Finding the most efficient way to solve a problem is difficult because some paths are long and complex without providing much useful information. It is hard to tell which steps are actually productive versus just being expensive to process.
It ranks different ways of solving a problem by measuring how much useful information each step provides compared to the amount of effort it takes to get there. It identifies the most efficient path by balancing information gain with cost.
It helps find the most direct and informative route to a solution without wasting resources on unnecessary steps.
It was run inside an isolated container with no network access. This is the exact command and the real output it produced — captured process output, not written by a model.
$ python3 rpv_tool.py Top paths from A to D ranked by RPVD score:
A screenshot of that run.
A clean run proves this does what is shown above, in a CPU-only sandbox. It is a small research demo — not a production tool, and nothing here was published anywhere.
All of it — 93 lines, one file, standard library only.
import math
from collections import defaultdict
class RPVDScorer:
def __init__(self, graph):
self.adjacency = defaultdict(list)
self.node_values = {}
self.node_costs = {}
for node, data in graph.items():
self.node_values[node] = data['value']
self.node_costs[node] = data['cost']
self.adjacency[node] = data.get('neighbors', [])
def pvd_score(self, path):
return sum(self.node_values[node] / self.node_costs[node] for node in path)
def its_factor(self, path):
return math.exp(-0.5 * len(path))
def rpvd_score(self, path):
return self.pvd_score(path) * self.its_factor(path)
def find_paths(self, start, end):
paths = []
visited = set()
def dfs(current, path):
visited.add(current)
path.append(current)
if current == end:
paths.append(list(path))
else:
for neighbor in self.adjacency.get(current, []):
if neighbor not in visited:
dfs(neighbor, path)
path.pop()
visited.remove(current)
dfs(start, [])
return paths
def rank_paths(self, start, end):
paths = self.find_paths(start, end)
scored = [(p, self.rpvd_score(p)) for p in paths]
return sorted(scored, key=lambda x: x[1], reverse=True)
def identify_bottleneck(self, path):
max_contrib = -1
bottleneck = None
for node in path:
contrib = self.node_values[node] / self.node_costs[node]
if contrib > max_contrib:
max_contrib = contrib
bottleneck = node
return bottleneck, max_contrib
def pbi_tool(graph, path):
scorer = RPVDScorer(graph)
rpvd = scorer.rpvd_score(path)
bottleneck, bottleneck_density = scorer.identify_bottleneck(path)
return {
'rpvd_score': rpvd,
'bottleneck': bottleneck,
'bottleneck_density': bottleneck_density,
'path': path,
}
if __name__ == "__main__":
graph = {
'A': {'value': 10, 'cost': 2, 'neighbors': ['B', 'C']},
'B': {'value': 6, 'cost': 1, 'neighbors': ['D']},
'C': {'value': 8, 'cost': 3, 'neighbors': ['D']},
'D': {'value': 5, 'cost': 1, 'neighbors': []}
}
scorer = RPVDScorer(graph)
start_node = 'A'
end_node = 'D'
ranked = scorer.rank_paths(start_node, end_node)
print("Top paths from A to D ranked by RPVD score:\n")
for path, score in ranked:
bottleneck, contrib = scorer.identify_bottleneck(path)
print(f' => {" -> ".join(path)} (RPVD: {score:.2f})')
print(f' Bottleneck: {bottleneck} (Contribution: {contrib:.2f})')
print("\npbi_tool direct call:\n")
result = pbi_tool(graph, ['A', 'B', 'D'])
print(f" RPVD: {result['rpvd_score']:.2f}")
print(f" Bottleneck: {result['bottleneck']} (Density: {result['bottleneck_density']:.2f})")