{
  "index": "1996-B-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "Define a \\textbf{selfish} set to be a set which has its own\ncardinality (number of elements) as an element. Find, with proof, the\nnumber of subsets of $\\{1, 2, \\ldots, n\\}$ which are \\textit{minimal}\nselfish sets, that is, selfish sets none of whose proper subsets is selfish.",
  "solution": "Let $[n]$ denote the set $\\{1,2,\\ldots,n\\}$, and let $f_n$ denote the\nnumber of minimal selfish subsets of $[n]$.  Then the number of\nminimal selfish subsets of $[n]$ not containing $n$ is equal to\n$f_{n-1}$.  On the other hand, for any minimal selfish subset of $[n]$\ncontaining $n$, by subtracting 1 from each element, and then taking\naway the element $n-1$ from the set, we obtain a minimal selfish\nsubset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish\nset).  Conversely, any minimal selfish subset of $[n-2]$ gives rise to\na minimal selfish subset of $[n]$ containing $n$ by the inverse\nprocedure.  Hence the number of minimal selfish subsets of $[n]$\ncontaining $n$ is $f_{n-2}$.  Thus we obtain $f_n=f_{n-1}+f_{n-2}$.\nSince $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th\nterm of the Fibonacci sequence.",
  "vars": [
    "f_n",
    "f_n-1",
    "f_n-2",
    "F_n"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f_n": "minimalcount",
        "f_n-1": "minimalcountprev",
        "f_n-2": "minimalcountprevprev",
        "F_n": "fiboterm",
        "n": "setsize"
      },
      "question": "Define a \\textbf{selfish} set to be a set which has its own\ncardinality (number of elements) as an element. Find, with proof, the\nnumber of subsets of $\\{1, 2, \\ldots, setsize\\}$ which are \\textit{minimal}\nselfish sets, that is, selfish sets none of whose proper subsets is selfish.",
      "solution": "Let $[setsize]$ denote the set $\\{1,2,\\ldots,setsize\\}$, and let minimalcount denote the\nnumber of minimal selfish subsets of $[setsize]$.  Then the number of\nminimal selfish subsets of $[setsize]$ not containing $setsize$ is equal to\nminimalcountprev.  On the other hand, for any minimal selfish subset of $[setsize]$\ncontaining $setsize$, by subtracting 1 from each element, and then taking\naway the element $setsize-1$ from the set, we obtain a minimal selfish\nsubset of $[setsize-2]$ (since $1$ and $setsize$ cannot both occur in a selfish\nset).  Conversely, any minimal selfish subset of $[setsize-2]$ gives rise to\na minimal selfish subset of $[setsize]$ containing $setsize$ by the inverse\nprocedure.  Hence the number of minimal selfish subsets of $[setsize]$\ncontaining $setsize$ is minimalcountprevprev.  Thus we obtain minimalcount=minimalcountprev+minimalcountprevprev.\nSince $f_1=f_2=1$, we have minimalcount=fiboterm, where fiboterm denotes the $setsize$th\nterm of the Fibonacci sequence."
    },
    "descriptive_long_confusing": {
      "map": {
        "f_n": "tortoise",
        "f_n-1": "armadillo",
        "f_n-2": "crocodile",
        "F_n": "kangaroo",
        "n": "sandstone"
      },
      "question": "Define a \\textbf{selfish} set to be a set which has its own\ncardinality (number of elements) as an element. Find, with proof, the\nnumber of subsets of $\\{1, 2, \\ldots, sandstone\\}$ which are \\textit{minimal}\nselfish sets, that is, selfish sets none of whose proper subsets is selfish.",
      "solution": "Let $[sandstone]$ denote the set $\\{1,2,\\ldots,sandstone\\}$, and let $tortoise$ denote the\nnumber of minimal selfish subsets of $[sandstone]$.  Then the number of\nminimal selfish subsets of $[sandstone]$ not containing $sandstone$ is equal to\n$armadillo$.  On the other hand, for any minimal selfish subset of $[sandstone]$\ncontaining $sandstone$, by subtracting $1$ from each element, and then taking\naway the element $sandstone-1$ from the set, we obtain a minimal selfish\nsubset of $[sandstone-2]$ (since $1$ and $sandstone$ cannot both occur in a selfish\nset).  Conversely, any minimal selfish subset of $[sandstone-2]$ gives rise to\na minimal selfish subset of $[sandstone]$ containing $sandstone$ by the inverse\nprocedure.  Hence the number of minimal selfish subsets of $[sandstone]$\ncontaining $sandstone$ is $crocodile$.  Thus we obtain $tortoise=armadillo+crocodile$.\nSince $f_1=f_2=1$, we have $tortoise=kangaroo$, where $kangaroo$ denotes the $sandstone$th\nterm of the Fibonacci sequence."
    },
    "descriptive_long_misleading": {
      "map": {
        "f_n": "altruisticcount",
        "f_n-1": "unselfishprev",
        "f_n-2": "unselfishprevtwo",
        "F_n": "randomsequence",
        "n": "undefinedindex"
      },
      "question": "Define a \\textbf{selfish} set to be a set which has its own\ncardinality (number of elements) as an element. Find, with proof, the\nnumber of subsets of $\\{1, 2, \\ldots, undefinedindex\\}$ which are \\textit{minimal}\nselfish sets, that is, selfish sets none of whose proper subsets is selfish.",
      "solution": "Let $[undefinedindex]$ denote the set $\\{1,2,\\ldots,undefinedindex\\}$, and let $altruisticcount$ denote the\nnumber of minimal selfish subsets of $[undefinedindex]$.  Then the number of\nminimal selfish subsets of $[undefinedindex]$ not containing $undefinedindex$ is equal to\n$unselfishprev$.  On the other hand, for any minimal selfish subset of $[undefinedindex]$\ncontaining $undefinedindex$, by subtracting 1 from each element, and then taking\naway the element $undefinedindex-1$ from the set, we obtain a minimal selfish\nsubset of $[undefinedindex-2]$ (since $1$ and $undefinedindex$ cannot both occur in a selfish\nset).  Conversely, any minimal selfish subset of $[undefinedindex-2]$ gives rise to\na minimal selfish subset of $[undefinedindex]$ containing $undefinedindex$ by the inverse\nprocedure.  Hence the number of minimal selfish subsets of $[undefinedindex]$\ncontaining $undefinedindex$ is $unselfishprevtwo$.  Thus we obtain $altruisticcount=unselfishprev+unselfishprevtwo$.\nSince $f_1=f_2=1$, we have $altruisticcount=randomsequence$, where $randomsequence$ denotes the $undefinedindex$th\nterm of the Fibonacci sequence."
    },
    "garbled_string": {
      "map": {
        "f_n": "mheldtuv",
        "f_n-1": "yahcijfq",
        "f_n-2": "wzptrokl",
        "F_n": "blsjdaow",
        "n": "qzxwvtnp"
      },
      "question": "Define a \\textbf{selfish} set to be a set which has its own\ncardinality (number of elements) as an element. Find, with proof, the\nnumber of subsets of $\\{1, 2, \\ldots, qzxwvtnp\\}$ which are \\textit{minimal}\nselfish sets, that is, selfish sets none of whose proper subsets is selfish.",
      "solution": "Let $[qzxwvtnp]$ denote the set $\\{1,2,\\ldots,qzxwvtnp\\}$, and let $mheldtuv$ denote the\nnumber of minimal selfish subsets of $[qzxwvtnp]$.  Then the number of\nminimal selfish subsets of $[qzxwvtnp]$ not containing $qzxwvtnp$ is equal to\n$yahcijfq$.  On the other hand, for any minimal selfish subset of $[qzxwvtnp]$\ncontaining $qzxwvtnp$, by subtracting 1 from each element, and then taking\naway the element $qzxwvtnp-1$ from the set, we obtain a minimal selfish\nsubset of $[qzxwvtnp-2]$ (since $1$ and $qzxwvtnp$ cannot both occur in a selfish\nset).  Conversely, any minimal selfish subset of $[qzxwvtnp-2]$ gives rise to\na minimal selfish subset of $[qzxwvtnp]$ containing $qzxwvtnp$ by the inverse\nprocedure.  Hence the number of minimal selfish subsets of $[qzxwvtnp]$\ncontaining $qzxwvtnp$ is $wzptrokl$.  Thus we obtain $mheldtuv=yahcijfq+wzptrokl$.\nSince $f_1=f_2=1$, we have $mheldtuv=blsjdaow$, where $blsjdaow$ denotes the $qzxwvtnp$th\nterm of the Fibonacci sequence."
    },
    "kernel_variant": {
      "question": "For a positive integer $n$ let\n$$O_n=\bigl\\{1,3,5,\\ldots,2n-1\\bigr\\}$$\nbe the set of the first $n$ odd positive integers.  Call a subset $S\\subseteq O_n$ \nself–aware if it contains its own cardinality, i.e.\n$|S|\\in S.$\nSuch a set is minimal if none of its proper subsets is self–aware.\nDetermine, with proof, the number $f_n$ of minimal self–aware subsets of $O_n$. ",
      "solution": "We wish to count, for each positive integer n, the number f_n of minimal self–aware subsets S of\n   O_n={1,3,5,…,2n−1},\nwhere “self–aware” means |S|∈S, and “minimal” means no proper subset of S is self–aware.\n\n1.  A self–aware S must have size k with k∈S.  Since S⊆O_n consists of odd integers, k must be odd and satisfy 1≤k≤n.\n\n2.  Minimality forces S to contain no smaller odd m<k.  Indeed, if m<k and m∈S, then since |S|=k>m we can pick any other m−1 elements of S to form a proper subset T of S of size m still containing m, so T would be self–aware—a contradiction.  Hence S∩{1,3,…,k−2}=∅.\n\n3.  Thus a minimal S of size k must consist of k itself together with k−1 elements chosen from the larger odds\n     {k+2, k+4,…,2n−1}.\n\n4.  How many choices are there?  The set {k+2,k+4,…,2n−1} has\n     ((2n−1)−(k+2))/2+1 = (2n−k−1)/2 = n − (k+1)/2\n   elements.  We must choose k−1 of these to complete S.  Hence for each odd k≤n the number of minimal self–aware S of size k is\n     C(n−(k+1)/2, k−1).\n\n5.  Summing over all odd k gives\n     f_n = ∑_{1≤k≤n, k odd} C\\bigl(n−(k+1)/2, k−1\\bigr).\n   Setting k=2j+1 (so j=0,1,…,⌊(n−1)/2⌋) we have (k+1)/2=j+1 and k−1=2j, giving the closed form\n     f_n = ∑_{j=0}^{⌊(n−1)/2⌋} C(n−(j+1), 2j)\n          = ∑_{j=0}^{⌊(n−1)/2⌋} C(n−j−1, 2j).\n\n6.  Quick checks:\n   n=1: j=0 only, C(1−0−1,0)=C(0,0)=1;\n   n=2: j=0 only, C(2−0−1,0)=1;\n   n=3: j=0,1 gives 1+0=1;\n   n=4: j=0,1 gives 1+1=2;\n   n=5: 1+3+0=4;\n   etc.\n\nHence the number of minimal self–aware subsets of O_n is\n   f_n = ∑_{j=0}^{⌊(n−1)/2⌋} C(n−j−1,2j)\n as claimed.",
      "_meta": {
        "core_steps": [
          "Split minimal-selfish subsets of [n] by whether they contain the element n.",
          "Note: absence of n ⇒ same object is a minimal-selfish subset of [n−1] (count f_{n−1}).",
          "Presence of n ⇒ via the bijection S ↦ (S−1) \\ {n−1}, relate them to minimal-selfish subsets of [n−2] (count f_{n−2}); this uses that 1 and n cannot coexist.",
          "Combine the two cases to obtain the recurrence f_n = f_{n−1} + f_{n−2} with bases f_1 = f_2 = 1, yielding Fibonacci numbers."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Which distinguished element is used for the initial case split (currently the maximum element).",
            "original": "n"
          },
          "slot2": {
            "description": "Indexing of the universal set; any contiguous block of integers of length n would serve the same purpose.",
            "original": "{1,2,…,n}"
          },
          "slot3": {
            "description": "Exact translation applied in the bijection (presently ‘subtract 1’ and delete the image of n); any fixed shift that removes the pivot element and shortens the universe by 2 keeps the argument intact.",
            "original": "subtract 1 from every element and delete n−1"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}