blog.pramp.com アルゴリズムイントロダクション 第三版 15章 動的計画法 2020.7.22 奥村圭祐 20年度 東工大Defago研・裏輪行 DYNAMIC PROGRAMMING 

/25 2 公開用の断り書き https://tanukifont.com/ フォントはたぬきフォントを使用させていただいています。 本資料は研究室内の非公式の輪読会で使用したもので、 対象書籍は「アルゴリズム イントロダクション」の第三版(通称CLRS)になります。 スライド中の擬似コード(英語)は英語版のスクリーンショットを使用しています。 厳密性に欠いているところはありますがご容赦ください。 間違い等ございましたら okumura.k[at]coord.c.titech.ac.jp まで、 ご連絡いただけると幸いです。 

/25 15.1 ロッド切出し 15.2 連鎖行列積 15.3 動的計画法の基本要素 15.4 最長共通部分列 15.5 最適2分探索木 15章の構成 目標: 動的計画法が適応できそうだ！と見積もりが立てられるようになること 3 

/25 Wikipediaより 復習: 分割統治法 部分問題の解を組合わせて元の解を得る 結合 部分問題を再帰的に解く 統治 問題をいくつかの部分問題に分割 分割 

/25 5 ケーススタディ1: ロッド切出し問題 1本の金属棒からいくつかの金属棒を切り出すとき 収益最大化を達成するような切断箇所を決定したい 長さ 1 2 3 4 5 6 7 8 9 10 価格 1 5 8 9 10 17 17 20 24 30 i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== pi AAAB6nicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48V7Qe0oWy2k3bpZhN2N0IJ/QlePCji1V/kzX/jts1BWx8MPN6bYWZekAiujet+O4W19Y3NreJ2aWd3b/+gfHjU0nGqGDZZLGLVCahGwSU2DTcCO4lCGgUC28H4dua3n1BpHstHM0nQj+hQ8pAzaqz0kPR5v1xxq+4cZJV4OalAjka//NUbxCyNUBomqNZdz02Mn1FlOBM4LfVSjQllYzrErqWSRqj9bH7qlJxZZUDCWNmShszV3xMZjbSeRIHtjKgZ6WVvJv7ndVMTXvsZl0lqULLFojAVxMRk9jcZcIXMiIkllClubyVsRBVlxqZTsiF4yy+vklat6l1Ua/eXlfpNHkcRTuAUzsGDK6jDHTSgCQyG8Ayv8OYI58V5dz4WrQUnnzmGP3A+fwBVro3U - 棒の長さ n AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlpuyXK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH2F+M9g== 入力 - 価格表 出力 切り出した棒を価格表に従って販売して得ることのできる収益の最大値 例 n = 4 AAAB6nicbVDLSgNBEOyNrxhfUY9eBoPgKezGgF6EoBePEc0DkiXMTmaTIbOzy0yvEEI+wYsHRbz6Rd78GyfJHjSxoKGo6qa7K0ikMOi6305ubX1jcyu/XdjZ3ds/KB4eNU2casYbLJaxbgfUcCkUb6BAyduJ5jQKJG8Fo9uZ33ri2ohYPeI44X5EB0qEglG00oO6rvaKJbfszkFWiZeREmSo94pf3X7M0ogrZJIa0/HcBP0J1SiY5NNCNzU8oWxEB7xjqaIRN/5kfuqUnFmlT8JY21JI5urviQmNjBlHge2MKA7NsjcT//M6KYZX/kSoJEWu2GJRmEqCMZn9TfpCc4ZybAllWthbCRtSTRnadAo2BG/55VXSrJS9i3Llvlqq3WRx5OEETuEcPLiEGtxBHRrAYADP8ApvjnRenHfnY9Gac7KZY/gD5/MHzpWNew== 9 1 8 5 5 1 5 1 最適解 1 1 1 1 

/25 6 アルゴリズムの設計指針 最適な分割を仮定 n = i1 + i2 + · · · + ik AAACBHicbVDLSgMxFM3UV62vUZfdBIsgCGWmCroRim5cVrAPaIeSSTNtaCYzJHeEMnThxl9x40IRt36EO//GdDoLbT2Qy8m595Dc48eCa3Ccb6uwsrq2vlHcLG1t7+zu2fsHLR0lirImjUSkOj7RTHDJmsBBsE6sGAl9wdr++GbWbz8wpXkk72ESMy8kQ8kDTgkYqW+XJb7CvO/iU1NrpvYGEejsNu7bFafqZMDLxM1JBeVo9O0v46ZJyCRQQbTuuk4MXkoUcCrYtNRLNIsJHZMh6xoqSci0l2ZLTPGxUQY4iJQ5EnCm/nakJNR6EvpmMiQw0ou9mfhfr5tAcOmlXMYJMEnnDwWJwBDhWSJ4wBWjICaGEKq4+SumI6IIBZNbyYTgLq68TFq1qntWrd2dV+rXeRxFVEZH6AS56ALV0S1qoCai6BE9o1f0Zj1ZL9a79TEfLVi55xD9gfX5AwFRlSU= 分割された棒の長さ rn = pi1 + pi2 + · · · + pik AAACEnicbZDLSgMxFIbPeK31VnXpJlgERSgzVdCNUHTjsoK9QDsMmTTThmYyQ5IRytBncOOruHGhiFtX7nwbM+0I2vpD4OM/5yQ5vx9zprRtf1kLi0vLK6uFteL6xubWdmlnt6miRBLaIBGPZNvHinImaEMzzWk7lhSHPqctf3id1Vv3VCoWiTs9iqkb4r5gASNYG8srHUtPoEsUeynznDE6mVI1o24v0urHGY69Utmu2BOheXByKEOuulf6NDeQJKRCE46V6jh2rN0US80Ip+NiN1E0xmSI+7RjUOCQKjedrDRGh8bpoSCS5giNJu7viRSHSo1C33SGWA/UbC0z/6t1Eh1cuCkTcaKpINOHgoQjHaEsH9RjkhLNRwYwkcz8FZEBlphok2LRhODMrjwPzWrFOa1Ub8/Ktas8jgLswwEcgQPnUIMbqEMDCDzAE7zAq/VoPVtv1vu0dcHKZ/bgj6yPbwt3m9c= 最適な収益 個に分割 k AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOl5rhfrrhVdw6ySrycVCBHo1/+6g1ilkYoDRNU667nJsbPqDKcCZyWeqnGhLIxHWLXUkkj1H42P3RKzqwyIGGsbElD5urviYxGWk+iwHZG1Iz0sjcT//O6qQmv/YzLJDUo2WJRmApiYjL7mgy4QmbExBLKFLe3EjaiijJjsynZELzll1dJu1b1Lqq15mWlfpPHUYQTOIVz8OAK6nAHDWgBA4RneIU359F5cd6dj0VrwclnjuEPnM8f09OM8w== rn = max (pn, r1 + rn 1, r2 + rn 2, . . . , rn 1 + r1) 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 部分構造最適性に着目 長さが と の2つの金属棒に切断, それぞれの金属棒を最適に 分割することで得られる総収入 n i AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4sSRV0GPRi8eK9gPaUDbbTbt0swm7E6GE/gQvHhTx6i/y5r9x2+agrQ8GHu/NMDMvSKQw6Lrfzsrq2vrGZmGruL2zu7dfOjhsmjjVjDdYLGPdDqjhUijeQIGStxPNaRRI3gpGt1O/9cS1EbF6xHHC/YgOlAgFo2ilB3UueqWyW3FnIMvEy0kZctR7pa9uP2ZpxBUySY3peG6CfkY1Cib5pNhNDU8oG9EB71iqaMSNn81OnZBTq/RJGGtbCslM/T2R0ciYcRTYzoji0Cx6U/E/r5NieO1nQiUpcsXmi8JUEozJ9G/SF5ozlGNLKNPC3krYkGrK0KZTtCF4iy8vk2a14l1UqveX5dpNHkcBjuEEzsCDK6jBHdShAQwG8Ayv8OZI58V5dz7mrStOPnMEf+B8/gAGqI2g i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== 切断なし 上式をもっと簡略化 rn = max 1in (pi + rn i) AAACHnicbVDLSgMxFM3UV62vqks3wSJUxDJTFd0IRTcuK9gHdMqQSTNtaCYzJnfEMvRL3PgrblwoIrjSvzF9LLR6INzDOfdyc48fC67Btr+szNz8wuJSdjm3srq2vpHf3KrrKFGU1WgkItX0iWaCS1YDDoI1Y8VI6AvW8PuXI79xx5TmkbyBQczaIelKHnBKwEhe/kR5Ep9jNyT3Xuq4gt1ijsdFDk0JoBh7HB9g5aXykA9dxbs92PfyBbtkj4H/EmdKCmiKqpf/cDsRTUImgQqidcuxY2inRAGngg1zbqJZTGifdFnLUElCptvp+Lwh3jNKBweRMk8CHqs/J1ISaj0IfdMZEujpWW8k/ue1EgjO2imXcQJM0smiIBEYIjzKCne4YhTEwBBCFTd/xbRHFKFgEs2ZEJzZk/+SernkHJXK18eFysU0jizaQbuoiBx0iiroClVRDVH0gJ7QC3q1Hq1n6816n7RmrOnMNvoF6/MbSKqhXA== 分割を左端から切り出された長さ の金属棒と 右側に残る長さ の部分とみなす i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== n i AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4sSRV0GPRi8eK9gPaUDbbTbt0swm7E6GE/gQvHhTx6i/y5r9x2+agrQ8GHu/NMDMvSKQw6Lrfzsrq2vrGZmGruL2zu7dfOjhsmjjVjDdYLGPdDqjhUijeQIGStxPNaRRI3gpGt1O/9cS1EbF6xHHC/YgOlAgFo2ilB3UueqWyW3FnIMvEy0kZctR7pa9uP2ZpxBUySY3peG6CfkY1Cib5pNhNDU8oG9EB71iqaMSNn81OnZBTq/RJGGtbCslM/T2R0ciYcRTYzoji0Cx6U/E/r5NieO1nQiUpcsXmi8JUEozJ9G/SF5ozlGNLKNPC3krYkGrK0KZTtCF4iy8vk2a14l1UqveX5dpNHkcBjuEEzsCDK6jBHdShAQwG8Ayv8OZI58V5dz7mrStOPnMEf+B8/gAGqI2g 

/25 7 再帰的トップダウン型実現 rn = max 1in (pi + rn i) AAACHnicbVDLSgMxFM3UV62vqks3wSJUxDJTFd0IRTcuK9gHdMqQSTNtaCYzJnfEMvRL3PgrblwoIrjSvzF9LLR6INzDOfdyc48fC67Btr+szNz8wuJSdjm3srq2vpHf3KrrKFGU1WgkItX0iWaCS1YDDoI1Y8VI6AvW8PuXI79xx5TmkbyBQczaIelKHnBKwEhe/kR5Ep9jNyT3Xuq4gt1ijsdFDk0JoBh7HB9g5aXykA9dxbs92PfyBbtkj4H/EmdKCmiKqpf/cDsRTUImgQqidcuxY2inRAGngg1zbqJZTGifdFnLUElCptvp+Lwh3jNKBweRMk8CHqs/J1ISaj0IfdMZEujpWW8k/ue1EgjO2imXcQJM0smiIBEYIjzKCne4YhTEwBBCFTd/xbRHFKFgEs2ZEJzZk/+SernkHJXK18eFysU0jizaQbuoiBx0iiroClVRDVH0gJ7QC3q1Hq1n6816n7RmrOnMNvoF6/MbSKqhXA== 棒の長さがゼロであれば最適解はゼロ このアルゴリズムは効率的だろうか？ 価格表 棒の長さ を返す rn = pi1 + pi2 + · · · + pik AAACEnicbZDLSgMxFIbPeK31VnXpJlgERSgzVdCNUHTjsoK9QDsMmTTThmYyQ5IRytBncOOruHGhiFtX7nwbM+0I2vpD4OM/5yQ5vx9zprRtf1kLi0vLK6uFteL6xubWdmlnt6miRBLaIBGPZNvHinImaEMzzWk7lhSHPqctf3id1Vv3VCoWiTs9iqkb4r5gASNYG8srHUtPoEsUeynznDE6mVI1o24v0urHGY69Utmu2BOheXByKEOuulf6NDeQJKRCE46V6jh2rN0US80Ip+NiN1E0xmSI+7RjUOCQKjedrDRGh8bpoSCS5giNJu7viRSHSo1C33SGWA/UbC0z/6t1Eh1cuCkTcaKpINOHgoQjHaEsH9RjkhLNRwYwkcz8FZEBlphok2LRhODMrjwPzWrFOa1Ub8/Ktas8jgLswwEcgQPnUIMbqEMDCDzAE7zAq/VoPVtv1vu0dcHKZ/bgj6yPbwt3m9c= 

/25 8 4 Cut-Rod(p, 4) AAACAXicbVDLSgNBEJyNrxhfq14EL4NBiKBhNwb0GMzFYxTzgCSE2dnZZMjsg5leMSzx4q948aCIV//Cm3/jJNmDJhY0FFXddHc5keAKLOvbyCwtr6yuZddzG5tb2zvm7l5DhbGkrE5DEcqWQxQTPGB14CBYK5KM+I5gTWdYnfjNeyYVD4M7GEWs65N+wD1OCWipZx50gD1A0lEUV2M4uw3dcSE6xeWTnpm3itYUeJHYKcmjFLWe+dVxQxr7LAAqiFJt24qgmxAJnAo2znVixSJCh6TP2poGxGeqm0w/GONjrbjYC6WuAPBU/T2REF+pke/oTp/AQM17E/E/rx2Dd9lNeBDFwAI6W+TFAkOIJ3Fgl0tGQYw0IVRyfSumAyIJBR1aTodgz7+8SBqlon1eLN2U85WrNI4sOkRHqIBsdIEq6BrVUB1R9Iie0St6M56MF+Pd+Ji1Zox0Zh/9gfH5A8amlck= 3 0 2 1 0 0 1 0 1 0 0 n = 4 AAAB6nicbVDLSgNBEOyNrxhfUY9eBoPgKezGgF6EoBePEc0DkiXMTmaTIbOzy0yvEEI+wYsHRbz6Rd78GyfJHjSxoKGo6qa7K0ikMOi6305ubX1jcyu/XdjZ3ds/KB4eNU2casYbLJaxbgfUcCkUb6BAyduJ5jQKJG8Fo9uZ33ri2ohYPeI44X5EB0qEglG00oO6rvaKJbfszkFWiZeREmSo94pf3X7M0ogrZJIa0/HcBP0J1SiY5NNCNzU8oWxEB7xjqaIRN/5kfuqUnFmlT8JY21JI5urviQmNjBlHge2MKA7NsjcT//M6KYZX/kSoJEWu2GJRmEqCMZn9TfpCc4ZybAllWthbCRtSTRnadAo2BG/55VXSrJS9i3Llvlqq3WRx5OEETuEcPLiEGtxBHRrAYADP8ApvjnRenHfnY9Gac7KZY/gD5/MHzpWNew== の場合 2 1 0 0 葉の数: 2n 1 AAAB7nicbVBNS8NAEJ34WetX1aOXxSJ4sSRV0GPRi8cK9gPaWDbbTbt0swm7E6GE/ggvHhTx6u/x5r9x2+agrQ8GHu/NMDMvSKQw6Lrfzsrq2vrGZmGruL2zu7dfOjhsmjjVjDdYLGPdDqjhUijeQIGStxPNaRRI3gpGt1O/9cS1EbF6wHHC/YgOlAgFo2ilVvUxU+fepFcquxV3BrJMvJyUIUe9V/rq9mOWRlwhk9SYjucm6GdUo2CST4rd1PCEshEd8I6likbc+Nns3Ak5tUqfhLG2pZDM1N8TGY2MGUeB7YwoDs2iNxX/8zophtd+JlSSIldsvihMJcGYTH8nfaE5Qzm2hDIt7K2EDammDG1CRRuCt/jyMmlWK95FpXp/Wa7d5HEU4BhO4Aw8uIIa3EEdGsBgBM/wCm9O4rw4787HvHXFyWeO4A+czx+dbI8Y 問題を解くのに に関して 指数時間必要とする n AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlpuyXK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH2F+M9g== 何度も同じ部分問題を解くのは無駄 各部分問題を”一度だけ”解くようにすれば効率的なはず！ 

/25 9 履歴管理を用いるトップダウン方式 保存された結果を利用 部分問題の解を保存 rn = max 1in (pi + rn i) AAACHnicbVDLSgMxFM3UV62vqks3wSJUxDJTFd0IRTcuK9gHdMqQSTNtaCYzJnfEMvRL3PgrblwoIrjSvzF9LLR6INzDOfdyc48fC67Btr+szNz8wuJSdjm3srq2vpHf3KrrKFGU1WgkItX0iWaCS1YDDoI1Y8VI6AvW8PuXI79xx5TmkbyBQczaIelKHnBKwEhe/kR5Ep9jNyT3Xuq4gt1ijsdFDk0JoBh7HB9g5aXykA9dxbs92PfyBbtkj4H/EmdKCmiKqpf/cDsRTUImgQqidcuxY2inRAGngg1zbqJZTGifdFnLUElCptvp+Lwh3jNKBweRMk8CHqs/J1ISaj0IfdMZEujpWW8k/ue1EgjO2imXcQJM0smiIBEYIjzKCne4YhTEwBBCFTd/xbRHFKFgEs2ZEJzZk/+SernkHJXK18eFysU0jizaQbuoiBx0iiroClVRDVH0gJ7QC3q1Hq1n6816n7RmrOnMNvoF6/MbSKqhXA== 履歴表を初期化 4 3 0 2 1 0 0 1 1 0 2 ⇥ n2 AAAB/3icbVDLSgNBEJyNrxhfq4IXL4tBiJewGwU9Br14jJAXZGOYnfQmQ2Znl5leIcQc/BUvHhTx6m9482+cPA6aWNBQVHXT3RUkgmt03W8rs7K6tr6R3cxtbe/s7tn7B3Udp4pBjcUiVs2AahBcQg05CmgmCmgUCGgEg5uJ33gApXksqzhMoB3RnuQhZxSN1LGP/GofkPoCQizI+5KveK+PZx077xbdKZxl4s1JnsxR6dhffjdmaQQSmaBatzw3wfaIKuRMwDjnpxoSyga0By1DJY1At0fT+8fOqVG6ThgrUxKdqfp7YkQjrYdRYDojin296E3E/7xWiuFVe8RlkiJINlsUpsLB2JmE4XS5AoZiaAhliptbHdanijI0keVMCN7iy8ukXip658XS3UW+fD2PI0uOyQkpEI9ckjK5JRVSI4w8kmfySt6sJ+vFerc+Zq0Zaz5zSP7A+vwBXiOVrQ== 計算量: 

/25 10 ボトムアップ方式 履歴表を初期化 3 2 0 1 4 ⇥ n2 AAAB/3icbVDLSgNBEJyNrxhfq4IXL4tBiJewGwU9Br14jJAXZGOYnfQmQ2Znl5leIcQc/BUvHhTx6m9482+cPA6aWNBQVHXT3RUkgmt03W8rs7K6tr6R3cxtbe/s7tn7B3Udp4pBjcUiVs2AahBcQg05CmgmCmgUCGgEg5uJ33gApXksqzhMoB3RnuQhZxSN1LGP/GofkPoCQizI+5KveK+PZx077xbdKZxl4s1JnsxR6dhffjdmaQQSmaBatzw3wfaIKuRMwDjnpxoSyga0By1DJY1At0fT+8fOqVG6ThgrUxKdqfp7YkQjrYdRYDojin296E3E/7xWiuFVe8RlkiJINlsUpsLB2JmE4XS5AoZiaAhliptbHdanijI0keVMCN7iy8ukXip658XS3UW+fD2PI0uOyQkpEI9ckjK5JRVSI4w8kmfySt6sJ+vFerc+Zq0Zaz5zSP7A+vwBXiOVrQ== 計算量: 部分問題グラフ と呼ばれる を計算 rj AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4KkkV9Fj04rGi/YA2lM12067dbMLuRCihP8GLB0W8+ou8+W/ctjlo64OBx3szzMwLEikMuu63s7K6tr6xWdgqbu/s7u2XDg6bJk414w0Wy1i3A2q4FIo3UKDk7URzGgWSt4LRzdRvPXFtRKwecJxwP6IDJULBKFrpXvcee6WyW3FnIMvEy0kZctR7pa9uP2ZpxBUySY3peG6CfkY1Cib5pNhNDU8oG9EB71iqaMSNn81OnZBTq/RJGGtbCslM/T2R0ciYcRTYzoji0Cx6U/E/r5NieOVnQiUpcsXmi8JUEozJ9G/SF5ozlGNLKNPC3krYkGrK0KZTtCF4iy8vk2a14p1XqncX5dp1HkcBjuEEzsCDS6jBLdShAQwG8Ayv8OZI58V5dz7mrStOPnMEf+B8/gBaPo3X 小さい問題から 順に解く 

/25 11 解の再構成 もともとのモチベーション 1本の金属棒からいくつかの金属棒を切り出すとき 収益最大化を達成するような切断箇所を決定したい 左端からの切出し位置 切出し位置の保存 長さ 1 2 3 4 5 6 7 価格 1 5 8 9 10 17 17 i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== pi AAAB6nicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48V7Qe0oWy2k3bpZhN2N0IJ/QlePCji1V/kzX/jts1BWx8MPN6bYWZekAiujet+O4W19Y3NreJ2aWd3b/+gfHjU0nGqGDZZLGLVCahGwSU2DTcCO4lCGgUC28H4dua3n1BpHstHM0nQj+hQ8pAzaqz0kPR5v1xxq+4cZJV4OalAjka//NUbxCyNUBomqNZdz02Mn1FlOBM4LfVSjQllYzrErqWSRqj9bH7qlJxZZUDCWNmShszV3xMZjbSeRIHtjKgZ6WVvJv7ndVMTXvsZl0lqULLFojAVxMRk9jcZcIXMiIkllClubyVsRBVlxqZTsiF4yy+vklat6l1Ua/eXlfpNHkcRTuAUzsGDK6jDHTSgCQyG8Ayv8OYI58V5dz4WrQUnnzmGP3A+fwBVro3U 練習. 下の価格表に対し の場合の 最適な切分けを求めよ n = 7 AAAB6nicbVDLSgNBEOyNrxhfUY9eBoPgKexGIV6EoBePEc0DkiXMTnqTIbOzy8ysEEI+wYsHRbz6Rd78GyfJHjSxoKGo6qa7K0gE18Z1v53c2vrG5lZ+u7Czu7d/UDw8auo4VQwbLBaxagdUo+ASG4Ybge1EIY0Cga1gdDvzW0+oNI/loxkn6Ed0IHnIGTVWepDX1V6x5JbdOcgq8TJSggz1XvGr249ZGqE0TFCtO56bGH9CleFM4LTQTTUmlI3oADuWShqh9ifzU6fkzCp9EsbKljRkrv6emNBI63EU2M6ImqFe9mbif14nNeGVP+EySQ1KtlgUpoKYmMz+Jn2ukBkxtoQyxe2thA2poszYdAo2BG/55VXSrJS9i3Ll/rJUu8niyMMJnMI5eFCFGtxBHRrAYADP8ApvjnBenHfnY9Gac7KZY/gD5/MH0yGNfg== 答え: i1 = 1, i2 = 6 AAAB+XicbZDLSsNAFIZP6q3WW9Slm8EiuJCStKJuhKIblxXsBdoQJtNJO3QyCTOTQgl9EzcuFHHrm7jzbZy2WWjrDwMf/zmHc+YPEs6Udpxvq7C2vrG5Vdwu7ezu7R/Yh0ctFaeS0CaJeSw7AVaUM0GbmmlOO4mkOAo4bQej+1m9PaZSsVg86UlCvQgPBAsZwdpYvm0z30W3yL1AzK8auPLtslNx5kKr4OZQhlwN3/7q9WOSRlRowrFSXddJtJdhqRnhdFrqpYommIzwgHYNChxR5WXzy6fozDh9FMbSPKHR3P09keFIqUkUmM4I66Fars3M/2rdVIc3XsZEkmoqyGJRmHKkYzSLAfWZpETziQFMJDO3IjLEEhNtwiqZENzlL69Cq1pxa5Xq42W5fpfHUYQTOIVzcOEa6vAADWgCgTE8wyu8WZn1Yr1bH4vWgpXPHMMfWZ8/b+KQ7w== 

/25 12 分割統治法っぽい手法で問題を解こうとするときに, 何度も同じ部分問題に遭遇する場合, その部分問題の解を保存しておき適宜利用すれば効率的 ここまでのハイライト 

/25 13 ケーススタディ2: 連鎖行列積 個の行列が与えられたとき を求めたい. A1A2 · · · An AAAB+3icbZDLSgMxFIYzXmu91bp0EyyCqzJTBV22unFZwV6gHYZMJtOGZpIhOSOW0ldx40IRt76IO9/GtJ2Ftv4Q+PjPOZyTP0wFN+C6387a+sbm1nZhp7i7t39wWDoqt43KNGUtqoTS3ZAYJrhkLeAgWDfVjCShYJ1wdDurdx6ZNlzJBxinzE/IQPKYUwLWCkrlRuDhRlDDfRopMBatWXGr7lx4FbwcKihXMyh99SNFs4RJoIIY0/PcFPwJ0cCpYNNiPzMsJXREBqxnUZKEGX8yv32Kz6wT4Vhp+yTguft7YkISY8ZJaDsTAkOzXJuZ/9V6GcTX/oTLNAMm6WJRnAkMCs+CwBHXjIIYWyBUc3srpkOiCQUbV9GG4C1/eRXatap3Ua3dX1bqN3kcBXSCTtE58tAVqqM71EQtRNETekav6M2ZOi/Ou/OxaF1z8plj9EfO5w+uvJLt n AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlpuyXK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH2F+M9g== (単純に)行列積を実行するのに必要なスカラー積の回数: 例 に対して の2通りの計算順序がある hA1, A2, A3 i AAACA3icbVDLSgMxFM3UV62vUXe6CRbBhZSZVtCFSKsblxXsA9oyZNJMG5rJDMkdoZSCG3/FjQtF3PoT7vwbM20X2nrgXg7n3Etyjx8LrsFxvq3M0vLK6lp2PbexubW9Y+/u1XWUKMpqNBKRavpEM8ElqwEHwZqxYiT0BWv4g5vUbzwwpXkk72EYs05IepIHnBIwkmcftAUL4LLiuae44hXTVmor3uvDlWfnnYIzAV4k7ozk0QxVz/5qdyOahEwCFUTrluvE0BkRBZwKNs61E81iQgekx1qGShIy3RlNbhjjY6N0cRApUxLwRP29MSKh1sPQN5Mhgb6e91LxP6+VQHDRGXEZJ8AknT4UJAJDhNNAcJcrRkEMDSFUcfNXTPtEEQomtpwJwZ0/eZHUiwW3VCjeneXL17M4sugQHaET5KJzVEa3qIpqiKJH9Ixe0Zv1ZL1Y79bHdDRjzXb20R9Ynz/ZbJW8 (A1(A2A3)) AAAB+HicbVDLSgMxFL3js9ZHR126CRah3ZSZVtBlqxuXFewD2mHIpJk2NJMZkoxQh36JGxeKuPVT3Pk3po+Fth64l8M595KbEyScKe0439bG5tb2zm5uL79/cHhUsI9P2ipOJaEtEvNYdgOsKGeCtjTTnHYTSXEUcNoJxrczv/NIpWKxeNCThHoRHgoWMoK1kXy7UGr4LjKtihp+rVz27aJTceZA68RdkiIs0fTtr/4gJmlEhSYcK9VznUR7GZaaEU6n+X6qaILJGA9pz1CBI6q8bH74FF0YZYDCWJoSGs3V3xsZjpSaRIGZjLAeqVVvJv7n9VIdXnsZE0mqqSCLh8KUIx2jWQpowCQlmk8MwUQycysiIywx0SarvAnBXf3yOmlXK26tUr2/LNZvlnHk4AzOoQQuXEEd7qAJLSCQwjO8wpv1ZL1Y79bHYnTDWu6cwh9Ynz/YJJCd ((A1A2)A3) AAAB+HicbVDLTgIxFL3jE/HBqEs3jcQENmQGTHQJunGJiTwSmEw6pUBDpzNpOyY44UvcuNAYt36KO//GArNQ8CT35uSce9PbE8ScKe0439bG5tb2zm5uL79/cHhUsI9P2ipKJKEtEvFIdgOsKGeCtjTTnHZjSXEYcNoJJrdzv/NIpWKReNDTmHohHgk2ZARrI/l2oVRq+C5q+NWyabWybxedirMAWiduRoqQoenbX/1BRJKQCk04VqrnOrH2Uiw1I5zO8v1E0RiTCR7RnqECh1R56eLwGbowygANI2lKaLRQf2+kOFRqGgZmMsR6rFa9ufif10v08NpLmYgTTQVZPjRMONIRmqeABkxSovnUEEwkM7ciMsYSE22yypsQ3NUvr5N2teLWKtX7y2L9JosjB2dwDiVw4QrqcAdNaAGBBJ7hFd6sJ+vFerc+lqMbVrZzCn9gff4A14SQnQ== , C AAAB6HicbVDLTgJBEOzFF+IL9ehlIjHxRHbRRI9ELh4hkUcCGzI79MLI7OxmZtaEEL7AiweN8eonefNvHGAPClbSSaWqO91dQSK4Nq777eQ2Nre2d/K7hb39g8Oj4vFJS8epYthksYhVJ6AaBZfYNNwI7CQKaRQIbAfj2txvP6HSPJYPZpKgH9Gh5CFn1FipUesXS27ZXYCsEy8jJchQ7xe/eoOYpRFKwwTVuuu5ifGnVBnOBM4KvVRjQtmYDrFrqaQRan+6OHRGLqwyIGGsbElDFurviSmNtJ5Ege2MqBnpVW8u/ud1UxPe+lMuk9SgZMtFYSqIicn8azLgCpkRE0soU9zeStiIKsqMzaZgQ/BWX14nrUrZuypXGtel6l0WRx7O4BwuwYMbqMI91KEJDBCe4RXenEfnxXl3PpatOSebOYU/cD5/AJczjMs= A 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AAAB/nicbVDLSgMxFL3js9bXqLhyEyyCq5Kpii6LblxWsA9oh5JJM21oJjMkGaEMBX/FjQtF3Pod7vwb03YEbT0QcjjnXu69J0gE1wbjL2dpeWV1bb2wUdzc2t7Zdff2GzpOFWV1GotYtQKimeCS1Q03grUSxUgUCNYMhjcTv/nAlOaxvDejhPkR6UseckqMlbruoYcx6hgeMY0ufn7cdUu4jKdAi8TLSQly1LruZ6cX0zRi0lBBtG57ODF+RpThVLBxsZNqlhA6JH3WtlQSO8fPpuuP0YlVeiiMlX3SoKn6uyMjkdajKLCVETEDPe9NxP+8dmrCKz/jMkkNk3Q2KEwFMjGaZIF6XDFqxMgSQhW3uyI6IIpQYxMr2hC8+ZMXSaNS9s7KlbvzUvU6j6MAR3AMp+DBJVThFmpQBwoZPMELvDqPzrPz5rzPSpecvOcA/sD5+AayR5QI 10 ⇥ 100 ⇥ 50 AAACAHicbZDLSgMxFIYzXmu9jbpw4SZYBFclUxVdFt24rGAv0A4lk2ba0ExmSM4IZejGV3HjQhG3PoY738a0HURbfwh8/OccTs4fJFIYIOTLWVpeWV1bL2wUN7e2d3bdvf2GiVPNeJ3FMtatgBouheJ1ECB5K9GcRoHkzWB4M6k3H7g2Ilb3MEq4H9G+EqFgFKzVdQ89gjsgIm6wR37wgnTdEimTqfAieDmUUK5a1/3s9GKWRlwBk9SYtkcS8DOqQTDJx8VOanhC2ZD2eduionaPn00PGOMT6/RwGGv7FOCp+3sio5ExoyiwnRGFgZmvTcz/au0Uwis/EypJgSs2WxSmEkOMJ2ngntCcgRxZoEwL+1fMBlRTBjazog3Bmz95ERqVsndWrtydl6rXeRwFdISO0Sny0CWqoltUQ3XE0Bg9oRf06jw6z86b8z5rXXLymQP0R87HN3r5lGg= 10 ⇥ 5 ⇥ 50 AAAB/HicbVDLSgNBEOz1GeNrNUcvg0HwFHajosegF48RzAOSJcxOJsmQ2QczvcKyxF/x4kERr36IN//GSbKCJhYMU1R1093lx1JodJwva2V1bX1js7BV3N7Z3du3Dw6bOkoU4w0WyUi1faq5FCFvoEDJ27HiNPAlb/njm6nfeuBKiyi8xzTmXkCHoRgIRtFIPbvkOl0UAdfk4ud3enbZqTgzkGXi5qQMOeo9+7Pbj1gS8BCZpFp3XCdGL6MKBZN8UuwmmseUjemQdwwNqZnjZbPlJ+TEKH0yiJR5IZKZ+rsjo4HWaeCbyoDiSC96U/E/r5Pg4MrLRBgnyEM2HzRIJMGITJMgfaE4Q5kaQpkSZlfCRlRRhiavognBXTx5mTSrFfesUr07L9eu8zgKcATHcAouXEINbqEODWCQwhO8wKv1aD1bb9b7vHTFyntK8AfWxzfoZpOk A.rows ⇥ A.columns ⇥ B.columns AAACLHicbVBdSwJBFJ21L7OvrR57GZKgJ9m1oB5NX3o0SA1UZHYcdXB2Zpm5W8jiD+qlvxJED0n02u9oVqUP7cKFwzn3cu49QSS4Ac+bOJmV1bX1jexmbmt7Z3fP3T+oGxVrympUCaXvAmKY4JLVgINgd5FmJAwEawTDSqo37pk2XMlbGEWsHZK+5D1OCViq41ZaIYEBh+SqoNWDGeMW8JAZ/ENbjziUy0r5W+m4ea/gTQsvA38O8mhe1Y770uoqGodMAhXEmKbvRdBOiAZOBRvnWrFhEaFD0mdNCyWxvu1k+uwYn1imi3tK25aAp+zvjYSExozCwE6ml5pFLSX/05ox9C7bCZdRDEzSmVEvFhgUTpPDXa4ZBTGygFDN7a2YDogmFGy+ORuCv/jyMqgXC/5ZoXhzni+V53Fk0RE6RqfIRxeohK5RFdUQRY/oGb2hifPkvDrvzsdsNOPMdw7Rn3I+vwBEe6l1 7500 AAAB63icbVBNSwMxEJ2tX7V+VT16CRbBU9mtlXosevFYwX5Au5Rsmm1Dk+ySZIWy9C948aCIV/+QN/+N2XYP2vpg4PHeDDPzgpgzbVz32ylsbG5t7xR3S3v7B4dH5eOTjo4SRWibRDxSvQBrypmkbcMMp71YUSwCTrvB9C7zu09UaRbJRzOLqS/wWLKQEWwyqXHtusNyxa26C6B14uWkAjlaw/LXYBSRRFBpCMda9z03Nn6KlWGE03lpkGgaYzLFY9q3VGJBtZ8ubp2jC6uMUBgpW9Kghfp7IsVC65kIbKfAZqJXvUz8z+snJrzxUybjxFBJlovChCMToexxNGKKEsNnlmCimL0VkQlWmBgbT8mG4K2+vE46tap3Va091CvN2zyOIpzBOVyCBw1owj20oA0EJvAMr/DmCOfFeXc+lq0FJ585hT9wPn8A1teNcg== 回 75000 AAAB7HicbVBNTwIxEJ3FL8Qv1KOXRmLiiXQRg0eiF4+YuEACG9ItXWjodjdt14Rs+A1ePGiMV3+QN/+NBfag4EsmeXlvJjPzgkRwbTD+dgobm1vbO8Xd0t7+weFR+fikreNUUebRWMSqGxDNBJfMM9wI1k0UI1EgWCeY3M39zhNTmsfy0UwT5kdkJHnIKTFW8hrXGONBuYKreAG0TtycVCBHa1D+6g9jmkZMGiqI1j0XJ8bPiDKcCjYr9VPNEkInZMR6lkoSMe1ni2Nn6MIqQxTGypY0aKH+nshIpPU0CmxnRMxYr3pz8T+vl5rwxs+4TFLDJF0uClOBTIzmn6MhV4waMbWEUMXtrYiOiSLU2HxKNgR39eV10q5V3atq7aFead7mcRThDM7hElxoQBPuoQUeUODwDK/w5kjnxXl3PpatBSefOYU/cD5/AEVcjaw= 回 10倍の差！ 問題: スカラー乗算回数を最小化するよう積 を括弧付けせよ A1A2 · · · An AAAB+3icbZDLSgMxFIYzXmu91bp0EyyCqzJTBV22unFZwV6gHYZMJtOGZpIhOSOW0ldx40IRt76IO9/GtJ2Ftv4Q+PjPOZyTP0wFN+C6387a+sbm1nZhp7i7t39wWDoqt43KNGUtqoTS3ZAYJrhkLeAgWDfVjCShYJ1wdDurdx6ZNlzJBxinzE/IQPKYUwLWCkrlRuDhRlDDfRopMBatWXGr7lx4FbwcKihXMyh99SNFs4RJoIIY0/PcFPwJ0cCpYNNiPzMsJXREBqxnUZKEGX8yv32Kz6wT4Vhp+yTguft7YkISY8ZJaDsTAkOzXJuZ/9V6GcTX/oTLNAMm6WJRnAkMCs+CwBHXjIIYWyBUc3srpkOiCQUbV9GG4C1/eRXatap3Ua3dX1bqN3kcBXSCTtE58tAVqqM71EQtRNETekav6M2ZOi/Ou/OxaF1z8plj9EfO5w+uvJLt 

/25 14 シラミツブシに調べる方法 個の行列の列に対する異なる括弧付けの個数 n AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlpuyXK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH2F+M9g== P(n) AAAB63icbVBNSwMxEJ2tX7V+VT16CRahXspuK+ix6MVjBfsB7VKyabYNTbJLkhXK0r/gxYMiXv1D3vw3Zts9aOuDgcd7M8zMC2LOtHHdb6ewsbm1vVPcLe3tHxwelY9POjpKFKFtEvFI9QKsKWeStg0znPZiRbEIOO0G07vM7z5RpVkkH80spr7AY8lCRrDJpFZVXg7LFbfmLoDWiZeTCuRoDctfg1FEEkGlIRxr3ffc2PgpVoYRTuelQaJpjMkUj2nfUokF1X66uHWOLqwyQmGkbEmDFurviRQLrWcisJ0Cm4le9TLxP6+fmPDGT5mME0MlWS4KE45MhLLH0YgpSgyfWYKJYvZWRCZYYWJsPCUbgrf68jrp1Gteo1Z/uKo0b/M4inAG51AFD66hCffQgjYQmMAzvMKbI5wX5935WLYWnHzmFP7A+fwBPOGNtQ== : P(n) = ( 1 if n = 1 Pn 1 k=1 P(k)P(n k) otherwise 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 どこで分割するか それぞれの部分積の括弧付けが とりうるパターンの組合せ ⌦ ⇣ 4n/n3/2 ⌘ AAACC3icbVA9SwNBEN3zM8avU0ubxSDEJrlLBC1FGzsVjAr5Ym8zlyzZ2zt254RwpLfxr9hYKGLrH7Dz37iJKTTxwcDjvZndmRckUhj0vC9nbn5hcWk5t5JfXVvf2HS3tm9MnGoONR7LWN8FzIAUCmooUMJdooFFgYTboH828m/vQRsRq2scJNCMWFeJUHCGVmq7e42LCLqsISHEYv6wpWiZqlZWLVeG+YYW3R4etN2CV/LGoLPEn5ACmeCy7X42OjFPI1DIJTOm7nsJNjOmUXAJ9t3UQMJ4n3WhbqliEZhmNr5lSPet0qFhrG0ppGP190TGImMGUWA7I4Y9M+2NxP+8eorhcTMTKkkRFP/5KEwlxZiOgqEdoYGjHFjCuBZ2V8p7TDOONr68DcGfPnmW3FRKfrVUuTosnJxO4siRXbJHisQnR+SEnJNLUiOcPJAn8kJenUfn2Xlz3n9a55zJzA75A+fjG8HMmPU= この数列は の速さで増加 シラミツブシに調べる戦略は連鎖行列の最適括弧付けには適さない. 部分問題の構造を利用する. 

/25 15 アルゴリズムの設計指針 最適な括弧付けを仮定 AiAi+1 · · · Aj AAACAHicbVDLSgMxFM3UV62vURcu3ASLIAhlpgq6bHXjsoJ9QDsMmUymjc0kQ5IRyjAbf8WNC0Xc+hnu/BvTdhZaPXDh5Jx7yb0nSBhV2nG+rNLS8srqWnm9srG5tb1j7+51lEglJm0smJC9ACnCKCdtTTUjvUQSFAeMdIPx9dTvPhCpqOB3epIQL0ZDTiOKkTaSbx80fQqbfkZP3RzCAQ6FVuZ979tVp+bMAP8StyBVUKDl25+DUOA0JlxjhpTqu06ivQxJTTEjeWWQKpIgPEZD0jeUo5goL5sdkMNjo4QwEtIU13Cm/pzIUKzUJA5MZ4z0SC16U/E/r5/q6NLLKE9STTiefxSlDGoBp2nAkEqCNZsYgrCkZleIR0girE1mFROCu3jyX9Kp19yzWv32vNq4KuIog0NwBE6ACy5AA9yAFmgDDHLwBF7Aq/VoPVtv1vu8tWQVM/vgF6yPb2K6lP4= を と分割 ((Ai · · · Ak)(Ak+1 · · · Aj)) AAACEHicbZDLSsNAFIYn9VbrLerSzWARW4SSVEGXrW5cVrAXaEOYTKbtmMkkzEyEEvoIbnwVNy4UcevSnW/jtA2orT8M/HznHM6c34sZlcqyvozc0vLK6lp+vbCxubW9Y+7utWSUCEyaOGKR6HhIEkY5aSqqGOnEgqDQY6TtBVeTevueCEkjfqtGMXFCNOC0TzFSGrnmcalUdynsYT9SEtbdoAw1SIMTe/wD78pl1yxaFWsquGjszBRBpoZrfvb8CCch4QozJGXXtmLlpEgoihkZF3qJJDHCARqQrrYchUQ66fSgMTzSxIf9SOjHFZzS3xMpCqUchZ7uDJEayvnaBP5X6yaqf+GklMeJIhzPFvUTBlUEJ+lAnwqCFRtpg7Cg+q8QD5FAWOkMCzoEe/7kRdOqVuzTSvXmrFi7zOLIgwNwCErABuegBq5BAzQBBg/gCbyAV+PReDbejPdZa87IZvbBHxkf30HTmjc= どちらも最適な括弧付けになっていなければならない 必要なスカラー積の回数 m[i, j] = m[i, k] + m[k + 1, j] + pi 1pkpj AAACFnicbZDLSsNAFIYnXmu9RV26GSyCUFuSKuhGKLpxWcFeIA1hMp2200ySYWYilNCncOOruHGhiFtx59s4TbPQ1h8GPv5zDmfO73NGpbKsb2NpeWV1bb2wUdzc2t7ZNff2WzJOBCZNHLNYdHwkCaMRaSqqGOlwQVDoM9L2g5tpvf1AhKRxdK/GnLghGkS0TzFS2vLMSujQUzhy4RXMKHBhWVNQtjO3DLmX0oo94V7AvZFnlqyqlQkugp1DCeRqeOZXtxfjJCSRwgxJ6dgWV26KhKKYkUmxm0jCEQ7QgDgaIxQS6abZWRN4rJ0e7MdCv0jBzP09kaJQynHo684QqaGcr03N/2pOovqXbkojnigS4dmifsKgiuE0I9ijgmDFxhoQFlT/FeIhEggrnWRRh2DPn7wIrVrVPqvW7s5L9es8jgI4BEfgBNjgAtTBLWiAJsDgETyDV/BmPBkvxrvxMWtdMvKZA/BHxucPmu6b8A== pk 1 ⇥ pk AAAB/HicbVDLSsNAFJ3UV62vaJduBovgxpJUQZdFNy4r2Ae0IUymk3bo5MHMjRBC/BU3LhRx64e482+ctFlo64GBc8+5l3vneLHgCizr26isrW9sblW3azu7e/sH5uFRT0WJpKxLIxHJgUcUEzxkXeAg2CCWjASeYH1vdlv4/UcmFY/CB0hj5gRkEnKfUwJacs167GazczsfAQ+YwkWVu2bDalpz4FVil6SBSnRc82s0jmgSsBCoIEoNbSsGJyMSOBUsr40SxWJCZ2TChpqGRK9ysvnxOT7Vyhj7kdQvBDxXf09kJFAqDTzdGRCYqmWvEP/zhgn4107GwzgBFtLFIj8RGCJcJIHHXDIKItWEUMn1rZhOiSQUdF41HYK9/OVV0ms17Ytm6/6y0b4p46iiY3SCzpCNrlAb3aEO6iKKUvSMXtGb8WS8GO/Gx6K1YpQzdfQHxucPsj+UzQ== Ak AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4KkkV9Fj14rGi/YA2lM120y7dbMLuRCihP8GLB0W8+ou8+W/ctjlo64OBx3szzMwLEikMuu63s7K6tr6xWdgqbu/s7u2XDg6bJk414w0Wy1i3A2q4FIo3UKDk7URzGgWSt4LR7dRvPXFtRKwecZxwP6IDJULBKFrp4bo36pXKbsWdgSwTLydlyFHvlb66/ZilEVfIJDWm47kJ+hnVKJjkk2I3NTyhbEQHvGOpohE3fjY7dUJOrdInYaxtKSQz9fdERiNjxlFgOyOKQ7PoTcX/vE6K4ZWfCZWkyBWbLwpTSTAm079JX2jOUI4toUwLeythQ6opQ5tO0YbgLb68TJrVindeqd5flGs3eRwFOIYTOAMPLqEGd1CHBjAYwDO8wpsjnRfn3fmYt644+cwR/IHz+QMRHI2n として のサイズを 部分問題のコストの和 に新規で必要なスカラー積の回数 ((Ai · · · Ak)(Ak+1 · · · Aj)) AAACEHicbZDLSsNAFIYn9VbrLerSzWARW4SSVEGXrW5cVrAXaEOYTKbtmMkkzEyEEvoIbnwVNy4UcevSnW/jtA2orT8M/HznHM6c34sZlcqyvozc0vLK6lp+vbCxubW9Y+7utWSUCEyaOGKR6HhIEkY5aSqqGOnEgqDQY6TtBVeTevueCEkjfqtGMXFCNOC0TzFSGrnmcalUdynsYT9SEtbdoAw1SIMTe/wD78pl1yxaFWsquGjszBRBpoZrfvb8CCch4QozJGXXtmLlpEgoihkZF3qJJDHCARqQrrYchUQ66fSgMTzSxIf9SOjHFZzS3xMpCqUchZ7uDJEayvnaBP5X6yaqf+GklMeJIhzPFvUTBlUEJ+lAnwqCFRtpg7Cg+q8QD5FAWOkMCzoEe/7kRdOqVuzTSvXmrFi7zOLIgwNwCErABuegBq5BAzQBBg/gCbyAV+PReDbejPdZa87IZvbBHxkf30HTmjc= m[i, j] = ( 0 i = j mini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 再帰的定義 

/25 16 履歴管理を用いるトップダウン方式 初期化 既知の値であれば結果を再利用 O(n 3) AAAB7XicbVBNSwMxEJ31s9avqkcvwSLUS9ltBT0WvXizgv2Adi3ZNNvGZpMlyQpl6X/w4kERr/4fb/4b03YP2vpg4PHeDDPzgpgzbVz321lZXVvf2Mxt5bd3dvf2CweHTS0TRWiDSC5VO8CaciZowzDDaTtWFEcBp61gdD31W09UaSbFvRnH1I/wQLCQEWys1LwtiYfqWa9QdMvuDGiZeBkpQoZ6r/DV7UuSRFQYwrHWHc+NjZ9iZRjhdJLvJprGmIzwgHYsFTii2k9n107QqVX6KJTKljBopv6eSHGk9TgKbGeEzVAvelPxP6+TmPDST5mIE0MFmS8KE46MRNPXUZ8pSgwfW4KJYvZWRIZYYWJsQHkbgrf48jJpVspetVy5Oy/WrrI4cnAMJ1ACDy6gBjdQhwYQeIRneIU3RzovzrvzMW9dcbKZI/gD5/MHY7yOWQ== 計算量: m[i, j] = ( 0 mini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 m[i, j] = ( 0 i = j mini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 

/25 17 ボトムアップ方式 l AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlpuiXK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH1VeM9A== i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== j AAAB6HicbVDLTgJBEOzFF+IL9ehlIjHxRHbRRI9ELx4hkUcCGzI79MLA7OxmZtaEEL7AiweN8eonefNvHGAPClbSSaWqO91dQSK4Nq777eQ2Nre2d/K7hb39g8Oj4vFJU8epYthgsYhVO6AaBZfYMNwIbCcKaRQIbAXj+7nfekKleSwfzSRBP6IDyUPOqLFSfdQrltyyuwBZJ15GSpCh1it+dfsxSyOUhgmqdcdzE+NPqTKcCZwVuqnGhLIxHWDHUkkj1P50ceiMXFilT8JY2ZKGLNTfE1MaaT2JAtsZUTPUq95c/M/rpCa89adcJqlByZaLwlQQE5P516TPFTIjJpZQpri9lbAhVZQZm03BhuCtvrxOmpWyd1Wu1K9L1bssjjycwTlcggc3UIUHqEEDGCA8wyu8OSPnxXl3PpatOSebOYU/cD5/ANJPjPI= k AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOl5rhfrrhVdw6ySrycVCBHo1/+6g1ilkYoDRNU667nJsbPqDKcCZyWeqnGhLIxHWLXUkkj1H42P3RKzqwyIGGsbElD5urviYxGWk+iwHZG1Iz0sjcT//O6qQmv/YzLJDUo2WJRmApiYjL7mgy4QmbExBLKFLe3EjaiijJjsynZELzll1dJu1b1Lqq15mWlfpPHUYQTOIVz8OAK6nAHDWgBA4RneIU359F5cd6dj0VrwclnjuEPnM8f09OM8w== A1A2 · · · An AAAB+3icbZDLSgMxFIYzXmu91bp0EyyCqzJTBV22unFZwV6gHYZMJtOGZpIhOSOW0ldx40IRt76IO9/GtJ2Ftv4Q+PjPOZyTP0wFN+C6387a+sbm1nZhp7i7t39wWDoqt43KNGUtqoTS3ZAYJrhkLeAgWDfVjCShYJ1wdDurdx6ZNlzJBxinzE/IQPKYUwLWCkrlRuDhRlDDfRopMBatWXGr7lx4FbwcKihXMyh99SNFs4RJoIIY0/PcFPwJ0cCpYNNiPzMsJXREBqxnUZKEGX8yv32Kz6wT4Vhp+yTguft7YkISY8ZJaDsTAkOzXJuZ/9V6GcTX/oTLNAMm6WJRnAkMCs+CwBHXjIIYWyBUc3srpkOiCQUbV9GG4C1/eRXatap3Ua3dX1bqN3kcBXSCTtE58tAVqqM71EQtRNETekav6M2ZOi/Ou/OxaF1z8plj9EfO5w+uvJLt 小さい問題から順に解く m[i, j] = ( 0 i = j mini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 括弧付けの位置 計算量: ⇥(n3) AAAB8nicbVBNSwMxEM3Wr1q/qh69BItQL2W3FfRY9OKxQr+gXUs2zbah2WRJZoWy9Gd48aCIV3+NN/+NabsHbX0w8Hhvhpl5QSy4Adf9dnIbm1vbO/ndwt7+weFR8fikbVSiKWtRJZTuBsQwwSVrAQfBurFmJAoE6wSTu7nfeWLacCWbMI2ZH5GR5CGnBKzU6zfHDEhZPtYuB8WSW3EXwOvEy0gJZWgMil/9oaJJxCRQQYzpeW4Mfko0cCrYrNBPDIsJnZAR61kqScSMny5OnuELqwxxqLQtCXih/p5ISWTMNApsZ0RgbFa9ufif10sgvPFTLuMEmKTLRWEiMCg8/x8PuWYUxNQSQjW3t2I6JppQsCkVbAje6svrpF2teLVK9eGqVL/N4sijM3SOyshD16iO7lEDtRBFCj2jV/TmgPPivDsfy9ack82coj9wPn8AON+Qjg== 

/25 18 行列 次元 A1 AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4KkkV9Fj14rGi/YA2lM120i7dbMLuRiihP8GLB0W8+ou8+W/ctjlo64OBx3szzMwLEsG1cd1vZ2V1bX1js7BV3N7Z3dsvHRw2dZwqhg0Wi1i1A6pRcIkNw43AdqKQRoHAVjC6nfqtJ1Sax/LRjBP0IzqQPOSMGis9XPe8XqnsVtwZyDLxclKGHPVe6avbj1kaoTRMUK07npsYP6PKcCZwUuymGhPKRnSAHUsljVD72ezUCTm1Sp+EsbIlDZmpvycyGmk9jgLbGVEz1IveVPzP66QmvPIzLpPUoGTzRWEqiInJ9G/S5wqZEWNLKFPc3krYkCrKjE2naEPwFl9eJs1qxTuvVO8vyrWbPI4CHMMJnIEHl1CDO6hDAxgM4Ble4c0Rzovz7nzMW1ecfOYI/sD5/AG5JY1t A2 AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4KkkV9Fj14rGi/YA2lM120y7dbMLuRCihP8GLB0W8+ou8+W/ctjlo64OBx3szzMwLEikMuu63s7K6tr6xWdgqbu/s7u2XDg6bJk414w0Wy1i3A2q4FIo3UKDk7URzGgWSt4LR7dRvPXFtRKwecZxwP6IDJULBKFrp4bpX7ZXKbsWdgSwTLydlyFHvlb66/ZilEVfIJDWm47kJ+hnVKJjkk2I3NTyhbEQHvGOpohE3fjY7dUJOrdInYaxtKSQz9fdERiNjxlFgOyOKQ7PoTcX/vE6K4ZWfCZWkyBWbLwpTSTAm079JX2jOUI4toUwLeythQ6opQ5tO0YbgLb68TJrVindeqd5flGs3eRwFOIYTOAMPLqEGd1CHBjAYwDO8wpsjnRfn3fmYt644+cwR/IHz+QO6qY1u A3 AAAB6nicbVDLTgJBEOzFF+IL9ehlIjHxRHbBRI+oF48Y5ZHAhswOszBhdnYz02tCCJ/gxYPGePWLvPk3DrAHBSvppFLVne6uIJHCoOt+O7m19Y3Nrfx2YWd3b/+geHjUNHGqGW+wWMa6HVDDpVC8gQIlbyea0yiQvBWMbmd+64lrI2L1iOOE+xEdKBEKRtFKD9e9aq9YcsvuHGSVeBkpQYZ6r/jV7ccsjbhCJqkxHc9N0J9QjYJJPi10U8MTykZ0wDuWKhpx40/mp07JmVX6JIy1LYVkrv6emNDImHEU2M6I4tAsezPxP6+TYnjlT4RKUuSKLRaFqSQYk9nfpC80ZyjHllCmhb2VsCHVlKFNp2BD8JZfXiXNStmrliv3F6XaTRZHHk7gFM7Bg0uowR3UoQEMBvAMr/DmSOfFeXc+Fq05J5s5hj9wPn8AvC2Nbw== A4 AAAB6nicbVDLTgJBEOzFF+IL9ehlIjHxRHaRRI+oF48Y5ZHAhswOvTBhdnYzM2tCCJ/gxYPGePWLvPk3DrAHBSvppFLVne6uIBFcG9f9dnJr6xubW/ntws7u3v5B8fCoqeNUMWywWMSqHVCNgktsGG4EthOFNAoEtoLR7cxvPaHSPJaPZpygH9GB5CFn1Fjp4bpX7RVLbtmdg6wSLyMlyFDvFb+6/ZilEUrDBNW647mJ8SdUGc4ETgvdVGNC2YgOsGOppBFqfzI/dUrOrNInYaxsSUPm6u+JCY20HkeB7YyoGeplbyb+53VSE175Ey6T1KBki0VhKoiJyexv0ucKmRFjSyhT3N5K2JAqyoxNp2BD8JZfXiXNStm7KFfuq6XaTRZHHk7gFM7Bg0uowR3UoQEMBvAMr/DmCOfFeXc+Fq05J5s5hj9wPn8AvbGNcA== A5 AAAB6nicbVDLTgJBEOzFF+IL9ehlIjHxRHZRo0fUi0eM8khgQ2aHBibMzm5mZk3Ihk/w4kFjvPpF3vwbB9iDgpV0UqnqTndXEAuujet+O7mV1bX1jfxmYWt7Z3evuH/Q0FGiGNZZJCLVCqhGwSXWDTcCW7FCGgYCm8Hoduo3n1BpHslHM47RD+lA8j5n1Fjp4bp70S2W3LI7A1kmXkZKkKHWLX51ehFLQpSGCap123Nj46dUGc4ETgqdRGNM2YgOsG2ppCFqP52dOiEnVumRfqRsSUNm6u+JlIZaj8PAdobUDPWiNxX/89qJ6V/5KZdxYlCy+aJ+IoiJyPRv0uMKmRFjSyhT3N5K2JAqyoxNp2BD8BZfXiaNStk7K1fuz0vVmyyOPBzBMZyCB5dQhTuoQR0YDOAZXuHNEc6L8+58zFtzTjZzCH/gfP4AvzWNcQ== A6 AAAB6nicbVDLTgJBEOzFF+IL9ehlIjHxRHbRqEfUi0eM8khgQ2aHBibMzm5mZk3Ihk/w4kFjvPpF3vwbB9iDgpV0UqnqTndXEAuujet+O7mV1bX1jfxmYWt7Z3evuH/Q0FGiGNZZJCLVCqhGwSXWDTcCW7FCGgYCm8Hoduo3n1BpHslHM47RD+lA8j5n1Fjp4bp70S2W3LI7A1kmXkZKkKHWLX51ehFLQpSGCap123Nj46dUGc4ETgqdRGNM2YgOsG2ppCFqP52dOiEnVumRfqRsSUNm6u+JlIZaj8PAdobUDPWiNxX/89qJ6V/5KZdxYlCy+aJ+IoiJyPRv0uMKmRFjSyhT3N5K2JAqyoxNp2BD8BZfXiaNStk7K1fuz0vVmyyOPBzBMZyCB5dQhTuoQR0YDOAZXuHNEc6L8+58zFtzTjZzCH/gfP4AwLmNcg== 30 ⇥ 35 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AAAB8XicbVDLTgJBEOzFF+IL9ehlIjHxRHZRokeiF4+YyCPChswOszBhdnYz02tCCH/hxYPGePVvvPk3DrAHBSvppFLVne6uIJHCoOt+O7m19Y3Nrfx2YWd3b/+geHjUNHGqGW+wWMa6HVDDpVC8gQIlbyea0yiQvBWMbmd+64lrI2L1gOOE+xEdKBEKRtFKj161iyLihlR7xZJbducgq8TLSAky1HvFr24/ZmnEFTJJjel4boL+hGoUTPJpoZsanlA2ogPesVRRu8afzC+ekjOr9EkYa1sKyVz9PTGhkTHjKLCdEcWhWfZm4n9eJ8Xw2p8IlaTIFVssClNJMCaz90lfaM5Qji2hTAt7K2FDqilDG1LBhuAtv7xKmpWyd1Gu3F+WajdZHHk4gVM4Bw+uoAZ3UIcGMFDwDK/w5hjnxXl3PhatOSebOYY/cD5/AG9EkBs= 5 ⇥ 10 AAAB8XicbVBNS8NAEJ34WetX1aOXxSJ4KklV9Fj04rGC/cA2lM120y7dbMLuRCih/8KLB0W8+m+8+W/ctjlo64OBx3szzMwLEikMuu63s7K6tr6xWdgqbu/s7u2XDg6bJk414w0Wy1i3A2q4FIo3UKDk7URzGgWSt4LR7dRvPXFtRKwecJxwP6IDJULBKFrp8bKLIuKGeG6vVHYr7gxkmXg5KUOOeq/01e3HLI24QiapMR3PTdDPqEbBJJ8Uu6nhCWUjOuAdSxW1e/xsdvGEnFqlT8JY21JIZurviYxGxoyjwHZGFIdm0ZuK/3mdFMNrPxMqSZErNl8UppJgTKbvk77QnKEcW0KZFvZWwoZUU4Y2pKINwVt8eZk0qxXvvFK9vyjXbvI4CnAMJ3AGHlxBDe6gDg1goOAZXuHNMc6L8+58zFtXnHzmCP7A+fwBaPuQFg== 10 ⇥ 20 AAAB8nicbVBNSwMxEJ31s9avqkcvwSJ4KrtV0GPRi8cK9gPapWTTbBuaTZZkViilP8OLB0W8+mu8+W9M2z1o64PA472ZzMyLUiks+v63t7a+sbm1Xdgp7u7tHxyWjo6bVmeG8QbTUpt2RC2XQvEGCpS8nRpOk0jyVjS6m/mtJ26s0OoRxykPEzpQIhaMopM6gd9FkXBLqn6vVPYr/hxklQQ5KUOOeq/01e1rliVcIZPUWvdZiuGEGhRM8mmxm1meUjaiA95xVFE3J5zMV56Sc6f0SayNewrJXP3dMaGJteMkcpUJxaFd9mbif14nw/gmnAiVZsgVWwyKM0lQk9n9pC8MZyjHjlBmhNuVsCE1lKFLqehCCJZPXiXNaiW4rFQfrsq12zyOApzCGVxAANdQg3uoQwMYaHiGV3jz0Hvx3r2PRemal/ecwB94nz/T45BN 20 ⇥ 25 AAAB8nicbVBNS8NAEN3Ur1q/qh69LBbBU0mioseiF48V7AekoWy2m3bpZjfsToQS+jO8eFDEq7/Gm//GbZuDtj4YeLw3w8y8KBXcgOt+O6W19Y3NrfJ2ZWd3b/+genjUNirTlLWoEkp3I2KY4JK1gINg3VQzkkSCdaLx3czvPDFtuJKPMElZmJCh5DGnBKwU+G4PeMIM9q/61Zpbd+fAq8QrSA0VaParX72BolnCJFBBjAk8N4UwJxo4FWxa6WWGpYSOyZAFlkpi94T5/OQpPrPKAMdK25KA5+rviZwkxkySyHYmBEZm2ZuJ/3lBBvFNmHOZZsAkXSyKM4FB4dn/eMA1oyAmlhCqub0V0xHRhIJNqWJD8JZfXiVtv+5d1P2Hy1rjtoijjE7QKTpHHrpGDXSPmqiFKFLoGb2iNwecF+fd+Vi0lpxi5hj9gfP5A90FkFM= i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== m AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3a3STsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgkfYMtwI7CQKqQwEPgTj25n/8IRK8zi6N5MEfUmHEQ85o8ZKTdkvV9yqOwdZJV5OKpCj0S9/9QYxSyVGhgmqdddzE+NnVBnOBE5LvVRjQtmYDrFraUQlaj+bHzolZ1YZkDBWtiJD5urviYxKrScysJ2SmpFe9mbif143NeG1n/EoSQ1GbLEoTAUxMZl9TQZcITNiYgllittbCRtRRZmx2ZRsCN7yy6ukXat6F9Va87JSv8njKMIJnMI5eHAFdbiDBrSAAcIzvMKb8+i8OO/Ox6K14OQzx/AHzucP1tuM9Q== スカラー積の回数 1 2 3 4 5 6 1 0 15750 7875 9375 11875 15125 2 0 2625 4375 7125 10500 3 0 750 2500 5375 4 0 1000 3500 5 0 5000 6 0 j AAAB6HicbVDLTgJBEOzFF+IL9ehlIjHxRHbRRI9ELx4hkUcCGzI79MLA7OxmZtaEEL7AiweN8eonefNvHGAPClbSSaWqO91dQSK4Nq777eQ2Nre2d/K7hb39g8Oj4vFJU8epYthgsYhVO6AaBZfYMNwIbCcKaRQIbAXj+7nfekKleSwfzSRBP6IDyUPOqLFSfdQrltyyuwBZJ15GSpCh1it+dfsxSyOUhgmqdcdzE+NPqTKcCZwVuqnGhLIxHWDHUkkj1P50ceiMXFilT8JY2ZKGLNTfE1MaaT2JAtsZUTPUq95c/M/rpCa89adcJqlByZaLwlQQE5P516TPFTIjJpZQpri9lbAhVZQZm03BhuCtvrxOmpWyd1Wu1K9L1bssjjycwTlcggc3UIUHqEEDGCA8wyu8OSPnxXl3PpatOSebOYU/cD5/ANJPjPI= i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== 1 2 3 4 5 6 1 1 1 3 3 3 2 2 3 3 3 3 3 3 3 4 4 5 5 5 6 j AAAB6HicbVDLTgJBEOzFF+IL9ehlIjHxRHbRRI9ELx4hkUcCGzI79MLA7OxmZtaEEL7AiweN8eonefNvHGAPClbSSaWqO91dQSK4Nq777eQ2Nre2d/K7hb39g8Oj4vFJU8epYthgsYhVO6AaBZfYMNwIbCcKaRQIbAXj+7nfekKleSwfzSRBP6IDyUPOqLFSfdQrltyyuwBZJ15GSpCh1it+dfsxSyOUhgmqdcdzE+NPqTKcCZwVuqnGhLIxHWDHUkkj1P50ceiMXFilT8JY2ZKGLNTfE1MaaT2JAtsZUTPUq95c/M/rpCa89adcJqlByZaLwlQQE5P516TPFTIjJpZQpri9lbAhVZQZm03BhuCtvrxOmpWyd1Wu1K9L1bssjjycwTlcggc3UIUHqEEDGCA8wyu8OSPnxXl3PpatOSebOYU/cD5/ANJPjPI= s AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlpu6XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH3/OM+w== 括弧付の位置 m[1, 2] + m[3, 3] + p0p2p3 =15750 + 0 + 30 · 15 · 5 =18000 AAACMXicbVBPS8MwHE39b/039eglOBRhMtLWoRdB9OJRwemgLSXNMg1L2pCkwih+JS9+E/HiQRGvfgmzrQedPvjB4733I/m9VHKmDUKvztT0zOzc/MKiu7S8srpWW9+41nmhCG2TnOeqk2JNOcto2zDDaUcqikXK6U3aPxv6N/dUaZZnV2YgaSzwbcZ6jGBjpaR2vitCb9+PGyIM9oO4IRMkE18mAXSj6Bjueq3DFoINOJwARaSbG+i14Ji0qswRQiip1VETjQD/Eq8idVDhIqk9R92cFIJmhnCsdeghaeISK8MIpw9uVGgqMenjWxpammFBdVyOLn6AO1bpwl6u7GQGjtSfGyUWWg9EapMCmzs96Q3F/7ywML2juGSZLAzNyPihXsGhyeGwPthlihLDB5Zgopj9KyR3WGFibMmuLcGbPPkvufabXtD0Lw/qJ6dVHQtgC2yDPeCBQ3ACzsEFaAMCHsELeAPvzpPz6nw4n+PolFPtbIJfcL6+AdcloX0= m[1, 1] + m[2, 3] + p0p1p3 =0 + 2625 + 30 · 35 · 5 =7875 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 ((A1(A2A3))((A4A5)A6)) AAACCHicbVC7TsMwFHXKq5RXgJEBiwqpWaqkLY+xhYWxSLRUaqPIcd3WquNEtoNURR1Z+BUWBhBi5RPY+BucNgO0HOlax+fcK/seP2JUKtv+NnIrq2vrG/nNwtb2zu6euX/QlmEsMGnhkIWi4yNJGOWkpahipBMJggKfkXt/fJ369w9ESBryOzWJiBugIacDipHSkmcel0oNz9FVgQ2valnptabpmaWPc8vyzKJdtmeAy8TJSBFkaHrmV68f4jggXGGGpOw6dqTcBAlFMSPTQi+WJEJ4jIakqylHAZFuMltkCk+10oeDUOjiCs7U3xMJCqScBL7uDJAayUUvFf/zurEaXLoJ5VGsCMfzhwYxgyqEaSqwTwXBik00QVhQ/VeIR0ggrHR2BR2Cs7jyMmlXyk61XLmtFetXWRx5cAROQAk44ALUwQ1oghbA4BE8g1fwZjwZL8a78TFvzRnZzCH4A+PzB+WAlM8= Q. 最適な括弧付けは?: 

/25 19 何度も同じ部分問題を解くのは無駄 各部分問題を一度だけ解くようにすれば効率的なはず！ 2つのケーススタディで共通している観点 

/25 20 動的計画法は分割統治法と同様, 部分問題の解を統合することによって問題を解く方法である. 計算結果を保存しておくことで, 部分部分問題を再計算する労を避ける. 部分構造最適性 (optimal substructure) 部分問題重複性 (overlapping subproblems) どんなときに適応できるのか? and ボトムアップ方式 (bottom-up method) 履歴管理を用いるトップダウン方式 (top-down with memorization) どのように実現するのか? or 

/25 21 部分構造最適性 連鎖行列積 AiAi+1 · · · Aj AAACAHicbVDLSgMxFM3UV62vURcu3ASLIAhlpgq6bHXjsoJ9QDsMmUymjc0kQ5IRyjAbf8WNC0Xc+hnu/BvTdhZaPXDh5Jx7yb0nSBhV2nG+rNLS8srqWnm9srG5tb1j7+51lEglJm0smJC9ACnCKCdtTTUjvUQSFAeMdIPx9dTvPhCpqOB3epIQL0ZDTiOKkTaSbx80fQqbfkZP3RzCAQ6FVuZ979tVp+bMAP8StyBVUKDl25+DUOA0JlxjhpTqu06ivQxJTTEjeWWQKpIgPEZD0jeUo5goL5sdkMNjo4QwEtIU13Cm/pzIUKzUJA5MZ4z0SC16U/E/r5/q6NLLKE9STTiefxSlDGoBp2nAkEqCNZsYgrCkZleIR0girE1mFROCu3jyX9Kp19yzWv32vNq4KuIog0NwBE6ACy5AA9yAFmgDDHLwBF7Aq/VoPVtv1vu8tWQVM/vgF6yPb2K6lP4= を と分割 ((Ai · · · Ak)(Ak+1 · · · Aj)) AAACEHicbZDLSsNAFIYn9VbrLerSzWARW4SSVEGXrW5cVrAXaEOYTKbtmMkkzEyEEvoIbnwVNy4UcevSnW/jtA2orT8M/HznHM6c34sZlcqyvozc0vLK6lp+vbCxubW9Y+7utWSUCEyaOGKR6HhIEkY5aSqqGOnEgqDQY6TtBVeTevueCEkjfqtGMXFCNOC0TzFSGrnmcalUdynsYT9SEtbdoAw1SIMTe/wD78pl1yxaFWsquGjszBRBpoZrfvb8CCch4QozJGXXtmLlpEgoihkZF3qJJDHCARqQrrYchUQ66fSgMTzSxIf9SOjHFZzS3xMpCqUchZ7uDJEayvnaBP5X6yaqf+GklMeJIhzPFvUTBlUEJ+lAnwqCFRtpg7Cg+q8QD5FAWOkMCzoEe/7kRdOqVuzTSvXmrFi7zOLIgwNwCErABuegBq5BAzQBBg/gCbyAV+PReDbejPdZa87IZvbBHxkf30HTmjc= どちらも最適な括弧付けになっていなければならない rn = max (pn, r1 + rn 1, r2 + rn 2, . . . , rn 1 + r1) 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 ロッド切出し問題 長さが と の2つの金属棒に切断, それぞれの金属棒を最適に 分割することで得られる総収入 n i AAAB6nicbVBNS8NAEJ34WetX1aOXxSJ4sSRV0GPRi8eK9gPaUDbbTbt0swm7E6GE/gQvHhTx6i/y5r9x2+agrQ8GHu/NMDMvSKQw6Lrfzsrq2vrGZmGruL2zu7dfOjhsmjjVjDdYLGPdDqjhUijeQIGStxPNaRRI3gpGt1O/9cS1EbF6xHHC/YgOlAgFo2ilB3UueqWyW3FnIMvEy0kZctR7pa9uP2ZpxBUySY3peG6CfkY1Cib5pNhNDU8oG9EB71iqaMSNn81OnZBTq/RJGGtbCslM/T2R0ciYcRTYzoji0Cx6U/E/r5NieO1nQiUpcsXmi8JUEozJ9G/SF5ozlGNLKNPC3krYkGrK0KZTtCF4iy8vk2a14l1UqveX5dpNHkcBjuEEzsCDK6jBHdShAQwG8Ayv8OZI58V5dz7mrStOPnMEf+B8/gAGqI2g i AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0mqoMeiF48t2FpoQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgpr6xubW8Xt0s7u3v5B+fCoreNUMWyxWMSqE1CNgktsGW4EdhKFNAoEPgTj25n/8IRK81jem0mCfkSHkoecUWOlJu+XK27VnYOsEi8nFcjR6Je/eoOYpRFKwwTVuuu5ifEzqgxnAqelXqoxoWxMh9i1VNIItZ/ND52SM6sMSBgrW9KQufp7IqOR1pMosJ0RNSO97M3E/7xuasJrP+MySQ1KtlgUpoKYmMy+JgOukBkxsYQyxe2thI2ooszYbEo2BG/55VXSrlW9i2qteVmp3+RxFOEETuEcPLiCOtxBA1rAAOEZXuHNeXRenHfnY9FacPKZY/gD5/MH0MuM8Q== 切断なし 問題の最適解をその部分問題の最適解から構成できるか? 1. ある部分問題の解が最適でないと仮定 2. その部分問題の最適解に貼替える 3. 元の問題に対するよりよい解を得ることができる -> 矛盾 切貼り法(cut-and-paste)を用いて妥当性を確認 

/25 22 あるグラフと2点 u, v が与えられたとき u->v への最短路を求めよ u->v への最長単純道を求めよ 部分構造最適性: YES 部分構造最適性: NO 解 u->w->v に対して 切出し法によるチェック 1. u->w への最短路を p とする 2. p より短い道 q が存在すると仮定 3. p を q に貼替える 4. 元の解より短い道が完成 -> 矛盾 u w v x uからvへの最長路: u->w->v uからwへの最長路: u->x->v->w wからvへの最長路: w->u->x->v 組合せても解にならない 部分問題が独立ではない 

/25 23 部分問題重複性 再帰アルゴリズムが同じ問題を繰返し訪れているか? 4 Cut-Rod(p, 4) AAACAXicbVDLSgNBEJyNrxhfq14EL4NBiKBhNwb0GMzFYxTzgCSE2dnZZMjsg5leMSzx4q948aCIV//Cm3/jJNmDJhY0FFXddHc5keAKLOvbyCwtr6yuZddzG5tb2zvm7l5DhbGkrE5DEcqWQxQTPGB14CBYK5KM+I5gTWdYnfjNeyYVD4M7GEWs65N+wD1OCWipZx50gD1A0lEUV2M4uw3dcSE6xeWTnpm3itYUeJHYKcmjFLWe+dVxQxr7LAAqiFJt24qgmxAJnAo2znVixSJCh6TP2poGxGeqm0w/GONjrbjYC6WuAPBU/T2REF+pke/oTp/AQM17E/E/rx2Dd9lNeBDFwAI6W+TFAkOIJ3Fgl0tGQYw0IVRyfSumAyIJBR1aTodgz7+8SBqlon1eLN2U85WrNI4sOkRHqIBsdIEq6BrVUB1R9Iie0St6M56MF+Pd+Ji1Zox0Zh/9gfH5A8amlck= 3 0 2 1 0 0 1 0 1 0 0 2 1 0 0 ロッド切出し問題 4 3 0 2 1 0 0 1 1 0 2 Memorized-Cut-Rod-Aux(p, 4, r) AAACEnicbVDLSgNBEJz1bXxFPXoZDIKCCbsa0KOPixdBxSRCEsLsbCcZnN1ZZnolcck3ePFXvHhQxKsnb/6Nk8dBowUNRVU33V1+LIVB1/1yJianpmdm5+YzC4tLyyvZ1bWyUYnmUOJKKn3jMwNSRFBCgRJuYg0s9CVU/NvTvl+5A22Eiq6xG0M9ZK1INAVnaKVGdqeG0MG0Zjg9h1BpcQ9B/jTB/JUK8sdJp7cd79LiLtU7jWzOLbgD0L/EG5EcGeGikf2sBYonIUTIJTOm6rkx1lOmUXAJvUwtMRAzfstaULU0YiGYejp4qUe3rBLQptK2IqQD9edEykJjuqFvO0OGbTPu9cX/vGqCzcN6KqI4QYj4cFEzkRQV7edDA6GBo+xawrgW9lbK20wzjjbFjA3BG3/5LynvFbz9wt5lMXd0MopjjmyQTbJNPHJAjsgZuSAlwskDeSIv5NV5dJ6dN+d92DrhjGbWyS84H98PfJxl 

/25 24 動的計画法の応用例 遺伝子コードの比較, 系統樹の作成など 配列アラインメント (The Needleman Wunsch algorithm) 時系列データの比較, 音声認識など 動的時間伸縮法 (dynamic time warping) グラフィカルモデルにおける推論手法 max-sum アルゴリズム 

/25 25 まとめ 大きい問題を分割, 部分問題を解いて得られた解を結合して 元の問題の答えを得る. 部分問題の解を保存しておき適宜利用する. 動的計画法 適用可能かの判断 解に部分構造最適性と部分問題重複性があるかどうか 実現方法 履歴管理を用いるトップダウン方式 or ボトムアップ方式