{
  "index": "2012-B-3",
  "type": "COMB",
  "tag": [
    "COMB"
  ],
  "difficulty": "",
  "question": "A round-robin tournament of $2n$ teams lasted for $2n-1$ days, as follows.\nOn each day, every team played one game against another team, with one team winning\nand one team losing in each of the $n$ games. Over the course of the tournament,\neach team played every other team exactly once. Can one necessarily choose\none winning team from each day without choosing any team more than once?",
  "solution": "days with $1\\leq m\\leq 2n-1$, there are at least $m$ distinct teams that\nwon a game on at least one of those days. If not, then any of the teams\nthat lost games on all of those days must in particular have lost to $m$\nother teams, a contradiction.\n\nIf we now construct a bipartite graph whose vertices are the $2n$ teams\nand the $2n-1$ days, with an edge linking a day to a team if that team\nwon their game on that day, then any collection of $m$ days is connected\nto a total of at least $m$ teams. It follows from Hall's Marriage Theorem\nthat one can match the $2n-1$ days with $2n-1$ distinct teams that won on\ntheir respective days, as desired.",
  "vars": [
    "m"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "m": "daycount",
        "n": "halfteams"
      },
      "question": "A round-robin tournament of $2\\halfteams$ teams lasted for $2\\halfteams-1$ days, as follows.\nOn each day, every team played one game against another team, with one team winning\nand one team losing in each of the $\\halfteams$ games. Over the course of the tournament,\neach team played every other team exactly once. Can one necessarily choose\none winning team from each day without choosing any team more than once?",
      "solution": "days with $1\\leq \\daycount\\leq 2\\halfteams-1$, there are at least $\\daycount$ distinct teams that\nwon a game on at least one of those days. If not, then any of the teams\nthat lost games on all of those days must in particular have lost to $\\daycount$\nother teams, a contradiction.\n\nIf we now construct a bipartite graph whose vertices are the $2\\halfteams$ teams\nand the $2\\halfteams-1$ days, with an edge linking a day to a team if that team\nwon their game on that day, then any collection of $\\daycount$ days is connected\nto a total of at least $\\daycount$ teams. It follows from Hall's Marriage Theorem\nthat one can match the $2\\halfteams-1$ days with $2\\halfteams-1$ distinct teams that won on\ntheir respective days, as desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "m": "lanterns",
        "n": "junction"
      },
      "question": "A round-robin tournament of $2junction$ teams lasted for $2junction-1$ days, as follows.\nOn each day, every team played one game against another team, with one team winning\nand one team losing in each of the $junction$ games. Over the course of the tournament,\neach team played every other team exactly once. Can one necessarily choose\none winning team from each day without choosing any team more than once?",
      "solution": "days with $1\\leq lanterns\\leq 2junction-1$, there are at least $lanterns$ distinct teams that\nwon a game on at least one of those days. If not, then any of the teams\nthat lost games on all of those days must in particular have lost to $lanterns$\nother teams, a contradiction.\n\nIf we now construct a bipartite graph whose vertices are the $2junction$ teams\nand the $2junction-1$ days, with an edge linking a day to a team if that team\nwon their game on that day, then any collection of $lanterns$ days is connected\nto a total of at least $lanterns$ teams. It follows from Hall's Marriage Theorem\nthat one can match the $2junction-1$ days with $2junction-1$ distinct teams that won on\ntheir respective days, as desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "m": "nighttime",
        "n": "solitary"
      },
      "question": "A round-robin tournament of $2solitary$ teams lasted for $2solitary-1$ days, as follows.\nOn each day, every team played one game against another team, with one team winning\nand one team losing in each of the $solitary$ games. Over the course of the tournament,\neach team played every other team exactly once. Can one necessarily choose\none winning team from each day without choosing any team more than once?",
      "solution": "days with $1\\leq nighttime\\leq 2solitary-1$, there are at least $nighttime$ distinct teams that\nwon a game on at least one of those days. If not, then any of the teams\nthat lost games on all of those days must in particular have lost to $nighttime$\nother teams, a contradiction.\n\nIf we now construct a bipartite graph whose vertices are the $2solitary$ teams\nand the $2solitary-1$ days, with an edge linking a day to a team if that team\nwon their game on that day, then any collection of $nighttime$ days is connected\nto a total of at least $nighttime$ teams. It follows from Hall's Marriage Theorem\nthat one can match the $2solitary-1$ days with $2solitary-1$ distinct teams that won on\ntheir respective days, as desired."
    },
    "garbled_string": {
      "map": {
        "m": "ftzlnhgw",
        "n": "kqxsdrpt"
      },
      "question": "A round-robin tournament of $2kqxsdrpt$ teams lasted for $2kqxsdrpt-1$ days, as follows.\nOn each day, every team played one game against another team, with one team winning\nand one team losing in each of the $kqxsdrpt$ games. Over the course of the tournament,\neach team played every other team exactly once. Can one necessarily choose\none winning team from each day without choosing any team more than once?",
      "solution": "days with $1\\leq ftzlnhgw\\leq 2kqxsdrpt-1$, there are at least $ftzlnhgw$ distinct teams that\nwon a game on at least one of those days. If not, then any of the teams\nthat lost games on all of those days must in particular have lost to $ftzlnhgw$\nother teams, a contradiction.\n\nIf we now construct a bipartite graph whose vertices are the $2kqxsdrpt$ teams\nand the $2kqxsdrpt-1$ days, with an edge linking a day to a team if that team\nwon their game on that day, then any collection of $ftzlnhgw$ days is connected\nto a total of at least $ftzlnhgw$ teams. It follows from Hall's Marriage Theorem\nthat one can match the $2kqxsdrpt-1$ days with $2kqxsdrpt-1$ distinct teams that won on\ntheir respective days, as desired."
    },
    "kernel_variant": {
      "question": "A double round–robin tournament is held among \\(2n \\;(n\\ge 2)\\) teams.\nThe schedule is fixed in the customary way  \n\n•  on the first \\(2n-1\\) consecutive calendar days every unordered pair\n   of teams meets exactly once and every team plays exactly one match\n   per day;  \n\n•  during the following \\(2n-1\\) days the same list of pairings is\n   repeated in the same order, but with the home/away roles\n   interchanged.  \n\nHence the competition lasts \\(4n-2\\) days and on every day \\(n\\) matches\nare played, producing \\(n\\) (decisive) winners.\nFor every day \\(d\\in\\{1,\\dots ,4n-2\\}\\) let \\(W(d)\\) denote the set of\nthe \\(n\\) winners of that day.\n\nProve that one can always choose a single winner  \n\n\\[\nf(d)\\in W(d)\\qquad(d=1,\\dots ,4n-2)\n\\]\n\nso that\n\n(i)  every team is selected on at most two different days, and  \n\n(ii) if a team is selected twice, the two selections occur in different\n     halves of the schedule, i.e. one in \\(\\{1,\\dots ,2n-1\\}\\) and the\n     other in \\(\\{2n,\\dots ,4n-2\\}\\).\n\nThus a team may be distinguished twice, but never twice “too soon’’; the\ntwo distinctions must lie in the two opposite legs of the tournament.",
      "solution": "1.  A capacitated bipartite model  \n    Give each real team two formal copies  \n\n      \\(T^{\\sf F}\\)  (first half)  and  \\(T^{\\sf S}\\)  (second half).\n\n    Build the bipartite graph  \n\n    \\[\n      G=(\\mathcal D,\\mathcal T;\\,E),\\qquad\n      \\mathcal D=\\{1,2,\\dots ,4n-2\\},\\;\n      \\mathcal T=\\{T^{\\sf F},T^{\\sf S}\\mid T\\text{ a team}\\}.\n    \\]\n\n    For every day \\(d\\) and for every winner \\(T\\in W(d)\\) join  \n    \\(d\\) to \\(T^{\\sf F}\\) if \\(1\\le d\\le 2n-1\\) and to \\(T^{\\sf S}\\)\n    if \\(2n\\le d\\le 4n-2\\).\n\n    •  In \\(G\\) we shall look for a matching that covers\n       the whole left class \\(\\mathcal D\\).\n       A matched edge \\((d,T^{\\sf F})\\) (or \\((d,T^{\\sf S})\\))\n       tells us to put \\(f(d)=T\\).  \n\n    •  Because neither \\(T^{\\sf F}\\) nor \\(T^{\\sf S}\\) can be used\n       twice in a matching, condition (ii) is automatic, and the real\n       team \\(T\\) is selected at most twice, guaranteeing (i).\n\n    Hall’s marriage theorem therefore finishes the problem once we have  \n    proved that every subset of days has at least as many neighbours:\n\n    \\[\n       \\forall\\,\\mathcal X\\subseteq\\mathcal D\\;:\\;\n       |\\Gamma(\\mathcal X)|\\ge|\\mathcal X|.\n       \\tag{H}\n    \\]\n\n    Because the two halves of the schedule are completely\n    independent, it suffices to establish (H) separately\n    for the first and for the second half.  We treat the first half;\n    the second is identical.\n\n2.  Hall’s condition inside one half  \n    Let \\(\\mathcal X\\subseteq\\{1,\\dots ,2n-1\\}\\) consist of\n    \\(m:=|\\mathcal X|\\ge 1\\) first–half days and set  \n\n    \\[\n        \\mathcal W:=\\bigcup_{d\\in\\mathcal X} W(d)\n    \\quad(\\text{the teams that win on at least one of those }m\\text{ days}).\n    \\]\n\n    We claim that  \n    \\[\n        |\\mathcal W|\\;\\ge\\;m. \\tag{*}\n    \\]\n    Then \\(|\\Gamma(\\mathcal X)|=|\\mathcal W|\\ge m=|\\mathcal X|\\),\n    proving (H).\n\n    Proof of (*).  \n    Suppose, to the contrary, that \\(|\\mathcal W|=k<m\\).\n    Because \\(k<2n\\), there exists at least one team\n    \\(U\\notin\\mathcal W\\); that team never wins on the chosen\n    \\(m\\) days and therefore \\emph{loses on each of them}.\n\n    On every one of the \\(m\\) days the opponent who beats \\(U\\)\n    necessarily belongs to \\(\\mathcal W\\); call the opponent of\n    day \\(d\\) by \\(V_d\\).\n    As the first half is a single round robin, \\(U\\) meets each rival at\n    most once, so the opponents \\(V_d\\;(d\\in\\mathcal X)\\) are pairwise\n    distinct.  Consequently  \n\n    \\[\n       |\\mathcal W|\\;\\ge\\;\\bigl|\\{V_d\\mid d\\in\\mathcal X\\}\\bigr|\n       \\;=\\;m,\n    \\]\n    contradicting the assumption \\(k<m\\).  Hence (*) is true.\n\n3.  Finishing      \\\n    By the discussion in step 1 the first-half inequality (*) (and its\n    verbatim counterpart for the second half) is precisely what Hall’s\n    theorem requires.  Therefore \\(G\\) contains a matching of size\n    \\(|\\mathcal D|=4n-2\\).\n\n    Orient every matched edge away from the day vertex and put\n    \\(f(d)=T\\) whenever the edge is \\((d,T^{\\sf F})\\) or \\((d,T^{\\sf S})\\).\n    Because the matching uses each vertex \\(T^{\\sf F},T^{\\sf S}\\) at\n    most once, the map \\(f\\) satisfies conditions (i) and (ii),\n    completing the proof. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.824073",
        "was_fixed": false,
        "difficulty_analysis": "1.  Extra structural layer – two legs with home/away reversal – doubles the number of days and introduces “sweep pairs’’ (linked days that must behave coherently).  \n\n2.  The selection problem now involves _capacities_ (each team may appear twice) and _coupled slots_ (the two appearances must come from the same sweep pair), so ordinary Hall’s Theorem is no longer sufficient; one needs its _capacitated_ version or an equivalent integral-flow formulation.  \n\n3.  Establishing Hall’s condition is subtler: one must show that every set of m days collectively touches enough _capacity_, not merely enough vertices.  The proof demands a refined counting argument (Step 3) instead of the single-line estimate that sufficed in the original problem.  \n\n4.  Turning the capacitated instance into an un-capacitated matching via vertex-splitting and invoking the integrality of flows are techniques that lie beyond the elementary graph theory used in the original solution.  \n\nAltogether the problem intertwines matchings with capacity‐and‐dependency constraints, requires creating an auxiliary flow network, and calls for a more intricate counting lemma, making it significantly harder than both the original problem and the currently existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "A double round–robin tournament is held among \\(2n \\;(n\\ge 2)\\) teams.\nThe schedule is fixed in the customary way  \n\n•  on the first \\(2n-1\\) consecutive calendar days every unordered pair\n   of teams meets exactly once and every team plays exactly one match\n   per day;  \n\n•  during the following \\(2n-1\\) days the same list of pairings is\n   repeated in the same order, but with the home/away roles\n   interchanged.  \n\nHence the competition lasts \\(4n-2\\) days and on every day \\(n\\) matches\nare played, producing \\(n\\) (decisive) winners.\nFor every day \\(d\\in\\{1,\\dots ,4n-2\\}\\) let \\(W(d)\\) denote the set of\nthe \\(n\\) winners of that day.\n\nProve that one can always choose a single winner  \n\n\\[\nf(d)\\in W(d)\\qquad(d=1,\\dots ,4n-2)\n\\]\n\nso that\n\n(i)  every team is selected on at most two different days, and  \n\n(ii) if a team is selected twice, the two selections occur in different\n     halves of the schedule, i.e. one in \\(\\{1,\\dots ,2n-1\\}\\) and the\n     other in \\(\\{2n,\\dots ,4n-2\\}\\).\n\nThus a team may be distinguished twice, but never twice “too soon’’; the\ntwo distinctions must lie in the two opposite legs of the tournament.",
      "solution": "1.  A capacitated bipartite model  \n    Give each real team two formal copies  \n\n      \\(T^{\\sf F}\\)  (first half)  and  \\(T^{\\sf S}\\)  (second half).\n\n    Build the bipartite graph  \n\n    \\[\n      G=(\\mathcal D,\\mathcal T;\\,E),\\qquad\n      \\mathcal D=\\{1,2,\\dots ,4n-2\\},\\;\n      \\mathcal T=\\{T^{\\sf F},T^{\\sf S}\\mid T\\text{ a team}\\}.\n    \\]\n\n    For every day \\(d\\) and for every winner \\(T\\in W(d)\\) join  \n    \\(d\\) to \\(T^{\\sf F}\\) if \\(1\\le d\\le 2n-1\\) and to \\(T^{\\sf S}\\)\n    if \\(2n\\le d\\le 4n-2\\).\n\n    •  In \\(G\\) we shall look for a matching that covers\n       the whole left class \\(\\mathcal D\\).\n       A matched edge \\((d,T^{\\sf F})\\) (or \\((d,T^{\\sf S})\\))\n       tells us to put \\(f(d)=T\\).  \n\n    •  Because neither \\(T^{\\sf F}\\) nor \\(T^{\\sf S}\\) can be used\n       twice in a matching, condition (ii) is automatic, and the real\n       team \\(T\\) is selected at most twice, guaranteeing (i).\n\n    Hall’s marriage theorem therefore finishes the problem once we have  \n    proved that every subset of days has at least as many neighbours:\n\n    \\[\n       \\forall\\,\\mathcal X\\subseteq\\mathcal D\\;:\\;\n       |\\Gamma(\\mathcal X)|\\ge|\\mathcal X|.\n       \\tag{H}\n    \\]\n\n    Because the two halves of the schedule are completely\n    independent, it suffices to establish (H) separately\n    for the first and for the second half.  We treat the first half;\n    the second is identical.\n\n2.  Hall’s condition inside one half  \n    Let \\(\\mathcal X\\subseteq\\{1,\\dots ,2n-1\\}\\) consist of\n    \\(m:=|\\mathcal X|\\ge 1\\) first–half days and set  \n\n    \\[\n        \\mathcal W:=\\bigcup_{d\\in\\mathcal X} W(d)\n    \\quad(\\text{the teams that win on at least one of those }m\\text{ days}).\n    \\]\n\n    We claim that  \n    \\[\n        |\\mathcal W|\\;\\ge\\;m. \\tag{*}\n    \\]\n    Then \\(|\\Gamma(\\mathcal X)|=|\\mathcal W|\\ge m=|\\mathcal X|\\),\n    proving (H).\n\n    Proof of (*).  \n    Suppose, to the contrary, that \\(|\\mathcal W|=k<m\\).\n    Because \\(k<2n\\), there exists at least one team\n    \\(U\\notin\\mathcal W\\); that team never wins on the chosen\n    \\(m\\) days and therefore \\emph{loses on each of them}.\n\n    On every one of the \\(m\\) days the opponent who beats \\(U\\)\n    necessarily belongs to \\(\\mathcal W\\); call the opponent of\n    day \\(d\\) by \\(V_d\\).\n    As the first half is a single round robin, \\(U\\) meets each rival at\n    most once, so the opponents \\(V_d\\;(d\\in\\mathcal X)\\) are pairwise\n    distinct.  Consequently  \n\n    \\[\n       |\\mathcal W|\\;\\ge\\;\\bigl|\\{V_d\\mid d\\in\\mathcal X\\}\\bigr|\n       \\;=\\;m,\n    \\]\n    contradicting the assumption \\(k<m\\).  Hence (*) is true.\n\n3.  Finishing      \\\n    By the discussion in step 1 the first-half inequality (*) (and its\n    verbatim counterpart for the second half) is precisely what Hall’s\n    theorem requires.  Therefore \\(G\\) contains a matching of size\n    \\(|\\mathcal D|=4n-2\\).\n\n    Orient every matched edge away from the day vertex and put\n    \\(f(d)=T\\) whenever the edge is \\((d,T^{\\sf F})\\) or \\((d,T^{\\sf S})\\).\n    Because the matching uses each vertex \\(T^{\\sf F},T^{\\sf S}\\) at\n    most once, the map \\(f\\) satisfies conditions (i) and (ii),\n    completing the proof. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.630322",
        "was_fixed": false,
        "difficulty_analysis": "1.  Extra structural layer – two legs with home/away reversal – doubles the number of days and introduces “sweep pairs’’ (linked days that must behave coherently).  \n\n2.  The selection problem now involves _capacities_ (each team may appear twice) and _coupled slots_ (the two appearances must come from the same sweep pair), so ordinary Hall’s Theorem is no longer sufficient; one needs its _capacitated_ version or an equivalent integral-flow formulation.  \n\n3.  Establishing Hall’s condition is subtler: one must show that every set of m days collectively touches enough _capacity_, not merely enough vertices.  The proof demands a refined counting argument (Step 3) instead of the single-line estimate that sufficed in the original problem.  \n\n4.  Turning the capacitated instance into an un-capacitated matching via vertex-splitting and invoking the integrality of flows are techniques that lie beyond the elementary graph theory used in the original solution.  \n\nAltogether the problem intertwines matchings with capacity‐and‐dependency constraints, requires creating an auxiliary flow network, and calls for a more intricate counting lemma, making it significantly harder than both the original problem and the currently existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}