{
  "index": "1996-B-5",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "Given a finite string $S$ of symbols $X$ and $O$, we write $\\Delta(S)$\nfor the number of $X$'s in $S$ minus the number of $O$'s. For example,\n$\\Delta(XOOXOOX) = -1$. We call a string $S$ \\textbf{balanced} if every\nsubstring $T$ of (consecutive symbols of) $S$ has $-2 \\leq \\Delta(T)\n\\leq 2$. Thus, $XOOXOOX$ is not balanced, since it contains the\nsubstring $OOXOO$. Find, with proof, the number of balanced strings of\nlength $n$.",
  "solution": "Consider a $1 \\times n$ checkerboard, in which we write an $n$-letter\nstring, one letter per square. If the string is balanced, we can cover\neach pair of adjacent squares containing the same letter with a $1\n\\times 2$ domino, and these will not overlap (because no three in a\nrow can be the same). Moreover, any domino is separated from the next\nby an even number of squares, since they must cover opposite letters,\nand the sequence must alternate in between.\n\nConversely, any arrangement of dominoes where adjacent dominoes are\nseparated by an even number of squares corresponds to a unique\nbalanced string, once we choose whether the string starts with $X$ or\n$O$. In other words, the number of balanced strings is twice the\nnumber of acceptable domino arrangements.\n\nWe count these arrangements by numbering the squares $0,1,\\dots,n-1$\nand distinguishing whether the dominoes start on even or odd numbers.\nOnce this is decided, one simply chooses whether or not to put a\ndomino in each eligible position. Thus\nwe have $2^{\\lfloor n/2 \\rfloor}$ arrangements in the first case and $2^{\\lfloor\n(n-1)/2 \\rfloor}$ in the second, but note that the case of no dominoes has\nbeen counted twice. Hence the number of balanced strings is\n\\[\n2^{\\lfloor (n+2)/2 \\rfloor} + 2^{\\lfloor (n+1)/2 \\rfloor} - 2.\n\\]",
  "vars": [
    "S",
    "T",
    "n",
    "\\\\Delta"
  ],
  "params": [
    "X",
    "O"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "stringvar",
        "T": "substrg",
        "n": "lengthn",
        "\\Delta": "diffcount",
        "X": "exsymbol",
        "O": "ohsymbol"
      },
      "question": "Given a finite string $stringvar$ of symbols $exsymbol$ and $ohsymbol$, we write $diffcount(stringvar)$\nfor the number of $exsymbol$'s in $stringvar$ minus the number of $ohsymbol$'s. For example,\n$diffcount(exsymbolohsymbolohsymbolexsymbolohsymbolohsymbolexsymbol) = -1$. We call a string $stringvar$ \\textbf{balanced} if every\nsubstring $substrg$ of (consecutive symbols of) $stringvar$ has $-2 \\leq diffcount(substrg)\n\\leq 2$. Thus, $exsymbolohsymbolohsymbolexsymbolohsymbolohsymbolexsymbol$ is not balanced, since it contains the\nsubstring $ohsymbolohsymbolexsymbolohsymbolohsymbol$. Find, with proof, the number of balanced strings of\nlength $lengthn$.",
      "solution": "Consider a $1 \\times lengthn$ checkerboard, in which we write an $lengthn$-letter\nstring, one letter per square. If the string is balanced, we can cover\neach pair of adjacent squares containing the same letter with a $1\n\\times 2$ domino, and these will not overlap (because no three in a\nrow can be the same). Moreover, any domino is separated from the next\nby an even number of squares, since they must cover opposite letters,\nand the sequence must alternate in between.\n\nConversely, any arrangement of dominoes where adjacent dominoes are\nseparated by an even number of squares corresponds to a unique\nbalanced string, once we choose whether the string starts with $exsymbol$ or\n$ohsymbol$. In other words, the number of balanced strings is twice the\nnumber of acceptable domino arrangements.\n\nWe count these arrangements by numbering the squares $0,1,\\dots,lengthn-1$\nand distinguishing whether the dominoes start on even or odd numbers.\nOnce this is decided, one simply chooses whether or not to put a\ndomino in each eligible position. Thus\nwe have $2^{\\lfloor lengthn/2 \\rfloor}$ arrangements in the first case and $2^{\\lfloor\n(lengthn-1)/2 \\rfloor}$ in the second, but note that the case of no dominoes has\nbeen counted twice. Hence the number of balanced strings is\n\\[\n2^{\\lfloor (lengthn+2)/2 \\rfloor} + 2^{\\lfloor (lengthn+1)/2 \\rfloor} - 2.\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "lighthouse",
        "T": "watermelon",
        "n": "hieroglyph",
        "\\\\Delta": "buttercup",
        "X": "porcupine",
        "O": "dandelion"
      },
      "question": "Given a finite string $lighthouse$ of symbols $porcupine$ and $dandelion$, we write $buttercup(lighthouse)$ for the number of $porcupine$'s in $lighthouse$ minus the number of $dandelion$'s. For example, $buttercup(porcupinedandeliondandelionporcupinedandeliondandelionporcupine) = -1$. We call a string $lighthouse$ \\textbf{balanced} if every substring $watermelon$ of (consecutive symbols of) $lighthouse$ has $-2 \\leq buttercup(watermelon) \\leq 2$. Thus, $porcupinedandeliondandelionporcupinedandeliondandelionporcupine$ is not balanced, since it contains the substring $dandeliondandelionporcupinedandeliondandelion$. Find, with proof, the number of balanced strings of length $hieroglyph$.",
      "solution": "Consider a $1 \\times hieroglyph$ checkerboard, in which we write an $hieroglyph$-letter string, one letter per square. If the string is balanced, we can cover each pair of adjacent squares containing the same letter with a $1 \\times 2$ domino, and these will not overlap (because no three in a row can be the same). Moreover, any domino is separated from the next by an even number of squares, since they must cover opposite letters, and the sequence must alternate in between.\n\nConversely, any arrangement of dominoes where adjacent dominoes are separated by an even number of squares corresponds to a unique balanced string, once we choose whether the string starts with $porcupine$ or $dandelion$. In other words, the number of balanced strings is twice the number of acceptable domino arrangements.\n\nWe count these arrangements by numbering the squares $0,1,\\dots,hieroglyph-1$ and distinguishing whether the dominoes start on even or odd numbers. Once this is decided, one simply chooses whether or not to put a domino in each eligible position. Thus\nwe have $2^{\\lfloor hieroglyph/2 \\rfloor}$ arrangements in the first case and $2^{\\lfloor\n(hieroglyph-1)/2 \\rfloor}$ in the second, but note that the case of no dominoes has\nbeen counted twice. Hence the number of balanced strings is\n\\[\n2^{\\lfloor (hieroglyph+2)/2 \\rfloor} + 2^{\\lfloor (hieroglyph+1)/2 \\rfloor} - 2.\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "voidsequence",
        "T": "wholesequence",
        "n": "boundless",
        "\\Delta": "equalityscore",
        "X": "circlesymbol",
        "O": "crosssymbol"
      },
      "question": "Given a finite string $voidsequence$ of symbols $circlesymbol$ and $crosssymbol$, we write $equalityscore(voidsequence)$ for the number of $circlesymbol$'s in $voidsequence$ minus the number of $crosssymbol$'s. For example, $equalityscore(circlesymbolcrosssymbolcrosssymbolcirclesymbolcrosssymbolcrosssymbolcirclesymbol) = -1$. We call a string $voidsequence$ \\textbf{balanced} if every substring $wholesequence$ of (consecutive symbols of) $voidsequence$ has $-2 \\leq equalityscore(wholesequence) \\leq 2$. Thus, $circlesymbolcrosssymbolcrosssymbolcirclesymbolcrosssymbolcrosssymbolcirclesymbol$ is not balanced, since it contains the substring $crosssymbolcrosssymbolcirclesymbolcrosssymbolcrosssymbol$. Find, with proof, the number of balanced strings of length $boundless$.",
      "solution": "Consider a $1 \\times boundless$ checkerboard, in which we write an boundless-letter string, one letter per square. If the string is balanced, we can cover each pair of adjacent squares containing the same letter with a $1 \\times 2$ domino, and these will not overlap (because no three in a row can be the same). Moreover, any domino is separated from the next by an even number of squares, since they must cover opposite letters, and the sequence must alternate in between.\n\nConversely, any arrangement of dominoes where adjacent dominoes are separated by an even number of squares corresponds to a unique balanced string, once we choose whether the string starts with $circlesymbol$ or $crosssymbol$. In other words, the number of balanced strings is twice the number of acceptable domino arrangements.\n\nWe count these arrangements by numbering the squares $0,1,\\dots,boundless-1$ and distinguishing whether the dominoes start on even or odd numbers. Once this is decided, one simply chooses whether or not to put a domino in each eligible position. Thus\nwe have $2^{\\lfloor boundless/2 \\rfloor}$ arrangements in the first case and $2^{\\lfloor\n(boundless-1)/2 \\rfloor}$ in the second, but note that the case of no dominoes has\nbeen counted twice. Hence the number of balanced strings is\n\\[\n2^{\\lfloor (boundless+2)/2 \\rfloor} + 2^{\\lfloor (boundless+1)/2 \\rfloor} - 2.\n\\]"
    },
    "garbled_string": {
      "map": {
        "S": "kmrzclam",
        "T": "vaznikpo",
        "n": "qewrtuio",
        "\\Delta": "ybqrstuv",
        "X": "fxjwpldk",
        "O": "zrmvchqa"
      },
      "question": "Given a finite string $kmrzclam$ of symbols $fxjwpldk$ and $zrmvchqa$, we write $ybqrstuv(kmrzclam)$\nfor the number of $fxjwpldk$'s in $kmrzclam$ minus the number of $zrmvchqa$'s. For example,\n$ybqrstuv(fxjwpldkzrmvchqazrmvchqafxjwpldkzrmvchqazrmvchqafxjwpldk) = -1$. We call a string $kmrzclam$ \\textbf{balanced} if every\nsubstring $vaznikpo$ of (consecutive symbols of) $kmrzclam$ has $-2 \\leq ybqrstuv(vaznikpo)\n\\leq 2$. Thus, $fxjwpldkzrmvchqazrmvchqafxjwpldkzrmvchqazrmvchqafxjwpldk$ is not balanced, since it contains the\nsubstring $zrmvchqazrmvchqafxjwpldkzrmvchqazrmvchqa$. Find, with proof, the number of balanced strings of\nlength $qewrtuio$.",
      "solution": "Consider a $1 \\times qewrtuio$ checkerboard, in which we write an $qewrtuio$-letter\nstring, one letter per square. If the string is balanced, we can cover\neach pair of adjacent squares containing the same letter with a $1\n\\times 2$ domino, and these will not overlap (because no three in a\nrow can be the same). Moreover, any domino is separated from the next\nby an even number of squares, since they must cover opposite letters,\nand the sequence must alternate in between.\n\nConversely, any arrangement of dominoes where adjacent dominoes are\nseparated by an even number of squares corresponds to a unique\nbalanced string, once we choose whether the string starts with $fxjwpldk$ or\n$zrmvchqa$. In other words, the number of balanced strings is twice the\nnumber of acceptable domino arrangements.\n\nWe count these arrangements by numbering the squares $0,1,\\dots,qewrtuio-1$\nand distinguishing whether the dominoes start on even or odd numbers.\nOnce this is decided, one simply chooses whether or not to put a\ndomino in each eligible position. Thus\nwe have $2^{\\lfloor qewrtuio/2 \\rfloor}$ arrangements in the first case and $2^{\\lfloor\n(qewrtuio-1)/2 \\rfloor}$ in the second, but note that the case of no dominoes has\nbeen counted twice. Hence the number of balanced strings is\n\\[\n2^{\\lfloor (qewrtuio+2)/2 \\rfloor} + 2^{\\lfloor (qewrtuio+1)/2 \\rfloor} - 2.\n\\]"
    },
    "kernel_variant": {
      "question": "For every non-negative integer n consider words  \n W = w₁w₂⋯wₙ of length n over the alphabet  \n 𝔄 = { N , S , E , W }.  \n\nAssociate to the four letters the planar unit steps  \n\n N ↦ (0,+1), S ↦ (0,–1), E ↦ (+1,0), W ↦ (–1,0).\n\nWrite  \n\n (xₜ , yₜ) = Σ_{i=1}^{t} wᵢ  (0 ≤ t ≤ n)\n\nfor the position of the walk after the first t steps and call the word W\n\n confined  if –2 ≤ xₜ ≤ 2 and –2 ≤ yₜ ≤ 2 for every t.\n\nLet Tₙ denote the number of confined words of length n that start and finish at the origin (0,0).  \nDetermine Tₙ in closed form.",
      "solution": "1.  A graph–theoretic model.  \n    The 25 lattice points (x,y) with –2 ≤ x,y ≤ 2 form the vertex set V(G) of the Cartesian product graph  \n  G = P₅ □ P₅,  \n    where P₅ is the path –2–––1––0––1––2.  \n    A letter of 𝔄 moves the walker to one of the four adjacent vertices of G, hence a confined word of length n is precisely an n-step walk in G that starts and ends at the centre  \n\n  c = (0,0).\n\n    With A the 25 × 25 adjacency matrix of G we therefore have  \n\n  Tₙ = (Aⁿ)_{c,c}.                                                              (1)\n\n2.  Separating the two dimensions.  \n    Label the vertices of P₅ by 1,…,5 corresponding to –2,…,2 and let A₁ be its 5 × 5 adjacency matrix.  \n    Because G = P₅ □ P₅ one has  \n\n  A = A₁ ⊗ I₅ + I₅ ⊗ A₁,\n\n    the Kronecker sum of A₁ with itself.\n\n    The path P₅ is well known to have eigen-values and an orthonormal eigen-basis  \n\n  λ_r = 2 cos(rπ/6),  u_{k,r} = √(2/6) sin(krπ/6),                   1 ≤ k,r ≤ 5.   (2)\n\n    (Here k indexes the vertex and r the eigen-value.)\n\n    The centre of P₅ is the vertex k = 3, and the diagonal entry of a power of A₁ at this vertex reads  \n\n  (A₁ⁿ)_{3,3} = Σ_{r=1}^{5} u_{3,r}² λ_rⁿ\n              = (2/6) Σ_{r=1}^{5} sin²(3rπ/6) λ_rⁿ.                     (3)\n\n    Since sin(3rπ/6)=0 for even r, only r = 1,3,5 contribute, giving\n\n  (A₁ⁿ)_{3,3} = (1/3)(λ₁ⁿ + λ₃ⁿ + λ₅ⁿ).                                 (4)\n\n    Numerical values: λ₁ = √3, λ₃ = 0, λ₅ = –√3.                        (5)\n\n3.  Spectrum of the Cartesian product.  \n   For a graph Cartesian product the spectrum adds: the 25 eigen-values of A are λ_r+λ_s (1 ≤ r,s ≤ 5) with eigen-vectors u_{·,r} ⊗ u_{·,s}.  Consequently,\n\n  (Aⁿ)_{c,c}\n  = Σ_{r,s=1}^{5} (u_{3,r}²)(u_{3,s}²) (λ_r+λ_s)ⁿ\n  = [(1/3)]² Σ_{r odd} Σ_{s odd} (λ_r+λ_s)ⁿ    (by (4))              (6)\n\n   because only r,s = 1,3,5 survive.  Using (5) we obtain the explicit nine summands\n\n  Tₙ = (1/9){(√3+√3)ⁿ+(√3+0)ⁿ+(√3–√3)ⁿ\n               + (0+√3)ⁿ+(0+0)ⁿ+(0–√3)ⁿ\n               + (–√3+√3)ⁿ+(–√3+0)ⁿ+(–√3–√3)ⁿ}.                       (7)\n\n   All terms containing a factor 0ⁿ vanish for n≥1, and grouping the remaining four equal pairs gives\n\n  Tₙ = (1/9)[(2√3)ⁿ + (–2√3)ⁿ + 2(√3)ⁿ + 2(–√3)ⁿ],  n≥1.             (8)\n\n4.  Parity considerations.  \n   The graph G is bipartite, hence no closed walk of odd length exists; equivalently, the right–hand side of (8) is 0 for odd n because the four non-zero summands cancel out.  Put n=2m (m≥0).  The signs disappear and\n\n  T_{2m} = (1/9)[(2√3)^{2m} + (–2√3)^{2m} + 2(√3)^{2m} + 2(–√3)^{2m}]\n         = (2/9)[(4·3)^{m} + 2·3^{m}]\n         = (2/9)(12^{m} + 2·3^{m}).                                    (9)\n\n5.  Initial value.  \n   The empty word is confined, so T₀ = 1.\n\n6.  Small-n check.  \n   m = 1: T₂ = (2/9)(12+6)=4.  \n   m = 2: T₄ = (2/9)(144+18)=36.  \n   m = 3: T₆ = (2/9)(1728+54)=396.  \n   These values agree with an exhaustive computer enumeration of all 4^{ n } words for n≤6.\n\nFinal closed form:\n\n T₀ = 1,  Tₙ = 0 for odd n,\n\n T_{2m} = (2/9)(12^{m} + 2·3^{m})   for m ≥ 1.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.748487",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension & more variables: the problem moves from one sign\n  difference to simultaneous horizontal and vertical imbalances, and\n  the state space grows from 2 kinds of letters to 4 and from 5 to 25\n  possible running–sum states.  \n• Sophisticated structures: the solution demands knowledge of Cartesian\n  graph products, tensor products of matrices, and spectral graph\n  theory; simple domino or colouring tricks no longer suffice.  \n• Deeper theory: diagonalising the adjacency matrix of P₅, using\n  orthonormal eigenbases, and exploiting the additive property of\n  eigen-values in Cartesian products are all linear–algebraic tools\n  well beyond the counting arguments of the original exercise.  \n• More steps & interactions: one has to (i) recast the word problem as\n  a walk on P₅□P₅; (ii) analyse one-dimensional spectra; (iii) lift the\n  result to two dimensions via Kronecker products; (iv) re-assemble the\n  data to obtain a closed formula, and (v) verify parity properties.  \n\nThese layers of algebraic and combinatorial reasoning make the enhanced\nvariant significantly harder than both the original problem and the\ncurrent kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "For every non-negative integer n consider words  \n W = w₁w₂⋯wₙ of length n over the alphabet  \n 𝔄 = { N , S , E , W }.  \n\nAssociate to the four letters the planar unit steps  \n\n N ↦ (0,+1), S ↦ (0,–1), E ↦ (+1,0), W ↦ (–1,0).\n\nWrite  \n\n (xₜ , yₜ) = Σ_{i=1}^{t} wᵢ  (0 ≤ t ≤ n)\n\nfor the position of the walk after the first t steps and call the word W\n\n confined  if –2 ≤ xₜ ≤ 2 and –2 ≤ yₜ ≤ 2 for every t.\n\nLet Tₙ denote the number of confined words of length n that start and finish at the origin (0,0).  \nDetermine Tₙ in closed form.",
      "solution": "1.  A graph–theoretic model.  \n    The 25 lattice points (x,y) with –2 ≤ x,y ≤ 2 form the vertex set V(G) of the Cartesian product graph  \n  G = P₅ □ P₅,  \n    where P₅ is the path –2–––1––0––1––2.  \n    A letter of 𝔄 moves the walker to one of the four adjacent vertices of G, hence a confined word of length n is precisely an n-step walk in G that starts and ends at the centre  \n\n  c = (0,0).\n\n    With A the 25 × 25 adjacency matrix of G we therefore have  \n\n  Tₙ = (Aⁿ)_{c,c}.                                                              (1)\n\n2.  Separating the two dimensions.  \n    Label the vertices of P₅ by 1,…,5 corresponding to –2,…,2 and let A₁ be its 5 × 5 adjacency matrix.  \n    Because G = P₅ □ P₅ one has  \n\n  A = A₁ ⊗ I₅ + I₅ ⊗ A₁,\n\n    the Kronecker sum of A₁ with itself.\n\n    The path P₅ is well known to have eigen-values and an orthonormal eigen-basis  \n\n  λ_r = 2 cos(rπ/6),  u_{k,r} = √(2/6) sin(krπ/6),                   1 ≤ k,r ≤ 5.   (2)\n\n    (Here k indexes the vertex and r the eigen-value.)\n\n    The centre of P₅ is the vertex k = 3, and the diagonal entry of a power of A₁ at this vertex reads  \n\n  (A₁ⁿ)_{3,3} = Σ_{r=1}^{5} u_{3,r}² λ_rⁿ\n              = (2/6) Σ_{r=1}^{5} sin²(3rπ/6) λ_rⁿ.                     (3)\n\n    Since sin(3rπ/6)=0 for even r, only r = 1,3,5 contribute, giving\n\n  (A₁ⁿ)_{3,3} = (1/3)(λ₁ⁿ + λ₃ⁿ + λ₅ⁿ).                                 (4)\n\n    Numerical values: λ₁ = √3, λ₃ = 0, λ₅ = –√3.                        (5)\n\n3.  Spectrum of the Cartesian product.  \n   For a graph Cartesian product the spectrum adds: the 25 eigen-values of A are λ_r+λ_s (1 ≤ r,s ≤ 5) with eigen-vectors u_{·,r} ⊗ u_{·,s}.  Consequently,\n\n  (Aⁿ)_{c,c}\n  = Σ_{r,s=1}^{5} (u_{3,r}²)(u_{3,s}²) (λ_r+λ_s)ⁿ\n  = [(1/3)]² Σ_{r odd} Σ_{s odd} (λ_r+λ_s)ⁿ    (by (4))              (6)\n\n   because only r,s = 1,3,5 survive.  Using (5) we obtain the explicit nine summands\n\n  Tₙ = (1/9){(√3+√3)ⁿ+(√3+0)ⁿ+(√3–√3)ⁿ\n               + (0+√3)ⁿ+(0+0)ⁿ+(0–√3)ⁿ\n               + (–√3+√3)ⁿ+(–√3+0)ⁿ+(–√3–√3)ⁿ}.                       (7)\n\n   All terms containing a factor 0ⁿ vanish for n≥1, and grouping the remaining four equal pairs gives\n\n  Tₙ = (1/9)[(2√3)ⁿ + (–2√3)ⁿ + 2(√3)ⁿ + 2(–√3)ⁿ],  n≥1.             (8)\n\n4.  Parity considerations.  \n   The graph G is bipartite, hence no closed walk of odd length exists; equivalently, the right–hand side of (8) is 0 for odd n because the four non-zero summands cancel out.  Put n=2m (m≥0).  The signs disappear and\n\n  T_{2m} = (1/9)[(2√3)^{2m} + (–2√3)^{2m} + 2(√3)^{2m} + 2(–√3)^{2m}]\n         = (2/9)[(4·3)^{m} + 2·3^{m}]\n         = (2/9)(12^{m} + 2·3^{m}).                                    (9)\n\n5.  Initial value.  \n   The empty word is confined, so T₀ = 1.\n\n6.  Small-n check.  \n   m = 1: T₂ = (2/9)(12+6)=4.  \n   m = 2: T₄ = (2/9)(144+18)=36.  \n   m = 3: T₆ = (2/9)(1728+54)=396.  \n   These values agree with an exhaustive computer enumeration of all 4^{ n } words for n≤6.\n\nFinal closed form:\n\n T₀ = 1,  Tₙ = 0 for odd n,\n\n T_{2m} = (2/9)(12^{m} + 2·3^{m})   for m ≥ 1.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.577786",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension & more variables: the problem moves from one sign\n  difference to simultaneous horizontal and vertical imbalances, and\n  the state space grows from 2 kinds of letters to 4 and from 5 to 25\n  possible running–sum states.  \n• Sophisticated structures: the solution demands knowledge of Cartesian\n  graph products, tensor products of matrices, and spectral graph\n  theory; simple domino or colouring tricks no longer suffice.  \n• Deeper theory: diagonalising the adjacency matrix of P₅, using\n  orthonormal eigenbases, and exploiting the additive property of\n  eigen-values in Cartesian products are all linear–algebraic tools\n  well beyond the counting arguments of the original exercise.  \n• More steps & interactions: one has to (i) recast the word problem as\n  a walk on P₅□P₅; (ii) analyse one-dimensional spectra; (iii) lift the\n  result to two dimensions via Kronecker products; (iv) re-assemble the\n  data to obtain a closed formula, and (v) verify parity properties.  \n\nThese layers of algebraic and combinatorial reasoning make the enhanced\nvariant significantly harder than both the original problem and the\ncurrent kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}