Spreadsheet dependency graph ($G=(V,E)$)

The directed graph whose vertices are the spreadsheet cells and whose directed edges $u\to v$ mean that the spreadsheet formula in $v$ references $u$. Input spreadsheet cells have no prerequisites in this model. Retain both incoming references and outgoing consumers: depth-first search for evaluation follows incoming references backwards, whereas downstream invalidation follows outgoing consumers forwards. A directed path then expresses indirect dependence, and a directed cycle expresses a spreadsheet circular reference.

A spreadsheet cell depends on itself along a nonempty directed path. For a separate formula-edit experiment, restore the initial inputs and replace only B1 by =A1+A2+C1. Now B1 $\to$ C1 $\to$ B1 is a directed cycle. Requesting D1 makes D1,C1,B1 active and then requests the still-active C1; the algorithm reports C1,B1,C1 rather than returning an old value. This witness follows references backwards, opposite to our dependency arrows; reversing it gives a directed cycle in the spreadsheet dependency graph.

D1 is not itself on that directed cycle, but it cannot receive a value from this evaluator because it needs C1. Independent requests for B2 and E1 still succeed. A topological order of the entire modified spreadsheet dependency graph is impossible: the two cyclic directed edges would require B1 before C1 and C1 before B1. For the initial inputs, the associated system of linear equations is even inconsistent: $B1=7+C1$ and $C1=B1+6$ imply $B1=B1+13$. Detecting a directed cycle is a structural check, not an attempt to solve these linear equations.

A different interpretation can treat cyclic spreadsheet formulas as simultaneous equations and seek a fixed point by repeated numerical updates. This requires an explicit update rule, initial values and stopping condition. It is not automatically the meaning of a spreadsheet formula. In particular, a directed cycle alone tells us neither whether a fixed point exists nor whether a chosen update rule approaches it.

Two original toy examples make the distinction exact. The cyclic pair X1=Y1+1, Y1=X1+1 has no solution over the real numbers: substitution gives $X1=X1+2$. For X1=Y1, Y1=X1, every equal pair is a fixed point, so there is no unique answer. Simultaneously updating both spreadsheet cells from the previous pair gives $(x_{n+1},y_{n+1})=(y_n,x_n)$. Starting at $(0,1)$ produces the periodic sequence $(0,1),(1,0),(0,1),\dots$, which does not approach a fixed point. Updating X1 first and immediately using its new value for Y1 instead gives $(1,1)$ in one sweep. That difference follows from the specified rules, not from a claim about any current spreadsheet product. Numerical iteration can be useful for a deliberately chosen model with a justification such as the contraction mapping theorem; ordinary acyclic spreadsheet dependency evaluation already has its answer from prerequisite order.