TAOCP 7.1.3 Exercise 147

The key correction is that Algorithm V must be followed literally: vertices are scanned in the prescribed external order $v_1,\dots,v_n$, and pointers $\pi_v,\beta_v,\alpha_v$ are updated only when th...

taocpmathematicsalgorithmsvolume-4math-medium

Section 7.1.3: Bitwise Tricks and Techniques

Exercise 147. ▶ [M20] [M20] What values of πv, βv, αv, and τj occur in Algorithm V when the forest is a) the empty digraph with vertices {v1, . . . , vn} and no arcs? b) the oriented path vn −−→· · · −−→v2 −−→v1?

Verified: no
Solve time: 3m

The key correction is that Algorithm V must be followed literally: vertices are scanned in the prescribed external order $v_1,\dots,v_n$, and pointers $\pi_v,\beta_v,\alpha_v$ are updated only when the algorithm actually discovers a vertex through an untraversed arc. Any interpretation that assumes a pre-existing rooted-tree structure is invalid.

We now restart from the algorithmic behavior.

a) Empty digraph on ${v_1,\dots,v_n}$

There are no arcs at all, so during the scan of vertices $v_1,\dots,v_n$, no recursive exploration is ever triggered.

When Algorithm V first encounters a vertex $v_i$, it is unmarked, so it is simply processed as an isolated vertex. Since there are no outgoing arcs to explore:

  • no child is ever created,
  • no sibling list is ever formed,
  • no ancestor updates propagate beyond the vertex itself.

Thus no structural links are created at all.

Parent pointers $\pi_v$

Since no vertex is ever discovered via another vertex, every vertex is a root of its own trivial tree. Hence

$$ \pi_{v_i} = 0 \quad \text{for all } i. $$

Sibling pointers $\beta_v$

No vertex acquires children, so there are no sibling chains:

$$ \beta_{v_i} = 0 \quad \text{for all } i. $$

Ancestor pointers $\alpha_v$

Each vertex is its own active component root at the moment of processing, and no later updates occur:

$$ \alpha_{v_i} = v_i \quad \text{for all } i. $$

Timestamps $\tau_v$

Each vertex is first encountered exactly when it is scanned, since no exploration ever reaches new vertices. Therefore the discovery times follow the scan order:

$$ \tau_{v_i} = i \quad \text{for } i=1,\dots,n. $$

b) Oriented path $v_n \to v_{n-1} \to \cdots \to v_1$

The crucial point is that Algorithm V still scans vertices in the fixed order $v_1, v_2, \dots, v_n$, and only traverses arcs from a vertex when that vertex is first processed.

We examine what happens during this scan.

Step 1: processing $v_1$

Vertex $v_1$ has no outgoing arcs. It is unmarked, so it is discovered immediately:

  • $\tau_{v_1} = 1$
  • $\pi_{v_1} = 0$
  • $\alpha_{v_1} = v_1$

Step 2: processing $v_2$

Vertex $v_2$ has a single outgoing arc $v_2 \to v_1$. But $v_1$ has already been marked, so the algorithm does not recurse or change any pointers.

Thus:

  • $v_2$ is discovered at scan time,
  • no tree edge is created.

So:

$$ \tau_{v_2} = 2, \quad \pi_{v_2} = 0, \quad \alpha_{v_2} = v_2. $$

General step

For each $v_i$, when it is scanned:

  • all vertices $v_{i-1}, \dots, v_1$ are already marked,
  • so every outgoing arc from $v_i$ points to already-visited vertices,
  • hence no arc causes a recursive discovery.

Therefore no vertex ever becomes a child of another.

Consequences for all pointers

Since no vertex is ever discovered via an arc:

$$ \pi_{v_i} = 0 \quad \text{for all } i. $$

No vertex ever gains children, so:

$$ \beta_{v_i} = 0 \quad \text{for all } i. $$

Each vertex is its own active root throughout its processing:

$$ \alpha_{v_i} = v_i \quad \text{for all } i. $$

Timestamps

Every vertex is discovered exactly at the moment it is scanned, so the timestamps again follow the scan order:

$$ \tau_{v_i} = i \quad \text{for } i=1,\dots,n. $$

Final conclusion

In both (a) and (b), the directed structure of the graph never triggers any recursive discovery, because every arc in the path points to a vertex that has already been processed by the time it is encountered.

Thus Algorithm V produces identical pointer and timestamp behavior in both cases:

$$ \boxed{ \pi_{v_i} = 0,\quad \beta_{v_i} = 0,\quad \alpha_{v_i} = v_i,\quad \tau_{v_i} = i \quad (1 \le i \le n) } $$

No other values occur.