Σ_{B≠∅} N(B)=360.

Now, the number of subsets B≠∅ with N(B)≥1 is X-1 (since X includes ∅ which has N=8). So,

Σ_{B≠∅} N(B) = Σ_{B≠∅, N(B)≥1} N(B) + Σ_{B≠∅, N(B)=0} 0 = Σ_{B≠∅, N(B)≥1} N(B) =360.

But the number of subsets with N(B)≥1 is X = (number with N≥1) = (number with N≥1 excluding ∅) +1.

Let me denote Z = number of non-empty subsets with N(B)≥1. Then,

Σ_{B≠∅} N(B) = Σ_{B in Z} N(B) =360.

But we need Z, the number of non-empty subsets with N(B)≥1. However, without knowing the distribution of N(B), it's hard to get Z directly.

Alternatively, if all non-empty subsets with N(B)≥1 had N(B)≥1, then Z ≤360. But this isn't helpful.

Alternatively, perhaps all non-empty subsets except the full set have N(B)≥1? But earlier we saw that subsets of size 4 can have N(B)=0, like the example {0,1,3,4} which couldn't be shifted to be disjoint.

Therefore, there are some subsets besides the full set with N(B)=0.

This approach might not be fruitful.

Perhaps I need to consider that the problem is from an Olympiad, so there might be a clever combinatorial trick.

Let me think of the problem in terms of the complement. Let me denote the complement of B as R. The condition is that there exists a rotation π such that π(B) ⊆ R. Since R is the complement of B, this is equivalent to π(B) ⊆ R = V \ B, so π(B) and B are disjoint.

Therefore, the existence of such a π means that B can be "shifted into its own complement". This is similar to saying that B is not a "fixed" set under any rotation, but not exactly.

Alternatively, consider that the octagon has 8 vertices. For a subset B, if it can be rotated into its complement, then the complement must have the same "shape" as B, just rotated. Therefore, the complement must be a rotation of B. Wait, but the problem states that π(B) must be a subset of the complement, not necessarily equal.

Wait, actually, if π(B) is a subset of the complement, then the complement must contain the rotated B, but the complement could be larger.

Alternatively, if we consider that the rotation maps B into R, then the size of B must be less than or equal to the size of R, which is 8 - |B|. So |B| ≤4, as previously noted.

Therefore, only subsets with |B| ≤4 can satisfy the condition. So we can restrict our attention to subsets of size 0 to4.

Now, for each subset size from0 to4, compute the number of subsets B where there exists a rotation π such that π(B) ⊆ R.

Let me compute this for each size:

Size0: Only the empty set. It trivially satisfies the condition (any rotation works). So count=1.

Size1: Any singleton set. As earlier, any singleton can be rotated to avoid itself. For a singleton {v}, shifting by any non-zero rotation that moves v to another position (which exists since the shift is cyclic). So all 8 subsets of size1 satisfy the condition. Count=8.

Size2: Any pair. As earlier examples showed, any pair can be shifted to avoid itself. Let me confirm with the earlier problematic subset {0,1,3,4} but that's size4. For pairs:

Take any two vertices. If they are adjacent, shifting by1 moves them to the next pair, which doesn't overlap. If they are separated, similar. For example, {0,1} shifted by2 becomes {2,3}, disjoint. {0,2} shifted by1 becomes {1,3}, disjoint. {0,4} (diametrically opposite) shifted by1 becomes {1,5}, disjoint. So yes, all pairs can be shifted to be disjoint. Therefore, all C(8,2)=28 subsets of size2 satisfy the condition. Count=28.

Size3: Triples. Here, some triples might not be shiftable. Let me see.

Take a triple like {0,1,2}. Shifting by3 would map it to {3,4,5}, which is disjoint. So that works.

Another triple {0,1,3}. Let's see:

Shift by2: {2,3,5}. Intersection with original is {3}. Not disjoint.

Shift by4: {4,5,7}. Disjoint from original {0,1,3}. So yes, shifting by4 works.

Another example: {0,2,4}. Shifting by1 gives {1,3,5}, disjoint. So works.

What about a triple that's spread out? {0,2,6}. Shifting by1: {1,3,7}, disjoint. Works.

Is there a triple that cannot be shifted to be disjoint?

Suppose we have a triple where every shift overlaps. Let me try {0,1,4}.

Shift by2: {2,3,6}. Disjoint? Original has 0,1,4. The shifted set has 2,3,6. No overlap. So that works.

Another example: {0,2,5}. Shift by3: {3,5,8 mod8=0}. The shifted set includes 0 and5, which are in original. Not good. But shift by1: {1,3,6}, which doesn't overlap with {0,2,5}. So that works.

Hmm, maybe all triples can be shifted?

Wait, let me try {0,1,5}. Shift by3: {3,4,8 mod8=0}. The shifted set includes0, which is in original. Not good. Shift by2: {2,3,7}, which is disjoint from {0,1,5}. So yes.

Another example: {0,3,6}. Shift by1: {1,4,7}, disjoint. Works.

Hmm, perhaps all triples can be shifted. Let me think of a potential problematic case.

Suppose the triple is {0,2,4}. Shift by 4: {4,6,0}, which overlaps with original at 0 and4. Not good. But shift by1: {1,3,5}, which is disjoint. So works.

Another example: {0,1,3}. Shift by5: {5,6,8 mod8=0}. Overlap at0. Not good. But shift by2: {2,4,5}, which is disjoint from {0,1,3}. So yes.

It's challenging to find a triple that can't be shifted. Maybe all triples can be shifted. Let me think of a triple with minimal spacing.

Take {0,1,2}: shifting by3 gives {3,4,5}, disjoint.

Take {0,1,7}: shifting by2 gives {2,3,1}, which overlaps at1. Not good. Shift by3: {3,4,2}, disjoint from {0,1,7}? Yes, {3,4,2} doesn't include0,1,7. So yes.

Hmm, perhaps all triples can be shifted. Let me think of a triple arranged such that every shift overlaps. Suppose {0,2,5}:

Shift by1: {1,3,6} – disjoint.

Shift by2: {2,4,7} – overlaps at2.

Shift by3: {3,5,0} – overlaps at0 and5.

Shift by4: {4,6,1} – disjoint.

So even if some shifts overlap, at least one shift works. Therefore, perhaps all triples can be shifted. So count= C(8,3)=56.

Size4: Here, as seen earlier, some subsets cannot be shifted. For example, the subset {0,1,3,4} cannot be shifted to be disjoint. So we need to count the number of 4-element subsets that can be shifted.

This is the tricky part. Let me try to find how many 4-element subsets can be shifted.

First, total 4-element subsets:70.

We need to subtract those 4-element subsets which cannot be shifted to be disjoint.

A subset B of size4 cannot be shifted if for every shift k, B and B +k share at least one common element.

This is equivalent to the difference set of B covering the entire group Z/8Z. Because if the difference set is the entire group, then for every shift k, there exists b, b' ∈ B such that b - b' ≡k mod8, so shifting by k would map b' to b, hence overlapping.

Therefore, the subsets B of size4 that cannot be shifted are exactly those whose difference sets are the entire Z/8Z.

Therefore, the number we need to subtract is the number of 4-element subsets whose difference sets cover Z/8Z.

So, how many 4-element subsets of Z/8Z have difference sets equal to Z/8Z?

This is equivalent to the subset being a "complete difference set".

In additive combinatorics, a complete difference set is a set such that every non-zero element of the group can be expressed as a difference of two elements in the set.

However, in our case, we need the difference set to cover all elements of Z/8Z, including zero? Wait, the difference set usually includes all differences b - b' for b, b' ∈ B, so it includes 0. But in our problem, the shift k=0 is allowed only for the empty set, but for the subset B of size4, the shift k=0 would require B disjoint from itself, which is only possible if B is empty. So, for non-empty B, we are concerned with non-zero shifts.

Therefore, the difference set needs to cover all non-zero elements of Z/8Z. So, the subset B is a complete difference set modulo8, meaning that every non-zero element of Z/8Z can be expressed as a difference of two elements of B.

Therefore, the number of 4-element subsets of Z/8Z that are complete difference sets is the number we need.

I need to find how many 4-element subsets of Z/8Z are complete difference sets.

This might be a known result. Let me think.

In Z/8Z, a complete difference set of size4 must have the property that every non-zero element is expressed as a difference of two elements in the set.

Let me consider known difference sets. For example, in Z/7Z, the set {0,1,3} is a difference set.

But in Z/8Z, perhaps the only complete difference sets of size4 are the affine planes or something?

Alternatively, let me try to construct such a set.

Suppose we have a subset B of size4 in Z/8Z such that every non-zero element is a difference.

Let me try the set {0,1,3,4} as in the earlier example.

Compute its differences:

0-0=0

0-1=7, 0-3=5, 0-4=4

1-0=1, 1-1=0, 1-3=6, 1-4=5

3-0=3, 3-1=2, 3-3=0, 3-4=7

4-0=4,4-1=3,4-3=1,4-4=0

So the differences are:

0,1,3,4,5,6,7,2. Wait, all elements except maybe check:

Looking at all differences:

From the above:

Differences obtained:

0,1,3,4,5,6,7,2. So all elements 0-7 are present. Therefore, the difference set is indeed Z/8Z. Therefore, this set {0,1,3,4} is a complete difference set.

Similarly, other such sets can be constructed by rotating this set. Since the group is cyclic, any rotation of this set would also be a complete difference set.

Therefore, the number of such subsets is equal to the number of orbits under rotation of such complete difference sets.

First, how many distinct complete difference sets of size4 are there in Z/8Z?

Let me see if there are others.

Take the set {0,1,4,5}.

Compute differences:

0-0=0

0-1=7, 0-4=4, 0-5=3

1-0=1,1-1=0,1-4=5,1-5=4

4-0=4,4-1=3,4-4=0,4-5=7

5-0=5,5-1=4,5-4=1,5-5=0

Differences: 0,1,3,4,5,7. Missing 2,6. So not complete.

Another example: {0,2,3,5}.

Differences:

0-0=0

0-2=6,0-3=5,0-5=3

2-0=2,2-2=0,2-3=7,2-5=5

3-0=3,3-2=1,3-3=0,3-5=6

5-0=5,5-2=3,5-3=2,5-5=0

Differences obtained:0,1,2,3,5,6,7. Missing 4. Not complete.

Another try: {0,1,2,4}.

Differences:

0-0=0

0-1=7,0-2=6,0-4=4

1-0=1,1-1=0,1-2=7,1-4=5

2-0=2,2-1=1,2-2=0,2-4=6

4-0=4,4-1=3,4-2=2,4-4=0

Differences:0,1,2,3,4,5,6,7. All elements! So this is also a complete difference set.

Therefore, {0,1,2,4} is another complete difference set.

Hmm, so there are at least two distinct complete difference sets under rotation.

Wait, let me check if {0,1,2,4} can be rotated to get {0,1,3,4}?

Rotating {0,1,2,4} by 2 gives {2,3,4,6}. Let's see if that's a complete difference set.

Differences of {2,3,4,6}:

2-2=0

2-3=7, 2-4=6, 2-6=4

3-2=1, 3-3=0, 3-4=7, 3-6=5

4-2=2,4-3=1,4-4=0,4-6=6

6-2=4,6-3=3,6-4=2,6-6=0

Differences obtained:0,1,2,3,4,5,6,7. Yes, complete.

Therefore, rotating a complete difference set gives another complete difference set.

Therefore, the number of distinct complete difference sets is equal to the number of orbits under rotation.

Let me see how many distinct complete difference sets there are.

We have two examples: {0,1,3,4} and {0,1,2,4}. Are these in the same orbit?

Let me see if rotating {0,1,2,4} can get to {0,1,3,4}.

Suppose we rotate {0,1,2,4} by k= something.

To get 3 in the set, we need to have an element 3 in the rotated set. The original set has 0,1,2,4. Rotating by 1 gives 1,2,3,5. Which includes3 but not4. Not the target.

Rotating by3: 3,4,5,7. Not the target.

Rotating by -1 (7): 7,0,1,3. Which is {0,1,3,7}. Not the target.

Hmm, so perhaps they are in different orbits.

Alternatively, let me see if there are multiple orbits.

Let me consider the two examples:

First type: {0,1,2,4} and its rotations.

Second type: {0,1,3,4} and its rotations.

Are these two orbits distinct?

Suppose there is a rotation or reflection that maps one to the other, but since we're only considering rotations, not reflections, let's see.

Suppose we have the set {0,1,3,4}. Rotating by1 gives {1,2,4,5}. Rotating by2 gives {2,3,5,6}, etc. None of these will give the other type.

Therefore, these are two distinct orbits.

Are there more?

Let me try another set: {0,1,2,5}.

Differences:

0-0=0

0-1=7,0-2=6,0-5=3

1-0=1,1-1=0,1-2=7,1-5=4

2-0=2,2-1=1,2-2=0,2-5=5

5-0=5,5-1=4,5-2=3,5-5=0

Differences:0,1,2,3,4,5,6,7. Complete.

So this is another complete difference set. Rotating this set:

Rotate by1: {1,2,3,6}. Let's check differences:

1-1=0,1-2=7,1-3=6,1-6=3,

2-1=1, etc. The differences will still be complete.

This is another orbit?

Wait, let me see if {0,1,2,5} can be rotated to {0,1,2,4}?

Rotating {0,1,2,5} by k= -1 (7): gives {7,0,1,4}, which is {0,1,4,7}. Not the same as {0,1,2,4}.

Therefore, this is a third orbit?

Hmm, this is getting complex. Maybe there are multiple orbits.

Alternatively, perhaps all complete difference sets of size4 in Z/8Z are in two orbits: one for each "type".

Alternatively, perhaps the number is 8. Wait, let me think of how many distinct complete difference sets there are.

Alternatively, perhaps each complete difference set corresponds to a "planar difference set", but I might be misremembering.

Alternatively, perhaps the number is 8. Since the first example {0,1,2,4} can be rotated into 8 positions, but some might overlap.

Wait, the set {0,1,2,4} rotated by4 gives {4,5,6,0}, which is the same as the original set shifted by4, so it's a different set but in the same orbit.

The orbit size would be 8 divided by the stabilizer. If the set is stabilized only by the identity, then the orbit has size8.

Similarly for the other type.

Assuming there are two distinct orbits, each of size8, then total number would be16. But I need to confirm.

Alternatively, let me count how many complete difference sets there are.

Let me consider the first type: {0,1,2,4} and its rotations. There are8 rotations.

Similarly, the second type {0,1,3,4} and its rotations: another8.

The third example {0,1,2,5} and its rotations: another8. Wait, but maybe {0,1,2,5} is in the same orbit as one of the previous?

Wait, rotating {0,1,2,5} by3 gives {3,4,5,0}, which is {0,3,4,5}. Let's see if this is a complete difference set.

Differences:

0-0=0,

0-3=5, 0-4=4,0-5=3,

3-0=3,3-3=0,3-4=7,3-5=6,

4-0=4,4-3=1,4-4=0,4-5=7,

5-0=5,5-3=2,5-4=1,5-5=0.

Differences obtained:0,1,2,3,4,5,6,7. Complete.

But this set {0,3,4,5} is different from the first two types. So perhaps this is a third orbit.

This suggests there are multiple orbits, making the count complicated.

Alternatively, perhaps the total number of complete difference sets of size4 in Z/8Z is 8. Let me see:

Each complete difference set must contain 4 elements. The number might be 8, corresponding to the 8 possible starting points, but with overlaps.

Alternatively, perhaps the total number is 8. Let me see:

Suppose that for each element, there is exactly one complete difference set containing it in a certain position, but I'm not sure.

Alternatively, perhaps the total number is 8. Let me see:

Suppose that the complete difference sets are exactly the sets of the form {0,1,2,4}, {1,2,3,5}, ..., rotated around, giving8 total.

Similarly, the other type {0,1,3,4}, rotated gives another8. So total16.

Alternatively, maybe there are 8.

Given the time I've spent and the complexity, perhaps I should look for another approach.

Wait, the problem asks for the probability, so maybe the total number of valid subsets is 1 (empty) +8 (size1)+28 (size2)+56 (size3)+ (70 - Y) (size4), where Y is the number of bad size4 subsets (those that can't be shifted). If I can find Y, then total is 1+8+28+56 + (70-Y)= 163 - Y.

The total valid subsets would then be 163 - Y, and the probability is (163 - Y)/256. Then, m+n would be accordingly.

To find Y, the number of 4-element subsets with complete difference sets.

From my earlier examples, each complete difference set has 8 rotations, so if there are N orbits, each contributing8 subsets, then Y=8*N.

Suppose there are two orbits, then Y=16. If three orbits, Y=24, etc.

Alternatively, perhaps the number is8. Let me see:

Take the set {0,1,2,4}. Its rotations give8 distinct subsets.

Similarly, the set {0,1,3,4} gives another8.

The set {0,1,2,5} gives another8, but maybe this is equivalent to one of the previous under rotation?

Wait, rotating {0,1,2,5} by4 gives {4,5,6,1}, which is {1,4,5,6}. Not the same as the first two.

Alternatively, perhaps there are four orbits, each with8 subsets, totaling32. But this is just guessing.

Alternatively, perhaps the total number of complete difference sets of size4 in Z/8Z is8. Let me see:

Suppose that the difference set must include all elements. For a subset of size4, the number of differences is4*3=12 (excluding zeros). Since Z/8Z has7 non-zero elements, each must be covered at least once. Since12 differences for7 elements, some elements are covered multiple times.

Perhaps the number is8. Let me assume Y=8 for now (though I'm not sure). Then total valid subsets would be163-8=155. 155/256 reduces to 155/256 (since 155=5×31, 256 is 2^8, so gcd(155,256)=1). Thus m+n=155+256=411. But I need to verify.

Alternatively, perhaps Y= 8. Let me see another way.

Suppose that each complete difference set corresponds to a "quadrangle" or something in the octagon. Alternatively, perhaps the number is 8.

Alternatively, perhaps the number of complete difference sets of size4 in Z/8Z is 8. Then Y=8, so total valid subsets would be1+8+28+56+ (70-8)=1+8=9; 9+28=37; 37+56=93; 93+62=155. So 155.

But I need to confirm Y=8.

Alternatively, perhaps the complete difference sets are exactly the 8 subsets obtained by rotating {0,1,2,4}. Let me check:

Rotating {0,1,2,4} by k=0 to7 gives8 distinct subsets. Each of these is a complete difference set. Similarly, rotating {0,1,3,4} gives another8. So total16.

If that's the case, then Y=16, so total valid subsets would be1+8+28+56 + (70-16)=1+8=9; 9+28=37; 37+56=93; 93+54=147. 147/256 reduces to 147/256 (gcd(147,256)=1), so m+n=147+256=403.

Alternatively, maybe there are 8 such subsets. Let me see an example:

Take the set {0,1,3,4}. Its rotations give8 subsets. Are there any others?

Suppose another set like {0,2,3,5}. Earlier, its differences missed 4, so not complete.

Another example: {0,1,4,5}. Differences missed 2 and6. Not complete.

Hmm, so maybe only two orbits: the {0,1,2,4} type and the {0,1,3,4} type, each with8 subsets. So total Y=16.

Thus total valid subsets would be1+8+28+56 + (70-16)=147. So probability147/256. 147 and256 are coprime? 147 factors into3×49=3×7². 256 is2^8. So yes, gcd is1. Thus m+n=147+256=403.

But I need to confirm if there are exactly16 such subsets.

Alternatively, let me think of the difference set must contain all elements. For a subset of size4, the differences must cover all8 elements (including0). But since we need the non-zero differences to cover Z/8Z, the non-zero differences must cover7 elements. Since there are12 non-zero differences (each pair in the subset), so each non-zero element must be represented at least once.

The two examples I found ({0,1,2,4} and {0,1,3,4}) both achieve this. Let me see if there are others.

Take the set {0,1,2,5}:

Differences include 5-0=5,5-1=4,5-2=3, etc. Let me list all non-zero differences:

From {0,1,2,5}:

Differences between elements:

1-0=1,

2-0=2,

5-0=5,

2-1=1,

5-1=4,

5-2=3,

0-1=7,

0-2=6,

0-5=3,

1-2=7,

1-5=6,

2-5=5.

Wait, non-zero differences are1,2,5,1,4,3,7,6,3,7,6,5.

So the non-zero elements covered are1,2,3,4,5,6,7. All seven non-zero elements. So yes, complete.

This set is different from the previous two types, so it must be in a third orbit. Rotating this gives another8 subsets. So now we have24 subsets.

Continuing, perhaps there are multiple such orbits.

Alternatively, perhaps the total number is 8×3=24. Then Y=24, so total valid subsets would be1+8+28+56 + (70-24)=1+8=9; 9+28=37; 37+56=93; 93+46=139. 139/256, which reduces to139+256=395.

This is getting too uncertain. Perhaps I need another approach.

Wait, let me think of the total number of complete difference sets of size4 in Z/8Z.

According to some references, in Z_8, the only known difference sets of size4 are the "complementary difference sets" which form a pair. For example, a difference set and its complement.

Wait, the complement of a complete difference set would also be a complete difference set?

The complement of a subset B is R=V\B. The difference set of R would be all differences of elements in R. If B's difference set is all of Z/8Z, then R's difference set would also include all differences, since R is also size4 in Z/8Z. Wait, no, the complement of a subset with a complete difference set might not necessarily have a complete difference set.

Alternatively, perhaps the number is8. Let me see:

Suppose that for each of the8 rotations of the set {0,1,2,4}, we have a complete difference set. That gives8 subsets.

Similarly, the set {0,1,3,4} and its rotations give another8.

The set {0,1,2,5} and its rotations give another8.

Total24.

But I can find another example:

Take {0,1,4,6}.

Differences:

0-0=0,

0-1=7,0-4=4,0-6=2,

1-0=1,1-1=0,1-4=5,1-6=3,

4-0=4,4-1=3,4-4=0,4-6=6,

6-0=6,6-1=5,6-4=2,6-6=0.

Non-zero differences:1,2,3,4,5,6,7. Complete.

This is another distinct set, so its rotations add another8. Now we have32.

This suggests that there are multiple such subsets, possibly32. But this is getting too time-consuming.

Given the time I've invested and the problem's context, perhaps the answer is 147/256 leading to 403, but I'm not sure.

Alternatively, perhaps the number of complete difference sets of size4 in Z/8Z is8. Let me assume Y=8 for now.

Then total valid subsets would be1+8+28+56+62=155. 155/256 reduces to155+256=411.

Alternatively, perhaps the correct answer is 147/256 (403), but I need to think differently.

Wait, let me think of the total number of subsets that can be shifted:

For size4 subsets, the number that can be shifted is70 minus the number of complete difference sets.

If the number of complete difference sets is8, then it's70-8=62.

Total valid subsets:1+8+28+56+62=155.

155/256=155/256. Since 155=5×31, and 256 is2^8, they are coprime. So m+n=155+256=411.

Alternatively, if the number is16, then 70-16=54, total 1+8+28+56+54=147, leading to403.

Given that in my examples I found at least two distinct orbits with8 each, I think Y=16 is more plausible, leading to403.

Alternatively, perhaps the correct answer is 147/256.

Wait, let me see an example from the problem's perspective.

The problem says "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".

This is equivalent to the existence of a rotation π such that π(B) ⊆ R = V \ B.

Which requires that π(B) and B are disjoint.

For subsets of size4, this requires that π(B) is contained in the other4 vertices.

Therefore, π(B) must be exactly the complement of B, since they are the same size.

Wait a second! If B has size4, then its complement also has size4. So for π(B) to be disjoint from B, π(B) must be exactly equal to the complement of B.

Because π(B) has size4, and the complement of B has size4. So π(B) must equal R=V\B.

Therefore, the condition for a size4 subset B is that there exists a rotation π such that π(B)=R.

Therefore, R must be a rotation of B.

Therefore, B and its complement are rotationally congruent.

Therefore, the subset B must be such that its complement is a rotation of itself.

Therefore, the number of such subsets B is equal to the number of subsets of size4 that are rotationally congruent to their complements.

Therefore, the number of size4 subsets where B and its complement are in the same rotation orbit.

Therefore, to count Y (the number of size4 subsets that cannot be rotated to their complement), we need the number of size4 subsets where B and its complement are in different orbits.

Wait, actually, the subsets B for which there exists a rotation π with π(B)=R are exactly those where B and R are in the same orbit under the rotation group.

Therefore, the number of such subsets B is equal to the number of orbits of4-element subsets under rotation, multiplied by the number of subsets in each orbit that satisfy the condition.

Wait, perhaps it's better to note that for each rotationally symmetric subset B where B is equal to its complement under some rotation, the number is equal to the number of such subsets.

Alternatively, the number of subsets B of size4 where B and its complement are in the same orbit.

The total number of such subsets is equal to the number of orbits of4-element subsets under rotation where the orbit of B intersects with the orbit of its complement.

Since the rotation group is acting, the orbit of the complement is the same as the orbit of B if and only if there exists a rotation π such that π(B)=R.

Therefore, the number of subsets B where such a π exists is equal to the number of orbits where the orbit of B equals the orbit of its complement.

Therefore, the total number is equal to the number of such orbits multiplied by the size of the orbit (but each subset in the orbit is counted once).

Alternatively, this is getting too abstract.

Alternatively, note that if B is rotationally congruent to its complement, then the entire set V can be partitioned into two rotated copies of B. Therefore, the subset B must be a "tile" of the octagon under rotation.

This is similar to a concept in combinatorics called a "rotational tiling".

For the octagon, such subsets B must be such that B and its rotation form a partition of the vertices.

This requires that the rotation by π shifts B to its complement, so applying the rotation twice would map B to itself.

Therefore, the rotation π must have order2, so that π^2 is the identity.

In Z/8Z, the rotations of order2 are those by4 steps (since 4*2=8≡0 mod8).

Therefore, the only rotation of order2 is rotation by4.

Therefore, the only way for a subset B to satisfy π(B)=R is if π is rotation by4, and rotating B by4 gives its complement.

Therefore, the condition is that rotating B by4 gives its complement.

Therefore, for a subset B of size4, the condition is that B +4 = V \ B.

Therefore, B +4 = complement of B.

This is equivalent to B being a "self-complementary" set under rotation by4.

Therefore, such subsets B must satisfy that for every element x in B, x+4 is not in B, and vice versa.

Therefore, B is a subset of4 elements such that it contains exactly one of each pair {x, x+4}.

Since the octagon has4 pairs of diametrically opposite vertices: {0,4}, {1,5}, {2,6}, {3,7}.

Therefore, to form such a subset B, we must choose exactly one element from each of the4 pairs.

Therefore, the number of such subsets is2^4=16.

Each pair has2 choices, so total16 subsets.

Therefore, the number of size4 subsets B where rotating by4 gives the complement is16.

These are exactly the subsets that are self-complementary under rotation by4.

Therefore, for these subsets, rotating by4 will map B to its complement, so they satisfy the condition.

Therefore, the number of size4 subsets that CAN be rotated to their complement is16.

Therefore, the number that cannot be rotated (Y) is70-16=54.

Wait, no. Wait, the question is asking for subsets that CAN be rotated so that all blue vertices end up at originally red positions. For size4 subsets, this is exactly the16 subsets that are self-complementary under some rotation (specifically rotation by4).

Wait, but rotation by4 is the only rotation that can achieve this, as per above.

Therefore, the number of size4 subsets that can be rotated to their complement is16.

Therefore, the total valid subsets are:

Size0:1,

Size1:8,

Size2:28,

Size3:56,

Size4:16.

Total=1+8=9; 9+28=37; 37+56=93; 93+16=109.

Wait, but this contradicts previous reasoning.

Wait, hold on. The problem states "can be rotated so that all of the blue vertices end up at positions where there were originally red vertices".

For a size4 subset B, this requires that there exists some rotation π where π(B) is a subset of R=V\B.

Since |π(B)|=4 and |R|=4, this requires π(B)=R.

Therefore, the subset B must be such that its complement is a rotation of itself.

The only rotations that can achieve this are those where the rotation is by4 steps, as above, leading to the16 subsets.

Therefore, only those16 subsets of size4 satisfy the condition.

Therefore, the total valid subsets are1+8+28+56+16=109.

Thus the probability is109/256. Since109 is prime (109 divides by... 109/13≈8.38, so yes, prime). So m=109, n=256, which are coprime. Thus m+n=365.

Wait, but this contradicts my earlier examples where some size4 subsets could be rotated to be disjoint.

Wait, for example, the subset {0,1,2,4} can be rotated by4 to get {4,5,6,0}, which is {0,4,5,6}. The original subset is {0,1,2,4}, so the rotated set {0,4,5,6} intersects with original at0 and4. So that's not disjoint.

Wait, but earlier I thought that {0,1,2,4} is a complete difference set, but according to this new approach, it cannot be rotated to be disjoint from itself.

Wait, there's a contradiction here.

Earlier, I thought that {0,1,2,4} could be rotated to be disjoint, but according to this new analysis, only the16 subsets that are self-complementary under rotation by4 can be rotated to their complement.

Let me check {0,1,2,4}:

Can this subset be rotated to be disjoint from itself?

Let me try rotating by3:

{3,4,5,7}. Original is {0,1,2,4}. The intersection is {4}. Not disjoint.

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

Rotate by1: {1,2,3,5}. Intersection {1,2}. Not disjoint.

Rotate by5: {5,6,7,1}. Intersection empty? Original is {0,1,2,4}; rotated set is {1,5,6,7}. The intersection is {1}. Not empty.

Rotate by6: {6,7,0,2}. Intersection {0,2}. Not empty.

Rotate by7: {7,0,1,3}. Intersection {0,1}. Not empty.

Rotate by4: {4,5,6,0}. Intersection {0,4}. Not empty.

So indeed, this subset cannot be rotated to be disjoint from itself. So my earlier assumption was wrong. Therefore, the only subsets of size4 that can be rotated to be disjoint are those16 self-complementary ones.

Therefore, the correct total is109.

Thus the probability is109/256, so m+n=109+256=365.

But let me confirm with another example of the16 subsets.

Take a subset that chooses one from each diametric pair:

Choose0 from {0,4},1 from {1,5},2 from {2,6},3 from {3,7}. So B={0,1,2,3}.

Rotating by4 gives {4,5,6,7}, which is the complement. So yes, this works.

Another example: choose0,1,2,7. Then complement is {4,5,6,3}. Rotating by4: {4,5,6,7+4=3 mod8}. So {4,5,6,3}, which is the complement. So yes.

Therefore, these16 subsets can be rotated (by4) to their complement, satisfying the condition.

Other subsets of size4 cannot be rotated to be disjoint.

Therefore, the total valid subsets are109.

Thus the probability is109/256, so m+n=365.

But let me check the count again:

Size0:1

Size1:8

Size2:28

Size3:56

Size4:16

Total:1+8=9; 9+28=37; 37+56=93; 93+16=109. Yes.

Therefore, the answer is365.

Wait, but I need to confirm that for size3 subsets, all can be rotated.

Earlier I thought so, but let me check a tricky case.

Take subset {0,1,2}. Rotating by4 gives {4,5,6}, which is disjoint from original. So yes.

Another subset {0,1,4}. Rotating by4 gives {4,5,0}, which intersects at0 and4. Not good. But rotating by3 gives {3,4,7}, which is disjoint from {0,1,4}? The rotated set {3,4,7} intersects with original at4. Not disjoint. Rotate by2: {2,3,6}, disjoint from {0,1,4}. Yes, so works.

Another example: {0,2,4}. Rotating by1 gives {1,3,5}, disjoint.

Yes, so all size3 subsets can be rotated.

Therefore, the total is indeed109.

Thus the final answer is365.

**Final Answer**
\boxed{365}
</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.

### 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**:
   - A subset \( B \) of blue vertices must be such that there exists a rotation \( \pi \) where \( \pi(B) \) is disjoint from \( B \).
   - This requires \( \pi(B) \subseteq V \setminus B \), meaning \( B \) and \( \pi(B) \) are disjoint.

3. **Size Considerations**:
   - Subsets of size 0 (empty set) trivially satisfy the condition.
   - Subsets of size 1, 2, and 3 can always be rotated to be disjoint from themselves.
   - Subsets of size 4 must be such that their complement is a rotation of themselves.

4. **Counting Valid Subsets**:
   - **Size 0**: 1 subset (empty set).
   - **Size 1**: 8 subsets (all single vertices).
   - **Size 2**: 28 subsets (all pairs).
   - **Size 3**: 56 subsets (all triples).
   - **Size 4**: 16 subsets (those that are self-complementary under rotation by 4 steps).

5. **Total Valid Subsets**:
   - Summing these, we get \( 1 + 8 + 28 + 56 + 16 = 109 \).

6. **Probability Calculation**:
   - The total number of subsets is \( 2^8 = 256 \).
   - The probability is \( \frac{109}{256} \).

7. **Final Result**:
   - The fraction \( \frac{109}{256} \) is in simplest form, so \( m = 109 \) and \( n = 256 \).
   - Thus, \( m + n = 109 + 256 = 365 \).

### Final Answer:
\boxed{365}