{
  "index": "2008-A-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Prove that there exists a constant $c>0$ such that in every\nnontrivial finite group $G$ there exists a sequence of length\nat most $c \\log |G|$ with the property that each element of $G$\nequals the product of some subsequence. (The elements of $G$ in the\nsequence are not required to be distinct. A \\emph{subsequence}\nof a sequence is obtained by selecting some of the terms,\nnot necessarily consecutive, without reordering them; for\nexample, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but\n$2, 2, 4$ is not.)",
  "solution": "For notational convenience, we will interpret the problem as\nallowing the empty subsequence, whose product is the identity element of\nthe group. To solve the problem in the interpretation where the empty\nsubsequence is not allowed, simply append the identity element to the sequence\ngiven by one of the following solutions.\n\n\\textbf{First solution:}\nPut $n = |G|$.\nWe will say that a sequence $S$ \\emph{produces}\nan element $g \\in G$ if $g$ occurs as the product of some subsequence\nof $S$.\nLet $H$ be the set of elements produced by the sequence $S$.\n\nStart with $S$ equal to the empty sequence. If at any point\nthe set $H^{-1}H = \\{h_1 h_2: h_1^{-1}, h_2 \\in H\\}$ fails to be\nall of $G$, extend $S$ by appending an element $g$ of $G$ not in\n$H^{-1} H$. Then $Hg \\cap H$ must be empty, otherwise there would\nbe an equation of the form $h_1 g= h_2 $ with $h_1, h_2 \\in G$,\nor $g = h_1^{-1} h_2$, a contradiction. Thus we can extend $S$ by one\nelement and double the size of $H$.\n\nAfter $k \\leq \\log_2 n$ steps, we must obtain a sequence $S\n= a_1,\\dots,a_k$ for which $H^{-1} H = G$. Then\nthe sequence $a_k^{-1}, \\dots, a_1^{-1}, a_1, \\dots, a_k$\nproduces all of $G$ and has length at most $(2/\\ln 2) \\ln n$.\n\n\\textbf{Second solution:}\n\nPut $m = |H|$. We will show that we can append one element\n$g$ to $S$ so that the resulting sequence of $k+1$ elements will produce\nat least $2m-m^2/n$ elements of $G$. To see this, we compute\n\\begin{align*}\n\\sum_{g \\in G} |H \\cup Hg|\n&= \\sum_{g \\in G} (|H| + |Hg| - |H \\cap Hg|) \\\\\n&= 2mn - \\sum_{g \\in G} |H \\cap Hg| \\\\\n&= 2mn - |\\{(g,h) \\in G^2: h \\in H \\cap Hg\\}| \\\\\n&= 2mn - \\sum_{h \\in H} |\\{g \\in G: h \\in Hg\\}| \\\\\n&= 2mn - \\sum_{h \\in H} |H^{-1} h| \\\\\n&= 2mn - m^2.\n\\end{align*}\nBy the pigeonhole principle, we have $|H \\cup Hg| \\geq 2m - m^2/n$ for\nsome choice of $g$, as claimed.\n\nIn other words, by extending the sequence by one element,\nwe can replace the ratio $s = 1-m/n$ (i.e., the fraction\nof elements of $G$ not generated by $S$)\nby a quantity no greater than\n\\[\n1-(2m-m^2/n)/n = s^2.\n\\]\nWe start out with $k = 0$ and $s = 1 - 1/n$;\nafter $k$ steps, we have $s \\leq (1-1/n)^{2^k}$.\nIt is enough to prove that for some $c > 0$, we can always find\nan integer $k \\leq c \\ln n$ such that\n\\[\n\\left(1 - \\frac{1}{n} \\right)^{2^k} < \\frac{1}{n},\n\\]\nas then we have $n-m < 1$ and hence $H = G$.\n\nTo obtain this last inequality, put\n\\[\nk = \\lfloor 2 \\log_2 n \\rfloor < (2/\\ln 2) \\ln n,\n\\]\nso that $2^{k+1} \\geq n^2$.\nFrom the facts that $\\ln n \\leq \\ln 2 + (n-2)/2 \\leq n/2$ and\n$\\ln (1-1/n) < -1/n$ for all $n \\geq 2$, we have\n\\[\n2^k \\ln \\left( 1 - \\frac{1}{n} \\right) < -\\frac{n^2}{2n} =  -\\frac{n}{2} < -\\ln n,\n\\]\nyielding the desired inequality.\n\n\\textbf{Remark:} An alternate approach in the second solution\nis to distinguish betwen the cases of $H$ small (i.e.,\n$m < n^{1/2}$, in which case $m$ can be replaced by a value\nno less than $2m-1$) and $H$ large.\nThis strategy is used in a number of recent results\nof Bourgain, Tao, Helfgott, and others on \\emph{small doubling}\nor \\emph{small tripling}\nof subsets of finite groups.\n\nIn the second solution, if we avoid the rather weak inequality\n$\\ln n \\leq n/2$, we instead get sequences of length\n$\\log_2 (n \\ln n) = \\log_2(n) + \\log_2 (\\ln n)$.\nThis is close to optimal: one cannot use fewer than $\\log_2 n$\nterms because the number of subsequences must be at least $n$.",
  "vars": [
    "G",
    "H",
    "S",
    "a",
    "a_1",
    "a_i",
    "a_k",
    "g",
    "h",
    "h_1",
    "h_2",
    "k",
    "m",
    "n",
    "s"
  ],
  "params": [
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "G": "groupglo",
        "H": "subgenset",
        "S": "seqwhole",
        "a": "singlet",
        "a_1": "singone",
        "a_i": "singith",
        "a_k": "singlast",
        "g": "elemvar",
        "h": "elemalt",
        "h_1": "elemone",
        "h_2": "elemtwo",
        "k": "stepcount",
        "m": "producedct",
        "n": "grouplsz",
        "s": "ratioleft",
        "c": "positvconst"
      },
      "question": "Prove that there exists a constant $positvconst>0$ such that in every\nnontrivial finite group $groupglo$ there exists a sequence of length\nat most $positvconst \\log |groupglo|$ with the property that each element of $groupglo$\nequals the product of some subsequence. (The elements of $groupglo$ in the\nsequence are not required to be distinct. A \\emph{subsequence}\nof a sequence is obtained by selecting some of the terms,\nnot necessarily consecutive, without reordering them; for\nexample, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but\n$2, 2, 4$ is not.)",
      "solution": "For notational convenience, we will interpret the problem as\nallowing the empty subsequence, whose product is the identity element of\nthe group. To solve the problem in the interpretation where the empty\nsubsequence is not allowed, simply append the identity element to the sequence\ngiven by one of the following solutions.\n\n\\textbf{First solution:}\nPut $grouplsz = |groupglo|$.\nWe will say that a sequence $seqwhole$ \\emph{produces}\nan element $elemvar \\in groupglo$ if $elemvar$ occurs as the product of some subsequence\nof $seqwhole$.\nLet $subgenset$ be the set of elements produced by the sequence $seqwhole$.\n\nStart with $seqwhole$ equal to the empty sequence. If at any point\nthe set $subgenset^{-1}subgenset = \\{\\elemone \\elemtwo: \\elemone^{-1}, \\elemtwo \\in subgenset\\}$ fails to be\nall of $groupglo$, extend $seqwhole$ by appending an element $elemvar$ of $groupglo$ not in\n$subgenset^{-1} subgenset$. Then $subgenset elemvar \\cap subgenset$ must be empty, otherwise there would\nbe an equation of the form $\\elemone elemvar= \\elemtwo $ with $\\elemone, \\elemtwo \\in groupglo$,\nor $elemvar = \\elemone^{-1} \\elemtwo$, a contradiction. Thus we can extend $seqwhole$ by one\nelement and double the size of $subgenset$.\n\nAfter $stepcount \\leq \\log_2 grouplsz$ steps, we must obtain a sequence $seqwhole\n= singone,\\dots,singlast$ for which $subgenset^{-1} subgenset = groupglo$. Then\nthe sequence $singlast^{-1}, \\dots, singone^{-1}, singone, \\dots, singlast$\nproduces all of $groupglo$ and has length at most $(2/\\ln 2) \\ln grouplsz$.\n\n\\textbf{Second solution:}\n\nPut $producedct = |subgenset|$. We will show that we can append one element\n$elemvar$ to $seqwhole$ so that the resulting sequence of $stepcount+1$ elements will produce\nat least $2producedct-producedct^2/grouplsz$ elements of $groupglo$. To see this, we compute\n\\begin{align*}\n\\sum_{elemvar \\in groupglo} |subgenset \\cup subgenset elemvar|\n&= \\sum_{elemvar \\in groupglo} (|subgenset| + |subgenset elemvar| - |subgenset \\cap subgenset elemvar|) \\\\\n&= 2producedct\\,grouplsz - \\sum_{elemvar \\in groupglo} |subgenset \\cap subgenset elemvar| \\\\\n&= 2producedct\\,grouplsz - |\\{(elemvar,\\elemalt) \\in groupglo^2: \\elemalt \\in subgenset \\cap subgenset elemvar\\}| \\\\\n&= 2producedct\\,grouplsz - \\sum_{\\elemalt \\in subgenset} |\\{elemvar \\in groupglo: \\elemalt \\in subgenset elemvar\\}| \\\\\n&= 2producedct\\,grouplsz - \\sum_{\\elemalt \\in subgenset} |subgenset^{-1} \\elemalt| \\\\\n&= 2producedct\\,grouplsz - producedct^2.\n\\end{align*}\nBy the pigeonhole principle, we have $|subgenset \\cup subgenset elemvar| \\geq 2producedct - producedct^2/grouplsz$ for\nsome choice of $elemvar$, as claimed.\n\nIn other words, by extending the sequence by one element,\nwe can replace the ratio $ratioleft = 1-producedct/grouplsz$ (i.e., the fraction\nof elements of $groupglo$ not generated by $seqwhole$)\nby a quantity no greater than\n\\[\n1-\\frac{2producedct-producedct^2/grouplsz}{grouplsz} = ratioleft^2.\n\\]\nWe start out with $stepcount = 0$ and $ratioleft = 1 - 1/grouplsz$;\nafter $stepcount$ steps, we have $ratioleft \\leq (1-1/grouplsz)^{2^{stepcount}}$.\nIt is enough to prove that for some $positvconst > 0$, we can always find\nan integer $stepcount \\leq positvconst \\ln grouplsz$ such that\n\\[\n\\left(1 - \\frac{1}{grouplsz} \\right)^{2^{stepcount}} < \\frac{1}{grouplsz},\n\\]\nas then we have $grouplsz-producedct < 1$ and hence $subgenset = groupglo$.\n\nTo obtain this last inequality, put\n\\[\nstepcount = \\lfloor 2 \\log_2 grouplsz \\rfloor < (2/\\ln 2) \\ln grouplsz,\n\\]\nso that $2^{stepcount+1} \\geq grouplsz^2$.\nFrom the facts that $\\ln grouplsz \\leq \\ln 2 + (grouplsz-2)/2 \\leq grouplsz/2$ and\n$\\ln (1-1/grouplsz) < -1/grouplsz$ for all $grouplsz \\geq 2$, we have\n\\[\n2^{stepcount} \\ln \\left( 1 - \\frac{1}{grouplsz} \\right) < -\\frac{grouplsz^2}{2grouplsz} =  -\\frac{grouplsz}{2} < -\\ln grouplsz,\n\\]\nyielding the desired inequality.\n\n\\textbf{Remark:} An alternate approach in the second solution\nis to distinguish betwen the cases of $subgenset$ small (i.e.,\n$producedct < grouplsz^{1/2}$, in which case $producedct$ can be replaced by a value\nno less than $2producedct-1$) and $subgenset$ large.\nThis strategy is used in a number of recent results\nof Bourgain, Tao, Helfgott, and others on \\emph{small doubling}\nor \\emph{small tripling}\nof subsets of finite groups.\n\nIn the second solution, if we avoid the rather weak inequality\n$\\ln grouplsz \\leq grouplsz/2$, we instead get sequences of length\n$\\log_2 (grouplsz \\ln grouplsz) = \\log_2(grouplsz) + \\log_2 (\\ln grouplsz)$.\nThis is close to optimal: one cannot use fewer than $\\log_2 grouplsz$\nterms because the number of subsequences must be at least $grouplsz$.",
      "}": "whoops"
    },
    "descriptive_long_confusing": {
      "map": {
        "G": "riverboat",
        "H": "clifftop",
        "S": "honeycomb",
        "a": "sandstorm",
        "a_1": "moonlight",
        "a_i": "raincloud",
        "a_k": "wildfire",
        "g": "hummingbird",
        "h": "snowflake",
        "h_1": "crosswind",
        "h_2": "driftwood",
        "k": "starlight",
        "m": "afterglow",
        "n": "gemstone",
        "s": "woodpecker",
        "c": "pinecone"
      },
      "question": "Prove that there exists a constant $pinecone>0$ such that in every\nnontrivial finite group $riverboat$ there exists a sequence of length\nat most $pinecone \\log |riverboat|$ with the property that each element of $riverboat$\nequals the product of some subsequence. (The elements of $riverboat$ in the\nsequence are not required to be distinct. A \\emph{subsequence}\nof a sequence is obtained by selecting some of the terms,\nnot necessarily consecutive, without reordering them; for\nexample, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but\n$2, 2, 4$ is not.)",
      "solution": "For notational convenience, we will interpret the problem as\nallowing the empty subsequence, whose product is the identity element of\nthe group. To solve the problem in the interpretation where the empty\nsubsequence is not allowed, simply append the identity element to the sequence\ngiven by one of the following solutions.\n\n\\textbf{First solution:}\nPut $gemstone = |riverboat|$.\nWe will say that a sequence $honeycomb$ \\emph{produces}\nan element $hummingbird \\in riverboat$ if $hummingbird$ occurs as the product of some subsequence\nof $honeycomb$.\nLet $clifftop$ be the set of elements produced by the sequence $honeycomb$.\n\nStart with $honeycomb$ equal to the empty sequence. If at any point\nthe set $clifftop^{-1}clifftop = \\{crosswind driftwood: crosswind^{-1}, driftwood \\in clifftop\\}$ fails to be\nall of $riverboat$, extend $honeycomb$ by appending an element $hummingbird$ of $riverboat$ not in\n$clifftop^{-1} clifftop$. Then $clifftophummingbird \\cap clifftop$ must be empty, otherwise there would\nbe an equation of the form $crosswind hummingbird= driftwood$ with $crosswind, driftwood \\in riverboat$,\nor $hummingbird = crosswind^{-1} driftwood$, a contradiction. Thus we can extend $honeycomb$ by one\nelement and double the size of $clifftop$.\n\nAfter $starlight \\leq \\log_2 gemstone$ steps, we must obtain a sequence $honeycomb\n= moonlight,\\dots,wildfire$ for which $clifftop^{-1} clifftop = riverboat$. Then\nthe sequence $wildfire^{-1}, \\dots, moonlight^{-1}, moonlight, \\dots, wildfire$\nproduces all of $riverboat$ and has length at most $(2/\\ln 2) \\ln gemstone$.\n\n\\textbf{Second solution:}\n\nPut $afterglow = |clifftop|$. We will show that we can append one element\n$hummingbird$ to $honeycomb$ so that the resulting sequence of $starlight+1$ elements will produce\nat least $2afterglow-afterglow^2/gemstone$ elements of $riverboat$. To see this, we compute\n\\begin{align*}\n\\sum_{hummingbird \\in riverboat} |clifftop \\cup clifftophummingbird|\n&= \\sum_{hummingbird \\in riverboat} (|clifftop| + |clifftophummingbird| - |clifftop \\cap clifftophummingbird|) \\\\\n&= 2afterglow gemstone - \\sum_{hummingbird \\in riverboat} |clifftop \\cap clifftophummingbird| \\\\\n&= 2afterglow gemstone - |\\{(hummingbird,snowflake) \\in riverboat^2: snowflake \\in clifftop \\cap clifftophummingbird\\}| \\\\\n&= 2afterglow gemstone - \\sum_{snowflake \\in clifftop} |\\{hummingbird \\in riverboat: snowflake \\in clifftophummingbird\\}| \\\\\n&= 2afterglow gemstone - \\sum_{snowflake \\in clifftop} |clifftop^{-1} snowflake| \\\\\n&= 2afterglow gemstone - afterglow^2.\n\\end{align*}\nBy the pigeonhole principle, we have $|clifftop \\cup clifftophummingbird| \\geq 2afterglow - afterglow^2/gemstone$ for\nsome choice of $hummingbird$, as claimed.\n\nIn other words, by extending the sequence by one element,\nwe can replace the ratio $woodpecker = 1-afterglow/gemstone$ (i.e., the fraction\nof elements of $riverboat$ not generated by $honeycomb$)\nby a quantity no greater than\n\\[\n1-(2afterglow-afterglow^2/gemstone)/gemstone = woodpecker^2.\n\\]\nWe start out with $starlight = 0$ and $woodpecker = 1 - 1/gemstone$;\nafter $starlight$ steps, we have $woodpecker \\leq (1-1/gemstone)^{2^{starlight}}$.\nIt is enough to prove that for some $pinecone > 0$, we can always find\nan integer $starlight \\leq pinecone \\ln gemstone$ such that\n\\[\n\\left(1 - \\frac{1}{gemstone} \\right)^{2^{starlight}} < \\frac{1}{gemstone},\n\\]\nas then we have $gemstone-afterglow < 1$ and hence $clifftop = riverboat$.\n\nTo obtain this last inequality, put\n\\[\nstarlight = \\lfloor 2 \\log_2 gemstone \\rfloor < (2/\\ln 2) \\ln gemstone,\n\\]\nso that $2^{starlight+1} \\geq gemstone^2$.\nFrom the facts that $\\ln gemstone \\leq \\ln 2 + (gemstone-2)/2 \\leq gemstone/2$ and\n$\\ln (1-1/gemstone) < -1/gemstone$ for all $gemstone \\geq 2$, we have\n\\[\n2^{starlight} \\ln \\left( 1 - \\frac{1}{gemstone} \\right) < -\\frac{gemstone^2}{2gemstone} =  -\\frac{gemstone}{2} < -\\ln gemstone,\n\\]\nyielding the desired inequality.\n\n\\textbf{Remark:} An alternate approach in the second solution\nis to distinguish betwen the cases of $clifftop$ small (i.e.,\n$afterglow < gemstone^{1/2}$, in which case $afterglow$ can be replaced by a value\nno less than $2afterglow-1$) and $clifftop$ large.\nThis strategy is used in a number of recent results\nof Bourgain, Tao, Helfgott, and others on \\emph{small doubling}\nor \\emph{small tripling}\nof subsets of finite groups.\n\nIn the second solution, if we avoid the rather weak inequality\n$\\ln gemstone \\leq gemstone/2$, we instead get sequences of length\n$\\log_2 (gemstone \\ln gemstone) = \\log_2(gemstone) + \\log_2 (\\ln gemstone)$.\nThis is close to optimal: one cannot use fewer than $\\log_2 gemstone$\nterms because the number of subsequences must be at least $gemstone$.}"
    },
    "descriptive_long_misleading": {
      "map": {
        "G": "ungrouped",
        "H": "universal",
        "S": "singleton",
        "a": "compound",
        "a_1": "compoundone",
        "a_i": "compoundvar",
        "a_k": "compoundend",
        "g": "insideone",
        "h": "outsider",
        "h_1": "outsiderone",
        "h_2": "outsidertwo",
        "k": "endlessn",
        "m": "emptiness",
        "n": "minisize",
        "s": "plentiful",
        "c": "variable"
      },
      "question": "Prove that there exists a constant $variable>0$ such that in every\nnontrivial finite group $ungrouped$ there exists a sequence of length\nat most $variable \\log |ungrouped|$ with the property that each element of $ungrouped$\nequals the product of some subsequence. (The elements of $ungrouped$ in the\nsequence are not required to be distinct. A \\emph{subsequence}\nof a sequence is obtained by selecting some of the terms,\nnot necessarily consecutive, without reordering them; for\nexample, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but\n$2, 2, 4$ is not.)",
      "solution": "For notational convenience, we will interpret the problem as\nallowing the empty subsequence, whose product is the identity element of\nthe group. To solve the problem in the interpretation where the empty\nsubsequence is not allowed, simply append the identity element to the sequence\ngiven by one of the following solutions.\n\n\\textbf{First solution:}\nPut $minisize = |ungrouped|$.\nWe will say that a sequence $singleton$ \\emph{produces}\nan element $insideone \\in ungrouped$ if $insideone$ occurs as the product of some subsequence\nof $singleton$.\nLet $universal$ be the set of elements produced by the sequence $singleton$.\n\nStart with $singleton$ equal to the empty sequence. If at any point\nthe set $universal^{-1}universal = \\{outsiderone outsidertwo: outsiderone^{-1}, outsidertwo \\in universal\\}$ fails to be\nall of $ungrouped$, extend $singleton$ by appending an element $insideone$ of $ungrouped$ not in\n$universal^{-1} universal$. Then $universal insideone \\cap universal$ must be empty, otherwise there would\nbe an equation of the form $outsiderone insideone= outsidertwo $ with $outsiderone, outsidertwo \\in ungrouped$,\nor $insideone = outsiderone^{-1} outsidertwo$, a contradiction. Thus we can extend $singleton$ by one\nelement and double the size of $universal$.\n\nAfter $endlessn \\leq \\log_2 minisize$ steps, we must obtain a sequence $singleton\n= compoundone,\\dots,compoundend$ for which $universal^{-1} universal = ungrouped$. Then\nthe sequence $compoundend^{-1}, \\dots, compoundone^{-1}, compoundone, \\dots, compoundend$\nproduces all of $ungrouped$ and has length at most $(2/\\ln 2) \\ln minisize$.\n\n\\textbf{Second solution:}\n\nPut $emptiness = |universal|$. We will show that we can append one element\n$insideone$ to $singleton$ so that the resulting sequence of $endlessn+1$ elements will produce\nat least $2 emptiness minisize-emptiness^2/minisize$ elements of $ungrouped$. To see this, we compute\n\\begin{align*}\n\\sum_{insideone \\in ungrouped} |universal \\cup universal insideone|\n&= \\sum_{insideone \\in ungrouped} (|universal| + |universal insideone| - |universal \\cap universal insideone|) \\\\\n&= 2 emptiness minisize - \\sum_{insideone \\in ungrouped} |universal \\cap universal insideone| \\\\\n&= 2 emptiness minisize - |\\{(insideone,outsider) \\in ungrouped^2: outsider \\in universal \\cap universal insideone\\}| \\\\\n&= 2 emptiness minisize - \\sum_{outsider \\in universal} |\\{insideone \\in ungrouped: outsider \\in universal insideone\\}| \\\\\n&= 2 emptiness minisize - \\sum_{outsider \\in universal} |universal^{-1} outsider| \\\\\n&= 2 emptiness minisize - emptiness^2.\n\\end{align*}\nBy the pigeonhole principle, we have $|universal \\cup universal insideone| \\geq 2emptiness - emptiness^2/minisize$ for\nsome choice of $insideone$, as claimed.\n\nIn other words, by extending the sequence by one element,\nwe can replace the ratio $plentiful = 1-emptiness/minisize$ (i.e., the fraction\nof elements of $ungrouped$ not generated by $singleton$)\nby a quantity no greater than\n\\[\n1-(2emptiness-emptiness^2/minisize)/minisize = plentiful^2.\n\\]\nWe start out with $endlessn = 0$ and $plentiful = 1 - 1/minisize$;\nafter $endlessn$ steps, we have $plentiful \\leq (1-1/minisize)^{2^{endlessn}}$.\nIt is enough to prove that for some $variable > 0$, we can always find\nan integer $endlessn \\leq variable \\ln minisize$ such that\n\\[\n\\left(1 - \\frac{1}{minisize} \\right)^{2^{endlessn}} < \\frac{1}{minisize},\n\\]\nas then we have $minisize-emptiness < 1$ and hence $universal = ungrouped$.\n\nTo obtain this last inequality, put\n\\[\nendlessn = \\lfloor 2 \\log_2 minisize \\rfloor < (2/\\ln 2) \\ln minisize,\n\\]\nso that $2^{endlessn+1} \\geq minisize^2$.\nFrom the facts that $\\ln minisize \\leq \\ln 2 + (minisize-2)/2 \\leq minisize/2$ and\n$\\ln (1-1/minisize) < -1/minisize$ for all $minisize \\geq 2$, we have\n\\[\n2^{endlessn} \\ln \\left( 1 - \\frac{1}{minisize} \\right) < -\\frac{minisize^2}{2minisize} =  -\\frac{minisize}{2} < -\\ln minisize,\n\\]\nyielding the desired inequality.\n\n\\textbf{Remark:} An alternate approach in the second solution\nis to distinguish betwen the cases of $universal$ small (i.e.,\n$emptiness < minisize^{1/2}$, in which case $emptiness$ can be replaced by a value\nno less than $2emptiness-1$) and $universal$ large.\nThis strategy is used in a number of recent results\nof Bourgain, Tao, Helfgott, and others on \\emph{small doubling}\nor \\emph{small tripling}\nof subsets of finite groups.\n\nIn the second solution, if we avoid the rather weak inequality\n$\\ln minisize \\leq minisize/2$, we instead get sequences of length\n$\\log_2 (minisize \\ln minisize) = \\log_2(minisize) + \\log_2 (\\ln minisize)$.\nThis is close to optimal: one cannot use fewer than $\\log_2 minisize$\nterms because the number of subsequences must be at least $minisize$.  }"
    },
    "garbled_string": {
      "map": {
        "G": "qzxwvtnp",
        "H": "hjgrksla",
        "S": "fmvqslre",
        "a": "vnmeiklp",
        "a_1": "yczstigb",
        "a_i": "dpqnmrtu",
        "a_k": "kngwyhza",
        "g": "wjslmpqe",
        "h": "rqltvnso",
        "h_1": "sdzmbpqa",
        "h_2": "lxvhzqwe",
        "k": "tbvysoam",
        "m": "pznrxvlu",
        "n": "gdjoukhr",
        "s": "vnrzqtla",
        "c": "jhvlktse"
      },
      "question": "Prove that there exists a constant $jhvlktse>0$ such that in every\nnontrivial finite group $qzxwvtnp$ there exists a sequence of length\nat most $jhvlktse \\log |qzxwvtnp|$ with the property that each element of $qzxwvtnp$\nequals the product of some subsequence. (The elements of $qzxwvtnp$ in the\nsequence are not required to be distinct. A \\emph{subsequence}\nof a sequence is obtained by selecting some of the terms,\nnot necessarily consecutive, without reordering them; for\nexample, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but\n$2, 2, 4$ is not.)",
      "solution": "For notational convenience, we will interpret the problem as\nallowing the empty subsequence, whose product is the identity element of\nthe group. To solve the problem in the interpretation where the empty\nsubsequence is not allowed, simply append the identity element to the sequence\ngiven by one of the following solutions.\n\n\\textbf{First solution:}\nPut $gdjoukhr = |qzxwvtnp|$.\nWe will say that a sequence $fmvqslre$ \\emph{produces}\nan element $wjslmpqe \\in qzxwvtnp$ if $wjslmpqe$ occurs as the product of some subsequence\nof $fmvqslre$.\nLet $hjgrksla$ be the set of elements produced by the sequence $fmvqslre$.\n\nStart with $fmvqslre$ equal to the empty sequence. If at any point\nthe set $hjgrksla^{-1}hjgrksla = \\{sdzmbpqa lxvhzqwe: sdzmbpqa^{-1}, lxvhzqwe \\in hjgrksla\\}$ fails to be\nall of $qzxwvtnp$, extend $fmvqslre$ by appending an element $wjslmpqe$ of $qzxwvtnp$ not in\n$hjgrksla^{-1} hjgrksla$. Then $hjgrksla wjslmpqe \\cap hjgrksla$ must be empty, otherwise there would\nbe an equation of the form $sdzmbpqa wjslmpqe= lxvhzqwe $ with $sdzmbpqa, lxvhzqwe \\in qzxwvtnp$,\nor $wjslmpqe = sdzmbpqa^{-1} lxvhzqwe$, a contradiction. Thus we can extend $fmvqslre$ by one\nelement and double the size of $hjgrksla$.\n\nAfter $tbvysoam \\leq \\log_2 gdjoukhr$ steps, we must obtain a sequence $fmvqslre\n= yczstigb,\\dots,kngwyhza$ for which $hjgrksla^{-1} hjgrksla = qzxwvtnp$. Then\nthe sequence $kngwyhza^{-1}, \\dots, yczstigb^{-1}, yczstigb, \\dots, kngwyhza$\nproduces all of $qzxwvtnp$ and has length at most $(2/\\ln 2) \\ln gdjoukhr$.\n\n\\textbf{Second solution:}\n\nPut $pznrxvlu = |hjgrksla|$. We will show that we can append one element\n$wjslmpqe$ to $fmvqslre$ so that the resulting sequence of $tbvysoam+1$ elements will produce\nat least $2pznrxvlu-pznrxvlu^2/gdjoukhr$ elements of $qzxwvtnp$. To see this, we compute\n\\begin{align*}\n\\sum_{wjslmpqe \\in qzxwvtnp} |hjgrksla \\cup hjgrksla wjslmpqe|\n&= \\sum_{wjslmpqe \\in qzxwvtnp} (|hjgrksla| + |hjgrksla wjslmpqe| - |hjgrksla \\cap hjgrksla wjslmpqe|) \\\\\n&= 2pznrxvlu gdjoukhr - \\sum_{wjslmpqe \\in qzxwvtnp} |hjgrksla \\cap hjgrksla wjslmpqe| \\\\\n&= 2pznrxvlu gdjoukhr - |\\{(wjslmpqe,rqltvnso) \\in qzxwvtnp^2: rqltvnso \\in hjgrksla \\cap hjgrksla wjslmpqe\\}| \\\\\n&= 2pznrxvlu gdjoukhr - \\sum_{rqltvnso \\in hjgrksla} |\\{wjslmpqe \\in qzxwvtnp: rqltvnso \\in hjgrksla wjslmpqe\\}| \\\\\n&= 2pznrxvlu gdjoukhr - \\sum_{rqltvnso \\in hjgrksla} |hjgrksla^{-1} rqltvnso| \\\\\n&= 2pznrxvlu gdjoukhr - pznrxvlu^2.\n\\end{align*}\nBy the pigeonhole principle, we have $|hjgrksla \\cup hjgrksla wjslmpqe| \\geq 2pznrxvlu - pznrxvlu^2/gdjoukhr$ for\nsome choice of $wjslmpqe$, as claimed.\n\nIn other words, by extending the sequence by one element,\nwe can replace the ratio $vnrzqtla = 1-pznrxvlu/gdjoukhr$ (i.e., the fraction\nof elements of $qzxwvtnp$ not generated by $fmvqslre$)\nby a quantity no greater than\n\\[\n1-(2pznrxvlu-pznrxvlu^2/gdjoukhr)/gdjoukhr = vnrzqtla^2.\n\\]\nWe start out with $tbvysoam = 0$ and $vnrzqtla = 1 - 1/gdjoukhr$;\nafter $tbvysoam$ steps, we have $vnrzqtla \\leq (1-1/gdjoukhr)^{2^{tbvysoam}}$.\nIt is enough to prove that for some $jhvlktse > 0$, we can always find\nan integer $tbvysoam \\leq jhvlktse \\ln gdjoukhr$ such that\n\\[\n\\left(1 - \\frac{1}{gdjoukhr} \\right)^{2^{tbvysoam}} < \\frac{1}{gdjoukhr},\n\\]\nas then we have $gdjoukhr-pznrxvlu < 1$ and hence $hjgrksla = qzxwvtnp$.\n\nTo obtain this last inequality, put\n\\[\ntbvysoam = \\lfloor 2 \\log_2 gdjoukhr \\rfloor < (2/\\ln 2) \\ln gdjoukhr,\n\\]\nso that $2^{tbvysoam+1} \\geq gdjoukhr^2$.\nFrom the facts that $\\ln gdjoukhr \\leq \\ln 2 + (gdjoukhr-2)/2 \\leq gdjoukhr/2$ and\n$\\ln (1-1/gdjoukhr) < -1/gdjoukhr$ for all $gdjoukhr \\geq 2$, we have\n\\[\n2^{tbvysoam} \\ln \\left( 1 - \\frac{1}{gdjoukhr} \\right) < -\\frac{gdjoukhr^2}{2gdjoukhr} =  -\\frac{gdjoukhr}{2} < -\\ln gdjoukhr,\n\\]\nyielding the desired inequality.\n\n\\textbf{Remark:} An alternate approach in the second solution\nis to distinguish betwen the cases of $hjgrksla$ small (i.e.,\n$pznrxvlu < gdjoukhr^{1/2}$, in which case $pznrxvlu$ can be replaced by a value\nno less than $2pznrxvlu-1$) and $hjgrksla$ large.\nThis strategy is used in a number of recent results\nof Bourgain, Tao, Helfgott, and others on \\emph{small doubling}\nor \\emph{small tripling}\nof subsets of finite groups.\n\nIn the second solution, if we avoid the rather weak inequality\n$\\ln gdjoukhr \\leq gdjoukhr/2$, we instead get sequences of length\n$\\log_2 (gdjoukhr \\ln gdjoukhr) = \\log_2(gdjoukhr) + \\log_2 (\\ln gdjoukhr)$.\nThis is close to optimal: one cannot use fewer than $\\log_2 gdjoukhr$\nterms because the number of subsequences must be at least $gdjoukhr$.}",
      "confidence": "0.12"
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 and let G be a finite group of order n.  Show that there is an absolute constant c = 7 for which one can find elements a₁, … , a_k of G with\n\tk ≤ c · log₃ n\nso that the sequence\n\t( 1 , a₁ , a₂ , … , a_k , a_k⁻¹ , a_{k-1}⁻¹ , … , a₁⁻¹ )\nhas the following property: every element of G can be written as the ordered product of the terms of some non-empty subsequence of that sequence (a subsequence is obtained by deleting any of the displayed terms, without changing the order of the remaining terms).\n\n(The constant 7 is not best possible; it is chosen only to keep the numerical statement simple.)",
      "solution": "Throughout we write log for the logarithm to base 3.  The argument has two parts.\n\n───────────────────────────────\n1.  Constructing a rapidly growing set H.\n───────────────────────────────\n\nSay that a sequence S of elements of G *produces* g ∈ G if g is the product of some subsequence of S.  Denote\n\tH(S) = { g ∈ G : g is produced by S }.\n\nStart with the one-term sequence S₀ = (1); then H₀ = {1}.  Suppose after some steps we have a sequence S whose produced set is H = H(S), and assume that H⁻¹H ≠ G.  Choose an element\n\tg ∉ H⁻¹H\nand append it to S, obtaining the longer sequence S′.  Every element obtained with the new final term is of the form h g with h ∈ H, so H(S′) = H ∪ H g.\n\nClaim:  H g ∩ H = ∅.  Indeed, were h₁ g = h₂ with h₁,h₂ ∈ H, then g = h₁⁻¹h₂ ∈ H⁻¹H, contradicting the choice of g.  Consequently\n\t|H(S′)| = |H| + |H g| = 2 |H|,\nso each extension *doubles* the size of H.\n\nAfter k steps we therefore have |H| = 2ᵏ.  We must stop when no element lies outside H⁻¹H, i.e. when H⁻¹H = G.  Since |H| ≤ n at every stage, doubling can occur at most ⌈log₂ n⌉ steps, so we obtain a sequence\n\tS_k = (a₁, … , a_k)          with k ≤ ⌈log₂ n⌉ ≤ 2 log n,\n(in base-3 units this is k ≤ (log₂3) log n < 1.6 log n).\nFor this S_k we have, by construction,\n\tH_k⁻¹ H_k = G  where H_k = H(S_k).\n\n───────────────────────────────\n2.  A short sequence producing the whole group.\n───────────────────────────────\n\nForm\n\tL = ( 1 , a₁ , … , a_k , a_k⁻¹ , … , a₁⁻¹ ).\nIts length is\n\tℓ = 1 + 2k ≤ 1 + 2 log₂ n < 1 + 3.2 log n < 7 log n,\nso the numerical bound (with base-3 logarithms) is satisfied.\n\nIt remains to show that every g ∈ G is produced by L.\nBecause H_k⁻¹ H_k = G, for each g there exist h₁ , h₂ ∈ H_k with\n\tg = h₁ h₂⁻¹.\nChoose subsequences of (a₁,…,a_k) whose products are h₁ and h₂; write the index sets\n\tI = { i₁ < … < i_r },      J = { j₁ < … < j_s }.\nThen\n\th₁ = a_{i₁} … a_{i_r},\n\th₂ = a_{j₁} … a_{j_s}.\n\nInside L the inverses appear in *reverse* order,\n\ta_k⁻¹ , … , a₁⁻¹ ,\nso the subsequence\n\ta_{i₁} … a_{i_r}   a_{j_s}⁻¹ … a_{j₁}⁻¹\noccurs in L (all direct a’s come first, all chosen inverses come later) and multiplies to h₁ h₂⁻¹ = g.  Thus every element of G is the product of some subsequence of L.\n\n───────────────────────────────\n3.  Conclusion.\n───────────────────────────────\n\nWe have exhibited a sequence of length at most 7·log₃|G| that begins with the identity and whose non-empty subsequences produce every element of the group.  Hence the desired constant works.",
      "_meta": {
        "core_steps": [
          "Let H be the set of products obtainable from the current sequence S.",
          "If H^{-1}H ≠ G, choose g ∉ H^{-1}H; this guarantees Hg ∩ H = ∅.",
          "Appending g therefore expands |H| to at least 2|H| (size ≥ double).",
          "After ≤ ⌈log₂|G|⌉ such extensions we reach H^{-1}H = G.",
          "Append the inverses (a_k^{-1},…,a_1^{-1}) to S = (a_1,…,a_k); the resulting sequence of length ≤ (2/ln2)·ln|G| produces every element of G."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Choice of set used to detect ‘missing’ elements (H^{-1}H can be replaced by HH^{-1}, HH, etc., any subset closed under taking quotients that still equals G when H is sufficiently large).",
            "original": "H^{-1}H"
          },
          "slot2": {
            "description": "Exact growth guarantee; the argument only needs a constant factor >1, not necessarily a full doubling.",
            "original": "factor 2 (|H| → at least 2|H|)"
          },
          "slot3": {
            "description": "Base of the logarithm used to count steps; changing the base merely rescales the constant c.",
            "original": "base 2 logarithm (log₂|G|)"
          },
          "slot4": {
            "description": "Precise concatenation that turns the ‘generating’ sequence into one whose subsequences produce all of G; any ordering that lists each a_i with its inverse works.",
            "original": "a_k^{-1},…,a_1^{-1},a_1,…,a_k"
          },
          "slot5": {
            "description": "Numerical upper bound on length: any constant ≥ 2/ln2 suffices for c.",
            "original": "c = 2/ln 2 ≈ 2.885"
          },
          "slot6": {
            "description": "Treatment of the identity element (allowing the empty subsequence vs. appending 1_G at the end).",
            "original": "Allow empty subsequence or add identity as final term"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}