El algoritmo de Bellman Ford es un algoritmo de ruta más corta de fuente única. Este algoritmo se utiliza para encontrar la distancia más corta desde un único vértice a todos los demás vértices de un gráfico ponderado. Hay varios otros algoritmos que se utilizan para encontrar el camino más corto, como el algoritmo de Dijkstra, etc. Si el gráfico ponderado contiene valores de peso negativos, entonces el algoritmo de Dijkstra no confirma si produce la respuesta correcta o no. A diferencia del algoritmo de Dijkstra, el algoritmo de Bellman Ford garantiza la respuesta correcta incluso si el gráfico ponderado contiene valores de peso negativos.
Regla de este algoritmo
We will go on relaxing all the edges (n - 1) times where, n = number of vertices
Considere el siguiente gráfico:
Como podemos observar en el gráfico anterior, algunos de los pesos son negativos. El gráfico anterior contiene 6 vértices, por lo que continuaremos relajándonos hasta los 5 vértices. Aquí relajaremos todos los bordes 5 veces. El ciclo se repetirá 5 veces para obtener la respuesta correcta. Si el bucle se repite más de 5 veces, la respuesta también será la misma, es decir, no habrá cambios en la distancia entre los vértices.
Relajante significa:
igualdad de objetos java
If (d(u) + c(u , v) <d(v)) d(v)="d(u)" + c(u , v) < pre> <p>To find the shortest path of the above graph, the first step is note down all the edges which are given below:</p> <p>(A, B), (A, C), (A, D), (B, E), (C, E), (D, C), (D, F), (E, F), (C, B)</p> <p>Let's consider the source vertex as 'A'; therefore, the distance value at vertex A is 0 and the distance value at all the other vertices as infinity shown as below:</p> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-2.webp" alt="Bellman-Ford Algorithm"> <p>Since the graph has six vertices so it will have five iterations.</p> <p> <strong>First iteration</strong> </p> <p>Consider the edge (A, B). Denote vertex 'A' as 'u' and vertex 'B' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 6</p> <p>Since (0 + 6) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 6 = 6</p> <p>Therefore, the distance of vertex B is 6.</p> <p>Consider the edge (A, C). Denote vertex 'A' as 'u' and vertex 'C' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 4</p> <p>Since (0 + 4) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 4 = 4</p> <p>Therefore, the distance of vertex C is 4.</p> <p>Consider the edge (A, D). Denote vertex 'A' as 'u' and vertex 'D' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 5</p> <p>Since (0 + 5) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 5 = 5</p> <p>Therefore, the distance of vertex D is 5.</p> <p>Consider the edge (B, E). Denote vertex 'B' as 'u' and vertex 'E' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 6</p> <p>d(v) = ∞</p> <p>c(u , v) = -1</p> <p>Since (6 - 1) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 6 - 1= 5</p> <p>Therefore, the distance of vertex E is 5.</p> <p>Consider the edge (C, E). Denote vertex 'C' as 'u' and vertex 'E' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 4</p> <p>d(v) = 5</p> <p>c(u , v) = 3</p> <p>Since (4 + 3) is greater than 5, so there will be no updation. The value at vertex E is 5.</p> <p>Consider the edge (D, C). Denote vertex 'D' as 'u' and vertex 'C' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 5</p> <p>d(v) = 4</p> <p>c(u , v) = -2</p> <p>Since (5 -2) is less than 4, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 5 - 2 = 3</p> <p>Therefore, the distance of vertex C is 3.</p> <p>Consider the edge (D, F). Denote vertex 'D' as 'u' and vertex 'F' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 5</p> <p>d(v) = ∞</p> <p>c(u , v) = -1</p> <p>Since (5 -1) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 5 - 1 = 4</p> <p>Therefore, the distance of vertex F is 4.</p> <p>Consider the edge (E, F). Denote vertex 'E' as 'u' and vertex 'F' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 5</p> <p>d(v) = ∞</p> <p>c(u , v) = 3</p> <p>Since (5 + 3) is greater than 4, so there would be no updation on the distance value of vertex F.</p> <p>Consider the edge (C, B). Denote vertex 'C' as 'u' and vertex 'B' as 'v'. Now use the relaxing formula:</p> <p>d(u) = 3</p> <p>d(v) = 6</p> <p>c(u , v) = -2</p> <p>Since (3 - 2) is less than 6, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 3 - 2 = 1</p> <p>Therefore, the distance of vertex B is 1.</p> <p>Now the first iteration is completed. We move to the second iteration.</p> <p> <strong>Second iteration:</strong> </p> <p>In the second iteration, we again check all the edges. The first edge is (A, B). Since (0 + 6) is greater than 1 so there would be no updation in the vertex B.</p> <p>The next edge is (A, C). Since (0 + 4) is greater than 3 so there would be no updation in the vertex C.</p> <p>The next edge is (A, D). Since (0 + 5) equals to 5 so there would be no updation in the vertex D.</p> <p>The next edge is (B, E). Since (1 - 1) equals to 0 which is less than 5 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(E) = d(B) +c(B , E)</p> <p>= 1 - 1 = 0</p> <p>The next edge is (C, E). Since (3 + 3) equals to 6 which is greater than 5 so there would be no updation in the vertex E.</p> <p>The next edge is (D, C). Since (5 - 2) equals to 3 so there would be no updation in the vertex C.</p> <p>The next edge is (D, F). Since (5 - 1) equals to 4 so there would be no updation in the vertex F.</p> <p>The next edge is (E, F). Since (5 + 3) equals to 8 which is greater than 4 so there would be no updation in the vertex F.</p> <p>The next edge is (C, B). Since (3 - 2) equals to 1` so there would be no updation in the vertex B.</p> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-3.webp" alt="Bellman-Ford Algorithm"> <p> <strong>Third iteration</strong> </p> <p>We will perform the same steps as we did in the previous iterations. We will observe that there will be no updation in the distance of vertices.</p> <pre> The following are the distances of vertices: A: 0 B: 1 C: 3 D: 5 E: 0 F: 3 </pre> <p> <strong>Time Complexity</strong> </p> <p>The time complexity of Bellman ford algorithm would be O(E|V| - 1).</p> <pre> function bellmanFord(G, S) for each vertex V in G distance[V] <- 0 infinite previous[v] <- null distance[s] for each vertex v in g edge (u,v) tempdistance distance[u] + edge_weight(u, v) if < distance[v] u distance[v} error: negative cycle exists return distance[], previous[] pre> <p> <strong>Drawbacks of Bellman ford algorithm</strong> </p> <ul> <li>The bellman ford algorithm does not produce a correct answer if the sum of the edges of a cycle is negative. Let's understand this property through an example. Consider the below graph. <br> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-4.webp" alt="Bellman-Ford Algorithm"></li> <li>In the above graph, we consider vertex 1 as the source vertex and provides 0 value to it. We provide infinity value to other vertices shown as below: <br> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-5.webp" alt="Bellman-Ford Algorithm"> <br>Edges can be written as: <br> (1, 3), (1, 2), (2, 4), (3, 2), (4, 3) </li> </ul> <p> <strong>First iteration</strong> </p> <p> <strong>Consider the edge (1, 3). Denote vertex '1' as 'u' and vertex '3' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 5</p> <p>Since (0 + 5) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 5 = 5</p> <p>Therefore, the distance of vertex 3 is 5.</p> <p> <strong>Consider the edge (1, 2). Denote vertex '1' as 'u' and vertex '2' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 4</p> <p>Since (0 + 4) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 4 = 4</p> <p>Therefore, the distance of vertex 2 is 4.</p> <p> <strong>Consider the edge (3, 2). Denote vertex '3' as 'u' and vertex '2' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 5</p> <p>d(v) = 4</p> <p>c(u , v) = 7</p> <p>Since (5 + 7) is greater than 4, so there would be no updation in the vertex 2.</p> <p> <strong>Consider the edge (2, 4). Denote vertex '2' as 'u' and vertex '4' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 4</p> <p>d(v) = ∞</p> <p>c(u , v) = 7</p> <p>Since (4 + 7) equals to 11 which is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 4 + 7 = 11</p> <p>Therefore, the distance of vertex 4 is 11.</p> <p> <strong>Consider the edge (4, 3). Denote vertex '4' as 'u' and vertex '3' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 11</p> <p>d(v) = 5</p> <p>c(u , v) = -15</p> <p>Since (11 - 15) equals to -4 which is less than 5, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 11 - 15 = -4</p> <p>Therefore, the distance of vertex 3 is -4.</p> <p>Now we move to the second iteration.</p> <p> <strong>Second iteration</strong> </p> <p>Now, again we will check all the edges. The first edge is (1, 3). Since (0 + 5) equals to 5 which is greater than -4 so there would be no updation in the vertex 3.</p> <p>The next edge is (1, 2). Since (0 + 4) equals to 4 so there would be no updation in the vertex 2.</p> <p>The next edge is (3, 2). Since (-4 + 7) equals to 3 which is less than 4 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(2) = d(3) +c(3, 2)</p> <p>= -4 + 7 = 3</p> <p>Therefore, the value at vertex 2 is 3.</p> <p>The next edge is (2, 4). Since ( 3+7) equals to 10 which is less than 11 so update</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(4) = d(2) +c(2, 4)</p> <p>= 3 + 7 = 10</p> <p>Therefore, the value at vertex 4 is 10.</p> <p>The next edge is (4, 3). Since (10 - 15) equals to -5 which is less than -4 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(3) = d(4) +c(4, 3)</p> <p>= 10 - 15 = -5</p> <p>Therefore, the value at vertex 3 is -5.</p> <p>Now we move to the third iteration.</p> <p> <strong>Third iteration</strong> </p> <p>Now again we will check all the edges. The first edge is (1, 3). Since (0 + 5) equals to 5 which is greater than -5 so there would be no updation in the vertex 3.</p> <p>The next edge is (1, 2). Since (0 + 4) equals to 4 which is greater than 3 so there would be no updation in the vertex 2.</p> <p>The next edge is (3, 2). Since (-5 + 7) equals to 2 which is less than 3 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(2) = d(3) +c(3, 2)</p> <p>= -5 + 7 = 2</p> <p>Therefore, the value at vertex 2 is 2.</p> <p>The next edge is (2, 4). Since (2 + 7) equals to 9 which is less than 10 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(4) = d(2) +c(2, 4)</p> <p>= 2 + 7 = 9</p> <p>Therefore, the value at vertex 4 is 9.</p> <p>The next edge is (4, 3). Since (9 - 15) equals to -6 which is less than -5 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(3) = d(4) +c(4, 3)</p> <p>= 9 - 15 = -6</p> <p>Therefore, the value at vertex 3 is -6.</p> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-6.webp" alt="Bellman-Ford Algorithm"> <p>Since the graph contains 4 vertices, so according to the bellman ford algorithm, there would be only 3 iterations. If we try to perform 4<sup>th</sup> iteration on the graph, the distance of the vertices from the given vertex should not change. If the distance varies, it means that the bellman ford algorithm is not providing the correct answer.</p> <p> <strong>4<sup>th</sup> iteration</strong> </p> <p>The first edge is (1, 3). Since (0 +5) equals to 5 which is greater than -6 so there would be no change in the vertex 3.</p> <p>The next edge is (1, 2). Since (0 + 4) is greater than 2 so there would be no updation.</p> <p>The next edge is (3, 2). Since (-6 + 7) equals to 1 which is less than 3 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(2) = d(3) +c(3, 2)</p> <p>= -6 + 7 = 1</p> <p>In this case, the value of the vertex is updated. So, we conclude that the bellman ford algorithm does not work when the graph contains the negative weight cycle.</p> <p>Therefore, the value at vertex 2 is 1.</p> <hr></-></pre></d(v))>
d(v) = 0 + 6 = 6
Por tanto, la distancia del vértice B es 6.
Considere el borde (A, C). Denota el vértice 'A' como 'u' y el vértice 'C' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 0
re(v) = ∞
c(tu, v) = 4
Dado que (0 + 4) es menor que ∞, actualice
d(v) = d(u) + c(u , v)
d(v) = 0 + 4 = 4
Por tanto, la distancia del vértice C es 4.
Considere el borde (A, D). Denota el vértice 'A' como 'u' y el vértice 'D' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 0
re(v) = ∞
c(tu, v) = 5
Dado que (0 + 5) es menor que ∞, actualice
d(v) = d(u) + c(u , v)
d(v) = 0 + 5 = 5
Por tanto, la distancia del vértice D es 5.
Considere el borde (B, E). Denota el vértice 'B' como 'u' y el vértice 'E' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 6
re(v) = ∞
c(tu, v) = -1
Dado que (6 - 1) es menor que ∞, actualice
d(v) = d(u) + c(u , v)
d(v) = 6 - 1= 5
Por tanto, la distancia del vértice E es 5.
Considere el borde (C, E). Denota el vértice 'C' como 'u' y el vértice 'E' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 4
d(v) = 5
c(tu, v) = 3
Dado que (4 + 3) es mayor que 5, no habrá actualización. El valor en el vértice E es 5.
Considere el borde (D, C). Denota el vértice 'D' como 'u' y el vértice 'C' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 5
d(v) = 4
c(tu, v) = -2
Dado que (5 -2) es menor que 4, actualice
d(v) = d(u) + c(u , v)
d(v) = 5 - 2 = 3
Por tanto, la distancia del vértice C es 3.
Considere el borde (D, F). Denota el vértice 'D' como 'u' y el vértice 'F' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 5
re(v) = ∞
c(tu, v) = -1
Dado que (5 -1) es menor que ∞, actualice
d(v) = d(u) + c(u , v)
d(v) = 5 - 1 = 4
Por tanto, la distancia del vértice F es 4.
Considere el borde (E, F). Denota el vértice 'E' como 'u' y el vértice 'F' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 5
re(v) = ∞
c(tu, v) = 3
Dado que (5 + 3) es mayor que 4, no habría actualización en el valor de distancia del vértice F.
Considere el borde (C, B). Denota el vértice 'C' como 'u' y el vértice 'B' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 3
d(v) = 6
c(tu, v) = -2
Dado que (3 - 2) es menor que 6, actualice
d(v) = d(u) + c(u , v)
d(v) = 3 - 2 = 1
Por tanto, la distancia al vértice B es 1.
Ahora se completa la primera iteración. Pasamos a la segunda iteración.
Segunda iteración:
En la segunda iteración, volvemos a comprobar todos los bordes. La primera arista es (A, B). Dado que (0 + 6) es mayor que 1, no habría actualización en el vértice B.
La siguiente arista es (A, C). Dado que (0 + 4) es mayor que 3, no habría actualización en el vértice C.
El siguiente borde es (A, D). Dado que (0 + 5) es igual a 5, no habría actualización en el vértice D.
El siguiente borde es (B, E). Dado que (1 - 1) es igual a 0, que es menor que 5, actualice:
creación de tabla de oráculo
d(v) = d(u) + c(u, v)
d(E) = d(B) +c(B , E)
= 1 - 1 = 0
El siguiente borde es (C, E). Dado que (3 + 3) es igual a 6, que es mayor que 5, no habría actualización en el vértice E.
El siguiente borde es (D, C). Dado que (5 - 2) es igual a 3, no habría actualización en el vértice C.
El siguiente borde es (D, F). Dado que (5 - 1) es igual a 4, no habría actualización en el vértice F.
El siguiente borde es (E, F). Dado que (5 + 3) es igual a 8, que es mayor que 4, no habría actualización en el vértice F.
El siguiente borde es (C, B). Dado que (3 - 2) es igual a 1`, no habría actualización en el vértice B.
Tercera iteración
Realizaremos los mismos pasos que hicimos en las iteraciones anteriores. Observaremos que no habrá actualización en la distancia de vértices.
The following are the distances of vertices: A: 0 B: 1 C: 3 D: 5 E: 0 F: 3
Complejidad del tiempo
La complejidad temporal del algoritmo de Bellman Ford sería O (E | V | - 1).
function bellmanFord(G, S) for each vertex V in G distance[V] <- 0 infinite previous[v] <- null distance[s] for each vertex v in g edge (u,v) tempdistance distance[u] + edge_weight(u, v) if < distance[v] u distance[v} error: negative cycle exists return distance[], previous[] pre> <p> <strong>Drawbacks of Bellman ford algorithm</strong> </p> <ul> <li>The bellman ford algorithm does not produce a correct answer if the sum of the edges of a cycle is negative. Let's understand this property through an example. Consider the below graph. <br> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-4.webp" alt="Bellman-Ford Algorithm"></li> <li>In the above graph, we consider vertex 1 as the source vertex and provides 0 value to it. We provide infinity value to other vertices shown as below: <br> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-5.webp" alt="Bellman-Ford Algorithm"> <br>Edges can be written as: <br> (1, 3), (1, 2), (2, 4), (3, 2), (4, 3) </li> </ul> <p> <strong>First iteration</strong> </p> <p> <strong>Consider the edge (1, 3). Denote vertex '1' as 'u' and vertex '3' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 5</p> <p>Since (0 + 5) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 5 = 5</p> <p>Therefore, the distance of vertex 3 is 5.</p> <p> <strong>Consider the edge (1, 2). Denote vertex '1' as 'u' and vertex '2' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 0</p> <p>d(v) = ∞</p> <p>c(u , v) = 4</p> <p>Since (0 + 4) is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 0 + 4 = 4</p> <p>Therefore, the distance of vertex 2 is 4.</p> <p> <strong>Consider the edge (3, 2). Denote vertex '3' as 'u' and vertex '2' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 5</p> <p>d(v) = 4</p> <p>c(u , v) = 7</p> <p>Since (5 + 7) is greater than 4, so there would be no updation in the vertex 2.</p> <p> <strong>Consider the edge (2, 4). Denote vertex '2' as 'u' and vertex '4' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 4</p> <p>d(v) = ∞</p> <p>c(u , v) = 7</p> <p>Since (4 + 7) equals to 11 which is less than ∞, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 4 + 7 = 11</p> <p>Therefore, the distance of vertex 4 is 11.</p> <p> <strong>Consider the edge (4, 3). Denote vertex '4' as 'u' and vertex '3' as 'v'. Now use the relaxing formula:</strong> </p> <p>d(u) = 11</p> <p>d(v) = 5</p> <p>c(u , v) = -15</p> <p>Since (11 - 15) equals to -4 which is less than 5, so update</p> <pre> d(v) = d(u) + c(u , v) </pre> <p>d(v) = 11 - 15 = -4</p> <p>Therefore, the distance of vertex 3 is -4.</p> <p>Now we move to the second iteration.</p> <p> <strong>Second iteration</strong> </p> <p>Now, again we will check all the edges. The first edge is (1, 3). Since (0 + 5) equals to 5 which is greater than -4 so there would be no updation in the vertex 3.</p> <p>The next edge is (1, 2). Since (0 + 4) equals to 4 so there would be no updation in the vertex 2.</p> <p>The next edge is (3, 2). Since (-4 + 7) equals to 3 which is less than 4 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(2) = d(3) +c(3, 2)</p> <p>= -4 + 7 = 3</p> <p>Therefore, the value at vertex 2 is 3.</p> <p>The next edge is (2, 4). Since ( 3+7) equals to 10 which is less than 11 so update</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(4) = d(2) +c(2, 4)</p> <p>= 3 + 7 = 10</p> <p>Therefore, the value at vertex 4 is 10.</p> <p>The next edge is (4, 3). Since (10 - 15) equals to -5 which is less than -4 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(3) = d(4) +c(4, 3)</p> <p>= 10 - 15 = -5</p> <p>Therefore, the value at vertex 3 is -5.</p> <p>Now we move to the third iteration.</p> <p> <strong>Third iteration</strong> </p> <p>Now again we will check all the edges. The first edge is (1, 3). Since (0 + 5) equals to 5 which is greater than -5 so there would be no updation in the vertex 3.</p> <p>The next edge is (1, 2). Since (0 + 4) equals to 4 which is greater than 3 so there would be no updation in the vertex 2.</p> <p>The next edge is (3, 2). Since (-5 + 7) equals to 2 which is less than 3 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(2) = d(3) +c(3, 2)</p> <p>= -5 + 7 = 2</p> <p>Therefore, the value at vertex 2 is 2.</p> <p>The next edge is (2, 4). Since (2 + 7) equals to 9 which is less than 10 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(4) = d(2) +c(2, 4)</p> <p>= 2 + 7 = 9</p> <p>Therefore, the value at vertex 4 is 9.</p> <p>The next edge is (4, 3). Since (9 - 15) equals to -6 which is less than -5 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(3) = d(4) +c(4, 3)</p> <p>= 9 - 15 = -6</p> <p>Therefore, the value at vertex 3 is -6.</p> <img src="//techcodeview.com/img/daa-tutorial/85/bellman-ford-algorithm-6.webp" alt="Bellman-Ford Algorithm"> <p>Since the graph contains 4 vertices, so according to the bellman ford algorithm, there would be only 3 iterations. If we try to perform 4<sup>th</sup> iteration on the graph, the distance of the vertices from the given vertex should not change. If the distance varies, it means that the bellman ford algorithm is not providing the correct answer.</p> <p> <strong>4<sup>th</sup> iteration</strong> </p> <p>The first edge is (1, 3). Since (0 +5) equals to 5 which is greater than -6 so there would be no change in the vertex 3.</p> <p>The next edge is (1, 2). Since (0 + 4) is greater than 2 so there would be no updation.</p> <p>The next edge is (3, 2). Since (-6 + 7) equals to 1 which is less than 3 so update:</p> <p>d(v) = d(u) + c(u, v)</p> <p>d(2) = d(3) +c(3, 2)</p> <p>= -6 + 7 = 1</p> <p>In this case, the value of the vertex is updated. So, we conclude that the bellman ford algorithm does not work when the graph contains the negative weight cycle.</p> <p>Therefore, the value at vertex 2 is 1.</p> <hr></->
d(v) = 0 + 5 = 5
Por tanto, la distancia del vértice 3 es 5.
java para bucle
Considere la arista (1, 2). Denota el vértice '1' como 'u' y el vértice '2' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 0
re(v) = ∞
c(tu, v) = 4
Dado que (0 + 4) es menor que ∞, actualice
d(v) = d(u) + c(u , v)
d(v) = 0 + 4 = 4
Por tanto, la distancia del vértice 2 es 4.
Considere la arista (3, 2). Denote el vértice '3' como 'u' y el vértice '2' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 5
d(v) = 4
c(tu, v) = 7
Dado que (5 + 7) es mayor que 4, no habría actualización en el vértice 2.
Considere la arista (2, 4). Denota el vértice '2' como 'u' y el vértice '4' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 4
re(v) = ∞
c(tu, v) = 7
Dado que (4 + 7) es igual a 11, que es menor que ∞, actualice
d(v) = d(u) + c(u , v)
d(v) = 4 + 7 = 11
Por tanto, la distancia del vértice 4 es 11.
Considere la arista (4, 3). Denote el vértice '4' como 'u' y el vértice '3' como 'v'. Ahora utiliza la fórmula relajante:
d(tu) = 11
d(v) = 5
c(u, v) = -15
Dado que (11 - 15) es igual a -4, que es menor que 5, actualice
d(v) = d(u) + c(u , v)
d(v) = 11 - 15 = -4
Por tanto, la distancia del vértice 3 es -4.
Ahora pasamos a la segunda iteración.
Segunda iteración
Ahora, nuevamente revisaremos todos los bordes. La primera arista es (1, 3). Dado que (0 + 5) es igual a 5, que es mayor que -4, no habría actualización en el vértice 3.
La siguiente arista es (1, 2). Dado que (0 + 4) es igual a 4, no habría actualización en el vértice 2.
La siguiente arista es (3, 2). Dado que (-4 + 7) es igual a 3, que es menor que 4, actualice:
d(v) = d(u) + c(u, v)
d(2) = d(3) +c(3, 2)
= -4 + 7 = 3
Por tanto, el valor en el vértice 2 es 3.
La siguiente arista es (2, 4). Dado que (3+7) es igual a 10, que es menor que 11, actualice
d(v) = d(u) + c(u, v)
d(4) = d(2) +c(2, 4)
= 3 + 7 = 10
Por tanto, el valor en el vértice 4 es 10.
matrices java
La siguiente arista es (4, 3). Dado que (10 - 15) es igual a -5, que es menor que -4, actualice:
d(v) = d(u) + c(u, v)
d(3) = d(4) +c(4, 3)
= 10 - 15 = -5
Por tanto, el valor en el vértice 3 es -5.
Ahora pasamos a la tercera iteración.
Tercera iteración
Ahora nuevamente revisaremos todos los bordes. La primera arista es (1, 3). Dado que (0 + 5) es igual a 5, que es mayor que -5, no habría actualización en el vértice 3.
La siguiente arista es (1, 2). Dado que (0 + 4) es igual a 4, que es mayor que 3, no habría actualización en el vértice 2.
La siguiente arista es (3, 2). Dado que (-5 + 7) es igual a 2, que es menor que 3, actualice:
d(v) = d(u) + c(u, v)
d(2) = d(3) +c(3, 2)
= -5 + 7 = 2
Por tanto, el valor en el vértice 2 es 2.
La siguiente arista es (2, 4). Dado que (2 + 7) es igual a 9, que es menor que 10, actualice:
d(v) = d(u) + c(u, v)
d(4) = d(2) +c(2, 4)
= 2 + 7 = 9
Por tanto, el valor en el vértice 4 es 9.
La siguiente arista es (4, 3). Dado que (9 - 15) es igual a -6, que es menor que -5, actualice:
d(v) = d(u) + c(u, v)
d(3) = d(4) +c(4, 3)
= 9 - 15 = -6
Por tanto, el valor en el vértice 3 es -6.
Dado que el gráfico contiene 4 vértices, según el algoritmo de Bellman Ford, solo habría 3 iteraciones. Si intentamos realizar 4thiteración en el gráfico, la distancia de los vértices desde el vértice dado no debería cambiar. Si la distancia varía, significa que el algoritmo de Bellman Ford no proporciona la respuesta correcta.
4thiteración
La primera arista es (1, 3). Dado que (0 +5) es igual a 5, que es mayor que -6, no habría ningún cambio en el vértice 3.
La siguiente arista es (1, 2). Dado que (0 + 4) es mayor que 2, no habría actualización.
La siguiente arista es (3, 2). Dado que (-6 + 7) es igual a 1, que es menor que 3, actualice:
d(v) = d(u) + c(u, v)
d(2) = d(3) +c(3, 2)
= -6 + 7 = 1
En este caso, se actualiza el valor del vértice. Entonces, concluimos que el algoritmo de Bellman Ford no funciona cuando el gráfico contiene el ciclo de peso negativo.
Por tanto, el valor en el vértice 2 es 1.