{
  "index": "1962-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO"
  ],
  "difficulty": "",
  "question": "4. The Euclidean plane is divided into regions by drawing a finite number of circles. Show that it is possible to color each of these regions either red or blue in such a way that no two adjacent regions have the same color. (Two such regions are said to be adjacent if and only if their boundaries have an arc of a circle in common.)",
  "solution": "Solution. For each region into which the plane is divided, count the number of circles within which that region lies. If this number is odd color the region red, while if the number is even color it blue.\n\nThen, if two regions share a common boundary arc, one of these regions is interior to one more of the given circles than is the other region. Hence these two regions have different colors.\n\nRemark. A number of coloring problems (from the simple up to the four-color problem) are discussed in the pamphlet by E. B. Dynkin and V. A. Uspenskii, Multicolor Problems, trans. from the Russian, Heath, Boston, 1963.",
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        "R": "radiant",
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        "h": "holistic",
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        "f": "fractal",
        "w": "wistful",
        "E": "electric",
        "m": "mineral",
        "u": "utopian",
        "I": "integral",
        "c": "crystal",
        "P": "polarize",
        "k": "kinetic",
        "A": "angular",
        "y": "youthful",
        "g": "genomic",
        "n": "nebular",
        "T": "thermal",
        "d": "dynamic",
        "j": "judicial",
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      "question": "4. The Euclidean plane is divided into regions by drawing a finite number of circles. Show that it is possible to color each of these regions either red or blue in such a way that no two adjacent regions have the same color. (Two such regions are said to be adjacent if and only if their boundaries have an arc of a circle in common.)",
      "solution": "Solution. For each region into which the plane is divided, count the number of circles within which that region lies. If this number is odd color the region red, while if the number is even color it blue.\n\nThen, if two regions share a common boundary arc, one of these regions is interior to one more of the given circles than is the other region. Hence these two regions have different colors.\n\nRemark. A number of coloring problems (from the simple up to the four-color problem) are discussed in the pamphlet by E. B. Dynkin and V. A. Uspenskii, Multicolor Problems, trans. from the Russian, Heath, Boston, 1963."
    },
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        "s": "blackbird",
        "R": "earthquake",
        "U": "goldmines",
        "B": "lighthouse",
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        "f": "flowerpot",
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      "question": "4. The Euclidean plane is divided into regions by drawing a finite number of circles. Show that it is possible to color each of these regions either red or blue in such a way that no two adjacent regions have the same color. (Two such regions are said to be adjacent if and only if their boundaries have an arc of a circle in common.)",
      "solution": "Solution. For each region into which the plane is divided, count the number of circles within which that region lies. If this number is odd color the region red, while if the number is even color it blue.\n\nThen, if two regions share a common boundary arc, one of these regions is interior to one more of the given circles than is the other region. Hence these two regions have different colors.\n\nRemark. A number of coloring problems (from the simple up to the four-color problem) are discussed in the pamphlet by E. B. Dynkin and V. A. Uspenskii, Multicolor Problems, trans. from the Russian, Heath, Boston, 1963."
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        "o": "outercore",
        "v": "verticalplane",
        "f": "falsetruth",
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        "E": "emptinessfull",
        "m": "maximumminimum",
        "u": "universalnone",
        "I": "invisiblesight",
        "c": "closedopen",
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        "k": "kineticrest",
        "A": "apexdepth",
        "y": "youngold",
        "g": "gravityvoid",
        "n": "negativepositive",
        "T": "transparentopaque",
        "d": "drynesswet",
        "j": "justicebias",
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      "question": "4. The Euclidean plane is divided into regions by drawing a finite number of circles. Show that it is possible to color each of these regions either red or blue in such a way that no two adjacent regions have the same color. (Two such regions are said to be adjacent if and only if their boundaries have an arc of a circle in common.)",
      "solution": "Solution. For each region into which the plane is divided, count the number of circles within which that region lies. If this number is odd color the region red, while if the number is even color it blue.\n\nThen, if two regions share a common boundary arc, one of these regions is interior to one more of the given circles than is the other region. Hence these two regions have different colors.\n\nRemark. A number of coloring problems (from the simple up to the four-color problem) are discussed in the pamphlet by E. B. Dynkin and V. A. Uspenskii, Multicolor Problems, trans. from the Russian, Heath, Boston, 1963."
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      "question": "4. The Euclidean plane is divided into regions by drawing a finite number of circles. Show that it is possible to color each of these regions either red or blue in such a way that no two adjacent regions have the same color. (Two such regions are said to be adjacent if and only if their boundaries have an arc of a circle in common.)",
      "solution": "Solution. For each region into which the plane is divided, count the number of circles within which that region lies. If this number is odd color the region red, while if the number is even color it blue.\n\nThen, if two regions share a common boundary arc, one of these regions is interior to one more of the given circles than is the other region. Hence these two regions have different colors.\n\nRemark. A number of coloring problems (from the simple up to the four-color problem) are discussed in the pamphlet by E. B. Dynkin and V. A. Uspenskii, Multicolor Problems, trans. from the Russian, Heath, Boston, 1963."
    },
    "kernel_variant": {
      "question": "Let $M^{n}$ be a closed, connected, orientable, smooth $n$–manifold with $n\\ge 2$.  \nA countable, locally–finite family  \n\\[\n\\mathcal H=\\{\\Sigma_{1},\\Sigma_{2},\\dots\\}\n\\]\nof pairwise distinct, connected, two–sided, smoothly embedded closed hypersurfaces is prescribed and satisfies  \n\n(1) (Pairwise transversality, no triple points)  \n    If $i\\neq j$ the intersection $\\Sigma_{i}\\cap\\Sigma_{j}$ (possibly empty) is a smooth closed $(n-2)$–submanifold, and no point of $M$ lies on three different hypersurfaces.  \n\n(2) (Separating property)  \n    For every $i$ the complement $M\\setminus\\Sigma_{i}$ has **exactly** two connected components.  \n    One of them is declared the positive side of $\\Sigma_{i}$, the other its negative side.  \n\nBecause of (1)–(2) the union $\\Sigma:=\\bigcup_{i}\\Sigma_{i}$ decomposes $M$ into connected components, called chambers.  \nTwo distinct chambers are said to be adjacent when their common boundary contains an open $(n-1)$–dimensional subset lying on **exactly one** hypersurface $\\Sigma_{k}$.\n\nFor every $i$ let $\\operatorname{PD}_{\\mathbb Z_{2}}\\bigl([\\Sigma_{i}]\\bigr)\\in H^{1}(M;\\mathbb Z_{2})$ be the Poincaré dual of the $\\mathbb Z_{2}$–homology class $[\\Sigma_{i}]$.  \nBecause of local finiteness the sum  \n\\[\n\\varphi\\;:=\\;\\sum_{i=1}^{\\infty} \\operatorname{PD}_{\\mathbb Z_{2}}\\bigl([\\Sigma_{i}]\\bigr)\\;\\in\\;H^{1}(M;\\mathbb Z_{2})\n\\tag{$\\ast$}\n\\]\nis well defined.\n\n(a)  Prove that the chambers admit a $2$–colouring (black/white) in which every two adjacent chambers receive opposite colours **iff** the cohomology class $\\varphi$ defined in $(\\ast)$ is zero.\n\n(b)  Show that for $M=S^{n}$ such a colouring is always possible, no matter what countable, locally–finite family $\\mathcal H$ satisfying (1)–(2) is given.",
      "solution": "Throughout all (co)homology groups carry $\\mathbb Z_{2}$–coefficients, so `$=$' means `$=$ mod $2$'.  Write  \n\\[\n\\Sigma=\\bigcup_{i}\\Sigma_{i},\\qquad M^{\\circ}=M\\setminus\\Sigma .\n\\]\nLocal finiteness implies: for every compact set $K\\subset M$ there are only finitely many indices $i$ with $\\Sigma_{i}\\cap K\\neq\\varnothing$.  Consequently the infinite sum $(\\ast)$ is a bona-fide $1$–cochain, hence a class in $H^{1}(M;\\mathbb Z_{2})$.\n\nStep 0.  The chamber graph.  \nChoose one interior point for every chamber; these points are the vertices of a graph $\\Gamma$.  \nFor every open $(n-1)$–piece $U\\subset\\Sigma_{k}$ that separates two chambers $C,C'$ pick a small segment transverse to $\\Sigma_{k}$, with interior disjoint from $\\Sigma$, joining the chosen points of $C$ and $C'$.  The segment represents an (unoriented) edge of $\\Gamma$.  \nBecause $M$ is connected, any two chamber points can be joined by a smooth path that meets $\\Sigma$ transversely in finitely many points; reading the chambers in the order met along the path shows that $\\Gamma$ is connected.  (The separating hypothesis is essential here.)\n\nStep 1.  Necessity of $\\varphi=0$.  \nAssume a $2$–colouring of the chambers is given and encode it as a $0$–cochain  \n\\[\nf\\in C^{0}(\\Gamma;\\mathbb Z_{2}),\\qquad f(C)=\n\\begin{cases}\n0 &\\text{if $C$ is white},\\\\\n1 &\\text{if $C$ is black}.\n\\end{cases}\n\\]\nBecause adjacent chambers have opposite colours, the coboundary $\\delta f$ evaluates to $1$ on every oriented edge of $\\Gamma$.  \n\nFix an index $k$.  Define a $1$–cochain $e_{k}$ on $\\Gamma$ by  \n\\[\ne_{k}(E)=\n\\begin{cases}\n1 &\\text{if the edge $E$ crosses }\\Sigma_{k},\\\\[2mm]\n0 &\\text{otherwise}.\n\\end{cases}\n\\]\nSince every $\\Sigma_{k}$ is separating, every connected component of  \n$\\Sigma_{k}\\setminus\\bigl(\\bigcup_{j\\neq k}\\Sigma_{j}\\bigr)$ appears as the common boundary\nof **exactly** two adjacent chambers, hence is met by some edge of $\\Gamma$; therefore $e_{k}\\not\\equiv 0$.  By construction  \n\\[\n\\delta f=\\sum_{k}e_{k}.\n\\tag{1}\n\\]\n\nClaim.  Each $e_{k}$ represents $\\operatorname{PD}_{\\mathbb Z_{2}}\\bigl([\\Sigma_{k}]\\bigr)\\in H^{1}(M;\\mathbb Z_{2})$.\n\nIndeed, let $\\gamma\\colon S^{1}\\to M$ be a smooth closed curve transverse to $\\Sigma_{k}$ and disjoint from the other hypersurfaces.  Homotope $\\gamma$ so that it meets $\\Gamma$ only in the mid-points of edges and meets each such edge once, transversely.  \nBecause $\\Sigma_{k}$ and the other hypersurfaces never coincide and triple points are excluded, $\\gamma$ crosses $\\Sigma_{k}$ precisely once for every edge of $\\Gamma$ on which $e_{k}$ takes the value $1$.  Hence the mod-$2$ intersection number of $\\gamma$ with $\\Sigma_{k}$ equals $e_{k}(\\gamma)$, showing that $e_{k}$ is a cocycle representing the desired Poincaré dual class.\n\nTaking cohomology classes of (1) gives  \n\\[\n0=[\\delta f]=\\sum_{k}\\operatorname{PD}_{\\mathbb Z_{2}}\\bigl([\\Sigma_{k}]\\bigr)=\\varphi ,\n\\]\nso $\\varphi=0$ is necessary.\n\nStep 2.  Sufficiency of $\\varphi=0$.  \nSuppose $\\varphi=0$.  Choose orientations on the edges of $\\Gamma$ once and for all.  Because $\\Gamma$ is a connected $1$–dimensional CW–complex,  \n\\[\nH^{1}(\\Gamma;\\mathbb Z_{2})\\;\\cong\\; \\operatorname{Hom}(H_{1}(\\Gamma),\\mathbb Z_{2}).\n\\]\nUnder the inclusion $\\iota\\colon\\Gamma\\hookrightarrow M$ every edge crosses a single hypersurface transversely and no other, hence  \n\\[\n\\iota^{*}\\Bigl(\\operatorname{PD}_{\\mathbb Z_{2}}\\bigl([\\Sigma_{k}]\\bigr)\\Bigr)=e_{k}.\n\\]\nConsequently \n\\[\n\\iota^{*}(\\varphi)=\\sum_{k}e_{k}\\in H^{1}(\\Gamma;\\mathbb Z_{2}).\n\\]\nBecause $\\varphi=0$ in $H^{1}(M;\\mathbb Z_{2})$, its pull-back to $\\Gamma$ is also zero, so the cocycle $\\sum_{k}e_{k}$ is a coboundary.  Hence there exists a $0$–cochain $g\\in C^{0}(\\Gamma;\\mathbb Z_{2})$ with  \n\\[\n\\delta g=\\sum_{k}e_{k}.\n\\tag{2}\n\\]\nFix a base chamber $C_{0}$ and add the normalising condition $g(C_{0})=0$.  \nDefine the colour of a chamber $C$ by  \n\\[\n\\chi(C):=g(C)\\in\\{0,1\\}.\n\\]\nBecause $\\Gamma$ is connected, $g$ is uniquely determined by (2) and the normalisation, so $\\chi$ is well defined.  \n\nLet $C,C'$ be adjacent chambers joined by an oriented edge $E$ crossing $\\Sigma_{k}$.  \nEvaluating (2) on $E$ yields  \n\\[\n\\chi(C')-\\chi(C)=\\delta g(E)=\\sum_{j}e_{j}(E)=e_{k}(E)=1 ,\n\\]\nso $C$ and $C'$ receive opposite colours.  Thus $\\chi$ is the sought $2$–colouring, completing the proof of (a).\n\nStep 3.  The sphere.  \nWhen $M=S^{n}$ we have $H^{1}(S^{n};\\mathbb Z_{2})=0$, hence $\\varphi=0$ for every locally-finite family $\\mathcal H$.  Therefore a $2$–colouring always exists by part (a), proving (b).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.540033",
        "was_fixed": false,
        "difficulty_analysis": "• Topological setting upgraded from ℝ² or S² to an arbitrary closed orientable n-manifold with n ≥ 2; the argument must now work in all dimensions.  \n• Instead of Jordan curves or spherical polygons we deal with smooth embedded hypersurfaces, so differential-topological notions (transversality, two-sidedness, local finiteness) appear.  \n• The key obstruction and criterion for 2-colourability is formulated in cohomological language: the sum of Poincaré duals in H¹(M;ℤ₂).  Hence algebraic topology (homology/cohomology with ℤ₂, Poincaré duality, Alexander duality implicitly for separation) is indispensable.  \n• Part (a) demands a two-way implication – one must both detect an obstruction and construct a colouring when the obstruction vanishes.  The construction involves path-independence arguments based on φ=0, not merely a naïve “counting circles’’ trick.  \n• Part (b) recovers the classical planar/spherical result as a corollary, but only after navigating the higher-dimensional framework.  \n• The necessity proof converts a hypothetical colouring into a non-trivial 1-cocycle; the sufficiency proof builds a colouring from cohomological nullity – both directions force competitors to manipulate cochains, integrals along loops, and intersection numbers mod 2.  \n\nThese layers of differential topology and algebraic topology, absent from the original and current kernels, make the enhanced variant significantly more technical and conceptually demanding."
      }
    },
    "original_kernel_variant": {
      "question": "Let Mⁿ be a closed, connected, orientable, smooth n–manifold (n ≥ 2).\nA countable, locally-finite family  \n\n  𝓗 = {Σ₁, Σ₂, … }  \n\nof pairwise distinct, connected, two-sided, smoothly embedded closed hypersurfaces is given and satisfies  \n\n(1) Pairwise transversality. Whenever i ≠ j the intersection Σᵢ∩Σⱼ (possibly empty) is a smooth closed (n−2)–submanifold, and no point of M lies on three different hypersurfaces.  \n\n(2) Sidedness choice. For every Σᵢ one of the two connected components of M\\Σᵢ is declared its positive side, the other its negative side.\n\nThe union ⋃ Σᵢ decomposes M into connected components, called chambers.\nTwo different chambers are adjacent if their intersection with some Σᵢ contains an (n−1)-dimensional open set that is contained in no other Σⱼ.\n\n(a) Prove that the chambers admit a 2-colouring (black/white) in which every two adjacent chambers receive opposite colours if and only if  \n\n  φ := Σᵢ PD_{ℤ₂}([Σᵢ]) ∈ H¹(M; ℤ₂)  \n\nis the zero class, where PD denotes the Poincaré dual with ℤ₂-coefficients.\n\n(b) Deduce that for M = Sⁿ such a colouring is always possible, independently of the locally-finite family 𝓗 verifying (1)–(2).",
      "solution": "Throughout, all (co)homology groups carry ℤ₂–coefficients; equality means equality mod 2.\n\nPreliminaries  \n• Because every Σᵢ is two-sided we may speak of crossing it from its negative to its positive side, but the corrected argument will only use the total number of crossings, disregarding this orientation.  \n• A generic path is a smooth path transverse to every Σᵢ and disjoint from all pairwise intersections Σᵢ∩Σⱼ.\n\nStep 1. Colouring ⇒ φ = 0 (necessity)  \nAssume a black/white colouring of the chambers exists.  \nChoose a base chamber C₀ and declare its colour to be 0 (white); colour 1 denotes black.\n\nExtend the colouring to a locally constant map\n  f : M\\⋃ Σᵢ → ℤ₂, f≡0 on C₀,\nand define the 1-cochain df by the usual difference along paths.  \nLet γ be any generic loop based at x₀∈C₀; each time γ crosses some Σᵢ the value of f flips, so\n\n  ∫_γ df = (number of intersections of γ with ⋃Σᵢ) = Σᵢ ⟨[Σᵢ],[γ]⟩.\n\nThus the cochain df equals Σᵢ PD([Σᵢ]); but df is a coboundary, hence its cohomology class is zero. Therefore φ=0.\n\nStep 2. φ=0 ⇒ existence of a well-defined parity function χ on chambers (sufficiency)  \nAssume φ=0.  \nFix a base chamber C₀ and set χ(C₀)=0.\n\nFor any other chamber C choose a generic path\n  γ: [0,1]→M\nwith γ(0)∈C₀, γ(1)∈C, and define\n\n  χ(C) := (# of intersection points of γ with ⋃ Σᵢ) (mod 2). (∗)\n\nWhy (∗) is well-defined.  \nLet γ₁, γ₂ be two such paths joining C₀ to C.  Consider the closed loop\n  ℓ := γ₁·γ̄₂,\nobtained by traversing γ₁ and returning along γ₂.  The parity difference produced by γ₁ and γ₂ equals the number of intersections of ℓ with ⋃Σᵢ, i.e.\n\n  χ_{γ₁}(C) − χ_{γ₂}(C) = Σᵢ ⟨[Σᵢ],[ℓ]⟩ = ⟨φ,[ℓ]⟩ = 0,\n\nbecause φ=0 by hypothesis.  Hence χ(C) does not depend on the chosen path γ.\n\nStep 3. Adjacent chambers receive opposite colours  \nLet C and C′ be adjacent across an (n−1)-dimensional open set U⊂Σ_k.  \nA sufficiently small path joining interior points of C and C′ crosses Σ_k exactly once and meets no other Σᵢ; consequently χ(C′)=χ(C)+1 (mod 2), so the colours differ.\n\nStep 4. Every chamber is coloured  \nLocal finiteness guarantees that from C₀ one can reach any chamber via a generic path meeting only finitely many Σᵢ, so χ is defined on all chambers; steps 2 and 3 show that χ furnishes the required 2-colouring.  This completes the proof of the equivalence in (a).\n\nStep 5. The special case M = Sⁿ  \nFor the n-sphere H¹(Sⁿ;ℤ₂)=0, hence φ=0 for any family 𝓗 satisfying the hypotheses.  Therefore a 2-colouring always exists, establishing part (b).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.448689",
        "was_fixed": false,
        "difficulty_analysis": "• Topological setting upgraded from ℝ² or S² to an arbitrary closed orientable n-manifold with n ≥ 2; the argument must now work in all dimensions.  \n• Instead of Jordan curves or spherical polygons we deal with smooth embedded hypersurfaces, so differential-topological notions (transversality, two-sidedness, local finiteness) appear.  \n• The key obstruction and criterion for 2-colourability is formulated in cohomological language: the sum of Poincaré duals in H¹(M;ℤ₂).  Hence algebraic topology (homology/cohomology with ℤ₂, Poincaré duality, Alexander duality implicitly for separation) is indispensable.  \n• Part (a) demands a two-way implication – one must both detect an obstruction and construct a colouring when the obstruction vanishes.  The construction involves path-independence arguments based on φ=0, not merely a naïve “counting circles’’ trick.  \n• Part (b) recovers the classical planar/spherical result as a corollary, but only after navigating the higher-dimensional framework.  \n• The necessity proof converts a hypothetical colouring into a non-trivial 1-cocycle; the sufficiency proof builds a colouring from cohomological nullity – both directions force competitors to manipulate cochains, integrals along loops, and intersection numbers mod 2.  \n\nThese layers of differential topology and algebraic topology, absent from the original and current kernels, make the enhanced variant significantly more technical and conceptually demanding."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}