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This overview explains how two branches can refer to one subexpression,

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and when execution can reuse its value.

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Shared storage alone does not guarantee that a later
traversal or call avoids computing the same value again.

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Shared storage alone does not guarantee that a later
traversal or call avoids computing the same value again.

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Let S be A and B, used by two roots.

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The left root combines S with C using or; the right root combines S

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with D using exclusive-or.

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The same mathematical subexpression occurs in both roots.

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For this input, S is one.

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The left root computes one or zero, giving one.

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The right root computes one exclusive-or one, giving zero.

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Both consumers need the same intermediate one, but they produce different final answers.

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An evaluation memo can store S during this evaluation.

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The first consumer computes it; the second reads the stored value.

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Sharing the representation makes that reuse possible, while the evaluator and its memo

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scope determine whether the calculation is actually reused.

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scope determine whether the calculation is actually reused.

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Let S equal A AND B.

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Our rule is S AND C AND D, OR S AND E.

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The written rule contains two occurrences of S, one in each branch.

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A shared graph can replace them with one S node and two outgoing uses.

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At A B C D E equal 11011, S is one.

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The left branch is zero and the right branch is one, so the final OR is one.

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A second root, G, takes S exclusive-or E.

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At this assignment, G returns zero.

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Both roots can point to the same S identity.

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In a deliberately memoized traversal over both roots,

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S is computed once and its saved value serves each consumer.

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A shared node makes reuse possible.

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Whether an executor actually reuses it across
roots or calls depends on its memo and lifecycle.

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The candidate companion checks a memoized teaching
executor separately from repository node sharing;

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neither demonstrates a speedup.

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Return to the opening four-input, two-root graph: S equals A and B, L equals S or C, and R equals S exclusive-or D.

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The intervening five-input graph is separate.

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Here set A, B, C and D to zero, one, one and zero.

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Compute S as zero and one, giving zero.

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The left consumer combines it with C one using or, giving one.

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The right consumer combines it with D zero using exclusive-or, giving zero.

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Sharing an intermediate does not merge the two root operations or their answers.

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This case establishes a specific contrast.

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One shared value S=0.

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Root outputs are still distinct.

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Keep the input, intermediate values, and conclusion together when comparing it with the previous case.

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The calculated answer belongs to this complete assignment under the stated rule.

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A later call can reuse the same compiled graph while beginning with an empty evaluation memo.

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Representation reuse and cached numerical results have separate lifetimes.

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Inspect the actual API and execution scope before claiming that the second call skips all shared-node calculations.

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First reset the four-input graph to A one, B one, C zero and D one.

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Its shared value S is one.

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Now change only B to zero, keeping A, C and D fixed.

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S becomes zero, L becomes zero, and R becomes one.

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An old memo entry containing S one is invalid unless its identity includes the input context.

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Three counts can differ: unique nodes in the stored graph, evaluations during one call, and cache hits across calls.

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A drawing with one S node establishes structure.

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A claim about saved execution must also establish how the node values are managed.

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Identify the shared node and its consumers.

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State the memo and lifecycle boundary before counting saved work.