{
  "index": "1959-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. Given the following matrix of 25 elements\n\\[\n\\left(\\begin{array}{rrrrr}\n11 & 17 & 25 & 19 & 16 \\\\\n24 & 10 & 13 & 15 & 3 \\\\\n12 & 5 & 14 & 2 & 18 \\\\\n23 & 4 & 1 & 8 & 22 \\\\\n6 & 20 & 7 & 21 & 9\n\\end{array}\\right),\n\\]\nchoose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct.",
  "solution": "Solution. Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central \\( 3 \\times 3 \\) submatrix. Since the largest element in this central submatrix is 15 , there is no admissible choice for which the minimum exceeds 15 . But the choice \\( 25,15,18,23,20 \\) satisfies the conditions and has minimum 15.\n\nRemark. It is easy to check that there is just one choice that realizes the minimum value of 15 .\n\nThe problem is a variation on what is called the assignment problem: Given an \\( n \\times n \\) matrix, choose \\( n \\) elements, one from each row and one from each column, so that the sum of these elements is as large as possible. The name derives from the following example. Suppose a company has \\( n \\) employees and \\( n \\) jobs. If \\( a_{i j} \\) is the utility of assigning the \\( i \\) th employee to the \\( j \\) th job, then the most desirable assignment is found by solving the assignment problem for the matrix ( \\( a_{i j} \\) ).\n\nFor the given matrix, it happens that the same choice is also the unique solution to the assignment problem.\n\nFor more information on the assignment problem, see G. Dantzig, Linear Programming and Extensions. Princeton University Press, 1963, page 316 ff ., or Reinfeld and Vogel, Mathematical Programming. PrenticeHall, Englewood Cliffs, N.J., 1958, page 238 ff.",
  "vars": [
    "n",
    "i",
    "j"
  ],
  "params": [
    "a_ij"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "sizevar",
        "i": "rowindex",
        "j": "colindex",
        "a_ij": "utilityentry"
      },
      "question": "4. Given the following matrix of 25 elements\n\\[\n\\left(\\begin{array}{rrrrr}\n11 & 17 & 25 & 19 & 16 \\\\\n24 & 10 & 13 & 15 & 3 \\\\\n12 & 5 & 14 & 2 & 18 \\\\\n23 & 4 & 1 & 8 & 22 \\\\\n6 & 20 & 7 & 21 & 9\n\\end{array}\\right),\n\\]\nchoose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct.",
      "solution": "Solution. Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central \\( 3 \\times 3 \\) submatrix. Since the largest element in this central submatrix is 15, there is no admissible choice for which the minimum exceeds 15. But the choice \\( 25,15,18,23,20 \\) satisfies the conditions and has minimum 15.\n\nRemark. It is easy to check that there is just one choice that realizes the minimum value of 15.\n\nThe problem is a variation on what is called the assignment problem: Given an \\( sizevar \\times sizevar \\) matrix, choose \\( sizevar \\) elements, one from each row and one from each column, so that the sum of these elements is as large as possible. The name derives from the following example. Suppose a company has \\( sizevar \\) employees and \\( sizevar \\) jobs. If \\( utilityentry \\) is the utility of assigning the \\( rowindex \\) th employee to the \\( colindex \\) th job, then the most desirable assignment is found by solving the assignment problem for the matrix ( \\( utilityentry \\) ).\n\nFor the given matrix, it happens that the same choice is also the unique solution to the assignment problem.\n\nFor more information on the assignment problem, see G. Dantzig, Linear Programming and Extensions. Princeton University Press, 1963, page 316 ff., or Reinfeld and Vogel, Mathematical Programming. PrenticeHall, Englewood Cliffs, N.J., 1958, page 238 ff."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "stonefruit",
        "i": "lanternfly",
        "j": "thunderfog",
        "a_ij": "meadowglass"
      },
      "question": "4. Given the following matrix of 25 elements\n\\[\n\\left(\\begin{array}{rrrrr}\n11 & 17 & 25 & 19 & 16 \\\\\n24 & 10 & 13 & 15 & 3 \\\\\n12 & 5 & 14 & 2 & 18 \\\\\n23 & 4 & 1 & 8 & 22 \\\\\n6 & 20 & 7 & 21 & 9\n\\end{array}\\right),\n\\]\nchoose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct.",
      "solution": "Solution. Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central \\( 3 \\times 3 \\) submatrix. Since the largest element in this central submatrix is 15 , there is no admissible choice for which the minimum exceeds 15 . But the choice \\( 25,15,18,23,20 \\) satisfies the conditions and has minimum 15.\n\nRemark. It is easy to check that there is just one choice that realizes the minimum value of 15 .\n\nThe problem is a variation on what is called the assignment problem: Given an \\( stonefruit \\times stonefruit \\) matrix, choose \\( stonefruit \\) elements, one from each row and one from each column, so that the sum of these elements is as large as possible. The name derives from the following example. Suppose a company has \\( stonefruit \\) employees and \\( stonefruit \\) jobs. If \\( meadowglass \\) is the utility of assigning the \\( lanternfly \\) th employee to the \\( thunderfog \\) th job, then the most desirable assignment is found by solving the assignment problem for the matrix ( \\( meadowglass \\) ).\n\nFor the given matrix, it happens that the same choice is also the unique solution to the assignment problem.\n\nFor more information on the assignment problem, see G. Dantzig, Linear Programming and Extensions. Princeton University Press, 1963, page 316 ff ., or Reinfeld and Vogel, Mathematical Programming. PrenticeHall, Englewood Cliffs, N.J., 1958, page 238 ff."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "zeroamount",
        "i": "collective",
        "j": "rowcounter",
        "a_ij": "displeasure"
      },
      "question": "4. Given the following matrix of 25 elements\n\\[\n\\left(\\begin{array}{rrrrr}\n11 & 17 & 25 & 19 & 16 \\\\\n24 & 10 & 13 & 15 & 3 \\\\\n12 & 5 & 14 & 2 & 18 \\\\\n23 & 4 & 1 & 8 & 22 \\\\\n6 & 20 & 7 & 21 & 9\n\\end{array}\\right),\n\\]\nchoose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct.",
      "solution": "Solution. Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central \\( 3 \\times 3 \\) submatrix. Since the largest element in this central submatrix is 15 , there is no admissible choice for which the minimum exceeds 15 . But the choice \\( 25,15,18,23,20 \\) satisfies the conditions and has minimum 15.\n\nRemark. It is easy to check that there is just one choice that realizes the minimum value of 15 .\n\nThe problem is a variation on what is called the assignment problem: Given an \\( zeroamount \\times zeroamount \\) matrix, choose \\( zeroamount \\) elements, one from each row and one from each column, so that the sum of these elements is as large as possible. The name derives from the following example. Suppose a company has \\( zeroamount \\) employees and \\( zeroamount \\) jobs. If \\( displeasure_{collective rowcounter} \\) is the utility of assigning the \\( collective \\) th employee to the \\( rowcounter \\) th job, then the most desirable assignment is found by solving the assignment problem for the matrix ( \\( displeasure_{collective rowcounter} \\) ).\n\nFor the given matrix, it happens that the same choice is also the unique solution to the assignment problem.\n\nFor more information on the assignment problem, see G. Dantzig, Linear Programming and Extensions. Princeton University Press, 1963, page 316 ff ., or Reinfeld and Vogel, Mathematical Programming. PrenticeHall, Englewood Cliffs, N.J., 1958, page 238 ff."
    },
    "garbled_string": {
      "map": {
        "n": "xbrtewpl",
        "j": "vnsqtgld",
        "a_ij": "rkmpsdqe"
      },
      "question": "4. Given the following matrix of 25 elements\n\\[\n\\left(\\begin{array}{rrrrr}\n11 & 17 & 25 & 19 & 16 \\\\\n24 & 10 & 13 & 15 & 3 \\\\\n12 & 5 & 14 & 2 & 18 \\\\\n23 & 4 & 1 & 8 & 22 \\\\\n6 & 20 & 7 & 21 & 9\n\\end{array}\\right),\n\\]\nchoose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct.",
      "solution": "Solution. Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central \\( 3 \\times 3 \\) submatrix. Since the largest element in this central submatrix is 15 , there is no admissible choice for which the minimum exceeds 15 . But the choice \\( 25,15,18,23,20 \\) satisfies the conditions and has minimum 15.\n\nRemark. It is easy to check that there is just one choice that realizes the minimum value of 15 .\n\nThe problem is a variation on what is called the assignment problem: Given an \\( xbrtewpl \\times xbrtewpl \\) matrix, choose \\( xbrtewpl \\) elements, one from each row and one from each column, so that the sum of these elements is as large as possible. The name derives from the following example. Suppose a company has \\( xbrtewpl \\) employees and \\( xbrtewpl \\) jobs. If \\( rkmpsdqe_{i vnsqtgld} \\) is the utility of assigning the \\( i \\) th employee to the \\( vnsqtgld \\) th job, then the most desirable assignment is found by solving the assignment problem for the matrix ( \\( rkmpsdqe_{i vnsqtgld} \\) ).\n\nFor the given matrix, it happens that the same choice is also the unique solution to the assignment problem.\n\nFor more information on the assignment problem, see G. Dantzig, Linear Programming and Extensions. Princeton University Press, 1963, page 316 ff ., or Reinfeld and Vogel, Mathematical Programming. PrenticeHall, Englewood Cliffs, N.J., 1958, page 238 ff."
    },
    "kernel_variant": {
      "question": "Let  \n\n  C = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝  \n  128  127  126  125  143  144  145  146  \n  124  123  122  121  142  141  140  139  \n  120  119  118  117  147  148  149  150  \n  116  115  114  113  132  152  151  134  \n\n  112  111  110  109  135  136  137  138  \n  108  107  106  105  133  154  155  156  \n  103  102  101   99  100  158  157  160  \n   95   94   93   92  161  162  163  164 ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠  \n\n(All sixty-four entries are distinct.)\n\nFor this 8 × 8 array we call  \n• the principal diagonal the eight cells (1,1), (2,2), …, (8,8);  \n• the anti-diagonal   the eight cells (1,8), (2,7), …, (8,1).\n\nChoose eight cells, no two taken from the same row or from the same column, in such a way that\n\n(1) Exactly three of the chosen cells lie on the principal diagonal;  \n(2) Exactly one of the chosen cells lies on the anti-diagonal.\n\nAmong all admissible 8-sets maximise the minimum of the eight chosen numbers.\n\n(a) What is this largest possible minimum?  \n(b) Exhibit an admissible 8-set whose minimum achieves the value found in (a).  \n(c) Prove rigorously that no admissible 8-set can have a larger minimum.",
      "solution": "Throughout, let m(S) denote the minimum of the eight numbers contained in an admissible set S.\n\n------------------------------------------------------------\nI A convenient partition of the rows\n------------------------------------------------------------\nFor every threshold T we split the rows into\n\n Low(T)  :=  { rows in which every entry of columns 1–4 is < T },  \n High(T) :=  the remaining rows.\n\nTable 1 shows, for rows 5–8, the largest entry occurring in columns 1–4.\n\n row 5 : 112 row 6 : 108 row 7 : 103 row 8 : 95               (1)\n\nConsequently\n\n Low(113) = {5, 6, 7, 8}, High(113) = {1, 2, 3, 4}.          (2)\n\nRows in Low(113) can reach a value ≥ 113 only in columns 5–8;  \nrows in High(113) can reach ≥ 113 already inside columns 1–4.\n\n------------------------------------------------------------\nII An upper bound m(S) ≤ 112\n------------------------------------------------------------\nAssume to the contrary that an admissible set S satisfies m(S) ≥ 113.\n\nStep 1. Where can the anti-diagonal element lie?  \nOn the anti-diagonal the four entries belonging to rows 5–8 are\n 95, 102, 106, 109  —all smaller than 113.\nTherefore the unique anti-diagonal element of S must lie in one of the\nrows 1–4, hence in a column belonging to {5, 6, 7, 8}.              (3)\n\nStep 2. Columns demanded by the rows 5–8.  \nBy (2) every row of {5, 6, 7, 8} lies in Low(113) and therefore must\ntake its element from columns 5–8.  Because the eight chosen cells are\ncolumn–wise disjoint, these four rows require four *different* columns\namong {5, 6, 7, 8}.                                                 (4)\n\nStep 3. Pigeon-hole contradiction.  \nFrom (3) one of the columns 5–8 is already occupied by the\nanti-diagonal cell coming from the *top* half of the board.  With four\nmore distinct columns needed for the bottom four rows by (4), we would\nneed five different columns inside the four-element set {5, 6, 7, 8}—an impossibility.\n\nHence no admissible 8-set can satisfy m(S) ≥ 113; whence\n\n m_max ≤ 112.                                                       (5)\n\n------------------------------------------------------------\nIII An admissible set with m = 112\n------------------------------------------------------------\nDefine\n\n S₀ = { (1,4), (2,3), (3,2), (4,5), (5,1), (6,6), (7,7), (8,8) }.   (6)\n\nThe associated numbers, row by row, are\n\n 125, 122, 119, 132, 112, 154, 157, 164.                            (7)\n\nTheir minimum equals 112, and the structural checks are\n\n • rows 1–8 and columns 1–8 each appear exactly once;  \n • principal-diagonal cells: (6,6), (7,7), (8,8)   = 3;  \n • anti-diagonal cell   : (4,5)           = 1.\n\nThus S₀ is admissible and\n\n m_max ≥ 112.                                                       (8)\n\n------------------------------------------------------------\nIV Conclusion\n------------------------------------------------------------\nFrom (5) and (8) we obtain\n\n m_max = 112.                                                       (9)\n\nThe 8-set S₀ given in (6) attains this value, completing parts (a) and\n(b); part (c) is the proof given in Sections I–II. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.516072",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original $5\\times5$ assignment problem and the $6\\times6$ kernel variant, the enhanced task is markedly harder for several independent reasons.\n\n1. Higher dimension: the matrix size rises from $5$ or $6$ to $8$, inflating the search space from $5!\\approx10^{2}$ to $8!\\approx4\\times10^{4}$ possible permutations even {\\em before} the extra conditions are imposed.\n\n2. Multiple interacting constraints: in addition to the standard “one per row and column”, solvers must respect\n   • four-quadrant coverage,  \n   • simultaneous representation on {\\em both} main diagonals.  \n   These requirements severely entangle positional choices, ruling out any naive greedy strategy.\n\n3. Two-stage bounding argument: the proof of optimality needs both\n   • a global numerical bound (arguing that one low quadrant caps the minimum at 28), and  \n   • a constructive combinatorial argument (producing a permutation satisfying all geometric constraints).  \n   The original problems required only one of these steps.\n\n4. Structural insight: discovering that the top-right quadrant is the “bottleneck’’ demands a global view of the matrix rather than local checking.  Conversely, building a compatible permutation afterwards involves an application of Hall’s marriage principle “by hand’’ to juggle rows and columns under the new diagonal/quadrant rules.\n\nAltogether these layers raise the problem well beyond the difficulty of the original assignment-style question, demanding deeper combinatorial reasoning and a sharper eye for hidden global restrictions."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  C = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝  \n  128  127  126  125  143  144  145  146  \n  124  123  122  121  142  141  140  139  \n  120  119  118  117  147  148  149  150  \n  116  115  114  113  132  152  151  134  \n\n  112  111  110  109  135  136  137  138  \n  108  107  106  105  133  154  155  156  \n  103  102  101   99  100  158  157  160  \n   95   94   93   92  161  162  163  164 ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠  \n\n(All sixty-four entries are distinct.)\n\nFor this 8 × 8 array we call  \n• the principal diagonal the eight cells (1,1), (2,2), …, (8,8);  \n• the anti-diagonal   the eight cells (1,8), (2,7), …, (8,1).\n\nChoose eight cells, no two taken from the same row or from the same column, in such a way that\n\n(1) Exactly three of the chosen cells lie on the principal diagonal;  \n(2) Exactly one of the chosen cells lies on the anti-diagonal.\n\nAmong all admissible 8-sets maximise the minimum of the eight chosen numbers.\n\n(a) What is this largest possible minimum?  \n(b) Exhibit an admissible 8-set whose minimum achieves the value found in (a).  \n(c) Prove rigorously that no admissible 8-set can have a larger minimum.",
      "solution": "Throughout, let m(S) denote the minimum of the eight numbers contained in an admissible set S.\n\n------------------------------------------------------------\nI A convenient partition of the rows\n------------------------------------------------------------\nFor every threshold T we split the rows into\n\n Low(T)  :=  { rows in which every entry of columns 1–4 is < T },  \n High(T) :=  the remaining rows.\n\nTable 1 shows, for rows 5–8, the largest entry occurring in columns 1–4.\n\n row 5 : 112 row 6 : 108 row 7 : 103 row 8 : 95               (1)\n\nConsequently\n\n Low(113) = {5, 6, 7, 8}, High(113) = {1, 2, 3, 4}.          (2)\n\nRows in Low(113) can reach a value ≥ 113 only in columns 5–8;  \nrows in High(113) can reach ≥ 113 already inside columns 1–4.\n\n------------------------------------------------------------\nII An upper bound m(S) ≤ 112\n------------------------------------------------------------\nAssume to the contrary that an admissible set S satisfies m(S) ≥ 113.\n\nStep 1. Where can the anti-diagonal element lie?  \nOn the anti-diagonal the four entries belonging to rows 5–8 are\n 95, 102, 106, 109  —all smaller than 113.\nTherefore the unique anti-diagonal element of S must lie in one of the\nrows 1–4, hence in a column belonging to {5, 6, 7, 8}.              (3)\n\nStep 2. Columns demanded by the rows 5–8.  \nBy (2) every row of {5, 6, 7, 8} lies in Low(113) and therefore must\ntake its element from columns 5–8.  Because the eight chosen cells are\ncolumn–wise disjoint, these four rows require four *different* columns\namong {5, 6, 7, 8}.                                                 (4)\n\nStep 3. Pigeon-hole contradiction.  \nFrom (3) one of the columns 5–8 is already occupied by the\nanti-diagonal cell coming from the *top* half of the board.  With four\nmore distinct columns needed for the bottom four rows by (4), we would\nneed five different columns inside the four-element set {5, 6, 7, 8}—an impossibility.\n\nHence no admissible 8-set can satisfy m(S) ≥ 113; whence\n\n m_max ≤ 112.                                                       (5)\n\n------------------------------------------------------------\nIII An admissible set with m = 112\n------------------------------------------------------------\nDefine\n\n S₀ = { (1,4), (2,3), (3,2), (4,5), (5,1), (6,6), (7,7), (8,8) }.   (6)\n\nThe associated numbers, row by row, are\n\n 125, 122, 119, 132, 112, 154, 157, 164.                            (7)\n\nTheir minimum equals 112, and the structural checks are\n\n • rows 1–8 and columns 1–8 each appear exactly once;  \n • principal-diagonal cells: (6,6), (7,7), (8,8)   = 3;  \n • anti-diagonal cell   : (4,5)           = 1.\n\nThus S₀ is admissible and\n\n m_max ≥ 112.                                                       (8)\n\n------------------------------------------------------------\nIV Conclusion\n------------------------------------------------------------\nFrom (5) and (8) we obtain\n\n m_max = 112.                                                       (9)\n\nThe 8-set S₀ given in (6) attains this value, completing parts (a) and\n(b); part (c) is the proof given in Sections I–II. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.431848",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original $5\\times5$ assignment problem and the $6\\times6$ kernel variant, the enhanced task is markedly harder for several independent reasons.\n\n1. Higher dimension: the matrix size rises from $5$ or $6$ to $8$, inflating the search space from $5!\\approx10^{2}$ to $8!\\approx4\\times10^{4}$ possible permutations even {\\em before} the extra conditions are imposed.\n\n2. Multiple interacting constraints: in addition to the standard “one per row and column”, solvers must respect\n   • four-quadrant coverage,  \n   • simultaneous representation on {\\em both} main diagonals.  \n   These requirements severely entangle positional choices, ruling out any naive greedy strategy.\n\n3. Two-stage bounding argument: the proof of optimality needs both\n   • a global numerical bound (arguing that one low quadrant caps the minimum at 28), and  \n   • a constructive combinatorial argument (producing a permutation satisfying all geometric constraints).  \n   The original problems required only one of these steps.\n\n4. Structural insight: discovering that the top-right quadrant is the “bottleneck’’ demands a global view of the matrix rather than local checking.  Conversely, building a compatible permutation afterwards involves an application of Hall’s marriage principle “by hand’’ to juggle rows and columns under the new diagonal/quadrant rules.\n\nAltogether these layers raise the problem well beyond the difficulty of the original assignment-style question, demanding deeper combinatorial reasoning and a sharper eye for hidden global restrictions."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}