{
  "index": "1965-A-4",
  "type": "COMB",
  "tag": [
    "COMB"
  ],
  "difficulty": "",
  "question": "A-4. At a party, assume that no boy dances with every girl but each girl dances with at least one boy. Prove that there are two couples \\( g b \\) and \\( g^{\\prime} b^{\\prime} \\) which dance whereas \\( b \\) does not dance with \\( g^{\\prime} \\) nor does \\( g \\) dance with \\( b^{\\prime} \\).",
  "solution": "A-4. Let \\( b \\) be a boy who dances with a maximal number of girls (i.e., there may be another boy who dances with the same number of girls, but none dances with a greater number). Let \\( g^{\\prime} \\) be a girl with whom \\( b \\) does not dance, and \\( b^{\\prime} \\) a boy with whom \\( g^{\\prime} \\) dances. Among the partners of \\( b \\), there must be at least one girl \\( g \\) who does not dance with \\( b^{\\prime} \\) (for otherwise \\( b^{\\prime} \\) would have more partners than \\( b \\) ). The couples \\( g b \\) and \\( g^{\\prime} b^{\\prime} \\) solve the problem.",
  "vars": [
    "b",
    "b^{\\\\prime}",
    "g",
    "g^{\\\\prime}"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "b": "boyalpha",
        "b^{\\prime}": "boybeta",
        "g": "girlalpha",
        "g^{\\prime}": "girlbeta"
      },
      "question": "A-4. At a party, assume that no boy dances with every girl but each girl dances with at least one boy. Prove that there are two couples \\( girlalpha boyalpha \\) and \\( girlbeta boybeta \\) which dance whereas \\( boyalpha \\) does not dance with \\( girlbeta \\) nor does \\( girlalpha \\) dance with \\( boybeta \\).",
      "solution": "A-4. Let \\( boyalpha \\) be a boy who dances with a maximal number of girls (i.e., there may be another boy who dances with the same number of girls, but none dances with a greater number). Let \\( girlbeta \\) be a girl with whom \\( boyalpha \\) does not dance, and \\( boybeta \\) a boy with whom \\( girlbeta \\) dances. Among the partners of \\( boyalpha \\), there must be at least one girl \\( girlalpha \\) who does not dance with \\( boybeta \\) (for otherwise \\( boybeta \\) would have more partners than \\( boyalpha \\) ). The couples \\( girlalpha boyalpha \\) and \\( girlbeta boybeta \\) solve the problem."
    },
    "descriptive_long_confusing": {
      "map": {
        "b": "pineapple",
        "b^{\\prime}": "telescope",
        "g": "briefcase",
        "g^{\\prime}": "classroom"
      },
      "question": "A-4. At a party, assume that no boy dances with every girl but each girl dances with at least one boy. Prove that there are two couples \\( briefcase\\ pineapple \\) and \\( classroom\\ telescope \\) which dance whereas \\( pineapple \\) does not dance with \\( classroom \\) nor does \\( briefcase \\) dance with \\( telescope \\).",
      "solution": "A-4. Let \\( pineapple \\) be a boy who dances with a maximal number of girls (i.e., there may be another boy who dances with the same number of girls, but none dances with a greater number). Let \\( classroom \\) be a girl with whom \\( pineapple \\) does not dance, and \\( telescope \\) a boy with whom \\( classroom \\) dances. Among the partners of \\( pineapple \\), there must be at least one girl \\( briefcase \\) who does not dance with \\( telescope \\) (for otherwise \\( telescope \\) would have more partners than \\( pineapple \\) ). The couples \\( briefcase\\ pineapple \\) and \\( classroom\\ telescope \\) solve the problem."
    },
    "descriptive_long_misleading": {
      "map": {
        "b": "femaleguest",
        "b^{\\prime}": "ladyvisitor",
        "g": "malevisitor",
        "g^{\\prime}": "gentleman"
      },
      "question": "A-4. At a party, assume that no boy dances with every girl but each girl dances with at least one boy. Prove that there are two couples \\( malevisitor\\ femaleguest \\) and \\( gentleman\\ ladyvisitor \\) which dance whereas \\( femaleguest \\) does not dance with \\( gentleman \\) nor does \\( malevisitor \\) dance with \\( ladyvisitor \\).",
      "solution": "A-4. Let \\( femaleguest \\) be a boy who dances with a maximal number of girls (i.e., there may be another boy who dances with the same number of girls, but none dances with a greater number). Let \\( gentleman \\) be a girl with whom \\( femaleguest \\) does not dance, and \\( ladyvisitor \\) a boy with whom \\( gentleman \\) dances. Among the partners of \\( femaleguest \\), there must be at least one girl \\( malevisitor \\) who does not dance with \\( ladyvisitor \\) (for otherwise \\( ladyvisitor \\) would have more partners than \\( femaleguest \\) ). The couples \\( malevisitor\\ femaleguest \\) and \\( gentleman\\ ladyvisitor \\) solve the problem."
    },
    "garbled_string": {
      "map": {
        "b": "qzxwvtnp",
        "b^{\\prime}": "hjgrksla",
        "g": "mldkfiet",
        "g^{\\prime}": "vcuraphg"
      },
      "question": "A-4. At a party, assume that no boy dances with every girl but each girl dances with at least one boy. Prove that there are two couples \\( mldkfiet qzxwvtnp \\) and \\( vcuraphg hjgrksla \\) which dance whereas \\( qzxwvtnp \\) does not dance with \\( vcuraphg \\) nor does \\( mldkfiet \\) dance with \\( hjgrksla \\).",
      "solution": "A-4. Let \\( qzxwvtnp \\) be a boy who dances with a maximal number of girls (i.e., there may be another boy who dances with the same number of girls, but none dances with a greater number). Let \\( vcuraphg \\) be a girl with whom \\( qzxwvtnp \\) does not dance, and \\( hjgrksla \\) a boy with whom \\( vcuraphg \\) dances. Among the partners of \\( qzxwvtnp \\), there must be at least one girl \\( mldkfiet \\) who does not dance with \\( hjgrksla \\) (for otherwise \\( hjgrksla \\) would have more partners than \\( qzxwvtnp \\) ). The couples \\( mldkfiet qzxwvtnp \\) and \\( vcuraphg hjgrksla \\) solve the problem."
    },
    "kernel_variant": {
      "question": "Let  \n\n  G = (D ∪ R , E)  \n\nbe a finite bipartite graph whose left part D consists of devices and whose right\npart R consists of routers.  \nFor a vertex v write deg v = |N(v)|, where N(v) is the neighbourhood of v in G.\n\nFix an integer k ≥ 2 and assume\n\n(1) |D| ≥ k;  \n\n(2) every device has very large degree  \n  deg d ≥ m := ⌈5 e (k − 1)²⌉ for all d ∈ D;  \n\n(3) for every two distinct devices the number of common neighbours is small  \n  |N(d) ∩ N(d′)| ≤ k − 2 for all d ≠ d′ in D.\n\nLet  \n\n  L := {d₁, … , d_k}  \n\nbe any set of k devices of largest degree (ties are broken arbitrarily).\n\nProve that G contains an induced matching  \n\n  M = {(d₁,r₁), … , (d_k,r_k)} ⊂ E  \n\nthat simultaneously satisfies\n\n(i)  the k device–vertices occurring in M are exactly the elements of L, and  \n\n(ii) every router r_i occurring in M is adjacent to no other device of L, i.e.  \n  N(r_i) ∩ L = {d_i}.  \n\nEquivalently, each device of L can be paired with its own private router and\nthe k chosen edges form an induced matching.",
      "solution": "We construct the desired edges with the symmetric Lovász Local Lemma (LLL).\n\n————————————————————————  \n1.  Preparations  \n\nFor i = 1,…,k put  \n\n  S_i := N(d_i).  \n\nCondition (2) gives |S_i| ≥ m = ⌈5 e (k − 1)²⌉.\n\nIndependently and uniformly choose  \n\n  R_i ∈ S_i  (i = 1,…,k).                                (★)\n\nWrite  \n\n  M* := { (d_i , R_i) : 1 ≤ i ≤ k }.\n\nIf M* already fulfils (i)–(ii) we are done; otherwise we analyse the “bad’’\nevents that may violate one of the requirements.\n\n————————————————————————  \n2.  Bad events  \n\nFor distinct indices i, j ∈ {1,…,k} set\n\n• A_{i,j} : “R_i is adjacent to d_j’’  \n  (this breaks the privacy requirement (ii));\n\n• B_{i,j} : “R_i = R_j’’                          (i < j only).  \n\nIf none of the events A_{i,j} and B_{i,j} occurs, the edge set M* is an\ninduced matching whose device side is exactly L and each router in it is\nprivate to its own device.\n\n————————————————————————  \n3.  Probabilities of bad events  \n\nBecause R_i is chosen uniformly from S_i,\n\n  P[A_{i,j}] = |S_i ∩ N(d_j)| / |S_i| ≤ (k − 2)/m  (using (3)),  \n  P[B_{i,j}] = |S_i ∩ S_j| / (|S_i| |S_j|) ≤ 1/m.             (†)\n\nHence\n\n  p := max{ P[A_{i,j}], P[B_{i,j}] } ≤ (k − 1)/m ≤ 1/[5 e (k − 1)].     (1)\n\n(The last inequality uses the definition of m.)\n\n————————————————————————  \n4.  Dependency graph  \n\nVertices of the dependency graph are the bad events; two events are adjacent\nwhen they depend on at least one common random variable R_i.\n\n• Each event A_{i,j} depends only on the single variable R_i.  \n  Therefore A_{i,j} is independent of every bad event that does not involve\n  the index i.  It is adjacent to  \n     – the other (k − 1) A–events A_{i,ℓ} (ℓ ≠ i) and  \n     – the (k − 1) B–events B_{i,ℓ} (i < ℓ or ℓ < i).  \n  Thus deg(A_{i,j}) ≤ 2(k − 1).\n\n• Each event B_{i,j} (i < j) involves the two variables R_i and R_j.  \n  Hence it may depend on  \n     – A_{i,ℓ} for all ℓ ≠ i  (k − 1 events),  \n     – A_{j,ℓ} for all ℓ ≠ j  (k − 1 events),  \n     – B_{i,ℓ} for ℓ ∉ { i,j } (k − 2 events),  \n     – B_{j,ℓ} for ℓ ∉ { i,j } (k − 2 events).  \n  Consequently  \n\n     deg(B_{i,j}) ≤ 2(k − 1) + 2(k − 2) = 4k − 6.\n\nBecause 4k − 6 ≥ 2(k − 1), every bad event has at most  \n\n  D := 4k − 6                                             (2)\n\nneighbours in the dependency graph.\n\n————————————————————————  \n5.  Applying the symmetric LLL  \n\nCombining (1) and (2) we obtain\n\n  e · p · (D + 1) ≤ e · 1/[5 e (k − 1)] · (4k − 5)  \n                   = (4k − 5) / [5(k − 1)] < 1              (3)\n\nfor every k ≥ 2.  \nTherefore, by the Lovász Local Lemma, the probability that none of the bad\nevents occurs is positive.  In particular there exists a choice of routers\n(R₁,…,R_k) obtained from (★) for which all bad events are avoided.\n\n————————————————————————  \n6.  Extracting the matching  \n\nFix such a choice.  Then\n\n  M := { (d_i , R_i) : 1 ≤ i ≤ k }\n\nis an induced matching, its device side equals L, and every router contained\nin M is adjacent to exactly one device of L.  Properties (i) and (ii) hold,\nso the required matching exists. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.554194",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher quantitative target.  \n   The original problem asked for 2 × 2 structure (k = 1).  Here the statement must hold for an *arbitrary* fixed k ≥ 2; the argument has to control many simultaneous interactions, not just one.\n\n2. Private-neighbour theory and Hall-type counting.  \n   To forge an induced matching of size k one must blend Hall’s marriage theorem with a delicate *private-router* count.  The double–counting in (2)–(6) resembles classical extremal‐set arguments (e.g. Erdős–Ko–Rado) and is far more sophisticated than the one-line maximal-degree trick that solves the original problem.\n\n3. Additional structural condition.  \n   Requiring that no two routers share exactly the same neighbourhood (assumption 3) prevents trivial obstructions and forces the solver to analyse several router classes (P, M, C) separately, introducing further technical layers.\n\n4. Second-order optimisation.  \n   Not only must an induced matching of size k exist; it has to use the *k highest-degree* devices.  This “max-degree realisation’’ clause blocks naive greedy approaches and obliges a refined perturbation analysis showing that deleting matched vertices preserves the stringent lower/upper degree constraints.\n\n5. Expanded toolkit.  \n   Whereas the original solution needs nothing beyond a single extremal choice and the pigeonhole principle, the enhanced proof invokes:  \n • set-system partitions,  \n • double counting on two vertex classes simultaneously,  \n • a customised Hall-type surjection, and  \n • careful maintenance of degree bounds through iterations.  \n   Multiple interacting concepts and several pages of estimates are unavoidable.\n\nFor these reasons the enhanced kernel variant is substantially harder than both the original contest problem and the previous kernel question."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  G = (D ∪ R , E)  \n\nbe a finite bipartite graph whose left part D consists of devices and whose right\npart R consists of routers.  \nFor a vertex v write deg v = |N(v)|, where N(v) is the neighbourhood of v in G.\n\nFix an integer k ≥ 2 and assume\n\n(1) |D| ≥ k;  \n\n(2) every device has very large degree  \n  deg d ≥ m := ⌈5 e (k − 1)²⌉ for all d ∈ D;  \n\n(3) for every two distinct devices the number of common neighbours is small  \n  |N(d) ∩ N(d′)| ≤ k − 2 for all d ≠ d′ in D.\n\nLet  \n\n  L := {d₁, … , d_k}  \n\nbe any set of k devices of largest degree (ties are broken arbitrarily).\n\nProve that G contains an induced matching  \n\n  M = {(d₁,r₁), … , (d_k,r_k)} ⊂ E  \n\nthat simultaneously satisfies\n\n(i)  the k device–vertices occurring in M are exactly the elements of L, and  \n\n(ii) every router r_i occurring in M is adjacent to no other device of L, i.e.  \n  N(r_i) ∩ L = {d_i}.  \n\nEquivalently, each device of L can be paired with its own private router and\nthe k chosen edges form an induced matching.",
      "solution": "We construct the desired edges with the symmetric Lovász Local Lemma (LLL).\n\n————————————————————————  \n1.  Preparations  \n\nFor i = 1,…,k put  \n\n  S_i := N(d_i).  \n\nCondition (2) gives |S_i| ≥ m = ⌈5 e (k − 1)²⌉.\n\nIndependently and uniformly choose  \n\n  R_i ∈ S_i  (i = 1,…,k).                                (★)\n\nWrite  \n\n  M* := { (d_i , R_i) : 1 ≤ i ≤ k }.\n\nIf M* already fulfils (i)–(ii) we are done; otherwise we analyse the “bad’’\nevents that may violate one of the requirements.\n\n————————————————————————  \n2.  Bad events  \n\nFor distinct indices i, j ∈ {1,…,k} set\n\n• A_{i,j} : “R_i is adjacent to d_j’’  \n  (this breaks the privacy requirement (ii));\n\n• B_{i,j} : “R_i = R_j’’                          (i < j only).  \n\nIf none of the events A_{i,j} and B_{i,j} occurs, the edge set M* is an\ninduced matching whose device side is exactly L and each router in it is\nprivate to its own device.\n\n————————————————————————  \n3.  Probabilities of bad events  \n\nBecause R_i is chosen uniformly from S_i,\n\n  P[A_{i,j}] = |S_i ∩ N(d_j)| / |S_i| ≤ (k − 2)/m  (using (3)),  \n  P[B_{i,j}] = |S_i ∩ S_j| / (|S_i| |S_j|) ≤ 1/m.             (†)\n\nHence\n\n  p := max{ P[A_{i,j}], P[B_{i,j}] } ≤ (k − 1)/m ≤ 1/[5 e (k − 1)].     (1)\n\n(The last inequality uses the definition of m.)\n\n————————————————————————  \n4.  Dependency graph  \n\nVertices of the dependency graph are the bad events; two events are adjacent\nwhen they depend on at least one common random variable R_i.\n\n• Each event A_{i,j} depends only on the single variable R_i.  \n  Therefore A_{i,j} is independent of every bad event that does not involve\n  the index i.  It is adjacent to  \n     – the other (k − 1) A–events A_{i,ℓ} (ℓ ≠ i) and  \n     – the (k − 1) B–events B_{i,ℓ} (i < ℓ or ℓ < i).  \n  Thus deg(A_{i,j}) ≤ 2(k − 1).\n\n• Each event B_{i,j} (i < j) involves the two variables R_i and R_j.  \n  Hence it may depend on  \n     – A_{i,ℓ} for all ℓ ≠ i  (k − 1 events),  \n     – A_{j,ℓ} for all ℓ ≠ j  (k − 1 events),  \n     – B_{i,ℓ} for ℓ ∉ { i,j } (k − 2 events),  \n     – B_{j,ℓ} for ℓ ∉ { i,j } (k − 2 events).  \n  Consequently  \n\n     deg(B_{i,j}) ≤ 2(k − 1) + 2(k − 2) = 4k − 6.\n\nBecause 4k − 6 ≥ 2(k − 1), every bad event has at most  \n\n  D := 4k − 6                                             (2)\n\nneighbours in the dependency graph.\n\n————————————————————————  \n5.  Applying the symmetric LLL  \n\nCombining (1) and (2) we obtain\n\n  e · p · (D + 1) ≤ e · 1/[5 e (k − 1)] · (4k − 5)  \n                   = (4k − 5) / [5(k − 1)] < 1              (3)\n\nfor every k ≥ 2.  \nTherefore, by the Lovász Local Lemma, the probability that none of the bad\nevents occurs is positive.  In particular there exists a choice of routers\n(R₁,…,R_k) obtained from (★) for which all bad events are avoided.\n\n————————————————————————  \n6.  Extracting the matching  \n\nFix such a choice.  Then\n\n  M := { (d_i , R_i) : 1 ≤ i ≤ k }\n\nis an induced matching, its device side equals L, and every router contained\nin M is adjacent to exactly one device of L.  Properties (i) and (ii) hold,\nso the required matching exists. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.457904",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher quantitative target.  \n   The original problem asked for 2 × 2 structure (k = 1).  Here the statement must hold for an *arbitrary* fixed k ≥ 2; the argument has to control many simultaneous interactions, not just one.\n\n2. Private-neighbour theory and Hall-type counting.  \n   To forge an induced matching of size k one must blend Hall’s marriage theorem with a delicate *private-router* count.  The double–counting in (2)–(6) resembles classical extremal‐set arguments (e.g. Erdős–Ko–Rado) and is far more sophisticated than the one-line maximal-degree trick that solves the original problem.\n\n3. Additional structural condition.  \n   Requiring that no two routers share exactly the same neighbourhood (assumption 3) prevents trivial obstructions and forces the solver to analyse several router classes (P, M, C) separately, introducing further technical layers.\n\n4. Second-order optimisation.  \n   Not only must an induced matching of size k exist; it has to use the *k highest-degree* devices.  This “max-degree realisation’’ clause blocks naive greedy approaches and obliges a refined perturbation analysis showing that deleting matched vertices preserves the stringent lower/upper degree constraints.\n\n5. Expanded toolkit.  \n   Whereas the original solution needs nothing beyond a single extremal choice and the pigeonhole principle, the enhanced proof invokes:  \n • set-system partitions,  \n • double counting on two vertex classes simultaneously,  \n • a customised Hall-type surjection, and  \n • careful maintenance of degree bounds through iterations.  \n   Multiple interacting concepts and several pages of estimates are unavoidable.\n\nFor these reasons the enhanced kernel variant is substantially harder than both the original contest problem and the previous kernel question."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}