{
  "index": "1958-2-B-6",
  "type": "COMB",
  "tag": [
    "COMB"
  ],
  "difficulty": "",
  "question": "6. Let a complete oriented graph on \\( n \\) points be given, i.e., a set of \\( n \\) points \\( 1,2,3, \\ldots, n \\), and between any two points \\( i \\) and \\( j \\) a direction, \\( i \\rightarrow j \\). Show that there exists a permutation of the points, \\( \\left[a_{1}, a_{2}, a_{3}, \\ldots, a_{n}\\right] \\), such that \\( a_{1} \\rightarrow a_{2} \\rightarrow a_{3} \\rightarrow \\cdots \\rightarrow a_{n} \\).",
  "solution": "First Solution. We shall prove this by induction on \\( n \\). It is obviously true for \\( n=2 \\) (and vacuously true for \\( n=1 \\) ). We assume the result for \\( n=1 \\), \\( 2, \\ldots, k \\) and consider a complete oriented graph on \\( k+1 \\) points. Pick any one of these points, say \\( b \\), and consider two subsets of the remaining \\( k \\) points\n\\[\nA=\\{x: x \\rightarrow b\\}, C=\\{x: b \\rightarrow x\\} .\n\\]\n\nThen \\( A \\) and \\( C \\), with the given directions for their pairs, are complete oriented graphs having, say, \\( p \\) and \\( q \\) points, respectively; \\( p+q=k \\). By the induction hypothesis \\( A \\) can be enumerated so that \\( a_{1} \\rightarrow a_{2} \\rightarrow \\cdots \\rightarrow a_{p} \\), and \\( C \\) can be enumerated so that \\( c_{1} \\rightarrow c_{2} \\rightarrow \\cdots \\rightarrow c_{q} \\). Then the required chain is given by\n\\[\na_{1} \\rightarrow a_{2} \\rightarrow \\cdots \\rightarrow a_{p} \\rightarrow b \\rightarrow c_{1} \\rightarrow c_{2} \\rightarrow \\cdots c_{q} .\n\\]\n[Note that either \\( A \\) or \\( C \\) might be empty, but this creates no difficulty.]\nSecond Solution. Again assume that the result is true for \\( n=k \\) points, and let a complete oriented graph \\( G \\) on \\( k+1 \\) points be given. Pick any point \\( b \\) of \\( G \\). By the inductive hypothesis, the remaining \\( k \\) points can be labeled \\( a_{1}, a_{2}, \\ldots, a_{k} \\) so that\n\\[\na_{1} \\rightarrow a_{2} \\rightarrow \\cdots \\rightarrow a_{k} .\n\\]\n\nThen \\( b \\) can be fitted into this sequence either just before \\( a_{i} \\), where \\( i \\) is the least index with \\( b \\rightarrow a_{i} \\), or at the end of the sequence if no such index exists.",
  "vars": [
    "i",
    "j",
    "b",
    "x",
    "A",
    "C",
    "G",
    "a_1",
    "a_2",
    "a_3",
    "a_n",
    "a_p",
    "a_k",
    "c_1",
    "c_2",
    "c_q"
  ],
  "params": [
    "n",
    "k",
    "p",
    "q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "i": "indexvar",
        "j": "secondindex",
        "b": "pivotpt",
        "x": "genericpt",
        "A": "setalpha",
        "C": "setgamma",
        "G": "fullgraph",
        "a_1": "seqfirst",
        "a_2": "seqsecond",
        "a_3": "seqthird",
        "a_n": "seqlast",
        "a_p": "seqppoint",
        "a_k": "seqkpoint",
        "c_1": "seqcfirst",
        "c_2": "seqcsecond",
        "c_q": "seqclast",
        "n": "totalpts",
        "k": "reducedpts",
        "p": "sizealpha",
        "q": "sizegamma"
      },
      "question": "6. Let a complete oriented graph on \\( totalpts \\) points be given, i.e., a set of \\( totalpts \\) points \\( 1,2,3, \\ldots, totalpts \\), and between any two points \\( indexvar \\) and \\( secondindex \\) a direction, \\( indexvar \\rightarrow secondindex \\). Show that there exists a permutation of the points, \\( \\left[seqfirst, seqsecond, seqthird, \\ldots, seqlast\\right] \\), such that \\( seqfirst \\rightarrow seqsecond \\rightarrow seqthird \\rightarrow \\cdots \\rightarrow seqlast \\).",
      "solution": "First Solution. We shall prove this by induction on \\( totalpts \\). It is obviously true for \\( totalpts=2 \\) (and vacuously true for \\( totalpts=1 \\) ). We assume the result for \\( totalpts=1 \\), \\( 2, \\ldots, reducedpts \\) and consider a complete oriented graph on \\( reducedpts+1 \\) points. Pick any one of these points, say pivotpt, and consider two subsets of the remaining \\( reducedpts \\) points\n\\[\nsetalpha=\\{genericpt: genericpt \\rightarrow pivotpt\\}, \\; setgamma=\\{genericpt: pivotpt \\rightarrow genericpt\\} .\n\\]\n\nThen \\( setalpha \\) and \\( setgamma \\), with the given directions for their pairs, are complete oriented graphs having, say, \\( sizealpha \\) and \\( sizegamma \\) points, respectively; \\( sizealpha+sizegamma=reducedpts \\). By the induction hypothesis \\( setalpha \\) can be enumerated so that \\( seqfirst \\rightarrow seqsecond \\rightarrow \\cdots \\rightarrow seqppoint \\), and \\( setgamma \\) can be enumerated so that \\( seqcfirst \\rightarrow seqcsecond \\rightarrow \\cdots \\rightarrow seqclast \\). Then the required chain is given by\n\\[\nseqfirst \\rightarrow seqsecond \\rightarrow \\cdots \\rightarrow seqppoint \\rightarrow pivotpt \\rightarrow seqcfirst \\rightarrow seqcsecond \\rightarrow \\cdots seqclast .\n\\]\n[Note that either \\( setalpha \\) or \\( setgamma \\) might be empty, but this creates no difficulty.]\n\nSecond Solution. Again assume that the result is true for \\( totalpts=reducedpts \\) points, and let a complete oriented graph \\( fullgraph \\) on \\( reducedpts+1 \\) points be given. Pick any point pivotpt of \\( fullgraph \\). By the inductive hypothesis, the remaining \\( reducedpts \\) points can be labeled \\( seqfirst, seqsecond, \\ldots, seqkpoint \\) so that\n\\[\nseqfirst \\rightarrow seqsecond \\rightarrow \\cdots \\rightarrow seqkpoint .\n\\]\n\nThen pivotpt can be fitted into this sequence either just before \\( a_{indexvar} \\), where \\( indexvar \\) is the least index with \\( pivotpt \\rightarrow a_{indexvar} \\), or at the end of the sequence if no such index exists."
    },
    "descriptive_long_confusing": {
      "map": {
        "j": "whimsyjar",
        "b": "lanternfox",
        "x": "cobblestone",
        "A": "rainbowkey",
        "C": "puzzlemint",
        "G": "alchemyrope",
        "a_1": "honeydew",
        "a_2": "blackwatch",
        "a_3": "silhouette",
        "a_n": "amberglow",
        "a_p": "waterlily",
        "a_k": "starlancer",
        "c_1": "pineconelm",
        "c_2": "dandelion",
        "c_q": "butterscotch",
        "n": "caterpillar",
        "k": "wanderlust",
        "p": "snowpebble",
        "q": "thunderclap",
        "i": "i"
      },
      "question": "6. Let a complete oriented graph on \\( caterpillar \\) points be given, i.e., a set of \\( caterpillar \\) points \\( 1,2,3, \\ldots, caterpillar \\), and between any two points \\( i \\) and \\( whimsyjar \\) a direction, \\( i \\rightarrow whimsyjar \\). Show that there exists a permutation of the points, \\( \\left[honeydew, blackwatch, silhouette, \\ldots, amberglow\\right] \\), such that \\( honeydew \\rightarrow blackwatch \\rightarrow silhouette \\rightarrow \\cdots \\rightarrow amberglow \\).",
      "solution": "First Solution. We shall prove this by induction on \\( caterpillar \\). It is obviously true for \\( caterpillar=2 \\) (and vacuously true for \\( caterpillar=1 \\) ). We assume the result for \\( caterpillar=1 \\), 2, \\ldots, wanderlust and consider a complete oriented graph on \\( wanderlust+1 \\) points. Pick any one of these points, say \\( lanternfox \\), and consider two subsets of the remaining \\( wanderlust \\) points\n\\[\nrainbowkey=\\{cobblestone: cobblestone \\rightarrow lanternfox\\}, \\quad puzzlemint=\\{cobblestone: lanternfox \\rightarrow cobblestone\\} .\n\\]\n\nThen \\( rainbowkey \\) and \\( puzzlemint \\), with the given directions for their pairs, are complete oriented graphs having, say, \\( snowpebble \\) and \\( thunderclap \\) points, respectively; \\( snowpebble+thunderclap=wanderlust \\). By the induction hypothesis \\( rainbowkey \\) can be enumerated so that \\( honeydew \\rightarrow blackwatch \\rightarrow \\cdots \\rightarrow waterlily \\), and \\( puzzlemint \\) can be enumerated so that \\( pineconelm \\rightarrow dandelion \\rightarrow \\cdots \\rightarrow butterscotch \\). Then the required chain is given by\n\\[\nhoneydew \\rightarrow blackwatch \\rightarrow \\cdots \\rightarrow waterlily \\rightarrow lanternfox \\rightarrow pineconelm \\rightarrow dandelion \\rightarrow \\cdots butterscotch .\n\\]\n[Note that either \\( rainbowkey \\) or \\( puzzlemint \\) might be empty, but this creates no difficulty.]\n\nSecond Solution. Again assume that the result is true for \\( caterpillar=wanderlust \\) points, and let a complete oriented graph \\( alchemyrope \\) on \\( wanderlust+1 \\) points be given. Pick any point \\( lanternfox \\) of \\( alchemyrope \\). By the inductive hypothesis, the remaining \\( wanderlust \\) points can be labeled \\( honeydew, blackwatch, \\ldots, starlancer \\) so that\n\\[\nhoneydew \\rightarrow blackwatch \\rightarrow \\cdots \\rightarrow starlancer .\n\\]\n\nThen \\( lanternfox \\) can be fitted into this sequence either just before \\( a_{i} \\), where \\( i \\) is the least index with \\( lanternfox \\rightarrow a_{i} \\), or at the end of the sequence if no such index exists."
    },
    "descriptive_long_misleading": {
      "map": {
        "i": "terminalindex",
        "j": "initialindex",
        "b": "peripheralpoint",
        "x": "specificpoint",
        "A": "outwardset",
        "C": "inwardset",
        "G": "incompletegraph",
        "a_1": "finalfirst",
        "a_2": "finalsecond",
        "a_3": "finalthird",
        "a_n": "finalgeneral",
        "a_p": "finalpindex",
        "a_k": "finalkindex",
        "c_1": "originfirst",
        "c_2": "originsecond",
        "c_q": "originqindex",
        "n": "fewpoints",
        "k": "extrapoints",
        "p": "missingcount",
        "q": "excesscount"
      },
      "question": "6. Let a complete oriented graph on \\( fewpoints \\) points be given, i.e., a set of \\( fewpoints \\) points 1,2,3, \\ldots, fewpoints, and between any two points \\( terminalindex \\) and \\( initialindex \\) a direction, \\( terminalindex \\rightarrow initialindex \\). Show that there exists a permutation of the points, \\( \\left[ finalfirst, finalsecond, finalthird, \\ldots, finalgeneral \\right] \\), such that \\( finalfirst \\rightarrow finalsecond \\rightarrow finalthird \\rightarrow \\cdots \\rightarrow finalgeneral \\).",
      "solution": "First Solution. We shall prove this by induction on \\( fewpoints \\). It is obviously true for \\( fewpoints=2 \\) (and vacuously true for \\( fewpoints=1 \\)). We assume the result for \\( fewpoints=1, 2, \\ldots, extrapoints \\) and consider a complete oriented graph on \\( extrapoints+1 \\) points. Pick any one of these points, say \\( peripheralpoint \\), and consider two subsets of the remaining \\( extrapoints \\) points\n\\[\noutwardset=\\{specificpoint: specificpoint \\rightarrow peripheralpoint\\}, \\quad inwardset=\\{specificpoint: peripheralpoint \\rightarrow specificpoint\\} .\n\\]\n\nThen \\( outwardset \\) and \\( inwardset \\), with the given directions for their pairs, are complete oriented graphs having, say, \\( missingcount \\) and \\( excesscount \\) points, respectively; \\( missingcount+excesscount=extrapoints \\). By the induction hypothesis \\( outwardset \\) can be enumerated so that \\( finalfirst \\rightarrow finalsecond \\rightarrow \\cdots \\rightarrow finalpindex \\), and \\( inwardset \\) can be enumerated so that \\( originfirst \\rightarrow originsecond \\rightarrow \\cdots \\rightarrow originqindex \\). Then the required chain is given by\n\\[\nfinalfirst \\rightarrow finalsecond \\rightarrow \\cdots \\rightarrow finalpindex \\rightarrow peripheralpoint \\rightarrow originfirst \\rightarrow originsecond \\rightarrow \\cdots originqindex .\n\\]\n[Note that either \\( outwardset \\) or \\( inwardset \\) might be empty, but this creates no difficulty.]\n\nSecond Solution. Again assume that the result is true for \\( fewpoints=extrapoints \\) points, and let a complete oriented graph \\( incompletegraph \\) on \\( extrapoints+1 \\) points be given. Pick any point \\( peripheralpoint \\) of \\( incompletegraph \\). By the inductive hypothesis, the remaining \\( extrapoints \\) points can be labeled \\( finalfirst, finalsecond, \\ldots, finalkindex \\) so that\n\\[\nfinalfirst \\rightarrow finalsecond \\rightarrow \\cdots \\rightarrow finalkindex .\n\\]\n\nThen \\( peripheralpoint \\) can be fitted into this sequence either just before \\( a_{terminalindex} \\), where \\( terminalindex \\) is the least index with \\( peripheralpoint \\rightarrow a_{terminalindex} \\), or at the end of the sequence if no such index exists."
    },
    "garbled_string": {
      "map": {
        "i": "qzxwvtnp",
        "j": "hjgrksla",
        "b": "plsnvwrq",
        "x": "mfgdrazt",
        "A": "tnqpslze",
        "C": "xwrtghqa",
        "G": "zmcxfrbe",
        "a_1": "ldkqmpse",
        "a_2": "wrvqkdsn",
        "a_3": "tvhzwlga",
        "a_n": "sdfjklqe",
        "a_p": "cvnrtxay",
        "a_k": "gltpwsue",
        "c_1": "bpqdrnfa",
        "c_2": "jzmbtvol",
        "c_q": "vxctlmow",
        "n": "fgrstlpa",
        "k": "whdlmpor",
        "p": "zxndtqre",
        "q": "kphqsdne"
      },
      "question": "6. Let a complete oriented graph on \\( fgrstlpa \\) points be given, i.e., a set of \\( fgrstlpa \\) points \\( 1,2,3, \\ldots, fgrstlpa \\), and between any two points \\( qzxwvtnp \\) and \\( hjgrksla \\) a direction, \\( qzxwvtnp \\rightarrow hjgrksla \\). Show that there exists a permutation of the points, \\( \\left[ldkqmpse, wrvqkdsn, tvhzwlga, \\ldots, sdfjklqe\\right] \\), such that \\( ldkqmpse \\rightarrow wrvqkdsn \\rightarrow tvhzwlga \\rightarrow \\cdots \\rightarrow sdfjklqe \\).",
      "solution": "First Solution. We shall prove this by induction on \\( fgrstlpa \\). It is obviously true for \\( fgrstlpa=2 \\) (and vacuously true for \\( fgrstlpa=1 \\) ). We assume the result for \\( fgrstlpa=1 \\), 2, \\ldots, whdlmpor and consider a complete oriented graph on \\( whdlmpor+1 \\) points. Pick any one of these points, say \\( plsnvwrq \\), and consider two subsets of the remaining \\( whdlmpor \\) points\n\\[\ntnqpslze=\\{mfgdrazt: mfgdrazt \\rightarrow plsnvwrq\\}, \\quad xwrtghqa=\\{mfgdrazt: plsnvwrq \\rightarrow mfgdrazt\\} .\n\\]\nThen \\( tnqpslze \\) and \\( xwrtghqa \\), with the given directions for their pairs, are complete oriented graphs having, say, \\( zxndtqre \\) and \\( kphqsdne \\) points, respectively; \\( zxndtqre+kphqsdne=whdlmpor \\). By the induction hypothesis \\( tnqpslze \\) can be enumerated so that \\( ldkqmpse \\rightarrow wrvqkdsn \\rightarrow \\cdots \\rightarrow cvnrtxay \\), and \\( xwrtghqa \\) can be enumerated so that \\( bpqdrnfa \\rightarrow jzmbtvol \\rightarrow \\cdots \\rightarrow vxctlmow \\). Then the required chain is given by\n\\[\nldkqmpse \\rightarrow wrvqkdsn \\rightarrow \\cdots \\rightarrow cvnrtxay \\rightarrow plsnvwrq \\rightarrow bpqdrnfa \\rightarrow jzmbtvol \\rightarrow \\cdots vxctlmow .\n\\]\n[Note that either \\( tnqpslze \\) or \\( xwrtghqa \\) might be empty, but this creates no difficulty.]\n\nSecond Solution. Again assume that the result is true for \\( fgrstlpa=whdlmpor \\) points, and let a complete oriented graph \\( zmcxfrbe \\) on \\( whdlmpor+1 \\) points be given. Pick any point \\( plsnvwrq \\) of \\( zmcxfrbe \\). By the inductive hypothesis, the remaining \\( whdlmpor \\) points can be labeled ldkqmpse, wrvqkdsn, \\ldots, gltpwsue so that\n\\[\nldkqmpse \\rightarrow wrvqkdsn \\rightarrow \\cdots \\rightarrow gltpwsue .\n\\]\nThen \\( plsnvwrq \\) can be fitted into this sequence either just before \\( a_{qzxwvtnp} \\), where \\( qzxwvtnp \\) is the least index with \\( plsnvwrq \\rightarrow a_{qzxwvtnp} \\), or at the end of the sequence if no such index exists."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 3$ be an integer.  \n\nAn airline serves $n$ distinct cities.  \nFor every unordered pair of different cities exactly one of the two possible {\\em directed} non-stop flights is scheduled; the resulting flight map is therefore a tournament $T$.  \nAssume moreover that $T$ is {\\em strongly connected}, i.e.\\ from every city one can reach any other city by some directed itinerary.  \n\na)  (Camion 1959)  Prove that $T$ possesses a Hamiltonian directed cycle; equivalently, the cities can be cyclically listed  \n\\[\n[c_{1},c_{2},\\dots ,c_{n}]\n\\]\nso that each of the flights  \n\\[\nc_{1}\\to c_{2}\\to\\dots\\to c_{n}\\to c_{1}\n\\]\nis scheduled.  \n\nb)  (Moon 1968)  Prove the {\\em vertex–pancyclic} strengthening:  \nfor every integer $\\ell$ with $3\\le\\ell\\le n$ and for every chosen city $v$ the tournament $T$ contains a directed cycle of length $\\ell$ that starts (and hence also ends) at $v$.  \n\nYou may freely quote Rédei’s Hamiltonian–path theorem, but every additional fact you use has to be proved within your solution.  \n\nIndices on cycles are read modulo the length of the cycle; hence on a $k$-cycle the successor of $c_{k}$ is $c_{1}$.",
      "solution": "Throughout “vertices’’ $=$ cities and “arcs’’ $=$ scheduled flights.  \n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n1.\\  A rotation tool for Hamiltonian paths  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\nLemma 1 (elementary rotation with the least out-neighbour).  \nLet  \n\\[\nP:=v_{1}\\to v_{2}\\to\\dots\\to v_{n}\\qquad (n\\ge 2)\n\\tag{1}\n\\]\nbe a Hamiltonian path in a tournament.  \nAssume that the last vertex $v_{n}$ has an arc to some $v_{i}$ ($1\\le i\\le n-1$) and that $i$ is the {\\em least}\nindex with this property.  Then the reordered list  \n\\[\nv_{1}\\to v_{2}\\to\\dots\\to v_{i-1}\\to v_{n}\\to v_{i}\\to v_{i+1}\\to\\dots\\to v_{n-1}\n\\tag{2}\n\\]\nis again a Hamiltonian path.  \n\nProof.  \nThe list (2) visits every vertex exactly once, so we only have to verify that the displayed arcs point forward.  \n\n$\\bullet$  The arc $v_{i-1}\\to v_{n}$ exists because, by the minimality of $i$, the opposite orientation $v_{n}\\to v_{i-1}$ is impossible.  \n\n$\\bullet$  The arc $v_{n}\\to v_{i}$ exists by assumption; every other arc is inherited from (1).  \n\nThus (2) is a Hamiltonian path.  \\hfill$\\square$  \n\nCorollary 2 (the last vertex moves left).  \nAfter a single rotation the old last vertex $v_{n}$ now occupies position $i(<n)$.  \\hfill$\\square$  \n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n2.\\  Camion’s Hamiltonian-cycle theorem – part (a)  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\nProposition 3.  Every strongly connected tournament on $n\\ge 3$ vertices contains a Hamiltonian directed cycle.  \n\nProof.  \n\nStep 1 – start with a Hamiltonian path.  \nRédei’s theorem supplies a Hamiltonian path  \n\\[\nP_{0}:=v_{1}\\to v_{2}\\to\\dots\\to v_{n}.\n\\tag{3}\n\\]\nIf the closing arc $v_{n}\\to v_{1}$ already exists we are done, so assume $v_{1}\\to v_{n}$.  \n\nStep 2 – rotate until the final vertex dominates the first one.  \nBecause $T$ is strongly connected, the last vertex $v_{n}$ has an out-neighbour among $\\{v_{1},\\dots ,v_{n-1}\\}$.  \nLet $i_{1}$ be the least such index and rotate once according to Lemma 1, obtaining a new Hamiltonian path whose last vertex is $v_{n-1}$ (Corollary 2).  \n\nIterate: after $k$ rotations the last vertex is $v_{n-k}$.  \nIf the procedure never stops, then for every $k$ we have $v_{1}\\to v_{n-k}$, hence $v_{1}$ sends arcs to all other vertices.  \nConsequently $\\operatorname{indeg}(v_{1})=0$, contradicting strong connectivity.  \nTherefore after at most $n-2$ rotations we arrive at a Hamiltonian path  \n\\[\nu_{1}\\to u_{2}\\to\\dots\\to u_{n}\n\\tag{4}\n\\]\nwith $u_{n}\\to u_{1}$.  \n\nStep 3 – close the cycle.  \nAdding the arc $u_{n}\\to u_{1}$ to (4) yields the Hamiltonian cycle  \n\\[\nu_{1}\\to u_{2}\\to\\dots\\to u_{n}\\to u_{1}.\n\\]\n\\hfill$\\square$  \n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n3.\\  Preparatory lemmas for part (b)  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\nLemma 4 (directed triangle through a prescribed vertex).  \nLet $v$ be any vertex of a strongly connected tournament $T$ with $n\\ge 3$.  \nThen $v$ lies on a directed $3$-cycle.  \n\nProof.  \nBecause $T$ is strong, $v$ has at least one out-neighbour $w$ and at least one in-neighbour $u$.  \nExactly one of $w\\to u$ or $u\\to w$ holds.  \n\n$\\bullet$  If $w\\to u$, then $v\\to w\\to u\\to v$ is a directed $3$-cycle containing $v$.  \n\n$\\bullet$  If $u\\to w$, then $u\\to v\\to w\\to u$ is a directed $3$-cycle containing $v$.  \n\n\\hfill$\\square$  \n\nLemma 5 (existence of a $4$-cycle through any vertex).  \nLet $v$ be any vertex of a strongly connected tournament $T$ with $n\\ge 4$.  \nThen $v$ lies on a directed $4$-cycle.  \n\nProof.  \nBy Lemma 4 choose a directed triangle  \n\\[\nC_{3}:=v\\to a\\to b\\to v.\n\\]\nLet $W:=V(T)\\setminus\\{v,a,b\\}$.  \n\nCase 1: Some vertex $x\\in W$ has both an in-neighbour and an out-neighbour on $C_{3}$.  \nBecause the vertices of $C_{3}$ appear in cyclic order, there are two {\\em consecutive} vertices $c_{j},c_{j+1}$ on $C_{3}$ with $c_{j}\\to x\\to c_{j+1}$ (a justification identical to that given at the start of Lemma 6(i) below).  \nInserting $x$ between $c_{j}$ and $c_{j+1}$ yields a $4$-cycle through $v$.  \n\nCase 2: Every vertex of $W$ is either dominated by all of $C_{3}$ or dominates all of $C_{3}$.  \nDefine  \n\\[\nR:=\\{x\\in W\\mid c\\to x\\ \\text{for all}\\ c\\in C_{3}\\},\\qquad \nS:=\\{x\\in W\\mid x\\to c\\ \\text{for all}\\ c\\in C_{3}\\}.\n\\]\n(The sets $R$ and $S$ are disjoint.)  \n\nSubcase 2a: $|W|=1$.  \nThen $W=\\{x\\}$ and either $x\\in R$ or $x\\in S$.  \nSuppose $x\\in R$.  \nThe triangle $C_{3}$ reaches $x$, but $x$ cannot reach $C_{3}$ because all three arcs are directed toward $x$.  \nThis violates strong connectivity.  \nAn identical contradiction arises when $x\\in S$.  \n\nSubcase 2b: $|W|\\ge 2$.  \nStrong connectivity implies that both $R$ and $S$ are non-empty.  \nBecause the subtournament induced by $R\\cup S$ is complete, there exist $r\\in R$ and $s\\in S$ with either $r\\to s$ or $s\\to r$.  \nSuppose $r\\to s$; we then have  \n\\[\nv\\to a\\to r\\to s\\to v,\n\\]\na $4$-cycle through $v$.  \nIf $s\\to r$ we instead use $v\\to s\\to r\\to b\\to v$.  \n\nTherefore a $4$-cycle containing $v$ exists in all cases. \\hfill$\\square$  \n\n\n\nLemma 6 (cycle extension by one or two vertices).  \nLet  \n\\[\nC:=c_{0}\\,(=v)\\to c_{1}\\to\\dots\\to c_{t-1}\\to c_{0}\n\\qquad (3\\le t<n)\n\\tag{5}\n\\]\nbe a directed cycle in a strongly connected tournament $T$.  Denote $V(C)$ by $C$ for brevity and put  \n\\[\nR:=\\{x\\notin C\\mid c\\to x\\ \\text{for every}\\ c\\in C\\},\\qquad\nS:=\\{x\\notin C\\mid x\\to c\\ \\text{for every}\\ c\\in C\\}.\n\\]\n(The sets $R$ and $S$ are clearly disjoint.)  \n\n(i)  If there exists a vertex $w\\notin C$ that has both an in-neighbour and an out-neighbour on $C$, then one may insert $w$ between suitable {\\em consecutive} vertices of $C$ and obtain a directed cycle of length $t+1$ still containing $v$.  \n\n(ii) If no such vertex exists (hence every vertex outside $C$ lies either in $R$ or in $S$), then $R$ and $S$ are both non-empty, and there exist $r\\in R$, $s\\in S$ and an index $j$ such that  \n\\[\nc_{j}\\to r\\to s\\to c_{j+1}.\n\\tag{6}\n\\]\nConsequently  \n\\[\nc_{0}\\to\\dots\\to c_{j}\\to r\\to s\\to c_{j+1}\\to\\dots\\to c_{t-1}\\to c_{0}\n\\]\nis a directed cycle of length $t+2$ that still contains $v$.  \n\nProof.  \n\n(i)  Let $w$ have an in-neighbour $c_{i}$ and an out-neighbour $c_{k}$ on $C$.  \nProceed clockwise round $C$ starting with $c_{i}$ until the first vertex  \n\\[\nc_{j}\\ \\text{such that}\\ w\\to c_{j}\n\\]\nis encountered.  \n(The vertex $c_{i}$ itself does {\\em not} satisfy $w\\to c_{i}$, so such a $c_{j}$ exists after at most $t$ steps.)  \nLet $c_{j-1}$ be the predecessor of $c_{j}$ on $C$.  \nThen $c_{j-1}\\to w$ and $w\\to c_{j}$, while all other adjacencies of $C$ are unchanged, giving the desired $(t+1)$-cycle.  \n\n(ii)  Assume every vertex outside $C$ belongs to $R\\cup S$.  \nBoth $R$ and $S$ must be non-empty; if, say, $R=\\varnothing$, then every vertex outside $C$ dominates $C$, so there is no directed path from $C$ to those vertices, contradicting strong connectivity.  \n\nBecause the subtournament induced by $R\\cup S$ is complete, there exists at least one ordered pair $(r,s)\\in R\\times S$ with $r\\to s$; otherwise every arc between $R$ and $S$ would be oriented $S\\to R$, again preventing vertices of $R$ from reaching $C$.  \nFix such an ordered pair $(r,s)$ and fix any index $j$ (indices are taken modulo $t$).  \nBy definition of $R$ and $S$ we have $c_{j}\\to r$ and $s\\to c_{j+1}$.  \nWith $r\\to s$ this yields the chain (6), completing the construction.  \\hfill$\\square$  \n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n4.\\  Vertex-pancyclicity – part (b)  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\nTheorem 7 (Moon 1968).  \nLet $T$ be a strongly connected tournament on $n$ vertices and let $v$ be any vertex of $T$.  \nFor every $\\ell$ with $3\\le\\ell\\le n$ there exists a directed cycle of length $\\ell$ that starts at $v$.  \n\nProof.  \n\nStep 0 – initial stock of cycles.  \nBy Proposition 3 the tournament contains a Hamiltonian cycle, so there exists a cycle of length $n$ through $v$.  \nLemmas 4 and 5 give cycles of lengths $3$ and $4$ through $v$.  \n\nStep 1 – an iterative extension procedure.  \nSuppose $C$ is a directed cycle through $v$ of length $t$ with $3\\le t\\le n-1$.  \nApply Lemma 6.\n\n$\\bullet$  If the situation (i) of Lemma 6 occurs, $C$ grows to length $t+1$.  \n\n$\\bullet$  Otherwise we are in situation (ii) and $C$ grows to length $t+2$.  \n\nThus starting from any available length, a $v$-cycle can be enlarged by $+1$ or $+2$ until it includes all vertices.  \n\nStep 2 – reaching all {\\em odd} lengths.  \nBeginning with length $3$ and repeatedly using the above procedure we obtain cycles of lengths  \n\\[\n3,\\,5,\\,7,\\,\\dots\n\\]\nuntil we reach $n$ or $n-1$ (depending on the parity of $n$).  \nA sporadic “\\,+1’’ extension, if it ever occurs, merely inserts an additional odd length and never creates a gap.  \nHence every odd $\\ell$ with $3\\le\\ell\\le n$ occurs.  \n\nStep 3 – reaching all {\\em even} lengths.  \nThe $4$-cycle obtained in Lemma 5 supplies the first even length.  \nApplying the same extension procedure to this cycle yields even lengths  \n\\[\n4,\\,6,\\,8,\\,\\dots\n\\]\nuntil we reach $n$ or $n-1$, and again any occasional “\\,+1’’ only fills extra values.  \nThus every even $\\ell$ with $4\\le\\ell\\le n$ occurs.  \n\nStep 4 – completeness.  \nCombining Steps 2 and 3 we have produced, for the prescribed vertex $v$, directed cycles of every length $\\ell$ with $3\\le\\ell\\le n$.  \\hfill$\\square$  \n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n5.\\  Concluding remarks on optimality  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\nMoon’s theorem is best possible in the sense that strong connectivity is indispensable.  \nA transitive tournament has no directed cycle at all, while a tournament that is strong but not $2$-strong need not be {\\em edge-}pancyclic (every arc contained in cycles of all lengths); both phenomena occur already for $n=4$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.500125",
        "was_fixed": false,
        "difficulty_analysis": "•  The original problem asks only for one Hamiltonian path; Rédei’s induction suffices.  \n•  Part (a) of the enhanced variant requires creating a Hamiltonian *cycle*.  Closing a path into a cycle forces one to exploit strong connectivity, develop an insertion algorithm, and prove its termination—Camion’s theorem.  \n•  Part (b) goes further, asking for cycles of *every* length through *every* vertex (pancyclicity).  This needs new structure (the chord lemma), iterative surgery on cycles, and careful bookkeeping to guarantee termination without repetitions.  \n•  These tasks integrate several advanced tools: Rédei’s ordering, strong–connectivity arguments, iterative edge-reversal (“insertion’’) techniques, and constructive shortening, all well beyond the elementary induction of the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 3$ be an integer.  \n\nAn airline serves $n$ distinct cities.  For every unordered pair of different cities exactly one of the two non–stop flight directions is scheduled; the resulting flight map is therefore a tournament $T$.  Assume moreover that $T$ is strongly connected – from every city one can reach any other one by a directed itinerary.\n\na)  (Camion 1959)  Prove that $T$ possesses a Hamiltonian directed cycle; equivalently, the cities can be cyclically listed  \n\\[\n[c_{1},c_{2},\\dots ,c_{n}]\n\\]\nso that the flights  \n\\[\nc_{1}\\to c_{2}\\to\\dots\\to c_{n}\\to c_{1}\n\\]\nare all scheduled.\n\nb)  (Moon 1968)  Prove the vertex–pancyclic strengthening:  \nfor every integer $\\ell$ with $3\\le\\ell\\le n$ and for every chosen city $v$ the tournament $T$ contains a directed cycle of length $\\ell$ that starts (and hence also ends) at $v$.\n\nYou may freely quote Rédei’s Hamiltonian–path theorem, but every additional fact you use has to be proved inside your solution.\n\nIndices on cycles are read modulo the length of the cycle, hence on a $k$–cycle the successor of $c_{k}$ is $c_{1}$.",
      "solution": "Throughout, “vertices’’ $=$ cities and “arcs’’ $=$ scheduled flights.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.\\  A rotation tool for Hamiltonian paths\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nLemma&nbsp;1 (elementary rotation with the least out–neighbour).  \nLet  \n\\[\nP:=v_{1}\\to v_{2}\\to\\dots\\to v_{n}\\qquad (n\\ge 2)\n\\tag{1}\n\\]\nbe a Hamiltonian path in a tournament.  \nAssume that the last vertex $v_{n}$ has an arc to some $v_{i}$ ($1\\le i\\le n-1$) and that $i$ is the least\nindex with this property.  Then the reordered list  \n\\[\nv_{1}\\to v_{2}\\to\\dots\\to v_{i-1}\\to v_{n}\\to v_{i}\\to v_{i+1}\\to\\dots\\to v_{n-1}\n\\tag{2}\n\\]\nis again a Hamiltonian path.\n\nProof.  \nThe list (2) visits every vertex exactly once, so we only have to verify that all displayed arcs point forward.\n\n$\\bullet$  The arc $v_{i-1}\\to v_{n}$ exists because, by the minimality of $i$, the orientation between $v_{i-1}$ and $v_{n}$ cannot be $v_{n}\\to v_{i-1}$.\n\n$\\bullet$  The arc $v_{n}\\to v_{i}$ exists by assumption; every other arc is inherited from (1).\n\nThus (2) is a Hamiltonian path.  $\\square$\n\n\n\nCorollary&nbsp;2 (the last vertex moves left).  \nAfter a single rotation the old last vertex $v_{n}$ now occupies position $i(<n)$.  $\\square$\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n2.\\  Camion’s Hamiltonian–cycle theorem – part (a)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nProposition&nbsp;3.  Every strongly connected tournament on $n\\ge 3$ vertices contains a Hamiltonian directed cycle.\n\nProof.  \n\nStep&nbsp;1 – start with a Hamiltonian path.  \nRédei’s theorem supplies a Hamiltonian path  \n\\[\nP_{0}:=v_{1}\\to v_{2}\\to\\dots\\to v_{n}.\n\\tag{3}\n\\]\nIf the closing arc $v_{n}\\to v_{1}$ already exists we are done, so assume $v_{1}\\to v_{n}$.\n\nStep&nbsp;2 – rotate until the final vertex dominates the first one.  \nBecause $T$ is strongly connected, the last vertex $v_{n}$ has an out–neighbour among $\\{v_{1},\\dots ,v_{n-1}\\}$.  \nLet $i_{1}$ be the least such index and rotate once according to Lemma&nbsp;1, obtaining a new Hamiltonian path whose last vertex is $v_{n-1}$ (Corollary&nbsp;2).  \n\nIterate the procedure: after $k$ rotations the last vertex is $v_{n-k}$, so after at most $n-2$ rotations we reach a Hamiltonian path  \n\\[\nu_{1}\\to u_{2}\\to\\dots\\to u_{n}\n\\tag{4}\n\\]\nsuch that $u_{n}\\to u_{1}$; otherwise all arcs between $u_{1}$ and earlier last vertices would have the fixed orientation $u_{1}\\to u_{i}$ for $i\\ge 2$, giving $u_{1}$\noutdegree $n-1$ and indegree $0$, contradicting strong connectivity.  Hence the process must stop in at most $n-2$ steps.\n\nStep&nbsp;3 – close the cycle.  \nAdding the arc $u_{n}\\to u_{1}$ to (4) yields the Hamiltonian cycle  \n\\[\nu_{1}\\to u_{2}\\to\\dots\\to u_{n}\\to u_{1}.\n\\]\n$\\square$\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n3.\\  Three preparatory lemmas for part (b)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nLemma&nbsp;4 (triangle through a prescribed vertex).  \nLet $v$ be any vertex of a strongly connected tournament $T$ with $n\\ge 3$.  \nThen $v$ lies on a directed $3$–cycle.\n\nProof.  \nBecause $T$ is strong, $v$ has at least one out–neighbour $w$ and at least one in–neighbour $u$.  \nExactly one of $w\\to u$ or $u\\to w$ holds.\n\n$\\bullet$  If $w\\to u$, then $v\\to w\\to u\\to v$ is a directed $3$–cycle containing $v$.\n\n$\\bullet$  If $u\\to w$, then $u\\to v\\to w\\to u$ is a directed $3$–cycle containing $v$.\n\n$\\square$\n\n\n\nLemma&nbsp;5 (vertex with mixed adjacency to a cycle).  \nLet $C:=c_{0}\\to c_{1}\\to\\dots\\to c_{t-1}\\to c_{0}$ be a directed cycle in a strongly connected tournament $T$ which does not cover all vertices.  \nThen there exists a vertex $z\\notin V(C)$ that has both an in–neighbour and an out–neighbour on $C$.\n\nProof.  \nBecause $T$ is strongly connected, there exists at least one vertex outside $C$ that is entered by some vertex of $C$.  \nAmong all vertices $x\\notin V(C)$ that receive at least one arc from $C$ choose $z$ so that the tail  \nindex of the {\\em last} such arc along $C$ is maximal.  Concretely, let  \n\\[\nJ(x):=\\{\\,j\\mid c_{j}\\to x\\,\\},\\qquad \n\\jmath(x):=\\max J(x)\\pmod{t},\n\\]\nand pick $z$ with $\\jmath(z)=\\max_{x\\notin C}\\jmath(x)$.\n\nBy construction $c_{\\jmath(z)}\\to z$.  Consider the successor $c_{\\jmath(z)+1}$ on the cycle.  \nIf $c_{\\jmath(z)+1}\\to z$, then $\\jmath(z)$ would not be maximal, so necessarily  \n\\[\nz\\to c_{\\jmath(z)+1}.\n\\]\nHence $z$ is entered by $c_{\\jmath(z)}\\in C$ and dominates $c_{\\jmath(z)+1}\\in C$, so $z$ has mixed adjacency to $C$.  $\\square$\n\n\n\nLemma&nbsp;6 (one–step insertion).  \nLet  \n\\[\nC:=c_{0}\\,(=v)\\to c_{1}\\to\\dots\\to c_{t-1}\\to c_{0}\n\\qquad (3\\le t<n)\n\\tag{5}\n\\]\nbe a directed cycle in a strongly connected tournament $T$.  \nThen one can find a vertex $w\\notin V(C)$ and an index $j$ such that  \n\\[\nc_{j}\\to w\\quad\\text{and}\\quad w\\to c_{j+1}.\n\\tag{6}\n\\]\nConsequently  \n\\[\nc_{0}\\to\\dots\\to c_{j}\\to w\\to c_{j+1}\\to\\dots\\to c_{t-1}\\to c_{0}\n\\]\nis a directed cycle of length $t+1$ that still contains $v$.\n\nProof.  \nLemma&nbsp;5 provides a vertex $w\\notin V(C)$ having mixed adjacency to $C$;  \nlet $c_{j}$ be an in–neighbour and $c_{j+1}$ a successor on $C$ that $w$ dominates.  \nThen (6) holds, so inserting $w$ between $c_{j}$ and $c_{j+1}$ lengthens the cycle as claimed.  $\\square$\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n4.\\  Vertex–pancyclicity – part (b)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nTheorem&nbsp;7 (Moon 1968).  \nLet $T$ be a strongly connected tournament on $n$ vertices and let $v$ be any vertex of $T$.  \nFor every $\\ell$ with $3\\le\\ell\\le n$ there exists a directed cycle of length $\\ell$ that starts at $v$.\n\nProof.  \n\nStep&nbsp;0 – initial supply of cycles.  \nBy Proposition&nbsp;3 the tournament contains a Hamiltonian cycle, in particular a cycle of length $n$ passing through $v$.  \nLemma&nbsp;4 furnishes a $3$–cycle through $v$.\n\nStep&nbsp;1 – grow a cycle by one vertex.  \nAssume inductively that for some $t$ with $3\\le t<n$ we already have a directed cycle $C_{t}$ of length $t$ starting at $v$.  \nLemma&nbsp;6 inserts a new vertex and yields a directed cycle $C_{t+1}$ of length $t+1$ still starting at $v$.\n\nStep&nbsp;2 – reach all lengths.  \nBeginning with the triangle ($t=3$) and iterating Step&nbsp;1 we obtain cycles of lengths $4,5,\\dots ,n$ through $v$.  \nHence $v$ is pancyclic.  As $v$ was arbitrary, every vertex of $T$ is pancyclic.  $\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.418726",
        "was_fixed": false,
        "difficulty_analysis": "•  The original problem asks only for one Hamiltonian path; Rédei’s induction suffices.  \n•  Part (a) of the enhanced variant requires creating a Hamiltonian *cycle*.  Closing a path into a cycle forces one to exploit strong connectivity, develop an insertion algorithm, and prove its termination—Camion’s theorem.  \n•  Part (b) goes further, asking for cycles of *every* length through *every* vertex (pancyclicity).  This needs new structure (the chord lemma), iterative surgery on cycles, and careful bookkeeping to guarantee termination without repetitions.  \n•  These tasks integrate several advanced tools: Rédei’s ordering, strong–connectivity arguments, iterative edge-reversal (“insertion’’) techniques, and constructive shortening, all well beyond the elementary induction of the original."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}