--- azemark: 2 title: Structured technical content author: - AzeForge examples --- # Structured technical content This document demonstrates the structured technical content family: typed tables, algorithms, theorem-family statements and worked examples. Each section introduces the directive, then walks from a minimal Block to the fullest form the directive registers. ## Typed table A typed table declares a closed column system, so every cell is validated, canonicalized and aligned by its declared type rather than by the author's spacing. The first table is a small three-column record of text, integer and boolean cells. :::: table id: pilot-line-yields number: true caption: Pilot line yields ---- columns: - key: batch name: Batch type: text - key: units name: Units type: integer - key: accepted name: Accepted type: boolean rows: - batch: B-101 units: 480 accepted: true - batch: B-102 units: 512 accepted: false - batch: B-103 units: 466 accepted: true :::: The second table mixes a `math` column, a `boolean` column and a numeric column whose alignment overrides the type default. :::: table id: control-loop-gains number: true caption: Control loop gains and their responses ---- columns: - key: loop name: Loop type: text - key: gain name: Gain type: decimal align: right - key: response name: Response type: math - key: stable name: Stable type: boolean rows: - loop: inner gain: 2.5 response: 1 / (1 + s * tau) stable: true - loop: outer gain: -1.75 response: k / (s * (s + 2)) stable: false :::: The third table is the fully typed form, with a column unit, a grouped header over two adjacent columns, a missing cell and a prose cell that keeps Markdown. :::: table id: cooling-measurements number: true caption: Cooling trials, ambient held at 22 C ---- columns: - key: trial name: Trial type: text - key: start-temp name: Start temp type: quantity unit: K - key: end-temp name: End temp type: quantity unit: K - key: interval name: Interval type: quantity unit: s - key: note name: Observation type: prose groups: - name: Temperature columns: - start-temp - end-temp rows: - trial: A1 start-temp: 344.2 end-temp: 338.4 interval: 300 note: lid on - trial: A2 start-temp: 344.5 interval: 300 note: lid off; **probe drifted late in run** :::: ## Algorithm An algorithm Block holds one procedure whose statements come from a closed six-form set and nest through the shared two-space record tree. The first procedure is a single bounded loop that accumulates a total. :::: algorithm id: dot-product number: true caption: Dot product of two equal-length vectors ---- procedure: DotProduct parameters: - A - B steps: - assign: total = 0 - for: i = 0 to length(A) - 1 do: - assign: total = total + A[i] * B[i] - return: total :::: The second procedure walks a vector backwards and resolves each element through an `if`/`else-if`/`else` ladder, opened by an authored invariant line. :::: algorithm id: classify-deviations number: true caption: Classify deviations against tolerance bands ---- procedure: ClassifyDeviations parameters: - deviations - tolerance steps: - text: invariant: tolerance is non-negative - for: k = length(deviations) - 1 downto 0 do: - if: abs(deviations[k]) <= tolerance then: - assign: verdict[k] = accept else-if: abs(deviations[k]) <= 2 * tolerance then: - assign: verdict[k] = rework else: - assign: verdict[k] = scrap - return: verdict :::: The third procedure is the reference search, with an invariant line, a `while` carrying an `if` ladder and an early return. :::: algorithm id: binary-search number: true caption: Binary search over a sorted array ---- procedure: BinarySearch parameters: - A - target steps: - text: invariant: A is sorted ascending - assign: lo = 0 - assign: hi = length(A) - 1 - while: lo <= hi do: - assign: mid = floor((lo + hi) / 2) - if: A[mid] == target then: - return: mid else-if: A[mid] < target then: - assign: lo = mid + 1 else: - assign: hi = mid - 1 - return: -1 :::: ## Statement A statement Block carries one closed `kind:` from the theorem family, the authored text and, when the form admits one, a proof whose QED mark the renderer derives. The first statement is a definition, the only form that carries no proof. :::: statement id: trace-definition number: true kind: definition caption: Trace of a square matrix ---- text: | The trace of a square matrix `M` is the sum of its diagonal entries, written `tr(M)`. :::: The second statement is a lemma whose proof nests an equation Block. :::: statement id: even-sum-lemma number: true kind: lemma caption: The sum of two even integers is even ---- text: | For any two even integers `m` and `n`, the sum `m + n` is even. proof: | Write `m = 2 a` and `n = 2 b` for integers `a` and `b`, then collect the common factor: :: equation ---- m + n = 2 a + 2 b = 2 (a + b) :: which exhibits `m + n` as twice the integer `a + b`. :::: The third statement is the plane triangle inequality, whose proof ends in the renderer-derived QED mark. :::: statement id: triangle-inequality number: true kind: theorem caption: Triangle inequality in the plane ---- text: | For any three points `A`, `B`, `C` in the Euclidean plane, the sum of the lengths of two sides of a triangle is at least the length of the third side. proof: | Place the points in a coordinate system. Let `d(X, Y)` denote the Euclidean distance between points `X` and `Y`. Then :: equation ---- abs(A - C) <= abs(A - B) + abs(B - C) :: by expanding each distance via the Pythagorean identity and comparing the squared lengths. :::: ## Worked example A worked example composes a problem, its givens, ordered steps and a result into one numbered object. The first example answers its question in a single text step. :::: example id: molar-dilution number: true caption: Diluting a stock solution to a working concentration ---- problem: | A `2.0 mol/L` stock solution must become `250 mL` at `0.05 mol/L`. givens: - stock concentration 2.0 mol/L - working concentration 0.05 mol/L - working volume 250 mL steps: - text: | The dilution relation `C1 V1 = C2 V2` solves directly for the stock volume, so `V1 = (0.05 * 250) / 2.0 = 6.25 mL`. result: | Measure `6.25 mL` of stock and dilute it to `250 mL`. :::: The second example nests an equation Block inside its single step, so the mathematics is composed rather than restated in prose. :::: example id: pendulum-period number: true caption: Period of a small-amplitude pendulum ---- problem: | A pendulum of length `1.2 m` swings with a small amplitude. Find its period. givens: - length 1.2 m - gravitational acceleration 9.81 m/s^2 - small-amplitude regime steps: - text: | The period depends only on the length and the gravitational acceleration: :: equation ---- T = 2 * pi * sqrt(L / g) :: Substituting the givens evaluates the period. result: | `T ~= 2.20 s`. :::: The third example is the full derivation, whose two steps compose a three-step derivation Block and an equation Block. :::: example id: cooling-model number: true caption: Deriving the exponential cooling model ---- problem: | Water cools from `344.2 K` to `338.4 K` in `300 s` in a `295 K` room. Fit Newton's law and predict the temperature at `600 s`. givens: - ambient temperature held constant - Newton cooling law with unknown constant k steps: - text: | Solve the law with the initial condition: :: derivation ---- - expression: T(t) = T_amb + (T_0 - T_amb) * exp(-k * t) annotation: general solution - expression: 338.4 = 295 + 49.2 * exp(-300 * k) annotation: substitute the first measurement - expression: k = -ln((338.4 - 295) / 49.2) / 300 annotation: solve for k :: - text: | Evaluate at `t = 600 s` using `k ~= 0.00334`: :: equation ---- T(600) = 295 + 49.2 * exp(-0.00334 * 600) :: result: | `T(600) ~= 333.1 K`. The fit uses two observations and claims nothing beyond them. ::::