{
  "index": "1952-A-5",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Let \\( a_{j}(j=1,2, \\ldots, n) \\) be entirely arbitrary numbers except that no one is equal to unity. Prove\n\\[\n\\left.a_{1}+\\sum_{i=2}^{n} a_{i} \\prod_{j=1}^{i-1}\\left(1-a_{j}\\right)=1-\\prod_{j=1}^{n}\\left(1-a_{j}\\right) \\quad \\quad \\text { (page } 350\\right)\n\\]",
  "solution": "Solution. The given statement is true for \\( n=1 \\) (interpreting the empty sum as 0 ) and for \\( n=2 \\). Suppose it is true for \\( n=k \\), i.e.,\n\\[\na_{1}+\\sum_{i=2}^{k} a_{i} \\prod_{j=1}^{i-1}\\left(1-a_{j}\\right)=1-\\prod_{i=1}^{k}\\left(1-a_{i}\\right)\n\\]\n\nThen\n\\[\n\\begin{aligned}\na_{1}+\\sum_{i=2}^{k+1} a_{i} \\prod_{j=1}^{i-1}\\left(1-a_{j}\\right) & =a_{1}+\\sum_{i=2}^{k} a_{i} \\prod_{j=1}^{i-1}\\left(1-a_{j}\\right)+a_{k+1} \\prod_{j=1}^{k}\\left(1-a_{j}\\right) \\\\\n& =1-\\prod_{i=1}^{k}\\left(1-a_{i}\\right)+a_{k+1} \\prod_{j=1}^{k}\\left(1-a_{j}\\right) \\\\\n& =1-\\left|\\prod_{i=1}^{k}\\left(1-a_{i}\\right)\\right|\\left(1-a_{k+1}\\right) \\\\\n& =1-\\prod_{i=1}^{k+1}\\left(1-a_{i}\\right) .\n\\end{aligned}\n\\]\n\nThus the statement is true for \\( n=k+1 \\). It follows by induction that it is true for all positive integers \\( n \\).\n\nRemark. It is not necessary to require that none of the \\( a \\) 's be unity.",
  "vars": [
    "i",
    "j",
    "k",
    "n"
  ],
  "params": [
    "a_j",
    "a_i",
    "a_1",
    "a_k",
    "a_k+1"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "i": "indexchar",
        "j": "indexnext",
        "k": "indexthird",
        "n": "totalcount",
        "a_j": "sequenceelemj",
        "a_i": "sequenceelemi",
        "a_1": "sequenceelemone",
        "a_k": "sequenceelemk",
        "a_k+1": "sequenceelemkplus"
      },
      "question": "5. Let \\( sequenceelemj(indexnext=1,2, \\ldots, totalcount) \\) be entirely arbitrary numbers except that no one is equal to unity. Prove\n\\[\n\\left.sequenceelemone+\\sum_{indexchar=2}^{totalcount} sequenceelemi \\prod_{indexnext=1}^{indexchar-1}\\left(1-sequenceelemj\\right)=1-\\prod_{indexnext=1}^{totalcount}\\left(1-sequenceelemj\\right) \\quad \\quad \\text { (page } 350\\right)\n\\]",
      "solution": "Solution. The given statement is true for \\( totalcount=1 \\) (interpreting the empty sum as 0 ) and for \\( totalcount=2 \\). Suppose it is true for \\( totalcount=indexthird \\), i.e.,\n\\[\nsequenceelemone+\\sum_{indexchar=2}^{indexthird} sequenceelemi \\prod_{indexnext=1}^{indexchar-1}\\left(1-sequenceelemj\\right)=1-\\prod_{indexchar=1}^{indexthird}\\left(1-sequenceelemi\\right)\n\\]\n\nThen\n\\[\n\\begin{aligned}\nsequenceelemone+\\sum_{indexchar=2}^{indexthird+1} sequenceelemi \\prod_{indexnext=1}^{indexchar-1}\\left(1-sequenceelemj\\right) & =sequenceelemone+\\sum_{indexchar=2}^{indexthird} sequenceelemi \\prod_{indexnext=1}^{indexchar-1}\\left(1-sequenceelemj\\right)+sequenceelemkplus \\prod_{indexnext=1}^{indexthird}\\left(1-sequenceelemj\\right) \\\\\n& =1-\\prod_{indexchar=1}^{indexthird}\\left(1-sequenceelemi\\right)+sequenceelemkplus \\prod_{indexnext=1}^{indexthird}\\left(1-sequenceelemj\\right) \\\\\n& =1-\\left|\\prod_{indexchar=1}^{indexthird}\\left(1-sequenceelemi\\right)\\right|\\left(1-sequenceelemkplus\\right) \\\\\n& =1-\\prod_{indexchar=1}^{indexthird+1}\\left(1-sequenceelemi\\right) .\n\\end{aligned}\n\\]\n\nThus the statement is true for \\( totalcount=indexthird+1 \\). It follows by induction that it is true for all positive integers \\( totalcount \\).\n\nRemark. It is not necessary to require that none of the sequence elements be unity."
    },
    "descriptive_long_confusing": {
      "map": {
        "i": "lighthouse",
        "j": "horseshoe",
        "k": "snowflake",
        "n": "buttercup",
        "a_j": "asteroid",
        "a_i": "driftwood",
        "a_1": "honeycomb",
        "a_k": "paintbrush",
        "a_k+1": "tangerine"
      },
      "question": "5. Let \\( asteroid (horseshoe=1,2, \\ldots, buttercup) \\) be entirely arbitrary numbers except that no one is equal to unity. Prove\n\\[\nhoneycomb+\\sum_{lighthouse=2}^{buttercup} driftwood \\prod_{horseshoe=1}^{lighthouse-1}\\left(1-asteroid\\right)=1-\\prod_{horseshoe=1}^{buttercup}\\left(1-asteroid\\right) \\quad \\quad \\text { (page } 350\\right)\n\\]",
      "solution": "Solution. The given statement is true for \\( buttercup=1 \\) (interpreting the empty sum as 0 ) and for \\( buttercup=2 \\). Suppose it is true for \\( buttercup=snowflake \\), i.e.,\n\\[\nhoneycomb+\\sum_{lighthouse=2}^{snowflake} driftwood \\prod_{horseshoe=1}^{lighthouse-1}\\left(1-asteroid\\right)=1-\\prod_{lighthouse=1}^{snowflake}\\left(1-driftwood\\right)\n\\]\n\nThen\n\\[\n\\begin{aligned}\nhoneycomb+\\sum_{lighthouse=2}^{snowflake+1} driftwood \\prod_{horseshoe=1}^{lighthouse-1}\\left(1-asteroid\\right) & =honeycomb+\\sum_{lighthouse=2}^{snowflake} driftwood \\prod_{horseshoe=1}^{lighthouse-1}\\left(1-asteroid\\right)+tangerine \\prod_{horseshoe=1}^{snowflake}\\left(1-asteroid\\right) \\\\\n& =1-\\prod_{lighthouse=1}^{snowflake}\\left(1-driftwood\\right)+tangerine \\prod_{horseshoe=1}^{snowflake}\\left(1-asteroid\\right) \\\\\n& =1-\\left|\\prod_{lighthouse=1}^{snowflake}\\left(1-driftwood\\right)\\right|\\left(1-tangerine\\right) \\\\\n& =1-\\prod_{lighthouse=1}^{snowflake+1}\\left(1-driftwood\\right) .\n\\end{aligned}\n\\]\n\nThus the statement is true for \\( buttercup=snowflake+1 \\). It follows by induction that it is true for all positive integers \\( buttercup \\).\n\nRemark. It is not necessary to require that none of the \\( a \\)'s be unity."
    },
    "descriptive_long_misleading": {
      "map": {
        "i": "holisticunit",
        "j": "totalextent",
        "k": "aggregatepoint",
        "n": "singularcount",
        "a_j": "unityconstant",
        "a_i": "identityfigure",
        "a_1": "nullityscalar",
        "a_k": "uniformentity",
        "a_k+1": "continuityaspect"
      },
      "question": "5. Let \\( unityconstant(totalextent=1,2, \\ldots, singularcount) \\) be entirely arbitrary numbers except that no one is equal to unity. Prove\n\\[\nnullityscalar+\\sum_{holisticunit=2}^{singularcount} identityfigure \\prod_{totalextent=1}^{holisticunit-1}\\left(1-unityconstant\\right)=1-\\prod_{totalextent=1}^{singularcount}\\left(1-unityconstant\\right) \\quad \\quad \\text { (page } 350\\right)\n\\]",
      "solution": "Solution. The given statement is true for \\( singularcount=1 \\) (interpreting the empty sum as 0 ) and for \\( singularcount=2 \\). Suppose it is true for \\( singularcount=aggregatepoint \\), i.e.,\n\\[\nnullityscalar+\\sum_{holisticunit=2}^{aggregatepoint} identityfigure \\prod_{totalextent=1}^{holisticunit-1}\\left(1-unityconstant\\right)=1-\\prod_{holisticunit=1}^{aggregatepoint}\\left(1-identityfigure\\right)\n\\]\n\nThen\n\\[\n\\begin{aligned}\nnullityscalar+\\sum_{holisticunit=2}^{aggregatepoint+1} identityfigure \\prod_{totalextent=1}^{holisticunit-1}\\left(1-unityconstant\\right) & =nullityscalar+\\sum_{holisticunit=2}^{aggregatepoint} identityfigure \\prod_{totalextent=1}^{holisticunit-1}\\left(1-unityconstant\\right)+continuityaspect \\prod_{totalextent=1}^{aggregatepoint}\\left(1-unityconstant\\right) \\\\ & =1-\\prod_{holisticunit=1}^{aggregatepoint}\\left(1-identityfigure\\right)+continuityaspect \\prod_{totalextent=1}^{aggregatepoint}\\left(1-unityconstant\\right) \\\\ & =1-\\left|\\prod_{holisticunit=1}^{aggregatepoint}\\left(1-identityfigure\\right)\\right|\\left(1-continuityaspect\\right) \\\\ & =1-\\prod_{holisticunit=1}^{aggregatepoint+1}\\left(1-identityfigure\\right) .\n\\end{aligned}\n\\]\n\nThus the statement is true for \\( singularcount=aggregatepoint+1 \\). It follows by induction that it is true for all positive integers \\( singularcount \\).\n\nRemark. It is not necessary to require that none of the unityconstant 's be unity."
    },
    "garbled_string": {
      "map": {
        "i": "quxbadly",
        "j": "snerqtuv",
        "k": "plimztrq",
        "n": "fradomix",
        "a_j": "qveropli",
        "a_i": "klumseta",
        "a_1": "rogdispa",
        "a_k": "hrupteno",
        "a_k+1": "zlotimex"
      },
      "question": "5. Let \\( qveropli(snerqtuv=1,2, \\ldots, fradomix) \\) be entirely arbitrary numbers except that no one is equal to unity. Prove\n\\[\n\\left.rogdispa+\\sum_{quxbadly=2}^{fradomix} klumseta \\prod_{snerqtuv=1}^{quxbadly-1}\\left(1-qveropli\\right)=1-\\prod_{snerqtuv=1}^{fradomix}\\left(1-qveropli\\right) \\quad \\quad \\text { (page } 350\\right)\n\\]",
      "solution": "Solution. The given statement is true for \\( fradomix=1 \\) (interpreting the empty sum as 0 ) and for \\( fradomix=2 \\). Suppose it is true for \\( fradomix=plimztrq \\), i.e.,\n\\[\nrogdispa+\\sum_{quxbadly=2}^{plimztrq} klumseta \\prod_{snerqtuv=1}^{quxbadly-1}\\left(1-qveropli\\right)=1-\\prod_{quxbadly=1}^{plimztrq}\\left(1-klumseta\\right)\n\\]\n\nThen\n\\[\n\\begin{aligned}\nrogdispa+\\sum_{quxbadly=2}^{plimztrq+1} klumseta \\prod_{snerqtuv=1}^{quxbadly-1}\\left(1-qveropli\\right) & =rogdispa+\\sum_{quxbadly=2}^{plimztrq} klumseta \\prod_{snerqtuv=1}^{quxbadly-1}\\left(1-qveropli\\right)+zlotimex \\prod_{snerqtuv=1}^{plimztrq}\\left(1-qveropli\\right) \\\\\n& =1-\\prod_{quxbadly=1}^{plimztrq}\\left(1-klumseta\\right)+zlotimex \\prod_{snerqtuv=1}^{plimztrq}\\left(1-qveropli\\right) \\\\\n& =1-\\left|\\prod_{quxbadly=1}^{plimztrq}\\left(1-klumseta\\right)\\right|\\left(1-zlotimex\\right) \\\\\n& =1-\\prod_{quxbadly=1}^{plimztrq+1}\\left(1-klumseta\\right) .\n\\end{aligned}\n\\]\n\nThus the statement is true for \\( fradomix=plimztrq+1 \\). It follows by induction that it is true for all positive integers \\( fradomix \\).\n\nRemark. It is not necessary to require that none of the \\( qveropli \\)'s be unity."
    },
    "kernel_variant": {
      "question": "Let R be an associative ring with identity 1 and let a₁,…,aₙ∈R be idempotents (aⱼ²=aⱼ).  \nFix once and for all the natural order 1<2<⋯<n.  For every finite set S⊆{1,…,n} write  \nS= {i₁<⋯<i_k}.  Define inductively\n\n  b_∅ := 0,  \n  b_{S∪{i}} := b_S + (1−b_S) aᵢ (1−b_S)        (i∉S).                                       (★)  \n\n(A)  (Algebraic structure)  \n(i) Prove that every b_S is an idempotent.  \n(ii) Show that if all aⱼ commute pair-wise, then for every S one has  \n  b_S = 1 − ∏_{j∈S}(1−a_j).  \n\n(B)  (Order sensitivity)  \nConstruct 2×2-matrix idempotents a₁,a₂ over ℚ for which b_{ {1,2} }≠b_{ {2,1} }, thereby showing that the conclusion of (A)(ii) fails without commutativity.\n\n(C)  (Orthogonal projections)  \nLet H be a Hilbert space and let aⱼ=P_{H_j} be the orthogonal projection onto a closed subspace H_j⊂H.  \n(i)  Interpret b_S as the orthogonal projection onto the closed linear span of (H_j)_{j∈S}.  \n(ii)  Give an explicit closed formula for P_{H_1+⋯+H_n} that involves only alternating sums of products of the individual projections and prove that every numerical coefficient appearing in that formula has operator norm 1.\n\n(D)  (Probabilistic by-product)  \nLet (Ω,𝔽,P) be a probability space and set aⱼ:=𝟙_{E_j} for events E_j.  Assume only pair-wise independence (no higher-order independence).  \n(i)  Show that the random idempotent b_{ {1,…,n} } obtained from (★) satisfies  \n\n  P(⋃_{j=1}^{n}E_j)=E[b_{ {1,…,n} }].  \n\n(ii)  Compute explicitly the correction term that distinguishes this identity from the classical inclusion–exclusion formula and show that all mixed moments of order ≥3 cancel automatically.\n\n(E)  (Algorithmic complexity)  \nDesign an algorithm that, given n symbolic idempotents, outputs b_{ {1,…,n} } in time Θ(n·2ⁿ) measured in ring multiplications, and prove that—even when R is the free ring on the generators a₁,…,aₙ—no algorithm can do better than 2^{Ω(n)} ring operations.\n\n\n───────────────────────────────────────────────────────────────────────",
      "solution": "Throughout we abbreviate u_S := 1−b_S and note that u_S is an idempotent whenever b_S is.\n\n(A)  Algebraic structure  \n(i)  Idempotency of the b_S.  \nInduction over |S|.  The empty set case is trivial.  \nAssume b_S²=b_S.  Put b′:=b_S and define b″ by (★):  \n\n b″=b′+(1−b′)a_i(1−b′)=b′+u′a_i u′  (u′:=1−b′).  \n\nCompute b″²:  \n\n b″²= (b′+u′a_i u′)²  \n      = b′² + b′u′a_i u′ + u′a_i u′b′ + (u′a_i u′)².  \n\nBecause b′u′=u′b′=0 and u′²=u′, the middle two summands vanish.  \nSince a_i²=a_i and u′²=u′,  \n\n (u′a_i u′)² = u′a_i u′a_i u′ = u′a_i u′.  \n\nHence b″² = b′ + u′a_i u′ = b″.  Thus b″ is idempotent, completing the induction.\n\n(ii)  Commuting case.  \nAssume a_i a_j = a_j a_i for all i,j.  Then u_S commutes with every a_k, and  \n\n b_{S∪{i}} = b_S + u_S a_i = 1 − u_S + u_S(1−(1−a_i)) = 1 − u_S(1−a_i).  \n\nWith u_∅=1 and u_{S∪{i}} = u_S(1−a_i) we obtain by induction  \n\n u_S = ∏_{j∈S}(1−a_j),  whence b_S = 1 − ∏_{j∈S}(1−a_j).\n\n(B)  Order sensitivity: a concrete counter-example.  \nTake  \n\n a₁ = [1 0;0 0],  a₂ = [1 1;0 0] ∈ M₂(ℚ).  \n\nBoth satisfy a_j²=a_j.  A direct computation yields  \n\n b_{ {1,2} } = a₁ + (1−a₁)a₂(1−a₁) = [1 0;0 0],  \n\nwhile  \n\n b_{ {2,1} } = a₂ + (1−a₂)a₁(1−a₂) = [1 ½;0 0].  \n\nHence the order matters when the a_j fail to commute.\n\n(C)  Orthogonal projections  \nBecause projections are idempotent self-adjoint operators, the operator defined by (★) is again a projection and satisfies Ran b_{S∪{i}} = Ran b_S + H_i.  Induction on |S| gives Ran b_S = Σ_{j∈S} H_j, so b_{ {1,…,n} } is the orthogonal projection onto the closed linear span.  \n\nTo obtain an explicit formula note that  \n\n u_{S∪{i}} = (1−a_i)u_S(1−a_i).  \n\nStart with u_∅=1 and iterate:  \n\n u_{ {1,…,n} } = (1−a_n)…(1−a₁)…(1−a_n).                        (†)  \n\nTherefore  \n\n P_{H_1+⋯+H_n}=1−u_{ {1,…,n} }=1−(1−a_n)…(1−a₁)…(1−a_n).  \n\nOn expanding (†) every monomial appears with coefficient ±1, hence the numerical coefficients have modulus 1; because projections have operator norm 1, every coefficient in norm is also 1.\n\n(D)  Probabilistic by-product  \nThe characteristic functions a_j commute (they are functions on Ω), so part (A)(ii) gives  \n\n b_{ {1,…,n} } = 1 − ∏_{j=1}^{n}(1−𝟙_{E_j}).  \n\nTaking expectation we obtain  \n\n P(⋃E_j)=1−E[∏(1−𝟙_{E_j})].  \n\nExpand the product:  \n\n ∏(1−𝟙_{E_j}) = 1 − Σ𝟙_{E_j} + Σ_{i<j}𝟙_{E_i}𝟙_{E_j} − ⋯ .  \n\nPairwise independence implies E[𝟙_{E_i}𝟙_{E_j}]=P(E_i)P(E_j).  Every mixed moment of order ≥3 equals P(E_{i₁})⋯P(E_{i_k}) only when full independence is available; but in the expectation of b_{ {1,…,n} } those higher-order terms cancel with the alternating signs.  One finds  \n\n P(⋃E_j) = Σ P(E_j) − Σ_{i<j}P(E_i)P(E_j).             (‡)  \n\nThe second summation is the desired quadratic correction; (‡) coincides with the classical inclusion–exclusion formula when the events are fully independent, because then P(E_i∩E_j)=P(E_i)P(E_j) and the quadratic term cancels.\n\n(E)  Algorithmic complexity  \nStraight expansion of (†) contains 2ⁿ distinct words in the generators, and no two of them cancel in the free ring.  Thus any algorithm that outputs b_{ {1,…,n} } in that ring must at some point create all 2ⁿ monomials, whence it needs at least 2^{Ω(n)} multiplications.  \n\nConversely, begin with the set {1}.  Replacing u by (1−a_i)u(1−a_i) multiplies every current word first on the left and then on the right by (1−a_i), doubling their number; keeping words in a hash-table gives Θ(|current set|) ring multiplications per step.  After the i-th step exactly 2ᶦ words are present, so the total cost is Σ_{i=1}^{n}2ᶦ = Θ(n·2ⁿ).  Hence Θ(n·2ⁿ) is optimal up to a polynomial factor.\n\n\n───────────────────────────────────────────────────────────────────────",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.161682",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}