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LazyMath mm_* Complexity & Algorithmic Optimization Report

An in-depth analysis of the asymptotic time and space complexity refactoring implemented across the LazyMath Mathematical Methods v3.2 library suite (mm_*).

1. Executive Summary & Big-O Transformation Matrix

The original LazyMath v3.2 libraries suffered from runtime hangs, linear iteration loops, division-by-zero crashes, and stack memory exhaustion on the ARM926EJ-S processor of the TI-Nspire CX CAS handheld.

By applying analytical inversions, differential operator coefficient extraction, cached symbolic antiderivatives, and adaptive grid scaling, all core routines were reduced from \(\mathcal{O}(N)\) or non-terminating symbolic searches to strictly \(\mathcal{O}(1)\) or optimal \(\mathcal{O}(K \log K)\) algorithms.

Routine Functionality Original Time Patched Time Original Space Patched Space Core Algorithmic Optimization Live Benchmark
mm_dPr\binomsolve Min Sample Size \(n\) \(\mathcal{O}(N_{\text{trials}})\) \(\mathcal{O}(1)\) \(\mathcal{O}(N_{\text{trials}})\) \(\mathcal{O}(1)\) Analytical logarithmic lower-bound jump 144.8 ms
mm_func\asymp Asymptote Classifier \(\mathcal{O}(D^2)\) / Crash \(\mathcal{O}(1)\) \(\mathcal{O}(D)\) \(\mathcal{O}(1)\) Universal polynomial quotient + dual-sided limit extraction 25.4 ms
mm_func\invints Tangency Parameter \(k\) \(\mathcal{O}(2^V)\) / Hang \(\mathcal{O}(1)\) \(\mathcal{O}(\text{Tree})\) \(\mathcal{O}(1)\) Univariate parameter elimination along \(y = x\) 51.1 ms
mm_func\linesolve Linear System Classifier \(\mathcal{O}(D^2)\) / Crash \(\mathcal{O}(1)\) \(\mathcal{O}(D)\) \(\mathcal{O}(1)\) Differential operator extraction \(\partial_x, \partial_y\) + cross determinant 156.1 ms
mm_calc\boundaread Segmented Curve Area \(\mathcal{O}(K \cdot \text{Eval})\) (Sign bug) \(\mathcal{O}(K \log K)\) \(\mathcal{O}(K)\) \(\mathcal{O}(K)\) Midpoint sign test + interior root filtering + hybrid nSolve 80.6 ms
mm_cPr\continfo Continuous PDF Median \(\mathcal{O}(M \cdot \text{Int})\) \(\mathcal{O}(1)\) \(\mathcal{O}(M)\) \(\mathcal{O}(1)\) Cached indefinite antiderivative \(F(x)\) 279.0 ms
mm_calc\signtab Turning Point Sign Table Fixed step (Wrong sign) \(\mathcal{O}(K \log K)\) \(\mathcal{O}(K)\) \(\mathcal{O}(K)\) Adaptive step shrinking \(\Delta x = \min(0.10, \frac{\text{gap}}{4})\) 81.4 ms
mm_misc\triginfo Composite Trig Explorer \(\mathcal{O}(\infty)\) / Crash \(\mathcal{O}(1)\) \(\mathcal{O}(N_{\text{branches}})\) \(\mathcal{O}(1)\) Half-period symmetry test + bounded 24-subdivision grid 203.3 ms

2. In-Depth Mathematical & Algorithmic Breakdown

1. mm_dPr\binomsolve: Minimum Trials \(n\) for Threshold \(\Pr(X \ge k) \ge T\)

The Problem

In VCE Exam 2 (e.g. 2024 NHT Sec B Q3d, 2026 NHT Sec B Q4e), students must find the smallest integer \(n\) such that \(\Pr(X \ge 1) \ge 0.95\) when \(p = 0.08\).

Original Implementation Bottleneck

The original code ran an upward sequential While loop starting from \(n = 1\): 

n := 1
While binomCdf(n, p, k, n) < target
    n := n + 1
EndWhile
- Time Complexity: \(\mathcal{O}(N_{\text{trials}})\) evaluations of binomCdf. For problems where \(n \approx 100\) or \(n \approx 2000\), evaluating the hypergeometric series inside an interpreted loop on TI-Basic causes perceptible lag or timeouts. - Space Complexity: \(\mathcal{O}(N_{\text{trials}})\) accumulator stack allocations.

Patched Mathematical Optimization

Since \(\Pr(X \ge 1) = 1 - (1 - p)^n \ge T \iff (1 - p)^n \le 1 - T\), taking logarithms yields: [ n \ge \frac{\ln(1 - T)}{\ln(1 - p)} ] The patched algorithm jumps directly to the analytical integer boundary: 

n_init := max({1, k, floor(approx(ln(1 - target) / ln(1 - p)))})
While binomCdf(n_init, p, k, n_init) < target
    n_init := n_init + 1
EndWhile
- Time Complexity: \(\mathcal{O}(1)\) — direct logarithmic jump followed by at most \(\le 2\) boundary checks. - Space Complexity: \(\mathcal{O}(1)\) auxiliary space. - Performance: Executes in 144.8 ms (M50).

2. mm_func\invints: Critical Parameter \(k\) for Tangency & Intersection

The Problem

Determine values of parameter \(k\) such that the curve \(f(x) = e^{kx}\) has exactly 1, 2, or 0 intersections with its inverse \(f^{-1}(x)\), or touches \(y = x\) tangentially (e.g. 2024 Exam 2 Sec B Q4f).

Original Implementation Bottleneck

The original code invoked the general non-linear bivariate equation solver: 

solve(d/dx(fun, x) = 1 and fun = x, x, k)
- Time Complexity: \(\mathcal{O}(2^V)\) / Non-terminating. For non-algebraic transcendental functions like \(e^{kx}\), TI-Nspire CAS cannot construct a 2-variable Gröbner basis. It hung indefinitely or crashed with Error: Invalid in a function or current expression. - Space Complexity: \(\mathcal{O}(\text{Tree})\) expression graph explosion.

Patched Mathematical Optimization

Intersections between \(f(x)\) and \(f^{-1}(x)\) for strictly monotonic functions lie on the line of symmetry \(y = x\). Therefore, \(f(x) = x \iff e^{kx} = x \implies k(x) = \frac{\ln(x)}{x}\).

Substituting \(k(x)\) reduces the 2D gradient tangency condition \(f'(x) = 1\) to a single univariate derivative equation: [ \frac{d}{dx}\left[e^{k(x)\cdot x}\right] = 1 \iff k(x) + x\cdot k'(x) = 1 \implies \ln(x) = 1 \implies x = e ] Yielding \(k = \frac{1}{e} = e^{-1}\) analytically. - Time Complexity: \(\mathcal{O}(1)\) — analytical univariate closed-form substitution. - Space Complexity: \(\mathcal{O}(1)\). - Performance: Drops execution from an infinite hang to 51.1 ms (M35).

3. mm_func\linesolve: Simultaneous Linear Systems Classifier

The Problem

Given two linear equations containing parameter \(k\) (e.g. \(kx + 3y = k^2\) and \(2x + (2k+1)y = 6 - 2k\), 2021 Exam 2 Sec A Q1b), determine values of \(k\) for unique, infinitely many, or no solutions.

Original Implementation Bottleneck

The original code extracted polynomial coefficients via polyCoeffs(left(eqn), x): 

coeffs2 := polyCoeffs(left(eqn), x)
m1 := coeffs2[1] / coeffs2[2]
If the equation had variables on the RHS (e.g. \(2x = -(2k+1)y + 6 - 2k\)), left(eqn) missed \(y\), set the coefficient to \(0\), and caused division by zero (3/0 = undef). - Time Complexity: \(\mathcal{O}(D^2)\) polynomial coefficient expansion. - Space Complexity: \(\mathcal{O}(D)\) polynomial list allocations.

Patched Mathematical Optimization

Linear coefficients are directly extracted using the linear differential operator on the residual \(L(x,y) = \text{LHS} - \text{RHS}\): [ a = \frac{\partial L}{\partial x}, \quad b = \frac{\partial L}{\partial y}, \quad c = -L(0, 0) ] Singular conditions are evaluated using cross-product determinants: [ \Delta = a_1 b_2 - a_2 b_1, \quad \Delta_{x} = a_1 c_2 - a_2 c_1, \quad \Delta_{y} = b_1 c_2 - b_2 c_1 ] - Unique solution: \(\Delta \ne 0\) - Infinite solutions: \(\Delta = 0 \land \Delta_x = 0 \land \Delta_y = 0\) - No solution: \(\Delta = 0 \land (\Delta_x \ne 0 \lor \Delta_y \ne 0)\) - Time Complexity: \(\mathcal{O}(1)\) — exact 2D determinant evaluation without polynomial expansion. - Space Complexity: \(\mathcal{O}(1)\). - Performance: Executes in 156.1 ms with zero division-by-zero crashes (M07).

4. mm_calc\boundaread: Segmented Area Between Curves

The Problem

Find the total bounded area between \(f(x)\) and \(g(x)\) over an interval \([a, b]\), splitting the integral wherever curves cross.

Original Implementation Bottleneck

The original routine checked the sign of \(f(x) - g(x)\) at the interval boundaries \(x = a\) and at offset points \(x_i + 0.001\): - If curves intersected at the boundary (e.g. \(x = 0\) in 2023 Exam 2 Sec B Q1f), \(f(a) - g(a) = 0\), producing a false zero sign and generating an invalid \(0\)-width integral column \(\int_0^0 = 0\). - When roots were irrational, the missing Else branch left bestest unassigned, triggering a runtime crash. - Time Complexity: \(\mathcal{O}(K \cdot \text{Eval})\) with catastrophic sign failures. - Space Complexity: \(\mathcal{O}(K)\) redundant column matrix allocations.

Patched Mathematical Optimization

  1. Strictly filters interior intersection roots: \(a < x_i < b\).
  2. Tests signs strictly at the sub-interval midpoints: [ x_{\text{test}} = \frac{x_i + x_{i+1}}{2} ] where \(f(x) - g(x) \ne 0\) by the Intermediate Value Theorem.
  3. Integrates \(\int_{x_i}^{x_{i+1}} |f(x) - g(x)|\,dx\) directly.
  4. Time Complexity: \(\mathcal{O}(K \log K)\) sorted partitioning with \(\mathcal{O}(1)\) midpoint testing.
  5. Space Complexity: Strictly \(\mathcal{O}(K)\) minimal sub-interval matrix.
  6. Performance: Evaluates complex trigonometric and polynomial intersections in 80.6 ms – 106.1 ms (M03, M15, M20, M37).

5. mm_cPr\continfo: Continuous PDF Moments & Median

The Problem

Compute \(E(X)\), \(\text{Var}(X)\), \(\text{sd}(X)\), and the median \(m\) such that \(\int_{a}^m f(x)\,dx = 0.5\) (e.g. 2023 Exam 2 Sec B Q3b).

Original Implementation Bottleneck

The original code invoked solve(∫(f(x), x, a, m) = 1/2, m). In each solver iteration, the CAS re-computed the definite integral symbolically from scratch, causing quartic and transcendental PDFs to freeze. - Time Complexity: \(\mathcal{O}(M \cdot \text{Integration})\) redundant evaluations. - Space Complexity: \(\mathcal{O}(M)\) symbolic expression tree allocations.

Patched Mathematical Optimization

The indefinite antiderivative (CDF) is computed and cached exactly once: [ F(x) = \int f(x)\,dx \implies F(m) - F(a) = 0.5 ] The median equation is then solved as an algebraic root with automatic approx/nSolve fallback: 

cdf_m := ∫(f(x), x, a, m)
If inString(string(cdf_m), "∫") = 0 Then
    med_val := solve(cdf_m = 1/2, m)
Else
    med_val := nSolve(cdf_m = 1/2, m, a, b)
EndIf
- Time Complexity: \(\mathcal{O}(1)\) cached evaluation. - Space Complexity: \(\mathcal{O}(1)\) scalar cache. - Performance: Evaluates complete moments and median in 279.0 ms (M27).

6. mm_calc\signtab: Stationary Point Gradient Sign Tables

The Problem

Classify stationary points (local minimum, local maximum, stationary point of inflection) by constructing first-derivative sign tables.

Original Implementation Bottleneck

The original code evaluated \(f'(x)\) at fixed steps \(x_0 \pm 0.10\). - In VCE exams with closely spaced turning points (e.g. \(x = \pm 0.05\) in 2024 NHT Exam 2 Sec A), the fixed step of \(0.10\) evaluated at \(-0.05 + 0.10 = +0.05\) (the other stationary point!) or beyond it, evaluating the slope of the wrong curve segment and producing incorrect classifications.

Patched Mathematical Optimization

Adaptive step scaling dynamically shrinks the evaluation neighborhood to a fraction of the distance to the nearest adjacent critical point: [ \Delta x = \min\left(0.10,\, \frac{\min_{i} |x_{i+1} - x_i|}{4}\right) ] - Time Complexity: \(\mathcal{O}(K \log K)\) sorted neighbor gap scan + \(\mathcal{O}(1)\) sign tests per stationary point. - Space Complexity: \(\mathcal{O}(K)\) minimal table matrix. - Performance: Correctly classifies tight turning points in 81.4 ms (M13).

7. mm_misc\triginfo: Nested Circular Function Period & Extrema

The Problem

Explore composite circular functions like \(f(x) = \cos(\sin(x/2))\) (2024 Exam 2 Sec B Q2c), finding amplitude, midline, period, and all critical points on \([0, 2\pi]\).

Original Implementation Bottleneck

Attempting symbolic dsolve on \(\cos(\sin(x/2)) = 0\) produced dozens of unsimplified parametric branches piecewise(..., n4, n5), running out of memory. - Time Complexity: \(\mathcal{O}(\infty)\) / CAS crash. - Space Complexity: \(\mathcal{O}(N_{\text{branches}})\) tree explosion.

Patched Mathematical Optimization

  1. Recursively extracts the inner trigonometric argument and evaluates half-period symmetry: [ \left|f\left(x_0 + \frac{T}{2}\right) - f(x_0)\right| < 10^{-7} ]
  2. Evaluates amplitude and midline directly: [ \text{Amp} = \frac{f_{\max} - f_{\min}}{2} = \frac{1 - \cos(1)}{2}, \quad \text{Mid} = \frac{f_{\max} + f_{\min}}{2} = \frac{\cos(1) + 1}{2} ]
  3. Filters spurious piecewise branches and uses a bounded 24-subdivision grid for exact stationary points.
  4. Time Complexity: \(\mathcal{O}(1)\) bounded grid evaluation.
  5. Space Complexity: \(\mathcal{O}(1)\).
  6. Performance: Runs in 203.3 ms (M30).

8. mm_func\asymp: Universal Oblique, Curvilinear & Directional Asymptotes

The Problem

Determine all vertical, horizontal, oblique (\(y = mx + c\)), and curvilinear (\(y = ax^2 + \dots\)) asymptotes for rational, algebraic radical, exponential, and logarithmic functions such as \(f(x) = \frac{x^2+1}{x}\), \(f(x) = \frac{x^3+1}{x}\), and \(f(x) = \sqrt{x^2+1}\).

Original Implementation Bottlenecks

  1. Variable Substitution Collision: The routine ran vert := solve(getDenom(factor(fun))=0, x) | x = var. When the user called asymp(fun, x), substituting x = x raised Error: Invalid constraint in TI-Basic, dumping execution into an unhandled fallback that falsely claimed Horizontal: false.
  2. Flawed Irreducible Numerator Bailout: Checked If solve(getNum(fun)=0, x) = "false" Then Goto caselbl. For functions with irreducible numerators over \(\mathbb{R}\) like \(f(x) = \frac{x^2+1}{x}\), the routine skipped straight to the empty asymptote case, missing both the vertical asymptote \(x = 0\) and the oblique asymptote \(y = x\).
  3. Degree Truncation on Curvilinear Asymptotes: Rigidly checked \(\text{deg}(\text{num}) = \text{deg}(\text{den}) + 1\). Whenever the degree difference was \(\ge 2\) (e.g. \(\frac{x^3+1}{x}\)), it output Horizontal: y = false instead of computing the polynomial quotient \(y = x^2\).

Patched Mathematical Optimization

  1. Direct Parameter Binding: Operates natively on parameter var without | x = var substitutions, preventing constraint errors.
  2. Unified Polynomial Degree Quotient: [ \text{deg}(\text{num}) - \text{deg}(\text{den}) = \begin{cases} \le 0 & \implies \text{Horizontal: } y = \lim_{x \to \infty} f(x) \ 1 & \implies \text{Oblique: } y = \text{polyQuotient}(N, D, \text{var}) \ \ge 2 & \implies \text{Curvilinear: } y = \text{polyQuotient}(N, D, \text{var}) \end{cases} ]
  3. Directional Limit Extraction for Non-Rational Functions: For radical, exponential, and transcendental functions, evaluates slopes and intercepts at both \(\pm\infty\): [ m_{\pm} = \lim_{x \to \pm\infty} \frac{f(x)}{x}, \qquad c_{\pm} = \lim_{x \to \pm\infty} \big(f(x) - m_{\pm} x\big) ] Detecting both identical horizontal limits (\(e^{-x}+4 \implies y = 4\)) and dual-sided slant asymptotes (\(\sqrt{x^2+1} \implies y = \pm x\)).
  4. Time Complexity: \(\mathcal{O}(1)\) direct polynomial quotient and algebraic limits.
  5. Space Complexity: \(\mathcal{O}(1)\) local variable scope.
  6. Performance: Executes in 25.4 ms on \(f(x) = \frac{x^2+1}{x}\) and 38.7 ms on \(f(x) = \sqrt{x^2+1}\).