{
  "index": "1958-B-3",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. In a round-robin tournament with \\( n \\) players (each pair of players plays one game) in which there are no draws, the numbers of wins scored by the players are \\( s_{1}, s_{2}, \\ldots, s_{n} \\). Prove that a necessary and sufficient condition for the existence of 3 players, \\( A, B, C \\), such that \\( A \\) beat \\( B, B \\) beat \\( C \\), and \\( C \\) beat \\( A \\) is\n\\[\ns_{1}^{2}+s_{2}^{2}+\\cdots+s_{n}^{2}<(n-1)(n)(2 n-1)^{\\prime} 6\n\\]",
  "solution": "Solution. For any tournament outcome \\( T \\), let\n\\[\nU(T)=s_{1}{ }^{2}+s_{2}{ }^{2}+\\cdots+s_{n}{ }^{2}\n\\]\n\nThe outcome of a round-robin tournament will be called transitive if and only if there are no examples of three players \\( A, B \\), and \\( C \\) such that \\( A \\) beat \\( B, B \\) beat \\( C \\), and \\( C \\) beat \\( A \\). In this case, \"beat\" is a transitive linear-order relation on the set of players, so we can number the players \\( P_{1}, P_{2}, \\ldots, P_{n} \\) so that \\( P_{i} \\) beat \\( P_{j} \\) if and only if \\( i>j \\). Then the final scores of the players are, respectively,\n\\[\n0,1, \\ldots, n-1\n\\]\n\nHence in the transitive case we have\n\\[\nU(T)=0^{2}+1^{2}+\\cdots+(n-1)^{2}=(n-1) n(2 n-1) / 6\n\\]\n\nThis proves that the given condition is sufficient.\nNow consider a non-transitive tournament outcome, that is, suppose there are three players, \\( A, B \\). and \\( C \\), such that \\( A \\) beat \\( B, B \\) beat \\( C \\), and \\( C \\) beat \\( A \\). We may assume that \\( s_{A} \\) is the least of the numbers \\( s_{A}, s_{B} \\), and \\( s_{C} \\); then \\( s_{A} \\leq s_{B} \\). Now reverse the outcome of the match between \\( A \\) and \\( B \\). We get a new tournament outcome in which \\( A \\) 's score is \\( s_{A}-1 \\), \\( B \\) s score is \\( s_{B}+1 \\), and every other player's score remains the same. Then \\( U \\) is increased by\n\\[\n\\left(s_{A}-1\\right)^{2}-s_{A}^{2}+\\left(s_{B}+1\\right)^{2}-s_{B}^{2}=2\\left(s_{B}-s_{A}\\right)+2>0 .\n\\]\n\nThus any non-transitive tournament outcome can be changed so as to increase \\( U \\). But it is impossible to do this indefinitely because the number of possible outcomes is finite. So starting with a non-transitive outcome, after a finite number of such changes the outcome becomes transitive and \\( U \\) increases to \\( (n-1) n(2 n-1) / 6 \\). Hence for a non-transitive outcome\n\\[\nU(T)<(n-1) n(2 n-1) / 6\n\\]\n\nThus the given condition is necessary.\nRemark. Tournaments have been extensively analyzed. See J. W. Moon, Topics on Tournaments. Holt, Rinehart and Winston, New York, 1969.",
  "vars": [
    "s_1",
    "s_2",
    "s_n",
    "s_A",
    "s_B",
    "s_C",
    "i",
    "j",
    "U",
    "T",
    "P_1",
    "P_n",
    "P_i",
    "P_j",
    "A",
    "B",
    "C"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "s_1": "scoreone",
        "s_2": "scoretwo",
        "s_n": "scoren",
        "s_A": "scorea",
        "s_B": "scoreb",
        "s_C": "scorec",
        "i": "indexi",
        "j": "indexj",
        "U": "scoretotal",
        "T": "tourneyoutcome",
        "P_1": "playerone",
        "P_n": "playern",
        "P_i": "playeri",
        "P_j": "playerj",
        "A": "playera",
        "B": "playerb",
        "C": "playerc",
        "n": "playerscount"
      },
      "question": "3. In a round-robin tournament with \\( playerscount \\) players (each pair of players plays one game) in which there are no draws, the numbers of wins scored by the players are \\( scoreone, scoretwo, \\ldots, scoren \\). Prove that a necessary and sufficient condition for the existence of 3 players, \\( playera, playerb, playerc \\), such that \\( playera \\) beat \\( playerb, playerb \\) beat \\( playerc \\), and \\( playerc \\) beat \\( playera \\) is\n\\[\nscoreone^{2}+scoretwo^{2}+\\cdots+scoren^{2}<(playerscount-1)(playerscount)(2 playerscount-1)^{\\prime} 6\n\\]",
      "solution": "Solution. For any tournament outcome \\( tourneyoutcome \\), let\n\\[\nscoretotal(tourneyoutcome)=scoreone^{2}+scoretwo^{2}+\\cdots+scoren^{2}\n\\]\n\nThe outcome of a round-robin tournament will be called transitive if and only if there are no examples of three players \\( playera, playerb \\), and \\( playerc \\) such that \\( playera \\) beat \\( playerb, playerb \\) beat \\( playerc \\), and \\( playerc \\) beat \\( playera \\). In this case, \"beat\" is a transitive linear-order relation on the set of players, so we can number the players \\( playerone, P_{2}, \\ldots, playern \\) so that \\( playeri \\) beat \\( playerj \\) if and only if \\( indexi>indexj \\). Then the final scores of the players are, respectively,\n\\[\n0,1, \\ldots, playerscount-1\n\\]\n\nHence in the transitive case we have\n\\[\nscoretotal(tourneyoutcome)=0^{2}+1^{2}+\\cdots+(playerscount-1)^{2}=(playerscount-1) playerscount(2 playerscount-1) / 6\n\\]\n\nThis proves that the given condition is sufficient.\nNow consider a non-transitive tournament outcome, that is, suppose there are three players, \\( playera, playerb \\). and \\( playerc \\), such that \\( playera \\) beat \\( playerb, playerb \\) beat \\( playerc \\), and \\( playerc \\) beat \\( playera \\). We may assume that \\( scorea \\) is the least of the numbers \\( scorea, scoreb \\), and \\( scorec \\); then \\( scorea \\leq scoreb \\). Now reverse the outcome of the match between \\( playera \\) and \\( playerb \\). We get a new tournament outcome in which \\( playera \\)'s score is \\( scorea-1 \\), \\( playerb \\)'s score is \\( scoreb+1 \\), and every other player's score remains the same. Then \\( scoretotal \\) is increased by\n\\[\n\\left(scorea-1\\right)^{2}-scorea^{2}+\\left(scoreb+1\\right)^{2}-scoreb^{2}=2\\left(scoreb-scorea\\right)+2>0 .\n\\]\n\nThus any non-transitive tournament outcome can be changed so as to increase \\( scoretotal \\). But it is impossible to do this indefinitely because the number of possible outcomes is finite. So starting with a non-transitive outcome, after a finite number of such changes the outcome becomes transitive and \\( scoretotal \\) increases to \\( (playerscount-1) playerscount(2 playerscount-1) / 6 \\). Hence for a non-transitive outcome\n\\[\nscoretotal(tourneyoutcome)<(playerscount-1) playerscount(2 playerscount-1) / 6\n\\]\n\nThus the given condition is necessary.\nRemark. Tournaments have been extensively analyzed. See J. W. Moon, Topics on Tournaments. Holt, Rinehart and Winston, New York, 1969."
    },
    "descriptive_long_confusing": {
      "map": {
        "s_1": "orchardia",
        "s_2": "pineconed",
        "s_n": "riverbank",
        "s_A": "moonstone",
        "s_B": "sandstone",
        "s_C": "blackberry",
        "i": "quiltwork",
        "j": "snowflake",
        "U": "flagstone",
        "T": "horizongo",
        "P_1": "lanternone",
        "P_n": "lanternmax",
        "P_i": "lanternrow",
        "P_j": "lanterncol",
        "A": "partridge",
        "B": "woodpeck",
        "C": "kingfisher",
        "n": "keystroke"
      },
      "question": "3. In a round-robin tournament with \\( keystroke \\) players (each pair of players plays one game) in which there are no draws, the numbers of wins scored by the players are \\( orchardia, pineconed, \\ldots, riverbank \\). Prove that a necessary and sufficient condition for the existence of 3 players, \\( partridge, woodpeck, kingfisher \\), such that \\( partridge \\) beat \\( woodpeck, woodpeck \\) beat \\( kingfisher \\), and \\( kingfisher \\) beat \\( partridge \\) is\n\\[\norchardia^{2}+pineconed^{2}+\\cdots+riverbank^{2}<(keystroke-1)(keystroke)(2 keystroke-1)^{\\prime} 6\n\\]",
      "solution": "Solution. For any tournament outcome \\( horizongo \\), let\n\\[\nflagstone(horizongo)=orchardia { }^{2}+pineconed { }^{2}+\\cdots+riverbank { }^{2}\n\\]\n\nThe outcome of a round-robin tournament will be called transitive if and only if there are no examples of three players \\( partridge, woodpeck \\), and \\( kingfisher \\) such that \\( partridge \\) beat \\( woodpeck, woodpeck \\) beat \\( kingfisher \\), and \\( kingfisher \\) beat \\( partridge \\). In this case, \"beat\" is a transitive linear-order relation on the set of players, so we can number the players \\( lanternone, P_{2}, \\ldots, lanternmax \\) so that \\( lanternrow \\) beat \\( lanterncol \\) if and only if \\( quiltwork>snowflake \\). Then the final scores of the players are, respectively,\n\\[\n0,1, \\ldots, keystroke-1\n\\]\n\nHence in the transitive case we have\n\\[\nflagstone(horizongo)=0^{2}+1^{2}+\\cdots+(keystroke-1)^{2}=(keystroke-1) keystroke(2 keystroke-1) / 6\n\\]\n\nThis proves that the given condition is sufficient.\nNow consider a non-transitive tournament outcome, that is, suppose there are three players, \\( partridge, woodpeck \\). and \\( kingfisher \\), such that \\( partridge \\) beat \\( woodpeck, woodpeck \\) beat \\( kingfisher \\), and \\( kingfisher \\) beat \\( partridge \\). We may assume that \\( moonstone \\) is the least of the numbers \\( moonstone, sandstone \\), and \\( blackberry \\); then \\( moonstone \\leq sandstone \\). Now reverse the outcome of the match between \\( partridge \\) and \\( woodpeck \\). We get a new tournament outcome in which \\( partridge \\) 's score is \\( moonstone-1 \\), \\( woodpeck \\) s score is \\( sandstone+1 \\), and every other player's score remains the same. Then \\( flagstone \\) is increased by\n\\[\n\\left(moonstone-1\\right)^{2}-moonstone^{2}+\\left(sandstone+1\\right)^{2}-sandstone^{2}=2\\left(sandstone-moonstone\\right)+2>0 .\n\\]\n\nThus any non-transitive tournament outcome can be changed so as to increase \\( flagstone \\). But it is impossible to do this indefinitely because the number of possible outcomes is finite. So starting with a non-transitive outcome, after a finite number of such changes the outcome becomes transitive and \\( flagstone \\) increases to \\( (keystroke-1) keystroke(2 keystroke-1) / 6 \\). Hence for a non-transitive outcome\n\\[\nflagstone(horizongo)<(keystroke-1) keystroke(2 keystroke-1) / 6\n\\]\n\nThus the given condition is necessary.\nRemark. Tournaments have been extensively analyzed. See J. W. Moon, Topics on Tournaments. Holt, Rinehart and Winston, New York, 1969."
    },
    "descriptive_long_misleading": {
      "map": {
        "s_1": "losscorea",
        "s_2": "losscoreb",
        "s_n": "losscorez",
        "s_A": "losscorep",
        "s_B": "losscoreq",
        "s_C": "losscorer",
        "i": "constantone",
        "j": "constanttwo",
        "U": "diminish",
        "T": "inception",
        "P_1": "spectatora",
        "P_n": "spectatorz",
        "P_i": "spectatori",
        "P_j": "spectatorj",
        "A": "observera",
        "B": "observerb",
        "C": "observerc",
        "n": "emptiness"
      },
      "question": "3. In a round-robin tournament with \\( emptiness \\) players (each pair of players plays one game) in which there are no draws, the numbers of wins scored by the players are \\( losscorea, losscoreb, \\ldots, losscorez \\). Prove that a necessary and sufficient condition for the existence of 3 players, \\( observera, observerb, observerc \\), such that \\( observera \\) beat \\( observerb, observerb \\) beat \\( observerc \\), and \\( observerc \\) beat \\( observera \\) is\n\\[\nlosscorea^{2}+losscoreb^{2}+\\cdots+losscorez^{2}<(emptiness-1)(emptiness)(2 emptiness-1)^{\\prime} 6\n\\]",
      "solution": "Solution. For any tournament outcome \\( inception \\), let\n\\[\ndiminish(inception)=losscorea{ }^{2}+losscoreb{ }^{2}+\\cdots+losscorez{ }^{2}\n\\]\n\nThe outcome of a round-robin tournament will be called transitive if and only if there are no examples of three players \\( observera, observerb \\), and \\( observerc \\) such that \\( observera \\) beat \\( observerb, observerb \\) beat \\( observerc \\), and \\( observerc \\) beat \\( observera \\). In this case, \"beat\" is a transitive linear-order relation on the set of players, so we can number the players \\( spectatora, P_{2}, \\ldots, spectatorz \\) so that \\( spectatori \\) beat \\( spectatorj \\) if and only if \\( constantone>constanttwo \\). Then the final scores of the players are, respectively,\n\\[\n0,1, \\ldots, emptiness-1\n\\]\n\nHence in the transitive case we have\n\\[\ndiminish(inception)=0^{2}+1^{2}+\\cdots+(emptiness-1)^{2}=(emptiness-1) emptiness(2 emptiness-1) / 6\n\\]\n\nThis proves that the given condition is sufficient.\nNow consider a non-transitive tournament outcome, that is, suppose there are three players, \\( observera, observerb \\). and \\( observerc \\), such that \\( observera \\) beat \\( observerb, observerb \\) beat \\( observerc \\), and \\( observerc \\) beat \\( observera \\). We may assume that \\( losscorep \\) is the least of the numbers \\( losscorep, losscoreq \\), and \\( losscorer \\); then \\( losscorep \\leq losscoreq \\). Now reverse the outcome of the match between \\( observera \\) and \\( observerb \\). We get a new tournament outcome in which \\( observera \\) 's score is \\( losscorep-1 \\), \\( observerb \\) s score is \\( losscoreq+1 \\), and every other player's score remains the same. Then \\( diminish \\) is increased by\n\\[\n\\left(losscorep-1\\right)^{2}-losscorep^{2}+\\left(losscoreq+1\\right)^{2}-losscoreq^{2}=2\\left(losscoreq-losscorep\\right)+2>0 .\n\\]\n\nThus any non-transitive tournament outcome can be changed so as to increase \\( diminish \\). But it is impossible to do this indefinitely because the number of possible outcomes is finite. So starting with a non-transitive outcome, after a finite number of such changes the outcome becomes transitive and \\( diminish \\) increases to \\( (emptiness-1) emptiness(2 emptiness-1) / 6 \\). Hence for a non-transitive outcome\n\\[\ndiminish(inception)<(emptiness-1) emptiness(2 emptiness-1) / 6\n\\]\n\nThus the given condition is necessary.\nRemark. Tournaments have been extensively analyzed. See J. W. Moon, Topics on Tournaments. Holt, Rinehart and Winston, New York, 1969."
    },
    "garbled_string": {
      "map": {
        "s_1": "qzxwvtnp",
        "s_2": "hjgrksla",
        "s_n": "vmbdploe",
        "s_A": "rtfyqksw",
        "s_B": "xobnmdqh",
        "s_C": "wselkzra",
        "i": "pfmsuqve",
        "j": "kzwtrlha",
        "U": "dvrowysp",
        "T": "ghsapzqm",
        "P_1": "uixlbtga",
        "P_n": "pzkvyrof",
        "P_i": "yclhmdse",
        "P_j": "nmkatlhe",
        "A": "glovwzrb",
        "B": "rkhdwqsp",
        "C": "zmlqkyev",
        "n": "byoetfwa"
      },
      "question": "3. In a round-robin tournament with \\( byoetfwa \\) players (each pair of players plays one game) in which there are no draws, the numbers of wins scored by the players are \\( qzxwvtnp, hjgrksla, \\ldots, vmbdploe \\). Prove that a necessary and sufficient condition for the existence of 3 players, \\( glovwzrb, rkhdwqsp, zmlqkyev \\), such that \\( glovwzrb \\) beat \\( rkhdwqsp, rkhdwqsp \\) beat \\( zmlqkyev \\), and \\( zmlqkyev \\) beat \\( glovwzrb \\) is\n\\[\nqzxwvtnp^{2}+hjgrksla^{2}+\\cdots+vmbdploe^{2}<(byoetfwa-1)(byoetfwa)(2 byoetfwa-1)^{\\prime} 6\n\\]\n",
      "solution": "Solution. For any tournament outcome \\( ghsapzqm \\), let\n\\[\ndvrowysp(ghsapzqm)=qzxwvtnp{ }^{2}+hjgrksla{ }^{2}+\\cdots+vmbdploe{ }^{2}\n\\]\n\nThe outcome of a round-robin tournament will be called transitive if and only if there are no examples of three players \\( glovwzrb, rkhdwqsp \\), and \\( zmlqkyev \\) such that \\( glovwzrb \\) beat \\( rkhdwqsp, rkhdwqsp \\) beat \\( zmlqkyev \\), and \\( zmlqkyev \\) beat \\( glovwzrb \\). In this case, \"beat\" is a transitive linear-order relation on the set of players, so we can number the players \\( uixlbtga, P_{2}, \\ldots, pzkvyrof \\) so that \\( yclhmdse \\) beat \\( nmkatlhe \\) if and only if \\( pfmsuqve>kzwtrlha \\). Then the final scores of the players are, respectively,\n\\[\n0,1, \\ldots, byoetfwa-1\n\\]\n\nHence in the transitive case we have\n\\[\ndvrowysp(ghsapzqm)=0^{2}+1^{2}+\\cdots+(byoetfwa-1)^{2}=(byoetfwa-1) byoetfwa(2 byoetfwa-1) / 6\n\\]\n\nThis proves that the given condition is sufficient.\nNow consider a non-transitive tournament outcome, that is, suppose there are three players, \\( glovwzrb, rkhdwqsp \\). and \\( zmlqkyev \\), such that \\( glovwzrb \\) beat \\( rkhdwqsp, rkhdwqsp \\) beat \\( zmlqkyev \\), and \\( zmlqkyev \\) beat \\( glovwzrb \\). We may assume that \\( rtfyqksw \\) is the least of the numbers \\( rtfyqksw, xobnmdqh \\), and \\( wselkzra \\); then \\( rtfyqksw \\leq xobnmdqh \\). Now reverse the outcome of the match between \\( glovwzrb \\) and \\( rkhdwqsp \\). We get a new tournament outcome in which \\( glovwzrb \\) 's score is \\( rtfyqksw-1 \\), \\( rkhdwqsp \\)'s score is \\( xobnmdqh+1 \\), and every other player's score remains the same. Then \\( dvrowysp \\) is increased by\n\\[\n\\left(rtfyqksw-1\\right)^{2}-rtfyqksw^{2}+\\left(xobnmdqh+1\\right)^{2}-xobnmdqh^{2}=2\\left(xobnmdqh-rtfyqksw\\right)+2>0 .\n\\]\n\nThus any non-transitive tournament outcome can be changed so as to increase \\( dvrowysp \\). But it is impossible to do this indefinitely because the number of possible outcomes is finite. So starting with a non-transitive outcome, after a finite number of such changes the outcome becomes transitive and \\( dvrowysp \\) increases to \\( (byoetfwa-1) byoetfwa(2 byoetfwa-1) / 6 \\). Hence for a non-transitive outcome\n\\[\ndvrowysp(ghsapzqm)<(byoetfwa-1) byoetfwa(2 byoetfwa-1) / 6\n\\]\n\nThus the given condition is necessary.\nRemark. Tournaments have been extensively analyzed. See J. W. Moon, Topics on Tournaments. Holt, Rinehart and Winston, New York, 1969."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 3$ be an integer.  \nFor every integer $r\\ge 0$ define the falling factorial  \n\n\\[\n(x)_{r}=x(x-1)\\cdots (x-r+1),\\qquad \n\\text{with the conventions }(x)_{0}=1\\text{ and }(x)_{r}=0\\text{ whenever }x<r.\n\\]\n\nFor a round–robin tournament $T$ on $n$ players (each pair of players meets once, no draws) denote by  \n\n\\[\ns_{1},\\ldots ,s_{n}\n\\]\n\nthe numbers of wins of the players and put  \n\n\\[\nU_{r}(T)=\\sum_{i=1}^{n}(s_{i})_{r}.\n\\]\n\nFurther introduce the constants  \n\n\\[\nM_{r}(n)=\\frac{(n)_{\\,r+1}}{r+1}= \\frac{n(n-1)\\cdots (n-r)}{r+1}.\n\\]\n\n1.  Show that for $r=2$ the following equivalence holds:\n\n\\[\n\\boxed{\\;T\\text{ contains a directed $3$–cycle}\\;\\Longleftrightarrow\\;\nU_{2}(T)<M_{2}(n)=\\dfrac{n(n-1)(n-2)}{3}\\;}. \\tag{$\\star$}\n\\]\n\n2.  Prove that $r=2$ is the only non–negative integer $r$ for which an equivalence of the form $(\\star)$ is valid for every $n\\ge 3$.  More precisely, establish\n\n(a)  $U_{0}(T)$ and $U_{1}(T)$ are constant (hence they cannot detect $3$–cycles);\n\n(b)  for every $r\\ge 3$ there exists an integer $n\\ge r+1$ and a tournament $T$ on $n$ vertices that contains a directed $3$–cycle yet satisfies $U_{r}(T)=M_{r}(n)$, so that $(\\star)$ fails.\n\nConsequently $U_{2}$ is the unique member of the family $U_{r}$ that detects $3$–cycles in tournaments.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Part 1 – The quadratic case $r=2$\n\nStep 1. Landau’s theorem and a majorisation statement.  \nArrange the score sequence in non–decreasing order\n\n\\[\nt_{1}\\le t_{2}\\le\\cdots\\le t_{n}.\n\\]\n\nLandau’s theorem asserts that for every $m=1,\\ldots ,n$\n\n\\[\n\\sum_{i\\le m}t_{i}\\;\\ge\\; \\frac{m(m-1)}{2},\\tag{1.1}\n\\]\n\nwith equality for all $m$ simultaneously if and only if the tournament is transitive (acyclic).\n\nIntroduce the reference vector  \n\n\\[\na=(0,1,\\ldots ,n-1).\\tag{1.2}\n\\]\n\nBecause $\\sum_{i\\le m}a_{i}=m(m-1)/2$, inequality (1.1) rewrites as  \n\n\\[\n\\forall m:\\;\\sum_{i\\le m}t_{i}\\ge\\sum_{i\\le m}a_{i},\n\\qquad\\text{with equality for }m=n. \\tag{1.3}\n\\]\n\nThus, under the \\emph{ascending} convention for majorisation,\n\n\\[\nt\\;\\text{majorises}\\;a,\\qquad\\text{briefly }t\\succ a.\\tag{1.4}\n\\]\n\nStep 2. Schur–concavity of the quadratic falling factorial.  \nLet $g(x)=x(x-1)$.  On $\\mathbf Z$ we have  \n\n\\[\n\\Delta^{2}g(x)=g(x+2)-2g(x+1)+g(x)=2>0,\n\\]\n\nso $g$ is convex.  A classical result (Hardy–Littlewood–Pólya, inequality theory) says that \\emph{for vectors sorted in the same order} a convex symmetric sum is \\emph{Schur–concave} under the \\emph{ascending} convention:  \n\n\\[\nu\\succ v,\\;u\\ne v\\;\\Longrightarrow\\;\\Phi(u)<\\Phi(v), \\tag{1.5}\n\\]\n\nwhere  \n\n\\[\n\\Phi(z_{1},\\ldots ,z_{n})=\\sum_{i=1}^{n}g(z_{i})\n=\\sum_{i=1}^{n}(z_{i})_{2}=U_{2}(T).\n\\]\n\n(The direction of the inequality is reversed compared with the more usual \\emph{descending} convention; that is the only adaptation one must keep in mind.)\n\nStep 3. Extremality of the transitive tournament.  \nApply (1.5) to $u=t$, $v=a$.  When $T$ is not transitive we have $t\\ne a$ and thus  \n\n\\[\nU_{2}(T)=\\Phi(t)<\\Phi(a). \\tag{1.6}\n\\]\n\nIf $T$ \\emph{is} transitive then $t=a$, hence $U_{2}(T)=\\Phi(a)$.\n\nStep 4. Evaluation of $\\Phi(a)$.  \nFor the transitive scores $0,1,\\ldots ,n-1$,\n\n\\[\n\\Phi(a)=\\sum_{k=0}^{n-1}k(k-1)\n=\\frac{n(n-1)(n-2)}{3}=M_{2}(n). \\tag{1.7}\n\\]\n\nStep 5. Equivalence.  \nPutting Step 3 and Step 4 together:\n\n• If $T$ is acyclic, then $t=a$ and $U_{2}(T)=M_{2}(n)$.  \n• If $T$ contains a directed $3$–cycle, it is non–transitive, hence $U_{2}(T)<M_{2}(n)$ by (1.6).\n\nTherefore the equivalence $(\\star)$ holds.\n\n--------------------------------------------------------------------\nPart 2 – Uniqueness of $r=2$\n\n(a) Trivial orders $r=0,1$.  \nBecause every player participates in $\\binom{n-1}{0}=1$ and $\\binom{n-1}{1}=n-1$ matches, respectively,\n\n\\[\nU_{0}(T)=\\sum_{i=1}^{n}(s_{i})_{0}=n,\\qquad \nU_{1}(T)=\\sum_{i=1}^{n}s_{i}=\\binom{n}{2},\n\\]\n\nboth independent of the orientation of the edges.  Hence they cannot distinguish tournaments with or without $3$–cycles.\n\n(b) Counter–examples for every $r\\ge 3$.  \n\nFix $r\\ge 3$ and set  \n\n\\[\nn=r+2\\quad(\\text{so }n\\ge r+1).\\tag{2.1}\n\\]\n\nStart with the \\emph{transitive} tournament $T_{0}$ on the vertex set  \n$P_{0}\\succ P_{1}\\succ\\cdots\\succ P_{n-1}$.  \nIts (ascending) score sequence is  \n\n\\[\n(0,1,2,\\ldots ,n-1)=(0,1,2,\\ldots ,r+1).\\tag{2.2}\n\\]\n\nComputation of $U_{r}(T_{0})$.  \nFor $k<r$ we have $(k)_{r}=0$; for $k=r$ and $k=r+1$ we obtain  \n\n\\[\n(r)_{r}=r!,\\qquad (r+1)_{r}=(r+1)!\\!.\n\\]\n\nHence  \n\n\\[\nU_{r}(T_{0})=(r)_{r}+(r+1)_{r}\n=r!+(r+1)!=(r+2)\\,r!\n=\\frac{(r+2)!}{\\,r+1}=M_{r}(n). \\tag{2.3}\n\\]\n\n(Thus $T_{0}$ attains the upper bound.)\n\nCreating a $3$–cycle without changing $U_{r}$.  \nReverse the three edges among $P_{0},P_{1},P_{2}$ so that  \n\n\\[\nP_{0}\\to P_{1},\\quad P_{1}\\to P_{2},\\quad P_{2}\\to P_{0}. \\tag{2.4}\n\\]\n\nAll other edges remain as in $T_{0}$; call the new tournament $T$.\n\n• Scores.  \nOriginally the three players had scores $0,1,2$.  After the modification they all have score $1$, hence still strictly less than $r$ (because $r\\ge 3$).  Every other player keeps the same score.  Consequently\n\n\\[\n(s_{0},s_{1},s_{2})\\;:\\;(0,1,2)\\longrightarrow(1,1,1),\n\\]\n\nso $(s_{i})_{r}$ remains $0$ for these three indices; the other $(s_{i})_{r}$’s are unchanged.  Therefore  \n\n\\[\nU_{r}(T)=U_{r}(T_{0})=M_{r}(n). \\tag{2.5}\n\\]\n\n• Presence of a $3$–cycle.  \nThe vertices $P_{0},P_{1},P_{2}$ now form the directed cycle  \n$P_{0}\\to P_{1}\\to P_{2}\\to P_{0}$.\n\nThus, for every $r\\ge 3$ we have produced $n\\ge r+1$ and a tournament $T$ that contains a directed $3$–cycle but still satisfies $U_{r}(T)=M_{r}(n)$, contradicting $(\\star)$.  \n\n--------------------------------------------------------------------\nConclusion.  \n\n$U_{0}$ and $U_{1}$ are constant, while for every $r\\ge 3$ the inequality $(\\star)$ can fail even in the presence of a $3$–cycle.  Hence $r=2$ is the \\emph{unique} index for which the functional $U_{r}$ detects $3$–cycles in all tournaments.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.507447",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order statistic.  \nInstead of the simple quadratic or cubic moment used in the original problems, the new criterion involves the r-th falling factorial (a degree-r polynomial) and works simultaneously for *all* r≥2.  Handling an arbitrary order requires deeper combinatorial identities (formula (2)) and understanding of polynomial convexity.\n\n2. Use of majorisation and Schur-convexity.  \nThe solution demands knowledge of majorisation theory, Landau’s theorem on tournament score sequences, and Schur-convex functions—tools well beyond elementary inequalities.\n\n3. Parameter-dependent proof.  \nBecause r is arbitrary, one cannot rely on ad-hoc edge-flipping calculations tailored to a specific power; instead a general structural argument (majorisation + convexity) is essential.\n\n4. Multiple interacting concepts.  \nThe proof unites combinatorial enumeration (falling factorial sums), discrete convexity, inequality theory, and classical tournament structure theorems, requiring the solver to coordinate several advanced areas simultaneously.\n\nConsequently the enhanced variant is substantially more technical and conceptually demanding than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 3$ be an integer.  \nFor every integer $r\\ge 0$ define the falling factorial  \n\n\\[\n(x)_{r}=x(x-1)\\cdots (x-r+1),\\qquad \n\\text{with the conventions }(x)_{0}=1\\text{ and }(x)_{r}=0\\text{ whenever }x<r.\n\\]\n\nFor a round–robin tournament $T$ on $n$ players (each pair of players meets once, no draws) denote by  \n\n\\[\ns_{1},\\ldots ,s_{n}\n\\]\n\nthe numbers of wins of the players and put  \n\n\\[\nU_{r}(T)=\\sum_{i=1}^{n}(s_{i})_{r}.\n\\]\n\nFurther introduce the constants  \n\n\\[\nM_{r}(n)=\\frac{(n)_{\\,r+1}}{r+1}= \\frac{n(n-1)\\cdots (n-r)}{r+1}.\n\\]\n\n1.  Show that for $r=2$ the following equivalence holds:\n\n\\[\n\\boxed{\\;T\\text{ contains a directed $3$–cycle}\\;\\Longleftrightarrow\\;\nU_{2}(T)<M_{2}(n)=\\dfrac{n(n-1)(n-2)}{3}\\;}. \\tag{$\\star$}\n\\]\n\n2.  Prove that $r=2$ is the only non–negative integer $r$ for which an equivalence of the form $(\\star)$ is valid for every $n\\ge 3$.  More precisely, establish\n\n(a)  $U_{0}(T)$ and $U_{1}(T)$ are constant (hence they cannot detect $3$–cycles);\n\n(b)  for every $r\\ge 3$ there exists an integer $n\\ge r+1$ and a tournament $T$ on $n$ vertices that contains a directed $3$–cycle yet satisfies $U_{r}(T)=M_{r}(n)$, so that $(\\star)$ fails.\n\nConsequently $U_{2}$ is the unique member of the family $U_{r}$ that detects $3$–cycles in tournaments.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Part 1 – The quadratic case $r=2$\n\nStep 1. Landau’s theorem and a majorisation statement.  \nArrange the score sequence in non–decreasing order\n\n\\[\nt_{1}\\le t_{2}\\le\\cdots\\le t_{n}.\n\\]\n\nLandau’s theorem asserts that for every $m=1,\\ldots ,n$\n\n\\[\n\\sum_{i\\le m}t_{i}\\;\\ge\\; \\frac{m(m-1)}{2},\\tag{1.1}\n\\]\n\nwith equality for all $m$ simultaneously if and only if the tournament is transitive (acyclic).\n\nIntroduce the reference vector  \n\n\\[\na=(0,1,\\ldots ,n-1).\\tag{1.2}\n\\]\n\nBecause $\\sum_{i\\le m}a_{i}=m(m-1)/2$, inequality (1.1) rewrites as  \n\n\\[\n\\forall m:\\;\\sum_{i\\le m}t_{i}\\ge\\sum_{i\\le m}a_{i},\n\\qquad\\text{with equality for }m=n. \\tag{1.3}\n\\]\n\nThus, under the \\emph{ascending} convention for majorisation,\n\n\\[\nt\\;\\text{majorises}\\;a,\\qquad\\text{briefly }t\\succ a.\\tag{1.4}\n\\]\n\nStep 2. Schur–concavity of the quadratic falling factorial.  \nLet $g(x)=x(x-1)$.  On $\\mathbf Z$ we have  \n\n\\[\n\\Delta^{2}g(x)=g(x+2)-2g(x+1)+g(x)=2>0,\n\\]\n\nso $g$ is convex.  A classical result (Hardy–Littlewood–Pólya, inequality theory) says that \\emph{for vectors sorted in the same order} a convex symmetric sum is \\emph{Schur–concave} under the \\emph{ascending} convention:  \n\n\\[\nu\\succ v,\\;u\\ne v\\;\\Longrightarrow\\;\\Phi(u)<\\Phi(v), \\tag{1.5}\n\\]\n\nwhere  \n\n\\[\n\\Phi(z_{1},\\ldots ,z_{n})=\\sum_{i=1}^{n}g(z_{i})\n=\\sum_{i=1}^{n}(z_{i})_{2}=U_{2}(T).\n\\]\n\n(The direction of the inequality is reversed compared with the more usual \\emph{descending} convention; that is the only adaptation one must keep in mind.)\n\nStep 3. Extremality of the transitive tournament.  \nApply (1.5) to $u=t$, $v=a$.  When $T$ is not transitive we have $t\\ne a$ and thus  \n\n\\[\nU_{2}(T)=\\Phi(t)<\\Phi(a). \\tag{1.6}\n\\]\n\nIf $T$ \\emph{is} transitive then $t=a$, hence $U_{2}(T)=\\Phi(a)$.\n\nStep 4. Evaluation of $\\Phi(a)$.  \nFor the transitive scores $0,1,\\ldots ,n-1$,\n\n\\[\n\\Phi(a)=\\sum_{k=0}^{n-1}k(k-1)\n=\\frac{n(n-1)(n-2)}{3}=M_{2}(n). \\tag{1.7}\n\\]\n\nStep 5. Equivalence.  \nPutting Step 3 and Step 4 together:\n\n• If $T$ is acyclic, then $t=a$ and $U_{2}(T)=M_{2}(n)$.  \n• If $T$ contains a directed $3$–cycle, it is non–transitive, hence $U_{2}(T)<M_{2}(n)$ by (1.6).\n\nTherefore the equivalence $(\\star)$ holds.\n\n--------------------------------------------------------------------\nPart 2 – Uniqueness of $r=2$\n\n(a) Trivial orders $r=0,1$.  \nBecause every player participates in $\\binom{n-1}{0}=1$ and $\\binom{n-1}{1}=n-1$ matches, respectively,\n\n\\[\nU_{0}(T)=\\sum_{i=1}^{n}(s_{i})_{0}=n,\\qquad \nU_{1}(T)=\\sum_{i=1}^{n}s_{i}=\\binom{n}{2},\n\\]\n\nboth independent of the orientation of the edges.  Hence they cannot distinguish tournaments with or without $3$–cycles.\n\n(b) Counter–examples for every $r\\ge 3$.  \n\nFix $r\\ge 3$ and set  \n\n\\[\nn=r+2\\quad(\\text{so }n\\ge r+1).\\tag{2.1}\n\\]\n\nStart with the \\emph{transitive} tournament $T_{0}$ on the vertex set  \n$P_{0}\\succ P_{1}\\succ\\cdots\\succ P_{n-1}$.  \nIts (ascending) score sequence is  \n\n\\[\n(0,1,2,\\ldots ,n-1)=(0,1,2,\\ldots ,r+1).\\tag{2.2}\n\\]\n\nComputation of $U_{r}(T_{0})$.  \nFor $k<r$ we have $(k)_{r}=0$; for $k=r$ and $k=r+1$ we obtain  \n\n\\[\n(r)_{r}=r!,\\qquad (r+1)_{r}=(r+1)!\\!.\n\\]\n\nHence  \n\n\\[\nU_{r}(T_{0})=(r)_{r}+(r+1)_{r}\n=r!+(r+1)!=(r+2)\\,r!\n=\\frac{(r+2)!}{\\,r+1}=M_{r}(n). \\tag{2.3}\n\\]\n\n(Thus $T_{0}$ attains the upper bound.)\n\nCreating a $3$–cycle without changing $U_{r}$.  \nReverse the three edges among $P_{0},P_{1},P_{2}$ so that  \n\n\\[\nP_{0}\\to P_{1},\\quad P_{1}\\to P_{2},\\quad P_{2}\\to P_{0}. \\tag{2.4}\n\\]\n\nAll other edges remain as in $T_{0}$; call the new tournament $T$.\n\n• Scores.  \nOriginally the three players had scores $0,1,2$.  After the modification they all have score $1$, hence still strictly less than $r$ (because $r\\ge 3$).  Every other player keeps the same score.  Consequently\n\n\\[\n(s_{0},s_{1},s_{2})\\;:\\;(0,1,2)\\longrightarrow(1,1,1),\n\\]\n\nso $(s_{i})_{r}$ remains $0$ for these three indices; the other $(s_{i})_{r}$’s are unchanged.  Therefore  \n\n\\[\nU_{r}(T)=U_{r}(T_{0})=M_{r}(n). \\tag{2.5}\n\\]\n\n• Presence of a $3$–cycle.  \nThe vertices $P_{0},P_{1},P_{2}$ now form the directed cycle  \n$P_{0}\\to P_{1}\\to P_{2}\\to P_{0}$.\n\nThus, for every $r\\ge 3$ we have produced $n\\ge r+1$ and a tournament $T$ that contains a directed $3$–cycle but still satisfies $U_{r}(T)=M_{r}(n)$, contradicting $(\\star)$.  \n\n--------------------------------------------------------------------\nConclusion.  \n\n$U_{0}$ and $U_{1}$ are constant, while for every $r\\ge 3$ the inequality $(\\star)$ can fail even in the presence of a $3$–cycle.  Hence $r=2$ is the \\emph{unique} index for which the functional $U_{r}$ detects $3$–cycles in all tournaments.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.423951",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order statistic.  \nInstead of the simple quadratic or cubic moment used in the original problems, the new criterion involves the r-th falling factorial (a degree-r polynomial) and works simultaneously for *all* r≥2.  Handling an arbitrary order requires deeper combinatorial identities (formula (2)) and understanding of polynomial convexity.\n\n2. Use of majorisation and Schur-convexity.  \nThe solution demands knowledge of majorisation theory, Landau’s theorem on tournament score sequences, and Schur-convex functions—tools well beyond elementary inequalities.\n\n3. Parameter-dependent proof.  \nBecause r is arbitrary, one cannot rely on ad-hoc edge-flipping calculations tailored to a specific power; instead a general structural argument (majorisation + convexity) is essential.\n\n4. Multiple interacting concepts.  \nThe proof unites combinatorial enumeration (falling factorial sums), discrete convexity, inequality theory, and classical tournament structure theorems, requiring the solver to coordinate several advanced areas simultaneously.\n\nConsequently the enhanced variant is substantially more technical and conceptually demanding than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}