Theory Pigeonhole

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theory Pigeonhole
imports EfficientNat
begin

(*  Title:      HOL/Extraction/Pigeonhole.thy
    ID:         $Id: Pigeonhole.thy,v 1.4 2005/09/23 14:05:42 nipkow Exp $
    Author:     Stefan Berghofer, TU Muenchen
*)

header {* The pigeonhole principle *}

theory Pigeonhole imports EfficientNat begin

text {*
We formalize two proofs of the pigeonhole principle, which lead
to extracted programs of quite different complexity. The original
formalization of these proofs in {\sc Nuprl} is due to
Aleksey Nogin \cite{Nogin-ENTCS-2000}.

We need decidability of equality on natural numbers:
*}

lemma nat_eq_dec: "!!n::nat. m = n ∨ m ≠ n"
  apply (induct m)
  apply (case_tac n)
  apply (case_tac [3] n)
  apply (simp only: nat.simps, iprover?)+
  done

text {*
We can decide whether an array @{term "f"} of length @{term "l"}
contains an element @{term "x"}.
*}

lemma search: "(∃j<(l::nat). (x::nat) = f j) ∨ ¬ (∃j<l. x = f j)"
proof (induct l)
  case 0
  have "¬ (∃j<0. x = f j)"
  proof
    assume "∃j<0. x = f j"
    then obtain j where j: "j < (0::nat)" by iprover
    thus "False" by simp
  qed
  thus ?case ..
next
  case (Suc l)
  thus ?case
  proof
    assume "∃j<l. x = f j"
    then obtain j where j: "j < l"
      and eq: "x = f j" by iprover
    from j have "j < Suc l" by simp
    with eq show ?case by iprover
  next
    assume nex: "¬ (∃j<l. x = f j)"
    from nat_eq_dec show ?case
    proof
      assume eq: "x = f l"
      have "l < Suc l" by simp
      with eq show ?case by iprover
    next
      assume neq: "x ≠ f l"
      have "¬ (∃j<Suc l. x = f j)"
      proof
        assume "∃j<Suc l. x = f j"
        then obtain j where j: "j < Suc l"
          and eq: "x = f j" by iprover
        show False
        proof cases
          assume "j = l"
          with eq have "x = f l" by simp
          with neq show False ..
        next
          assume "j ≠ l"
          with j have "j < l" by simp
          with nex and eq show False by iprover
        qed
      qed
      thus ?case ..
    qed
  qed
qed

text {*
This proof yields a polynomial program.
*}

theorem pigeonhole:
  "!!f. (!!i. i ≤ Suc n ==> f i ≤ n) ==> ∃i j. i ≤ Suc n ∧ j < i ∧ f i = f j"
proof (induct n)
  case 0
  hence "Suc 0 ≤ Suc 0 ∧ 0 < Suc 0 ∧ f (Suc 0) = f 0" by simp
  thus ?case by iprover
next
  case (Suc n)
  {
    fix k
    have
      "k ≤ Suc (Suc n) ==>
      (!!i j. Suc k ≤ i ==> i ≤ Suc (Suc n) ==> j < i ==> f i ≠ f j) ==>
      (∃i j. i ≤ k ∧ j < i ∧ f i = f j)"
    proof (induct k)
      case 0
      let ?f = "λi. if f i = Suc n then f (Suc (Suc n)) else f i"
      have "¬ (∃i j. i ≤ Suc n ∧ j < i ∧ ?f i = ?f j)"
      proof
        assume "∃i j. i ≤ Suc n ∧ j < i ∧ ?f i = ?f j"
        then obtain i j where i: "i ≤ Suc n" and j: "j < i"
          and f: "?f i = ?f j" by iprover
        from j have i_nz: "Suc 0 ≤ i" by simp
        from i have iSSn: "i ≤ Suc (Suc n)" by simp
        have S0SSn: "Suc 0 ≤ Suc (Suc n)" by simp
        show False
        proof cases
          assume fi: "f i = Suc n"
          show False
          proof cases
            assume fj: "f j = Suc n"
            from i_nz and iSSn and j have "f i ≠ f j" by (rule 0)
            moreover from fi have "f i = f j"
              by (simp add: fj [symmetric])
            ultimately show ?thesis ..
          next
            from i and j have "j < Suc (Suc n)" by simp
            with S0SSn and le_refl have "f (Suc (Suc n)) ≠ f j"
              by (rule 0)
            moreover assume "f j ≠ Suc n"
            with fi and f have "f (Suc (Suc n)) = f j" by simp
            ultimately show False ..
          qed
        next
          assume fi: "f i ≠ Suc n"
          show False
          proof cases
            from i have "i < Suc (Suc n)" by simp
            with S0SSn and le_refl have "f (Suc (Suc n)) ≠ f i"
              by (rule 0)
            moreover assume "f j = Suc n"
            with fi and f have "f (Suc (Suc n)) = f i" by simp
            ultimately show False ..
          next
            from i_nz and iSSn and j
            have "f i ≠ f j" by (rule 0)
            moreover assume "f j ≠ Suc n"
            with fi and f have "f i = f j" by simp
            ultimately show False ..
          qed
        qed
      qed
      moreover have "!!i. i ≤ Suc n ==> ?f i ≤ n"
      proof -
        fix i assume "i ≤ Suc n"
        hence i: "i < Suc (Suc n)" by simp
        have "f (Suc (Suc n)) ≠ f i"
          by (rule 0) (simp_all add: i)
        moreover have "f (Suc (Suc n)) ≤ Suc n"
          by (rule Suc) simp
        moreover from i have "i ≤ Suc (Suc n)" by simp
        hence "f i ≤ Suc n" by (rule Suc)
        ultimately show "?thesis i"
          by simp
      qed
      hence "∃i j. i ≤ Suc n ∧ j < i ∧ ?f i = ?f j"
        by (rule Suc)
      ultimately show ?case ..
    next
      case (Suc k)
      from search show ?case
      proof
        assume "∃j<Suc k. f (Suc k) = f j"
        thus ?case by (iprover intro: le_refl)
      next
        assume nex: "¬ (∃j<Suc k. f (Suc k) = f j)"
        have "∃i j. i ≤ k ∧ j < i ∧ f i = f j"
        proof (rule Suc)
          from Suc show "k ≤ Suc (Suc n)" by simp
          fix i j assume k: "Suc k ≤ i" and i: "i ≤ Suc (Suc n)"
            and j: "j < i"
          show "f i ≠ f j"
          proof cases
            assume eq: "i = Suc k"
            show ?thesis
            proof
              assume "f i = f j"
              hence "f (Suc k) = f j" by (simp add: eq)
              with nex and j and eq show False by iprover
            qed
          next
            assume "i ≠ Suc k"
            with k have "Suc (Suc k) ≤ i" by simp
            thus ?thesis using i and j by (rule Suc)
          qed
        qed
        thus ?thesis by (iprover intro: le_SucI)
      qed
    qed
  }
  note r = this
  show ?case by (rule r) simp_all
qed

text {*
The following proof, although quite elegant from a mathematical point of view,
leads to an exponential program:
*}

theorem pigeonhole_slow:
  "!!f. (!!i. i ≤ Suc n ==> f i ≤ n) ==> ∃i j. i ≤ Suc n ∧ j < i ∧ f i = f j"
proof (induct n)
  case 0
  have "Suc 0 ≤ Suc 0" ..
  moreover have "0 < Suc 0" ..
  moreover from 0 have "f (Suc 0) = f 0" by simp
  ultimately show ?case by iprover
next
  case (Suc n)
  from search show ?case
  proof
    assume "∃j < Suc (Suc n). f (Suc (Suc n)) = f j"
    thus ?case by (iprover intro: le_refl)
  next
    assume "¬ (∃j < Suc (Suc n). f (Suc (Suc n)) = f j)"
    hence nex: "∀j < Suc (Suc n). f (Suc (Suc n)) ≠ f j" by iprover
    let ?f = "λi. if f i = Suc n then f (Suc (Suc n)) else f i"
    have "!!i. i ≤ Suc n ==> ?f i ≤ n"
    proof -
      fix i assume i: "i ≤ Suc n"
      show "?thesis i"
      proof (cases "f i = Suc n")
        case True
        from i and nex have "f (Suc (Suc n)) ≠ f i" by simp
        with True have "f (Suc (Suc n)) ≠ Suc n" by simp
        moreover from Suc have "f (Suc (Suc n)) ≤ Suc n" by simp
        ultimately have "f (Suc (Suc n)) ≤ n" by simp
        with True show ?thesis by simp
      next
        case False
        from Suc and i have "f i ≤ Suc n" by simp
        with False show ?thesis by simp
      qed
    qed
    hence "∃i j. i ≤ Suc n ∧ j < i ∧ ?f i = ?f j" by (rule Suc)
    then obtain i j where i: "i ≤ Suc n" and ji: "j < i" and f: "?f i = ?f j"
      by iprover
    have "f i = f j"
    proof (cases "f i = Suc n")
      case True
      show ?thesis
      proof (cases "f j = Suc n")
        assume "f j = Suc n"
        with True show ?thesis by simp
      next
        assume "f j ≠ Suc n"
        moreover from i ji nex have "f (Suc (Suc n)) ≠ f j" by simp
        ultimately show ?thesis using True f by simp
      qed
    next
      case False
      show ?thesis
      proof (cases "f j = Suc n")
        assume "f j = Suc n"
        moreover from i nex have "f (Suc (Suc n)) ≠ f i" by simp
        ultimately show ?thesis using False f by simp
      next
        assume "f j ≠ Suc n"
        with False f show ?thesis by simp
      qed
    qed
    moreover from i have "i ≤ Suc (Suc n)" by simp
    ultimately show ?thesis using ji by iprover
  qed
qed

extract pigeonhole pigeonhole_slow

text {*
The programs extracted from the above proofs look as follows:
@{thm [display] pigeonhole_def}
@{thm [display] pigeonhole_slow_def}
The program for searching for an element in an array is
@{thm [display,eta_contract=false] search_def}
The correctness statement for @{term "pigeonhole"} is
@{thm [display] pigeonhole_correctness [no_vars]}

In order to analyze the speed of the above programs,
we generate ML code from them.
*}

consts_code
  arbitrary :: "nat × nat" ("{* (0::nat, 0::nat) *}")

code_module PH
contains
  test = "λn. pigeonhole n (λm. m - 1)"
  test' = "λn. pigeonhole_slow n (λm. m - 1)"
  sel = "op !"

ML "timeit (fn () => PH.test 10)"
ML "timeit (fn () => PH.test' 10)"
ML "timeit (fn () => PH.test 20)"
ML "timeit (fn () => PH.test' 20)"
ML "timeit (fn () => PH.test 25)"
ML "timeit (fn () => PH.test' 25)"
ML "timeit (fn () => PH.test 500)"

ML "PH.pigeonhole 8 (PH.sel [0,1,2,3,4,5,6,3,7,8])"

end

lemma nat_eq_dec:

  m = nmn

lemma search:

  (∃j<l. x = f j) ∨ ¬ (∃j<l. x = f j)

theorem pigeonhole:

  (!!i. i ≤ Suc n ==> f in) ==> ∃i j. i ≤ Suc nj < if i = f j

theorem pigeonhole_slow:

  (!!i. i ≤ Suc n ==> f in) ==> ∃i j. i ≤ Suc nj < if i = f j