{
  "index": "1990-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "$b$. Prove or disprove: there is a sequence\n\\[\ng_1, g_2, g_3, \\dots, g_{2n}\n\\]\nsuch that\n\\begin{itemize}\n   \\item[(1)] every element of $G$ occurs exactly twice, and\n   \\item[(2)] $g_{i+1}$ equals $g_i a$ or $g_i b$ for $i = 1, 2, \\dots,\n   2n$. (Interpret $g_{2n+1}$ as $g_1$.)\n\\end{itemize}",
  "solution": "Solution. We use graph theory terminology; see the remark below. Construct a directed multigraph \\( \\mathcal{D} \\) whose vertices are the elements of \\( G \\), and whose arcs are indexed by \\( G \\times\\{a, b\\} \\), such that the arc corresponding to the pair \\( (g, x) \\) goes from vertex \\( g \\) to vertex \\( g x \\). (See Figure 16 for an example, with \\( G \\) equal to the symmetric group \\( S_{3}, a \\) equal to the transposition (12), and \\( b \\) equal to the 3 -cycle (123).) At the vertex \\( g \\), there are two arcs going out (to \\( g a \\) and to \\( g b \\) ), and two arcs coming in (from \\( g a^{-1} \\) and from \\( g b^{-1} \\) ). Also, \\( \\mathcal{D} \\) is weakly connected, since \\( a \\) and \\( b \\) generate \\( G \\). Hence, by the first theorem in the remark below, \\( \\mathcal{D} \\) has an Eulerian circuit. Take \\( g_{1} \\), \\( g_{2}, \\ldots, g_{m} \\) to be the the startpoints of the arcs in this circuit, in order. Each element of \\( G \\) occurs exactly twice in this sequence, since each vertex of \\( \\mathcal{D} \\) has outdegree 2 ; in particular \\( m=2 n \\). Also, for \\( 1 \\leq i \\leq 2 n \\), the element \\( g_{i+1} \\) equals either \\( g_{i} a \\) or \\( g_{i} b \\), because the two outgoing arcs from \\( g_{i} \\) end at \\( g_{i} a \\) and \\( g_{i} b \\).\n\nRemark (Eulerian paths and circuits). A directed multigraph \\( \\mathcal{D} \\) consists of a set \\( V \\) (whose elements are called vertices), and a set \\( E \\) (whose elements are called arcs or sometimes edges or directed edges), with a map \\( E \\rightarrow V \\times V \\) (thought of as sending an arc to the pair consisting of its startpoint and the endpoint). Typically one draws each element of \\( V \\) as a point, and each arc of \\( E \\) as an arc from the startpoint to the endpoint, with an arrow to indicate the direction. What makes it a multigraph is that for some vertices \\( v, w \\in V \\), there may be more than one arc from \\( v \\) to \\( w \\). Also, there may be loops: arcs from a vertex to itself. Call \\( \\mathcal{D} \\) finite if \\( V \\) and \\( E \\) are both finite sets.\n\nThe outdegree of a vertex \\( v \\) is the number of arcs in \\( E \\) having \\( v \\) as startpoint. Similarly the indegree of \\( v \\) is the number of arcs in \\( E \\) having \\( v \\) as endpoint. If \\( v, w \\in V \\) then a path from \\( v \\) to \\( w \\) in \\( \\mathcal{D} \\) is a finite sequence of arcs in \\( E \\), not necessarily distinct, such that the startpoint of the first arc is \\( v \\), the endpoint of each arc (other than the last) is the startpoint of the next arc, and the endpoint of the last arc is \\( w \\). Such a path is called a circuit or cycle if \\( v=w \\). An Eulerian path in \\( \\mathcal{D} \\) is a path in which each arc in \\( E \\) occurs exactly once. An Eulerian circuit is a circuit in which each arc in \\( E \\) occurs exactly once. We say that \\( \\mathcal{D} \\) is strongly connected if for every two distinct vertices \\( v, w \\in V \\), there is a path from \\( v \\) to \\( w \\) in \\( \\mathcal{D} \\). On the other hand, \\( \\mathcal{D} \\) is weakly connected if for every two distinct vertices \\( v, w \\in V \\), there is a path from \\( v \\) to \\( w \\) in the corresponding undirected graph, that is, a path consisting of arcs some of which may be the reverses of the arcs in \\( E \\).\n\nThen one can prove the following two theorems.\n- A finite directed multigraph with at least one arc has an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex.\n- A finite directed multigraph with at least one arc has an Eulerian path but not an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex, except at one vertex at which indegree is 1 larger than outdegree, and one other vertex at which outdegree is 1 larger than indegree.\nSee Chapter 7 of [Ros2], especially \\( \\S 7.4 \\) and \\( \\S 7.5 \\).\nRemark. The directed multigraph constructed in the solution is called the Cayley digraph or Cayley diagram associated to \\( G \\) and its set of generators \\( \\{a, b\\} \\). See \\( [\\mathrm{Fr} \\), pp. 87-91].",
  "vars": [
    "g",
    "g_1",
    "g_2",
    "g_3",
    "g_2n",
    "g_2n+1",
    "g_i",
    "g_i+1",
    "i",
    "v",
    "w",
    "x"
  ],
  "params": [
    "a",
    "b",
    "n",
    "D",
    "G",
    "E",
    "V",
    "m",
    "S_3"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "g": "groupelt",
        "g_1": "groupone",
        "g_2": "grouptwo",
        "g_3": "groupthree",
        "g_2n": "grouptwoen",
        "g_2n+1": "groupplus",
        "g_i": "groupi",
        "g_i+1": "groupiplus",
        "i": "indexvar",
        "v": "vertexv",
        "w": "vertexw",
        "x": "elementx",
        "a": "generatora",
        "b": "generatorb",
        "n": "integern",
        "D": "digraphd",
        "G": "groupbig",
        "E": "arcsetall",
        "V": "vertexcoll",
        "m": "lengthm",
        "S_3": "symmetricthree"
      },
      "question": "$b$. Prove or disprove: there is a sequence\n\\[\ngroupone, grouptwo, groupthree, \\dots, grouptwoen\n\\]\nsuch that\n\\begin{itemize}\n   \\item[(1)] every element of $groupbig$ occurs exactly twice, and\n   \\item[(2)] $groupiplus$ equals $groupi \\, generatora$ or $groupi \\, generatorb$ for $indexvar = 1, 2, \\dots, 2integern$. (Interpret $groupplus$ as $groupone$.)\n\\end{itemize}",
      "solution": "Solution. We use graph theory terminology; see the remark below. Construct a directed multigraph \\( \\mathcal{digraphd} \\) whose vertices are the elements of \\( groupbig \\), and whose arcs are indexed by \\( groupbig \\times\\{generatora, generatorb\\} \\), such that the arc corresponding to the pair \\( (groupelt, elementx) \\) goes from vertex \\( groupelt \\) to vertex \\( groupelt \\, elementx \\). (See Figure 16 for an example, with \\( groupbig \\) equal to the symmetric group \\( symmetricthree, generatora \\) equal to the transposition (12), and \\( generatorb \\) equal to the 3 -cycle (123).) At the vertex \\( groupelt \\), there are two arcs going out (to \\( groupelt \\, generatora \\) and to \\( groupelt \\, generatorb \\) ), and two arcs coming in (from \\( groupelt \\, generatora^{-1} \\) and from \\( groupelt \\, generatorb^{-1} \\) ). Also, \\( \\mathcal{digraphd} \\) is weakly connected, since \\( generatora \\) and \\( generatorb \\) generate \\( groupbig \\). Hence, by the first theorem in the remark below, \\( \\mathcal{digraphd} \\) has an Eulerian circuit. Take \\( groupone, grouptwo, \\ldots, groupelt_{lengthm} \\) to be the startpoints of the arcs in this circuit, in order. Each element of \\( groupbig \\) occurs exactly twice in this sequence, since each vertex of \\( \\mathcal{digraphd} \\) has outdegree 2; in particular \\( lengthm = 2 integern \\). Also, for \\( 1 \\leq indexvar \\leq 2 integern \\), the element \\( groupiplus \\) equals either \\( groupi \\, generatora \\) or \\( groupi \\, generatorb \\), because the two outgoing arcs from \\( groupi \\) end at \\( groupi \\, generatora \\) and \\( groupi \\, generatorb \\).\n\nRemark (Eulerian paths and circuits). A directed multigraph \\( \\mathcal{digraphd} \\) consists of a set \\( vertexcoll \\) (whose elements are called vertices), and a set \\( arcsetall \\) (whose elements are called arcs or sometimes edges or directed edges), with a map \\( arcsetall \\rightarrow vertexcoll \\times vertexcoll \\) (thought of as sending an arc to the pair consisting of its startpoint and the endpoint). Typically one draws each element of \\( vertexcoll \\) as a point, and each arc of \\( arcsetall \\) as an arc from the startpoint to the endpoint, with an arrow to indicate the direction. What makes it a multigraph is that for some vertices \\( vertexv, vertexw \\in vertexcoll \\), there may be more than one arc from \\( vertexv \\) to \\( vertexw \\). Also, there may be loops: arcs from a vertex to itself. Call \\( \\mathcal{digraphd} \\) finite if \\( vertexcoll \\) and \\( arcsetall \\) are both finite sets.\n\nThe outdegree of a vertex \\( vertexv \\) is the number of arcs in \\( arcsetall \\) having \\( vertexv \\) as startpoint. Similarly the indegree of \\( vertexv \\) is the number of arcs in \\( arcsetall \\) having \\( vertexv \\) as endpoint. If \\( vertexv, vertexw \\in vertexcoll \\) then a path from \\( vertexv \\) to \\( vertexw \\) in \\( \\mathcal{digraphd} \\) is a finite sequence of arcs in \\( arcsetall \\), not necessarily distinct, such that the startpoint of the first arc is \\( vertexv \\), the endpoint of each arc (other than the last) is the startpoint of the next arc, and the endpoint of the last arc is \\( vertexw \\). Such a path is called a circuit or cycle if \\( vertexv = vertexw \\). An Eulerian path in \\( \\mathcal{digraphd} \\) is a path in which each arc in \\( arcsetall \\) occurs exactly once. An Eulerian circuit is a circuit in which each arc in \\( arcsetall \\) occurs exactly once. We say that \\( \\mathcal{digraphd} \\) is strongly connected if for every two distinct vertices \\( vertexv, vertexw \\in vertexcoll \\), there is a path from \\( vertexv \\) to \\( vertexw \\) in \\( \\mathcal{digraphd} \\). On the other hand, \\( \\mathcal{digraphd} \\) is weakly connected if for every two distinct vertices \\( vertexv, vertexw \\in vertexcoll \\), there is a path from \\( vertexv \\) to \\( vertexw \\) in the corresponding undirected graph, that is, a path consisting of arcs some of which may be the reverses of the arcs in \\( arcsetall \\).\n\nThen one can prove the following two theorems.\n- A finite directed multigraph with at least one arc has an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex.\n- A finite directed multigraph with at least one arc has an Eulerian path but not an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex, except at one vertex at which indegree is 1 larger than outdegree, and one other vertex at which outdegree is 1 larger than indegree.\nSee Chapter 7 of [Ros2], especially \\( \\S 7.4 \\) and \\( \\S 7.5 \\).\nRemark. The directed multigraph constructed in the solution is called the Cayley digraph or Cayley diagram associated to \\( groupbig \\) and its set of generators \\{generatora, generatorb\\}. See \\( [\\mathrm{Fr} \\), pp. 87-91]."
    },
    "descriptive_long_confusing": {
      "map": {
        "g": "whistling",
        "g_1": "partridge",
        "g_2": "backstory",
        "g_3": "campfire",
        "g_2n": "lullabies",
        "g_2n+1": "steamboat",
        "g_i": "megapixel",
        "g_i+1": "harmonica",
        "i": "lanterns",
        "v": "snowflake",
        "w": "pinecones",
        "x": "tangerine",
        "a": "rainstorm",
        "b": "buttercup",
        "n": "honeycomb",
        "D": "locomotive",
        "G": "waterfall",
        "E": "gallantry",
        "V": "millinery",
        "m": "scarecrow",
        "S_3": "dragonfly"
      },
      "question": "$b$. Prove or disprove: there is a sequence\n\\[\npartridge, backstory, campfire, \\dots, lullabies\n\\]\nsuch that\n\\begin{itemize}\n   \\item[(1)] every element of waterfall occurs exactly twice, and\n   \\item[(2)] harmonica equals megapixel rainstorm or megapixel buttercup for lanterns = 1, 2, \\dots,\n   2 honeycomb. (Interpret steamboat as partridge.)\n\\end{itemize}",
      "solution": "Solution. We use graph theory terminology; see the remark below. Construct a directed multigraph \\( \\mathcal{locomotive} \\) whose vertices are the elements of \\( waterfall \\), and whose arcs are indexed by \\( waterfall \\times\\{rainstorm, buttercup\\} \\), such that the arc corresponding to the pair \\( (whistling, tangerine) \\) goes from vertex \\( whistling \\) to vertex \\( whistling tangerine \\). (See Figure 16 for an example, with \\( waterfall \\) equal to the symmetric group \\( dragonfly, rainstorm \\) equal to the transposition (12), and \\( buttercup \\) equal to the 3 -cycle (123).) At the vertex \\( whistling \\), there are two arcs going out (to \\( whistling rainstorm \\) and to \\( whistling buttercup \\) ), and two arcs coming in (from \\( whistling rainstorm^{-1} \\) and from \\( whistling buttercup^{-1} \\) ). Also, \\( \\mathcal{locomotive} \\) is weakly connected, since \\( rainstorm \\) and \\( buttercup \\) generate \\( waterfall \\). Hence, by the first theorem in the remark below, \\( \\mathcal{locomotive} \\) has an Eulerian circuit. Take \\( partridge \\), \\( backstory, \\ldots, scarecrow \\) to be the the startpoints of the arcs in this circuit, in order. Each element of \\( waterfall \\) occurs exactly twice in this sequence, since each vertex of \\( \\mathcal{locomotive} \\) has outdegree 2 ; in particular \\( scarecrow=2 honeycomb \\). Also, for \\( 1 \\leq lanterns \\leq 2 honeycomb \\), the element \\( harmonica \\) equals either \\( megapixel rainstorm \\) or \\( megapixel buttercup \\), because the two outgoing arcs from \\( megapixel \\) end at \\( megapixel rainstorm \\) and \\( megapixel buttercup \\).\n\nRemark (Eulerian paths and circuits). A directed multigraph \\( \\mathcal{locomotive} \\) consists of a set \\( millinery \\) (whose elements are called vertices), and a set \\( gallantry \\) (whose elements are called arcs or sometimes edges or directed edges), with a map \\( gallantry \\rightarrow millinery \\times millinery \\) (thought of as sending an arc to the pair consisting of its startpoint and the endpoint). Typically one draws each element of \\( millinery \\) as a point, and each arc of \\( gallantry \\) as an arc from the startpoint to the endpoint, with an arrow to indicate the direction. What makes it a multigraph is that for some vertices \\( snowflake, pinecones \\in millinery \\), there may be more than one arc from \\( snowflake \\) to \\( pinecones \\). Also, there may be loops: arcs from a vertex to itself. Call \\( \\mathcal{locomotive} \\) finite if \\( millinery \\) and \\( gallantry \\) are both finite sets.\n\nThe outdegree of a vertex \\( snowflake \\) is the number of arcs in \\( gallantry \\) having \\( snowflake \\) as startpoint. Similarly the indegree of \\( snowflake \\) is the number of arcs in \\( gallantry \\) having \\( snowflake \\) as endpoint. If \\( snowflake, pinecones \\in millinery \\) then a path from \\( snowflake \\) to \\( pinecones \\) in \\( \\mathcal{locomotive} \\) is a finite sequence of arcs in \\( gallantry \\), not necessarily distinct, such that the startpoint of the first arc is \\( snowflake \\), the endpoint of each arc (other than the last) is the startpoint of the next arc, and the endpoint of the last arc is \\( pinecones \\). Such a path is called a circuit or cycle if \\( snowflake=pinecones \\). An Eulerian path in \\( \\mathcal{locomotive} \\) is a path in which each arc in \\( gallantry \\) occurs exactly once. An Eulerian circuit is a circuit in which each arc in \\( gallantry \\) occurs exactly once. We say that \\( \\mathcal{locomotive} \\) is strongly connected if for every two distinct vertices \\( snowflake, pinecones \\in millinery \\), there is a path from \\( snowflake \\) to \\( pinecones \\) in \\( \\mathcal{locomotive} \\). On the other hand, \\( \\mathcal{locomotive} \\) is weakly connected if for every two distinct vertices \\( snowflake, pinecones \\in millinery \\), there is a path from \\( snowflake \\) to \\( pinecones \\) in the corresponding undirected graph, that is, a path consisting of arcs some of which may be the reverses of the arcs in \\( gallantry \\).\n\nThen one can prove the following two theorems.\n- A finite directed multigraph with at least one arc has an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex.\n- A finite directed multigraph with at least one arc has an Eulerian path but not an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex, except at one vertex at which indegree is 1 larger than outdegree, and one other vertex at which outdegree is 1 larger than indegree.\nSee Chapter 7 of [Ros2], especially \\( \\S 7.4 \\) and \\( \\S 7.5 \\).\nRemark. The directed multigraph constructed in the solution is called the Cayley digraph or Cayley diagram associated to \\( waterfall \\) and its set of generators \\{rainstorm, buttercup\\}. See \\( [\\mathrm{Fr} \\), pp. 87-91]."
    },
    "descriptive_long_misleading": {
      "map": {
        "g": "outsider",
        "g_1": "outsiderone",
        "g_{1}": "outsiderone",
        "g_2": "outsidertwo",
        "g_{2}": "outsidertwo",
        "g_3": "outsiderthree",
        "g_{3}": "outsiderthree",
        "g_{2n}": "outsidermany",
        "g_{2n+1}": "outsiderextra",
        "g_i": "outsiderindex",
        "g_{i}": "outsiderindex",
        "g_{i+1}": "outsidernext",
        "i": "constant",
        "v": "scalarval",
        "w": "edgevalue",
        "x": "knownval",
        "a": "follower",
        "b": "laggards",
        "n": "infinite",
        "D": "forestgraph",
        "G": "nongroup",
        "E": "vertices",
        "V": "edgeslist",
        "m": "expansion",
        "S_{3}": "asymmetric"
      },
      "question": "$b$. Prove or disprove: there is a sequence\n\\[\noutsiderone, outsidertwo, outsiderthree, \\dots, outsidermany\n\\]\nsuch that\n\\begin{itemize}\n   \\item[(1)] every element of $nongroup$ occurs exactly twice, and\n   \\item[(2)] $outsidernext$ equals $outsiderindex$ follower or $outsiderindex$ laggards for $constant = 1, 2, \\dots,\n   2infinite$. (Interpret $outsiderextra$ as $outsiderone$.)\n\\end{itemize}",
      "solution": "Solution. We use graph theory terminology; see the remark below. Construct a directed multigraph \\( \\mathcal{forestgraph} \\) whose vertices are the elements of \\( nongroup \\), and whose arcs are indexed by \\( nongroup \\times\\{follower, laggards\\} \\), such that the arc corresponding to the pair \\( (outsider, knownval) \\) goes from vertex \\( outsider \\) to vertex \\( outsider knownval \\). (See Figure 16 for an example, with \\( nongroup \\) equal to the symmetric group \\( asymmetric, follower \\) equal to the transposition (12), and \\( laggards \\) equal to the 3 -cycle (123).) At the vertex \\( outsider \\), there are two arcs going out (to \\( outsider follower \\) and to \\( outsider laggards \\) ), and two arcs coming in (from \\( outsider follower^{-1} \\) and from \\( outsider laggards^{-1} \\) ). Also, \\( \\mathcal{forestgraph} \\) is weakly connected, since \\( follower \\) and \\( laggards \\) generate \\( nongroup \\). Hence, by the first theorem in the remark below, \\( \\mathcal{forestgraph} \\) has an Eulerian circuit. Take \\( outsider_{1} \\), \\( outsider_{2}, \\ldots, outsider_{expansion} \\) to be the the startpoints of the arcs in this circuit, in order. Each element of \\( nongroup \\) occurs exactly twice in this sequence, since each vertex of \\( \\mathcal{forestgraph} \\) has outdegree 2 ; in particular \\( expansion=2 infinite \\). Also, for \\( 1 \\leq constant \\leq 2 infinite \\), the element \\( outsider_{constant+1} \\) equals either \\( outsider_{constant} follower \\) or \\( outsider_{constant} laggards \\), because the two outgoing arcs from \\( outsider_{constant} \\) end at \\( outsider_{constant} follower \\) and \\( outsider_{constant} laggards \\).\n\nRemark (Eulerian paths and circuits). A directed multigraph \\( \\mathcal{forestgraph} \\) consists of a set \\( edgeslist \\) (whose elements are called vertices), and a set \\( vertices \\) (whose elements are called arcs or sometimes edges or directed edges), with a map \\( vertices \\rightarrow edgeslist \\times edgeslist \\) (thought of as sending an arc to the pair consisting of its startpoint and the endpoint). Typically one draws each element of \\( edgeslist \\) as a point, and each arc of \\( vertices \\) as an arc from the startpoint to the endpoint, with an arrow to indicate the direction. What makes it a multigraph is that for some vertices \\( scalarval, edgevalue \\in edgeslist \\), there may be more than one arc from \\( scalarval \\) to \\( edgevalue \\). Also, there may be loops: arcs from a vertex to itself. Call \\( \\mathcal{forestgraph} \\) finite if \\( edgeslist \\) and \\( vertices \\) are both finite sets.\n\nThe outdegree of a vertex \\( scalarval \\) is the number of arcs in \\( vertices \\) having \\( scalarval \\) as startpoint. Similarly the indegree of \\( scalarval \\) is the number of arcs in \\( vertices \\) having \\( scalarval \\) as endpoint. If \\( scalarval, edgevalue \\in edgeslist \\) then a path from \\( scalarval \\) to \\( edgevalue \\) in \\( \\mathcal{forestgraph} \\) is a finite sequence of arcs in \\( vertices \\), not necessarily distinct, such that the startpoint of the first arc is \\( scalarval \\), the endpoint of each arc (other than the last) is the startpoint of the next arc, and the endpoint of the last arc is \\( edgevalue \\). Such a path is called a circuit or cycle if \\( scalarval=edgevalue \\). An Eulerian path in \\( \\mathcal{forestgraph} \\) is a path in which each arc in \\( vertices \\) occurs exactly once. An Eulerian circuit is a circuit in which each arc in \\( vertices \\) occurs exactly once. We say that \\( \\mathcal{forestgraph} \\) is strongly connected if for every two distinct vertices \\( scalarval, edgevalue \\in edgeslist \\), there is a path from \\( scalarval \\) to \\( edgevalue \\) in \\( \\mathcal{forestgraph} \\). On the other hand, \\( \\mathcal{forestgraph} \\) is weakly connected if for every two distinct vertices \\( scalarval, edgevalue \\in edgeslist \\), there is a path from \\( scalarval \\) to \\( edgevalue \\) in the corresponding undirected graph, that is, a path consisting of arcs some of which may be the reverses of the arcs in \\( vertices \\).\n\nThen one can prove the following two theorems.\n- A finite directed multigraph with at least one arc has an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex.\n- A finite directed multigraph with at least one arc has an Eulerian path but not an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex, except at one vertex at which indegree is 1 larger than outdegree, and one other vertex at which outdegree is 1 larger than indegree.\nSee Chapter 7 of [Ros2], especially \\( \\S 7.4 \\) and \\( \\S 7.5 \\).\nRemark. The directed multigraph constructed in the solution is called the Cayley digraph or Cayley diagram associated to \\( nongroup \\) and its set of generators \\{follower, laggards\\}. See \\( [\\mathrm{Fr} \\), pp. 87-91]."
    },
    "garbled_string": {
      "map": {
        "g": "qzxwvtnp",
        "g_1": "hjgrksla",
        "g_2": "lmpqzvwr",
        "g_3": "ndfksbli",
        "g_2n": "tqrplmns",
        "g_2n+1": "fghsadzx",
        "g_i": "kjnmpqrs",
        "g_i+1": "vclotxzi",
        "i": "bcsnmdlf",
        "v": "wquxyvmp",
        "w": "pocnqwer",
        "x": "asdjklre",
        "a": "zqplmnbv",
        "b": "ghynmktr",
        "n": "mnbvcxzq",
        "D": "kzpmivqz",
        "G": "rtyuioas",
        "E": "qwopzxcv",
        "V": "lkjhgfds",
        "m": "poiuytre",
        "S_3": "qjwrlxzn"
      },
      "question": "$b$. Prove or disprove: there is a sequence\n\\[\nhjgrksla, lmpqzvwr, ndfksbli, \\dots, tqrplmns\n\\]\nsuch that\n\\begin{itemize}\n   \\item[(1)] every element of $rtyuioas$ occurs exactly twice, and\n   \\item[(2)] $vclotxzi$ equals $kjnmpqrs zqplmnbv$ or $kjnmpqrs ghynmktr$ for $bcsnmdlf = 1, 2, \\dots,\n   2 mnbvcxzq$. (Interpret $fghsadzx$ as $hjgrksla$.)\n\\end{itemize}",
      "solution": "Solution. We use graph theory terminology; see the remark below. Construct a directed multigraph \\( \\mathcal{kzpmivqz} \\) whose vertices are the elements of \\( rtyuioas \\), and whose arcs are indexed by \\( rtyuioas \\times\\{zqplmnbv, ghynmktr\\} \\), such that the arc corresponding to the pair \\( (qzxwvtnp, asdjklre) \\) goes from vertex \\( qzxwvtnp \\) to vertex \\( qzxwvtnp asdjklre \\). (See Figure 16 for an example, with \\( rtyuioas \\) equal to the symmetric group \\( qjwrlxzn, zqplmnbv \\) equal to the transposition (12), and \\( ghynmktr \\) equal to the 3 -cycle (123).) At the vertex \\( qzxwvtnp \\), there are two arcs going out (to \\( qzxwvtnp zqplmnbv \\) and to \\( qzxwvtnp ghynmktr \\) ), and two arcs coming in (from \\( qzxwvtnp zqplmnbv^{-1} \\) and from \\( qzxwvtnp ghynmktr^{-1} \\) ). Also, \\( \\mathcal{kzpmivqz} \\) is weakly connected, since \\( zqplmnbv \\) and \\( ghynmktr \\) generate \\( rtyuioas \\). Hence, by the first theorem in the remark below, \\( \\mathcal{kzpmivqz} \\) has an Eulerian circuit. Take \\( hjgrksla \\), \\( lmpqzvwr \\), \\ldots, \\( qzxwvtnp_{poiuytre} \\) to be the the startpoints of the arcs in this circuit, in order. Each element of \\( rtyuioas \\) occurs exactly twice in this sequence, since each vertex of \\( \\mathcal{kzpmivqz} \\) has outdegree 2 ; in particular \\( poiuytre=2 mnbvcxzq \\). Also, for \\( 1 \\leq bcsnmdlf \\leq 2 mnbvcxzq \\), the element \\( vclotxzi \\) equals either \\( kjnmpqrs zqplmnbv \\) or \\( kjnmpqrs ghynmktr \\), because the two outgoing arcs from \\( kjnmpqrs \\) end at \\( kjnmpqrs zqplmnbv \\) and \\( kjnmpqrs ghynmktr \\).\n\nRemark (Eulerian paths and circuits). A directed multigraph \\( \\mathcal{kzpmivqz} \\) consists of a set \\( lkjhgfds \\) (whose elements are called vertices), and a set \\( qwopzxcv \\) (whose elements are called arcs or sometimes edges or directed edges), with a map \\( qwopzxcv \\rightarrow lkjhgfds \\times lkjhgfds \\) (thought of as sending an arc to the pair consisting of its startpoint and the endpoint). Typically one draws each element of \\( lkjhgfds \\) as a point, and each arc of \\( qwopzxcv \\) as an arc from the startpoint to the endpoint, with an arrow to indicate the direction. What makes it a multigraph is that for some vertices \\( wquxyvmp, pocnqwer \\in lkjhgfds \\), there may be more than one arc from \\( wquxyvmp \\) to \\( pocnqwer \\). Also, there may be loops: arcs from a vertex to itself. Call \\( \\mathcal{kzpmivqz} \\) finite if \\( lkjhgfds \\) and \\( qwopzxcv \\) are both finite sets.\n\nThe outdegree of a vertex \\( wquxyvmp \\) is the number of arcs in \\( qwopzxcv \\) having \\( wquxyvmp \\) as startpoint. Similarly the indegree of \\( wquxyvmp \\) is the number of arcs in \\( qwopzxcv \\) having \\( wquxyvmp \\) as endpoint. If \\( wquxyvmp, pocnqwer \\in lkjhgfds \\) then a path from \\( wquxyvmp \\) to \\( pocnqwer \\) in \\( \\mathcal{kzpmivqz} \\) is a finite sequence of arcs in \\( qwopzxcv \\), not necessarily distinct, such that the startpoint of the first arc is \\( wquxyvmp \\), the endpoint of each arc (other than the last) is the startpoint of the next arc, and the endpoint of the last arc is \\( pocnqwer \\). Such a path is called a circuit or cycle if \\( wquxyvmp=pocnqwer \\). An Eulerian path in \\( \\mathcal{kzpmivqz} \\) is a path in which each arc in \\( qwopzxcv \\) occurs exactly once. An Eulerian circuit is a circuit in which each arc in \\( qwopzxcv \\) occurs exactly once. We say that \\( \\mathcal{kzpmivqz} \\) is strongly connected if for every two distinct vertices \\( wquxyvmp, pocnqwer \\in lkjhgfds \\), there is a path from \\( wquxyvmp \\) to \\( pocnqwer \\) in \\( \\mathcal{kzpmivqz} \\). On the other hand, \\( \\mathcal{kzpmivqz} \\) is weakly connected if for every two distinct vertices \\( wquxyvmp, pocnqwer \\in lkjhgfds \\), there is a path from \\( wquxyvmp \\) to \\( pocnqwer \\) in the corresponding undirected graph, that is, a path consisting of arcs some of which may be the reverses of the arcs in \\( qwopzxcv \\).\n\nThen one can prove the following two theorems.\n- A finite directed multigraph with at least one arc has an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex.\n- A finite directed multigraph with at least one arc has an Eulerian path but not an Eulerian circuit if and only if it is weakly connected and the indegree and outdegree are equal at each vertex, except at one vertex at which indegree is 1 larger than outdegree, and one other vertex at which outdegree is 1 larger than indegree.\nSee Chapter 7 of [Ros2], especially \\( \\S 7.4 \\) and \\( \\S 7.5 \\).\nRemark. The directed multigraph constructed in the solution is called the Cayley digraph or Cayley diagram associated to \\( rtyuioas \\) and its set of generators \\{zqplmnbv, ghynmktr\\}. See \\( [\\mathrm{Fr} \\), pp. 87-91]."
    },
    "kernel_variant": {
      "question": "Let $G$ be a finite group of order $m$ which is generated by the $k\\,( \\ge 3)$ elements  \n\\[\nS=\\{\\,s_{1},s_{2},\\dots ,s_{k}\\,\\}.\n\\]\nProve that one can arrange the elements of $G$ in a cyclic list  \n\\[\ng_{1},g_{2},\\dots ,g_{k(k-1)m}\\qquad \n(\\text{indices understood modulo }k(k-1)m)\n\\]\ntogether with a map $\\sigma:\\{1,\\dots ,k(k-1)m\\}\\longrightarrow S$ such that\n\n(1)  every element of $G$ occurs exactly $k(k-1)$ times in the list;\n\n(2)  for every index $i$ we have \n\\[\ng_{i+1}=\\,\\sigma(i)\\,g_{i};\n\\]\n\n(3)  no generator acts twice in a row, i.e. $\\sigma(i+1)\\ne\\sigma(i)$ for all $i$;\n\n(4)  (uniform ordered–pair distribution) for every ordered pair $(s_{p},s_{q})$ of\ndistinct generators exactly $m$ indices $i$ satisfy  \n\\[\n\\sigma(i)=s_{p}\\quad\\text{and}\\quad \\sigma(i+1)=s_{q}.\n\\]\n\nShow that such a cyclic sequence always exists.",
      "solution": "Multiplication is written on the \\emph{left}; the identity of $G$ is denoted by $1$.\n\n──────────────────────────────────\nStep 1.  Construction of the auxiliary digraph $\\mathfrak D$.\n──────────────────────────────────\nDefine the directed multigraph\n\\[\nV := G\\times S,\\qquad \n(g,s_{p})\\longrightarrow (s_{q}g,s_{q}) \\quad (s_{q}\\ne s_{p}).\n\\tag{1}\n\\]\nIn words, an edge is labelled by the left multiplier that brings its first\ncoordinate to the first coordinate of the head.\n\n──────────────────────────────────\nStep 2.  Basic parameters.\n──────────────────────────────────\nEvery vertex has indegree $k-1$ and outdegree $k-1$; consequently\n\\[\n\\lvert V\\rvert=mk,\\qquad\n\\lvert E\\rvert=mk(k-1).\n\\tag{2}\n\\]\nThus $\\mathfrak D$ is finite, balanced and $(k-1)$–regular.\n\n──────────────────────────────────\nStep 3.  Strong connectivity of $\\mathfrak D$.\n──────────────────────────────────\nIt suffices to show that for arbitrary\n\\[\n(g,a),(h,b)\\in G\\times S\n\\]\nthere exists a directed path from $(g,a)$ to $(h,b)$ in $\\mathfrak D$.\n\n─────────────────\n(3.1)   Repeat–free words.\n─────────────────\nA word $t_{1}t_{2}\\dots t_{r}$ on the alphabet $S$ is \\emph{repeat–free} if\n$t_{i+1}\\ne t_{i}$ for every $i$.\n\n─────────────────\n(3.2)   Identity gadgets (all repeat–free and equal $1$ in $G$).\n─────────────────\nBecause $k\\ge 3$, any singleton or pair of generators can be complemented by\na third one.\n\n(i)  One–sided gadget  \nFor distinct $x,y\\in S$ choose $z\\in S\\setminus\\{x,y\\}$ and define\n\\[\nJ(x\\!\\to\\! y):=(zx)^{\\operatorname{ord}(zx)}\\,(zy)^{\\operatorname{ord}(zy)}.\n\\tag{3}\n\\]\nThe two blocks have different terminal and initial letters ($x\\ne z$ and\n$z\\ne y$), hence their concatenation is repeat–free.  As each block is a power\nof an element of $G$, $J(x\\!\\to\\! y)$ evaluates to $1$.\n\n(ii)  Separation gadget  \nFor $x\\in S$ pick $u,v\\in S\\setminus\\{x\\}$ with $u\\ne v$ and set\n\\[\nK(x):=(uv)^{\\operatorname{ord}(uv)}.\n\\tag{4}\n\\]\nConsecutive letters inside the power alternate $u,v$, so the word is\nrepeat–free; its value is $1$.\n\n─────────────────\n(3.3)   Flexible repeat–free representation.\n─────────────────\nLemma 3.1  \nLet $a,b\\in S$ (not necessarily distinct) and $g,h\\in G$.  \nThere exists a repeat–free word $w=t_{1}\\dots t_{r}$ satisfying\n\\[\nt_{1}\\ne a,\\qquad t_{r}=b,\\qquad \nt_{r}\\dots t_{2}t_{1}=h\\,g^{-1}.\n\\tag{5}\n\\]\n\nProof.  \nPut $\\delta:=h\\,g^{-1}$ and select an arbitrary word\n\\[\nv=q_{1}q_{2}\\dots q_{s}\\quad(q_{i}\\in S)\n\\]\nwith group value $q_{s}\\dots q_{1}=\\delta$ (existence follows from the fact\nthat $S$ generates $G$).  The word $v$ is transformed in three repair stages.\n\nStage A – forcing the terminal letter $b$.  \nIf the last letter of $v$ is already $b$, do nothing.  Otherwise append the\none–sided gadget $J(q_{s}\\!\\to\\! b)$.  The new word $v^{(A)}$ still evaluates\nto $\\delta$, ends with $b$ and remains repeat–free except possibly at the\njunction $q_{s}\\,(\\text{first letter of }J)$, where the two letters differ by\nconstruction.\n\nStage B – avoiding the initial letter $a$.  \nIf the first letter of $v^{(A)}$ is different from $a$, leave the word\nunchanged.  Otherwise \\emph{prepend} the separation gadget $K(a)$ to obtain\n$v^{(B)}$.  The prefix $K(a)$ starts with a letter $u\\ne a$, so\n$t_{1}\\ne a$.  Because $K(a)$ evaluates to $1$, the total value is still\n$\\delta$.  Again, all potential repetitions at the new junction are prevented\nby $u\\ne a$.\n\nStage C – removing internal repetitions.  \nScan the current word from left to right.  Whenever two consecutive equal\nletters $xx$ are encountered, insert the gadget $K(x)$ between them.\nInsertion keeps the overall value, and $K(x)$ starts with $u\\ne x$ and ends\nwith $v\\ne x$, so the double $xx$ disappears and no new repetition is\ncreated.  As the word is finite, the process terminates after finitely many\ninsertions; the output is the desired word $w$.\n\nBecause none of the stages affects the value, $w$ fulfils (5).\\hfill$\\square$\n\n─────────────────\n(3.4)   Translation of $w$ into a directed path.\n─────────────────\nGiven $w=t_{1}\\dots t_{r}$ of Lemma 3.1 set\n\\[\nv_{i}:=\\bigl(t_{i}\\dots t_{1}g,\\; t_{i}\\bigr)\\qquad(0\\le i\\le r).\n\\]\nThen $v_{0}=(g,a)$, $v_{r}=(h,b)$ and for $1\\le i\\le r$\n\\[\nv_{i-1}\\longrightarrow v_{i}\n\\]\nis an edge of type (1), since $t_{i}\\ne t_{i-1}$.\nHence $\\mathfrak D$ is strongly connected. \\hfill$\\square$\n\n──────────────────────────────────\nStep 4.  An Eulerian circuit.\n──────────────────────────────────\nA finite, balanced, strongly connected digraph admits an Euler tour.\nFix such a circuit\n\\[\nC=(e_{0},e_{1},\\dots ,e_{L-1}),\\qquad L:=k(k-1)m.\n\\]\nWrite the tail of $e_{i}$ as $(g_{i},\\sigma_{i})$ and the head as\n$(g_{i+1},\\sigma_{i+1})$; indices are modulo $L$.\nDiscarding the second coordinates gives the cyclic list\n\\[\ng_{0},g_{1},\\dots ,g_{L-1}.\n\\tag{6}\n\\]\nFor conformity with the statement relabel $g_{0}=g_{1},\\dots,g_{L-1}=g_{L}$ and define\n\\[\n\\sigma(i):=\\sigma_{i+1}\\qquad(1\\le i\\le L).\n\\tag{7}\n\\]\n\n──────────────────────────────────\nStep 5.  Verification of the four requirements.\n──────────────────────────────────\n(1)  Frequency of group elements.  \nFix $g\\in G$.  The $k$ vertices $(g,s_{1}),\\dots,(g,s_{k})$ are the tails of\nexactly $k(k-1)$ distinct edges (each vertex contributes $k-1$ outgoing\nedges).  An Euler circuit traverses every edge \\emph{exactly once}, hence $g$\nappears $k(k-1)=L/m$ times in the list (6).\n\n(2)  Transition rule.  \nEdge $e_{i}$ goes from $(g_{i},\\sigma_{i})$ to $(g_{i+1},\\sigma_{i+1})$, so\n$g_{i+1}=\\sigma_{i+1}g_{i}$.  With the shift (7) this is\n$g_{i+1}= \\sigma(i)\\,g_{i}$.\n\n(3)  No consecutive repetition.  \nBecause $e_{i}$ leaves $(g_{i},\\sigma_{i})$ toward a vertex whose second\ncoordinate differs, $\\sigma_{i+1}\\ne\\sigma_{i}$ for all $i$, and consequently\n$\\sigma(i+1)\\ne\\sigma(i)$.\n\n(4)  Uniform ordered–pair distribution.  \nFix $p\\ne q$.  For every $g\\in G$ the edge\n\\[\n(g,s_{p})\\longrightarrow (s_{q}g,s_{q})\n\\]\noccurs exactly once in $\\mathfrak D$, so precisely $m$ edges carry the ordered\npair $(s_{p},s_{q})$ of generators.  The Euler circuit uses every edge once,\ntherefore exactly $m$ indices $j$ satisfy\n$(\\sigma_{j},\\sigma_{j+1})=(s_{p},s_{q})$.\nBy definition (7) this is equivalent to\n$(\\sigma(j-1),\\sigma(j))=(s_{p},s_{q})$, i.e. precisely the $m$ indices\nrequired in property (4).\n\nAll four conditions are fulfilled; hence the desired cyclic list exists.\\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.715272",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original and the current kernel variant, the enhanced\nproblem adds a second-order uniformity condition (Property 4):  not only\nmust each generator occur equally often, but every ordered pair of\ndistinct generators must also occur equally often in consecutive\npositions.  This forces simultaneous control of vertex frequencies and\nedge-pair frequencies, raising the combinatorial load from first-order\nto second-order statistics.\n\nTechnically, meeting this stronger requirement demands replacing the\nordinary Cayley digraph (whose vertices are the group elements) with a\nmuch larger “state” digraph of size |G|·k whose vertices keep track of\nboth a group element and the last generator used.  Only after analyzing\nthis higher-dimensional object and proving it is Eulerian can one\nproject back to a sequence in G.  The solver must therefore:\n\n• Identify the correct augmented state space (G×S) that encodes the\nsecond-order constraint.  \n• Show that this larger digraph is balanced and connected, invoking a\nnon-trivial generalization of the standard Cayley-graph argument.  \n• Construct and interpret an Eulerian circuit in that digraph, then\nprove that the projected walk satisfies the original four properties.\n\nThese additional layers of abstraction, size blow-up (m → m·k), and\nhigher-order combinatorial constraints render the enhanced variant\nsignificantly more intricate and conceptually demanding than either of\nthe predecessors."
      }
    },
    "original_kernel_variant": {
      "question": "Let $G$ be a finite group of order $m$ which is generated by the $k\\,( \\ge 3)$ elements  \n\\[\nS=\\{\\,s_{1},s_{2},\\dots ,s_{k}\\,\\}.\n\\]\nProve that one can arrange the elements of $G$ in a cyclic list  \n\\[\ng_{1},g_{2},\\dots ,g_{k(k-1)m}\\qquad \n(\\text{indices understood modulo }k(k-1)m)\n\\]\ntogether with a map $\\sigma:\\{1,\\dots ,k(k-1)m\\}\\longrightarrow S$ such that\n\n(1)  every element of $G$ occurs exactly $k(k-1)$ times in the list;\n\n(2)  for every index $i$ we have \n\\[\ng_{i+1}=\\,\\sigma(i)\\,g_{i};\n\\]\n\n(3)  no generator acts twice in a row, i.e. $\\sigma(i+1)\\ne\\sigma(i)$ for all $i$;\n\n(4)  (uniform ordered–pair distribution) for every ordered pair $(s_{p},s_{q})$ of\ndistinct generators exactly $m$ indices $i$ satisfy  \n\\[\n\\sigma(i)=s_{p}\\quad\\text{and}\\quad \\sigma(i+1)=s_{q}.\n\\]\n\nShow that such a cyclic sequence always exists.",
      "solution": "Multiplication is written on the \\emph{left}; the identity of $G$ is denoted by $1$.\n\n──────────────────────────────────\nStep 1.  Construction of the auxiliary digraph $\\mathfrak D$.\n──────────────────────────────────\nDefine the directed multigraph\n\\[\nV := G\\times S,\\qquad \n(g,s_{p})\\longrightarrow (s_{q}g,s_{q}) \\quad (s_{q}\\ne s_{p}).\n\\tag{1}\n\\]\nIn words, an edge is labelled by the left multiplier that brings its first\ncoordinate to the first coordinate of the head.\n\n──────────────────────────────────\nStep 2.  Basic parameters.\n──────────────────────────────────\nEvery vertex has indegree $k-1$ and outdegree $k-1$; consequently\n\\[\n\\lvert V\\rvert=mk,\\qquad\n\\lvert E\\rvert=mk(k-1).\n\\tag{2}\n\\]\nThus $\\mathfrak D$ is finite, balanced and $(k-1)$–regular.\n\n──────────────────────────────────\nStep 3.  Strong connectivity of $\\mathfrak D$.\n──────────────────────────────────\nIt suffices to show that for arbitrary\n\\[\n(g,a),(h,b)\\in G\\times S\n\\]\nthere exists a directed path from $(g,a)$ to $(h,b)$ in $\\mathfrak D$.\n\n─────────────────\n(3.1)   Repeat–free words.\n─────────────────\nA word $t_{1}t_{2}\\dots t_{r}$ on the alphabet $S$ is \\emph{repeat–free} if\n$t_{i+1}\\ne t_{i}$ for every $i$.\n\n─────────────────\n(3.2)   Identity gadgets (all repeat–free and equal $1$ in $G$).\n─────────────────\nBecause $k\\ge 3$, any singleton or pair of generators can be complemented by\na third one.\n\n(i)  One–sided gadget  \nFor distinct $x,y\\in S$ choose $z\\in S\\setminus\\{x,y\\}$ and define\n\\[\nJ(x\\!\\to\\! y):=(zx)^{\\operatorname{ord}(zx)}\\,(zy)^{\\operatorname{ord}(zy)}.\n\\tag{3}\n\\]\nThe two blocks have different terminal and initial letters ($x\\ne z$ and\n$z\\ne y$), hence their concatenation is repeat–free.  As each block is a power\nof an element of $G$, $J(x\\!\\to\\! y)$ evaluates to $1$.\n\n(ii)  Separation gadget  \nFor $x\\in S$ pick $u,v\\in S\\setminus\\{x\\}$ with $u\\ne v$ and set\n\\[\nK(x):=(uv)^{\\operatorname{ord}(uv)}.\n\\tag{4}\n\\]\nConsecutive letters inside the power alternate $u,v$, so the word is\nrepeat–free; its value is $1$.\n\n─────────────────\n(3.3)   Flexible repeat–free representation.\n─────────────────\nLemma 3.1  \nLet $a,b\\in S$ (not necessarily distinct) and $g,h\\in G$.  \nThere exists a repeat–free word $w=t_{1}\\dots t_{r}$ satisfying\n\\[\nt_{1}\\ne a,\\qquad t_{r}=b,\\qquad \nt_{r}\\dots t_{2}t_{1}=h\\,g^{-1}.\n\\tag{5}\n\\]\n\nProof.  \nPut $\\delta:=h\\,g^{-1}$ and select an arbitrary word\n\\[\nv=q_{1}q_{2}\\dots q_{s}\\quad(q_{i}\\in S)\n\\]\nwith group value $q_{s}\\dots q_{1}=\\delta$ (existence follows from the fact\nthat $S$ generates $G$).  The word $v$ is transformed in three repair stages.\n\nStage A – forcing the terminal letter $b$.  \nIf the last letter of $v$ is already $b$, do nothing.  Otherwise append the\none–sided gadget $J(q_{s}\\!\\to\\! b)$.  The new word $v^{(A)}$ still evaluates\nto $\\delta$, ends with $b$ and remains repeat–free except possibly at the\njunction $q_{s}\\,(\\text{first letter of }J)$, where the two letters differ by\nconstruction.\n\nStage B – avoiding the initial letter $a$.  \nIf the first letter of $v^{(A)}$ is different from $a$, leave the word\nunchanged.  Otherwise \\emph{prepend} the separation gadget $K(a)$ to obtain\n$v^{(B)}$.  The prefix $K(a)$ starts with a letter $u\\ne a$, so\n$t_{1}\\ne a$.  Because $K(a)$ evaluates to $1$, the total value is still\n$\\delta$.  Again, all potential repetitions at the new junction are prevented\nby $u\\ne a$.\n\nStage C – removing internal repetitions.  \nScan the current word from left to right.  Whenever two consecutive equal\nletters $xx$ are encountered, insert the gadget $K(x)$ between them.\nInsertion keeps the overall value, and $K(x)$ starts with $u\\ne x$ and ends\nwith $v\\ne x$, so the double $xx$ disappears and no new repetition is\ncreated.  As the word is finite, the process terminates after finitely many\ninsertions; the output is the desired word $w$.\n\nBecause none of the stages affects the value, $w$ fulfils (5).\\hfill$\\square$\n\n─────────────────\n(3.4)   Translation of $w$ into a directed path.\n─────────────────\nGiven $w=t_{1}\\dots t_{r}$ of Lemma 3.1 set\n\\[\nv_{i}:=\\bigl(t_{i}\\dots t_{1}g,\\; t_{i}\\bigr)\\qquad(0\\le i\\le r).\n\\]\nThen $v_{0}=(g,a)$, $v_{r}=(h,b)$ and for $1\\le i\\le r$\n\\[\nv_{i-1}\\longrightarrow v_{i}\n\\]\nis an edge of type (1), since $t_{i}\\ne t_{i-1}$.\nHence $\\mathfrak D$ is strongly connected. \\hfill$\\square$\n\n──────────────────────────────────\nStep 4.  An Eulerian circuit.\n──────────────────────────────────\nA finite, balanced, strongly connected digraph admits an Euler tour.\nFix such a circuit\n\\[\nC=(e_{0},e_{1},\\dots ,e_{L-1}),\\qquad L:=k(k-1)m.\n\\]\nWrite the tail of $e_{i}$ as $(g_{i},\\sigma_{i})$ and the head as\n$(g_{i+1},\\sigma_{i+1})$; indices are modulo $L$.\nDiscarding the second coordinates gives the cyclic list\n\\[\ng_{0},g_{1},\\dots ,g_{L-1}.\n\\tag{6}\n\\]\nFor conformity with the statement relabel $g_{0}=g_{1},\\dots,g_{L-1}=g_{L}$ and define\n\\[\n\\sigma(i):=\\sigma_{i+1}\\qquad(1\\le i\\le L).\n\\tag{7}\n\\]\n\n──────────────────────────────────\nStep 5.  Verification of the four requirements.\n──────────────────────────────────\n(1)  Frequency of group elements.  \nFix $g\\in G$.  The $k$ vertices $(g,s_{1}),\\dots,(g,s_{k})$ are the tails of\nexactly $k(k-1)$ distinct edges (each vertex contributes $k-1$ outgoing\nedges).  An Euler circuit traverses every edge \\emph{exactly once}, hence $g$\nappears $k(k-1)=L/m$ times in the list (6).\n\n(2)  Transition rule.  \nEdge $e_{i}$ goes from $(g_{i},\\sigma_{i})$ to $(g_{i+1},\\sigma_{i+1})$, so\n$g_{i+1}=\\sigma_{i+1}g_{i}$.  With the shift (7) this is\n$g_{i+1}= \\sigma(i)\\,g_{i}$.\n\n(3)  No consecutive repetition.  \nBecause $e_{i}$ leaves $(g_{i},\\sigma_{i})$ toward a vertex whose second\ncoordinate differs, $\\sigma_{i+1}\\ne\\sigma_{i}$ for all $i$, and consequently\n$\\sigma(i+1)\\ne\\sigma(i)$.\n\n(4)  Uniform ordered–pair distribution.  \nFix $p\\ne q$.  For every $g\\in G$ the edge\n\\[\n(g,s_{p})\\longrightarrow (s_{q}g,s_{q})\n\\]\noccurs exactly once in $\\mathfrak D$, so precisely $m$ edges carry the ordered\npair $(s_{p},s_{q})$ of generators.  The Euler circuit uses every edge once,\ntherefore exactly $m$ indices $j$ satisfy\n$(\\sigma_{j},\\sigma_{j+1})=(s_{p},s_{q})$.\nBy definition (7) this is equivalent to\n$(\\sigma(j-1),\\sigma(j))=(s_{p},s_{q})$, i.e. precisely the $m$ indices\nrequired in property (4).\n\nAll four conditions are fulfilled; hence the desired cyclic list exists.\\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.557088",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original and the current kernel variant, the enhanced\nproblem adds a second-order uniformity condition (Property 4):  not only\nmust each generator occur equally often, but every ordered pair of\ndistinct generators must also occur equally often in consecutive\npositions.  This forces simultaneous control of vertex frequencies and\nedge-pair frequencies, raising the combinatorial load from first-order\nto second-order statistics.\n\nTechnically, meeting this stronger requirement demands replacing the\nordinary Cayley digraph (whose vertices are the group elements) with a\nmuch larger “state” digraph of size |G|·k whose vertices keep track of\nboth a group element and the last generator used.  Only after analyzing\nthis higher-dimensional object and proving it is Eulerian can one\nproject back to a sequence in G.  The solver must therefore:\n\n• Identify the correct augmented state space (G×S) that encodes the\nsecond-order constraint.  \n• Show that this larger digraph is balanced and connected, invoking a\nnon-trivial generalization of the standard Cayley-graph argument.  \n• Construct and interpret an Eulerian circuit in that digraph, then\nprove that the projected walk satisfies the original four properties.\n\nThese additional layers of abstraction, size blow-up (m → m·k), and\nhigher-order combinatorial constraints render the enhanced variant\nsignificantly more intricate and conceptually demanding than either of\nthe predecessors."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}