{
  "index": "2004-A-5",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "An $m \\times n$ checkerboard is colored randomly: each square is\nindependently\nassigned red or black with probability $1/2$. We say that two squares,\n$p$ and $q$,  are in the same connected monochromatic region if there is\na sequence of squares, all of the same color, starting at $p$ and ending\nat $q$, in which successive squares in the sequence share a common side.\nShow that the expected number of connected monochromatic regions is\ngreater than $m n / 8$.",
  "solution": "\\textbf{First solution:}\nFirst recall that any graph with $n$ vertices and $e$ edges has at\nleast $n-e$ connected components (add each edge one at a time, and note\nthat it reduces the number of components by at most 1). Now\nimagine the squares of the checkerboard as a graph, whose vertices are\nconnected if the corresponding squares share a side and are\nthe same color. Let\n$A$ be the number of edges in the graph, and let $B$ be the number of\n4-cycles (formed by monochromatic $2 \\times 2$ squares).\nIf we remove the bottom edge of each 4-cycle, the resulting graph has\nthe same number of connected components as the original one;\nhence this number is at least\n\\[\nmn - A+B.\n\\]\nBy the linearity of expectation, the expected number of connected\ncomponents is at least\n\\[\nmn - E(A) + E(B).\n\\]\nMoreover, we may compute $E(A)$ by summing over the individual\npairs of adjacent squares, and we may compute $E(B)$ by summing over\nthe individual $2 \\times 2$ squares. Thus\n\\begin{align*}\nE(A) &= \\frac{1}{2}(m(n-1) + (m-1)n), \\\\\nE(B) &= \\frac{1}{8}(m-1)(n-1),\n\\end{align*}\nand so the expected number of components is at least\n\\begin{align*}\n& mn - \\frac{1}{2}(m(n-1)+(m-1)n) + \\frac{1}{8}(m-1)(n-1) \\\\\n&= \\frac{mn + 3m + 3n + 1}{8} >\n\\frac{mn}{8}.\n\\end{align*}\n\n\\textbf{Remark:}\nA ``dual'' approach is to consider\nthe graph whose vertices are the corners of the squares of the checkerboard,\nwith two vertices joined if they are adjacent and the edge between then\ndoes not separate two squares of the same color. In this approach,\nthe 4-cycles become isolated vertices, and the bound on components\nis replaced by a call to Euler's formula relating the vertices, edges\nand faces of a planar figure. (One must be careful, however, to correctly\nhandle faces which are not simply connected.)\n\n\\textbf{Second solution:} (by Noam Elkies)\nNumber the squares of the checkerboard $1, \\dots, mn$ by numbering the\nfirst row from left to right, then the second row, and so on. We prove\nby induction on $i$ that if we just consider the figure formed by the\nfirst $i$ squares, its expected number of monochromatic components is\nat least $i/8$. For $i=1$, this is clear.\n\nSuppose the $i$-th square does not abut the left edge or the top row\nof the board.\nThen we may divide into three cases.\n\\begin{itemize}\n\\item\nWith probability $1/4$, the $i$-th square is opposite in color from\nthe adjacent squares directly above and to the left of it. In this case\nadding the $i$-th square adds one component.\n\\item\nWith probability $1/8$, the $i$-th square is the same in color as the\nadjacent squares directly above and to the left of it, but opposite in\ncolor from its diagonal neighbor above and to the left. In this case,\nadding the $i$-th square either removes a component or leaves the number\nunchanged.\n\\item\nIn all other cases, the number of components remains unchanged upon\nadding the $i$-th square.\n\\end{itemize}\nHence adding the $i$-th square increases the expected number of components\nby $1/4 - 1/8 = 1/8$.\n\nIf the $i$-th square does abut the left edge of the board, the situation\nis even simpler: if the $i$-th square differs in color from the square\nabove it, one component is added, otherwise the number does not change. Hence\nadding the $i$-th square increases the expected number of components\nby $1/2$; likewise if the $i$-th square abuts the top edge of the board.\nThus the expected number of components is at least $i/8$ by induction,\nas desired.\n\n\\textbf{Remark:} Some solvers attempted to consider adding one row\nat a time, rather than one square; this must be handled with great\ncare, as it is possible that the number of components can drop\nrather precipitously upon adding an entire row.",
  "vars": [
    "p",
    "q",
    "A",
    "B",
    "e",
    "i"
  ],
  "params": [
    "m",
    "n",
    "E"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "p": "squareone",
        "q": "squaretwo",
        "A": "samecoloredges",
        "B": "squarecycles",
        "e": "edgequantity",
        "i": "indexsquare",
        "m": "rownumber",
        "n": "columnnumber",
        "E": "expectationop"
      },
      "question": "An $rownumber \\times columnnumber$ checkerboard is colored randomly: each square is independently assigned red or black with probability $1/2$. We say that two squares, $squareone$ and $squaretwo$, are in the same connected monochromatic region if there is a sequence of squares, all of the same color, starting at $squareone$ and ending at $squaretwo$, in which successive squares in the sequence share a common side. Show that the expected number of connected monochromatic regions is greater than $rownumber columnnumber / 8$.",
      "solution": "\\textbf{First solution:}\nFirst recall that any graph with $columnnumber$ vertices and $edgequantity$ edges has at least $columnnumber-edgequantity$ connected components (add each edge one at a time, and note that it reduces the number of components by at most 1). Now imagine the squares of the checkerboard as a graph, whose vertices are connected if the corresponding squares share a side and are the same color. Let $samecoloredges$ be the number of edges in the graph, and let $squarecycles$ be the number of 4-cycles (formed by monochromatic $2 \\times 2$ squares). If we remove the bottom edge of each 4-cycle, the resulting graph has the same number of connected components as the original one; hence this number is at least\n\\[\nrownumber columnnumber - samecoloredges + squarecycles.\n\\]\nBy the linearity of expectation, the expected number of connected components is at least\n\\[\nrownumber columnnumber - \\expectationop(samecoloredges) + \\expectationop(squarecycles).\n\\]\nMoreover, we may compute $\\expectationop(samecoloredges)$ by summing over the individual pairs of adjacent squares, and we may compute $\\expectationop(squarecycles)$ by summing over the individual $2 \\times 2$ squares. Thus\n\\begin{align*}\n\\expectationop(samecoloredges) &= \\frac{1}{2}\\bigl(rownumber(columnnumber-1) + (rownumber-1)columnnumber\\bigr), \\\\\n\\expectationop(squarecycles) &= \\frac{1}{8}(rownumber-1)(columnnumber-1),\n\\end{align*}\nand so the expected number of components is at least\n\\begin{align*}\n&\\, rownumber columnnumber - \\frac{1}{2}\\bigl(rownumber(columnnumber-1)+(rownumber-1)columnnumber\\bigr) + \\frac{1}{8}(rownumber-1)(columnnumber-1) \\\\\n&= \\frac{rownumber columnnumber + 3\\,rownumber + 3\\,columnnumber + 1}{8} >\n\\frac{rownumber columnnumber}{8}.\n\\end{align*}\n\n\\textbf{Remark:}\nA ``dual'' approach is to consider the graph whose vertices are the corners of the squares of the checkerboard, with two vertices joined if they are adjacent and the edge between then does not separate two squares of the same color. In this approach, the 4-cycles become isolated vertices, and the bound on components is replaced by a call to Euler's formula relating the vertices, edges and faces of a planar figure. (One must be careful, however, to correctly handle faces which are not simply connected.)\n\n\\textbf{Second solution:} (by Noam Elkies)\nNumber the squares of the checkerboard $1, \\dots, rownumber columnnumber$ by numbering the first row from left to right, then the second row, and so on. We prove by induction on $indexsquare$ that if we just consider the figure formed by the first $indexsquare$ squares, its expected number of monochromatic components is at least $indexsquare/8$. For $indexsquare=1$, this is clear.\n\nSuppose the $indexsquare$-th square does not abut the left edge or the top row of the board.\nThen we may divide into three cases.\n\\begin{itemize}\n\\item\nWith probability $1/4$, the $indexsquare$-th square is opposite in color from the adjacent squares directly above and to the left of it. In this case adding the $indexsquare$-th square adds one component.\n\\item\nWith probability $1/8$, the $indexsquare$-th square is the same in color as the adjacent squares directly above and to the left of it, but opposite in color from its diagonal neighbor above and to the left. In this case, adding the $indexsquare$-th square either removes a component or leaves the number unchanged.\n\\item\nIn all other cases, the number of components remains unchanged upon adding the $indexsquare$-th square.\n\\end{itemize}\nHence adding the $indexsquare$-th square increases the expected number of components by $1/4 - 1/8 = 1/8$.\n\nIf the $indexsquare$-th square does abut the left edge of the board, the situation is even simpler: if the $indexsquare$-th square differs in color from the square above it, one component is added, otherwise the number does not change. Hence adding the $indexsquare$-th square increases the expected number of components by $1/2$; likewise if the $indexsquare$-th square abuts the top edge of the board. Thus the expected number of components is at least $indexsquare/8$ by induction, as desired.\n\n\\textbf{Remark:} Some solvers attempted to consider adding one row at a time, rather than one square; this must be handled with great care, as it is possible that the number of components can drop rather precipitously upon adding an entire row."
    },
    "descriptive_long_confusing": {
      "map": {
        "p": "candidate",
        "q": "insulator",
        "A": "porcelain",
        "B": "saxophone",
        "e": "dandelion",
        "i": "lighthouse",
        "m": "watermelon",
        "n": "buttercup",
        "E": "hummingbird"
      },
      "question": "An $watermelon \\times buttercup$ checkerboard is colored randomly: each square is\nindependently\nassigned red or black with probability $1/2$. We say that two squares,\n$candidate$ and $insulator$,  are in the same connected monochromatic region if there is\na sequence of squares, all of the same color, starting at $candidate$ and ending\nat $insulator$, in which successive squares in the sequence share a common side.\nShow that the expected number of connected monochromatic regions is\ngreater than $watermelon buttercup / 8$.",
      "solution": "\\textbf{First solution:}\nFirst recall that any graph with $buttercup$ vertices and $dandelion$ edges has at\nleast $buttercup-dandelion$ connected components (add each edge one at a time, and note\nthat it reduces the number of components by at most 1). Now\nimagine the squares of the checkerboard as a graph, whose vertices are\nconnected if the corresponding squares share a side and are\nthe same color. Let\n$porcelain$ be the number of edges in the graph, and let $saxophone$ be the number of\n4-cycles (formed by monochromatic $2 \\times 2$ squares).\nIf we remove the bottom edge of each 4-cycle, the resulting graph has\nthe same number of connected components as the original one;\nhence this number is at least\n\\[\nwatermelon buttercup - porcelain+saxophone.\n\\]\nBy the linearity of expectation, the expected number of connected\ncomponents is at least\n\\[\nwatermelon buttercup - hummingbird(porcelain) + hummingbird(saxophone).\n\\]\nMoreover, we may compute $hummingbird(porcelain)$ by summing over the individual\npairs of adjacent squares, and we may compute $hummingbird(saxophone)$ by summing over\nthe individual $2 \\times 2$ squares. Thus\n\\begin{align*}\nhummingbird(porcelain) &= \\frac{1}{2}(watermelon(buttercup-1) + (watermelon-1)buttercup), \\\\\nhummingbird(saxophone) &= \\frac{1}{8}(watermelon-1)(buttercup-1),\n\\end{align*}\nand so the expected number of components is at least\n\\begin{align*}\n& watermelon buttercup - \\frac{1}{2}(watermelon(buttercup-1)+(watermelon-1)buttercup) + \\frac{1}{8}(watermelon-1)(buttercup-1) \\\\\n&= \\frac{watermelon buttercup + 3watermelon + 3buttercup + 1}{8} >\n\\frac{watermelon buttercup}{8}.\n\\end{align*}\n\n\\textbf{Remark:}\nA ``dual'' approach is to consider\nthe graph whose vertices are the corners of the squares of the checkerboard,\nwith two vertices joined if they are adjacent and the edge between then\ndoes not separate two squares of the same color. In this approach,\nthe 4-cycles become isolated vertices, and the bound on components\nis replaced by a call to Euler's formula relating the vertices, edges\nand faces of a planar figure. (One must be careful, however, to correctly\nhandle faces which are not simply connected.)\n\n\\textbf{Second solution:} (by Noam Elkies)\nNumber the squares of the checkerboard $1, \\dots, watermelon buttercup$ by numbering the\nfirst row from left to right, then the second row, and so on. We prove\nby induction on $lighthouse$ that if we just consider the figure formed by the\nfirst $lighthouse$ squares, its expected number of monochromatic components is\nat least $lighthouse/8$. For $lighthouse=1$, this is clear.\n\nSuppose the $lighthouse$-th square does not abut the left edge or the top row\nof the board.\nThen we may divide into three cases.\n\\begin{itemize}\n\\item\nWith probability $1/4$, the $lighthouse$-th square is opposite in color from\nthe adjacent squares directly above and to the left of it. In this case\nadding the $lighthouse$-th square adds one component.\n\\item\nWith probability $1/8$, the $lighthouse$-th square is the same in color as the\nadjacent squares directly above and to the left of it, but opposite in\ncolor from its diagonal neighbor above and to the left. In this case,\nadding the $lighthouse$-th square either removes a component or leaves the number\nunchanged.\n\\item\nIn all other cases, the number of components remains unchanged upon\nadding the $lighthouse$-th square.\n\\end{itemize}\nHence adding the $lighthouse$-th square increases the expected number of components\nby $1/4 - 1/8 = 1/8$.\n\nIf the $lighthouse$-th square does abut the left edge of the board, the situation\nis even simpler: if the $lighthouse$-th square differs in color from the square\nabove it, one component is added, otherwise the number does not change. Hence\nadding the $lighthouse$-th square increases the expected number of components\nby $1/2$; likewise if the $lighthouse$-th square abuts the top edge of the board.\nThus the expected number of components is at least $lighthouse/8$ by induction,\nas desired.\n\n\\textbf{Remark:} Some solvers attempted to consider adding one row\nat a time, rather than one square; this must be handled with great\ncare, as it is possible that the number of components can drop\nrather precipitously upon adding an entire row."
    },
    "descriptive_long_misleading": {
      "map": {
        "p": "wholeset",
        "q": "fragment",
        "A": "cornerless",
        "B": "acyclicset",
        "e": "vertexless",
        "i": "wholeindex",
        "m": "thinness",
        "n": "shortness",
        "E": "surprise"
      },
      "question": "An $thinness \\times shortness$ checkerboard is colored randomly: each square is independently\nassigned red or black with probability $1/2$. We say that two squares,\n$wholeset$ and $fragment$,  are in the same connected monochromatic region if there is\na sequence of squares, all of the same color, starting at $wholeset$ and ending\nat $fragment$, in which successive squares in the sequence share a common side.\nShow that the expected number of connected monochromatic regions is\ngreater than $thinness shortness / 8$.",
      "solution": "\\textbf{First solution:}\nFirst recall that any graph with $shortness$ vertices and $vertexless$ edges has at\nleast $shortness-vertexless$ connected components (add each edge one at a time, and note\nthat it reduces the number of components by at most 1). Now\nimagine the squares of the checkerboard as a graph, whose vertices are\nconnected if the corresponding squares share a side and are\nthe same color. Let\n$cornerless$ be the number of edges in the graph, and let $acyclicset$ be the number of\n4-cycles (formed by monochromatic $2 \\times 2$ squares).\nIf we remove the bottom edge of each 4-cycle, the resulting graph has\nthe same number of connected components as the original one;\nhence this number is at least\n\\[\nthinness shortness - cornerless+acyclicset.\n\\]\nBy the linearity of expectation, the expected number of connected\ncomponents is at least\n\\[\nthinness shortness - surprise(cornerless) + surprise(acyclicset).\n\\]\nMoreover, we may compute $surprise(cornerless)$ by summing over the individual\npairs of adjacent squares, and we may compute $surprise(acyclicset)$ by summing over\nthe individual $2 \\times 2$ squares. Thus\n\\begin{align*}\nsurprise(cornerless) &= \\frac{1}{2}(thinness(shortness-1) + (thinness-1)shortness), \\\\\nsurprise(acyclicset) &= \\frac{1}{8}(thinness-1)(shortness-1),\n\\end{align*}\nand so the expected number of components is at least\n\\begin{align*}\n& thinness shortness - \\frac{1}{2}(thinness(shortness-1)+(thinness-1)shortness) + \\frac{1}{8}(thinness-1)(shortness-1) \\\\\n&= \\frac{thinness shortness + 3thinness + 3shortness + 1}{8} >\n\\frac{thinness shortness}{8}.\n\\end{align*}\n\n\\textbf{Remark:}\nA ``dual'' approach is to consider\nthe graph whose vertices are the corners of the squares of the checkerboard,\nwith two vertices joined if they are adjacent and the edge between then\ndoes not separate two squares of the same color. In this approach,\nthe 4-cycles become isolated vertices, and the bound on components\nis replaced by a call to Euler's formula relating the vertices, edges\nand faces of a planar figure. (One must be careful, however, to correctly\nhandle faces which are not simply connected.)\n\n\\textbf{Second solution:} (by Noam Elkies)\nNumber the squares of the checkerboard $1, \\dots, thinness shortness$ by numbering the\nfirst row from left to right, then the second row, and so on. We prove\nby induction on $wholeindex$ that if we just consider the figure formed by the\nfirst $wholeindex$ squares, its expected number of monochromatic components is\nat least $wholeindex/8$. For $wholeindex=1$, this is clear.\n\nSuppose the $wholeindex$-th square does not abut the left edge or the top row\nof the board.\nThen we may divide into three cases.\n\\begin{itemize}\n\\item\nWith probability $1/4$, the $wholeindex$-th square is opposite in color from\nthe adjacent squares directly above and to the left of it. In this case\nadding the $wholeindex$-th square adds one component.\n\\item\nWith probability $1/8$, the $wholeindex$-th square is the same in color as the\nadjacent squares directly above and to the left of it, but opposite in\ncolor from its diagonal neighbor above and to the left. In this case,\nadding the $wholeindex$-th square either removes a component or leaves the number\nunchanged.\n\\item\nIn all other cases, the number of components remains unchanged upon\nadding the $wholeindex$-th square.\n\\end{itemize}\nHence adding the $wholeindex$-th square increases the expected number of components\nby $1/4 - 1/8 = 1/8$.\n\nIf the $wholeindex$-th square does abut the left edge of the board, the situation\nis even simpler: if the $wholeindex$-th square differs in color from the square\nabove it, one component is added, otherwise the number does not change. Hence\nadding the $wholeindex$-th square increases the expected number of components\nby $1/2$; likewise if the $wholeindex$-th square abuts the top edge of the board.\nThus the expected number of components is at least $wholeindex/8$ by induction,\nas desired.\n\n\\textbf{Remark:} Some solvers attempted to consider adding one row\nat a time, rather than one square; this must be handled with great\ncare, as it is possible that the number of components can drop\nrather precipitously upon adding an entire row."
    },
    "garbled_string": {
      "map": {
        "p": "rufgktha",
        "q": "zolvydqe",
        "A": "vnasjyht",
        "B": "kuytlmse",
        "e": "e",
        "i": "i",
        "m": "glatbens",
        "n": "qorvixad",
        "E": "udmikona"
      },
      "question": "An $glatbens \\times qorvixad$ checkerboard is colored randomly: each square is\nindependently\nassigned red or black with probability $1/2$. We say that two squares,\n$rufgktha$ and $zolvydqe$,  are in the same connected monochromatic region if there is\na sequence of squares, all of the same color, starting at $rufgktha$ and ending\nat $zolvydqe$, in which successive squares in the sequence share a common side.\nShow that the expected number of connected monochromatic regions is\ngreater than $glatbens qorvixad / 8$.",
      "solution": "\\textbf{First solution:}\nFirst recall that any graph with $qorvixad$ vertices and $e$ edges has at\nleast $qorvixad-e$ connected components (add each edge one at a time, and note\nthat it reduces the number of components by at most 1). Now\nimagine the squares of the checkerboard as a graph, whose vertices are\nconnected if the corresponding squares share a side and are\nthe same color. Let\n$vnasjyht$ be the number of edges in the graph, and let $kuytlmse$ be the number of\n4-cycles (formed by monochromatic $2 \\times 2$ squares).\nIf we remove the bottom edge of each 4-cycle, the resulting graph has\nthe same number of connected components as the original one;\nhence this number is at least\n\\[\nglatbensqorvixad - vnasjyht+kuytlmse.\n\\]\nBy the linearity of expectation, the expected number of connected\ncomponents is at least\n\\[\nglatbensqorvixad - udmikona(vnasjyht) + udmikona(kuytlmse).\n\\]\nMoreover, we may compute $udmikona(vnasjyht)$ by summing over the individual\npairs of adjacent squares, and we may compute $udmikona(kuytlmse)$ by summing over\nthe individual $2 \\times 2$ squares. Thus\n\\begin{align*}\nudmikona(vnasjyht) &= \\frac{1}{2}(glatbens(qorvixad-1) + (glatbens-1)qorvixad), \\\\\nudmikona(kuytlmse) &= \\frac{1}{8}(glatbens-1)(qorvixad-1),\n\\end{align*}\nand so the expected number of components is at least\n\\begin{align*}\n& glatbensqorvixad - \\frac{1}{2}(glatbens(qorvixad-1)+(glatbens-1)qorvixad) + \\frac{1}{8}(glatbens-1)(qorvixad-1) \\\\\n&= \\frac{glatbensqorvixad + 3glatbens + 3qorvixad + 1}{8} >\n\\frac{glatbensqorvixad}{8}.\n\\end{align*}\n\n\\textbf{Remark:}\nA ``dual'' approach is to consider\nthe graph whose vertices are the corners of the squares of the checkerboard,\nwith two vertices joined if they are adjacent and the edge between then\ndoes not separate two squares of the same color. In this approach,\nthe 4-cycles become isolated vertices, and the bound on components\nis replaced by a call to Euler's formula relating the vertices, edges\nand faces of a planar figure. (One must be careful, however, to correctly\nhandle faces which are not simply connected.)\n\n\\textbf{Second solution:} (by Noam Elkies)\nNumber the squares of the checkerboard $1, \\dots, glatbensqorvixad$ by numbering the\nfirst row from left to right, then the second row, and so on. We prove\nby induction on $i$ that if we just consider the figure formed by the\nfirst $i$ squares, its expected number of monochromatic components is\nat least $i/8$. For $i=1$, this is clear.\n\nSuppose the $i$-th square does not abut the left edge or the top row\nof the board.\nThen we may divide into three cases.\n\\begin{itemize}\n\\item\nWith probability $1/4$, the $i$-th square is opposite in color from\nthe adjacent squares directly above and to the left of it. In this case\nadding the $i$-th square adds one component.\n\\item\nWith probability $1/8$, the $i$-th square is the same in color as the\nadjacent squares directly above and to the left of it, but opposite in\ncolor from its diagonal neighbor above and to the left. In this case,\nadding the $i$-th square either removes a component or leaves the number\nunchanged.\n\\item\nIn all other cases, the number of components remains unchanged upon\nadding the $i$-th square.\n\\end{itemize}\nHence adding the $i$-th square increases the expected number of components\nby $1/4 - 1/8 = 1/8$.\n\nIf the $i$-th square does abut the left edge of the board, the situation\nis even simpler: if the $i$-th square differs in color from the square\nabove it, one component is added, otherwise the number does not change. Hence\nadding the $i$-th square increases the expected number of components\nby $1/2$; likewise if the $i$-th square abuts the top edge of the board.\nThus the expected number of components is at least $i/8$ by induction,\nas desired.\n\n\\textbf{Remark:} Some solvers attempted to consider adding one row\nat a time, rather than one square; this must be handled with great\ncare, as it is possible that the number of components can drop\nrather precipitously upon adding an entire row."
    },
    "kernel_variant": {
      "question": "Let $m,n,k\\in\\mathbb Z_{\\ge 1}$ and put  \n\\[\nB:=\\{1,\\dots ,m\\}\\times\\{1,\\dots ,n\\}\\times\\{1,\\dots ,k\\},\n\\qquad N:=mnk .\n\\]\n\nEvery unit cube (voxel) $\\xi\\in B$ is coloured independently with one of the three colours  \n\n\\[\n\\text{crimson},\\qquad \\text{jade},\\qquad \\text{obsidian}\n\\]\n\nwith respective probabilities  \n\n\\[\np_{\\text{cr}}=\\tfrac12,\\qquad \np_{\\text{ja}}=\\tfrac13,\\qquad \np_{\\text{ob}}=\\tfrac16,\\qquad \np_{\\text{cr}}+p_{\\text{ja}}+p_{\\text{ob}}=1 .\n\\]\n\nTwo cubes are \\emph{face–adjacent} if they share an entire $(1\\times1)$ face.  \nA (monochromatic) \\emph{face–connected component} is a maximal set of cubes of one fixed colour that can be linked by a chain of face–adjacent cubes of that colour.  \nDenote by $C$ the (random) number of monochromatic face–connected components in the whole block.\n\nFor $\\xi=(x,y,z)\\in B$ define the set of (existing) neighbours  \n\n\\[\n\\mathcal N(\\xi)=\\bigl\\{\n(x\\!-\\!1,y,z),\\,(x\\!+\\!1,y,z),\\,\n(x,y\\!-\\!1,z),\\,(x,y\\!+\\!1,z),\\,\n(x,y,z\\!-\\!1),\\,(x,y,z\\!+\\!1)\n\\bigr\\}\\cap B .\n\\]\n\nThe cube $\\xi$ is called \\emph{solitary} if none of the cubes in $\\mathcal N(\\xi)$ has the same colour as $\\xi$.  \nPut  \n\n\\[\nJ_{\\xi}:=\n\\begin{cases}\n1,&\\text{if }\\xi\\text{ is solitary},\\\\[4pt]\n0,&\\text{otherwise},\n\\end{cases}\n\\qquad \nS:=\\displaystyle\\sum_{\\xi\\in B}J_{\\xi}\\quad\n(\\text{number of solitary cubes}).\n\\]\n\nThroughout write  \n\n\\[\np_{1}=p_{\\text{cr}},\\quad\np_{2}=p_{\\text{ja}},\\quad\np_{3}=p_{\\text{ob}},\\qquad\n\\tau:=\\sum_{i=1}^{3}p_{i}(1-p_{i})^{6}\n      =\\frac{2167}{23328}\\approx0.09290.\n\\]\n\nAnswer the following.\n\n(a)  Prove the uniform lower bound  \n\\[\n\\boxed{\\;\n\\mathbb P\\!\\bigl[J_{\\xi}=1\\bigr]\\;\\ge\\;\\tau\\quad\n\\text{for all }\\xi\\in B\\;}\n\\tag{1}\n\\]\nand compute the exact value of $\\tau$ given above.\n\n(b)  Show that $C\\ge S$ and hence  \n\n\\[\n\\boxed{\\;\n\\mathbb E[C]\\;\\ge\\;\\tau\\,N\\;}\n\\tag{2}\n\\]\n\n(c)  A single recolouring of one voxel may affect $J_{\\xi}$ for the voxel itself and for each of its (up to) six face–neighbours.\n\n\\quad(i)  Prove that for every pair of colourings $\\omega,\\omega'$ differing in the colour of exactly one voxel,\n\\[\n       |S(\\omega)-S(\\omega')|\\le 7 .\n\\]\n\n\\quad(ii)  Using McDiarmid’s bounded–difference inequality,  \nshow that for every $t>0$\n\\[\n       \\boxed{\\;\n       \\mathbb P\\bigl[S\\le\\mathbb E[S]-t\\bigr]\n       \\;\\le\\;\n       \\exp\\!\\Bigl(\n          -\\dfrac{2t^{2}}{49N}\n       \\Bigr)\\;}\n       \\tag{3}\n\\]\n\n(d)  Put $\\displaystyle\\delta:=\\tau-\\dfrac1{12}=\\dfrac{223}{23328}\\approx0.00956$ and define  \n\n\\[\nt_{0}:=\\delta\\,N\\quad(>0).\n\\]\n\nCombine (2) and (3) to show the high–probability lower bound  \n\n\\[\n\\boxed{\\;\n     \\mathbb P\\!\\Bigl[C<\\dfrac{N}{12}\\Bigr]\n     \\;\\le\\;\n     \\exp\\!\\Bigl(-\\gamma\\,N\\Bigr),\\qquad\n     \\gamma:=\\frac{2\\delta^{2}}{49}\n     =\\frac{49729}{13332791808}\\approx3.73\\times10^{-6}\\;}\n     \\tag{4}\n\\]\n\nThat is, with probability at least $1-\\exp(-\\gamma mnk)$ the block contains more than $\\dfrac{mnk}{12}$ monochromatic face–connected components.\n\n(The constant $\\gamma$ is still not optimal; any positive constant obtained from a valid argument earns full credit.)\n\n(e)  Briefly discuss why the technique in (d) cannot be replaced by the elementary “solitary–cube” argument used in (b), and indicate a family of $(m,n,k)$ for which the exponential estimate in (4) is essentially sharp up to the value of $\\gamma$.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "(a)  Fix $\\xi\\in B$ and let $d(\\xi):=|\\mathcal N(\\xi)|\\in\\{3,4,5,6\\}$.  \nCondition on the event that $\\xi$ receives colour $i\\in\\{1,2,3\\}$.  \nIndependence implies  \n\\[\n\\mathbb P\\!\\bigl[J_{\\xi}=1\\,\\bigl|\\,\\text{$\\xi$ has colour }i\\bigr]\n =(1-p_{i})^{d(\\xi)}\\ge(1-p_{i})^{6}.\n\\]\nRemoving the conditioning and summing over the three colours gives  \n\\[\n\\mathbb P\\!\\bigl[J_{\\xi}=1\\bigr]\n =\\sum_{i=1}^{3}p_{i}(1-p_{i})^{d(\\xi)}\n \\ge\\sum_{i=1}^{3}p_{i}(1-p_{i})^{6}\n =\\tau,\n\\]\nestablishing (1).  A direct calculation shows  \n\\[\n\\tau\n =\\tfrac12\\!\\left(\\tfrac12\\right)^{6}\n +\\tfrac13\\!\\left(\\tfrac23\\right)^{6}\n +\\tfrac16\\!\\left(\\tfrac56\\right)^{6}\n =\\frac{2167}{23328}\\approx0.09290 .\n\\]\n\n(b)  A solitary cube forms a monochromatic component of size $1$, hence $C\\ge S$.  \nTaking expectations and using (1) we obtain\n\\[\n\\mathbb E[C]\\ge\\mathbb E[S]\n          =\\sum_{\\xi\\in B}\\mathbb P\\!\\bigl[J_{\\xi}=1\\bigr]\n          \\ge\\tau\\,N,\n\\]\nwhich is (2).\n\n(c-i)  Recolouring a single voxel $\\zeta$ can influence $J_{\\zeta}$ itself and at most the six indicators $J_{\\xi}$ with $\\xi\\in\\mathcal N(\\zeta)$.  \nTherefore $|S(\\omega)-S(\\omega')|\\le 7$ whenever $\\omega,\\omega'$ differ only at $\\zeta$.\n\n(c-ii)  Reveal the colours of the $N$ voxels in an arbitrary but fixed order and form the Doob martingale  \n\\[\nM_{j}:=\\mathbb E\\!\\bigl[S\\mid\\text{first $j$ colours revealed}\\bigr],\n\\qquad j=0,\\dots ,N.\n\\]\nChanging the $j$th colour alters $M_{j}$ by at most $7$, so  \n$|M_{j}-M_{j-1}|\\le7$ for every $j$.  \nAzuma–Hoeffding (or McDiarmid’s bounded–difference) inequality yields\n\\[\n\\mathbb P\\!\\bigl[S-\\mathbb E[S]\\le -t\\bigr]\n \\le\\exp\\!\\Bigl(\n      -\\frac{2t^{2}}{\\sum_{j=1}^{N}7^{2}}\n    \\Bigr)\n =\\exp\\!\\Bigl(\n      -\\frac{2t^{2}}{49N}\n    \\Bigr),\n\\]\nwhich is (3).\n\n(d)  From (2) we have $\\mathbb E[S]\\ge\\tau N$.  Put  \n\\[\nt_{0}:=\\delta N,\n\\qquad\\text{where}\\quad \n\\delta=\\tau-\\frac1{12}=\\frac{223}{23328}.\n\\]\n\n\\textbf{Threshold comparison.}  \nBecause $\\delta>0$,  \n\\[\n\\tau N-t_{0}=(\\tau-\\delta)N=\\frac{N}{12}.\n\\]\nHence the implication\n\\[\n\\Bigl\\{\\,S\\le\\frac{N}{12}\\Bigr\\}\n\\;\\subseteq\\;\n\\bigl\\{\\,S\\le\\mathbb E[S]-t_{0}\\bigr\\}\n\\]\nholds.\n\n\\textbf{Applying (3).}  Taking $t=t_{0}$ in (3) gives  \n\\[\n\\mathbb P\\!\\bigl[S\\le\\mathbb E[S]-t_{0}\\bigr]\n\\le\n\\exp\\!\\Bigl(\n  -\\frac{2t_{0}^{2}}{49N}\n\\Bigr)\n=\\exp\\!\\Bigl(\n  -\\frac{2\\delta^{2}}{49}N\n\\Bigr)\n=\\exp\\!\\bigl(-\\gamma N\\bigr),\n\\]\nwhere  \n\\[\n\\gamma=\\frac{2\\delta^{2}}{49}\n       =\\frac{2\\cdot 49729}{49\\cdot 23328^{2}}\n       =\\frac{49729}{13332791808}\n       \\approx3.73\\times10^{-6}.\n\\]\nSince $C\\ge S$, the same bound holds for $C$, completing the proof of (4).\n\n(e)  The estimate in (b) involves only the \\emph{expectation}.  Although the indicators $J_{\\xi}$ are individually simple, they are highly dependent, so straightforward Chernoff or Bernstein bounds cannot be applied directly to $S$.  The martingale approach of McDiarmid exploits the \\emph{global $7$–Lipschitz property} exhibited in (c) and gives exponential concentration around the mean.\n\nFor rectangular solids with $m=n=k\\to\\infty$ one has $\\operatorname{Var}(S)=\\Theta(N)$ (a routine though tedious calculation).  Consequently no tail bound of the form $\\exp(-\\Theta(N^{1+\\varepsilon}))$ can be valid; up to the constant $\\gamma$, (4) is therefore essentially sharp for this family.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.787028",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-dimensional setting – the 3-dimensional grid forces the solver to control three independent edge families and three families of 2×2 squares instead of one, and to juggle non-trivial relations between them.\n\n2.  Multiple colours with unequal probabilities introduce higher-order moment sums p₂, p₄, demanding careful probability bookkeeping; the solver must track Σpi² and Σpi⁴ rather than a single ½ or ⅔.\n\n3.  The argument now uses a miniature version of 3-dimensional cell-complex topology (deleting edges from 4-cycles scattered through three orthogonal planes) instead of planar graph ideas only.\n\n4.  A non-obvious optimisation problem appears: the bound must hold for every triple (m,n,k) and therefore requires analysing a three-variable function, locating its minimum, and proving that it occurs at an interior limit point (a=b=c=0).\n\n5.  The resulting algebra is longer and more intricate: expressions (5)–(6) contain nine monomials in a,b,c, each with coefficients depending on p₂ and p₄; verifying positivity of all coefficients is essential and cannot be skipped.\n\n6.  Constants are no longer “nice” fractions such as 1/8 or 1/11; reaching the sharp factor 13/216 forces exact arithmetic with rational numbers and careful bounding of edge and face counts.\n\nThese layers of extra structure and computation place the enhanced variant well beyond the scope of the original problem and its simpler kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $m,n,k\\in\\mathbb Z_{\\ge 1}$ and put  \n\\[\nB:=\\{1,\\dots ,m\\}\\times\\{1,\\dots ,n\\}\\times\\{1,\\dots ,k\\},\n\\qquad N:=mnk .\n\\]\n\nEvery unit cube (voxel) $\\xi\\in B$ is coloured independently with one of the three colours  \n\n\\[\n\\text{crimson},\\qquad \\text{jade},\\qquad \\text{obsidian}\n\\]\n\nwith respective probabilities  \n\n\\[\np_{\\text{cr}}=\\tfrac12,\\qquad \np_{\\text{ja}}=\\tfrac13,\\qquad \np_{\\text{ob}}=\\tfrac16,\\qquad \np_{\\text{cr}}+p_{\\text{ja}}+p_{\\text{ob}}=1 .\n\\]\n\nTwo cubes are \\emph{face–adjacent} if they share an entire $(1\\times1)$ face.  \nA (monochromatic) \\emph{face–connected component} is a maximal set of cubes of one fixed colour that can be linked by a chain of face–adjacent cubes of that colour.  \nDenote by $C$ the (random) number of monochromatic face–connected components in the whole block.\n\nFor $\\xi=(x,y,z)\\in B$ define the set of (existing) neighbours  \n\n\\[\n\\mathcal N(\\xi)=\\bigl\\{\n(x\\!-\\!1,y,z),\\,(x\\!+\\!1,y,z),\\,\n(x,y\\!-\\!1,z),\\,(x,y\\!+\\!1,z),\\,\n(x,y,z\\!-\\!1),\\,(x,y,z\\!+\\!1)\n\\bigr\\}\\cap B .\n\\]\n\nThe cube $\\xi$ is called \\emph{solitary} if none of the cubes in $\\mathcal N(\\xi)$ has the same colour as $\\xi$.  \nPut  \n\n\\[\nJ_{\\xi}:=\n\\begin{cases}\n1,&\\text{if }\\xi\\text{ is solitary},\\\\[4pt]\n0,&\\text{otherwise},\n\\end{cases}\n\\qquad \nS:=\\displaystyle\\sum_{\\xi\\in B}J_{\\xi}\\quad\n(\\text{number of solitary cubes}).\n\\]\n\nThroughout write  \n\n\\[\np_{1}=p_{\\text{cr}},\\quad\np_{2}=p_{\\text{ja}},\\quad\np_{3}=p_{\\text{ob}},\\qquad\n\\tau:=\\sum_{i=1}^{3}p_{i}(1-p_{i})^{6}\n      =\\frac{2167}{23328}\\approx0.09290.\n\\]\n\nAnswer the following.\n\n(a)  Prove the uniform lower bound  \n\\[\n\\boxed{\\;\n\\mathbb P\\!\\bigl[J_{\\xi}=1\\bigr]\\;\\ge\\;\\tau\\quad\n\\text{for all }\\xi\\in B\\;}\n\\tag{1}\n\\]\nand compute the exact value of $\\tau$ given above.\n\n(b)  Show that $C\\ge S$ and hence  \n\n\\[\n\\boxed{\\;\n\\mathbb E[C]\\;\\ge\\;\\tau\\,N\\;}\n\\tag{2}\n\\]\n\n(c)  A single recolouring of one voxel may affect $J_{\\xi}$ for the voxel itself and for each of its (up to) six face–neighbours.\n\n\\quad(i)  Prove that for every pair of colourings $\\omega,\\omega'$ differing in the colour of exactly one voxel,\n\\[\n       |S(\\omega)-S(\\omega')|\\le 7 .\n\\]\n\n\\quad(ii)  Using McDiarmid’s bounded–difference inequality,  \nshow that for every $t>0$\n\\[\n       \\boxed{\\;\n       \\mathbb P\\bigl[S\\le\\mathbb E[S]-t\\bigr]\n       \\;\\le\\;\n       \\exp\\!\\Bigl(\n          -\\dfrac{2t^{2}}{49N}\n       \\Bigr)\\;}\n       \\tag{3}\n\\]\n\n(d)  Put $\\displaystyle\\delta:=\\tau-\\dfrac1{12}=\\dfrac{223}{23328}\\approx0.00956$ and define  \n\n\\[\nt_{0}:=\\delta\\,N\\quad(>0).\n\\]\n\nCombine (2) and (3) to show the high–probability lower bound  \n\n\\[\n\\boxed{\\;\n     \\mathbb P\\!\\Bigl[C<\\dfrac{N}{12}\\Bigr]\n     \\;\\le\\;\n     \\exp\\!\\Bigl(-\\gamma\\,N\\Bigr),\\qquad\n     \\gamma:=\\frac{2\\delta^{2}}{49}\n     =\\frac{49729}{13332791808}\\approx3.73\\times10^{-6}\\;}\n     \\tag{4}\n\\]\n\nThat is, with probability at least $1-\\exp(-\\gamma mnk)$ the block contains more than $\\dfrac{mnk}{12}$ monochromatic face–connected components.\n\n(The constant $\\gamma$ is still not optimal; any positive constant obtained from a valid argument earns full credit.)\n\n(e)  Briefly discuss why the technique in (d) cannot be replaced by the elementary “solitary–cube” argument used in (b), and indicate a family of $(m,n,k)$ for which the exponential estimate in (4) is essentially sharp up to the value of $\\gamma$.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "(a)  Fix $\\xi\\in B$ and let $d(\\xi):=|\\mathcal N(\\xi)|\\in\\{3,4,5,6\\}$.  \nCondition on the event that $\\xi$ receives colour $i\\in\\{1,2,3\\}$.  \nIndependence implies  \n\\[\n\\mathbb P\\!\\bigl[J_{\\xi}=1\\,\\bigl|\\,\\text{$\\xi$ has colour }i\\bigr]\n =(1-p_{i})^{d(\\xi)}\\ge(1-p_{i})^{6}.\n\\]\nRemoving the conditioning and summing over the three colours gives  \n\\[\n\\mathbb P\\!\\bigl[J_{\\xi}=1\\bigr]\n =\\sum_{i=1}^{3}p_{i}(1-p_{i})^{d(\\xi)}\n \\ge\\sum_{i=1}^{3}p_{i}(1-p_{i})^{6}\n =\\tau,\n\\]\nestablishing (1).  A direct calculation shows  \n\\[\n\\tau\n =\\tfrac12\\!\\left(\\tfrac12\\right)^{6}\n +\\tfrac13\\!\\left(\\tfrac23\\right)^{6}\n +\\tfrac16\\!\\left(\\tfrac56\\right)^{6}\n =\\frac{2167}{23328}\\approx0.09290 .\n\\]\n\n(b)  A solitary cube forms a monochromatic component of size $1$, hence $C\\ge S$.  \nTaking expectations and using (1) we obtain\n\\[\n\\mathbb E[C]\\ge\\mathbb E[S]\n          =\\sum_{\\xi\\in B}\\mathbb P\\!\\bigl[J_{\\xi}=1\\bigr]\n          \\ge\\tau\\,N,\n\\]\nwhich is (2).\n\n(c-i)  Recolouring a single voxel $\\zeta$ can influence $J_{\\zeta}$ itself and at most the six indicators $J_{\\xi}$ with $\\xi\\in\\mathcal N(\\zeta)$.  \nTherefore $|S(\\omega)-S(\\omega')|\\le 7$ whenever $\\omega,\\omega'$ differ only at $\\zeta$.\n\n(c-ii)  Reveal the colours of the $N$ voxels in an arbitrary but fixed order and form the Doob martingale  \n\\[\nM_{j}:=\\mathbb E\\!\\bigl[S\\mid\\text{first $j$ colours revealed}\\bigr],\n\\qquad j=0,\\dots ,N.\n\\]\nChanging the $j$th colour alters $M_{j}$ by at most $7$, so  \n$|M_{j}-M_{j-1}|\\le7$ for every $j$.  \nAzuma–Hoeffding (or McDiarmid’s bounded–difference) inequality yields\n\\[\n\\mathbb P\\!\\bigl[S-\\mathbb E[S]\\le -t\\bigr]\n \\le\\exp\\!\\Bigl(\n      -\\frac{2t^{2}}{\\sum_{j=1}^{N}7^{2}}\n    \\Bigr)\n =\\exp\\!\\Bigl(\n      -\\frac{2t^{2}}{49N}\n    \\Bigr),\n\\]\nwhich is (3).\n\n(d)  From (2) we have $\\mathbb E[S]\\ge\\tau N$.  Put  \n\\[\nt_{0}:=\\delta N,\n\\qquad\\text{where}\\quad \n\\delta=\\tau-\\frac1{12}=\\frac{223}{23328}.\n\\]\n\n\\textbf{Threshold comparison.}  \nBecause $\\delta>0$,  \n\\[\n\\tau N-t_{0}=(\\tau-\\delta)N=\\frac{N}{12}.\n\\]\nHence the implication\n\\[\n\\Bigl\\{\\,S\\le\\frac{N}{12}\\Bigr\\}\n\\;\\subseteq\\;\n\\bigl\\{\\,S\\le\\mathbb E[S]-t_{0}\\bigr\\}\n\\]\nholds.\n\n\\textbf{Applying (3).}  Taking $t=t_{0}$ in (3) gives  \n\\[\n\\mathbb P\\!\\bigl[S\\le\\mathbb E[S]-t_{0}\\bigr]\n\\le\n\\exp\\!\\Bigl(\n  -\\frac{2t_{0}^{2}}{49N}\n\\Bigr)\n=\\exp\\!\\Bigl(\n  -\\frac{2\\delta^{2}}{49}N\n\\Bigr)\n=\\exp\\!\\bigl(-\\gamma N\\bigr),\n\\]\nwhere  \n\\[\n\\gamma=\\frac{2\\delta^{2}}{49}\n       =\\frac{2\\cdot 49729}{49\\cdot 23328^{2}}\n       =\\frac{49729}{13332791808}\n       \\approx3.73\\times10^{-6}.\n\\]\nSince $C\\ge S$, the same bound holds for $C$, completing the proof of (4).\n\n(e)  The estimate in (b) involves only the \\emph{expectation}.  Although the indicators $J_{\\xi}$ are individually simple, they are highly dependent, so straightforward Chernoff or Bernstein bounds cannot be applied directly to $S$.  The martingale approach of McDiarmid exploits the \\emph{global $7$–Lipschitz property} exhibited in (c) and gives exponential concentration around the mean.\n\nFor rectangular solids with $m=n=k\\to\\infty$ one has $\\operatorname{Var}(S)=\\Theta(N)$ (a routine though tedious calculation).  Consequently no tail bound of the form $\\exp(-\\Theta(N^{1+\\varepsilon}))$ can be valid; up to the constant $\\gamma$, (4) is therefore essentially sharp for this family.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.601809",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-dimensional setting – the 3-dimensional grid forces the solver to control three independent edge families and three families of 2×2 squares instead of one, and to juggle non-trivial relations between them.\n\n2.  Multiple colours with unequal probabilities introduce higher-order moment sums p₂, p₄, demanding careful probability bookkeeping; the solver must track Σpi² and Σpi⁴ rather than a single ½ or ⅔.\n\n3.  The argument now uses a miniature version of 3-dimensional cell-complex topology (deleting edges from 4-cycles scattered through three orthogonal planes) instead of planar graph ideas only.\n\n4.  A non-obvious optimisation problem appears: the bound must hold for every triple (m,n,k) and therefore requires analysing a three-variable function, locating its minimum, and proving that it occurs at an interior limit point (a=b=c=0).\n\n5.  The resulting algebra is longer and more intricate: expressions (5)–(6) contain nine monomials in a,b,c, each with coefficients depending on p₂ and p₄; verifying positivity of all coefficients is essential and cannot be skipped.\n\n6.  Constants are no longer “nice” fractions such as 1/8 or 1/11; reaching the sharp factor 13/216 forces exact arithmetic with rational numbers and careful bounding of edge and face counts.\n\nThese layers of extra structure and computation place the enhanced variant well beyond the scope of the original problem and its simpler kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}