{
  "index": "1994-A-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $f_1, \\dots, f_{10}$ be bijections of the set of integers such that for\neach integer $n$, there is some composition $f_{i_1} \\circ f_{i_2}\n\\circ \\cdots \\circ f_{i_m}$ of these functions (allowing repetitions)\nwhich maps 0 to $n$. Consider the set of 1024 functions\n\\[\n\\mathcal{F} = \\{f_1^{e_1} \\circ f_2^{e_2} \\circ \\cdots \\circ f_{10}^{e_{10}}\\},\n\\]\n$e_i = 0$ or 1 for $1 \\leq i \\leq 10$. ($f_i^0$ is the identity function\nand $f_i^1 = f_i$.) Show that if $A$ is any nonempty finite set of\nintegers, then at most 512 of the functions in $\\mathcal{F}$ map $A$ to\nitself.",
  "solution": "Solution. We say that a bijection of the integers to itself preserves a subset \\( A \\) if it restricts to a bijection of \\( A \\).\n\nLet \\( G \\) be the group of permutations of \\( \\mathbb{Z} \\) generated by the \\( f_{i} \\). Then \\( G \\) has elements mapping any integer \\( m \\) to 0 , and elements mapping 0 to any integer \\( n \\), so \\( G \\) acts transitively on \\( \\mathbb{Z} \\). Hence no nonempty proper subset \\( A \\) can be preserved by all the \\( f_{i} \\). It remains to prove the following lemma.\n\nLemma. Let \\( f_{1}, \\ldots, f_{n} \\) be bijections \\( \\mathbb{Z} \\rightarrow \\mathbb{Z} \\), and let \\( A \\) be a subset of \\( \\mathbb{Z} \\). Suppose that some \\( f_{i} \\) does not preserve \\( A \\). Then at most \\( 2^{n-1} \\) elements of\n\\[\n\\mathcal{F}_{n}=\\left\\{f_{1}^{e_{1}} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ f_{n}^{e_{n}}: e_{i}=0 \\text { or } 1\\right\\}\n\\]\npreserve \\( A \\).\nProof 1 of Lemma (inductive). The lemma is true for \\( n=1 \\), so assume it it is true for all \\( n<k \\) (for some \\( k>1 \\) ), and false for \\( n=k \\). Then by the Pigeonhole Principle, there are \\( e_{1}, \\ldots, e_{k-1} \\in\\{0,1\\} \\) such that both\n\\[\nf_{1}^{e_{1}} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ f_{k-1}^{e_{k-1}} \\quad \\text { and } \\quad f_{1}^{e_{1}} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ f_{k-1}^{e_{k-1}} \\circ f_{k}\n\\]\nfix \\( A \\). Hence \\( f_{k} \\) preserves \\( A \\) as well. By the inductive hypothesis, at most \\( 2^{k-2} \\) elements of \\( \\mathcal{F}_{k-1} \\) fix \\( A \\); thus at most \\( 2^{k-1} \\) elements of \\( \\mathcal{F}_{k} \\) fix \\( A \\), giving a contradiction.\n\nProof 2 of Lemma (noninductive). Let \\( k \\) be the largest integer such that \\( f_{k} \\) does not map \\( A \\) to itself, and suppose that more than \\( 2^{n-1} \\) of the functions \\( \\mathcal{F}_{n} \\operatorname{map} A \\) to itself. By the Pigeonhole Principle, there are\n\\[\ne_{1}, \\ldots, e_{k-1}, e_{k+1}, \\ldots, e_{n} \\in\\{0,1\\}\n\\]\nsuch that both\n\\[\nf_{1}^{e_{1}} \\circ \\cdots \\circ f_{k-1}^{e_{k-1}} \\circ f_{k+1}^{e_{k+1}} \\circ \\cdots \\circ f_{n}^{e_{n}} \\quad \\text { and } \\quad f_{1}^{e_{1}} \\circ \\cdots \\circ f_{k-1}^{e_{k-1}} \\circ f_{k} \\circ f_{k+1}^{e_{k+1}} \\circ \\cdots \\circ f_{n}^{e_{n}}\n\\]\nboth fix \\( A \\). Hence both \\( F_{1}=f_{1}^{e_{1}} \\circ \\cdots \\circ f_{k-1}^{e_{k-1}} \\) and \\( F_{1}=f_{1}^{e_{1}} \\circ \\cdots \\circ f_{k-1}^{e_{k-1}} \\circ f_{k} \\) both map \\( A \\) to itself. But then \\( F_{1}^{-1} \\circ F_{2}=f_{k} \\) also maps \\( A \\) to itself, giving a contradiction.",
  "vars": [
    "n",
    "m",
    "k",
    "i",
    "e_i",
    "e_1"
  ],
  "params": [
    "A",
    "G",
    "F",
    "F_n",
    "F_1",
    "F_2",
    "f_1",
    "f_10",
    "f_i",
    "f_i_1",
    "f_i_2",
    "f_i_m",
    "f_k-1",
    "f_k",
    "f_n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "targetint",
        "m": "sourceint",
        "k": "pivotalidx",
        "i": "indexvar",
        "e_i": "exponentidx",
        "e_1": "exponentone",
        "A": "intset",
        "G": "permgroup",
        "F": "composition",
        "F_n": "compositionn",
        "F_1": "compositionone",
        "F_2": "compositiontwo",
        "f_1": "bijectionone",
        "f_10": "bijectionten",
        "f_i": "bijectionidx",
        "f_i_1": "bijectionfirst",
        "f_i_2": "bijectionsecond",
        "f_i_m": "bijectionmth",
        "f_k-1": "bijectionprev",
        "f_k": "bijectionkth",
        "f_n": "bijectionnth"
      },
      "question": "Let $bijectionone, \\dots, bijectionten$ be bijections of the set of integers such that for\neach integer $targetint$, there is some composition $bijectionfirst \\circ bijectionsecond\n\\circ \\cdots \\circ bijectionmth$ of these functions (allowing repetitions)\nwhich maps $0$ to $targetint$. Consider the set of $1024$ functions\n\\[\n\\mathcal{composition} = \\{bijectionone^{exponentone} \\circ f_2^{e_2} \\circ \\cdots \\circ bijectionten^{e_{10}}\\},\n\\]\n$exponentidx = 0$ or $1$ for $1 \\leq indexvar \\leq 10$. (bijectionidx$^{0}$ is the identity function\nand bijectionidx$^{1} = $bijectionidx.) Show that if $intset$ is any nonempty finite set of\nintegers, then at most $512$ of the functions in $\\mathcal{composition}$ map $intset$ to\nitself.",
      "solution": "Solution. We say that a bijection of the integers to itself preserves a subset \\( intset \\) if it restricts to a bijection of \\( intset \\).\n\nLet \\( permgroup \\) be the group of permutations of \\( \\mathbb{Z} \\) generated by the \\( bijectionidx \\). Then \\( permgroup \\) has elements mapping any integer \\( sourceint \\) to 0, and elements mapping 0 to any integer \\( targetint \\), so \\( permgroup \\) acts transitively on \\( \\mathbb{Z} \\). Hence no nonempty proper subset \\( intset \\) can be preserved by all the \\( bijectionidx \\). It remains to prove the following lemma.\n\nLemma. Let \\( bijectionone, \\ldots, bijectionnth \\) be bijections \\( \\mathbb{Z} \\rightarrow \\mathbb{Z} \\), and let \\( intset \\) be a subset of \\( \\mathbb{Z} \\). Suppose that some \\( bijectionidx \\) does not preserve \\( intset \\). Then at most \\( 2^{targetint-1} \\) elements of\n\\[\n\\mathcal{composition}_{targetint}=\\left\\{bijectionone^{exponentone} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ bijectionnth^{e_{targetint}}: exponentidx=0 \\text { or } 1\\right\\}\n\\]\npreserve \\( intset \\).\n\nProof 1 of Lemma (inductive). The lemma is true for \\( targetint=1 \\), so assume it is true for all \\( targetint<pivotalidx \\) (for some \\( pivotalidx>1 \\)), and false for \\( targetint=pivotalidx \\). Then by the Pigeonhole Principle, there are \\( exponentone, e_{2}, \\ldots, e_{pivotalidx-1} \\in\\{0,1\\} \\) such that both\n\\[\nbijectionone^{exponentone} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ bijectionprev^{e_{pivotalidx-1}}\n\\quad \\text { and } \\quad\nbijectionone^{exponentone} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ bijectionprev^{e_{pivotalidx-1}} \\circ bijectionkth\n\\]\nfix \\( intset \\). Hence \\( bijectionkth \\) preserves \\( intset \\) as well. By the inductive hypothesis, at most \\( 2^{pivotalidx-2} \\) elements of \\( \\mathcal{composition}_{pivotalidx-1} \\) fix \\( intset \\); thus at most \\( 2^{pivotalidx-1} \\) elements of \\( \\mathcal{composition}_{pivotalidx} \\) fix \\( intset \\), giving a contradiction.\n\nProof 2 of Lemma (noninductive). Let \\( pivotalidx \\) be the largest integer such that \\( bijectionkth \\) does not map \\( intset \\) to itself, and suppose that more than \\( 2^{targetint-1} \\) of the functions \\( \\mathcal{composition}_{targetint} \\operatorname{map} intset \\) to itself. By the Pigeonhole Principle, there are\n\\[\nexponentone, e_{2}, \\ldots, e_{pivotalidx-1}, e_{pivotalidx+1}, \\ldots, e_{targetint} \\in\\{0,1\\}\n\\]\nsuch that both\n\\[\nbijectionone^{exponentone} \\circ \\cdots \\circ bijectionprev^{e_{pivotalidx-1}} \\circ f_{pivotalidx+1}^{e_{pivotalidx+1}} \\circ \\cdots \\circ bijectionnth^{e_{targetint}}\n\\quad \\text { and } \\quad\nbijectionone^{exponentone} \\circ \\cdots \\circ bijectionprev^{e_{pivotalidx-1}} \\circ bijectionkth \\circ f_{pivotalidx+1}^{e_{pivotalidx+1}} \\circ \\cdots \\circ bijectionnth^{e_{targetint}}\n\\]\nboth fix \\( intset \\). Hence both \\( compositionone = bijectionone^{exponentone} \\circ \\cdots \\circ bijectionprev^{e_{pivotalidx-1}} \\) and \\( compositionone = bijectionone^{exponentone} \\circ \\cdots \\circ bijectionprev^{e_{pivotalidx-1}} \\circ bijectionkth \\) both map \\( intset \\) to itself. But then \\( compositionone^{-1} \\circ compositiontwo = bijectionkth \\) also maps \\( intset \\) to itself, giving a contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "seashells",
        "m": "pinecones",
        "k": "sandpiper",
        "i": "lighthouse",
        "e_i": "rainwater",
        "e_1": "moonlight",
        "A": "blueberry",
        "G": "dragonfly",
        "F": "riverbank",
        "F_n": "riverdelta",
        "F_1": "rivermouth",
        "F_2": "riversource",
        "f_1": "sunflower",
        "f_10": "windswept",
        "f_i": "stargazer",
        "f_i_1": "stargazeone",
        "f_i_2": "stargazetwo",
        "f_i_m": "stargazemany",
        "f_k-1": "woodland",
        "f_k": "treetop",
        "f_n": "hillcrest"
      },
      "question": "Let $sunflower, \\dots, windswept$ be bijections of the set of integers such that for\neach integer $seashells$, there is some composition $stargazeone \\circ stargazetwo\n\\circ \\cdots \\circ stargazemany$ of these functions (allowing repetitions)\nwhich maps 0 to $seashells$. Consider the set of 1024 functions\n\\[\n\\mathcal{F} = \\{sunflower^{moonlight} \\circ f_2^{e_2} \\circ \\cdots \\circ windswept^{e_{10}}\\},\n\\]\n$rainwater = 0$ or 1 for $1 \\leq lighthouse \\leq 10$. ($stargazer^0$ is the identity function\nand $stargazer^1 = stargazer$.) Show that if $blueberry$ is any nonempty finite set of\nintegers, then at most 512 of the functions in $\\mathcal{F}$ map $blueberry$ to\nitself.",
      "solution": "Solution. We say that a bijection of the integers to itself preserves a subset \\( blueberry \\) if it restricts to a bijection of \\( blueberry \\).\n\nLet \\( dragonfly \\) be the group of permutations of \\( \\mathbb{Z} \\) generated by the \\( stargazer \\). Then \\( dragonfly \\) has elements mapping any integer \\( pinecones \\) to 0 , and elements mapping 0 to any integer \\( seashells \\), so \\( dragonfly \\) acts transitively on \\( \\mathbb{Z} \\). Hence no nonempty proper subset \\( blueberry \\) can be preserved by all the \\( stargazer \\). It remains to prove the following lemma.\n\nLemma. Let \\( sunflower, \\ldots, hillcrest \\) be bijections \\( \\mathbb{Z} \\rightarrow \\mathbb{Z} \\), and let \\( blueberry \\) be a subset of \\( \\mathbb{Z} \\). Suppose that some \\( stargazer \\) does not preserve \\( blueberry \\). Then at most \\( 2^{seashells-1} \\) elements of\n\\[\n\\mathcal{F}_{seashells}=\\left\\{sunflower^{moonlight} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ hillcrest^{rainwater}: rainwater=0 \\text { or } 1\\right\\}\n\\]\npreserve \\( blueberry \\).\n\nProof 1 of Lemma (inductive). The lemma is true for \\( seashells=1 \\), so assume it is true for all \\( seashells<sandpiper \\) (for some \\( sandpiper>1 \\) ), and false for \\( seashells=sandpiper \\). Then by the Pigeonhole Principle, there are \\( e_{1}, \\ldots, e_{sandpiper-1} \\in\\{0,1\\} \\) such that both\n\\[\nsunflower^{moonlight} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ woodland^{e_{sandpiper-1}} \\quad \\text { and } \\quad sunflower^{moonlight} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ woodland^{e_{sandpiper-1}} \\circ treetop\n\\]\nfix \\( blueberry \\). Hence \\( treetop \\) preserves \\( blueberry \\) as well. By the inductive hypothesis, at most \\( 2^{sandpiper-2} \\) elements of \\( \\mathcal{F}_{sandpiper-1} \\) fix \\( blueberry \\); thus at most \\( 2^{sandpiper-1} \\) elements of \\( \\mathcal{F}_{sandpiper} \\) fix \\( blueberry \\), giving a contradiction.\n\nProof 2 of Lemma (noninductive). Let \\( sandpiper \\) be the largest integer such that \\( treetop \\) does not map \\( blueberry \\) to itself, and suppose that more than \\( 2^{seashells-1} \\) of the functions \\( \\mathcal{F}_{seashells} \\operatorname{map} blueberry \\) to itself. By the Pigeonhole Principle, there are\n\\[\ne_{1}, \\ldots, e_{sandpiper-1}, e_{sandpiper+1}, \\ldots, e_{seashells} \\in\\{0,1\\}\n\\]\nsuch that both\n\\[\nsunflower^{moonlight} \\circ \\cdots \\circ woodland^{e_{sandpiper-1}} \\circ f_{sandpiper+1}^{e_{sandpiper+1}} \\circ \\cdots \\circ hillcrest^{e_{seashells}} \\quad \\text { and } \\quad sunflower^{moonlight} \\circ \\cdots \\circ woodland^{e_{sandpiper-1}} \\circ treetop \\circ f_{sandpiper+1}^{e_{sandpiper+1}} \\circ \\cdots \\circ hillcrest^{e_{seashells}}\n\\]\nboth fix \\( blueberry \\). Hence both \\( rivermouth=sunflower^{moonlight} \\circ \\cdots \\circ woodland^{e_{sandpiper-1}} \\) and \\( rivermouth=sunflower^{moonlight} \\circ \\cdots \\circ woodland^{e_{sandpiper-1}} \\circ treetop \\) both map \\( blueberry \\) to itself. But then \\( rivermouth^{-1} \\circ riversource=treetop \\) also maps \\( blueberry \\) to itself, giving a contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "irrational",
        "m": "fractional",
        "k": "minimalist",
        "i": "aggregate",
        "e_i": "basevalue",
        "e_1": "tipvalue",
        "A": "wholeset",
        "G": "antigroup",
        "F": "emptiness",
        "F_n": "voidfamily",
        "F_1": "voidfirst",
        "F_2": "voidsecond",
        "f_1": "constantone",
        "f_10": "constantten",
        "f_i": "constantfunc",
        "f_i_1": "constantalpha",
        "f_i_2": "constantbeta",
        "f_i_m": "constantomega",
        "f_k-1": "constantprev",
        "f_k": "constantpeak",
        "f_n": "constantend"
      },
      "question": "Let $constantone, \\dots, constantten$ be bijections of the set of integers such that for\neach integer $irrational$, there is some composition $constantalpha \\circ constantbeta\n\\circ \\cdots \\circ constantomega$ of these functions (allowing repetitions)\nwhich maps 0 to $irrational$. Consider the set of 1024 functions\n\\[\n\\mathcal{emptiness} = \\{constantone^{tipvalue} \\circ f_2^{e_2} \\circ \\cdots \\circ constantten^{e_{10}}\\},\n\\]\n$basevalue = 0$ or 1 for $1 \\leq aggregate \\leq 10$. ($constantfunc^0$ is the identity function\nand $constantfunc^1 = constantfunc$.) Show that if $wholeset$ is any nonempty finite set of\nintegers, then at most 512 of the functions in $\\mathcal{emptiness}$ map $wholeset$ to\nitself.",
      "solution": "Solution. We say that a bijection of the integers to itself preserves a subset \\( wholeset \\) if it restricts to a bijection of \\( wholeset \\).\n\nLet \\( antigroup \\) be the group of permutations of \\( \\mathbb{Z} \\) generated by the \\( constantfunc \\). Then \\( antigroup \\) has elements mapping any integer \\( fractional \\) to 0 , and elements mapping 0 to any integer \\( irrational \\), so \\( antigroup \\) acts transitively on \\( \\mathbb{Z} \\). Hence no nonempty proper subset \\( wholeset \\) can be preserved by all the \\( constantfunc \\). It remains to prove the following lemma.\n\nLemma. Let \\( constantone, \\ldots, constantend \\) be bijections \\( \\mathbb{Z} \\rightarrow \\mathbb{Z} \\), and let \\( wholeset \\) be a subset of \\( \\mathbb{Z} \\). Suppose that some \\( constantfunc \\) does not preserve \\( wholeset \\). Then at most \\( 2^{irrational-1} \\) elements of\n\\[\n\\mathcal{emptiness}_{irrational}=\\left\\{constantone^{tipvalue} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ constantend^{basevalue}: basevalue=0 \\text { or } 1\\right\\}\n\\]\npreserve \\( wholeset \\).\nProof 1 of Lemma (inductive). The lemma is true for \\( irrational=1 \\), so assume it it is true for all \\( irrational<minimalist \\) (for some \\( minimalist>1 \\) ), and false for \\( irrational=minimalist \\). Then by the Pigeonhole Principle, there are \\( e_{1}, \\ldots, e_{minimalist-1} \\in\\{0,1\\} \\) such that both\n\\[\nconstantone^{e_{1}} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ constantprev^{e_{minimalist-1}} \\quad \\text { and } \\quad constantone^{e_{1}} \\circ f_{2}^{e_{2}} \\circ \\cdots \\circ constantprev^{e_{minimalist-1}} \\circ constantpeak\n\\]\nfix \\( wholeset \\). Hence \\( constantpeak \\) preserves \\( wholeset \\) as well. By the inductive hypothesis, at most \\( 2^{minimalist-2} \\) elements of \\( \\mathcal{emptiness}_{minimalist-1} \\) fix \\( wholeset \\); thus at most \\( 2^{minimalist-1} \\) elements of \\( \\mathcal{emptiness}_{minimalist} \\) fix \\( wholeset \\), giving a contradiction.\n\nProof 2 of Lemma (noninductive). Let \\( minimalist \\) be the largest integer such that \\( constantpeak \\) does not map \\( wholeset \\) to itself, and suppose that more than \\( 2^{irrational-1} \\) of the functions \\( \\mathcal{emptiness}_{irrational} \\operatorname{map} wholeset \\) to itself. By the Pigeonhole Principle, there are\n\\[\ne_{1}, \\ldots, e_{minimalist-1}, e_{minimalist+1}, \\ldots, e_{irrational} \\in\\{0,1\\}\n\\]\nsuch that both\n\\[\nconstantone^{e_{1}} \\circ \\cdots \\circ constantprev^{e_{minimalist-1}} \\circ f_{minimalist+1}^{e_{minimalist+1}} \\circ \\cdots \\circ constantend^{e_{irrational}} \\quad \\text { and } \\quad constantone^{e_{1}} \\circ \\cdots \\circ constantprev^{e_{minimalist-1}} \\circ constantpeak \\circ f_{minimalist+1}^{e_{minimalist+1}} \\circ \\cdots \\circ constantend^{e_{irrational}}\n\\]\nboth fix \\( wholeset \\). Hence both \\( voidfirst=constantone^{e_{1}} \\circ \\cdots \\circ constantprev^{e_{minimalist-1}} \\) and \\( voidfirst=constantone^{e_{1}} \\circ \\cdots \\circ constantprev^{e_{minimalist-1}} \\circ constantpeak \\) both map \\( wholeset \\) to itself. But then \\( voidfirst^{-1} \\circ voidsecond=constantpeak \\) also maps \\( wholeset \\) to itself, giving a contradiction."
    },
    "garbled_string": {
      "map": {
        "n": "zyqplmne",
        "m": "hrxovtka",
        "k": "psdlenji",
        "i": "gvatkour",
        "e_i": "wskmzypl",
        "e_1": "njrfouks",
        "A": "xmqnfadz",
        "G": "lepzovuq",
        "F": "vidorgaz",
        "F_n": "gozpitld",
        "F_1": "kucebyhz",
        "F_2": "ravlikum",
        "f_1": "thqmsnwa",
        "f_10": "becalvud",
        "f_i": "pyhontcz",
        "f_i_1": "tapkleri",
        "f_i_2": "ygtrumed",
        "f_i_m": "bsevukon",
        "f_k-1": "zodithap",
        "f_k": "alqorime",
        "f_n": "hstafwer"
      },
      "question": "Let $thqmsnwa, \\dots, becalvud$ be bijections of the set of integers such that for\neach integer $zyqplmne$, there is some composition $tapkleri \\circ ygtrumed\n\\circ \\cdots \\circ bsevukon$ of these functions (allowing repetitions)\nwhich maps 0 to $zyqplmne$. Consider the set of 1024 functions\n\\[\n\\mathcal{vidorgaz} = \\{thqmsnwa^{njrfouks} \\circ f_2^{e_2} \\circ \\cdots \\circ becalvud^{e_{10}}\\},\n\\]\n$wskmzypl = 0$ or 1 for $1 \\leq gvatkour \\leq 10$. ($pyhontcz^0$ is the identity function\nand $pyhontcz^1 = pyhontcz$.) Show that if $xmqnfadz$ is any nonempty finite set of\nintegers, then at most 512 of the functions in $\\mathcal{vidorgaz}$ map $xmqnfadz$ to\nitself.",
      "solution": "Solution. We say that a bijection of the integers to itself preserves a subset \\( xmqnfadz \\) if it restricts to a bijection of \\( xmqnfadz \\).\n\nLet \\( lepzovuq \\) be the group of permutations of \\( \\mathbb{Z} \\) generated by the \\( pyhontcz \\). Then \\( lepzovuq \\) has elements mapping any integer \\( hrxovtka \\) to 0 , and elements mapping 0 to any integer \\( zyqplmne \\), so \\( lepzovuq \\) acts transitively on \\( \\mathbb{Z} \\). Hence no nonempty proper subset \\( xmqnfadz \\) can be preserved by all the \\( pyhontcz \\). It remains to prove the following lemma.\n\nLemma. Let \\( thqmsnwa, \\ldots, hstafwer \\) be bijections \\( \\mathbb{Z} \\rightarrow \\mathbb{Z} \\), and let \\( xmqnfadz \\) be a subset of \\( \\mathbb{Z} \\). Suppose that some \\( pyhontcz \\) does not preserve \\( xmqnfadz \\). Then at most \\( 2^{zyqplmne-1} \\) elements of\n\\[\n\\mathcal{vidorgaz}_{zyqplmne}=\\left\\{thqmsnwa^{njrfouks} \\circ f_2^{e_2} \\circ \\cdots \\circ hstafwer^{e_{zyqplmne}}: wskmzypl=0 \\text { or } 1\\right\\}\n\\]\npreserve \\( xmqnfadz \\).\n\nProof 1 of Lemma (inductive). The lemma is true for \\( zyqplmne=1 \\), so assume it is true for all \\( zyqplmne<psdlenji \\) (for some \\( psdlenji>1 \\) ), and false for \\( zyqplmne=psdlenji \\). Then by the Pigeonhole Principle, there are \\( njrfouks, \\ldots, e_{psdlenji-1} \\in\\{0,1\\} \\) such that both\n\\[\nthqmsnwa^{njrfouks} \\circ f_2^{e_2} \\circ \\cdots \\circ zodithap^{e_{psdlenji-1}} \\quad \\text { and } \\quad thqmsnwa^{njrfouks} \\circ f_2^{e_2} \\circ \\cdots \\circ zodithap^{e_{psdlenji-1}} \\circ alqorime\n\\]\nfix \\( xmqnfadz \\). Hence \\( alqorime \\) preserves \\( xmqnfadz \\) as well. By the inductive hypothesis, at most \\( 2^{psdlenji-2} \\) elements of \\( \\mathcal{vidorgaz}_{psdlenji-1} \\) fix \\( xmqnfadz \\); thus at most \\( 2^{psdlenji-1} \\) elements of \\( \\mathcal{vidorgaz}_{psdlenji} \\) fix \\( xmqnfadz \\), giving a contradiction.\n\nProof 2 of Lemma (noninductive). Let \\( psdlenji \\) be the largest integer such that \\( alqorime \\) does not map \\( xmqnfadz \\) to itself, and suppose that more than \\( 2^{zyqplmne-1} \\) of the functions \\( \\mathcal{vidorgaz}_{zyqplmne} \\operatorname{map} xmqnfadz \\) to itself. By the Pigeonhole Principle, there are\n\\[\nnjrfouks, \\ldots, e_{psdlenji-1}, e_{psdlenji+1}, \\ldots, e_{zyqplmne} \\in\\{0,1\\}\n\\]\nsuch that both\n\\[\nthqmsnwa^{njrfouks} \\circ \\cdots \\circ zodithap^{e_{psdlenji-1}} \\circ f_{psdlenji+1}^{e_{psdlenji+1}} \\circ \\cdots \\circ hstafwer^{e_{zyqplmne}} \\quad \\text { and } \\quad thqmsnwa^{njrfouks} \\circ \\cdots \\circ zodithap^{e_{psdlenji-1}} \\circ alqorime \\circ f_{psdlenji+1}^{e_{psdlenji+1}} \\circ \\cdots \\circ hstafwer^{e_{zyqplmne}}\n\\]\nboth fix \\( xmqnfadz \\). Hence both \\( kucebyhz=thqmsnwa^{njrfouks} \\circ \\cdots \\circ zodithap^{e_{psdlenji-1}} \\) and \\( kucebyhz=thqmsnwa^{njrfouks} \\circ \\cdots \\circ zodithap^{e_{psdlenji-1}} \\circ alqorime \\) both map \\( xmqnfadz \\) to itself. But then \\( kucebyhz^{-1} \\circ ravlikum=alqorime \\) also maps \\( xmqnfadz \\) to itself, giving a contradiction."
    },
    "kernel_variant": {
      "question": "Let $g_{1},\\dots ,g_{12}$ be permutations of the set $\\mathbb Q$ of rational numbers with the property that for every $q\\in\\mathbb Q$ there exists a finite composition of the $g_i$’s (repetitions allowed) that sends the rational number $5$ to $q$.  For each $i$ put $g_i^{0}=\\operatorname{id}_{\\mathbb Q}$ and $g_i^{1}=g_i$, and form the family of $2^{12}=4096$ permutations\n\\[\n\\mathcal H\\;=\\;\\Bigl\\{\\;g_1^{e_1}\\circ g_2^{e_2}\\circ\\cdots\\circ g_{12}^{e_{12}}\n : e_i\\in\\{0,1\\}\\;\\Bigr\\}.\n\\]\nShow that if $B\\subset\\mathbb Q$ is an infinite proper subset, then at most $2048$ elements of $\\mathcal H$ map $B$ to itself (i.e. restrict to a bijection $B\\to B$).",
      "solution": "Proof.  Call a permutation σ of Q ‘preserving B’ if σ(B)=B.  Let G=⟨g₁,…,g₁₂⟩.  By hypothesis for every q∈Q some composition of the gᵢ’s sends 5 to q, so G acts transitively on Q.  Hence no nonempty proper subset of Q can be fixed by all the gᵢ, and in particular there exists at least one index i for which gᵢ does not preserve B.  Let k be the largest such index; then for every j>k the generator g_j does preserve B.\n\nFor each binary vector e=(e₁,…,e₁₂) with eᵢ∈{0,1}, set\n  h_e = g₁^{e₁} ∘ g₂^{e₂} ∘ ⋯ ∘ g_{12}^{e_{12}}.\nWe claim at most 2¹¹ of these 2¹² maps can preserve B.  Suppose to the contrary that more than 2¹¹ of the h_e do preserve B.  Pair up all 2¹² vectors by flipping the kᵗʰ bit: e↔e′ where e′_i=e_i for i≠k and e′_k=1−e_k.  There are 2¹¹ disjoint pairs, so by pigeonhole at least one pair {e,e′} has both h_e and h_{e′} preserving B.\n\nWrite\n  U  = g₁^{e₁} ∘ ⋯ ∘ g_{k−1}^{e_{k−1}},\n  T  = g_{k+1}^{e_{k+1}} ∘ ⋯ ∘ g_{12}^{e_{12}}.\nThen\n  h_e  = U ∘ g_k^{e_k} ∘ T,\n  h_{e′}= U ∘ g_k^{1−e_k} ∘ T.\nSince every g_j with j>k preserves B, the ‘tail’ T does as well.  Hence for i=0,1,\n  h_e (resp. h_{e′}) preserves B  ⇔  U∘g_k^{e_k} preserves B.\nBecause both h_e and h_{e′} preserve B, we conclude that both U and U∘g_k preserve B.  But then\n  g_k = U^{-1} ∘ (U ∘ g_k)\nwould preserve B as well, contradicting our choice of k.  This contradiction shows that at most 2¹¹=2048 of the h_e preserve B, as required.  □",
      "_meta": {
        "core_steps": [
          "Use the reachability hypothesis to show the group generated by the f_i acts transitively on ℤ.",
          "Conclude that at least one generator f_k does not preserve the given subset A.",
          "Apply the pigeonhole argument: among the 2^n binary words in the generators, two that differ only by f_k would both preserve A if more than half preserved A, forcing f_k itself to preserve A—a contradiction.",
          "Hence no more than 2^{n-1} of the 2^n words preserve A (Lemma).",
          "Substitute n = 10 to obtain the numeric bound 2^9 = 512 for the problem."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Total number of generators/bijections under consideration.",
            "original": 10
          },
          "slot2": {
            "description": "Chosen reference element that can be moved to every other element (presently 0).",
            "original": 0
          },
          "slot3": {
            "description": "Cardinality of the set of binary-word compositions (2^{num_generators}).",
            "original": 1024
          },
          "slot4": {
            "description": "Numeric bound on the number of compositions that can preserve A (2^{num_generators−1}).",
            "original": 512
          },
          "slot5": {
            "description": "Underlying set on which the generators act; only transitivity is needed.",
            "original": "ℤ (the integers)"
          },
          "slot6": {
            "description": "Stipulation that A be finite; the reasoning works for any non-empty proper subset.",
            "original": "finite"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}