{
  "index": "2018-A-2",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $S_1, S_2, \\dots, S_{2^n-1}$ be the nonempty subsets of $\\{1,2,\\dots,n\\}$ in some order, and let\n$M$ be the $(2^n-1) \\times (2^n-1)$ matrix whose $(i,j)$ entry is\n\\[\nm_{ij} = \\begin{cases} 0 & \\mbox{if }S_i \\cap S_j = \\emptyset; \\\\\n1 & \\mbox{otherwise.}\n\\end{cases}\n\\]\nCalculate the determinant of $M$.",
  "solution": "The answer is $1$ if $n=1$ and $-1$ if $n>1$. Write $M_n$ for a $(2^n-1) \\times (2^n-1)$ matrix of the given form, and note that $\\det M_n$ does not depend on the ordering of the subsets: transposing two subsets has the effect of transposing two rows and then transposing two columns in $M_n$, and this does not change the determinant.\n\nClearly $\\det M_1 = 1$. We claim that for $n>1$, $\\det M_n = -(\\det M_{n-1})^2$, and the desired answer will follow by induction. Let $S_1',\\ldots,S_{2^{n-1}-1}'$ denote the nonempty subsets of $\\{1,\\ldots,n-1\\}$ in any order, with resulting matrix $M_{n-1}$. Let $m_{ij}'$ denote the $(i,j)$ entry of $M_{n-1}$. Now order the nonempty subsets $S_1,\\ldots,S_{2^n-1}$ of $\\{1,\\ldots,n\\}$ as follows: \n\\[\nS_i= \\begin{cases} S_i' &  i \\leq 2^{n-1}-1 \\\\\nS_{i-2^{n-1}+1}' \\cup \\{n\\} & 2^{n-1} \\leq i \\leq 2^n-2 \\\\\n\\{n\\} & i=2^{n}-1.\n\\end{cases}\n\\]\n(For example, if $S_1', \\dots, S_{2^{n-1}-1}'$ are ordered in lexicographic order\nas binary strings, then so are $S_1,\\dots,S_{2^n-1}$.)\nLet $M_n$ be the resulting matrix. Then we have:\n\\[\nM_n = \\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& M_{n-1} &&& M_{n-1} && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&1 & \\cdots & 1 & 1 \\\\\n& M_{n-1} && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&1 &\\cdots & 1 & 1 \\\\ \\hline\n0 & \\cdots & 0 & 1 & \\cdots & 1 & 1\n\\end{array} \\right).\n\\]\n\nIn $M_n$, perform the following operations, which do not change the determinant: subtract the final row from rows $2^{n-1}$ through $2^n-2$, and then subtract the final column from columns $2^{n-1}$ through $2^n-2$. The result is the matrix\n\\[\n\\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& M_{n-1} &&& M_{n-1} && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&0 & \\cdots & 0 & 0 \\\\\n& M_{n-1} && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&0 &\\cdots & 0 & 0 \\\\ \\hline\n0 & \\cdots & 0 & 0 & \\cdots & 0 & 1\n\\end{array} \\right).\n\\]\n\nWe can remove the final row and column without changing the determinant. Now swap the first $2^{n-1}-1$ rows with the final $2^{n-1}-1$ rows: this changes the determinant by an overall factor of $(-1)^{(2^{n-1}-1)^2} = -1$. The result is the block-diagonal matrix $\\left( \\begin{matrix} M_{n-1} & 0 \\\\ M_{n-1} & M_{n-1} \\end{matrix} \\right)$, whose determinant is $(\\det M_{n-1})^2$. Thus $\\det M_n = -(\\det M_{n-1})^2$ as desired.",
  "vars": [
    "S_i",
    "S_j",
    "m_ij",
    "M",
    "M_n",
    "M_n-1",
    "i",
    "j"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S_i": "subsetfirst",
        "S_j": "subsetsecond",
        "m_ij": "matrixentry",
        "M": "bigmatrix",
        "M_n": "currentmatrix",
        "M_n-1": "prevmatrix",
        "i": "rowindex",
        "j": "colindex",
        "n": "setsize"
      },
      "question": "Let $S_1, S_2, \\dots, S_{2^{setsize}-1}$ be the nonempty subsets of \\{1,2,\\dots,setsize\\} in some order, and let $bigmatrix$ be the $(2^{setsize}-1) \\times (2^{setsize}-1)$ matrix whose $(rowindex,colindex)$ entry is\n\\[\nmatrixentry = \\begin{cases} 0 & \\mbox{if }subsetfirst \\cap subsetsecond = \\emptyset; \\\\ 1 & \\mbox{otherwise.} \\end{cases}\n\\]\nCalculate the determinant of $bigmatrix$.",
      "solution": "The answer is $1$ if $setsize=1$ and $-1$ if $setsize>1$. Write $currentmatrix$ for a $(2^{setsize}-1) \\times (2^{setsize}-1)$ matrix of the given form, and note that $\\det currentmatrix$ does not depend on the ordering of the subsets: transposing two subsets has the effect of transposing two rows and then transposing two columns in $currentmatrix$, and this does not change the determinant.\n\nClearly $\\det currentmatrix = 1$ when $setsize=1$. We claim that for $setsize>1$, $\\det currentmatrix = -(\\det prevmatrix)^2$, and the desired answer will follow by induction. Let $S_1',\\ldots,S_{2^{setsize-1}-1}'$ denote the nonempty subsets of \\{1,\\ldots,setsize-1\\} in any order, with resulting matrix $prevmatrix$. Let $matrixentry'$ denote the $(rowindex,colindex)$ entry of $prevmatrix$. Now order the nonempty subsets $S_1,\\ldots,S_{2^{setsize}-1}$ of \\{1,\\ldots,setsize\\} as follows:\n\\[\nsubsetfirst_{rowindex}= \\begin{cases} subsetfirst_{rowindex}' &  rowindex \\leq 2^{setsize-1}-1 \\\\ subsetfirst_{rowindex-2^{setsize-1}+1}' \\cup \\{setsize\\} & 2^{setsize-1} \\leq rowindex \\leq 2^{setsize}-2 \\\\ \\{setsize\\} & rowindex=2^{setsize}-1. \\end{cases}\n\\]\n(For example, if $S_1', \\dots, S_{2^{setsize-1}-1}'$ are ordered in lexicographic order as binary strings, then so are $S_1,\\dots,S_{2^{setsize}-1}$.) Let $currentmatrix$ be the resulting matrix. Then we have:\n\\[\ncurrentmatrix = \\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& prevmatrix &&& prevmatrix && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&1 & \\cdots & 1 & 1 \\\\\n& prevmatrix && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&1 &\\cdots & 1 & 1 \\\\ \\hline\n0 & \\cdots & 0 & 1 & \\cdots & 1 & 1\n\\end{array} \\right).\n\\]\n\nIn $currentmatrix$, perform the following operations, which do not change the determinant: subtract the final row from rows $2^{setsize-1}$ through $2^{setsize}-2$, and then subtract the final column from columns $2^{setsize-1}$ through $2^{setsize}-2$. The result is the matrix\n\\[\n\\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& prevmatrix &&& prevmatrix && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&0 & \\cdots & 0 & 0 \\\\\n& prevmatrix && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&0 &\\cdots & 0 & 0 \\\\ \\hline\n0 & \\cdots & 0 & 0 & \\cdots & 0 & 1\n\\end{array} \\right).\n\\]\n\nWe can remove the final row and column without changing the determinant. Now swap the first $2^{setsize-1}-1$ rows with the final $2^{setsize-1}-1$ rows: this changes the determinant by an overall factor of $(-1)^{(2^{setsize-1}-1)^2} = -1$. The result is the block-diagonal matrix $\\left( \\begin{matrix} prevmatrix & 0 \\\\ prevmatrix & prevmatrix \\end{matrix} \\right)$, whose determinant is $(\\det prevmatrix)^2$. Thus $\\det currentmatrix = -(\\det prevmatrix)^2$ as desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "S_i": "thistledown",
        "S_j": "afterglow",
        "m_ij": "buttermint",
        "M": "riverstone",
        "M_n": "candlewick",
        "M_n-1": "applecart",
        "i": "dragonfly",
        "j": "crocodile",
        "n": "caterpillar"
      },
      "question": "Let $S_1, S_2, \\dots, S_{2^{caterpillar}-1}$ be the nonempty subsets of $\\{1,2,\\dots,caterpillar\\}$ in some order, and let\n$riverstone$ be the $(2^{caterpillar}-1) \\times (2^{caterpillar}-1)$ matrix whose $(dragonfly,crocodile)$ entry is\n\\[\nbuttermint_{ij} = \\begin{cases} 0 & \\mbox{if }thistledown_{dragonfly} \\cap afterglow_{crocodile} = \\emptyset; \\\\ 1 & \\mbox{otherwise.}\\end{cases}\n\\]\nCalculate the determinant of $riverstone$.",
      "solution": "The answer is $1$ if $caterpillar=1$ and $-1$ if $caterpillar>1$. Write $candlewick$ for a $(2^{caterpillar}-1) \\times (2^{caterpillar}-1)$ matrix of the given form, and note that $\\det candlewick$ does not depend on the ordering of the subsets: transposing two subsets has the effect of transposing two rows and then transposing two columns in $candlewick$, and this does not change the determinant.\n\nClearly $\\det M_1 = 1$. We claim that for $caterpillar>1$, $\\det candlewick = -(\\det applecart)^2$, and the desired answer will follow by induction. Let $S_1',\\ldots,S_{2^{caterpillar-1}-1}'$ denote the nonempty subsets of $\\{1,\\ldots,caterpillar-1\\}$ in any order, with resulting matrix $applecart$. Let $buttermint_{ij}'$ denote the $(dragonfly,crocodile)$ entry of $applecart$. Now order the nonempty subsets $S_1,\\ldots,S_{2^{caterpillar}-1}$ of $\\{1,\\ldots,caterpillar\\}$ as follows:\n\\[\nthistledown_{dragonfly}= \\begin{cases} S_i' &  \\ dragonfly \\leq 2^{caterpillar-1}-1 \\\\ S_{dragonfly-2^{caterpillar-1}+1}' \\cup \\{caterpillar\\} & 2^{caterpillar-1} \\leq dragonfly \\leq 2^{caterpillar}-2 \\\\ \\{caterpillar\\} & dragonfly=2^{caterpillar}-1.\\end{cases}\n\\]\n(For example, if $S_1', \\dots, S_{2^{caterpillar-1}-1}'$ are ordered in lexicographic order as binary strings, then so are $S_1,\\dots,S_{2^{caterpillar}-1}$.)\nLet $candlewick$ be the resulting matrix. Then we have:\n\\[\n\\candlewick = \\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& applecart &&& applecart && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&1 & \\cdots & 1 & 1 \\\\\n& applecart && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&1 &\\cdots & 1 & 1 \\\\ \\hline\n0 & \\cdots & 0 & 1 & \\cdots & 1 & 1\n\\end{array} \\right).\n\\]\n\nIn $candlewick$, perform the following operations, which do not change the determinant: subtract the final row from rows $2^{caterpillar-1}$ through $2^{caterpillar}-2$, and then subtract the final column from columns $2^{caterpillar-1}$ through $2^{caterpillar}-2$. The result is the matrix\n\\[\n\\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& applecart &&& applecart && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&0 & \\cdots & 0 & 0 \\\\\n& applecart && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&0 &\\cdots & 0 & 0 \\\\ \\hline\n0 & \\cdots & 0 & 0 & \\cdots & 0 & 1\n\\end{array} \\right).\n\\]\n\nWe can remove the final row and column without changing the determinant. Now swap the first $2^{caterpillar-1}-1$ rows with the final $2^{caterpillar-1}-1$ rows: this changes the determinant by an overall factor of $(-1)^{(2^{caterpillar-1}-1)^2} = -1$. The result is the block-diagonal matrix $\\left( \\begin{matrix} applecart & 0 \\\\ applecart & applecart \\end{matrix} \\right)$, whose determinant is $(\\det applecart)^2$. Thus $\\det candlewick = -(\\det applecart)^2$ as desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "S_i": "supersetone",
        "S_j": "supersettwo",
        "m_ij": "wholematrix",
        "M": "vectorial",
        "M_n": "vectorblock",
        "M_n-1": "vectorprior",
        "i": "wholeset",
        "j": "fragment",
        "n": "emptiness"
      },
      "question": "Let $S_1, S_2, \\dots, S_{2^{emptiness}-1}$ be the nonempty subsets of $\\{1,2,\\dots,emptiness\\}$ in some order, and let\n$vectorial$ be the $(2^{emptiness}-1) \\times (2^{emptiness}-1)$ matrix whose $(wholeset,fragment)$ entry is\n\\[\nwholematrix = \\begin{cases} 0 & \\mbox{if } supersetone \\cap supersettwo = \\emptyset; \\\\\n1 & \\mbox{otherwise.}\n\\end{cases}\n\\]\nCalculate the determinant of $vectorial$.",
      "solution": "The answer is $1$ if $emptiness=1$ and $-1$ if $emptiness>1$. Write $vectorblock$ for a $(2^{emptiness}-1) \\times (2^{emptiness}-1)$ matrix of the given form, and note that $\\det vectorblock$ does not depend on the ordering of the subsets: transposing two subsets has the effect of transposing two rows and then transposing two columns in $vectorblock$, and this does not change the determinant.\n\nClearly $\\det vectorial_1 = 1$. We claim that for $emptiness>1$, $\\det vectorblock = -(\\det vectorprior)^2$, and the desired answer will follow by induction. Let $S_1',\\ldots,S_{2^{emptiness-1}-1}'$ denote the nonempty subsets of $\\{1,\\ldots,emptiness-1\\}$ in any order, with resulting matrix $vectorprior$. Let $wholematrix'$ denote the $(wholeset,fragment)$ entry of $vectorprior$. Now order the nonempty subsets $S_1,\\ldots,S_{2^{emptiness}-1}$ of $\\{1,\\ldots,emptiness\\}$ as follows: \n\\[\nsupersetone= \\begin{cases} supersetone' &  wholeset \\leq 2^{emptiness-1}-1 \\\\\nS_{wholeset-2^{emptiness-1}+1}' \\cup \\{emptiness\\} & 2^{emptiness-1} \\leq wholeset \\leq 2^{emptiness}-2 \\\\\n\\{emptiness\\} & wholeset=2^{emptiness}-1.\n\\end{cases}\n\\]\n(For example, if $S_1', \\dots, S_{2^{emptiness-1}-1}'$ are ordered in lexicographic order\nas binary strings, then so are $S_1,\\dots,S_{2^{emptiness}-1}$.)\nLet $vectorblock$ be the resulting matrix. Then we have:\n\\[\nvectorblock = \\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& vectorprior &&& vectorprior && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&1 & \\cdots & 1 & 1 \\\\\n& vectorprior && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&1 &\\cdots & 1 & 1 \\\\ \\hline\n0 & \\cdots & 0 & 1 & \\cdots & 1 & 1\n\\end{array} \\right).\n\\]\n\nIn $vectorblock$, perform the following operations, which do not change the determinant: subtract the final row from rows $2^{emptiness-1}$ through $2^{emptiness}-2$, and then subtract the final column from columns $2^{emptiness-1}$ through $2^{emptiness}-2$. The result is the matrix\n\\[\n\\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& vectorprior &&& vectorprior && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&0 & \\cdots & 0 & 0 \\\\\n& vectorprior && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&0 &\\cdots & 0 & 0 \\\\ \\hline\n0 & \\cdots & 0 & 0 & \\cdots & 0 & 1\n\\end{array} \\right).\n\\]\n\nWe can remove the final row and column without changing the determinant. Now swap the first $2^{emptiness-1}-1$ rows with the final $2^{emptiness-1}-1$ rows: this changes the determinant by an overall factor of $(-1)^{(2^{emptiness-1}-1)^2} = -1$. The result is the block-diagonal matrix $\\left( \\begin{matrix} vectorprior & 0 \\\\ vectorprior & vectorprior \\end{matrix} \\right)$, whose determinant is $(\\det vectorprior)^2$. Thus $\\det vectorblock = -(\\det vectorprior)^2$ as desired."
    },
    "garbled_string": {
      "map": {
        "S_i": "qzxwvtnp",
        "S_j": "hjgrksla",
        "m_ij": "vbnmqwer",
        "M": "plokijuh",
        "M_n": "asdfghjk",
        "M_n-1": "zxcvbnml",
        "i": "rtyuiofg",
        "j": "dfghjklm",
        "n": "qwerasdf"
      },
      "question": "Let $S_1, S_2, \\dots, S_{2^{qwerasdf}-1}$ be the nonempty subsets of $\\{1,2,\\dots,qwerasdf\\}$ in some order, and let\n$plokijuh$ be the $(2^{qwerasdf}-1) \\times (2^{qwerasdf}-1)$ matrix whose $(rtyuiofg,dfghjklm)$ entry is\n\\[\nvbnmqwer = \\begin{cases} 0 & \\mbox{if }qzxwvtnp \\cap hjgrksla = \\emptyset; \\\\\n1 & \\mbox{otherwise.}\n\\end{cases}\n\\]\nCalculate the determinant of $plokijuh$.",
      "solution": "The answer is $1$ if $qwerasdf=1$ and $-1$ if $qwerasdf>1$. Write $asdfghjk$ for a $(2^{qwerasdf}-1) \\times (2^{qwerasdf}-1)$ matrix of the given form, and note that $\\det asdfghjk$ does not depend on the ordering of the subsets: transposing two subsets has the effect of transposing two rows and then transposing two columns in $asdfghjk$, and this does not change the determinant.\n\nClearly $\\det M_1 = 1$. We claim that for $qwerasdf>1$, $\\det asdfghjk = -(\\det zxcvbnml)^2$, and the desired answer will follow by induction. Let $S_1',\\ldots,S_{2^{qwerasdf-1}-1}'$ denote the nonempty subsets of $\\{1,\\ldots,qwerasdf-1\\}$ in any order, with resulting matrix $zxcvbnml$. Let $vbnmqwer'$ denote the $(rtyuiofg,dfghjklm)$ entry of $zxcvbnml$. Now order the nonempty subsets $S_1,\\ldots,S_{2^{qwerasdf}-1}$ of $\\{1,\\ldots,qwerasdf\\}$ as follows: \n\\[\nqzxwvtnp= \\begin{cases} qzxwvtnp' &  rtyuiofg \\leq 2^{qwerasdf-1}-1 \\\\\nS_{rtyuiofg-2^{qwerasdf-1}+1}' \\cup \\{qwerasdf\\} & 2^{qwerasdf-1} \\leq rtyuiofg \\leq 2^{qwerasdf}-2 \\\\\n\\{qwerasdf\\} & rtyuiofg=2^{qwerasdf}-1.\n\\end{cases}\n\\]\n(For example, if $S_1', \\dots, S_{2^{qwerasdf-1}-1}'$ are ordered in lexicographic order\nas binary strings, then so are $S_1,\\dots,S_{2^{qwerasdf}-1}$.)\nLet $asdfghjk$ be the resulting matrix. Then we have:\n\\[\nasdfghjk = \\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& zxcvbnml &&& zxcvbnml && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&1 & \\cdots & 1 & 1 \\\\\n& zxcvbnml && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&1 &\\cdots & 1 & 1 \\\\ \\hline\n0 & \\cdots & 0 & 1 & \\cdots & 1 & 1\n\\end{array} \\right).\n\\]\n\nIn $asdfghjk$, perform the following operations, which do not change the determinant: subtract the final row from rows $2^{qwerasdf-1}$ through $2^{qwerasdf}-2$, and then subtract the final column from columns $2^{qwerasdf-1}$ through $2^{qwerasdf}-2$. The result is the matrix\n\\[\n\\left( \\begin{array}{ccc|ccc|c}\n&&&&&& 0 \\\\\n& zxcvbnml &&& zxcvbnml && \\vdots \\\\\n&&&&&& 0 \\\\ \\hline\n&&&0 & \\cdots & 0 & 0 \\\\\n& zxcvbnml && \\vdots & \\ddots & \\vdots & \\vdots \\\\\n&&&0 &\\cdots & 0 & 0 \\\\ \\hline\n0 & \\cdots & 0 & 0 & \\cdots & 0 & 1\n\\end{array} \\right).\n\\]\n\nWe can remove the final row and column without changing the determinant. Now swap the first $2^{qwerasdf-1}-1$ rows with the final $2^{qwerasdf-1}-1$ rows: this changes the determinant by an overall factor of $(-1)^{(2^{qwerasdf-1}-1)^2} = -1$. The result is the block-diagonal matrix $\\left( \\begin{matrix} zxcvbnml & 0 \\\\ zxcvbnml & zxcvbnml \\end{matrix} \\right)$, whose determinant is $(\\det zxcvbnml)^2$. Thus $\\det asdfghjk = -(\\det zxcvbnml)^2$ as desired."
    },
    "kernel_variant": {
      "question": "For every integer $n\\ge 1$ list the $2^{\\,n}-1$ non-empty subsets  \n\\[\nR_1,R_2,\\dots ,R_{\\,2^{\\,n}-1}\\subset\\{1,2,\\dots ,n\\}\n\\]\nin the following order:\n\n• first all subsets that do not contain $n$;  \n• next all subsets that do contain $n$ but are different from the singleton $\\{n\\}$;  \n• finally the singleton $\\{n\\}$ itself.\n\nDefine the $(2^{\\,n}-1)\\times(2^{\\,n}-1)$ matrix $B_n=(b_{ij})$ by  \n\\[\nb_{ij}=\\begin{cases}\n0 &\\text{if }R_i\\cap R_j=\\varnothing,\\\\[3pt]\n(-1)^{|R_i\\cap R_j|} &\\text{if }R_i\\cap R_j\\ne\\varnothing .\n\\end{cases}\n\\]\n\nFind a closed formula for $\\displaystyle\\det B_n$ for every $n\\ge 1$.",
      "solution": "Throughout, $\\det M$ denotes the determinant of a square matrix $M$.\n\nStep 1.  The initial case $n=1$  \n\nThere is only one non–empty subset, namely $\\{1\\}$; hence $B_1=(-1)$ and  \n\\[\n\\det B_1=-1.\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 2.  Block decomposition for $n\\ge 2$  \n\nSet  \n\\[\nm:=2^{\\,n-1}-1\\quad\\bigl(\\text{so that $m$ is odd}\\bigr),\n\\qquad C:=B_{\\,n-1}\\ \\text{(size $m$)},\\qquad\nJ_m:=\\begin{pmatrix}1\\\\\\vdots\\\\1\\end{pmatrix}\\in\\Bbb R^{m}.\n\\]\n\nWrite the non–empty subsets of $\\{1,\\dots ,n-1\\}$ as  \n\\[\n\\mathcal S=\\{S_1,S_2,\\dots ,S_m\\},\n\\]\nalready ordered as prescribed for $B_{\\,n-1}$.  \nWith this notation the ordered list of the $R_k$ is\n\n1. $R_i=S_i\\quad(1\\le i\\le m)$;  \n2. $R_{m+i}=S_i\\cup\\{n\\}\\quad(1\\le i\\le m)$;  \n3. $R_{2m+1}=\\{n\\}$.\n\nA case distinction on the four possible intersection patterns gives the\nblock representation  \n\n\\[\nB_n=\n\\begin{pmatrix}\nC & C & 0\\\\[4pt]\nC & -C & -J_m\\\\[4pt]\n0 & -J_m^{\\!\\mathsf T} & -1\n\\end{pmatrix}.\n\\tag{1}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 3.  Two elementary operations preserving the determinant  \n\nApply the $m$ row operations  \n\\[\nR_{m+i}\\;\\longleftarrow\\;R_{m+i}-R_i\\qquad(1\\le i\\le m).\n\\]\nThey annihilate the left block of the second group of rows, giving  \n\n\\[\nB_n'=\n\\begin{pmatrix}\nC & C & 0\\\\\n0 & -2C & -J_m\\\\\n0 & -J_m^{\\!\\mathsf T} & -1\n\\end{pmatrix}.\n\\]\n\nNext perform the $m$ column operations  \n\\[\nC_{m+i}\\;\\longleftarrow\\;C_{m+i}-C_i\\qquad(1\\le i\\le m),\n\\]\nwhich kill the upper-right $m\\times m$ block and produce a block-upper-triangular\nmatrix  \n\n\\[\nB_n''=\n\\begin{pmatrix}\nC & 0 & 0\\\\\n0 & -2C & -J_m\\\\\n0 & -J_m^{\\!\\mathsf T} & -1\n\\end{pmatrix}.\n\\tag{2}\n\\]\n\nBoth types “row/column $+\\lambda$ (other row/column)” leave the determinant\nunchanged, so\n\\[\n\\det B_n=\\det B_n''.\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 4.  A Schur-complement reduction  \n\nWrite\n\\[\nD:=\n\\begin{pmatrix}-2C & -J_m\\\\[2pt]-J_m^{\\!\\mathsf T}&-1\\end{pmatrix}\n\\qquad(\\text{size }m+1).\n\\]\nBecause $-2C$ is invertible, the Schur complement formula yields  \n\n\\[\n\\det D\n=\\det(-2C)\\;\\Bigl(-1-J_m^{\\!\\mathsf T}(-2C)^{-1}J_m\\Bigr).\n\\tag{3}\n\\]\n\nThus\n\\[\n\\det B_n\n=\\det C \\;\\det D.\n\\tag{4}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 5.  The scalar $J_m^{\\!\\mathsf T}C^{-1}J_m$  \n\nFor any fixed non-empty subset $S\\subset\\{1,\\dots ,n-1\\}$,\n\\[\n\\sum_{T\\subset\\{1,\\dots ,n-1\\},\\,T\\ne\\varnothing}\n      (-1)^{|S\\cap T|}\n=\\Bigl(1+(-1)\\Bigr)^{|S|}-1\n=0-1=-1,\n\\]\nso every row-sum of $C$ equals $-1$.  Consequently\n\\[\nCJ_m=-J_m\\quad\\Longrightarrow\\quad C^{-1}J_m=-J_m\n\\]\nand\n\\[\nJ_m^{\\!\\mathsf T}C^{-1}J_m\n=-J_m^{\\!\\mathsf T}J_m\n=-m.\n\\tag{5}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 6.  Back to the Schur complement  \n\nSince $(-2C)^{-1}= -\\dfrac12\\,C^{-1}$, we have  \n\\[\nJ_m^{\\!\\mathsf T}(-2C)^{-1}J_m\n=-\\frac12\\,J_m^{\\!\\mathsf T}C^{-1}J_m\n=-\\frac12\\,(-m)=\\frac m2.\n\\]\nSubstituting into (3) gives\n\\[\n\\det D\n=\\det(-2C)\\;\\Bigl(-1-\\tfrac m2\\Bigr)\n=(-2)^m\\,\\det C\\;\\Bigl(-1-\\tfrac m2\\Bigr).\n\\tag{6}\n\\]\n\nBecause $m$ is odd, $(-2)^m=-2^{\\,m}$.  Multiplying the two negative\nfactors in (6) we obtain the positive simplification\n\\[\n(-2)^m\\Bigl(-1-\\tfrac m2\\Bigr)\n=2^{\\,m}\\Bigl(1+\\tfrac m2\\Bigr)\n=2^{\\,m-1}(m+2).\n\\tag{7}\n\\]\nHence\n\\[\n\\det D\n=2^{\\,m-1}(m+2)\\,\\det C.\n\\tag{8}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 7.  The fundamental recurrence for $\\det B_n$  \n\nCombining (4) with (8) yields\n\\[\n\\boxed{\\;\n\\det B_n\n=2^{\\,m-1}(m+2)\\,(\\det B_{\\,n-1})^{2}\n\\;}\n\\qquad\\bigl(m=2^{\\,n-1}-1\\bigr).\n\\tag{9}\n\\]\n\nNoting that $m+2=2^{\\,n-1}+1$, write (9) as\n\\[\n\\det B_n\n=2^{\\,2^{\\,n-1}-2}\\bigl(2^{\\,n-1}+1\\bigr)\\,(\\det B_{\\,n-1})^{2}.\n\\tag{10}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 8.  Solving the recurrence  \n\nDefine $\\Delta_n:=|\\det B_n|$.  Since (10) is positive for $n\\ge 2$ and\n$\\det B_1=-1$, one has  \n\n\\[\n\\Delta_1=1,\\qquad\n\\Delta_n\n=\\Bigl[\\,2^{\\,2^{\\,n-1}-2}\\bigl(2^{\\,n-1}+1\\bigr)\\Bigr]\\,\\Delta_{\\,n-1}^{\\,2}\n\\quad(n\\ge 2).\n\\]\n\nUnfolding the recursion gives  \n\\[\n\\Delta_n\n=\\prod_{k=2}^{n}\n     \\Bigl[\\,2^{\\,2^{\\,k-1}-2}\\bigl(2^{\\,k-1}+1\\bigr)\\Bigr]^{2^{\\,n-k}}\n =2^{\\,2^{\\,n-1}(n-3)+2}\\;\n  \\prod_{k=2}^{n}\\bigl(2^{\\,k-1}+1\\bigr)^{2^{\\,n-k}}.\n\\]\n\nConsequently  \n\n\\[\n\\boxed{\\;\n\\det B_1=-1,\\qquad\n\\det B_n=\n2^{\\,2^{\\,n-1}(n-3)+2}\\,\n\\prod_{k=2}^{n}\\bigl(2^{\\,k-1}+1\\bigr)^{2^{\\,n-k}}\n\\quad(n\\ge 2).\n\\;}\n\\tag{11}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 9.  Quick numerical check  \n\n$n=2$ : (11) yields $2^{0}(2+1)=3$, matching a direct computation.  \n$n=3$ : (11) gives $2^{2}(3^{2}\\cdot5)=4\\cdot45=180$, again correct.\n\nHence the formula (11) is rigorously established.\n\n────────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.849401",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional structure.  Unlike the original problem (entries $0$\nor a single constant), the enhanced matrix distinguishes the parity\nof the intersection, so every non-zero entry is either $+1$ or $-1$.\n\n2.  More intricate algebra.  \n   • The new sign pattern destroys the simple “all-ones rank 1”\n     behaviour exploited in the original; a non-trivial eigen-vector\n     analysis (Step 5) is now necessary.  \n   • The Schur-complement computation (Step 4) introduces\n     a non-constant scalar factor that depends on $n$ in an\n     exponential way, yielding a multiplicative cascade rather than\n     the simple alternating sign of the kernel problem.\n\n3.  Exponential-tower recursion.  \n   The recurrence (6) mixes an exponential power of 2 with a quadratic\n   squaring of the previous determinant, producing double-exponential\n   growth that must be tamed carefully to obtain the closed product.\n\n4.  Length and depth of proof.  \n   To reach the final formula one must perform:  \n   • a three-block decomposition;  \n   • two coordinated elementary transformations;  \n   • an eigen-value argument to evaluate a quadratic form;  \n   • a Schur-complement determinant;  \n   • an induction that needs telescoping of both\n     product and exponent sums.  \n   Each of these steps is conceptually more sophisticated than the\n   single block-swap that solved the original variant.\n\n5.  Final expression.  The determinant is no longer $\\pm1$ but an\n   explicit positive integer whose prime-factorisation involves both\n   powers of 2 and a geometric product $\\prod(2^{k-1}+1)^{2^{n-k}}$,\n   reflecting the much richer combinatorial information encoded in\n   $B_n$."
      }
    },
    "original_kernel_variant": {
      "question": "For every integer $n\\ge 1$ list the $2^{\\,n}-1$ non-empty subsets  \n\\[\nR_1,R_2,\\dots ,R_{\\,2^{\\,n}-1}\\subset\\{1,2,\\dots ,n\\}\n\\]\nin the following order:\n\n• first all subsets that do not contain $n$;  \n• next all subsets that do contain $n$ but are different from the singleton $\\{n\\}$;  \n• finally the singleton $\\{n\\}$ itself.\n\nDefine the $(2^{\\,n}-1)\\times(2^{\\,n}-1)$ matrix $B_n=(b_{ij})$ by  \n\\[\nb_{ij}=\\begin{cases}\n0 &\\text{if }R_i\\cap R_j=\\varnothing,\\\\[3pt]\n(-1)^{|R_i\\cap R_j|} &\\text{if }R_i\\cap R_j\\ne\\varnothing .\n\\end{cases}\n\\]\n\nFind a closed formula for $\\displaystyle\\det B_n$ for every $n\\ge 1$.",
      "solution": "Throughout, $\\det M$ denotes the determinant of a square matrix $M$.\n\nStep 1.  The initial case $n=1$  \n\nThere is only one non–empty subset, namely $\\{1\\}$; hence $B_1=(-1)$ and  \n\\[\n\\det B_1=-1.\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 2.  Block decomposition for $n\\ge 2$  \n\nSet  \n\\[\nm:=2^{\\,n-1}-1\\quad\\bigl(\\text{so that $m$ is odd}\\bigr),\n\\qquad C:=B_{\\,n-1}\\ \\text{(size $m$)},\\qquad\nJ_m:=\\begin{pmatrix}1\\\\\\vdots\\\\1\\end{pmatrix}\\in\\Bbb R^{m}.\n\\]\n\nWrite the non–empty subsets of $\\{1,\\dots ,n-1\\}$ as  \n\\[\n\\mathcal S=\\{S_1,S_2,\\dots ,S_m\\},\n\\]\nalready ordered as prescribed for $B_{\\,n-1}$.  \nWith this notation the ordered list of the $R_k$ is\n\n1. $R_i=S_i\\quad(1\\le i\\le m)$;  \n2. $R_{m+i}=S_i\\cup\\{n\\}\\quad(1\\le i\\le m)$;  \n3. $R_{2m+1}=\\{n\\}$.\n\nA case distinction on the four possible intersection patterns gives the\nblock representation  \n\n\\[\nB_n=\n\\begin{pmatrix}\nC & C & 0\\\\[4pt]\nC & -C & -J_m\\\\[4pt]\n0 & -J_m^{\\!\\mathsf T} & -1\n\\end{pmatrix}.\n\\tag{1}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 3.  Two elementary operations preserving the determinant  \n\nApply the $m$ row operations  \n\\[\nR_{m+i}\\;\\longleftarrow\\;R_{m+i}-R_i\\qquad(1\\le i\\le m).\n\\]\nThey annihilate the left block of the second group of rows, giving  \n\n\\[\nB_n'=\n\\begin{pmatrix}\nC & C & 0\\\\\n0 & -2C & -J_m\\\\\n0 & -J_m^{\\!\\mathsf T} & -1\n\\end{pmatrix}.\n\\]\n\nNext perform the $m$ column operations  \n\\[\nC_{m+i}\\;\\longleftarrow\\;C_{m+i}-C_i\\qquad(1\\le i\\le m),\n\\]\nwhich kill the upper-right $m\\times m$ block and produce a block-upper-triangular\nmatrix  \n\n\\[\nB_n''=\n\\begin{pmatrix}\nC & 0 & 0\\\\\n0 & -2C & -J_m\\\\\n0 & -J_m^{\\!\\mathsf T} & -1\n\\end{pmatrix}.\n\\tag{2}\n\\]\n\nBoth types “row/column $+\\lambda$ (other row/column)” leave the determinant\nunchanged, so\n\\[\n\\det B_n=\\det B_n''.\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 4.  A Schur-complement reduction  \n\nWrite\n\\[\nD:=\n\\begin{pmatrix}-2C & -J_m\\\\[2pt]-J_m^{\\!\\mathsf T}&-1\\end{pmatrix}\n\\qquad(\\text{size }m+1).\n\\]\nBecause $-2C$ is invertible, the Schur complement formula yields  \n\n\\[\n\\det D\n=\\det(-2C)\\;\\Bigl(-1-J_m^{\\!\\mathsf T}(-2C)^{-1}J_m\\Bigr).\n\\tag{3}\n\\]\n\nThus\n\\[\n\\det B_n\n=\\det C \\;\\det D.\n\\tag{4}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 5.  The scalar $J_m^{\\!\\mathsf T}C^{-1}J_m$  \n\nFor any fixed non-empty subset $S\\subset\\{1,\\dots ,n-1\\}$,\n\\[\n\\sum_{T\\subset\\{1,\\dots ,n-1\\},\\,T\\ne\\varnothing}\n      (-1)^{|S\\cap T|}\n=\\Bigl(1+(-1)\\Bigr)^{|S|}-1\n=0-1=-1,\n\\]\nso every row-sum of $C$ equals $-1$.  Consequently\n\\[\nCJ_m=-J_m\\quad\\Longrightarrow\\quad C^{-1}J_m=-J_m\n\\]\nand\n\\[\nJ_m^{\\!\\mathsf T}C^{-1}J_m\n=-J_m^{\\!\\mathsf T}J_m\n=-m.\n\\tag{5}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 6.  Back to the Schur complement  \n\nSince $(-2C)^{-1}= -\\dfrac12\\,C^{-1}$, we have  \n\\[\nJ_m^{\\!\\mathsf T}(-2C)^{-1}J_m\n=-\\frac12\\,J_m^{\\!\\mathsf T}C^{-1}J_m\n=-\\frac12\\,(-m)=\\frac m2.\n\\]\nSubstituting into (3) gives\n\\[\n\\det D\n=\\det(-2C)\\;\\Bigl(-1-\\tfrac m2\\Bigr)\n=(-2)^m\\,\\det C\\;\\Bigl(-1-\\tfrac m2\\Bigr).\n\\tag{6}\n\\]\n\nBecause $m$ is odd, $(-2)^m=-2^{\\,m}$.  Multiplying the two negative\nfactors in (6) we obtain the positive simplification\n\\[\n(-2)^m\\Bigl(-1-\\tfrac m2\\Bigr)\n=2^{\\,m}\\Bigl(1+\\tfrac m2\\Bigr)\n=2^{\\,m-1}(m+2).\n\\tag{7}\n\\]\nHence\n\\[\n\\det D\n=2^{\\,m-1}(m+2)\\,\\det C.\n\\tag{8}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 7.  The fundamental recurrence for $\\det B_n$  \n\nCombining (4) with (8) yields\n\\[\n\\boxed{\\;\n\\det B_n\n=2^{\\,m-1}(m+2)\\,(\\det B_{\\,n-1})^{2}\n\\;}\n\\qquad\\bigl(m=2^{\\,n-1}-1\\bigr).\n\\tag{9}\n\\]\n\nNoting that $m+2=2^{\\,n-1}+1$, write (9) as\n\\[\n\\det B_n\n=2^{\\,2^{\\,n-1}-2}\\bigl(2^{\\,n-1}+1\\bigr)\\,(\\det B_{\\,n-1})^{2}.\n\\tag{10}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 8.  Solving the recurrence  \n\nDefine $\\Delta_n:=|\\det B_n|$.  Since (10) is positive for $n\\ge 2$ and\n$\\det B_1=-1$, one has  \n\n\\[\n\\Delta_1=1,\\qquad\n\\Delta_n\n=\\Bigl[\\,2^{\\,2^{\\,n-1}-2}\\bigl(2^{\\,n-1}+1\\bigr)\\Bigr]\\,\\Delta_{\\,n-1}^{\\,2}\n\\quad(n\\ge 2).\n\\]\n\nUnfolding the recursion gives  \n\\[\n\\Delta_n\n=\\prod_{k=2}^{n}\n     \\Bigl[\\,2^{\\,2^{\\,k-1}-2}\\bigl(2^{\\,k-1}+1\\bigr)\\Bigr]^{2^{\\,n-k}}\n =2^{\\,2^{\\,n-1}(n-3)+2}\\;\n  \\prod_{k=2}^{n}\\bigl(2^{\\,k-1}+1\\bigr)^{2^{\\,n-k}}.\n\\]\n\nConsequently  \n\n\\[\n\\boxed{\\;\n\\det B_1=-1,\\qquad\n\\det B_n=\n2^{\\,2^{\\,n-1}(n-3)+2}\\,\n\\prod_{k=2}^{n}\\bigl(2^{\\,k-1}+1\\bigr)^{2^{\\,n-k}}\n\\quad(n\\ge 2).\n\\;}\n\\tag{11}\n\\]\n\n────────────────────────────────────────────────────────────────\nStep 9.  Quick numerical check  \n\n$n=2$ : (11) yields $2^{0}(2+1)=3$, matching a direct computation.  \n$n=3$ : (11) gives $2^{2}(3^{2}\\cdot5)=4\\cdot45=180$, again correct.\n\nHence the formula (11) is rigorously established.\n\n────────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.650131",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional structure.  Unlike the original problem (entries $0$\nor a single constant), the enhanced matrix distinguishes the parity\nof the intersection, so every non-zero entry is either $+1$ or $-1$.\n\n2.  More intricate algebra.  \n   • The new sign pattern destroys the simple “all-ones rank 1”\n     behaviour exploited in the original; a non-trivial eigen-vector\n     analysis (Step 5) is now necessary.  \n   • The Schur-complement computation (Step 4) introduces\n     a non-constant scalar factor that depends on $n$ in an\n     exponential way, yielding a multiplicative cascade rather than\n     the simple alternating sign of the kernel problem.\n\n3.  Exponential-tower recursion.  \n   The recurrence (6) mixes an exponential power of 2 with a quadratic\n   squaring of the previous determinant, producing double-exponential\n   growth that must be tamed carefully to obtain the closed product.\n\n4.  Length and depth of proof.  \n   To reach the final formula one must perform:  \n   • a three-block decomposition;  \n   • two coordinated elementary transformations;  \n   • an eigen-value argument to evaluate a quadratic form;  \n   • a Schur-complement determinant;  \n   • an induction that needs telescoping of both\n     product and exponent sums.  \n   Each of these steps is conceptually more sophisticated than the\n   single block-swap that solved the original variant.\n\n5.  Final expression.  The determinant is no longer $\\pm1$ but an\n   explicit positive integer whose prime-factorisation involves both\n   powers of 2 and a geometric product $\\prod(2^{k-1}+1)^{2^{n-k}}$,\n   reflecting the much richer combinatorial information encoded in\n   $B_n$."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}