Return-Path: sakane@mist.wide.ydc.co.jp Received: from orange.kame.net by galilei.v6.hitachi.co.jp (8.8.5/3.4W-EBINA) id MAA10353 for ; Fri, 28 Aug 1998 12:41:40 +0900 (JST) Received: from mist.wide.ydc.co.jp (mach117.xnet.com [207.227.19.117]) by orange.kame.net (8.8.8+3.0Wbeta13/3.6W/smtpfeed 0.63) with ESMTP id MAA02040 for ; Fri, 28 Aug 1998 12:41:38 +0900 (JST) Received: (from sakane@localhost) by mist.wide.ydc.co.jp (8.8.8/3.6Wbeta7) id MAA00399; Fri, 28 Aug 1998 12:39:49 +0900 (JST) Date: Fri, 28 Aug 1998 12:39:49 +0900 (JST) Message-Id: <199808280339.MAA00399@mist.wide.ydc.co.jp> From: Ne To: itojun@iijlab.net Cc: onoe@sm.sony.co.jp, sumikawa@ebina.hitachi.co.jp, core@kame.net, itojun@itojun.org Subject: Re: KQC In-Reply-To: Your message of "Fri, 28 Aug 1998 09:59:18 JST". <346.904265958@turmeric.itojun.org> Mime-Version: 1.0 X-Mailer: mnews [version 1.21PL3] 1998-04/12(Sun) X-UIDL: beed814df629c33ba24a0ee0f592aa1f Content-Type: text/plain; charset=ISO-2022-JP :>> 2) 同様に、nを用いて一般解を求めなさい :> 2*n-3 (n >= 2 の場合) :> 全員の記憶を集めるのに最低 (n-1) 回必要 :> この時点で全部保持している人は 2人 :> 残り (n-2) 人は記憶が欠落しているので、充当するのに (n-2) 回必要 :> 合計 (n-1) + (n-2) = 2*n-3 : : 6人の場合、以下のようにすると8回(2 * n - 4)でいけます。 4回の様な気も。まちがったらごめんなさい？ A(1) 0 A(2) 1 A(3) a b c -> ab ab c -> ab abc abc -> abc abc abc A(3) A(4) a b -> ab ab -> abcd abcd c d ab ab abcd abcd A(4) -> A(2) -> A(1) A(5) a b -> ab ab -> abcd abcd -> abcd abcd -> abcde abcde c d e cd cd e abcd abcd e abcd abcde abcde abcd abcde abcde -> abcde abcde abcde abcde abcde A(6) a b c -> ad be cf -> abde abde cf -> abde abdecf abdecf d e f ad be cf abde abde cf abde abdecf abdecf A(6) -> A(3) -> abdecf abdecf abdecf abdecf abdecf abdecf A(7) a b c -> ad be cf -> abde abde cf -> abde abdecf abdecf d e f g ad be cf g abde abde cf g abde abdecf abdecf g -> abdecf abdecf abdecf -> .. -> .. -> .. abdecf abdecf abdecf g A(8) a b c d -> ae bf cg dh -> aebf aebf cgdh cgdh -> aebfcgdh aebfcgdh ... e f g h ae bf cg dh aebf aebf cgdh cgdh aebfcgdh aebfcgdh ... A(8) -> A(4) -> A(2) -> (1) A(10) a b c d e -> ag bh ci dj ek -> agbh agbh cidj cidj ek g h i j k ag bh ci dj ek agbh agbh cidj cidj ek A(5) -> agbhcidj agbhcidj agbhcidj agbhcidj ek -> .. -> .. agbhcidj agbhcidj agbhcidj agbhcidj ek A(12) a b c d e f -> ag bh ci dj ek fl -> agbh agbh cidj cidj ekfl ekfl g h i j k l ag bh ci dj ek fl agbh agbh cidj cidj ekfl ekfl A(6) -> A(3) -> agbhcidj abdhcidj agbhcidj agbhcidj ekfl ekfl agbhcidj abdhcidj agbhcidj agbhcidj ekfl ekfl -> .. -> .. n&1==0 a 0 0 2 1 4 2 6 4 8 3 10 6 12 5 : : n&1==1 a 1 0 3 3 5 5 7 7 : : n n