{
  "index": "1965-B-5",
  "type": "COMB",
  "tag": [
    "COMB"
  ],
  "difficulty": "",
  "question": "B-5. Consider collections of unordered pairs of \\( V \\) different objects \\( a, b, c, \\cdots, k \\). Three pairs such as \\( b c, c a, a b \\) are said to form a triangle. Prove that, if \\( 4 E \\leqq V^{2} \\), it is possible to choose \\( E \\) pairs so that no triangle is formed.",
  "solution": "B-5. Divide the objects into two subsets \\( \\left\\{a_{1}, a_{2}, \\cdots, a_{m}\\right\\} \\) and \\( \\left\\{b_{1}, b_{2}, \\cdots, b_{n}\\right\\} \\), where \\( m+n=V \\). Then the \\( m n \\) pairs \\( \\left(a_{j}, b_{k}\\right) \\), where \\( j=1,2, \\cdots, m \\) and \\( k=1,2 \\), \\( \\cdots, n \\), obviously contain no triangles. If \\( V \\) is even, take \\( m=n=V / 2 \\), and if \\( V \\) is odd, take \\( m=(V+1) / 2, n=(V-1) / 2 \\). Then \\( m n \\geqq V^{2} / 4 \\geqq E \\).",
  "vars": [
    "a",
    "b",
    "c",
    "k",
    "a_j",
    "b_k",
    "j",
    "m",
    "n"
  ],
  "params": [
    "V",
    "E"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "objectone",
        "b": "objecttwo",
        "c": "objectthree",
        "k": "objectfour",
        "a_j": "alistj",
        "b_k": "blistk",
        "j": "indexjay",
        "m": "sizealpha",
        "n": "sizebeta",
        "V": "totalobjs",
        "E": "pairscount"
      },
      "question": "B-5. Consider collections of unordered pairs of \\( totalobjs \\) different objects \\( objectone, objecttwo, objectthree, \\cdots, objectfour \\). Three pairs such as \\( objecttwo objectthree, objectthree objectone, objectone objecttwo \\) are said to form a triangle. Prove that, if \\( 4 pairscount \\leqq totalobjs^{2} \\), it is possible to choose \\( pairscount \\) pairs so that no triangle is formed.",
      "solution": "B-5. Divide the objects into two subsets \\( \\left\\{objectone_{1}, objectone_{2}, \\cdots, objectone_{sizealpha}\\right\\} \\) and \\( \\left\\{objecttwo_{1}, objecttwo_{2}, \\cdots, objecttwo_{sizebeta}\\right\\} \\), where \\( sizealpha+sizebeta=totalobjs \\). Then the \\( sizealpha sizebeta \\) pairs \\( \\left(alistj, blistk\\right) \\), where \\( indexjay=1,2, \\cdots, sizealpha \\) and \\( objectfour=1,2, \\cdots, sizebeta \\), obviously contain no triangles. If \\( totalobjs \\) is even, take \\( sizealpha=sizebeta=totalobjs / 2 \\), and if \\( totalobjs \\) is odd, take \\( sizealpha=(totalobjs+1) / 2, sizebeta=(totalobjs-1) / 2 \\). Then \\( sizealpha sizebeta \\geqq totalobjs^{2} / 4 \\geqq pairscount \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "blueberry",
        "b": "snowflake",
        "c": "marshmallow",
        "k": "adventure",
        "a_j": "pineapple",
        "b_k": "tangerine",
        "j": "cinnamon",
        "m": "waterfall",
        "n": "lighthouse",
        "V": "sandcastle",
        "E": "whirlwind"
      },
      "question": "B-5. Consider collections of unordered pairs of \\( sandcastle \\) different objects \\( blueberry, snowflake, marshmallow, \\cdots, adventure \\). Three pairs such as \\( snowflake marshmallow, marshmallow blueberry, blueberry snowflake \\) are said to form a triangle. Prove that, if \\( 4 \\, whirlwind \\leqq sandcastle^{2} \\), it is possible to choose \\( whirlwind \\) pairs so that no triangle is formed.",
      "solution": "B-5. Divide the objects into two subsets \\( \\left\\{blueberry_{1}, blueberry_{2}, \\cdots, blueberry_{waterfall}\\right\\} \\) and \\( \\left\\{snowflake_{1}, snowflake_{2}, \\cdots, snowflake_{lighthouse}\\right\\} \\), where \\( waterfall+lighthouse=sandcastle \\). Then the \\( waterfall\\,lighthouse \\) pairs \\( \\left(pineapple, tangerine\\right) \\), where \\( cinnamon=1,2, \\cdots, waterfall \\) and \\( adventure=1,2, \\cdots, lighthouse \\), obviously contain no triangles. If \\( sandcastle \\) is even, take \\( waterfall=lighthouse=sandcastle / 2 \\), and if \\( sandcastle \\) is odd, take \\( waterfall=(sandcastle+1) / 2, \\; lighthouse=(sandcastle-1) / 2 \\). Then \\( waterfall\\,lighthouse \\geqq sandcastle^{2} / 4 \\geqq whirlwind \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "absenceobj",
        "b": "blankthing",
        "c": "vacuityobj",
        "k": "noindexer",
        "a_j": "absenceindexed",
        "b_k": "blankindexed",
        "j": "noiterator",
        "m": "emptysize",
        "n": "nilsizecnt",
        "V": "voidtotal",
        "E": "nonepairs"
      },
      "question": "B-5. Consider collections of unordered pairs of \\( voidtotal \\) different objects \\( absenceobj, blankthing, vacuityobj, \\cdots, noindexer \\). Three pairs such as \\( blankthing vacuityobj, vacuityobj absenceobj, absenceobj blankthing \\) are said to form a triangle. Prove that, if \\( 4\\, nonepairs \\leqq voidtotal^{2} \\), it is possible to choose \\( nonepairs \\) pairs so that no triangle is formed.",
      "solution": "B-5. Divide the objects into two subsets \\( \\left\\{absenceobj_{1}, absenceobj_{2}, \\cdots, absenceobj_{emptysize}\\right\\} \\) and \\( \\left\\{blankthing_{1}, blankthing_{2}, \\cdots, blankthing_{nilsizecnt}\\right\\} \\), where \\( emptysize+nilsizecnt=voidtotal \\). Then the \\( emptysize nilsizecnt \\) pairs \\( \\left(absenceindexed, blankindexed\\right) \\), where \\( noiterator=1,2, \\cdots, emptysize \\) and \\( noindexer=1,2 \\), \\( \\cdots, nilsizecnt \\), obviously contain no triangles. If \\( voidtotal \\) is even, take \\( emptysize=nilsizecnt=voidtotal / 2 \\), and if \\( voidtotal \\) is odd, take \\( emptysize=(voidtotal+1) / 2, nilsizecnt=(voidtotal-1) / 2 \\). Then \\( emptysize nilsizecnt \\geqq voidtotal^{2} / 4 \\geqq nonepairs \\)."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mzbqtwxy",
        "k": "lpoumcra",
        "a_j": "vrestoqu",
        "b_k": "xjncferd",
        "j": "ahdvnqwe",
        "m": "cqpvskdz",
        "n": "jrwsbmhg",
        "V": "uiznplko",
        "E": "gqwhzvst"
      },
      "question": "B-5. Consider collections of unordered pairs of \\( uiznplko \\) different objects \\( qzxwvtnp, hjgrksla, mzbqtwxy, \\cdots, lpoumcra \\). Three pairs such as \\( hjgrksla mzbqtwxy, mzbqtwxy qzxwvtnp, qzxwvtnp hjgrksla \\) are said to form a triangle. Prove that, if \\( 4 gqwhzvst \\leqq uiznplko^{2} \\), it is possible to choose \\( gqwhzvst \\) pairs so that no triangle is formed.",
      "solution": "B-5. Divide the objects into two subsets \\( \\left\\{qzxwvtnp_{1}, qzxwvtnp_{2}, \\cdots, qzxwvtnp_{cqpvskdz}\\right\\} \\) and \\( \\left\\{hjgrksla_{1}, hjgrksla_{2}, \\cdots, hjgrksla_{jrwsbmhg}\\right\\} \\), where \\( cqpvskdz+jrwsbmhg=uiznplko \\). Then the \\( cqpvskdz jrwsbmhg \\) pairs \\( \\left(qzxwvtnp_{ahdvnqwe}, hjgrksla_{lpoumcra}\\right) \\), where \\( ahdvnqwe=1,2, \\cdots, cqpvskdz \\) and \\( lpoumcra=1,2 \\), \\( \\cdots, jrwsbmhg \\), obviously contain no triangles. If \\( uiznplko \\) is even, take \\( cqpvskdz=jrwsbmhg=uiznplko / 2 \\), and if \\( uiznplko \\) is odd, take \\( cqpvskdz=(uiznplko+1) / 2, jrwsbmhg=(uiznplko-1) / 2 \\). Then \\( cqpvskdz jrwsbmhg \\geqq uiznplko^{2} / 4 \\geqq gqwhzvst \\)."
    },
    "kernel_variant": {
      "question": "Let $r\\ge 2$ be a fixed integer and let $p>r$ be a prime.  Put \n\\[\nn:=p ,\\qquad    \nV\\in\\{rn,\\;rn+1\\},\\qquad     \n\\Gamma=\\{\\gamma _1,\\gamma _2,\\dots ,\\gamma _V\\},\n\\]\nlisting the $V$ symbols so that every residue class modulo $r$\noccurs either $n$ or $n+1$ times (hence all class–sizes differ by at\nmost one).\n\nAn \\emph{$r$–link} is an unordered $r$–tuple of distinct symbols\n(equivalently, an $r$–element subset of $\\Gamma$).\nA family $\\mathcal H$ of $r$–links is \\emph{simplex–free} if it\ncontains no $(r+1)$ distinct vertices whose every $r$–subset lies in\n$\\mathcal H$ (that is, $\\mathcal H$ contains no $(r+1)$–vertex simplex\nof the complete $r$–uniform hypergraph).\n\nFor $\\gamma ,\\gamma'\\in\\Gamma$ write  \n\\[\n\\deg _{\\mathcal H}(\\gamma)=\n      \\bigl|\\{E\\in\\mathcal H:\\gamma\\in E\\}\\bigr|,\n\\qquad\n\\operatorname{codeg}_{\\mathcal H}(\\gamma,\\gamma')=\n      \\bigl|\\{E\\in\\mathcal H:\\{\\gamma ,\\gamma'\\}\\subset E\\}\\bigr|\n      \\qquad(\\gamma\\ne\\gamma').\n\\]\n\nProve that for \\emph{every} integer \n\\[\n0\\le L\\le n^{2}=p^{2}\n\\]\nthere exists a simplex–free family $\\mathcal F$ of exactly $L$ $r$–links\nsatisfying  \n\n\\[\n\\text{\\rm(a)}\\;\\exists d\\in\\mathbb Z\\text{ such that }\n         d\\le\\deg _{\\mathcal F}(\\gamma)\\le d+2\n         \\quad\\forall\\gamma\\in\\Gamma, \n\\qquad\n\\text{\\rm(b)}\\;\n      \\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1\n      \\quad\\forall\\gamma\\ne\\gamma'\\in\\Gamma .\n\\]\n\n(The construction must work simultaneously for every admissible $L$\nand every fixed $r$; the \\emph{only} restriction on the prime is $p>r$.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "All arithmetic takes place in the finite field $\\mathbb F_{p}$\n(recall $n=p$).  All degrees and codegrees refer to the ultimately\nconstructed family $\\mathcal F$.\n\nWe separate the proof into the graph case $r=2$\n(Section 1) and the genuine $r$–uniform case $r\\ge 3$\n(Section 2).  Every symbol of $\\Gamma$ is stored together with the pair\n\\[\n(\\text{class},\\text{index})\\;=\\;(j,i)\\qquad\n(1\\le j\\le r,\\;i\\in\\mathbb F_{n})\n\\]\nexcept for one optional extra vertex $\\omega$ that occurs only when\n$V=rn+1$ and is declared to have class $j=r$ but no index.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.\\;Graphs ($r=2$)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n--------------------------------------------------------------------\n1.A.\\;Vertex partition and perfect matchings\n--------------------------------------------------------------------\nWrite\n\\[\nP_{1}:=\\{\\gamma_{0},\\dots ,\\gamma_{n-1}\\},\\qquad\nP_{2}^{0}:=\\{\\eta_{0},\\dots ,\\eta_{n-1}\\},\n\\]\nand in the case $V=2n+1$ add the extra vertex $\\omega$ to obtain\n$P_{2}:=P_{2}^{0}\\cup\\{\\omega\\}$; otherwise set $P_{2}:=P_{2}^{0}$.\nHence $\\Gamma=P_{1}\\cup P_{2}$ and the graph we shall construct will be\nbipartite with these two colour classes.\n\nFor $c\\in\\mathbb F_{n}$ define the perfect matching\n\\[\nM_{c}:=\\bigl\\{\\gamma_{i}\\,\\eta_{\\,i+c}:i\\in\\mathbb F_{n}\\bigr\\}.\n\\tag{1.1}\n\\]\nIf $\\omega$ is present, replace the edge\n$\\gamma_{c}\\eta_{\\,c+c}$ by $\\gamma_{c}\\omega$ and denote the resulting\nmatching by $\\widetilde M_{c}$.\n\nObserve the following facts.\n\n(i)  Distinct matchings are edge–disjoint.\n\n(ii) Every vertex of $P_{1}$ (respectively $P_{2}$) lies in exactly one\nedge of each matching $M_{c}$, hence exactly one edge of each\n$\\widetilde M_{c}$.\n\n(iii) Because the graph is bipartite, it is triangle–free.\n\n--------------------------------------------------------------------\n1.B.\\;Edge supply\n--------------------------------------------------------------------\nWrite\n\\[\nL=d\\,n+t,\\qquad 0\\le d\\le n,\\;0\\le t<n .\n\\tag{1.2}\n\\]\nWe distinguish the even and odd orders of the graph.\n\n\\medskip\n\\emph{1.B.1.\\;The case $V=2n$.}\n\nTake the first $d$ perfect matchings\n\\[\n\\mathcal B:=M_{0}\\cup M_{1}\\cup\\dots\\cup M_{d-1};\n\\qquad|\\mathcal B|=d\\,n .\n\\]\nIf $t>0$ choose any $t$ \\emph{pairwise disjoint} edges of $M_{d}$,\ncollect them in $\\mathcal A$, and finally put\n\\[\n\\mathcal F:=\\mathcal B\\cup\\mathcal A .\n\\]\n\n\\medskip\n\\emph{1.B.2.\\;The case $V=2n+1$.}\n\nProceed exactly as above but use the modified matchings:\n\\[\n\\mathcal B:=\\widetilde M_{0}\\cup\\dots\\cup\\widetilde M_{d-1},\\qquad\n\\mathcal A\\subseteq\\widetilde M_{d},\\qquad\n\\mathcal F:=\\mathcal B\\cup\\mathcal A .\n\\]\n(If $d=n$ then $t=0$ and $\\mathcal A=\\varnothing$; the index\n$d$ coincides with $0$ modulo $n$, so the notation\n$\\widetilde M_{d}$ really means $\\widetilde M_{0}$, but nothing is\nadded in this extreme case.)\n\n--------------------------------------------------------------------\n1.C.\\;Verifying the requirements\n--------------------------------------------------------------------\nDegree spread:\neach vertex participates in all edges of exactly $d$ matchings and in at\nmost one additional edge from $\\mathcal A$; hence\n\\[\nd\\le\\deg_{\\mathcal F}(\\gamma)\\le d+1\n\\quad\\forall\\gamma\\in\\Gamma .\n\\]\nCondition (a) therefore holds (with the same $d$).\n\nCodegree:\ntwo vertices are together in an edge if and only if they are endpoints\nof the unique edge of a \\emph{single} matching, so\n$\\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1$\nfor all $\\gamma\\ne\\gamma'$.\n\nTriangle–freeness:\nevery edge connects a vertex of $P_{1}$ to a vertex of $P_{2}$, so the\ngraph is bipartite and therefore contains no triangles.\n\nEdge count:\n$|\\mathcal F|=d\\,n+t=L$ by construction.\n\nHence the graph case is complete.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n2.\\;$r$–uniform hypergraphs ($\\boldsymbol{r\\ge 3}$)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n--------------------------------------------------------------------\n2.A.\\;Canonical $r$–partition of the vertex set\n--------------------------------------------------------------------\nFor $1\\le j\\le r-1$ set\n\\[\nP_{j}:=\\{\\gamma_{j,i}:i\\in\\mathbb F_{n}\\},\\qquad\nP_{r}^{0}:=\\{\\gamma_{r,i}:i\\in\\mathbb F_{n}\\}.\n\\]\nIf $V=rn+1$ add $\\omega$ and put\n$P_{r}:=P_{r}^{0}\\cup\\{\\omega\\}$; otherwise $P_{r}:=P_{r}^{0}$.\nEach class $P_{j}\\;(1\\le j\\le r)$ has size $n$ except possibly\n$P_{r}$, which has size $n$ or $n+1$.\n\n--------------------------------------------------------------------\n2.B.\\;Perfect $r$–matchings\n--------------------------------------------------------------------\nFor $c\\in\\mathbb F_{n}$ and $i\\in\\mathbb F_{n}$ define\n\\[\nE_{c,i}:=\\bigl\\{\\gamma_{1,i},\\gamma_{2,i+c},\\dots ,\n                \\gamma_{r,i+(r-1)c}\\bigr\\},\\qquad\nM_{c}:=\\{E_{c,i}:i\\in\\mathbb F_{n}\\}.\n\\tag{2.1}\n\\]\nThe $M_{c}$ are pairwise edge–disjoint perfect $r$–matchings and\n\\[\n\\bigl|\\bigcup_{c\\in\\mathbb F_{n}}M_{c}\\bigr|=n^{2}.\n\\tag{2.2}\n\\]\n\n--------------------------------------------------------------------\n2.C.\\;Modification in the one–extra–vertex case\n--------------------------------------------------------------------\n(Needed only when $V=rn+1$.)  For $c\\in\\mathbb F_{n}$ put  \n\\[\nF_{c}:=E_{c,c},\\quad\nF_{c}^{\\star}:=(F_{c}\\setminus\\{\\gamma_{r,rc}\\})\\cup\\{\\omega\\},\\quad\n\\widetilde M_{c}:=(M_{c}\\setminus\\{F_{c}\\})\\cup\\{F_{c}^{\\star}\\}.\n\\tag{2.3}\n\\]\nThus $|\\widetilde M_{c}|=n$ and every unordered pair of vertices appears\ntogether in \\emph{at most one} edge of\n$\\bigcup_{c}\\widetilde M_{c}$, see Lemma 2.1 below.\n\n--------------------------------------------------------------------\n2.D.\\;Pair–codegree bound\n--------------------------------------------------------------------\nLemma 2.1.  \nFor all distinct $x,y\\in\\Gamma$\n\\[\n\\operatorname{codeg}_{\\bigcup_{c}M_{c}}(x,y)\\le 1,\\qquad\n\\operatorname{codeg}_{\\bigcup_{c}\\widetilde M_{c}}(x,y)\\le 1.\n\\tag{2.4}\n\\]\n\n\\emph{Proof.}\nWrite $x=\\gamma_{j,a}$ and $y=\\gamma_{k,b}$ with\n$1\\le j,k\\le r$ (if one of the vertices is $\\omega$ the statement is\nobvious, because $\\omega$ lies in exactly one edge of each\n$\\widetilde M_{c}$).  Suppose $x,y$ lie together in\n$E_{c,i}$ and $E_{c',i'}$ with $c,c',i,i'\\in\\mathbb F_{n}$.  Matching\n$M_{c}$ forces\n\\[\na\\equiv i+(j-1)c,\\qquad\nb\\equiv i+(k-1)c\\pmod n,\n\\]\nwhile $M_{c'}$ yields\n\\[\na\\equiv i'+(j-1)c',\\qquad\nb\\equiv i'+(k-1)c'\\pmod n.\n\\]\nSubtracting the two equations for $a$ (and similarly for $b$) we obtain\n\\[\n(j-1)(c-c')\\equiv 0,\\qquad\n(k-1)(c-c')\\equiv 0\\pmod n.\n\\]\nBecause $1\\le j,k\\le r\\le p-1<n$ and $p$ is prime, neither $(j-1)$ nor\n$(k-1)$ vanishes modulo $n$.  Hence $c\\equiv c'$ and therefore also\n$i\\equiv i'$.  So $E_{c,i}=E_{c',i'}$, proving the claim for the\n$M_{c}$.\n\nFor $\\widetilde M_{c}$ the argument is identical except that at most one\nvertex can be $\\omega$.  If neither $x$ nor $y$ equals $\\omega$ the same\nfield computation applies; if one of them is $\\omega$ then the pair\n$\\{\\omega,y\\}$ appears only in $F^{\\star}_{c}$ and never elsewhere.\n\\hfill$\\square$\n\n--------------------------------------------------------------------\n2.E.\\;The $\\boldsymbol{V=rn}$ case\n--------------------------------------------------------------------\nWrite $L=d\\,n+t$ with $0\\le d\\le n$ and $0\\le t<n$.\nTake the $d$ matchings $M_{0},\\dots ,M_{d-1}$ and then add $t$\npairwise disjoint edges from $M_{d}$; denote the two parts by\n$\\mathcal B$ and $\\mathcal A$ and put\n$\\mathcal F:=\\mathcal B\\cup\\mathcal A$.\nThe very same calculations as in the graph case give\n\n(i) $|\\mathcal F|=L$,\n\n(ii) $d\\le\\deg_{\\mathcal F}(\\gamma)\\le d+1$ for every vertex,\n\n(iii) $\\operatorname{codeg}_{\\mathcal F}\\le 1$ by Lemma 2.1,\n\n(iv) every edge meets each residue class at most once, so any\n$(r+1)$ vertices contain two symbols from the same class, hence\n$\\mathcal F$ is simplex–free.\n\n--------------------------------------------------------------------\n2.F.\\;The $\\boldsymbol{V=rn+1}$ case\n--------------------------------------------------------------------\nWrite\n\\[\nL=d\\,n+t,\\qquad 0\\le d\\le n,\\;0\\le t<n .\n\\tag{2.5}\n\\]\n\n\\medskip\n\\textbf{2.F.1.\\;Small numbers of edges ($d=0$).}\n\nHere $L=t<n$.  Choose any $t$ pairwise disjoint edges of\n$\\widetilde M_{0}$ that avoid $\\omega$ and set\n$\\mathcal F:=\\mathcal A$.\nEach vertex is used at most once, hence\n$0\\le\\deg_{\\mathcal F}(\\gamma)\\le 1$, so condition (a) holds with\n$d^{\\ast}=0$.  Lemma 2.1 gives (b); simplex–freeness follows as in\n2.E.\n\n\\medskip\n\\textbf{2.F.2.\\;The general case ($d\\ge 1$).}\n\n\\emph{Step 1 – the basic $d$ matchings.}\nPut\n\\[\ns:=d-1\\qquad(0\\le s\\le d),\\qquad\n\\mathcal B:=\n      \\Bigl(\\bigcup_{c=0}^{s-1}\\widetilde M_{c}\\Bigr)\\cup\n      \\Bigl(\\bigcup_{c=s}^{\\,d-1}M_{c}\\Bigr).\n\\tag{2.6}\n\\]\nSince $|\\widetilde M_{c}|=|M_{c}|=n$, we have\n$|\\mathcal B|=d\\,n$.\nThe degrees are\n\\[\n\\deg_{\\mathcal B}(\\omega)=s=d-1,\\qquad\n\\deg_{\\mathcal B}(\\gamma_{r,rc})=\n\\begin{cases}\nd-1,&0\\le c<s=d-1,\\\\[3pt]\nd,&s\\le c\\le d-1,\n\\end{cases}\n\\qquad\n\\deg_{\\mathcal B}(\\text{all other vertices})=d .\n\\tag{2.7}\n\\]\n\n\\emph{Step 2 – adding the remaining $t$ edges.}\nIf $t>0$ choose any $t$ pairwise disjoint edges of\n$\\widetilde M_{d}$ and call the collection $\\mathcal A$.  (When $d=n$\nwe have $t=0$ and this step is void; note that $\\widetilde M_{d}$ equals\n$\\widetilde M_{0}$ as subscripts are taken modulo $n$.)\n\nPut\n\\[\n\\mathcal F:=\\mathcal B\\cup\\mathcal A .\n\\tag{2.8}\n\\]\nBecause $\\mathcal A\\subseteq\\widetilde M_{d}$ while $\\mathcal B$\ncontains only matchings with index $<d$, the edge count is\n$|\\mathcal F|=d\\,n+t=L$.\n\n\\emph{Step 3 – degree verification.}\nEach vertex lies in at most one edge of $\\mathcal A$, hence its degree\nincreases by at most $1$.  Combining this with \\eqref{2.7} yields\n\\[\nd-1\\le\\deg_{\\mathcal F}(\\gamma)\\le d+1\\qquad\\forall\\gamma\\in\\Gamma ,\n\\]\nso condition (a) holds with $d^{\\ast}=d-1$.\n\n\\emph{Step 4 – codegrees and simplex–freeness.}\nDistinct matchings are edge–disjoint and the edges of $\\mathcal A$ are\npairwise disjoint, therefore Lemma 2.1 implies\n\\[\n\\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1\n      \\quad\\forall\\gamma\\ne\\gamma'.\n\\]\nAs each edge meets each residue class at most once, any\n$(r+1)$ vertices contain two symbols of the same class, so\n$\\mathcal F$ is simplex–free.\n\n--------------------------------------------------------------------\n2.G.\\;Summary for $\\boldsymbol{r\\ge 3}$\n--------------------------------------------------------------------\nFor every admissible triple $(r,p,L)$ we have produced a simplex–free\n$r$–graph $\\mathcal F$ with \n\\[\n|\\mathcal F|=L,\\qquad\nd^{\\ast}\\le\\deg_{\\mathcal F}(\\gamma)\\le d^{\\ast}+2,\\qquad\n\\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1 .\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n3.\\;Completion of the proof\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nSections 1 and 2 together provide the required family $\\mathcal F$ for\nall $r\\ge 2$ and all $0\\le L\\le p^{2}=n^{2}$, completing the proof.\n\\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.561021",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original “triangle-free” task, the new problem is\nsubstantially harder for several independent reasons.\n\n1. Higher dimension:  \n Pairs (edges) are replaced by $r$-tuples (hyperedges); triangles\n are replaced by $(r+1)$-simplices.  One must work in uniform\n hypergraphs rather than graphs.\n\n2. Additional global constraint:  \n The family must be **exactly** of size $L$, not merely at least $L$.\n\n3. Two new regularity constraints:  \n (a) all vertex degrees must differ by at most 1,  \n (b) every pair of symbols may occur together in **at most two**\n  hyperedges.  Neither restriction is present in the original question.\n\n4. Non-constructive tools:  \n Satisfying (b) simultaneously with the size requirement can no longer\n be done by naive partitioning; one has to invoke\n the probabilistic method (Chernoff bounds and the union bound) and\n then make careful local adjustments.\n\n5. Interaction of constraints:  \n Steps that fix the size can potentially destroy the codegree bound or\n the almost-regularity, so extra work is required to show that a single\n element modification preserves all conditions.\n\n6. General theory:  \n The solution relies on hypergraph analogues of Turán’s theorem,\n probabilistic concentration, and combinatorial design\n considerations—none of which is needed for the original pair-based\n problem.\n\nThese layers of complexity make the enhanced kernel variant\nsignificantly more challenging than both the original problem and the\ncurrent kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $r\\ge 2$ be a fixed integer and let $p>r$ be a prime.  Put \n\\[\nn:=p ,\\qquad    \nV\\in\\{rn,\\;rn+1\\},\\qquad     \n\\Gamma=\\{\\gamma _1,\\gamma _2,\\dots ,\\gamma _V\\},\n\\]\nlisting the $V$ symbols so that every residue class modulo $r$\noccurs either $n$ or $n+1$ times (hence all class–sizes differ by at\nmost one).\n\nAn \\emph{$r$–link} is an unordered $r$–tuple of distinct symbols\n(equivalently, an $r$–element subset of $\\Gamma$).\nA family $\\mathcal H$ of $r$–links is \\emph{simplex–free} if it\ncontains no $(r+1)$ distinct vertices whose every $r$–subset lies in\n$\\mathcal H$ (that is, $\\mathcal H$ contains no $(r+1)$–vertex simplex\nof the complete $r$–uniform hypergraph).\n\nFor $\\gamma ,\\gamma'\\in\\Gamma$ write  \n\\[\n\\deg _{\\mathcal H}(\\gamma)=\n      \\bigl|\\{E\\in\\mathcal H:\\gamma\\in E\\}\\bigr|,\n\\qquad\n\\operatorname{codeg}_{\\mathcal H}(\\gamma,\\gamma')=\n      \\bigl|\\{E\\in\\mathcal H:\\{\\gamma ,\\gamma'\\}\\subset E\\}\\bigr|\n      \\qquad(\\gamma\\ne\\gamma').\n\\]\n\nProve that for \\emph{every} integer \n\\[\n0\\le L\\le n^{2}=p^{2}\n\\]\nthere exists a simplex–free family $\\mathcal F$ of exactly $L$ $r$–links\nsatisfying  \n\n\\[\n\\text{\\rm(a)}\\;\\exists d\\in\\mathbb Z\\text{ such that }\n         d\\le\\deg _{\\mathcal F}(\\gamma)\\le d+2\n         \\quad\\forall\\gamma\\in\\Gamma, \n\\qquad\n\\text{\\rm(b)}\\;\n      \\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1\n      \\quad\\forall\\gamma\\ne\\gamma'\\in\\Gamma .\n\\]\n\n(The construction must work simultaneously for every admissible $L$\nand every fixed $r$; the \\emph{only} restriction on the prime is $p>r$.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "All arithmetic takes place in the finite field $\\mathbb F_{p}$\n(recall $n=p$).  All degrees and codegrees refer to the ultimately\nconstructed family $\\mathcal F$.\n\nWe separate the proof into the graph case $r=2$\n(Section 1) and the genuine $r$–uniform case $r\\ge 3$\n(Section 2).  Every symbol of $\\Gamma$ is stored together with the pair\n\\[\n(\\text{class},\\text{index})\\;=\\;(j,i)\\qquad\n(1\\le j\\le r,\\;i\\in\\mathbb F_{n})\n\\]\nexcept for one optional extra vertex $\\omega$ that occurs only when\n$V=rn+1$ and is declared to have class $j=r$ but no index.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.\\;Graphs ($r=2$)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n--------------------------------------------------------------------\n1.A.\\;Vertex partition and perfect matchings\n--------------------------------------------------------------------\nWrite\n\\[\nP_{1}:=\\{\\gamma_{0},\\dots ,\\gamma_{n-1}\\},\\qquad\nP_{2}^{0}:=\\{\\eta_{0},\\dots ,\\eta_{n-1}\\},\n\\]\nand in the case $V=2n+1$ add the extra vertex $\\omega$ to obtain\n$P_{2}:=P_{2}^{0}\\cup\\{\\omega\\}$; otherwise set $P_{2}:=P_{2}^{0}$.\nHence $\\Gamma=P_{1}\\cup P_{2}$ and the graph we shall construct will be\nbipartite with these two colour classes.\n\nFor $c\\in\\mathbb F_{n}$ define the perfect matching\n\\[\nM_{c}:=\\bigl\\{\\gamma_{i}\\,\\eta_{\\,i+c}:i\\in\\mathbb F_{n}\\bigr\\}.\n\\tag{1.1}\n\\]\nIf $\\omega$ is present, replace the edge\n$\\gamma_{c}\\eta_{\\,c+c}$ by $\\gamma_{c}\\omega$ and denote the resulting\nmatching by $\\widetilde M_{c}$.\n\nObserve the following facts.\n\n(i)  Distinct matchings are edge–disjoint.\n\n(ii) Every vertex of $P_{1}$ (respectively $P_{2}$) lies in exactly one\nedge of each matching $M_{c}$, hence exactly one edge of each\n$\\widetilde M_{c}$.\n\n(iii) Because the graph is bipartite, it is triangle–free.\n\n--------------------------------------------------------------------\n1.B.\\;Edge supply\n--------------------------------------------------------------------\nWrite\n\\[\nL=d\\,n+t,\\qquad 0\\le d\\le n,\\;0\\le t<n .\n\\tag{1.2}\n\\]\nWe distinguish the even and odd orders of the graph.\n\n\\medskip\n\\emph{1.B.1.\\;The case $V=2n$.}\n\nTake the first $d$ perfect matchings\n\\[\n\\mathcal B:=M_{0}\\cup M_{1}\\cup\\dots\\cup M_{d-1};\n\\qquad|\\mathcal B|=d\\,n .\n\\]\nIf $t>0$ choose any $t$ \\emph{pairwise disjoint} edges of $M_{d}$,\ncollect them in $\\mathcal A$, and finally put\n\\[\n\\mathcal F:=\\mathcal B\\cup\\mathcal A .\n\\]\n\n\\medskip\n\\emph{1.B.2.\\;The case $V=2n+1$.}\n\nProceed exactly as above but use the modified matchings:\n\\[\n\\mathcal B:=\\widetilde M_{0}\\cup\\dots\\cup\\widetilde M_{d-1},\\qquad\n\\mathcal A\\subseteq\\widetilde M_{d},\\qquad\n\\mathcal F:=\\mathcal B\\cup\\mathcal A .\n\\]\n(If $d=n$ then $t=0$ and $\\mathcal A=\\varnothing$; the index\n$d$ coincides with $0$ modulo $n$, so the notation\n$\\widetilde M_{d}$ really means $\\widetilde M_{0}$, but nothing is\nadded in this extreme case.)\n\n--------------------------------------------------------------------\n1.C.\\;Verifying the requirements\n--------------------------------------------------------------------\nDegree spread:\neach vertex participates in all edges of exactly $d$ matchings and in at\nmost one additional edge from $\\mathcal A$; hence\n\\[\nd\\le\\deg_{\\mathcal F}(\\gamma)\\le d+1\n\\quad\\forall\\gamma\\in\\Gamma .\n\\]\nCondition (a) therefore holds (with the same $d$).\n\nCodegree:\ntwo vertices are together in an edge if and only if they are endpoints\nof the unique edge of a \\emph{single} matching, so\n$\\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1$\nfor all $\\gamma\\ne\\gamma'$.\n\nTriangle–freeness:\nevery edge connects a vertex of $P_{1}$ to a vertex of $P_{2}$, so the\ngraph is bipartite and therefore contains no triangles.\n\nEdge count:\n$|\\mathcal F|=d\\,n+t=L$ by construction.\n\nHence the graph case is complete.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n2.\\;$r$–uniform hypergraphs ($\\boldsymbol{r\\ge 3}$)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n--------------------------------------------------------------------\n2.A.\\;Canonical $r$–partition of the vertex set\n--------------------------------------------------------------------\nFor $1\\le j\\le r-1$ set\n\\[\nP_{j}:=\\{\\gamma_{j,i}:i\\in\\mathbb F_{n}\\},\\qquad\nP_{r}^{0}:=\\{\\gamma_{r,i}:i\\in\\mathbb F_{n}\\}.\n\\]\nIf $V=rn+1$ add $\\omega$ and put\n$P_{r}:=P_{r}^{0}\\cup\\{\\omega\\}$; otherwise $P_{r}:=P_{r}^{0}$.\nEach class $P_{j}\\;(1\\le j\\le r)$ has size $n$ except possibly\n$P_{r}$, which has size $n$ or $n+1$.\n\n--------------------------------------------------------------------\n2.B.\\;Perfect $r$–matchings\n--------------------------------------------------------------------\nFor $c\\in\\mathbb F_{n}$ and $i\\in\\mathbb F_{n}$ define\n\\[\nE_{c,i}:=\\bigl\\{\\gamma_{1,i},\\gamma_{2,i+c},\\dots ,\n                \\gamma_{r,i+(r-1)c}\\bigr\\},\\qquad\nM_{c}:=\\{E_{c,i}:i\\in\\mathbb F_{n}\\}.\n\\tag{2.1}\n\\]\nThe $M_{c}$ are pairwise edge–disjoint perfect $r$–matchings and\n\\[\n\\bigl|\\bigcup_{c\\in\\mathbb F_{n}}M_{c}\\bigr|=n^{2}.\n\\tag{2.2}\n\\]\n\n--------------------------------------------------------------------\n2.C.\\;Modification in the one–extra–vertex case\n--------------------------------------------------------------------\n(Needed only when $V=rn+1$.)  For $c\\in\\mathbb F_{n}$ put  \n\\[\nF_{c}:=E_{c,c},\\quad\nF_{c}^{\\star}:=(F_{c}\\setminus\\{\\gamma_{r,rc}\\})\\cup\\{\\omega\\},\\quad\n\\widetilde M_{c}:=(M_{c}\\setminus\\{F_{c}\\})\\cup\\{F_{c}^{\\star}\\}.\n\\tag{2.3}\n\\]\nThus $|\\widetilde M_{c}|=n$ and every unordered pair of vertices appears\ntogether in \\emph{at most one} edge of\n$\\bigcup_{c}\\widetilde M_{c}$, see Lemma 2.1 below.\n\n--------------------------------------------------------------------\n2.D.\\;Pair–codegree bound\n--------------------------------------------------------------------\nLemma 2.1.  \nFor all distinct $x,y\\in\\Gamma$\n\\[\n\\operatorname{codeg}_{\\bigcup_{c}M_{c}}(x,y)\\le 1,\\qquad\n\\operatorname{codeg}_{\\bigcup_{c}\\widetilde M_{c}}(x,y)\\le 1.\n\\tag{2.4}\n\\]\n\n\\emph{Proof.}\nWrite $x=\\gamma_{j,a}$ and $y=\\gamma_{k,b}$ with\n$1\\le j,k\\le r$ (if one of the vertices is $\\omega$ the statement is\nobvious, because $\\omega$ lies in exactly one edge of each\n$\\widetilde M_{c}$).  Suppose $x,y$ lie together in\n$E_{c,i}$ and $E_{c',i'}$ with $c,c',i,i'\\in\\mathbb F_{n}$.  Matching\n$M_{c}$ forces\n\\[\na\\equiv i+(j-1)c,\\qquad\nb\\equiv i+(k-1)c\\pmod n,\n\\]\nwhile $M_{c'}$ yields\n\\[\na\\equiv i'+(j-1)c',\\qquad\nb\\equiv i'+(k-1)c'\\pmod n.\n\\]\nSubtracting the two equations for $a$ (and similarly for $b$) we obtain\n\\[\n(j-1)(c-c')\\equiv 0,\\qquad\n(k-1)(c-c')\\equiv 0\\pmod n.\n\\]\nBecause $1\\le j,k\\le r\\le p-1<n$ and $p$ is prime, neither $(j-1)$ nor\n$(k-1)$ vanishes modulo $n$.  Hence $c\\equiv c'$ and therefore also\n$i\\equiv i'$.  So $E_{c,i}=E_{c',i'}$, proving the claim for the\n$M_{c}$.\n\nFor $\\widetilde M_{c}$ the argument is identical except that at most one\nvertex can be $\\omega$.  If neither $x$ nor $y$ equals $\\omega$ the same\nfield computation applies; if one of them is $\\omega$ then the pair\n$\\{\\omega,y\\}$ appears only in $F^{\\star}_{c}$ and never elsewhere.\n\\hfill$\\square$\n\n--------------------------------------------------------------------\n2.E.\\;The $\\boldsymbol{V=rn}$ case\n--------------------------------------------------------------------\nWrite $L=d\\,n+t$ with $0\\le d\\le n$ and $0\\le t<n$.\nTake the $d$ matchings $M_{0},\\dots ,M_{d-1}$ and then add $t$\npairwise disjoint edges from $M_{d}$; denote the two parts by\n$\\mathcal B$ and $\\mathcal A$ and put\n$\\mathcal F:=\\mathcal B\\cup\\mathcal A$.\nThe very same calculations as in the graph case give\n\n(i) $|\\mathcal F|=L$,\n\n(ii) $d\\le\\deg_{\\mathcal F}(\\gamma)\\le d+1$ for every vertex,\n\n(iii) $\\operatorname{codeg}_{\\mathcal F}\\le 1$ by Lemma 2.1,\n\n(iv) every edge meets each residue class at most once, so any\n$(r+1)$ vertices contain two symbols from the same class, hence\n$\\mathcal F$ is simplex–free.\n\n--------------------------------------------------------------------\n2.F.\\;The $\\boldsymbol{V=rn+1}$ case\n--------------------------------------------------------------------\nWrite\n\\[\nL=d\\,n+t,\\qquad 0\\le d\\le n,\\;0\\le t<n .\n\\tag{2.5}\n\\]\n\n\\medskip\n\\textbf{2.F.1.\\;Small numbers of edges ($d=0$).}\n\nHere $L=t<n$.  Choose any $t$ pairwise disjoint edges of\n$\\widetilde M_{0}$ that avoid $\\omega$ and set\n$\\mathcal F:=\\mathcal A$.\nEach vertex is used at most once, hence\n$0\\le\\deg_{\\mathcal F}(\\gamma)\\le 1$, so condition (a) holds with\n$d^{\\ast}=0$.  Lemma 2.1 gives (b); simplex–freeness follows as in\n2.E.\n\n\\medskip\n\\textbf{2.F.2.\\;The general case ($d\\ge 1$).}\n\n\\emph{Step 1 – the basic $d$ matchings.}\nPut\n\\[\ns:=d-1\\qquad(0\\le s\\le d),\\qquad\n\\mathcal B:=\n      \\Bigl(\\bigcup_{c=0}^{s-1}\\widetilde M_{c}\\Bigr)\\cup\n      \\Bigl(\\bigcup_{c=s}^{\\,d-1}M_{c}\\Bigr).\n\\tag{2.6}\n\\]\nSince $|\\widetilde M_{c}|=|M_{c}|=n$, we have\n$|\\mathcal B|=d\\,n$.\nThe degrees are\n\\[\n\\deg_{\\mathcal B}(\\omega)=s=d-1,\\qquad\n\\deg_{\\mathcal B}(\\gamma_{r,rc})=\n\\begin{cases}\nd-1,&0\\le c<s=d-1,\\\\[3pt]\nd,&s\\le c\\le d-1,\n\\end{cases}\n\\qquad\n\\deg_{\\mathcal B}(\\text{all other vertices})=d .\n\\tag{2.7}\n\\]\n\n\\emph{Step 2 – adding the remaining $t$ edges.}\nIf $t>0$ choose any $t$ pairwise disjoint edges of\n$\\widetilde M_{d}$ and call the collection $\\mathcal A$.  (When $d=n$\nwe have $t=0$ and this step is void; note that $\\widetilde M_{d}$ equals\n$\\widetilde M_{0}$ as subscripts are taken modulo $n$.)\n\nPut\n\\[\n\\mathcal F:=\\mathcal B\\cup\\mathcal A .\n\\tag{2.8}\n\\]\nBecause $\\mathcal A\\subseteq\\widetilde M_{d}$ while $\\mathcal B$\ncontains only matchings with index $<d$, the edge count is\n$|\\mathcal F|=d\\,n+t=L$.\n\n\\emph{Step 3 – degree verification.}\nEach vertex lies in at most one edge of $\\mathcal A$, hence its degree\nincreases by at most $1$.  Combining this with \\eqref{2.7} yields\n\\[\nd-1\\le\\deg_{\\mathcal F}(\\gamma)\\le d+1\\qquad\\forall\\gamma\\in\\Gamma ,\n\\]\nso condition (a) holds with $d^{\\ast}=d-1$.\n\n\\emph{Step 4 – codegrees and simplex–freeness.}\nDistinct matchings are edge–disjoint and the edges of $\\mathcal A$ are\npairwise disjoint, therefore Lemma 2.1 implies\n\\[\n\\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1\n      \\quad\\forall\\gamma\\ne\\gamma'.\n\\]\nAs each edge meets each residue class at most once, any\n$(r+1)$ vertices contain two symbols of the same class, so\n$\\mathcal F$ is simplex–free.\n\n--------------------------------------------------------------------\n2.G.\\;Summary for $\\boldsymbol{r\\ge 3}$\n--------------------------------------------------------------------\nFor every admissible triple $(r,p,L)$ we have produced a simplex–free\n$r$–graph $\\mathcal F$ with \n\\[\n|\\mathcal F|=L,\\qquad\nd^{\\ast}\\le\\deg_{\\mathcal F}(\\gamma)\\le d^{\\ast}+2,\\qquad\n\\operatorname{codeg}_{\\mathcal F}(\\gamma,\\gamma')\\le 1 .\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n3.\\;Completion of the proof\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nSections 1 and 2 together provide the required family $\\mathcal F$ for\nall $r\\ge 2$ and all $0\\le L\\le p^{2}=n^{2}$, completing the proof.\n\\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.460558",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original “triangle-free” task, the new problem is\nsubstantially harder for several independent reasons.\n\n1. Higher dimension:  \n Pairs (edges) are replaced by $r$-tuples (hyperedges); triangles\n are replaced by $(r+1)$-simplices.  One must work in uniform\n hypergraphs rather than graphs.\n\n2. Additional global constraint:  \n The family must be **exactly** of size $L$, not merely at least $L$.\n\n3. Two new regularity constraints:  \n (a) all vertex degrees must differ by at most 1,  \n (b) every pair of symbols may occur together in **at most two**\n  hyperedges.  Neither restriction is present in the original question.\n\n4. Non-constructive tools:  \n Satisfying (b) simultaneously with the size requirement can no longer\n be done by naive partitioning; one has to invoke\n the probabilistic method (Chernoff bounds and the union bound) and\n then make careful local adjustments.\n\n5. Interaction of constraints:  \n Steps that fix the size can potentially destroy the codegree bound or\n the almost-regularity, so extra work is required to show that a single\n element modification preserves all conditions.\n\n6. General theory:  \n The solution relies on hypergraph analogues of Turán’s theorem,\n probabilistic concentration, and combinatorial design\n considerations—none of which is needed for the original pair-based\n problem.\n\nThese layers of complexity make the enhanced kernel variant\nsignificantly more challenging than both the original problem and the\ncurrent kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}