(* Title: HOL/Proofs/Extraction/Pigeonhole.thy Author: Stefan Berghofer, TU Muenchen *) section ‹The pigeonhole principle› theory Pigeonhole imports Util "HOL-Library.Realizers" "HOL-Library.Code_Target_Numeral" 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"}. 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 then have "Suc 0 ≤ Suc 0 ∧ 0 < Suc 0 ∧ f (Suc 0) = f 0" by simp then show ?case by iprover next case (Suc n) have r: "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)" for k 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 "?f i ≤ n" if "i ≤ Suc n" for i proof - from that have 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 then have "f i ≤ Suc n" by (rule Suc) ultimately show ?thesis by simp qed then have "∃i j. i ≤ Suc n ∧ j < i ∧ ?f i = ?f j" by (rule Suc) ultimately show ?case .. next case (Suc k) from search [OF nat_eq_dec] show ?case proof assume "∃j<Suc k. f (Suc k) = f j" then show ?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" then have "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 then show ?thesis using i and j by (rule Suc) qed qed then show ?thesis by (iprover intro: le_SucI) qed qed 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 [OF nat_eq_dec] show ?case proof assume "∃j < Suc (Suc n). f (Suc (Suc n)) = f j" then show ?case by (iprover intro: le_refl) next assume "¬ (∃j < Suc (Suc n). f (Suc (Suc n)) = f j)" then have 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 then have "∃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. › instantiation nat :: default begin definition "default = (0::nat)" instance .. end instantiation prod :: (default, default) default begin definition "default = (default, default)" instance .. end definition "test n u = pigeonhole (nat_of_integer n) (λm. m - 1)" definition "test' n u = pigeonhole_slow (nat_of_integer n) (λm. m - 1)" definition "test'' u = pigeonhole 8 (List.nth [0, 1, 2, 3, 4, 5, 6, 3, 7, 8])" ML_val "timeit (@{code test} 10)" ML_val "timeit (@{code test'} 10)" ML_val "timeit (@{code test} 20)" ML_val "timeit (@{code test'} 20)" ML_val "timeit (@{code test} 25)" ML_val "timeit (@{code test'} 25)" ML_val "timeit (@{code test} 500)" ML_val "timeit @{code test''}" end