PRH | Essay | 7.20 • Fifty–Plus Exercises with Solutions on Two-Channel Arithmetic

Fifty–Plus Exercises with Solutions on Two-Channel Arithmetic Block Operators, Mixed Expressions, Quantum-Style Amplitudes, and Sharp/Blur Layer Bookkeeping Aleksandar Periˇsi´c November 2025 Abstract We present a compact exercise set (60+ problems with solutions) that trains the twochannel view of arithmetic—an additive channel and a multiplicative channel—equipped with blurs ( B+ τ, B× ε ), a bridge U , and honest switches ( S×→+, S+→× ). Part I develops the 2 × 2 block-operator calculus. Part II practices the mixed-expression word notation and its blur budget. Part III casts the same moves as quantum-style amplitudes on a two-component state. Between the preliminaries and Part I we insert a short section on the sharp/blur notation x♯, x♭ and a generic blur operator Bσ , so that an entire equation (or whole proof) can be declared to live at a given blur “lens”. The final mini-drills touch the Collatz accelerated step and show how the per-step budget ρ drops out immediately from the count of switches. 1 Preliminaries (one page) Two channels. We carry a two-component state add mult , with native blocks A+ (additive) and M× (multiplicative). The bridge U implements the Mellin ↔ Fourier move on the log-line. Blurs B+ τ and B× ε are Gaussian-type regularizers; we keep the honesty constraint ετ ≳ 1 (no over-concentration in a conjugate lens). The switches S×→+:= B+ τU B× ε, S+→× := B× εU−1B+ τ are the only off-diagonal actions in our 2 × 2 calculus. We read out numbers only at the end. Budget rule: ρ= #(switches) ·ρswitch(ε, τ), ετ ≳1. Soft arithmetic. A soft addition x⊞y means: act natively in the +-channel, blur once, and read out. Similarly x⊠yworks natively in the ×-channel. 2 Sharp and Blur Notation for Two Channels (6 exercises) To make the blur explicit and portable across problems, we adopt a very light “musical” notation. Sharp vs. blurred objects. •x♯ denotes the ideal (model) value of x , as if we worked with infinite precision and no blur. •x♭denotes a blurred value seen at some finite resolution. • When we want to remember the blur scale σ , we can write x♭ σ and relate the two by a blur operator x♭ σ=Bσ(x♯),Bσ: state space →state space. In our two-channel setting it is often convenient to write x♭,+ τ:= B+ τ(x♯), x♭,× ε:= B× ε(x♯), 1 for additive vs. multiplicative blur, always with the honesty constraint ετ ≳ 1 in the background. Equations at a given lens. Once the symbols have their layer, we may also mark entire equations as sharp or blurred. For instance, the schematic Collatz drift inequality can be written as (·)♯: log2T(n)♯+ϕ♯(r′)≤log2n♯+ϕ♯(r)−δstruct, and the same statement read at a blurred lens as (·)♭: log2T(n)♭+ϕ♭(r′)≤log2n♭+ϕ♭(r)−(δstruct −ρ) + ε2(n), where ρ is the per-step blur budget and ε2 ( n ) is the residual analytic error. One may think of the superscript ( · ) ♯ or ( · ) ♭ as a small marker above the whole equation: “this line lives at the sharp/blurred level.” The next six exercises let you practice this bookkeeping in toy situations. Sharp/blur drills (1–6) Exercise 2.1 (Layered equality).Consider the scalar identity x+y=z. (a) Write the sharp version using x♯, y♯, z♯. (b) Write a blurred version using x♭ σ, y♭ σ, z♭ σand a generic blur Bσ. (c) Explain in words what it means to mark the equality as (·)♭. Solution. (a) x♯ + y♯ = z♯ . (b) x♭ σ + y♭ σ = z♭ σ with x♭ σ = Bσ ( x♯ ), etc. (c) It means we are not claiming anything about the ideal sharp values; only the blurred versions at scale σare being related. □ Exercise 2.2 (Lane-aware blur).Let a♯ be a number you want to treat in the multiplicative lane. Define a♭,× εand describe its meaning. What is the corresponding a♭,+ τ? Solution. a♭,× ε := B× ε ( a♯ ) is the multiplicative blur of a♯ at scale ε (e.g. Gaussian on the log-line). The additive blur is a♭,+ τ:= B+ τ(a♯). □ Exercise 2.3 (Budgeted drift).Suppose a sharp inequality (·)♯:X♯≤Y♯−δstruct is known, and that in the blurred picture the budget cost per step is ρ > 0. Define a blurred inequality with a margin δ > 0that absorbs the budget, and state the condition on δstruct. Solution. Set δ:= δstruct −ρand assume δ > 0. Then (·)♭:X♭≤Y♭−δ is the blurred version with the budget accounted for. □ Exercise 2.4 (Commuting natives with blur).Argue why it is natural to assume that A+ commutes with B+ τ (up to small technicalities), but not necessarily with B× ε . Express this using x♯and x♭,+ τ. Solution. A+ is native to the additive lane, as is B+ τ , so it is natural to model A+B+ τx♯≈B+ τA+x♯ . In contrast, B× ε acts in the multiplicative lens; composing it with A+ crosses modes and should be mediated by a switch. In layered notation: ( A+x ) ♭,+ τ≈A+ ( x♭,+ τ ), but ( A+x ) ♭,× ε is ill-typed without an explicit switch. □ Exercise 2.5 (Interpreting n♭≈n♯ ).In the Collatz setting, n♯ is the ideal odd integer, and n♭ σ is a blurred copy. Explain what a statement of the form n♭ σ≈n♯ means operationally, and why we do not require equality. 2 Solution. It means the blurred reading at scale σ is consistent with n♯ up to the blur tolerance (for instance, it lies in the same blur bin or interval). We do not require equality because blur explicitly hides micro-differences; only the bin index matters. □ Exercise 2.6 (Marking entire formulas).Take the soft addition identity x⊞y = ( x + y ) ♭ τ . Rewrite it so that the entire formula is marked as blurred, and explain how this helps avoid writing ♭on every symbol. Solution. One can write (·)♭:x⊞y=x+y with the understanding that both x, y and the result are seen at blur τ in the +-lane. This way the superscript (·)♭is a global lens label for the whole equation. □ 3 Part I: Block-Operator Calculus (20 exercises) Throughout, one beat (one composite move) acts by a 2 ×2 block operator B="A+S×→+ S+→× M×#. When native blocks act alone, the off-diagonal cost is 0. Each switch contributes ρswitch. Warmups (1–6) Exercise 3.1 (Diagonal dominance).Let D = diag ( A+, M× )and assume  A+ ,∥M×∥ ≤ 1. Show ∥D∥ ≤ 1on the two-channel Hilbert space. Solution. ∥D∥ = max{∥A+∥,∥M×∥} ≤ 1 because D is block-diagonal and the operator norm on a direct sum is the maximum of the block norms. □ Exercise 3.2 (One switch).Consider S="IS×→+ 0I#. Count the blur budget. Solution. Exactly one switch is used, hence ρ=ρswitch.□ Exercise 3.3 (Two switches).T="I0 S+→× I#"IS×→+ 0I#. How many switches? Solution. Two off-diagonal calls: one S×→+and one S+→×, so ρ= 2ρswitch.□ Exercise 3.4 (Budget additivity).Show that concatenating block moves adds the switch counts (hence budgets). Solution. Off-diagonal uses add under composition; each appearance of S×→+ or S+→× contributes ρswitch. Thus counts (and budgets) add. □ Exercise 3.5 (No fake in-channel).Explain why it is illegal to apply A+ to a multiplicative component without a preceding switch. Solution. Type-mismatch: A+ is a native +-channel operator. Acting on the × -channel must be mediated by S×→+or S+→×; otherwise we silently smuggle an unbooked blur. □ Exercise 3.6 (Energy bound).If B+ τand B× εare contractions, show ∥S×→+∥≤∥U∥. Solution. ∥S×→+∥=∥B+ τU B× ε∥≤∥B+ τ∥ ∥U∥ ∥B× ε∥≤∥U∥.□ Composition and powers (7–12) Exercise 3.7 (Square of a beat).Compute B2 symbolically and list how many switches appear in each block. 3 Solution. B2=(A+)2+S×→+S+→× A+S×→++S×→+M× S+→×A++M×S+→× S+→×S×→++ (M×)2. The diagonal gains 2 switches via S×→+S+→× and S+→×S×→+ ; each off-diagonal block has one switch per summand. □ Exercise 3.8 (Geometric series).Assume ∥B∥ < 1. Write Pk≥0Bk and interpret its switch budget. Solution. (I −B ) −1 = Pk≥0Bk . Each Bk contains at most k switches per path; total budget is bounded by the expected switch count of paths under the norm-weighted series. □ Exercise 3.9 (Block resolvent).Give the Schur complement for (I− B)−1. Solution. Schur on the additive block: (I −A+−S×→+ (I −M× ) −1S+→× ) −1 ; similarly for the multiplicative block. Off-diagonal resolvents insert explicit switch pairs. □ Exercise 3.10 (Commuting natives).If A+ and M× commute with U (idealization), simplify S×→+S+→×. Solution. S×→+S+→× = B+ τUB× εB× εU−1B+ τ = B+ τU B× ε 2U−1B+ τ , a diagonal regularizer in the +-channel. □ Exercise 3.11 (Minimal switch normal form).Show any block product can be rewritten with all switches grouped, without changing count. Solution. Use associativity and slide diagonal blocks past switches (they act in their native lanes) to bundle S×→+, S+→×; occurrences are preserved, so the count and budget remain. □ Exercise 3.12 (Budget lower bound).Prove any nontrivial off-diagonal entry in Bk requires at least one switch. Solution. Off-diagonal maps between channels; the only such maps are S×→+ or S+→× . Hence minimal count ≥1. □ Numerical cartoons (13–20) In these, treat  B+ τ =∥B× ε∥= 1, ∥U∥= 1, and ρswitch =αfor some fixed α > 0. Exercise 3.13. Budget for B? Solution. Budget is bookkeeping, not an operator norm: one potential switch per off-diagonal use. For a single beat if used once, worst-case ρ=αper off-diagonal access. □ Exercise 3.14. Budget for B3on a path that alternates channels every step. Solution. Each step requires a switch, so 3 switches ⇒ρ= 3α.□ Exercise 3.15. Path: +native, switch to ×, native, switch back to +, read out. Budget? Solution. Two switches ⇒ρ= 2α.□ Exercise 3.16. If a policy forbids two consecutive switches, what is the maximum switch count in ksteps? Solution. At most ⌈k/2⌉.□ Exercise 3.17. Show any k-step composition has ρ≤k α. Solution. At most one switch per step. □ Exercise 3.18. When can ρ= 0 in ksteps? Solution. If we remain in one native channel (all-diagonal). □ Exercise 3.19. Explain why grouping all natives first, then switching once, is budget-optimal for a single cross-channel evaluation. 4 Solution. Any interleaving adds extra switches without benefit. □ Exercise 3.20. Give an example where two switches are unavoidable. Solution. Compute +-native, then must evaluate a × -native, then return to + to read out: + →×→ + requires two switches. □ 4 Part II: Mixed-Expression (Word) Calculus (20 exercises) Alphabet: [ ×a ], [+ b ], S×→+ , S+→× , and optional [ ÷ 2 m ] as a × -native move. Budget is the number of switches times ρswitch. Counting and normal forms (1–10) Exercise 4.1. Word W= [+1] [+2] [+3]. Budget? Solution. All additive; ρ= 0. □ Exercise 4.2. W= [×5] [×7] [×2]. Budget? Solution. All multiplicative; ρ= 0. □ Exercise 4.3. W= [×3] S×→+[+1]. Budget? Solution. One switch; ρ=ρswitch.□ Exercise 4.4. W= [×3] S×→+[+1] S+→× [×2]. Budget? Solution. Two switches; ρ= 2ρswitch.□ Exercise 4.5. Move all S×→+, S+→× to the rightmost possible positions while preserving count. Solution. Slide natives left (they are lane-diagonal). Switch count unchanged. □ Exercise 4.6. Show any word with schannel alternations has budget s ρswitch. Solution. Alternation ⇔a switch; count equals s.□ Exercise 4.7. Minimal budget to evaluate [×a] [+b]in that order and read out additively. Solution. × -native then +-native: must switch once for [+ b ] plus possibly none to start if we begin in × ; but we read out in +, so if starting in × , we need one switch (enter +) used once—still one switch total. □ Exercise 4.8. Budget for [×a] [+b] [×c]reading out multiplicatively. Solution. Two alternations ⇒2ρswitch.□ Exercise 4.9. Show [×a] [+b] [+c]has same budget as [×a] [+b+c]. Solution. Both require one switch × → +; combining + natives doesn’t add switches. □ Exercise 4.10. Argue [×a] [+b] [÷2m]has budget 2ρswitch if read out in +. Solution. × native, switch to + for [+ b ], then back to × for [ ÷ 2 m ], then (if reading out in +) another switch. If the final readout is in ×, budget is 2ρswitch; if in +, budget is 3ρswitch.□ Tiny identities and estimates (11–20) Exercise 4.11. Show the budget of Wrev (reverse order) equals the alternation count of W. Solution. Reversing preserves the number of channel toggles. □ Exercise 4.12. Prove any insertion of a native block in the current lane keeps budget unchanged. Solution. No switch is introduced. □ Exercise 4.13. If [×a]and [+b]commute as maps, does the budget vanish? 5 Solution. No. Budget accounts for lane changes, not algebraic commutativity. □ Exercise 4.14. Bound the budget of a word with ksymbols by k ρswitch. Solution. At most one switch per boundary. □ Exercise 4.15. Give a budget-optimal parenthesization strategy. Solution. Evaluate maximal native blocks before switching; group all + parts and all ×parts. □ Exercise 4.16. Show merging adjacent +blocks never increases budget; likewise for ×. Solution. Merging reduces boundaries; switches occur only across lane boundaries. □ Exercise 4.17. Explain why a normal form is: [native in lane 1]→(switch) →[native in lane 2]→(switch) → · · · Solution. All budget is at switches; natives compact inside lanes. □ Exercise 4.18. If we restrict to at most one switch total, which words are admissible? Solution. Words with at most one alternation: all natives in a lane, then one switch, then possibly readout in the new lane. □ Exercise 4.19. Show any word that starts and ends in the same lane has even switch count. Solution. Each switch toggles the lane; returning requires an even number. □ Exercise 4.20 (Mini Collatz word).Consider the accelerated odd step: [×3] S×→+[+1] S+→× [÷2v2(3n+1)]. Count switches and write ρ. Solution. Two switches ⇒ρ= 2ρswitch.□ 5 Part III: Quantum-Style Amplitudes (20 exercises) State vector ψ = "ψ+ ψ×# with ∥ψ∥2 = ∥ψ+∥2 + ∥ψ×∥2 . Natives act diagonally; switches are unitary bridges dressed by blurs. Amplitudes and probabilities (1–8) Exercise 5.1. Let ψ0="1 0#. Apply A+(norm-preserving). What is ∥ψ∥? Solution. Unchanged: 1. □ Exercise 5.2. Apply S×→+to ψ0. Where is the mass? Solution. Moves amplitude from + to × (up to blur); norm preserved if S×→+ is (near) unitary. The budget records the switch. □ Exercise 5.3. Sequence S×→+then S+→× to ψ0. End in which lane? Solution. Back to the + lane. □ Exercise 5.4. If readout happens in +, what is the probability to be in +after S×→+? Solution. Zero (ideal unitary case), as amplitude sits in ×.□ Exercise 5.5. Argue that inserting a native block between two identical switches does not change the switch count. Solution. Switch count depends only on the number of off-diagonal maps. □ Exercise 5.6. Explain why ετ ≳1protects norm stability under switching. 6 Solution. It prevents over-concentration in conjugate lenses, keeping the dressed bridge near-isometric. □ Exercise 5.7. Show that two consecutive switches S×→+S+→× act effectively like a diagonal smoothing in +. Solution. By definition: B+ τU B× εB× εU−1B+ τ=B+ τU B× ε 2U−1B+ τ.□ Exercise 5.8. Why is it legitimate to postpone all readouts to the end? Solution. Amplitudes evolve linearly; intermediate measurements would collapse. The calculus models “compute first, measure once”. □ Short computations (9–16) Exercise 5.9. Compute ψ′=S×→+A+ψ0and identify the lane. Solution. Lane: ×. First A+keeps us in +, then S×→+moves to ×.□ Exercise 5.10. Given ψ="a b#with ∥ψ∥= 1, what is ∥S+→×ψ∥? Solution. ≈1 under near-unitary switches; exactly 1 if the dressing blurs are normalized. □ Exercise 5.11. Find the probability to read out in +after S+→×S×→+ψ0. Solution. Near 1, since we returned to + (modulo blur leakage if non-ideal). □ Exercise 5.12. Budget for (A+)mS×→+(M×)nS+→×. Solution. Two switches ⇒ρ= 2ρswitch.□ Exercise 5.13. Which sequence has lower budget: (S×→+M×S+→×)A+ or S×→+(M×A+)S+→×? Solution. Equal budgets (two switches). The difference is purely where natives sit. □ Exercise 5.14. Show any closed loop +→×→+has even switch number. Solution. Each switch toggles the lane; returning requires an even count. □ Exercise 5.15. If one insists on measuring after each native block, how does the expected budget change? Solution. Budget is unaffected (switch count unchanged), but variance of outcomes rises; this is a pedagogical note on “compute-then-measure”. □ Exercise 5.16. Relate the amplitude picture to the word picture. Solution. Words describe the same operator product acting on amplitudes; switch count is invariant. □ Mini-bridges to Collatz (17–20) Exercise 5.17. Write the odd-step operator as an amplitude word and state ρ. Solution. [×3] S×→+[+1] S+→× [÷2v2(3n+1)]; budget 2ρswitch.□ Exercise 5.18. Why is it natural to read out in ×after the division by 2v2(3n+1)? Solution. Division is ×-native; reading out there avoids an extra switch. □ Exercise 5.19. If we must report an additive potential, what extra budget appears? Solution. One more switch × → +, so add ρswitch.□ Exercise 5.20. Explain how the per-step Collatz drift bound absorbs ρ. 7 Solution. The certificate margin ( δ + ρ ) is chosen so that ρ covers the total switch budget while preserving a net drop. □ 6 From Drills to Collatz: Three Short Capstones These final mini-exercises show how the bookkeeping plugs straight into a residue certificate. Exercise 6.1 (Capstone A: Minimal odd-step budget).Confirm the minimal switch budget for one accelerated odd step when read out multiplicatively, then when read out additively. Solution. Multiplicative readout: two switches ( ρ = 2 ρswitch ). Additive readout: three switches (ρ= 3ρswitch). □ Exercise 6.2 (Capstone B: Two-step pattern).Consider two consecutive odd steps (ignoring the even descent between). Give a budget upper bound. Solution. Each odd step costs 2 ρswitch (multiplicative readout). If we thread them multiplicatively, still 2ρswitch per step, so ≤4ρswitch for two steps. □ Exercise 6.3 (Capstone C: Plug to residue drift).Let ϕ be a residue potential mod 2 k . Show that with ρ= #switches ·ρswitch the inequality log nt+1 +ϕ(r′)≤log nt+ϕ(r)−(δ+ρ)+ε(nt) yields a net drop for all tlarge enough. Solution. ε ( nt ) ↓ 0 while ( δ + ρ ) is a fixed positive margin; for nt large, ε ( nt ) < δ/ 2 so the right-hand side drops by at least δ/2 per accelerated step. □ Definition of the per–step margin δ .Fix k≥ 1 and a residue potential ϕ : S→R , S=Z/2kZ. For r∈Sset a(r) := log23−v2(3r+ 1), r′:= Fk(r), and, for the (unique) exceptional class r∗ , use a conservative bound a ( r∗ ) := amin ( r∗ ). Define the structural slack ∆struct(r) := ϕ(r)−ϕ(r′)−a(r), δstruct := min r∈S∆struct(r). Let ρ≥ 0 denote the fixed blur budget per accelerated odd step ( ρ = 2 ρswitch ) under the two–switch word [×3] S×→+[+1] S+→× [÷2v2(3n+1)]). We set the per–step margin δ:= δstruct −ρand require δ > 0. Then for every odd nwith residue r=nmod 2k, drift♭: log2T(n)♭+ϕ♭(r′)≤log2n♭+ϕ♭(r)−(δ+ρ)+ε2(n), ε2(n) = log21 + 1 3n. Since ε2 ( n ) ↓ 0, there exists N0 such that ε2 ( n ) ≤δ/ 2 for all n≥N0 , yielding a uniform drop log2T(n)♭+ϕ♭(r′)≤log2n♭+ϕ♭(r)−δ/2 for all such n. 7 Complex Numbers as a Special Case of Blur (and Why We Do Not Use ias the Universal Notation) At several points it is tempting to identify “blur” with the imaginary direction in C , and to treat expressions of the form x+iy ≈x♯+i·(uncertainty) 8 as a ready-made blur calculus. After all, complex analysis has been spectacularly successful wherever oscillations, phases and interference appear, and our bridge operator U is explicitly Mellin/Fourier in spirit. In that sense, complex numbers do point in the right direction. However, as a universal notation for blur this identification is misleading. The sharp/blur distinction we actually need is both simpler and more brutal: blur is lost or unrecoverable information, tracked additively as a budget, not an orthogonal degree of freedom that can be rotated away. The musical notation x♯, x♭ and the explicit blur operators B+ τ, B× ε serve that purpose more honestly. Here are the key differences. 1. Blur adds; it does not cancel In the complex plane, the algebra is designed so that i2=−1, eiθ1eiθ2=ei(θ1+θ2), and destructive interference is a central feature: two coherent contributions can sum to zero. Our blur behaves differently: • Blur from independent sources adds up as a budget ( ρ = # switches ·ρswitch ), or as some monotone aggregate (e.g. an L1or L2norm). • There is no notion of “negative blur” that cancels positive blur; the best one can do is reduce blur by changing the measurement protocol (moving to a sharper lens), but not by combining two noisy histories in a clever way. If we encoded blur into an imaginary part, the algebra would falsely suggest that uncertainties can be engineered to annihilate each other via phase, which contradicts the intended “no free information” interpretation. Blur is about non-recoverable loss, not about hidden structure that can be brought back by the right rotation. 2. Complex numbers presuppose a rigid geometry Working over Cautomatically installs a very strong geometric structure: •a fixed Euclidean metric |z|=px2+y2, •a canonical notion of rotation (phases eiθ), •and, in the quantum setting, a Hilbert space with unitary evolution. Our blur is deliberately more general: • It may live on discrete state spaces, residue classes, trees or phase spaces where no canonical complex structure exists. • The honesty constraint ετ ≳ 1 is about conjugate lenses (additive vs. multiplicative views), not about a fixed circular symmetry. • The two-channel bookkeeping (additive vs. multiplicative) is about which algebra is native at each step, not about a single orthogonal direction in a plane. If we forced all blur into an imaginary axis, we would be smuggling in Hilbert-space structure and unitary symmetry precisely where we want to remain agnostic. 3. Lane semantics vs. anonymous orthogonal directions In the present calculus we care a lot about which lane we are in: lane ∈ {+-channel,×-channel}, and about the typed nature of each operation. Our switches S×→+, S+→× are expensive precisely because they jump between two logically distinct regimes: addition and multiplication. By contrast, the imaginary unit i is anonymous: it is “the” orthogonal direction, independent of any choice of algebra. Complex notation does not remember whether a contribution came from an additive step or a multiplicative step; it only remembers magnitude and phase. 9