Acronym  ... 
Name  hyperbolic order 4 octagon tiling 
©  
Circumradius  sqrt[sqrt(2)]/2 = 0.594604 i 
Vertex figure  [8^{4}] 
Dual  x4o8o 
Confer 

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Note that this tiling allows for a consistent quartering of octagons (connecting opposite edge midpoints), such as to derive a combinatorical variant of x5o4o. While for that one the edge lengths of the regular pentagons aaaaa can be calculated as given below for comparison, in that decomposition the edge lengths of these nonregular pentagons ccbab would calculate (using the formulas provided here) as:
x8o4o (quartering each octagon gives a topological variant of x5o4o): P0 o+ P1 P1 /  P0 o  P1 ++ P2 edges a: hypdist(P0,P0) = 2 arcosh[cos(π/8)/sin(π/4)] = 1.528571 edges b: hypdist(P0,P1) = arcosh[cos(π/8)/sin(π/4)] = 0.764285 edges c: hypdist(P1,P2) = arcosh[cos(π/4)/sin(π/8)] = 1.224226 deformed pentagons ccbab (to be conpared to: x5o4o all edges alike: hypdist(P0,P0) = 2 arcosh[cos(π/5)/sin(π/4)] = 1.061275 regular pentagons aaaaa )
There exists also a series of regular modwraps of this tiling, obtained by identifying every kth vertex on each hole (q.e. straight edge sequence here). Then it allows a representation as (possibly) infinite regular skew polyhedron, which happens to be the facial subset of octagons of the spherical tetracontoctachoron (k=3: finite modwrap), of the hyperbolic o4x4x4o tesselation (k=4), etc. This series of modwraps generally is being denoted as x8o4ok (with according value of k).
Incidence matrix according to Dynkin symbol
o4o8x (N → ∞) . . .  2N  4  4 +++ . . x  2  4N  2 +++ . o8x  8  8  N snubbed forms: o4o8s
o8x8o (N → ∞) . . .  4N  4  2 2 +++ . x .  2  8N  1 1 +++ o8x .  8  8  N * . x8o  8  8  * N snubbed forms: o8s8o
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