Read the syntax. Carry the context. Make control explicit.

Answer Numbers evaluate directly. A compound expression asks evaluate to interpret both nested operands before combining their values. The evaluator follows the expression tree one node at a time.

Answer evaluate does not attach one permanent meaning to x or y. It receives an environment with the expression, so the same expression tree can be reused with different bindings.

Answer In the first example the alternative divides by zero. The program still returns 60 because the true predicate selects the addition branch and the invalid alternative remains expression data.

Answer The resulting plan accepts an environment. Running it only looks up variables, runs the stored operand plans, and applies the already selected operator. One analyzed plan can therefore serve many environments.

Answer The evaluator does not need a second implementation of local binding. Its let case rewrites the expression and sends the result back through the ordinary lambda and application cases in the same environment.

Answer force-it inspects the tag. On the first demand it calls the stored computation, changes the tag to evaluated-thunk, discards the computation, and stores the value. Later demands select the cached slot and perform no computation again. This exposes the representation change that a lazy evaluator normally hides behind argument handling.

Answer The second program makes two choice positions explicit. scan-y tests every y for one x, and scan-x repeats that work for every x. Returning all pairs whose squared components sum to 25 exposes a finite nondeterministic search as ordinary list-producing control.

Answer A later occurrence looks up the existing binding before extending anything. Equal data preserves the frame; conflicting data returns failed, which every remaining recursive step propagates. This is one-way matching with variables in the pattern, not full bidirectional unification or database search.

Answer The second program binds grand from the rule head, then solve-goals processes the two parent goals in order. The first goal produces middle values ben and dia. Each frame becomes input to the second goal, which finds cy and eli. The lesson evaluator processes direct parent facts and one ordered parent-rule body while preserving every intermediate frame.

Answer The second program uses a three-person cycle. Expanding ada reaches ben, ben reaches cy, and cy reaches ada again. The evaluator spends exactly one work unit per removed frontier item and returns truncated with the remaining frontier when the budget reaches zero. This breadth-first lesson model covers parent expansion, an explicit frontier, repeated answers, and complete or truncated status.

Answer The second program gives married a symmetric operational rule: when a direct fact is absent, swap the arguments and search again. This proves (married mickey minnie) after one swap because the reverse fact exists. The same rule alternates forever for an unrelated pair unless the evaluator detects repetition or spends a visible work budget. The logical statement may be symmetric, but the direction and control of the rule still determine the behavior of this executable search.

Answer The second program executes both forms. classify-original uses internal definitions directly. classify-scanned writes the transformed let and set! structure explicitly. Both return the same parity observations for 7 and 8, and equal? reports true. The finite comparison checks this transformation and makes its explicit intermediate structure visible.

Answer run-program is the driver. It receives quoted top-level forms, one explicit global environment, and a form budget. Every completed form contributes one transcript value while definitions and assignments remain visible to later forms. The full run defines square, make-adder, add-five, and base; add-five keeps the local x value 5 even after the global base changes. The second run stops after five forms and reports four forms still pending. This form budget does not bound recursion or work inside a single form; the packaged browser runtime still supplies the lower-level execution limits.

Answer require evaluates its predicate in the same continuation system. A true predicate succeeds with ok; a false predicate invokes the next predicate alternative, which ultimately returns to the most recent amb choice. all-values repeatedly calls the next alternative supplied by each success. The first run exhausts the selected finite search and reports complete. The second deliberately stops after four solutions and reports truncated. The lesson evaluator covers ordered amb choices, require, procedure application, and success or failure continuation transfer.

Answer apply-procedure receives already classified procedure values. A host procedure receives the argument list through apply. A compound-procedure record supplies parameters, body expressions, and its creation environment; application extends that environment with a fresh frame and evaluates the body sequence. The reused canonical driver exercises every core form in one persistent guest environment and makes the educational eval/apply loop visible.

Answer The second program tags a host + procedure as primitive data and applies it after unwrapping its implementation. list-of-values records left-to-right operand evaluation with explicit probes. The guest truth predicate returns false for #f, so zero and a quoted symbol both count as true. Together these records define the lesson evaluator protocol for frames, procedures, arguments, and truth tests.

Answer The second program treats a quadratic expression as data and recursively replaces the symbol x with 3. The transformation preserves arithmetic as symbolic data. arithmetic-eval then interprets + and * and produces 16. The lesson evaluator processes the listed arithmetic forms through the same explicit representation shared by the producer and transformer.

Answer In the second expression the body uses x twice. Strict evaluation computes probe once and binds the resulting 10. The non-memoized normal evaluator forces the saved expression for each lookup, so probe runs twice. Both return 20, but the work differs. These two evaluators are explicit guest models; they do not switch the evaluation order of the host Lispex program running them.

Answer force-it evaluates a fresh thunk once, rewrites the same mutable record to evaluated-thunk, stores the result, and discards the saved expression and environment. The unused argument therefore performs no probe. The duplicated x expression is demanded twice but probes once. The false branch of the selected if is never evaluated. This evaluator implements call by need for the listed forms under an explicit work budget.

Answer integers-from and lazy-map use the same ordinary recursive syntax. Their recursive calls appear as delayed y operands to lazy-cons, so the infinite remainder is represented by memoized evaluator thunks. take is the demand boundary: primitive cons needs actual argument values, so it forces exactly the requested number of heads and tails into a finite host list. This example integrates lazy lists with the lesson evaluator and makes the consumer demand explicit.

Answer The Pythagorean program reuses the same continuations with numeric choices. Ordering constraints remove permutations and the square equality accepts two triples in the finite range 1 through 10. Both examples use depth-first finite lists and record every choice reached within that range.

Answer qeval recognizes and as a pipeline. The first query creates possible who bindings for all programmers. The second query receives each frame separately and keeps only the frame whose same who also supervises under bob. This lesson materializes the finite frame stream as a list while preserving the pipeline protocol.

Answer simple-query combines direct assertion matches with rule results. A rule result unifies the query pattern with the renamed conclusion, then sends the resulting frame through qeval on the rule body. and pipelines frames, or appends alternatives, and not keeps a frame only when its subquery has no result under that frame. The explicit depth is the exact finite resource boundary for rule expansion.