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For an all-to-one shortest path problem with root $r$, initialize $d_r^{(0)}=0$ and all other labels to infinity. Set $d_i^{(k)}=\min (d_i^{(k-1)},{\min }_{(i,j)\in A}\{c_{ij}+d_j^{(k-1)}\})$, keeping the root at zero. Induction identifies the labels with the shortest directed paths using at most $k$ directed edges. Without a reachable negative directed cycle, $|V|-1$ rounds suffice after removing directed cycles from a shortest walk. In-place or queue variants repeatedly relax incoming directed edges to an improved label. Labels may decrease multiple times, so this is a label-correcting shortest-path algorithm.

The Bellman inequalities turn local directed edge checks into a global certificate for the shortest path problem. Here the convention is all-to-one: every journey ends at a fixed vertex $r$, and $c_{ij}$ is the cost of the original directed edge $i\to j$. Work in a finite directed graph with finite real costs. Write $L_i$ for candidate labels, normalized by $L_r=0$, on the vertices that can reach $r$. A feasible potential means precisely that these finite labels satisfyfor every directed edge within that set. These labels give lower bounds on costs to $r$. A discovered directed path of cost $U_i$ instead gives an upper bound. The following constructed graph shows both bounds and the exact distances between them; its weights are illustrative costs, not measured travel data.

The arrows in the picture point toward the destination, whereas edge relaxation propagates an improved label in the opposite direction: an improved $d_j$ can lower $d_i$. In the pictured graph, the directed paths from $a$ to $r$ cost $7$, $2+3=5$, and $2-2+4=4$. Thus $\delta (a,r)=4$. The candidate $L=(3,1,3,0)$ in vertex order $(a,b,c,r)$ is feasible, but underestimates the exact distances $(4,2,4,0)$. Merely satisfying the Bellman inequalities does not make a candidate exact.

Why do they give lower bounds? Along any directed path $i=v_0\to v_1\to \cdots \to v_k=r$, successive substitution givesThe same calculation works for a directed walk, with repeated vertices allowed. Taking the infimum of route costs gives $L_i\le \delta (i,r)$. A discovered route gives $\delta (i,r)\le U_i$. If the bounds meet, $L_i=U_i$, they certify an optimum without enumerating routes. Equivalently, check the Bellman inequalities for the proposed distances and exhibit directed paths realizing them. The authored Bellman-Ford algorithm implementation checks the corresponding single-source inequalities and equality on recorded route edges; reversing every directed edge gives this all-to-one convention.^[ref]

For the exact labels in the picture, the slack $c_{ij}+d_j-d_i$ is respectively $0,3,1,0,0$ on $a\to b$, $a\to r$, $b\to r$, $b\to c$, $c\to r$. Zero slack means equality; the zero-slack directed path $a\to b\to c\to r$ realizes the cost $4$. This is the useful beauty of the certificate: a few local inequalities rule out every cheaper journey, while one explicit journey attains the bound.

Keep unreachable vertices separate: $\delta (i,r)=\infty$ when no directed path to $r$ exists, and a finite lower bound alone cannot certify that fact. Extended labels can use $\infty +c=\infty$ for finite $c$, but never compute a slack as $\infty -\infty$. If $i$ can reach a negative-cost directed cycle and then reach $r$, repeatedly traversing the directed cycle makes route costs unbounded below: their infimum is $-\infty$, with no finite shortest directed walk. A finite feasible potential on such a directed cycle is impossible, since summing its Bellman inequalities would give $0\le \sum c_{ij}<0$. A negative-cost directed cycle that cannot reach $r$ does not affect distances to $r$. These are statements about unrestricted directed walks; requiring a directed path with no repeated vertices is a different optimization problem when negative-cost directed cycles exist.^[ref]

For the single-source convention, the same original directed edge $u\to v$ instead gives $L(v)\le L(u)+w_{uv}$ with $L(s)=0$, a lower bound on distance from $s$. The endpoint carrying the label on the left therefore changes with the direction of the problem.

A numerical value retained for reuse by memoization. It is usable only while the spreadsheet formula and all its relevant inputs remain valid. After changing A1 from 3 to 5, following outgoing directed edges finds exactly B1,B2,C1,D1. Invalidate their cached cell values before evaluating another requested result; install the new input value for A1. The cached cell values of A2 and E1 remain valid because neither has a directed path from A1.

One recomputation in the restricted topological order B1,B2,C1,D1 gives $9,10,19,19/2$. Four spreadsheet formulas are recomputed, while E1 stays at 16. Removing invalid values from the cache prevents a subtle error: a finished marker from the previous evaluation does not prove that the old value is correct after an edit. The affected set is conservative; a changed prerequisite might leave a consumer's value numerically equal, but this simple scheme still invalidates it. For a formula edit, update the incoming directed edges and corresponding outgoing consumer lists as well as invalidating the edited spreadsheet cell and its downstream consumers. Microsoft's documentation explicitly describes both direct and indirect dependents being marked for recalculation.^[ref]

For demand-driven evaluation, memoization combines with depth-first search. Each spreadsheet cell has one of three states: UNSEEN means no valid cached cell value; ACTIVE means its evaluation is on the current call stack; FINISHED means a valid value is available. Initialize valid input spreadsheet cells as FINISHED. Start with all spreadsheet formulas UNSEEN, or retain only valid FINISHED entries after invalidation. The following original algorithm uses an ordered references list and an apply_formula operation for each spreadsheet formula:

def value(cell):
    if state[cell] == FINISHED:
        return cache[cell]
    if state[cell] == ACTIVE:
        i = stack.index(cell)
        raise CircularReference(stack[i:] + [cell])
    state[cell] = ACTIVE
    stack.append(cell)
    try:
        args = [value(p) for p in references[cell]]
        cache[cell] = apply_formula(cell, args)
        state[cell] = FINISHED
        return cache[cell]
    except Exception:
        state[cell] = UNSEEN
        raise
    finally:
        stack.pop()

Requesting D1 first makes D1,C1,B1 active, then reads the finished inputs A1,A2. B1 finishes at 7. Evaluating B2 reuses the finished value of A1 and finishes at 6; C1 finishes at 13 and D1 at $13/2$. A later request for D1 immediately returns its cached cell value. The shared prerequisite A1 is not a directed cycle: it was finished before the second request. The crucial distinction is between a previously visited vertex and a still-active vertex. The exceptional path restores unfinished states while unwinding the call stack, so failed evaluations are not cached as successful results. MIT's depth-first search notes use stack-duration marks to distinguish an edge to an active ancestor and to detect a directed cycle.^[ref]

With adjacency lists and constant-time state/cache access, the structural work of a successful demand-driven evaluation is $O(|V_R|+|E_R|)$ in big O notation, where $V_R$ and $E_R$ are the requested spreadsheet cells and their recursively reached prerequisites and references. Each reached spreadsheet formula is computed once; each reference is inspected once. This time complexity counts dependency bookkeeping separately from the cost of the arithmetic operations. Detecting a spreadsheet circular reference can additionally scan the current call stack once to report its witness.

A cyclic ordering of a directed graph is considered unchanged by rotations or swaps of consecutive nonadjacent vertices. A representative is coherent when every backward edge $v_j\to v_i$ has a forward directed path from $v_i$ to $v_j$. Equivalently every edge is on a directed circuit that winds exactly once around the order. The number of backward edges on a circuit is invariant under the allowed changes. Every nontrivial strongly connected directed graph admits such an order.

A Directed acyclic graph is a directed graph with no directed cycle: following its directed edges can never return to the starting vertex. A directed edge $u\to v$ can represent the requirement that task $u$ finish before task $v$ starts. A directed path then records a chain of prerequisites. The essential prerequisite for topological sorting is that these requirements contain no circular dependency. For finite directed graphs, this is equivalent to the existence of a topological order. ^[ref]

Consider an invented publishing workflow with five tasks: $A$ drafts the text, $B$ draws a figure, $C$ reviews the text, $D$ lays out the text and figure, and $E$ publishes the result. Its directed edges are $A\to C$, $A\to D$, $B\to D$, $C\to E$, and $D\to E$. Reviewing needs the draft; layout needs both the draft and figure; publication needs both review and layout. The drawing below encodes only these prerequisites, with no measured durations or resource constraints.

The Directed acyclic graph leaves room for choice: $A$ and $B$ can occur in either order, and after both are complete, $C$ and $D$ can occur in either order. Task $C$ may even precede $B$, since the review does not need the figure. A topological order turns these local prerequisites into a complete executable list while retaining this freedom. It supplies a feasible sequential plan; task durations, available workers, and earliest finishing times require additional information.

In bond percolation on a directed graph, independently open each directed edge with probability $p$. The out-cluster of $x$ consists of vertices reachable by open directed paths. In site percolation, independently open vertices and require every vertex of the directed path, including $x$, to be open. The same definitions allow a multigraph. The orientation matters: an infinite underlying connected graph need not contain an infinite out-cluster from a specified root.

Suppose every vertex of a directed graph has at most $\Delta$ incoming bonds. In bond percolation with parameter $q$, open a site whenever at least one incoming bond is open. Incoming bond sets have different heads and hence are disjoint, so these site indicators are independent random variables. Supplement their probabilities to $r=1-(1-q{)}^{\Delta }$. Every vertex reached by a nontrivial open directed path is now open. Conditioning the root to be open gives ${\theta }_s(r)\ge r{\theta }_b(q)$, since a simple directed path from the root never uses a bond entering the root. Thus $p_H^s\le 1-(1-p_H^b{)}^{\Delta }$.

An algorithm whose tentative distance labels may be revised repeatedly until all relevant Bellman inequalities hold. Bellman-Ford algorithm uses this approach: a label is an upper bound supplied by a discovered directed path, and relaxing an directed edge can replace it with a shorter discovered directed path. A finite label is not declared permanent merely because it has been processed.

Choose a strict total order $v_1<\cdots <v_n$ of the vertices of a finite simple graph. Direct every edge $v_i{v}_j$ from $v_i$ to $v_j$ when $i<j$. Along every directed path, the index strictly increases. Returning to the starting vertex would require an impossible strict increase back to the initial index, so this is an acyclic orientation. The displayed $D_1$ comes from $A<B<C$.

The chosen strict total order is a topological order of the resulting Directed acyclic graph. Conversely, every acyclic orientation of a finite simple graph has a topological order, and orienting its edges from earlier to later in that order recovers it. Different topological orders can induce the same acyclic orientation, since vertices with no edge between them need not impose a choice of arrow.^[ref]

An order respects reachability but does not create it: $u<v$ in a chosen topological order does not imply a directed path from $u$ to $v$. For example, orient the two edges of the graph path $A,B,C$ using $A<C<B$. The arrows are $A\to B$ and $C\to B$; there is no directed path from $A$ to $C$, despite $A<C$. In $D_1$, however, $A$ reaches $C$ both directly and via $B$. Following outgoing directed edges with graph search decides this directional reachability.^[ref]