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feat: python impl for 0-1 BFS #1380

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109 changes: 71 additions & 38 deletions src/graph/01_bfs.md
Original file line number Diff line number Diff line change
Expand Up @@ -17,27 +17,44 @@ In this article we demonstrate how we can use BFS to solve the SSSP (single-sour
We can develop the algorithm by closely studying Dijkstra's algorithm and thinking about the consequences that our special graph implies.
The general form of Dijkstra's algorithm is (here a `set` is used for the priority queue):

```cpp
d.assign(n, INF);
d[s] = 0;
set<pair<int, int>> q;
q.insert({0, s});
while (!q.empty()) {
int v = q.begin()->second;
q.erase(q.begin());

for (auto edge : adj[v]) {
int u = edge.first;
int w = edge.second;

if (d[v] + w < d[u]) {
q.erase({d[u], u});
d[u] = d[v] + w;
q.insert({d[u], u});
=== "C++"
```cpp
d.assign(n, INF);
d[s] = 0;
set<pair<int, int>> q;
q.insert({0, s});
while (!q.empty()) {
int v = q.begin()->second;
q.erase(q.begin());

for (auto edge : adj[v]) {
int u = edge.first;
int w = edge.second;

if (d[v] + w < d[u]) {
q.erase({d[u], u});
d[u] = d[v] + w;
q.insert({d[u], u});
}
}
}
Comment on lines +20 to 40
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@adamant-pwn adamant-pwn Oct 24, 2024

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Suggested change
=== "C++"
```cpp
d.assign(n, INF);
d[s] = 0;
set<pair<int, int>> q;
q.insert({0, s});
while (!q.empty()) {
int v = q.begin()->second;
q.erase(q.begin());
for (auto edge : adj[v]) {
int u = edge.first;
int w = edge.second;
if (d[v] + w < d[u]) {
q.erase({d[u], u});
d[u] = d[v] + w;
q.insert({d[u], u});
}
}
}
=== "C++"
```cpp
d.assign(n, INF);
d[s] = 0;
priority_queue<pair<int, int>> q = {{0, s}};
while (!empty(q)) {
int [dv, v] = q.top();
q.pop();
if (-dv == d[v]) {
for (auto [u, w] : adj[v]) {
if (d[v] + w < d[u]) {
d[u] = d[v] + w;
q.insert({-d[u], u});
}
}
}
}

Probably best to make this consistent with Python then. Also remove the comment about "set" above if this is incorporated.

}
```
```
=== "Python"
```py
from heapq import heappop, heappush


d = [float("inf")] * n
d[s] = 0
q = [(0, s)]
while q:
dv, v = heappop(q)
if dv == d[v]:
for u, w in adj[v]:
if d[v] + w < d[u]:
d[u] = d[v] + w
heappush(q, (d[u], u))
```

We can notice that the difference between the distances between the source `s` and two other vertices in the queue differs by at most one.
Especially, we know that $d[v] \le d[u] \le d[v] + 1$ for each $u \in Q$.
Expand All @@ -53,27 +70,43 @@ This structure is so simple, that we don't need an actual priority queue, i.e. u
We can simply use a normal queue, and append new vertices at the beginning if the corresponding edge has weight $0$, i.e. if $d[u] = d[v]$, or at the end if the edge has weight $1$, i.e. if $d[u] = d[v] + 1$.
This way the queue still remains sorted at all time.

```cpp
vector<int> d(n, INF);
d[s] = 0;
deque<int> q;
q.push_front(s);
while (!q.empty()) {
int v = q.front();
q.pop_front();
for (auto edge : adj[v]) {
int u = edge.first;
int w = edge.second;
if (d[v] + w < d[u]) {
d[u] = d[v] + w;
if (w == 1)
q.push_back(u);
else
q.push_front(u);
=== "C++"
```cpp
vector<int> d(n, INF);
d[s] = 0;
deque<int> q;
q.push_front(s);
while (!q.empty()) {
int v = q.front();
q.pop_front();
for (auto edge : adj[v]) {
int u = edge.first;
int w = edge.second;
if (d[v] + w < d[u]) {
d[u] = d[v] + w;
if (w == 1)
q.push_back(u);
else
q.push_front(u);
}
}
}
Comment on lines +75 to 93
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Suggested change
vector<int> d(n, INF);
d[s] = 0;
deque<int> q;
q.push_front(s);
while (!q.empty()) {
int v = q.front();
q.pop_front();
for (auto edge : adj[v]) {
int u = edge.first;
int w = edge.second;
if (d[v] + w < d[u]) {
d[u] = d[v] + w;
if (w == 1)
q.push_back(u);
else
q.push_front(u);
}
}
}
vector<int> d(n, INF);
d[s] = 0;
deque<int> q = {s};
while (!empty(q)) {
int v = q.front();
q.pop_front();
for (auto [u, w] : adj[v]) {
if (d[v] + w < d[u]) {
d[u] = d[v] + w;
if (w == 1)
q.push_back(u);
else
q.push_front(u);
}
}
}

}
```
```
=== "Python"
```py
d = [float("inf")] * n
d[s] = 0
q = deque([s])
while q:
v = q.popleft()
for u, w in adj[v]:
if d[v] + w < d[u]:
d[u] = d[v] + w
if w == 1:
q.append(u)
else:
q.appendleft(u)
```

## Dial's algorithm

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