Question textConsider the directed network below. Each arc label has the form arc current flow/capacity. For example, sa 3/8 means that arc sa currently carries 3 units of flow and has capacity 8. The displayed flow has value 10. Optional working area, not marked: the boxes below correspond to the current flow on each arc. Each column lists arcs with the same starting vertex: s first, then the remaining starting vertices in alphabetical order. They are initially the numerators shown on the graph. You may update them while working. [table]
from s | from a | from b | from c | from d | from e
sa 3/7 | ad 3/6 | bt 6/6 | ct 1/5 | dt 3/8 | eb 3/3 sb 3/6 | | | | | ec 1/3 se 4/4 | | | | |
[/table] Marked area below. Answer both parts. Maximum value of an s-t flow: Answer 1 Question 3[input] Vertices on the side of s in the minimum cut: [table]
s | a | b | c | d | e
[/table] Check Question 3

Question

Question textConsider the directed network below. Each arc label has the form arc current flow/capacity. For example, sa 3/8 means that arc sa currently carries 3 units of flow and has capacity 8. The displayed flow has value 10. Optional working area, not marked: the boxes below correspond to the current flow on each arc. Each column lists arcs with the same starting vertex: s first, then the remaining starting vertices in alphabetical order. They are initially the numerators shown on the graph. You may update them while working. [table]
from s | from a | from b | from c | from d | from e
sa 3/7 | ad 3/6 | bt 6/6 | ct 1/5 | dt 3/8 | eb 3/3 sb 3/6 |  |  |  |  | ec 1/3 se 4/4 |  |  |  |  | 
[/table] Marked area below. Answer both parts. Maximum value of an s-t flow: Answer 1 Question 3[input] Vertices on the side of s in the minimum cut: [table]
s | a | b | c | d | e
[/table] Check Question 3

BlackTom AI · Accepted Answer

Max-flow/min-cut theorem: the value of a maximum s–t flow equals the capacity of a minimum s–t cut. A cut is a partition (S-side, T-side) with s in S-side and t in T-side; its capacity is the sum of capacities of arcs leaving S-side to T-side.

List relevant arc capacities (from the diagram):
- s→a: 7
- s→b: 6
- s→e: 4
- a→d: 6
- b→t: 6
- c→t: 5
- d→t: 8
- e→b: 3
- e→c: 3

Step-by-step reasoning and cut-capacity calculations:
1. Consider S-side = {s} (only s). Arcs leaving S-side are s→a (7), s→b (6), s→e (4). Capacity = 7 + 6 + 4 = 17.
2. Consider S-side = {s,a}. Arcs leaving are s→b (6), s→e (4) and a→d (6) (s→a is internal). Capacity = 6 + 4 + 6 = 16.
3. Check S-side = {s,e}. Arcs leaving are s→a (7), s→b (6), and e→b or e→c if those go to T-side. Here e→b (3) and e→c (3) go to T-side; but only those from S to T count — with S-side {s,e}, e→b (3) and e→c (3) both leave, but s→a and s→b also leave. The smallest of these reasonable partitions sums at least 16 (for example counting e→c only gives 7+6+3 = 16).
4. Consider S-side = {s,a,b}. Arcs leaving include s→e (4), a→d (6), b→t (6). Capacity = 4 + 6 + 6 = 16.
5. Any cut that omits a but keeps only s has capacity 17, and any cut that contains a but not some other vertices we checked gives capacity ≥16.

Why the correct answer is correct:
- The cut S-side = {s,a} has capacity 16, and we cannot find any cut with capacity less than 16 (other natural candidates give 16 or larger). By the max-flow/min-cut theorem, the maximum s–t flow equals this minimum cut capacity, so the maximum flow value is 16.
- The specified minimum cut can be chosen so that the vertices on the s-side are exactly s and a.

Why other choices are incorrect:
- S alone gives capacity 17, so it is not minimum.
- Any choice that omits a while including s yields capacity ≥17, so a must be on the s-side for a minimum cut of value 16.
- Some other S-side sets (like {s,a,b} or {s,e}) also achieve capacity 16, so a minimum cut is not unique; however the particular minimum cut asked for in the answer has s and a on the s-side.

Conclusion: the maximum s–t flow equals 16, and one minimum cut has S-side = {s, a}. This matches the provided answers: 16 and s,a.

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