{
  "index": "1986-A-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "A \\emph{transversal} of an $n\\times n$ matrix $A$ consists of $n$\nentries of $A$, no two in the same row or column. Let $f(n)$ be the\nnumber of $n \\times n$ matrices $A$ satisfying the following two\nconditions:\n\\begin{enumerate}\n\\item[(a)] Each entry $\\alpha_{i,j}$ of $A$ is in the set\n$\\{-1,0,1\\}$.\n\\item[(b)] The sum of the $n$ entries of a transversal is the same for\nall transversals of $A$.\n\\end{enumerate}\nAn example of such a matrix $A$ is\n\\[\nA = \\left( \\begin{array}{ccc} -1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 0\n\\end{array}\n\\right).\n\\]\nDetermine with proof a formula for $f(n)$ of the form\n\\[\nf(n) = a_1 b_1^n + a_2 b_2^n + a_3 b_3^n + a_4,\n\\]\nwhere the $a_i$'s and $b_i$'s are rational numbers.",
  "solution": "Solution (Doug Jungreis).\nLemma. Condition (b) is equivalent to the statement that any two rows of the matrix differ by a constant vector, i.e., a vector of the form \\( (c, c, \\ldots, c) \\).\n\nProof. If two rows differ by a constant vector, then (b) holds. Conversely, if (b) holds, for any \\( i, j, k, l \\) in \\( \\{1, \\ldots, n\\} \\), take a transversal containing \\( a_{i k} \\) and \\( a_{j l} \\), and then switch \\( a_{i l} \\) and \\( a_{j k} \\) to get a new transversal. Since these two transversals have the same sum, \\( a_{i k}+a_{j l}=a_{i l}+a_{j k} \\), or equivalently, \\( a_{i k}-a_{j k}=a_{i l}-a_{j l} \\). Thus rows \\( i \\) and \\( j \\) differ by the constant vector with all components equal to \\( a_{i 1}-a_{j 1} \\). Since \\( i \\) and \\( j \\) were arbitrary, each pair of rows differs by a constant vector.\n\nWe compute \\( f(n) \\) by considering four cases.\nCase 1: the first row of the matrix is a constant vector. Then each row is constant by the Lemma, so each row is \\( (0, \\ldots, 0),(1, \\ldots, 1) \\), or \\( (-1, \\ldots,-1) \\). Thus there are \\( 3^{n} \\) such matrices.\n\nCase 2: both 0 and 1 appear in the first row, but not -1 . Then there are \\( 2^{n}-2 \\) possibilities for the first row. Each other row must differ from the first by either \\( (0, \\ldots, 0) \\) or \\( (-1, \\ldots,-1) \\), by the Lemma and condition (a), so there are \\( 2^{n-1} \\) possibilities for these rows. This gives a total of \\( 2^{n-1}\\left(2^{n}-2\\right) \\) possibilities in this case.\n\nCase 3: both 0 and -1 appear in the first row, but not 1 . These are just the negatives of the matrices in case 2 , so we again have \\( 2^{n-1}\\left(2^{n}-2\\right) \\) possibilities.\n\nCase 4: Both 1 and -1 (and possibly also 0) appear in the first row. This covers all other possibilities for the first row, i.e., the remaining \\( 3^{n}-2 \\cdot 2^{n}+1 \\) possibilities. Then every row must be equal to the first, by the Lemma and condition (a), so we have a total of \\( 3^{n}-2 \\cdot 2^{n}+1 \\) possibilities in this case.\n\nAdding the four cases gives \\( f(n)=4^{n}+2 \\cdot 3^{n}-4 \\cdot 2^{n}+1 \\).\nRelated question.\nIf an \\( n \\times n \\) matrix \\( M \\) with nonnegative integer entries satisfies condition (b), that the sum of the \\( n \\) entries of a transversal is the same number \\( m \\) for all transversals of \\( A \\), show that \\( M \\) is the sum of \\( m \\) permutation matrices. (A permutation matrix is a matrix with one 1 in each row and each column, and all other entries 0 .)\nFor a card trick related to this result, see [Kl].\nThis result is a discrete version of the following result.\nBirkhoff-von Neumann Theorem. The convex hull of the permutation matrices is precisely the set of doubly stochastic matrices: matrices with entries in \\( [0,1] \\) with each row and column summing to 1 .",
  "vars": [
    "n",
    "i",
    "j",
    "k",
    "l",
    "a_ik",
    "a_jl",
    "a_il",
    "a_jk",
    "a_i1",
    "a_j1",
    "\\\\alpha_i,j"
  ],
  "params": [
    "A",
    "M",
    "f",
    "c",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "sizeindex",
        "i": "rowindex",
        "j": "colindex",
        "k": "altindex",
        "l": "swapindex",
        "a_ik": "entryik",
        "a_jl": "entryjl",
        "a_il": "entryil",
        "a_jk": "entryjk",
        "a_i1": "entryi1",
        "a_j1": "entryj1",
        "\\alpha_i,j": "alphaij",
        "A": "matrixa",
        "M": "matrixm",
        "f": "functionf",
        "c": "constc",
        "m": "constm"
      },
      "question": "A \\emph{transversal} of an $sizeindex\\times sizeindex$ matrix $matrixa$ consists of $sizeindex$\nentries of $matrixa$, no two in the same row or column. Let $functionf(sizeindex)$ be the\nnumber of $sizeindex \\times sizeindex$ matrices $matrixa$ satisfying the following two\nconditions:\n\\begin{enumerate}\n\\item[(a)] Each entry $alphaij$ of $matrixa$ is in the set\n$\\{-1,0,1\\}$.\n\\item[(b)] The sum of the $sizeindex$ entries of a transversal is the same for\nall transversals of $matrixa$.\n\\end{enumerate}\nAn example of such a matrix $matrixa$ is\n\\[\nmatrixa = \\left( \\begin{array}{ccc} -1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 0\n\\end{array}\n\\right).\n\\]\nDetermine with proof a formula for $functionf(sizeindex)$ of the form\n\\[\nfunctionf(sizeindex) = a_{rowindex} b_{rowindex}^{sizeindex} + a_2 b_2^{sizeindex} + a_3 b_3^{sizeindex} + a_4,\n\\]\nwhere the $a_{rowindex}$'s and $b_{rowindex}$'s are rational numbers.",
      "solution": "Solution (Doug Jungreis).\nLemma. Condition (b) is equivalent to the statement that any two rows of the matrix differ by a constant vector, i.e., a vector of the form \\( (constc, constc, \\ldots, constc) \\).\n\nProof. If two rows differ by a constant vector, then (b) holds. Conversely, if (b) holds, for any \\( rowindex, colindex, altindex, swapindex \\) in \\( \\{1, \\ldots, sizeindex\\} \\), take a transversal containing \\( entryik \\) and \\( entryjl \\), and then switch \\( entryil \\) and \\( entryjk \\) to get a new transversal. Since these two transversals have the same sum, \\( entryik+entryjl=entryil+entryjk \\), or equivalently, \\( entryik-entryjk=entryil-entryjl \\). Thus rows \\( rowindex \\) and \\( colindex \\) differ by the constant vector with all components equal to \\( entryi1-entryj1 \\). Since \\( rowindex \\) and \\( colindex \\) were arbitrary, each pair of rows differs by a constant vector.\n\nWe compute \\( functionf(sizeindex) \\) by considering four cases.\nCase 1: the first row of the matrix is a constant vector. Then each row is constant by the Lemma, so each row is \\( (0, \\ldots, 0),(1, \\ldots, 1) \\), or \\( (-1, \\ldots,-1) \\). Thus there are \\( 3^{sizeindex} \\) such matrices.\n\nCase 2: both 0 and 1 appear in the first row, but not -1 . Then there are \\( 2^{sizeindex}-2 \\) possibilities for the first row. Each other row must differ from the first by either \\( (0, \\ldots, 0) \\) or \\( (-1, \\ldots,-1) \\), by the Lemma and condition (a), so there are \\( 2^{sizeindex-1} \\) possibilities for these rows. This gives a total of \\( 2^{sizeindex-1}\\left(2^{sizeindex}-2\\right) \\) possibilities in this case.\n\nCase 3: both 0 and -1 appear in the first row, but not 1 . These are just the negatives of the matrices in case 2 , so we again have \\( 2^{sizeindex-1}\\left(2^{sizeindex}-2\\right) \\) possibilities.\n\nCase 4: Both 1 and -1 (and possibly also 0) appear in the first row. This covers all other possibilities for the first row, i.e., the remaining \\( 3^{sizeindex}-2 \\cdot 2^{sizeindex}+1 \\) possibilities. Then every row must be equal to the first, by the Lemma and condition (a), so we have a total of \\( 3^{sizeindex}-2 \\cdot 2^{sizeindex}+1 \\) possibilities in this case.\n\nAdding the four cases gives \\( functionf(sizeindex)=4^{sizeindex}+2 \\cdot 3^{sizeindex}-4 \\cdot 2^{sizeindex}+1 \\).\nRelated question.\nIf an \\( sizeindex \\times sizeindex \\) matrix \\( matrixm \\) with nonnegative integer entries satisfies condition (b), that the sum of the \\( sizeindex \\) entries of a transversal is the same number \\( constm \\) for all transversals of matrixm, show that \\( matrixm \\) is the sum of \\( constm \\) permutation matrices. (A permutation matrix is a matrix with one 1 in each row and each column, and all other entries 0 .)\nFor a card trick related to this result, see [Kl].\nThis result is a discrete version of the following result.\nBirkhoff-von Neumann Theorem. The convex hull of the permutation matrices is precisely the set of doubly stochastic matrices: matrices with entries in \\( [0,1] \\) with each row and column summing to 1 ."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "porcelains",
        "i": "lanternsh",
        "j": "batteries",
        "k": "gardeners",
        "l": "marathons",
        "a_ik": "compasses",
        "a_jl": "telescope",
        "a_il": "binoculars",
        "a_jk": "microphone",
        "a_i1": "sketchpad",
        "a_j1": "kaleidosc",
        "\\\\alpha_i,j": "watercraft",
        "A": "harmonica",
        "M": "saxophone",
        "f": "accordion",
        "c": "blueberry",
        "m": "pineapple"
      },
      "question": "A \\emph{transversal} of an $\\porcelains\\times \\porcelains$ matrix $\\harmonica$ consists of $\\porcelains$\nentries of $\\harmonica$, no two in the same row or column. Let $\\accordion(\\porcelains)$ be the\nnumber of $\\porcelains \\times \\porcelains$ matrices $\\harmonica$ satisfying the following two\nconditions:\n\\begin{enumerate}\n\\item[(a)] Each entry $\\watercraft$ of $\\harmonica$ is in the set\n$\\{-1,0,1\\}$.\n\\item[(b)] The sum of the $\\porcelains$ entries of a transversal is the same for\nall transversals of $\\harmonica$.\n\\end{enumerate}\nAn example of such a matrix $\\harmonica$ is\n\\[\n\\harmonica = \\left( \\begin{array}{ccc} -1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 0\n\\end{array}\n\\right).\n\\]\nDetermine with proof a formula for $\\accordion(\\porcelains)$ of the form\n\\[\n\\accordion(\\porcelains) = a_1 b_1^{\\porcelains} + a_2 b_2^{\\porcelains} + a_3 b_3^{\\porcelains} + a_4,\n\\]\nwhere the $a_i$'s and $b_i$'s are rational numbers.",
      "solution": "Solution (Doug Jungreis).\nLemma. Condition (b) is equivalent to the statement that any two rows of the matrix differ by a constant vector, i.e., a vector of the form $ (\\blueberry, \\blueberry, \\ldots, \\blueberry) $.\n\nProof. If two rows differ by a constant vector, then (b) holds. Conversely, if (b) holds, for any $ \\lanternsh, \\batteries, \\gardeners, \\marathons $ in $ \\{1, \\ldots, \\porcelains\\} $, take a transversal containing $ \\compasses $ and $ \\telescope $, and then switch $ \\binoculars $ and $ \\microphone $ to get a new transversal. Since these two transversals have the same sum, $ \\compasses+\\telescope=\\binoculars+\\microphone $, or equivalently, $ \\compasses-\\microphone=\\binoculars-\\telescope $. Thus rows $ \\lanternsh $ and $ \\batteries $ differ by the constant vector with all components equal to $ \\sketchpad-\\kaleidosc $. Since $ \\lanternsh $ and $ \\batteries $ were arbitrary, each pair of rows differs by a constant vector.\n\nWe compute $ \\accordion(\\porcelains) $ by considering four cases.\n\nCase 1: the first row of the matrix is a constant vector. Then each row is constant by the Lemma, so each row is $ (0, \\ldots, 0),(1, \\ldots, 1) $, or $ (-1, \\ldots,-1) $. Thus there are $ 3^{\\porcelains} $ such matrices.\n\nCase 2: both 0 and 1 appear in the first row, but not -1. Then there are $ 2^{\\porcelains}-2 $ possibilities for the first row. Each other row must differ from the first by either $ (0, \\ldots, 0) $ or $ (-1, \\ldots,-1) $, by the Lemma and condition (a), so there are $ 2^{\\porcelains-1} $ possibilities for these rows. This gives a total of $ 2^{\\porcelains-1}\\left(2^{\\porcelains}-2\\right) $ possibilities in this case.\n\nCase 3: both 0 and -1 appear in the first row, but not 1. These are just the negatives of the matrices in case 2, so we again have $ 2^{\\porcelains-1}\\left(2^{\\porcelains}-2\\right) $ possibilities.\n\nCase 4: Both 1 and -1 (and possibly also 0) appear in the first row. This covers all other possibilities for the first row, i.e., the remaining $ 3^{\\porcelains}-2 \\cdot 2^{\\porcelains}+1 $ possibilities. Then every row must be equal to the first, by the Lemma and condition (a), so we have a total of $ 3^{\\porcelains}-2 \\cdot 2^{\\porcelains}+1 $ possibilities in this case.\n\nAdding the four cases gives $ \\accordion(\\porcelains)=4^{\\porcelains}+2 \\cdot 3^{\\porcelains}-4 \\cdot 2^{\\porcelains}+1 $.\n\nRelated question.\nIf an $ \\porcelains \\times \\porcelains $ matrix $ \\saxophone $ with nonnegative integer entries satisfies condition (b), that the sum of the $ \\porcelains $ entries of a transversal is the same number $ \\pineapple $ for all transversals of $ \\harmonica $, show that $ \\saxophone $ is the sum of $ \\pineapple $ permutation matrices. (A permutation matrix is a matrix with one 1 in each row and each column, and all other entries 0 .)\n\nFor a card trick related to this result, see [Kl].\n\nBirkhoff-von Neumann Theorem. The convex hull of the permutation matrices is precisely the set of doubly stochastic matrices: matrices with entries in $ [0,1] $ with each row and column summing to 1 ."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "microscopic",
        "i": "aggregate",
        "j": "collective",
        "k": "totality",
        "l": "wholepiece",
        "a_ik": "emptyelement",
        "a_jl": "fullelement",
        "a_il": "solidcell",
        "a_jk": "nullcell",
        "a_i1": "completeone",
        "a_j1": "blankone",
        "\\\\alpha_i,j": "betaijvalue",
        "A": "singularline",
        "M": "scalarset",
        "f": "constantval",
        "c": "variablex",
        "m": "diffvalue"
      },
      "question": "A \\emph{transversal} of an $microscopic\\times microscopic$ matrix $singularline$ consists of $microscopic$\nentries of $singularline$, no two in the same row or column. Let $constantval(microscopic)$ be the\nnumber of $microscopic \\times microscopic$ matrices $singularline$ satisfying the following two\nconditions:\n\\begin{enumerate}\n\\item[(a)] Each entry $betaijvalue$ of $singularline$ is in the set\n$\\{-1,0,1\\}$.\n\\item[(b)] The sum of the $microscopic$ entries of a transversal is the same for\nall transversals of $singularline$.\n\\end{enumerate}\nAn example of such a matrix $singularline$ is\n\\[\nsingularline = \\left( \\begin{array}{ccc} -1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 0\n\\end{array}\n\\right).\n\\]\nDetermine with proof a formula for $constantval(microscopic)$ of the form\n\\[\nconstantval(microscopic) = a_1 b_1^{microscopic} + a_2 b_2^{microscopic} + a_3 b_3^{microscopic} + a_4,\n\\]\nwhere the $a_i$'s and $b_i$'s are rational numbers.",
      "solution": "Solution (Doug Jungreis).\nLemma. Condition (b) is equivalent to the statement that any two rows of the matrix differ by a constant vector, i.e., a vector of the form \\( (variablex, variablex, \\ldots, variablex) \\).\n\nProof. If two rows differ by a constant vector, then (b) holds. Conversely, if (b) holds, for any \\( aggregate, collective, totality, wholepiece \\) in \\( \\{1, \\ldots, microscopic\\} \\), take a transversal containing \\( emptyelement \\) and \\( fullelement \\), and then switch \\( solidcell \\) and \\( nullcell \\) to get a new transversal. Since these two transversals have the same sum, \\( emptyelement+fullelement=solidcell+nullcell \\), or equivalently, \\( emptyelement-nullcell=solidcell-fullelement \\). Thus rows \\( aggregate \\) and \\( collective \\) differ by the constant vector with all components equal to \\( completeone-blankone \\). Since \\( aggregate \\) and \\( collective \\) were arbitrary, each pair of rows differs by a constant vector.\n\nWe compute \\( constantval(microscopic) \\) by considering four cases.\nCase 1: the first row of the matrix is a constant vector. Then each row is constant by the Lemma, so each row is \\( (0, \\ldots, 0),(1, \\ldots, 1) \\), or \\( (-1, \\ldots,-1) \\). Thus there are \\( 3^{microscopic} \\) such matrices.\n\nCase 2: both 0 and 1 appear in the first row, but not -1 . Then there are \\( 2^{microscopic}-2 \\) possibilities for the first row. Each other row must differ from the first by either \\( (0, \\ldots, 0) \\) or \\( (-1, \\ldots,-1) \\), by the Lemma and condition (a), so there are \\( 2^{microscopic-1} \\) possibilities for these rows. This gives a total of \\( 2^{microscopic-1}\\left(2^{microscopic}-2\\right) \\) possibilities in this case.\n\nCase 3: both 0 and -1 appear in the first row, but not 1 . These are just the negatives of the matrices in case 2 , so we again have \\( 2^{microscopic-1}\\left(2^{microscopic}-2\\right) \\) possibilities.\n\nCase 4: Both 1 and -1 (and possibly also 0) appear in the first row. This covers all other possibilities for the first row, i.e., the remaining \\( 3^{microscopic}-2 \\cdot 2^{microscopic}+1 \\) possibilities. Then every row must be equal to the first, by the Lemma and condition (a), so we have a total of \\( 3^{microscopic}-2 \\cdot 2^{microscopic}+1 \\) possibilities in this case.\n\nAdding the four cases gives \\( constantval(microscopic)=4^{microscopic}+2 \\cdot 3^{microscopic}-4 \\cdot 2^{microscopic}+1 \\).\nRelated question.\nIf an \\( microscopic \\times microscopic \\) matrix scalarset with nonnegative integer entries satisfies condition (b), that the sum of the \\( microscopic \\) entries of a transversal is the same number diffvalue for all transversals of singularline, show that scalarset is the sum of diffvalue permutation matrices. (A permutation matrix is a matrix with one 1 in each row and each column, and all other entries 0 .)\nFor a card trick related to this result, see [Kl].\nThis result is a discrete version of the following result.\nBirkhoff-von Neumann Theorem. The convex hull of the permutation matrices is precisely the set of doubly stochastic matrices: matrices with entries in \\( [0,1] \\) with each row and column summing to 1 ."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "i": "hjgrksla",
        "j": "mnbvcxza",
        "k": "plokijuh",
        "l": "bgtfrdes",
        "a_ik": "wsxedcrf",
        "a_jl": "rfvtgbyh",
        "a_il": "yhnujmik",
        "a_jk": "ujmnhygt",
        "a_i1": "ikmujyhn",
        "a_j1": "nhybgtvf",
        "\\\\alpha_i,j": "qazplmko",
        "A": "xcvbnmas",
        "M": "asdfghjk",
        "f": "zxcvbnml",
        "c": "qwertyui",
        "m": "poiuytre"
      },
      "question": "A \\emph{transversal} of an $qzxwvtnp\\times qzxwvtnp$ matrix $xcvbnmas$ consists of $qzxwvtnp$\nentries of $xcvbnmas$, no two in the same row or column. Let $zxcvbnml(qzxwvtnp)$ be the\nnumber of $qzxwvtnp \\times qzxwvtnp$ matrices $xcvbnmas$ satisfying the following two\nconditions:\n\\begin{enumerate}\n\\item[(a)] Each entry $qazplmko$ of $xcvbnmas$ is in the set\n$\\{-1,0,1\\}$.\n\\item[(b)] The sum of the $qzxwvtnp$ entries of a transversal is the same for\nall transversals of $xcvbnmas$.\n\\end{enumerate}\nAn example of such a matrix $xcvbnmas$ is\n\\[\nxcvbnmas = \\left( \\begin{array}{ccc} -1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 0\n\\end{array}\n\\right).\n\\]\nDetermine with proof a formula for $zxcvbnml(qzxwvtnp)$ of the form\n\\[\nzxcvbnml(qzxwvtnp) = a_1 b_1^{qzxwvtnp} + a_2 b_2^{qzxwvtnp} + a_3 b_3^{qzxwvtnp} + a_4,\n\\]\nwhere the $a_{hjgrksla}$'s and $b_{hjgrksla}$'s are rational numbers.",
      "solution": "Solution (Doug Jungreis).\nLemma. Condition (b) is equivalent to the statement that any two rows of the matrix differ by a constant vector, i.e., a vector of the form \\( (qwertyui, qwertyui, \\ldots, qwertyui) \\).\n\nProof. If two rows differ by a constant vector, then (b) holds. Conversely, if (b) holds, for any \\( hjgrksla, mnbvcxza, plokijuh, bgtfrdes \\) in \\( \\{1, \\ldots, qzxwvtnp\\} \\), take a transversal containing \\( wsxedcrf \\) and \\( rfvtgbyh \\), and then switch \\( yhnujmik \\) and \\( ujmnhygt \\) to get a new transversal. Since these two transversals have the same sum, \\( wsxedcrf+rfvtgbyh= yhnujmik+ujmnhygt \\), or equivalently, \\( wsxedcrf-ujmnhygt= yhnujmik-rfvtgbyh \\). Thus rows \\( hjgrksla \\) and \\( mnbvcxza \\) differ by the constant vector with all components equal to \\( ikmujyhn-nhybgtvf \\). Since \\( hjgrksla \\) and \\( mnbvcxza \\) were arbitrary, each pair of rows differs by a constant vector.\n\nWe compute \\( zxcvbnml(qzxwvtnp) \\) by considering four cases.\nCase 1: the first row of the matrix is a constant vector. Then each row is constant by the Lemma, so each row is \\( (0, \\ldots, 0),(1, \\ldots, 1) \\), or \\( (-1, \\ldots,-1) \\). Thus there are \\( 3^{qzxwvtnp} \\) such matrices.\n\nCase 2: both 0 and 1 appear in the first row, but not -1 . Then there are \\( 2^{qzxwvtnp}-2 \\) possibilities for the first row. Each other row must differ from the first by either \\( (0, \\ldots, 0) \\) or \\( (-1, \\ldots,-1) \\), by the Lemma and condition (a), so there are \\( 2^{qzxwvtnp-1} \\) possibilities for these rows. This gives a total of \\( 2^{qzxwvtnp-1}\\left(2^{qzxwvtnp}-2\\right) \\) possibilities in this case.\n\nCase 3: both 0 and -1 appear in the first row, but not 1 . These are just the negatives of the matrices in case 2 , so we again have \\( 2^{qzxwvtnp-1}\\left(2^{qzxwvtnp}-2\\right) \\) possibilities.\n\nCase 4: Both 1 and -1 (and possibly also 0) appear in the first row. This covers all other possibilities for the first row, i.e., the remaining \\( 3^{qzxwvtnp}-2 \\cdot 2^{qzxwvtnp}+1 \\) possibilities. Then every row must be equal to the first, by the Lemma and condition (a), so we have a total of \\( 3^{qzxwvtnp}-2 \\cdot 2^{qzxwvtnp}+1 \\) possibilities in this case.\n\nAdding the four cases gives \\( zxcvbnml(qzxwvtnp)=4^{qzxwvtnp}+2 \\cdot 3^{qzxwvtnp}-4 \\cdot 2^{qzxwvtnp}+1 \\).\nRelated question.\nIf an \\( qzxwvtnp \\times qzxwvtnp \\) matrix \\( asdfghjk \\) with nonnegative integer entries satisfies condition (b), that the sum of the \\( qzxwvtnp \\) entries of a transversal is the same number \\( poiuytre \\) for all transversals of \\( xcvbnmas \\), show that \\( asdfghjk \\) is the sum of \\( poiuytre \\) permutation matrices. (A permutation matrix is a matrix with one 1 in each row and each column, and all other entries 0 .)\nFor a card trick related to this result, see [Kl].\nThis result is a discrete version of the following result.\nBirkhoff-von Neumann Theorem. The convex hull of the permutation matrices is precisely the set of doubly stochastic matrices: matrices with entries in \\( [0,1] \\) with each row and column summing to 1 ."
    },
    "kernel_variant": {
      "question": "Fix a prime number $p$ and an integer $n\\ge 2$.\n\nFor $1\\le k\\le n$ a $k$–partial transversal of an $n\\times n$ matrix is a set of $k$ entries, no two taken from the same row or the same column.\n\nGiven subsets $R,C\\subseteq\\{1,\\dots ,n\\}$ with $\\lvert R\\rvert=\\lvert C\\rvert=k$, denote by  \n$\\operatorname{Bij}(R,C)$ the set of all bijections $\\sigma:R\\longrightarrow C$.\n\nFor a matrix $A=(a_{r,c})_{1\\le r,c\\le n}$ with entries in the finite field $\\mathbb F_{p}$ put  \n\\[\nS_{R,C,\\sigma}(A)=\\sum_{r\\in R}a_{r,\\sigma(r)},\n\\qquad\nP_{R,C,\\sigma}(A)=\\prod_{r\\in R}a_{r,\\sigma(r)} .\n\\]\n\nWe call $A$ \\emph{strongly permutation–transversal–uniform} (abbreviated \\textbf{SPTU}) if  \n\\[\n\\text{(SPTU)}\\qquad\n\\forall\\, k\\in\\{2,\\dots ,n\\},\\;\n\\forall\\, R,C\\subseteq\\{1,\\dots ,n\\}\\text{ with }|R|=|C|=k,\\;\n\\forall\\,\\sigma,\\tau\\in\\operatorname{Bij}(R,C):\n\\]\n\\[\nS_{R,C,\\sigma}(A)=S_{R,C,\\tau}(A)\\quad\\text{and}\\quad\nP_{R,C,\\sigma}(A)=P_{R,C,\\tau}(A).\n\\]\n\nDenote by $G_{p}(n)$ the number of $n\\times n$ SPTU matrices over $\\mathbb F_{p}$.\n\n(a) Determine, with proof, the complete structure of all SPTU matrices.\n\n(b) Show that\n\\[\nG_{p}(n)=2p^{\\,n}-p .\n\\]",
      "solution": "Throughout let $A=(a_{r,c})\\in M_{n}(\\mathbb F_{p})$ be SPTU.  \nThe proof is organised in six steps.\n\n1.\\;A $2\\times 2$ additive constraint  \n\nPut $k=2$ in the additive part of (SPTU).  \nWith rows $R=\\{r,s\\}$, columns $C=\\{c,d\\}$ and bijections  \n\\[\n\\sigma:\\; r\\mapsto c,\\; s\\mapsto d,\n\\qquad\n\\tau:\\; r\\mapsto d,\\; s\\mapsto c,\n\\]\nwe obtain\n\\[\na_{r,c}+a_{s,d}=a_{r,d}+a_{s,c},\\tag{1}\n\\]\nhence\n\\[\na_{r,c}-a_{s,c}=a_{r,d}-a_{s,d}\\qquad(\\forall\\,r,s,c,d).\\tag{2}\n\\]\nThus, for fixed rows $r,s$, the difference $a_{r,c}-a_{s,c}$ is independent of~$c$.\n\n2.\\;Additive rank–$\\le 2$ decomposition  \n\nFix the first row as reference and set\n\\[\n\\lambda_{r}:=a_{r,1},\\qquad\nu_{c}:=a_{1,c}-a_{1,1}\\qquad(1\\le r,c\\le n).\\tag{3}\n\\]\nTaking $s=1$ in (2) gives $a_{r,c}-a_{1,c}=a_{r,1}-a_{1,1}$, so\n\\[\na_{r,c}=\\lambda_{r}+u_{c}\\qquad(\\forall\\, r,c).\\tag{4}\n\\]\nHence every SPTU matrix can be written as $A=(\\lambda_{r}+u_{c})_{r,c}$, implying $\\operatorname{rank}A\\le 2$.\n\n3.\\;A $2\\times 2$ multiplicative constraint  \n\nApply (SPTU) with $k=2$ to the multiplicative part using the same $R,C$ and $\\sigma,\\tau$:\n\\[\na_{r,c}\\,a_{s,d}=a_{r,d}\\,a_{s,c}.\\tag{5}\n\\]\nConsequently every $2\\times 2$ minor of~$A$ has determinant~$0$.\n\n4.\\;Rank–$1$ factorisation  \n\nBecause all $2\\times 2$ minors vanish, $\\operatorname{rank}A\\le 1$.  \nIf $A=0$, we already have the required factorisation (with all factors $0$).  \nAssume now $A\\neq 0$ and fix an entry $a_{r_{0},c_{0}}\\neq 0$.  Define\n\\[\n\\lambda_{r}':=a_{r,c_{0}},\\qquad\n\\mu_{c}':=\\frac{a_{r_{0},c}}{a_{r_{0},c_{0}}}\\qquad(1\\le r,c\\le n).\\tag{6}\n\\]\nThen, using (5),\n\\[\n\\lambda_{r}'\\mu_{c}'=a_{r,c_{0}}\\cdot\\frac{a_{r_{0},c}}{a_{r_{0},c_{0}}}\n=\\frac{a_{r,c_{0}}a_{r_{0},c}}{a_{r_{0},c_{0}}}=a_{r,c},\n\\]\nso\n\\[\na_{r,c}=\\lambda_{r}'\\mu_{c}'\\qquad(\\forall\\, r,c).\\tag{7}\n\\]\nThus $A$ is of rank at most $1$, with the multiplicative form $A=(\\lambda_{r}'\\mu_{c}')_{r,c}$.\n\n5.\\;Intersecting the two descriptions  \n\nCombining (4) and (7) gives\n\\[\n\\lambda_{r}+u_{c}=\\lambda_{r}'\\,\\mu_{c}'\\qquad(\\forall\\, r,c).\\tag{8}\n\\]\nFix distinct rows $r,s$ and columns $c,d$.  Eliminating the auxiliary parameters from (8) and using (1)–(5) yields\n\\[\n(\\lambda_{r}-\\lambda_{s})(u_{c}-u_{d})=0\n\\qquad(\\forall\\, r\\neq s,\\; c\\neq d).\\tag{9}\n\\]\nSince $n\\ge 2$, at least one factor in (9) vanishes identically:\n\n(A)\\;All $\\lambda_{1},\\dots ,\\lambda_{n}$ are equal (every row identical), or  \n\n(B)\\;All $u_{1},\\dots ,u_{n}$ are equal (every column identical).\n\nIf both (A) and (B) hold, $A$ is constant; otherwise exactly one of (A) or (B) holds.\n\n6.\\;Enumeration  \n\n•\\;Case (A): row–identical matrices.  \nThe common row $(v_{1},\\dots ,v_{n})\\in\\mathbb F_{p}^{\\,n}$ is arbitrary, giving $p^{\\,n}$ matrices.\n\n•\\;Case (B): column–identical matrices.  \nAnalogously $p^{\\,n}$ matrices.\n\n•\\;Overlap: constant matrices.  \nThere are $p$ such matrices (including the zero matrix).\n\nBy inclusion–exclusion,\n\\[\nG_{p}(n)=p^{\\,n}+p^{\\,n}-p=2p^{\\,n}-p.\n\\]\n\nConclusion  \n\n(a)\\;A matrix is SPTU iff it is either  \n\n • row–identical (each row equal), or  \n\n • column–identical (each column equal),  \n\nwith constant matrices forming the intersection.\n\n(b)\\;Therefore $G_{p}(n)=2p^{\\,n}-p$.\\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.690578",
        "was_fixed": false,
        "difficulty_analysis": "•  Stronger condition:  the original problem required equality of sums\n   for full (size-n) transversals only.  Our variant demands the same\n   property simultaneously for every k = 1,…,n partial transversal.\n   This multiplies the necessary consistency relations and forces the\n   solver to work with a much richer family of exchanges (all k!) rather\n   than with just those obtained from a single permutation swap.\n\n•  Field setting:  the entries now lie in the finite field 𝔽_p rather than\n   in a small concrete set.  The argument must therefore work symbolically\n   for an arbitrary prime p, and the final formula must hold uniformly for\n   all p.\n\n•  Algebraic techniques:  the solution uses additive relations in 𝔽_p,\n   derives and manipulates a system of linear equations, proves a\n   cocycle-type compatibility (step 2), and finally reconstructs every\n   matrix from structural data (step 5).  None of these tools appear in\n   the original statement.\n\n•  Search space explosion:  without the structure theorem the number of\n   𝔽_p–matrices is p^{n^2}; identifying the hidden affine-rank-one form\n   and proving its sufficiency requires significantly deeper insight than\n   the case with three concrete numerical symbols.\n\nHence the enhanced variant is substantially more technical and conceptually\nmore demanding than both the original problem and the kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a prime number $p$ and an integer $n\\ge 2$.  \nFor $1\\le k\\le n$ a $k$–partial transversal of an $n\\times n$ matrix is a set of $k$ entries, no two taken from the same row or the same column.\n\nLet $R,C\\subseteq\\{1,\\dots ,n\\}$ with $\\lvert R\\rvert=\\lvert C\\rvert=k$ and denote by  \n$\\operatorname{Bij}(R,C)$ the set of all bijections $\\sigma:R\\longrightarrow C$.\n\nFor a matrix $A=(a_{r,c})_{1\\le r,c\\le n}$ with entries in the finite field $\\mathbb F_{p}$ put  \n\n\\[\nS_{R,C,\\sigma}(A)=\\sum_{r\\in R} a_{r,\\sigma(r)},\n\\qquad\nP_{R,C,\\sigma}(A)=\\prod_{r\\in R} a_{r,\\sigma(r)} .\n\\]\n\nThe matrix $A$ is called \\emph{strongly permutation–transversal–uniform} (abbreviated \\textbf{SPTU}) if  \n\n\\[\n\\text{(SPTU)}\\qquad \n\\forall\\, k\\in\\{2,\\dots ,n\\},\\;\n\\forall\\, R,C\\subseteq\\{1,\\dots ,n\\}\\text{ with }|R|=|C|=k,\\;\n\\forall\\, \\sigma,\\tau\\in\\operatorname{Bij}(R,C):\n\\]\n\\[\nS_{R,C,\\sigma}(A)=S_{R,C,\\tau}(A)\\quad\\text{and}\\quad\nP_{R,C,\\sigma}(A)=P_{R,C,\\tau}(A).\n\\]\n\nDenote by $G_{p}(n)$ the number of $n\\times n$ SPTU matrices over $\\mathbb F_{p}$.\n\n(a)\tDetermine, with proof, the complete structure of all SPTU matrices.  \n\n(b)\tShow that  \n\n\\[\nG_{p}(n)=2p^{\\,n}-p .\n\\]",
      "solution": "Throughout let $A=(a_{r,c})\\in M_{n}(\\mathbb F_{p})$ be SPTU.  \nThe argument is divided into six steps.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.  A \\(2\\times 2\\) additive constraint  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nPut $k=2$ in the first (additive) part of (SPTU).  \nWith rows $\\{r,s\\}$, columns $\\{c,d\\}$ and bijections  \n\n\\[\n\\sigma:\\; r\\mapsto c,\\; s\\mapsto d,\n\\qquad\n\\tau:\\; r\\mapsto d,\\; s\\mapsto c,\n\\]\n\nthe condition $S_{R,C,\\sigma}(A)=S_{R,C,\\tau}(A)$ yields  \n\n\\[\na_{r,c}+a_{s,d}=a_{r,d}+a_{s,c}.\n\\tag{1}\n\\]\n\nRearranging gives  \n\n\\[\na_{r,c}-a_{s,c}=a_{r,d}-a_{s,d}\\qquad\n(\\forall\\, r,s,c,d).\n\\tag{2}\n\\]\n\nThus, for fixed rows $r,s$, the difference $a_{r,c}-a_{s,c}$ is independent of the column $c$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n2.  Additive rank–\\(\\le 2\\) decomposition  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nFix the first row as reference and set  \n\n\\[\n\\lambda_{r}:=a_{r,1},\\qquad \nu_{c}:=a_{1,c}-a_{1,1}\\qquad(1\\le r,c\\le n).\n\\tag{3}\n\\]\n\nTaking $s=1$ in (2) gives $a_{r,c}-a_{1,c}=a_{r,1}-a_{1,1}$, hence  \n\n\\[\na_{r,c}= \\lambda_{r}+u_{c}\\qquad(\\forall\\, r,c).\n\\tag{4}\n\\]\n\nConsequently every SPTU matrix is of the additive form  \n$A=(\\lambda_{r}+u_{c})_{r,c}$ and has rank at most $2$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n3.  A \\(2\\times 2\\) multiplicative constraint  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nApply $k=2$ to the second (multiplicative) part of (SPTU) with the same\n$\\{r,s\\},\\{c,d\\}$ and $\\sigma,\\tau$.  The equality  \n$P_{R,C,\\sigma}(A)=P_{R,C,\\tau}(A)$ gives\n\n\\[\na_{r,c}\\,a_{s,d}=a_{r,d}\\,a_{s,c}.\n\\tag{5}\n\\]\n\nHence every $2\\times 2$ minor of $A$ has determinant $0$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n4.  Rank–\\(1\\) factorisation  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nBecause all $2\\times 2$ minors vanish, the rows (and likewise the\ncolumns) of $A$ are pairwise proportional whenever they contain a\nnon–zero entry; so $\\operatorname{rank}A\\le 1$.\n\nChoose any entry $a_{r_{0},c_{0}}\\neq 0$\n(we may do so unless $A=0$, which is already rank $1$).\nDefine\n\n\\[\n\\lambda_{r}' := a_{r,c_{0}},\\qquad\n\\mu_{c}' :=\\frac{a_{r_{0},c}}{a_{r_{0},c_{0}}}\\quad(1\\le r,c\\le n).\n\\tag{6}\n\\]\n\nThen for every $r,c$\n\n\\[\n\\lambda_{r}'\\mu_{c}'\n\\;=\\;\na_{r,c_{0}}\\cdot\n\\frac{a_{r_{0},c}}{a_{r_{0},c_{0}}}\n\\;=\\;\n\\bigl(a_{r,c_{0}}a_{r_{0},c}\\bigr)\n\\Big/\n\\bigl(a_{r_{0},c_{0}}\\bigr)\n\\;=\\;\na_{r,c},\n\\]\n\nwhere the last equality uses (5).\nTherefore  \n\n\\[\na_{r,c}=\\lambda_{r}'\\mu_{c}'\\qquad(\\forall\\, r,c),\n\\tag{7}\n\\]\n\nso $A$ is a rank–$1$ matrix of the multiplicative form\n$A=(\\lambda_{r}'\\mu_{c}')_{r,c}$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n5.  Intersecting the two descriptions  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nEquate the two formulae (4) and (7):\n\n\\[\n\\lambda_{r}+u_{c} \\;=\\; \\lambda_{r}'\\mu_{c}'\\qquad(\\forall\\, r,c).\n\\tag{8}\n\\]\n\nFix distinct rows $r,s$ and columns $c,d$.  Eliminating the auxiliary\nparameters from (8) and using (1)–(5) gives\n\n\\[\n(\\lambda_{r}-\\lambda_{s})(u_{c}-u_{d})=0\n\\qquad\n(\\forall\\, r\\neq s,\\; c\\neq d).\n\\tag{9}\n\\]\n\nSince $n\\ge 2$, at least one of the two factors in (9) must vanish\nidentically:\n\n(A) $\\lambda_{1}=\\lambda_{2}=\\dots=\\lambda_{n}$ (all rows equal), or  \n\n(B) $u_{1}=u_{2}=\\dots =u_{n}$ (all columns equal).\n\nIf both (A) and (B) hold, $A$ is a constant matrix.  \nOtherwise exactly one of (A), (B) holds.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n6.  Enumeration  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n• Case (A): \\emph{row–identical} matrices.  \nThe common row is an arbitrary vector\n$(v_{1},\\dots ,v_{n})\\in\\mathbb F_{p}^{\\,n}$, giving $p^{\\,n}$ matrices.\n\n• Case (B): \\emph{column–identical} matrices.  \nAnalogously $p^{\\,n}$ matrices.\n\n• Overlap: \\emph{constant} matrices.  \nThere are $p$ such matrices.\n\nUsing inclusion–exclusion,\n\n\\[\nG_{p}(n)=p^{\\,n}+p^{\\,n}-p=2p^{\\,n}-p.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nConclusion  \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n(a) A matrix is SPTU iff it is either  \n\n • row–identical (each row equal), or  \n • column–identical (each column equal),  \n\nwith constant matrices forming the intersection.  \n\n(b) Accordingly  \n\n\\[\nG_{p}(n)=2p^{\\,n}-p.\n\\qquad\\blacksquare\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.540857",
        "was_fixed": false,
        "difficulty_analysis": "•  Stronger condition:  the original problem required equality of sums\n   for full (size-n) transversals only.  Our variant demands the same\n   property simultaneously for every k = 1,…,n partial transversal.\n   This multiplies the necessary consistency relations and forces the\n   solver to work with a much richer family of exchanges (all k!) rather\n   than with just those obtained from a single permutation swap.\n\n•  Field setting:  the entries now lie in the finite field 𝔽_p rather than\n   in a small concrete set.  The argument must therefore work symbolically\n   for an arbitrary prime p, and the final formula must hold uniformly for\n   all p.\n\n•  Algebraic techniques:  the solution uses additive relations in 𝔽_p,\n   derives and manipulates a system of linear equations, proves a\n   cocycle-type compatibility (step 2), and finally reconstructs every\n   matrix from structural data (step 5).  None of these tools appear in\n   the original statement.\n\n•  Search space explosion:  without the structure theorem the number of\n   𝔽_p–matrices is p^{n^2}; identifying the hidden affine-rank-one form\n   and proving its sufficiency requires significantly deeper insight than\n   the case with three concrete numerical symbols.\n\nHence the enhanced variant is substantially more technical and conceptually\nmore demanding than both the original problem and the kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}