{
  "index": "1966-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "B-4. Let \\( 0<a_{1}<a_{2}<\\cdots<a_{m n+1} \\) be \\( m n+1 \\) integers. Prove that you can select either \\( m+1 \\) of them no one of which divides any other, or \\( n+1 \\) of them each dividing the following one.",
  "solution": "B-4 Let, for each \\( 1 \\leqq i \\leqq m n+1, n_{i} \\) denote the length of the longest chain, starting with \\( a_{i} \\) and each dividing the following one, that we can select out of \\( a_{i}, a_{i+1}, \\cdots, a_{m n+1} \\). If no \\( n_{i} \\) is greater than \\( n \\) then there are at least \\( m+1 n_{i} \\) 's that are the same. However, the integers \\( a_{i} \\) corresponding to these \\( n_{i} \\) 's cannot divide each other, because \\( a_{i} \\mid a_{j} \\) implies that \\( n_{i} \\geqq n_{j}+1 \\).",
  "vars": [
    "a_1",
    "a_2",
    "a_i",
    "a_i+1",
    "a_mn+1",
    "a_j",
    "n_i",
    "n_j",
    "i",
    "j"
  ],
  "params": [
    "m",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_1": "seqelemone",
        "a_2": "seqelemtwo",
        "a_i": "seqelemi",
        "a_i+1": "seqelemiplus",
        "a_mn+1": "seqelemlast",
        "a_j": "seqelemj",
        "n_i": "chainleni",
        "n_j": "chainlenj",
        "i": "indexi",
        "j": "indexj",
        "m": "groupcount",
        "n": "chainlimit"
      },
      "question": "B-4. Let \\( 0<seqelemone<seqelemtwo<\\cdots<seqelemlast \\) be \\( groupcount chainlimit+1 \\) integers. Prove that you can select either \\( groupcount+1 \\) of them no one of which divides any other, or \\( chainlimit+1 \\) of them each dividing the following one.",
      "solution": "B-4 Let, for each \\( 1 \\leqq indexi \\leqq groupcount chainlimit+1, chainleni \\) denote the length of the longest chain, starting with \\( seqelemi \\) and each dividing the following one, that we can select out of \\( seqelemi, seqelemiplus, \\cdots, seqelemlast \\). If no \\( chainleni \\) is greater than \\( chainlimit \\) then there are at least \\( groupcount+1 chainleni \\)'s that are the same. However, the integers \\( seqelemi \\) corresponding to these \\( chainleni \\)'s cannot divide each other, because \\( seqelemi \\mid seqelemj \\) implies that \\( chainleni \\geqq chainlenj+1 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_1": "sunflower",
        "a_2": "buttercup",
        "a_i": "daffodil",
        "a_i+1": "gardenia",
        "a_mn+1": "hydrangea",
        "a_j": "chrysalis",
        "n_i": "marigold",
        "n_j": "bluebells",
        "i": "snowdrop",
        "j": "mellower",
        "m": "cornstalk",
        "n": "rainstorm"
      },
      "question": "B-4. Let \\( 0<sunflower<buttercup<\\cdots<hydrangea \\) be \\( cornstalk rainstorm+1 \\) integers. Prove that you can select either \\( cornstalk+1 \\) of them no one of which divides any other, or \\( rainstorm+1 \\) of them each dividing the following one.",
      "solution": "B-4 Let, for each \\( 1 \\leqq snowdrop \\leqq cornstalk rainstorm+1, marigold \\) denote the length of the longest chain, starting with \\( daffodil \\) and each dividing the following one, that we can select out of \\( daffodil, gardenia, \\cdots, hydrangea \\). If no \\( marigold \\) is greater than \\( rainstorm \\) then there are at least \\( cornstalk+1 marigold \\) 's that are the same. However, the integers \\( daffodil \\) corresponding to these \\( marigold \\) 's cannot divide each other, because \\( daffodil \\mid chrysalis \\) implies that \\( marigold \\geqq bluebells+1 \\ )."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_1": "fractional",
        "a_2": "irrational",
        "a_i": "composite",
        "a_i+1": "descending",
        "a_mn+1": "minimally",
        "a_j": "imaginary",
        "n_i": "shortness",
        "n_j": "briefness",
        "i": "aggregate",
        "j": "complete",
        "m": "limitless",
        "n": "limitedly"
      },
      "question": "B-4. Let \\( 0<fractional<irrational<\\cdots<minimally \\) be \\( limitless limitedly+1 \\) integers. Prove that you can select either \\( limitless+1 \\) of them no one of which divides any other, or \\( limitedly+1 \\) of them each dividing the following one.",
      "solution": "B-4 Let, for each \\( 1 \\leqq aggregate \\leqq limitless limitedly+1, shortness \\) denote the length of the longest chain, starting with \\( composite \\) and each dividing the following one, that we can select out of \\( composite, descending, \\cdots, minimally \\). If no \\( shortness \\) is greater than \\( limitedly \\) then there are at least \\( limitless+1 shortness \\) 's that are the same. However, the integers \\( composite \\) corresponding to these \\( shortness \\) 's cannot divide each other, because \\( composite \\mid imaginary \\) implies that \\( shortness \\geqq briefness+1 \\)."
    },
    "garbled_string": {
      "map": {
        "a_1": "qzxwvtnp",
        "a_2": "hjgrksla",
        "a_i": "tbvrcyuk",
        "a_i+1": "lfdkezno",
        "a_mn+1": "wqmsylpa",
        "a_j": "pzfdxeno",
        "n_i": "kqhrtlva",
        "n_j": "cvgpropn",
        "i": "sblmqwzk",
        "j": "ncrxtdye",
        "m": "gbltrqse",
        "n": "zmxqpvdu"
      },
      "question": "B-4. Let \\( 0<qzxwvtnp<hjgrksla<\\cdots<wqmsylpa \\) be \\( gbltrqse zmxqpvdu+1 \\) integers. Prove that you can select either \\( gbltrqse+1 \\) of them no one of which divides any other, or \\( zmxqpvdu+1 \\) of them each dividing the following one.",
      "solution": "B-4 Let, for each \\( 1 \\leqq sblmqwzk \\leqq gbltrqse zmxqpvdu+1, kqhrtlva \\) denote the length of the longest chain, starting with \\( tbvrcyuk \\) and each dividing the following one, that we can select out of \\( tbvrcyuk, lfdkezno, \\cdots, wqmsylpa \\). If no \\( kqhrtlva \\) is greater than \\( zmxqpvdu \\) then there are at least \\( gbltrqse+1 kqhrtlva \\) 's that are the same. However, the integers \\( tbvrcyuk \\) corresponding to these \\( kqhrtlva \\) 's cannot divide each other, because \\( tbvrcyuk \\mid pzfdxeno \\) implies that \\( kqhrtlva \\geqq cvgpropn+1 \\)."
    },
    "kernel_variant": {
      "question": "Fix positive integers $k$ and $r$ and an integer $d\\ge 1$.  \nLet $\\Sigma$ be a finite alphabet with $|\\Sigma|\\ge 2$ and write $\\Sigma^{\\ast}$ for the set of all finite words over $\\Sigma$.  \nConsider the product poset  \n\n\\[\nP\\ :=\\ \\Sigma^{\\ast}\\ \\times\\ \\mathbb N_{>0}^{\\,d},\n\\]\n\nendowed with the partial order  \n\n\\[\n\\bigl(w,v\\bigr)\\;\\preceq\\;\\bigl(w',v'\\bigr)\n\\quad\\Longleftrightarrow\\quad\n\\text{$w$ is a prefix of $w'$ and $v$ divides $v'$ coordinate-wise}.\n\\]\n\nFor a finite family $F\\subseteq P$ put  \n\n\\[\nh(x)\\ :=\\ \\max\\{\\ell : x=x_{\\ell}\\preceq\\cdots\\preceq x_{1}\\text{ in }F\\},\\qquad\nH(F)\\ :=\\ \\max_{x\\in F}h(x),\\qquad\nw(F)\\ :=\\ \\bigl|\\text{largest $\\preceq$–antichain in }F\\bigr|.\n\\]\n\nA family $F$ is called $(k,r)$-avoiding if it contains\nneither a $\\preceq$–chain of length $r+1$ nor a $\\preceq$–antichain of size $k+1$.\n\n1. Determine, with proof, the extremal number  \n\n\\[\nM(k,r)\\ :=\\ \\max\\bigl\\{|F| : F\\subseteq P\\text{ is $(k,r)$-avoiding}\\bigr\\}.\n\\]\n\n2. (Erd\\H{o}s–Szekeres type) Show that every family of size $M(k,r)+1$\nnecessarily contains either a $\\preceq$–chain of length $r+1$ or a $\\preceq$–antichain of size $k+1$.\n\n3. Give a complete description of all extremal families, i.e.\\ of all\n$F\\subseteq P$ that are $(k,r)$-avoiding and satisfy $|F|=M(k,r)$.  \nProvide necessary and sufficient structural conditions.\n\n\n\n────────────────────────────────────────────────────",
      "solution": "Part 0.  Notation.  \nFor $t=1,\\dots ,r$ write  \n\n\\[\nA_{t}(F)\\ :=\\ \\bigl\\{\\,x\\in F : h(x)=t \\bigr\\}.\n\\]\n\nEach $A_{t}(F)$ is an antichain, hence $|A_{t}(F)|\\le w(F)$.\n\n────────────────────────────────────────────────────\n1.  The extremal number $M(k,r)$.\n\nUpper bound.  \nFor every finite poset and every finite $F$ one has  \n\n\\[\n\\lvert F\\rvert \\;=\\;\\sum_{t=1}^{H(F)}\\lvert A_{t}(F)\\rvert\n\\;\\le\\;H(F)\\cdot w(F). \\tag{1}\n\\]\n\nIf $F$ is $(k,r)$-avoiding then $H(F)\\le r$ and $w(F)\\le k$, so  \n\n\\[\n\\lvert F\\rvert\\;\\le\\;kr,\n\\qquad\\text{hence}\\qquad\nM(k,r)\\;\\le\\;kr.\n\\]\n\nLower bound – a construction of size $kr$.  \nChoose\n\n• $k$ words $u_{1},\\dots ,u_{k}$ that are pairwise prefix-incomparable,  \n• $r$ vectors $v_{1}\\mid v_{2}\\mid\\cdots\\mid v_{r}$ (a divisibility chain).\n\nPut  \n\n\\[\nF_{0}\\ :=\\ \\bigl\\{(u_{i},v_{j}) : 1\\le i\\le k,\\;1\\le j\\le r\\bigr\\}. \\tag{2}\n\\]\n\nFor each fixed $i$ the $r$ elements $(u_{i},v_{j})_{j=1}^{r}$ form a $\\preceq$-chain,\nwhile the $k$ columns form a $\\preceq$-antichain; hence $F_{0}$ is $(k,r)$-avoiding and $|F_{0}|=kr$.\nThus $M(k,r)=kr$. ∎\n\n\n\n────────────────────────────────────────────────────\n2.  The Erd\\H{o}s–Szekeres type statement.\n\nLet $|F|=kr+1$.  Inequality (1) forces either $H(F)\\ge r+1$ or $w(F)\\ge k+1$,\ni.e.\\ $F$ contains the required chain or antichain. ∎\n\n\n\n────────────────────────────────────────────────────\n3.  All extremal families.\n\nFix an extremal family $F\\subseteq P$, i.e.\\ $|F|=kr$ and $F$ is $(k,r)$-avoiding.\n\nStep 3.1.  Height and width are forced.  \nEquality in (1) implies  \n\n\\[\nH(F)=r,\\qquad w(F)=k,\\qquad\\text{and}\\qquad\n|A_{t}(F)|=k\\ \\text{for every }t=1,\\dots ,r. \\tag{3}\n\\]\n\nStep 3.2.  A saturated chain decomposition.  \nBy Dilworth’s theorem $F$ can be partitioned into \\emph{some} $k$ disjoint\n$\\preceq$-chains; with $|F|=kr$ and $H(F)=r$ each of them must have length $r$.\nFix one such decomposition and write  \n\n\\[\nC_{i} :\\ x_{i,1}\\;\\preceq\\;x_{i,2}\\;\\preceq\\;\\cdots\\;\\preceq\\;x_{i,r}\n\\qquad (1\\le i\\le k). \\tag{4}\n\\]\n\nRemark.  The label $p=h(x_{i,p})$ is intrinsic, but the chain index $i$\n\\emph{does} depend on the chosen decomposition; different decompositions\nare possible and will differ by a permutation of the chains if $r\\ge 2$,\nand by \\emph{any} permutation of the singletons when $r=1$.\n\nStep 3.3.  The crucial “strict-level’’ property.\n\nLemma 3.1.  \nFor two distinct chains $C_{i},C_{j}$ and indices $p,q\\in\\{1,\\dots ,r\\}$ we have  \n\n\\[\nx_{i,p}\\;\\preceq\\;x_{j,q}\\quad\\Longrightarrow\\quad p<q. \\tag{5}\n\\]\n\nProof.  \nAssume $x_{i,p}\\preceq x_{j,q}$ with $p\\ge q$.  Then  \n\n\\[\nx_{i,1},\\dots ,x_{i,p},\\,x_{j,q},\\dots ,x_{j,r}  \\tag{6}\n\\]\n\nis a $\\preceq$-chain of length $p+(r-q+1)\\ge q+(r-q+1)=r+1$,\ncontradicting the $(k,r)$-avoidance of $F$. ∎\n\n\n\nDefinition 3.2 (upper-triangular family).  \nA family $F$ satisfying (3) that admits \\emph{some} decomposition (4)\nfulfilling (5) is called \\emph{upper-triangular}.\n\n\n\nStep 3.4.  Sufficiency: upper-triangular $\\Longrightarrow$ $(k,r)$-avoiding.  \nLet $F$ be upper-triangular and take any $\\preceq$-chain  \n\n\\[\ny_{1}\\;\\preceq\\;y_{2}\\;\\preceq\\;\\dots\\;\\preceq\\;y_{s}\\subseteq F,\\qquad\ny_{t}=x_{i_{t},p_{t}}.\n\\]\n\nIf $i_{t}\\ne i_{t+1}$ then (5) gives $p_{t}<p_{t+1}$; if $i_{t}=i_{t+1}$\nwe also have $p_{t}<p_{t+1}$ because the $x$’s lie in the same chain.\nHence $p_{1}<p_{2}<\\dots <p_{s}\\le r$, so $s\\le r$.\nTherefore no chain of length $r+1$ exists, while $w(F)=k$ by (3);\nconsequently $F$ is $(k,r)$-avoiding. ∎\n\n\n\nStep 3.5.  Necessity: every extremal family is upper-triangular.  \nWe have already fixed a saturated chain decomposition (4).  \nLemma 3.1 shows that (5) holds for this decomposition, so $F$ is\nupper-triangular.\n\nStep 3.6.  A complete but non-redundant parametrisation.\n\nAdmissible data.  \nFor every $1\\le i\\le k$ choose two $r$-tuples\n\n\\[\nW_{i}:=\\bigl(w_{i,1},\\dots ,w_{i,r}\\bigr)\\in(\\Sigma^{\\ast})^{r},\n\\qquad\nV_{i}:=\\bigl(v_{i,1},\\dots ,v_{i,r}\\bigr)\\in(\\mathbb N_{>0}^{\\,d})^{r},\n\\]\n\nsubject to  \n\n\\[\n\\begin{array}{ll}\n\\text{(A1)} & w_{i,1}\\preceq w_{i,2}\\preceq\\cdots\\preceq w_{i,r}\\quad\\text{(prefix chain)},\\\\[2pt]\n\\text{(A2)} & v_{i,1}\\mid v_{i,2}\\mid\\cdots\\mid v_{i,r}\\quad\\text{(divisibility chain)},\\\\[2pt]\n\\text{(A3)} & \\text{for all }i\\ne j\\text{ and }p,q\\text{ with }p\\ge q:\\\\\n            & \\quad\\bigl(w_{i,p}\\text{ prefix of }w_{j,q}\\ \\text{and}\\ v_{i,p}\\mid v_{j,q}\\bigr)\n              \\ \\Longrightarrow\\ p<q.\n\\end{array}\\tag{7}\n\\]\n\nDefine  \n\n\\[\nF(W,V)\\ :=\\ \\bigl\\{(w_{i,p},v_{i,p}) : 1\\le i\\le k,\\;1\\le p\\le r\\bigr\\}. \\tag{8}\n\\]\n\nProposition 3.3.  \n\n(a) $F(W,V)$ is an extremal family: it is $(k,r)$-avoiding and $|F(W,V)|=kr$.  \n\n(b) Conversely, every extremal family equals $F(W,V)$ for some admissible pair $(W,V)$.  \n\n(c) Two admissible pairs $(W,V)$ and $(\\widehat W,\\widehat V)$ yield the\n\\emph{same} family iff one can be obtained from the other by a permutation\nof the $k$ index blocks $(W_{i},V_{i})$.\n\nProof.  \n\n(a) Inside each fixed $i$ the $r$ elements form a chain by (A1)–(A2);\nproperty (A3) is exactly Lemma 3.1, hence $F(W,V)$ is upper-triangular\nand therefore $(k,r)$-avoiding with size $kr$ (Steps 3.3–3.4).\n\n(b) Let $F$ be extremal.  Choose an arbitrary saturated chain\ndecomposition (4) and set $w_{i,p}$, $v_{i,p}$ to be the two coordinates\nof $x_{i,p}$.  Conditions (A1)–(A2) hold by construction, while (A3) is\nnothing but Lemma 3.1.\n\n(c) If we start from the \\emph{same} extremal family but pick a\n\\emph{different} saturated chain decomposition, the $r$-level\npartitioning is fixed whereas the $k$ chains may be permuted.\nHence any two arrays describing $F$ differ only by such a permutation,\nand conversely permuting whole blocks obviously keeps the underlying\nset unchanged. ∎\n\n\n\nTheorem 3.4 (complete characterisation).  \nA family $F\\subseteq P$ has size $kr$ and is $(k,r)$-avoiding iff\n\n(i) $F$ can be partitioned into $k$ disjoint $\\preceq$-chains of length $r$, and  \n\n(ii) whenever $x$ belongs to an earlier chain than $y$ in that partition,\ncomparability $x\\preceq y$ forces $h(x)<h(y)$.\n\nEquivalently, $F=F(W,V)$ for some admissible pair $(W,V)$, the pair being\nunique up to permutation of the $k$ chains.\n\n\n\nConcrete example (showing no extra restriction is necessary).  \nFix a symbol $\\sigma\\in\\Sigma$ and $r$ distinct primes\n$\\pi_{1},\\dots ,\\pi_{r}$.  Pick $k$ arbitrary pairwise prefix-incomparable\nwords $u_{1},\\dots ,u_{k}$ and set  \n\n\\[\nw_{i,p}\\ :=\\ u_{i}\\sigma^{p},\\qquad\nv_{i,p}\\ :=\\ \\bigl(\\pi_{p}^{\\,i},\\dots ,\\pi_{p}^{\\,i}\\bigr)\\in\\mathbb N_{>0}^{\\,d}.\n\\]\n\nConditions (A1)–(A3) are readily verified, so we obtain\nan extremal family distinct from $F_{0}$ whenever $r\\ge 2$.\n\n\n\n────────────────────────────────────────────────────\nSummary.  \n\nExtremal size: $M(k,r)=kr$.  \n\nErd\\H{o}s–Szekeres property: every set of $kr+1$ elements in $P$\ncontains either a $\\preceq$-chain of length $r+1$ or a\n$\\preceq$-antichain of size $k+1$.  \n\nStructure: extremal families are precisely the upper-triangular ones\ndescribed in Theorem 3.4; the admissible pair $(W,V)$ that encodes such a\nfamily is unique up to permuting the $k$ chains. ∎\n\n\n\n────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.567469",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional structure – Elements now carry a d-component vector, and comparability demands simultaneous prefix inclusion and component-wise divisibility.  This turns the order into the product of two non-trivial partial orders, sharply increasing the combinatorial complexity.\n\n2. Stronger comparability condition – For two pairs to be comparable, every coordinate of one vector must divide the corresponding coordinate of the other; failure in any single coordinate breaks comparability.  This makes chains much rarer and antichains subtler to detect.\n\n3. Abstract poset methods indispensable – The solution needs the height/width dichotomy of posets (Mirsky/Dilworth ideas) together with a new “rank number” argument; elementary pigeon-hole or Erdős–Szekeres tricks no longer suffice.\n\n4. Parameter interplay – The bound N = k·r +1 is tight for this much richer order; showing tightness amid the added divisibility constraints requires careful counting, not present in the original problems.\n\n5. Multiple advanced concepts – The problem blends string theory (prefix order), number theory (divisibility in every coordinate), and high-dimensional partial order theory, demanding fluency with each and their interaction.\n\nAll these features markedly exceed the technical and conceptual load of both the original B-4 problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix positive integers $k$ and $r$ and an integer $d\\ge 1$.  \nLet $\\Sigma$ be a finite alphabet with $|\\Sigma|\\ge 2$ and write $\\Sigma^{\\ast}$ for the set of all finite words over $\\Sigma$.  \nConsider the product poset  \n\n\\[\nP\\ :=\\ \\Sigma^{\\ast}\\ \\times\\ \\mathbb N_{>0}^{\\,d},\n\\]\n\nendowed with the partial order  \n\n\\[\n\\bigl(w,v\\bigr)\\;\\preceq\\;\\bigl(w',v'\\bigr)\n\\quad\\Longleftrightarrow\\quad\n\\text{$w$ is a prefix of $w'$ and $v$ divides $v'$ coordinate-wise}.\n\\]\n\nFor a finite family $F\\subseteq P$ put  \n\n\\[\nh(x)\\ :=\\ \\max\\{\\ell : x=x_{\\ell}\\preceq\\cdots\\preceq x_{1}\\text{ in }F\\},\\qquad\nH(F)\\ :=\\ \\max_{x\\in F}h(x),\\qquad\nw(F)\\ :=\\ \\bigl|\\text{largest $\\preceq$–antichain in }F\\bigr|.\n\\]\n\nA family $F$ is called $(k,r)$-avoiding if it contains\nneither a $\\preceq$–chain of length $r+1$ nor a $\\preceq$–antichain of size $k+1$.\n\n1. Determine, with proof, the extremal number  \n\n\\[\nM(k,r)\\ :=\\ \\max\\bigl\\{|F| : F\\subseteq P\\text{ is $(k,r)$-avoiding}\\bigr\\}.\n\\]\n\n2. (Erd\\H{o}s–Szekeres type) Show that every family of size $M(k,r)+1$\nnecessarily contains either a $\\preceq$–chain of length $r+1$ or a $\\preceq$–antichain of size $k+1$.\n\n3. Give a complete description of all extremal families, i.e.\\ of all\n$F\\subseteq P$ that are $(k,r)$-avoiding and satisfy $|F|=M(k,r)$.  \nProvide necessary and sufficient structural conditions.\n\n\n\n────────────────────────────────────────────────────",
      "solution": "Part 0.  Notation.  \nFor $t=1,\\dots ,r$ write  \n\n\\[\nA_{t}(F)\\ :=\\ \\bigl\\{\\,x\\in F : h(x)=t \\bigr\\}.\n\\]\n\nEach $A_{t}(F)$ is an antichain, hence $|A_{t}(F)|\\le w(F)$.\n\n────────────────────────────────────────────────────\n1.  The extremal number $M(k,r)$.\n\nUpper bound.  \nFor every finite poset and every finite $F$ one has  \n\n\\[\n\\lvert F\\rvert \\;=\\;\\sum_{t=1}^{H(F)}\\lvert A_{t}(F)\\rvert\n\\;\\le\\;H(F)\\cdot w(F). \\tag{1}\n\\]\n\nIf $F$ is $(k,r)$-avoiding then $H(F)\\le r$ and $w(F)\\le k$, so  \n\n\\[\n\\lvert F\\rvert\\;\\le\\;kr,\n\\qquad\\text{hence}\\qquad\nM(k,r)\\;\\le\\;kr.\n\\]\n\nLower bound – a construction of size $kr$.  \nChoose\n\n• $k$ words $u_{1},\\dots ,u_{k}$ that are pairwise prefix-incomparable,  \n• $r$ vectors $v_{1}\\mid v_{2}\\mid\\cdots\\mid v_{r}$ (a divisibility chain).\n\nPut  \n\n\\[\nF_{0}\\ :=\\ \\bigl\\{(u_{i},v_{j}) : 1\\le i\\le k,\\;1\\le j\\le r\\bigr\\}. \\tag{2}\n\\]\n\nFor each fixed $i$ the $r$ elements $(u_{i},v_{j})_{j=1}^{r}$ form a $\\preceq$-chain,\nwhile the $k$ columns form a $\\preceq$-antichain; hence $F_{0}$ is $(k,r)$-avoiding and $|F_{0}|=kr$.\nThus $M(k,r)=kr$. ∎\n\n\n\n────────────────────────────────────────────────────\n2.  The Erd\\H{o}s–Szekeres type statement.\n\nLet $|F|=kr+1$.  Inequality (1) forces either $H(F)\\ge r+1$ or $w(F)\\ge k+1$,\ni.e.\\ $F$ contains the required chain or antichain. ∎\n\n\n\n────────────────────────────────────────────────────\n3.  All extremal families.\n\nFix an extremal family $F\\subseteq P$, i.e.\\ $|F|=kr$ and $F$ is $(k,r)$-avoiding.\n\nStep 3.1.  Height and width are forced.  \nEquality in (1) implies  \n\n\\[\nH(F)=r,\\qquad w(F)=k,\\qquad\\text{and}\\qquad\n|A_{t}(F)|=k\\ \\text{for every }t=1,\\dots ,r. \\tag{3}\n\\]\n\nStep 3.2.  A saturated chain decomposition.  \nBy Dilworth’s theorem $F$ can be partitioned into \\emph{some} $k$ disjoint\n$\\preceq$-chains; with $|F|=kr$ and $H(F)=r$ each of them must have length $r$.\nFix one such decomposition and write  \n\n\\[\nC_{i} :\\ x_{i,1}\\;\\preceq\\;x_{i,2}\\;\\preceq\\;\\cdots\\;\\preceq\\;x_{i,r}\n\\qquad (1\\le i\\le k). \\tag{4}\n\\]\n\nRemark.  The label $p=h(x_{i,p})$ is intrinsic, but the chain index $i$\n\\emph{does} depend on the chosen decomposition; different decompositions\nare possible and will differ by a permutation of the chains if $r\\ge 2$,\nand by \\emph{any} permutation of the singletons when $r=1$.\n\nStep 3.3.  The crucial “strict-level’’ property.\n\nLemma 3.1.  \nFor two distinct chains $C_{i},C_{j}$ and indices $p,q\\in\\{1,\\dots ,r\\}$ we have  \n\n\\[\nx_{i,p}\\;\\preceq\\;x_{j,q}\\quad\\Longrightarrow\\quad p<q. \\tag{5}\n\\]\n\nProof.  \nAssume $x_{i,p}\\preceq x_{j,q}$ with $p\\ge q$.  Then  \n\n\\[\nx_{i,1},\\dots ,x_{i,p},\\,x_{j,q},\\dots ,x_{j,r}  \\tag{6}\n\\]\n\nis a $\\preceq$-chain of length $p+(r-q+1)\\ge q+(r-q+1)=r+1$,\ncontradicting the $(k,r)$-avoidance of $F$. ∎\n\n\n\nDefinition 3.2 (upper-triangular family).  \nA family $F$ satisfying (3) that admits \\emph{some} decomposition (4)\nfulfilling (5) is called \\emph{upper-triangular}.\n\n\n\nStep 3.4.  Sufficiency: upper-triangular $\\Longrightarrow$ $(k,r)$-avoiding.  \nLet $F$ be upper-triangular and take any $\\preceq$-chain  \n\n\\[\ny_{1}\\;\\preceq\\;y_{2}\\;\\preceq\\;\\dots\\;\\preceq\\;y_{s}\\subseteq F,\\qquad\ny_{t}=x_{i_{t},p_{t}}.\n\\]\n\nIf $i_{t}\\ne i_{t+1}$ then (5) gives $p_{t}<p_{t+1}$; if $i_{t}=i_{t+1}$\nwe also have $p_{t}<p_{t+1}$ because the $x$’s lie in the same chain.\nHence $p_{1}<p_{2}<\\dots <p_{s}\\le r$, so $s\\le r$.\nTherefore no chain of length $r+1$ exists, while $w(F)=k$ by (3);\nconsequently $F$ is $(k,r)$-avoiding. ∎\n\n\n\nStep 3.5.  Necessity: every extremal family is upper-triangular.  \nWe have already fixed a saturated chain decomposition (4).  \nLemma 3.1 shows that (5) holds for this decomposition, so $F$ is\nupper-triangular.\n\nStep 3.6.  A complete but non-redundant parametrisation.\n\nAdmissible data.  \nFor every $1\\le i\\le k$ choose two $r$-tuples\n\n\\[\nW_{i}:=\\bigl(w_{i,1},\\dots ,w_{i,r}\\bigr)\\in(\\Sigma^{\\ast})^{r},\n\\qquad\nV_{i}:=\\bigl(v_{i,1},\\dots ,v_{i,r}\\bigr)\\in(\\mathbb N_{>0}^{\\,d})^{r},\n\\]\n\nsubject to  \n\n\\[\n\\begin{array}{ll}\n\\text{(A1)} & w_{i,1}\\preceq w_{i,2}\\preceq\\cdots\\preceq w_{i,r}\\quad\\text{(prefix chain)},\\\\[2pt]\n\\text{(A2)} & v_{i,1}\\mid v_{i,2}\\mid\\cdots\\mid v_{i,r}\\quad\\text{(divisibility chain)},\\\\[2pt]\n\\text{(A3)} & \\text{for all }i\\ne j\\text{ and }p,q\\text{ with }p\\ge q:\\\\\n            & \\quad\\bigl(w_{i,p}\\text{ prefix of }w_{j,q}\\ \\text{and}\\ v_{i,p}\\mid v_{j,q}\\bigr)\n              \\ \\Longrightarrow\\ p<q.\n\\end{array}\\tag{7}\n\\]\n\nDefine  \n\n\\[\nF(W,V)\\ :=\\ \\bigl\\{(w_{i,p},v_{i,p}) : 1\\le i\\le k,\\;1\\le p\\le r\\bigr\\}. \\tag{8}\n\\]\n\nProposition 3.3.  \n\n(a) $F(W,V)$ is an extremal family: it is $(k,r)$-avoiding and $|F(W,V)|=kr$.  \n\n(b) Conversely, every extremal family equals $F(W,V)$ for some admissible pair $(W,V)$.  \n\n(c) Two admissible pairs $(W,V)$ and $(\\widehat W,\\widehat V)$ yield the\n\\emph{same} family iff one can be obtained from the other by a permutation\nof the $k$ index blocks $(W_{i},V_{i})$.\n\nProof.  \n\n(a) Inside each fixed $i$ the $r$ elements form a chain by (A1)–(A2);\nproperty (A3) is exactly Lemma 3.1, hence $F(W,V)$ is upper-triangular\nand therefore $(k,r)$-avoiding with size $kr$ (Steps 3.3–3.4).\n\n(b) Let $F$ be extremal.  Choose an arbitrary saturated chain\ndecomposition (4) and set $w_{i,p}$, $v_{i,p}$ to be the two coordinates\nof $x_{i,p}$.  Conditions (A1)–(A2) hold by construction, while (A3) is\nnothing but Lemma 3.1.\n\n(c) If we start from the \\emph{same} extremal family but pick a\n\\emph{different} saturated chain decomposition, the $r$-level\npartitioning is fixed whereas the $k$ chains may be permuted.\nHence any two arrays describing $F$ differ only by such a permutation,\nand conversely permuting whole blocks obviously keeps the underlying\nset unchanged. ∎\n\n\n\nTheorem 3.4 (complete characterisation).  \nA family $F\\subseteq P$ has size $kr$ and is $(k,r)$-avoiding iff\n\n(i) $F$ can be partitioned into $k$ disjoint $\\preceq$-chains of length $r$, and  \n\n(ii) whenever $x$ belongs to an earlier chain than $y$ in that partition,\ncomparability $x\\preceq y$ forces $h(x)<h(y)$.\n\nEquivalently, $F=F(W,V)$ for some admissible pair $(W,V)$, the pair being\nunique up to permutation of the $k$ chains.\n\n\n\nConcrete example (showing no extra restriction is necessary).  \nFix a symbol $\\sigma\\in\\Sigma$ and $r$ distinct primes\n$\\pi_{1},\\dots ,\\pi_{r}$.  Pick $k$ arbitrary pairwise prefix-incomparable\nwords $u_{1},\\dots ,u_{k}$ and set  \n\n\\[\nw_{i,p}\\ :=\\ u_{i}\\sigma^{p},\\qquad\nv_{i,p}\\ :=\\ \\bigl(\\pi_{p}^{\\,i},\\dots ,\\pi_{p}^{\\,i}\\bigr)\\in\\mathbb N_{>0}^{\\,d}.\n\\]\n\nConditions (A1)–(A3) are readily verified, so we obtain\nan extremal family distinct from $F_{0}$ whenever $r\\ge 2$.\n\n\n\n────────────────────────────────────────────────────\nSummary.  \n\nExtremal size: $M(k,r)=kr$.  \n\nErd\\H{o}s–Szekeres property: every set of $kr+1$ elements in $P$\ncontains either a $\\preceq$-chain of length $r+1$ or a\n$\\preceq$-antichain of size $k+1$.  \n\nStructure: extremal families are precisely the upper-triangular ones\ndescribed in Theorem 3.4; the admissible pair $(W,V)$ that encodes such a\nfamily is unique up to permuting the $k$ chains. ∎\n\n\n\n────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.462708",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional structure – Elements now carry a d-component vector, and comparability demands simultaneous prefix inclusion and component-wise divisibility.  This turns the order into the product of two non-trivial partial orders, sharply increasing the combinatorial complexity.\n\n2. Stronger comparability condition – For two pairs to be comparable, every coordinate of one vector must divide the corresponding coordinate of the other; failure in any single coordinate breaks comparability.  This makes chains much rarer and antichains subtler to detect.\n\n3. Abstract poset methods indispensable – The solution needs the height/width dichotomy of posets (Mirsky/Dilworth ideas) together with a new “rank number” argument; elementary pigeon-hole or Erdős–Szekeres tricks no longer suffice.\n\n4. Parameter interplay – The bound N = k·r +1 is tight for this much richer order; showing tightness amid the added divisibility constraints requires careful counting, not present in the original problems.\n\n5. Multiple advanced concepts – The problem blends string theory (prefix order), number theory (divisibility in every coordinate), and high-dimensional partial order theory, demanding fluency with each and their interaction.\n\nAll these features markedly exceed the technical and conceptual load of both the original B-4 problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}