{
  "index": "1996-A-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $S$ be the set of ordered triples $(a, b, c)$ of distinct elements\nof a finite set $A$. Suppose that\n\\begin{enumerate}\n\\item $(a,b,c) \\in S$ if and only if $(b,c,a) \\in S$;\n\\item $(a,b,c) \\in S$ if and only if $(c,b,a) \\notin S$;\n\\item $(a,b,c)$ and $(c,d,a)$ are both in $S$ if and only if $(b,c,d)$\nand $(d,a,b)$ are both in $S$.\n\\end{enumerate}\nProve that there exists a one-to-one function $g$ from $A$ to $R$ such\nthat $g(a) < g(b) < g(c)$ implies $(a,b,c) \\in S$. Note: $R$ is the\nset of real numbers.",
  "solution": "In fact, we will show that such a function $g$ exists with the\nproperty that $(a,b,c) \\in S$ if and only if $g(d) < g(e) < g(f)$ for\nsome cyclic permutation $(d,e,f)$ of $(a,b,c)$. We proceed by\ninduction on the number of elements in $A$. If $A =\n\\{a,b,c\\}$ and $(a,b,c) \\in S$, then choose $g$ with $g(a) < g(b) <\ng(c)$, otherwise choose $g$ with $g(a) > g(b) > g(c)$.\n\nNow let $z$ be an element of $A$ and $B = A - \\{z\\}$.\nLet $a_{1}, \\dots, a_{n}$ be the elements of $B$ labeled such that\n$g(a_{1}) < g(a_{2}) < \\cdots < g(a_{n})$. We claim that there exists\na unique $i \\in \\{1, \\dots, n\\}$ such that $(a_{i}, z, a_{i+1})\n\\in S$, where hereafter $a_{n+k} = a_{k}$.\n\nWe show existence first. Suppose no such $i$ exists; then for all\n$i,k \\in \\{1, \\dots, n\\}$, we have $(a_{i+k}, z, a_{i}) \\notin S$.\nThis holds by property 1 for $k=1$ and by induction on $k$ in\ngeneral, noting that\n\\begin{align*}\n(a_{i+k+1}, z, a_{i+k}), &(a_{i+k}, z, a_{i})  \\in S \\\\\n&\\Rightarrow (a_{i+k}, a_{i+k+1}, z), (z, a_{i}, a_{i+k}) \\in S \\\\\n&\\Rightarrow (a_{i+k+1},z,a_{i}) \\in S.\n\\end{align*}\nApplying this when $k=n$, we get $(a_{i-1}, z, a_{i}) \\in S$,\ncontradicting the fact that $(a_{i}, z, a_{i-1}) \\in S$. Hence\nexistence follows.\n\nNow we show uniqueness. Suppose $(a_{i}, z, a_{i+1}) \\in S$; then for\nany $j \\neq i-1, i, i+1$, we have $(a_{i}, a_{i+1}, a_{j}), (a_{j},\na_{j+1}, a_{i}) \\in S$ by the\nassumption on $G$. Therefore\n\\begin{align*}\n(a_{i}, z, a_{i+1}), (a_{i+1}, a_{j}, a_{i}) \\in S\n&\\Rightarrow (a_{j}, a_{i}, z) \\in S \\\\\n(a_{i}, z, a_{j}), (a_{j}, a_{j+1}, a_{i}) \\in S\n&\\Rightarrow (z, a_{j}, a_{j+1}),\n\\end{align*}\nso $(a_{j}, z, a_{j+1}) \\notin S$. The case $j =i+1$ is ruled out by\n\\[\n(a_{i}, z, a_{i+1}), (a_{i+1}, a_{i+2}, a_{i}) \\in S \\Rightarrow (z,\na_{i+1}, a_{i+2}) \\in S\n\\]\nand the case $j=i-1$ is similar.\n\nFinally, we put $g(z)$ in $(g(a_{n}), + \\infty)$ if $i = n$, and\n$(g(a_{i}), g(a_{i+1}))$ otherwise; an analysis similar to that above\nshows that $g$ has the desired property.",
  "vars": [
    "a",
    "b",
    "c",
    "d",
    "e",
    "f",
    "z",
    "i",
    "j",
    "k",
    "n",
    "a_1",
    "a_2",
    "a_n",
    "a_i",
    "a_j"
  ],
  "params": [
    "S",
    "A",
    "B",
    "g",
    "R",
    "G"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "elementa",
        "b": "elementb",
        "c": "elementc",
        "d": "elementd",
        "e": "elemente",
        "f": "elementf",
        "z": "elementz",
        "i": "indexi",
        "j": "indexj",
        "k": "indexk",
        "n": "sizen",
        "a_1": "elementone",
        "a_2": "elementtwo",
        "a_n": "elementn",
        "a_i": "elementi",
        "a_j": "elementj",
        "S": "tripleset",
        "A": "bigseta",
        "B": "subsetb",
        "g": "ordering",
        "R": "realspace",
        "G": "assumgraph"
      },
      "question": "Let $tripleset$ be the set of ordered triples $(elementa, elementb, elementc)$ of distinct elements of a finite set $bigseta$. Suppose that\n\\begin{enumerate}\n\\item $(elementa,elementb,elementc) \\in tripleset$ if and only if $(elementb,elementc,elementa) \\in tripleset$;\n\\item $(elementa,elementb,elementc) \\in tripleset$ if and only if $(elementc,elementb,elementa) \\notin tripleset$;\n\\item $(elementa,elementb,elementc)$ and $(elementc,elementd,elementa)$ are both in $tripleset$ if and only if $(elementb,elementc,elementd)$ and $(elementd,elementa,elementb)$ are both in $tripleset$.\n\\end{enumerate}\nProve that there exists a one-to-one function $ordering$ from $bigseta$ to $realspace$ such that $ordering(elementa) < ordering(elementb) < ordering(elementc)$ implies $(elementa,elementb,elementc) \\in tripleset$. Note: $realspace$ is the set of real numbers.",
      "solution": "In fact, we will show that such a function $ordering$ exists with the property that $(elementa,elementb,elementc) \\in tripleset$ if and only if $ordering(elementd) < ordering(elemente) < ordering(elementf)$ for some cyclic permutation $(elementd,elemente,elementf)$ of $(elementa,elementb,elementc)$. We proceed by induction on the number of elements in $bigseta$. If $bigseta = \\{elementa,elementb,elementc\\}$ and $(elementa,elementb,elementc) \\in tripleset$, then choose $ordering$ with $ordering(elementa) < ordering(elementb) < ordering(elementc)$; otherwise choose $ordering$ with $ordering(elementa) > ordering(elementb) > ordering(elementc)$.\n\nNow let $elementz$ be an element of $bigseta$ and set $subsetb = bigseta - \\{elementz\\}$. Let $elementone, \\dots, elementn$ be the elements of $subsetb$ labeled so that\n$ordering(elementone) < ordering(elementtwo) < \\cdots < ordering(elementn)$. We claim that there exists a unique $indexi \\in \\{1, \\dots, sizen\\}$ such that $(elementa_{indexi}, elementz, elementa_{indexi+1}) \\in tripleset$, where hereafter we agree that $elementa_{sizen+indexk} = elementa_{indexk}$.\n\nExistence.  Suppose no such $indexi$ exists; then for all $indexi,indexk \\in \\{1, \\dots, sizen\\}$ we have $(elementa_{indexi+indexk}, elementz, elementa_{indexi}) \\notin tripleset$.  This holds for $indexk=1$ by property~1 and for general $indexk$ by induction, noting that\n\\begin{align*}\n(elementa_{indexi+indexk+1}, elementz, elementa_{indexi+indexk}), &(elementa_{indexi+indexk}, elementz, elementa_{indexi}) \\in tripleset\\\\\n&\\Rightarrow (elementa_{indexi+indexk}, elementa_{indexi+indexk+1}, elementz), (elementz, elementa_{indexi}, elementa_{indexi+indexk}) \\in tripleset\\\\\n&\\Rightarrow (elementa_{indexi+indexk+1}, elementz, elementa_{indexi}) \\in tripleset.\n\\end{align*}\nTaking $indexk = sizen$ yields $(elementa_{indexi-1}, elementz, elementa_{indexi}) \\in tripleset$, contradicting $(elementa_{indexi}, elementz, elementa_{indexi-1}) \\in tripleset$. Hence existence is proved.\n\nUniqueness.  Suppose $(elementa_{indexi}, elementz, elementa_{indexi+1}) \\in tripleset$. For any $indexj \\neq indexi-1, indexi, indexi+1$ we have $(elementa_{indexi}, elementa_{indexi+1}, elementa_{indexj}), (elementa_{indexj}, elementa_{indexj+1}, elementa_{indexi}) \\in tripleset$ by the assumption on $assumgraph$. Therefore\n\\begin{align*}\n(elementa_{indexi}, elementz, elementa_{indexi+1}), (elementa_{indexi+1}, elementa_{indexj}, elementa_{indexi}) &\\in tripleset \\;\\Rightarrow\\; (elementa_{indexj}, elementa_{indexi}, elementz) \\in tripleset,\\\\\n(elementa_{indexi}, elementz, elementa_{indexj}), (elementa_{indexj}, elementa_{indexj+1}, elementa_{indexi}) &\\in tripleset \\;\\Rightarrow\\; (elementz, elementa_{indexj}, elementa_{indexj+1}) \\in tripleset,\n\\end{align*}\nso $(elementa_{indexj}, elementz, elementa_{indexj+1}) \\notin tripleset$.  The case $indexj = indexi+1$ is ruled out by\n\\[\n(elementa_{indexi}, elementz, elementa_{indexi+1}), (elementa_{indexi+1}, elementa_{indexi+2}, elementa_{indexi}) \\in tripleset \\Rightarrow (elementz, elementa_{indexi+1}, elementa_{indexi+2}) \\in tripleset,\n\\]\nand the case $indexj = indexi-1$ is analogous. Hence the required $indexi$ is unique.\n\nFinally, set $ordering(elementz)$ in $(ordering(elementn), +\\infty)$ if $indexi = sizen$, and in $(ordering(elementa_{indexi}),\\, ordering(elementa_{indexi+1}))$ otherwise.  An argument parallel to the one above shows that the resulting $ordering$ has the desired property."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "whaleroad",
        "b": "sunlitpath",
        "c": "moondapple",
        "d": "breezehill",
        "e": "stoneglade",
        "f": "riverbloom",
        "z": "embercloak",
        "i": "foxglovep",
        "j": "crimsonark",
        "k": "velvetdawn",
        "n": "dovetailr",
        "a_1": "azurefinch",
        "a_2": "amberlynx",
        "a_n": "obsidianr",
        "a_i": "marigolds",
        "a_j": "silverfern",
        "S": "lanternset",
        "A": "willowbank",
        "B": "cedargrove",
        "g": "lanternmap",
        "R": "pebblestream",
        "G": "frosthaven"
      },
      "question": "Let $lanternset$ be the set of ordered triples $(whaleroad, sunlitpath, moondapple)$ of distinct elements\nof a finite set $willowbank$. Suppose that\n\\begin{enumerate}\n\\item $(whaleroad,sunlitpath,moondapple) \\in lanternset$ if and only if $(sunlitpath,moondapple,whaleroad) \\in lanternset$;\n\\item $(whaleroad,sunlitpath,moondapple) \\in lanternset$ if and only if $(moondapple,sunlitpath,whaleroad) \\notin lanternset$;\n\\item $(whaleroad,sunlitpath,moondapple)$ and $(moondapple,breezehill,whaleroad)$ are both in $lanternset$ if and only if $(sunlitpath,moondapple,breezehill)$\nand $(breezehill,whaleroad,sunlitpath)$ are both in $lanternset$.\n\\end{enumerate}\nProve that there exists a one-to-one function $lanternmap$ from $willowbank$ to $pebblestream$ such\nthat $lanternmap(whaleroad) < lanternmap(sunlitpath) < lanternmap(moondapple)$ implies $(whaleroad,sunlitpath,moondapple) \\in lanternset$. Note: $pebblestream$ is the\nset of real numbers.",
      "solution": "In fact, we will show that such a function $lanternmap$ exists with the\nproperty that $(whaleroad,sunlitpath,moondapple) \\in lanternset$ if and only if $lanternmap(breezehill) < lanternmap(stoneglade) < lanternmap(riverbloom)$ for\nsome cyclic permutation $(breezehill,stoneglade,riverbloom)$ of $(whaleroad,sunlitpath,moondapple)$. We proceed by\ninduction on the number of elements in $willowbank$. If $willowbank =\n\\{whaleroad,sunlitpath,moondapple\\}$ and $(whaleroad,sunlitpath,moondapple) \\in lanternset$, then choose $lanternmap$ with $lanternmap(whaleroad) < lanternmap(sunlitpath) <\nlanternmap(moondapple)$, otherwise choose $lanternmap$ with $lanternmap(whaleroad) > lanternmap(sunlitpath) > lanternmap(moondapple)$.\n\nNow let $embercloak$ be an element of $willowbank$ and $cedargrove = willowbank - \\{embercloak\\}$.\nLet $whaleroad_{1}, \\dots, whaleroad_{dovetailr}$ be the elements of $cedargrove$ labeled such that\n$lanternmap(whaleroad_{1}) < lanternmap(whaleroad_{2}) < \\cdots < lanternmap(whaleroad_{dovetailr})$. We claim that there exists\na unique $foxglovep \\in \\{1, \\dots, dovetailr\\}$ such that $(whaleroad_{foxglovep}, embercloak, whaleroad_{foxglovep+1})\n\\in lanternset$, where hereafter $whaleroad_{dovetailr+velvetdawn} = whaleroad_{velvetdawn}$.\n\nWe show existence first. Suppose no such $foxglovep$ exists; then for all\n$foxglovep,velvetdawn \\in \\{1, \\dots, dovetailr\\}$, we have $(whaleroad_{foxglovep+velvetdawn}, embercloak, whaleroad_{foxglovep}) \\notin lanternset$.\nThis holds by property 1 for $velvetdawn=1$ and by induction on $velvetdawn$ in\ngeneral, noting that\n\\begin{align*}\n(whaleroad_{foxglovep+velvetdawn+1}, embercloak, whaleroad_{foxglovep+velvetdawn}), &(whaleroad_{foxglovep+velvetdawn}, embercloak, whaleroad_{foxglovep})  \\in lanternset \\\\\n&\\Rightarrow (whaleroad_{foxglovep+velvetdawn}, whaleroad_{foxglovep+velvetdawn+1}, embercloak), (embercloak, whaleroad_{foxglovep}, whaleroad_{foxglovep+velvetdawn}) \\in lanternset \\\\\n&\\Rightarrow (whaleroad_{foxglovep+velvetdawn+1},embercloak,whaleroad_{foxglovep}) \\in lanternset.\n\\end{align*}\nApplying this when $velvetdawn=dovetailr$, we get $(whaleroad_{foxglovep-1}, embercloak, whaleroad_{foxglovep}) \\in lanternset$,\ncontradicting the fact that $(whaleroad_{foxglovep}, embercloak, whaleroad_{foxglovep-1}) \\in lanternset$. Hence\nexistence follows.\n\nNow we show uniqueness. Suppose $(whaleroad_{foxglovep}, embercloak, whaleroad_{foxglovep+1}) \\in lanternset$; then for\nany $crimsonark \\neq foxglovep-1, foxglovep, foxglovep+1$, we have $(whaleroad_{foxglovep}, whaleroad_{foxglovep+1}, whaleroad_{crimsonark}), (whaleroad_{crimsonark},\nwhaleroad_{crimsonark+1}, whaleroad_{foxglovep}) \\in lanternset$ by the\nassumption on $frosthaven$. Therefore\n\\begin{align*}\n(whaleroad_{foxglovep}, embercloak, whaleroad_{foxglovep+1}), (whaleroad_{foxglovep+1}, whaleroad_{crimsonark}, whaleroad_{foxglovep}) \\in lanternset\n&\\Rightarrow (whaleroad_{crimsonark}, whaleroad_{foxglovep}, embercloak) \\in lanternset \\\\\n(whaleroad_{foxglovep}, embercloak, whaleroad_{crimsonark}), (whaleroad_{crimsonark}, whaleroad_{crimsonark+1}, whaleroad_{foxglovep}) \\in lanternset\n&\\Rightarrow (embercloak, whaleroad_{crimsonark}, whaleroad_{crimsonark+1}),\n\\end{align*}\nso $(whaleroad_{crimsonark}, embercloak, whaleroad_{crimsonark+1}) \\notin lanternset$. The case $crimsonark =foxglovep+1$ is ruled out by\n\\[\n(whaleroad_{foxglovep}, embercloak, whaleroad_{foxglovep+1}), (whaleroad_{foxglovep+1}, whaleroad_{foxglovep+2}, whaleroad_{foxglovep}) \\in lanternset \\Rightarrow (embercloak,\nwhaleroad_{foxglovep+1}, whaleroad_{foxglovep+2}) \\in lanternset\n\\]\nand the case $crimsonark=foxglovep-1$ is similar.\n\nFinally, we put $lanternmap(embercloak)$ in $(lanternmap(whaleroad_{dovetailr}), + \\infty)$ if $foxglovep = dovetailr$, and\n$(lanternmap(whaleroad_{foxglovep}), lanternmap(whaleroad_{foxglovep+1}))$ otherwise; an analysis similar to that above\nshows that $lanternmap$ has the desired property."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "omegaelem",
        "b": "alphachar",
        "c": "xenovalue",
        "d": "terminalid",
        "e": "westfield",
        "f": "nullvalue",
        "z": "firstentry",
        "i": "outsider",
        "j": "insidera",
        "k": "centered",
        "n": "zerobase",
        "a_1": "lastfirst",
        "a_2": "lastsecond",
        "a_n": "lastzero",
        "a_i": "lastindex",
        "a_j": "lastinside",
        "S": "emptysup",
        "A": "infinitepool",
        "B": "wholesetb",
        "g": "disorderfn",
        "R": "imaginaryset",
        "G": "phantomset"
      },
      "question": "Let $emptysup$ be the set of ordered triples $(omegaelem, alphachar, xenovalue)$ of distinct elements\nof a finite set $infinitepool$. Suppose that\n\\begin{enumerate}\n\\item $(omegaelem,alphachar,xenovalue) \\in emptysup$ if and only if $(alphachar,xenovalue,omegaelem) \\in emptysup$;\n\\item $(omegaelem,alphachar,xenovalue) \\in emptysup$ if and only if $(xenovalue,alphachar,omegaelem) \\notin emptysup$;\n\\item $(omegaelem,alphachar,xenovalue)$ and $(xenovalue,terminalid,omegaelem)$ are both in $emptysup$ if and only if $(alphachar,xenovalue,terminalid)$\nand $(terminalid,omegaelem,alphachar)$ are both in $emptysup$.\n\\end{enumerate}\nProve that there exists a one-to-one function $disorderfn$ from $infinitepool$ to $imaginaryset$ such\nthat $disorderfn(omegaelem) < disorderfn(alphachar) < disorderfn(xenovalue)$ implies $(omegaelem,alphachar,xenovalue) \\in emptysup$. Note: $imaginaryset$ is the\nset of real numbers.",
      "solution": "In fact, we will show that such a function $disorderfn$ exists with the\nproperty that $(omegaelem,alphachar,xenovalue) \\in emptysup$ if and only if $disorderfn(terminalid) < disorderfn(westfield) < disorderfn(nullvalue)$ for\nsome cyclic permutation $(terminalid,westfield,nullvalue)$ of $(omegaelem,alphachar,xenovalue)$. We proceed by\ninduction on the number of elements in $infinitepool$. If $infinitepool =\n\\{omegaelem,alphachar,xenovalue\\}$ and $(omegaelem,alphachar,xenovalue) \\in emptysup$, then choose $disorderfn$ with $disorderfn(omegaelem) < disorderfn(alphachar) <\ndisorderfn(xenovalue)$, otherwise choose $disorderfn$ with $disorderfn(omegaelem) > disorderfn(alphachar) > disorderfn(xenovalue)$.\n\nNow let $firstentry$ be an element of $infinitepool$ and $wholesetb = infinitepool - \\{firstentry\\}$.\nLet $omegaelem_{1}, \\dots, omegaelem_{zerobase}$ be the elements of $wholesetb$ labeled such that\n$disorderfn(omegaelem_{1}) < disorderfn(omegaelem_{2}) < \\cdots < disorderfn(omegaelem_{zerobase})$. We claim that there exists\na unique $outsider \\in \\{1, \\dots, zerobase\\}$ such that $(omegaelem_{outsider}, firstentry, omegaelem_{outsider+1})\n\\in emptysup$, where hereafter $omegaelem_{zerobase+centered} = omegaelem_{centered}$.\n\nWe show existence first. Suppose no such $outsider$ exists; then for all\n$outsider,centered \\in \\{1, \\dots, zerobase\\}$, we have $(omegaelem_{outsider+centered}, firstentry, omegaelem_{outsider}) \\notin emptysup$.\nThis holds by property 1 for $centered=1$ and by induction on $centered$ in\ngeneral, noting that\n\\begin{align*}\n(omegaelem_{outsider+centered+1}, firstentry, omegaelem_{outsider+centered}), &(omegaelem_{outsider+centered}, firstentry, omegaelem_{outsider})  \\in emptysup \\\\\n&\\Rightarrow (omegaelem_{outsider+centered}, omegaelem_{outsider+centered+1}, firstentry), (firstentry, omegaelem_{outsider}, omegaelem_{outsider+centered}) \\in emptysup \\\\\n&\\Rightarrow (omegaelem_{outsider+centered+1},firstentry,omegaelem_{outsider}) \\in emptysup.\n\\end{align*}\nApplying this when $centered=zerobase$, we get $(omegaelem_{outsider-1}, firstentry, omegaelem_{outsider}) \\in emptysup$,\ncontradicting the fact that $(omegaelem_{outsider}, firstentry, omegaelem_{outsider-1}) \\in emptysup$. Hence\nexistence follows.\n\nNow we show uniqueness. Suppose $(omegaelem_{outsider}, firstentry, omegaelem_{outsider+1}) \\in emptysup$; then for\nany $insidera \\neq outsider-1, outsider, outsider+1$, we have $(omegaelem_{outsider}, omegaelem_{outsider+1}, omegaelem_{insidera}), (omegaelem_{insidera},\nomegaelem_{insidera+1}, omegaelem_{outsider}) \\in emptysup$ by the\nassumption on $phantomset$. Therefore\n\\begin{align*}\n(omegaelem_{outsider}, firstentry, omegaelem_{outsider+1}), (omegaelem_{outsider+1}, omegaelem_{insidera}, omegaelem_{outsider}) \\in emptysup\n&\\Rightarrow (omegaelem_{insidera}, omegaelem_{outsider}, firstentry) \\in emptysup \\\\\n(omegaelem_{outsider}, firstentry, omegaelem_{insidera}), (omegaelem_{insidera}, omegaelem_{insidera+1}, omegaelem_{outsider}) \\in emptysup\n&\\Rightarrow (firstentry, omegaelem_{insidera}, omegaelem_{insidera+1}),\n\\end{align*}\nso $(omegaelem_{insidera}, firstentry, omegaelem_{insidera+1}) \\notin emptysup$. The case $insidera =outsider+1$ is ruled out by\n\\[\n(omegaelem_{outsider}, firstentry, omegaelem_{outsider+1}), (omegaelem_{outsider+1}, omegaelem_{outsider+2}, omegaelem_{outsider}) \\in emptysup \\Rightarrow (firstentry,\nomegaelem_{outsider+1}, omegaelem_{outsider+2}) \\in emptysup\n\\]\nand the case $insidera=outsider-1$ is similar.\n\nFinally, we put $disorderfn(firstentry)$ in $(disorderfn(omegaelem_{zerobase}), + \\infty)$ if $outsider = zerobase$, and\n$(disorderfn(omegaelem_{outsider}), disorderfn(omegaelem_{outsider+1}))$ otherwise; an analysis similar to that above\nshows that $disorderfn$ has the desired property."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "ptvfqion",
        "d": "lmsrkdbe",
        "e": "vfwyrzna",
        "f": "kgzlqmtp",
        "z": "tnjgwepo",
        "i": "mczvkdrl",
        "j": "nubksyaf",
        "k": "fhzqvemi",
        "n": "drxqplou",
        "a_1": "xkdjshla",
        "a_2": "sbvcltno",
        "a_n": "wqtmpdse",
        "a_i": "plnxryud",
        "a_j": "gdjqvrka",
        "S": "lqmgidest",
        "A": "qfztmxsa",
        "B": "rnszvpqa",
        "g": "xmtrawob",
        "R": "szpyghum",
        "G": "bwcrlnaq"
      },
      "question": "Let $lqmgidest$ be the set of ordered triples $(qzxwvtnp, hjgrksla, ptvfqion)$ of distinct elements\nof a finite set $qfztmxsa$. Suppose that\n\\begin{enumerate}\n\\item $(qzxwvtnp,hjgrksla,ptvfqion) \\in lqmgidest$ if and only if $(hjgrksla,ptvfqion,qzxwvtnp) \\in lqmgidest$;\n\\item $(qzxwvtnp,hjgrksla,ptvfqion) \\in lqmgidest$ if and only if $(ptvfqion,hjgrksla,qzxwvtnp) \\notin lqmgidest$;\n\\item $(qzxwvtnp,hjgrksla,ptvfqion)$ and $(ptvfqion,lmsrkdbe,qzxwvtnp)$ are both in $lqmgidest$ if and only if $(hjgrksla,ptvfqion,lmsrkdbe)$\nand $(lmsrkdbe,qzxwvtnp,hjgrksla)$ are both in $lqmgidest$.\n\\end{enumerate}\nProve that there exists a one-to-one function $xmtrawob$ from $qfztmxsa$ to $szpyghum$ such\nthat $xmtrawob(qzxwvtnp) < xmtrawob(hjgrksla) < xmtrawob(ptvfqion)$ implies $(qzxwvtnp,hjgrksla,ptvfqion) \\in lqmgidest$. Note: $szpyghum$ is the\nset of real numbers.",
      "solution": "In fact, we will show that such a function $xmtrawob$ exists with the property that $(qzxwvtnp,hjgrksla,ptvfqion) \\in lqmgidest$ if and only if $xmtrawob(lmsrkdbe) < xmtrawob(vfwyrzna) < xmtrawob(kgzlqmtp)$ for some cyclic permutation $(lmsrkdbe,vfwyrzna,kgzlqmtp)$ of $(qzxwvtnp,hjgrksla,ptvfqion)$. We proceed by induction on the number of elements in $qfztmxsa$. If $qfztmxsa = \\{qzxwvtnp,hjgrksla,ptvfqion\\}$ and $(qzxwvtnp,hjgrksla,ptvfqion) \\in lqmgidest$, then choose $xmtrawob$ with $xmtrawob(qzxwvtnp) < xmtrawob(hjgrksla) < xmtrawob(ptvfqion)$, otherwise choose $xmtrawob$ with $xmtrawob(qzxwvtnp) > xmtrawob(hjgrksla) > xmtrawob(ptvfqion)$.\n\nNow let $tnjgwepo$ be an element of $qfztmxsa$ and $rnszvpqa = qfztmxsa - \\{tnjgwepo\\}$.\nLet $xkdjshla, \\dots, wqtmpdse$ be the elements of $rnszvpqa$ labeled such that\n$xmtrawob(xkdjshla) < xmtrawob(sbvcltno) < \\cdots < xmtrawob(wqtmpdse)$. We claim that there exists\na unique $mczvkdrl \\in \\{1, \\dots, drxqplou\\}$ such that $(plnxryud, tnjgwepo, qzxwvtnp_{mczvkdrl+1})\n\\in lqmgidest$, where hereafter $qzxwvtnp_{drxqplou+fhzqvemi} = qzxwvtnp_{fhzqvemi}$.\n\nWe show existence first. Suppose no such $mczvkdrl$ exists; then for all\n$mczvkdrl,fhzqvemi \\in \\{1, \\dots, drxqplou\\}$, we have $(qzxwvtnp_{mczvkdrl+fhzqvemi}, tnjgwepo, plnxryud) \\notin lqmgidest$.\nThis holds by property 1 for $fhzqvemi=1$ and by induction on $fhzqvemi$ in\ngeneral, noting that\n\\begin{align*}\n(qzxwvtnp_{mczvkdrl+fhzqvemi+1}, tnjgwepo, qzxwvtnp_{mczvkdrl+fhzqvemi}), &(qzxwvtnp_{mczvkdrl+fhzqvemi}, tnjgwepo, plnxryud)  \\in lqmgidest \\\\\n&\\Rightarrow (qzxwvtnp_{mczvkdrl+fhzqvemi}, qzxwvtnp_{mczvkdrl+fhzqvemi+1}, tnjgwepo), (tnjgwepo, plnxryud, qzxwvtnp_{mczvkdrl+fhzqvemi}) \\in lqmgidest \\\\\n&\\Rightarrow (qzxwvtnp_{mczvkdrl+fhzqvemi+1},tnjgwepo,plnxryud) \\in lqmgidest.\n\\end{align*}\nApplying this when $fhzqvemi=drxqplou$, we get $(qzxwvtnp_{mczvkdrl-1}, tnjgwepo, plnxryud) \\in lqmgidest$,\ncontradicting the fact that $(plnxryud, tnjgwepo, qzxwvtnp_{mczvkdrl-1}) \\in lqmgidest$. Hence\nexistence follows.\n\nNow we show uniqueness. Suppose $(plnxryud, tnjgwepo, qzxwvtnp_{mczvkdrl+1}) \\in lqmgidest$; then for\nany $nubksyaf \\neq mczvkdrl-1, mczvkdrl, mczvkdrl+1$, we have $(plnxryud, qzxwvtnp_{mczvkdrl+1}, qzxwvtnp_{nubksyaf}), (qzxwvtnp_{nubksyaf},\nqzxwvtnp_{nubksyaf+1}, plnxryud) \\in lqmgidest$ by the\nassumption on $bwcrlnaq$. Therefore\n\\begin{align*}\n(plnxryud, tnjgwepo, qzxwvtnp_{mczvkdrl+1}), (qzxwvtnp_{mczvkdrl+1}, qzxwvtnp_{nubksyaf}, plnxryud) \\in lqmgidest\n&\\Rightarrow (qzxwvtnp_{nubksyaf}, plnxryud, tnjgwepo) \\in lqmgidest \\\\\n(plnxryud, tnjgwepo, qzxwvtnp_{nubksyaf}), (qzxwvtnp_{nubksyaf}, qzxwvtnp_{nubksyaf+1}, plnxryud) \\in lqmgidest\n&\\Rightarrow (tnjgwepo, qzxwvtnp_{nubksyaf}, qzxwvtnp_{nubksyaf+1}),\n\\end{align*}\nso $(qzxwvtnp_{nubksyaf}, tnjgwepo, qzxwvtnp_{nubksyaf+1}) \\notin lqmgidest$. The case $nubksyaf = mczvkdrl+1$ is ruled out by\n\\[\n(plnxryud, tnjgwepo, qzxwvtnp_{mczvkdrl+1}), (qzxwvtnp_{mczvkdrl+1}, qzxwvtnp_{mczvkdrl+2}, plnxryud) \\in lqmgidest \\Rightarrow (tnjgwepo,\nqzxwvtnp_{mczvkdrl+1}, qzxwvtnp_{mczvkdrl+2}) \\in lqmgidest\n\\]\nand the case $nubksyaf=mczvkdrl-1$ is similar.\n\nFinally, we put $xmtrawob(tnjgwepo)$ in $(xmtrawob(wqtmpdse), + \\infty)$ if $mczvkdrl = drxqplou$, and\n$(xmtrawob(plnxryud), xmtrawob(qzxwvtnp_{mczvkdrl+1}))$ otherwise; an analysis similar to that above\nshows that $xmtrawob$ has the desired property."
    },
    "kernel_variant": {
      "question": "Let A be a finite set with |A| \\ge 3 and let S be a collection of ordered triples (a,b,c) of pairwise distinct elements of A such that\n\n1. (cyclic invariance) (a,b,c) \\in S \\;\\Longleftrightarrow\\; (b,c,a) \\in S;\n2. (reversal / asymmetry) (a,b,c) \\in S \\;\\Longrightarrow\\; (c,b,a) \\notin S;\n3. (four–term exchange) for every four distinct a,b,c,d \\in A\n   (a,b,c),(c,d,a) \\in S \\;\\Longleftrightarrow\\; (b,c,d),(d,a,b) \\in S;\n4. (totality) for every three distinct a,b,c \\in A exactly one of the two triples (a,b,c) and (c,b,a) belongs to S.\n\nProve that there exists an injective map g:A \\to \\mathbb R such that for all distinct a,b,c \\in A\n\ng(a)<g(b)<g(c)\\qquad\\Longrightarrow\\qquad(a,b,c)\\in S.\n(The converse implication is not required.)",
      "solution": "Throughout write n:=|A| \\ge 3.\n\n─────────────────────────────────────────────\n1.  The axioms make S a (complete) cyclic order\n─────────────────────────────────────────────\n\nA ternary relation T on a set X is called a cyclic order if it fulfils\n(C1)  (a,b,c)\\in T \\Longrightarrow (b,c,a)\\in T;\n(C2)  (a,b,c)\\in T \\Longrightarrow (c,b,a)\\notin T; and\n(C3)  whenever (a,b,c),(a,c,d)\\in T with a,b,c,d pairwise distinct, then (a,b,d)\\in T.\n\nIt is classical (and easy to check) that condition (C3) is equivalent to the\nfour–term exchange used in the statement.  Items 1–3 therefore already yield\n(C1)–(C3).  Item 4 additionally guarantees that for every distinct a,b,c\nexactly one of the two possible orientations occurs, i.e. the cyclic order is\ntotal (sometimes called **complete**).  From now on we treat S as a fixed\ncomplete cyclic order on A and freely use the standard facts about cyclic\norders.\n\n─────────────────────────────────────────────\n2.  Deleting one point produces a linear order\n─────────────────────────────────────────────\n\nFix an element z\\in A and put B:=A\\setminus\\{z\\} (so |B|=n-1).  Define a\nbinary relation\n\n    x <_{z} y \\;\\Longleftrightarrow\\; (z,x,y) \\in S\\qquad(x,y\\in B).   (2.1)\n\nLemma 2.1  The relation <_{z} is a strict linear (total) order on B.\n\nProof.\nAnti–symmetry and transitivity follow exactly as in the original write–up; we\nonly repeat the argument for totality because this is where the new condition\n(4) is used.\n\n• (totality)  Let x\\ne y in B.  Applying (4) to the three distinct elements\n  z,x,y we know that precisely one of (z,x,y) and (z,y,x) is in S.  Hence\n  exactly one of x<_{z}y or y<_{z}x holds.\n∎\n\nBecause <_{z} is strict and total we may list the elements of B as\n\n    b_{1} <_{z} b_{2} <_{z} \\dots <_{z} b_{m}\\qquad(m:=n-1).   (2.2)\n\n─────────────────────────────────────────────\n3.  Chains in the linear order are compatible with S\n─────────────────────────────────────────────\n\nLemma 3.1  If x<_{z}y<_{z}w in B, then (x,y,w)\\in S.\n\nProof.\nExactly as before: (z,x,y),(z,y,w)\\in S yields (y,w,z)\\in S by cyclicity, and\nthe four–term rule for z,x,y,w gives (x,y,w)\\in S. ∎\n\nCorollary 3.2  For any indices 1\\le i<j<k\\le m we have (b_{i},b_{j},b_{k})\\in S.\n(The corollary follows by repeatedly applying Lemma 3.1.)\n\n─────────────────────────────────────────────\n4.  A convenient linear embedding of A into \\mathbb R\n─────────────────────────────────────────────\n\nDefine g:A\\to\\mathbb R by\n\n    g(b_{i}) := i \\quad(1\\le i\\le m), \\qquad g(z):=m+1.   (4.1)\n\nThe map g is clearly injective.  We verify that it has the required property.\n\nLet a,b,c\\in A be distinct with g(a)<g(b)<g(c).  Three situations can occur.\n\nCase 1.  None of a,b,c equals z.  Then a=b_{i}, b=b_{j}, c=b_{k} with i<j<k,\nand Corollary 3.2 gives (a,b,c)\\in S.\n\nCase 2.  c=z.  Then a,b\\in B and g(a)<g(b) implies a<_{z}b, i.e. (z,a,b)\\in S\nby (2.1); cyclicity yields (a,b,z)=(a,b,c)\\in S.\n\nCase 3.  The element z would have to be the smallest or middle value, but\nby construction g(z)=m+1 is the largest value among all g–images, so this case\ncannot occur.\n\nHence in every possible configuration\n\ng(a)<g(b)<g(c)\\;\\Rightarrow\\;(a,b,c)\\in S.\n\nThis finishes the proof. ∎",
      "_meta": {
        "core_steps": [
          "Regard properties (1) & (2) as defining a cyclic orientation on every triple.",
          "Do induction on n = |A| (base n = 3 handled by choosing either increasing or decreasing order).",
          "Inductive lemma: for the new element z there is a UNIQUE adjacent pair (a_i , a_{i+1}) in the current linear order with (a_i , z , a_{i+1}) ∈ S (proved by contradiction using the three axioms).",
          "Insert z by picking g(z) strictly between g(a_i) and g(a_{i+1}) (or to the right of all of them when i = n).",
          "Check that this extended g keeps ↔ between ‘cyclicly increasing’ and membership in S; induction completes."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The target set only needs to be a totally ordered set large enough to host |A| distinct values; any such set works.",
            "original": "R (the real numbers)"
          },
          "slot2": {
            "description": "The direction of the inequality criterion can be reversed (use g' = −g, or decree that decreasing triples lie in S).",
            "original": "g(a) < g(b) < g(c) ⇒ (a,b,c) ∈ S"
          },
          "slot3": {
            "description": "The specific choice in the 3-element base case (increasing vs. decreasing assignment) is arbitrary; either orientation starts the induction.",
            "original": "For |A| = 3 pick g(a)<g(b)<g(c) when (a,b,c)∈S, otherwise g(a)>g(b)>g(c)"
          },
          "slot4": {
            "description": "Any consistent labelling/ordering of B = A\\{z} (not necessarily a_1,…,a_n) suffices; only the relative order matters.",
            "original": "a_1,…,a_n with g(a_1)<⋯<g(a_n)"
          },
          "slot5": {
            "description": "The modular indexing convention (a_{n+k}=a_k) is inessential; one could work with indices mod n in any equivalent notation.",
            "original": "a_{n+k} = a_k"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}