How is this calculated?
The math continuously tracks how strongly this post is connected to the rest of the network. Every tag forms a 2-way link. The base stats determine personal node strength, and the pie charts below show this node's share against its direct neighbours.
// 1. Base variables (floored at 1 to prevent zero-multiplication math errors)
$inbound = max(1, 1) = 1
$outbound = max(1, 1) = 1
// 2. Node Base Values (Local connection strength)
Base_Strength (PV) = $inbound * $outbound = 1 * 1 = 1
Base_Influence (IV) = $inbound / $outbound = 1 / 1 = 1
// 3. Exponential Network Values (accumulating 20 direct neighbours)
Network_Strength (CV) = Base_PV * (Neighbour_1_PV * Neighbour_2_PV * ...)
= 1 *
( 6 [Mattias Eklund] *
42 [Charlize Theron] *
90 [John Goodman] *
1 [Attila Árpa] *
100 [Eddie Marsan] *
25 [Til Schweiger] *
4 [Roland Møller] *
81 [Toby Jones] *
16 [Daniel Bernhardt] *
25 [James McAvoy] *
400 [Atomic Blonde] *
16 [Barbara Sukowa] *
16 [Declan Hannigan] *
1 [James Faulkner] *
1 [Lili Gesler] *
1 [Martin Angerbauer] *
16 [Sam Hargrave] *
1 [Sara Natasa Szonda] *
1 [Bill Skarsgård] *
1 [Jóhannes Haukur Jóhannesson]
)
= 1.20 x 10^19
Network_Influence (TV) = Base_IV * (Neighbour_1_IV * Neighbour_2_IV * ...)
= 1 *
( 1.5 [Mattias Eklund] *
1.17 [Charlize Theron] *
1.11 [John Goodman] *
1 [Attila Árpa] *
1 [Eddie Marsan] *
1 [Til Schweiger] *
1 [Roland Møller] *
1 [Toby Jones] *
1 [Daniel Bernhardt] *
1 [James McAvoy] *
1 [Atomic Blonde] *
1 [Barbara Sukowa] *
1 [Declan Hannigan] *
1 [James Faulkner] *
1 [Lili Gesler] *
1 [Martin Angerbauer] *
1 [Sam Hargrave] *
1 [Sara Natasa Szonda] *
1 [Bill Skarsgård] *
1 [Jóhannes Haukur Jóhannesson]
)
= 1.94
Outbound
1
Tags on post
Inbound
1
Posts tagging this
Connections
20
Total nodes
Base Node Strength
1
Base Node Influence
1
Strength Share (vs Direct Neighbours)
0.12%
(1.20 × 1019 overall)
Dominant nodes (excluded from chart)Atomic Blonde 47.34%
Influence Share (vs Direct Neighbours)
Connected Network Hierarchy
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Connection Health Audit (Red = broken 1-way link)