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 19 direct neighbours)
Network_Strength (CV) = Base_PV * (Neighbour_1_PV * Neighbour_2_PV * ...)
= 1 *
( 56 [Joan Cusack] *
225 [Paul Rudd] *
361 [The Perks of Being a Wallflower] *
25 [Tom Savini] *
16 [Mae Whitman] *
16 [Brian Balzerini] *
16 [Chelsea Zhang] *
36 [Dylan McDermott] *
1 [Emily Marie Callaway] *
25 [Emma Watson] *
16 [Ezra Miller] *
9 [Johnny Simmons] *
16 [Julia Garner] *
1 [Justine Nicole Schaefer] *
16 [Logan Lerman] *
4 [Nicholas Braun] *
1 [Nina Dobrev] *
9 [Patrick de Ledebur] *
1 [Tom Kruszewski]
)
= 5.56 x 10^20
Network_Influence (TV) = Base_IV * (Neighbour_1_IV * Neighbour_2_IV * ...)
= 1 *
( 1.14 [Joan Cusack] *
1 [Paul Rudd] *
1 [The Perks of Being a Wallflower] *
1 [Tom Savini] *
1 [Mae Whitman] *
1 [Brian Balzerini] *
1 [Chelsea Zhang] *
1 [Dylan McDermott] *
1 [Emily Marie Callaway] *
1 [Emma Watson] *
1 [Ezra Miller] *
1 [Johnny Simmons] *
1 [Julia Garner] *
1 [Justine Nicole Schaefer] *
1 [Logan Lerman] *
1 [Nicholas Braun] *
1 [Nina Dobrev] *
1 [Patrick de Ledebur] *
1 [Tom Kruszewski]
)
= 1.14
Outbound
1
Tags on post
Inbound
1
Posts tagging this
Connections
19
Total nodes
Base Node Strength
1
Base Node Influence
1
Strength Share (vs Direct Neighbours)
0.12%
(5.56 × 1020 overall)
Dominant nodes (excluded from chart)The Perks of Being a Wallflower 42.42%Paul Rudd 26.44%
Influence Share (vs Direct Neighbours)
Connected Network Hierarchy
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Connection Health Audit (Red = broken 1-way link)