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backwardPropagate wrapper: simulate backwardPropagateInternal calculating all paths in one go

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Heiko Schaefer 2023-07-10 14:17:18 +02:00
parent 4583b0a742
commit 7b7601326c
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1 changed files with 48 additions and 5 deletions
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53
wot-dijkstra/src/main/kotlin/org/pgpainless/wot/dijkstra/Query.kt
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@ -194,13 +194,56 @@ class Query(
* FIXME: public for unit tests (undo!)
*/
fun backwardPropagate(targetFpr: Fingerprint, targetUserid: String, filter: CertificationFilter): HashMap<Fingerprint, Pair<Path, Int>> {
// XXX: this is an experiment, calculating both variants is possibly not the most efficient approach.
// Prefer paths where the target User ID is self-signed as long as possible. (But .. Why?)
val authPaths = backwardPropagateInternal(targetFpr, targetUserid, true, filter)
if (authPaths.isNotEmpty()) return authPaths
// However, in terms of semantics, this function now simulates what would happen if
// backwardPropagateInternal() were generalized to find both types of paths:
// If we find no "self-signed" paths, return any others
return backwardPropagateInternal(targetFpr, targetUserid, false, filter)
// It would prefer shorter over longer paths, or higher trust amount - so this is what we do here.
val a = backwardPropagateInternal(targetFpr, targetUserid, true, filter)
val b = backwardPropagateInternal(targetFpr, targetUserid, false, filter)
val c = HashMap<Fingerprint, Pair<Path, Int>>()
val keys = a.keys.toMutableSet()
keys.addAll(b.keys)
for (fp in keys) {
val x = a[fp]
val y = b[fp]
if (x != null && y != null) {
// Pick the path we like better, first by length, then by amount
println("x: $x")
println("y: $y")
if (y.first.length < x.first.length) { // prefer smaller length
println("XX found two for $fp! -> picking y by length")
c[fp] = y
} else if (x.first.length < y.first.length) {
println("XX found two for $fp! -> picking x by length")
c[fp] = x
} else {
// length is the same, pick by amount
if (y.second > x.second) { // prefer bigger amount
println("XX found two for $fp! -> picking y by amount")
c[fp] = y
} else {
println("XX found two for $fp! -> picking x by amount")
c[fp] = x
}
}
} else if (x != null) {
println("XX found one for $fp, in x")
c[fp] = x
} else if (y != null) {
println("XX found one for $fp, in y")
c[fp] = y
}
}
return c
}
/**