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一種可調整分支度之新型互連網路設計及其拓樸性質分析(2/2)

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(1)行政院國家科學委員會專題研究計畫. 成果報告. 一種可調整分支度之新型互連網路設計及其拓樸性質分析 (2/2). 計畫類別: 個別型計畫 計畫編號: NSC94-2213-E-006-025執行期間: 94 年 08 月 01 日至 95 年 07 月 31 日 執行單位: 國立成功大學資訊工程學系(所). 計畫主持人: 謝孫源 計畫參與人員: 張乃文(博士生)、伍昶德(碩士生)、畢文豪(碩士生)、莊宗諺 (碩士生). 報告類型: 完整報告 處理方式: 本計畫可公開查詢. 中. 華 民 國 95 年 8 月 23 日.

(2) 行政院國家科學委員會補助專題研究計畫.  成 果 報 告 □ 期中進度報告. 一種可調整分支度之新型互連網路設計及其拓樸性質分析(2/2). 計畫類別: 個別型計畫. □ 整合型計畫. 計畫編號:NSC 94-2213-E-006 -025 執行期間:. 94 年. 8 月 1 日至. 95 年. 7 月. 31 日. 計畫主持人:謝孫源 共同主持人: 計畫參與人員:張乃文(博士生)、伍昶德(碩士生)、畢文豪(碩士生)、 莊宗諺(碩士生). 成果報告類型(依經費核定清單規定繳交):□精簡報告. 完整報告. 本成果報告包括以下應繳交之附件: □赴國外出差或研習心得報告一份 □赴大陸地區出差或研習心得報告一份 □出席國際學術會議心得報告及發表之論文各一份 □國際合作研究計畫國外研究報告書一份. 處理方式:除產學合作研究計畫、提升產業技術及人才培育研究計畫、 列管計畫及下列情形者外,得立即公開查詢 □涉及專利或其他智慧財產權,□一年二年後可公開查詢 執行單位:國立成功大學資訊工程系 中 華 民 國 95 年. 8. 月. 23. 日.

(3) (二) 研究背景及目的: 對平行處理或分散式系統而言,互連網路(interconnection networks)的設計 與分析是一個相當重要的主題。許多重要的文獻在探討互連網路之架構[1, 2, 5, 7, 8, 12, 13, 21, 22, 23, 25, 27, 28, 31, 33]。選擇何種網路互連方式將會決定系統 的效能。互連網路有許多分析的參數,例如:連結性(connectivity)、分支度 (degree)、直徑(diameter)、對稱性(symmetry)、容錯(fault-tolerance) 等[36]。無 向圖(undirected graph)常被用來作為互連網路的模型,其中,節點(node)代表處 理機(processor),邊(edge)則代表到通訊管道(communication channels)。因此, 本計畫中,網路(network)和圖(graph)、節點(node)和處理器(processor)、邊(edge) 和通訊管道(communication channels)將交互使用。 Cayle 圖 已 被 證 明 非 常 適 合 用 來 設 計 互 連 網 路 [2]。 它 們 是 以 排 列 群 (permutation groups)為基礎所發展出來的一重要理論,目前許多重要的互連網 路,包括 hypercubes[5],star graphs[1, 2],pancake graphs[2] 等,都是 Cayle 圖。這些圖是正則(rugular)且對稱的(symmetric)。然而,這些節點的分支度 (degree)會隨著圖的大小而增加。從 VLSI 實作[7]的觀點來看,固定節點分支 度之網路(the fixed node-degree networks)是相當重要的,其中一原因是在互連 網路的研究領域中,有些計算只適用於有固定數目的 I/O 埠上(即固定分支度) [7]。 P. Vadapalli and P. K. Srimany[31]提出 trivalent Cayley 圖,該圖已被證明出 有正則性、直徑與節點數目對數成正比、以及最大容錯。此外,論文[26, 30, 31, 32, 34]中探討 trivalent Cayley 圖各種拓樸屬性(topological properties)和最短路 1.

(4) 徑繞徑演算法。然而,trivalent Cayley 圖的連結度只能被侷限在分支度為 3, 而且當網路擴增時,損毀的機率也會隨之增加。 本計畫提出一個新的 Cayley 圖族,用於建構互連網路,我們稱之為. k-valent 圖 Gk,n,k3 and n2。k-valent 圖跟一般 Cayley 圖族最大的不同 是它具有可視需要而調整的分支度 k;當 k 值固定之後,我們可持續增加. k-valent 圖的節點(node),而仍能保持其連結度。 本計畫證明 k-valent 圖許多重要的拓撲特性,例如: 一、 節點對稱性(node-symmetric); 二、 節點正規(regular)性:每一節點對外連接度為 k; 三、 邊(edge)的無向性(bi-directional) ; 四、 最大的容錯度等於連結度為 k; 五、 直徑(diameter)相對於 k-valent 圖節點數目對數成正比; 六、 證明 3-valent 圖與 trivalent 圖為同構(isomorphism); 七、 設計最短路徑之繞徑演算法(shortest path routing algorithm),用以建構 任兩節點間之最短路徑; 八、 探討 k-valent 圖的嵌入(embedding)性質。我們嵌入下列結構於. k-valent 圖 : 互 斥 環 路 ( node-disjoint cycles ) 和 互 斥 cliques (node-disjoint cliques)。. (三) 研究方法、進行步驟及執行進度: 3-1 研究方法 本計畫建構一種新型可調整分支度 (degree)的互連網路(interconnection 2.

(5) network):k-valent 圖 Gk ,n 。k-valent 圖屬於 Cayley 圖族(Cayley graphs family)。 我们證明 Gk ,n 具有許多優異的拓樸性質(topological properties),並設計一個最短 路徑繞徑演算法(shortest path routing algorithm)。另外,我们也探討 Gk ,n 圖的直 徑(diameter)、環路的嵌入( cycles embedding )、cliques embedding、容錯(fault tolerance)等性質。 方 法 一 ︰ 我 們 將 互 連 網 路 以 圖 形 (graph) 來 表 示 ; 節 點 (node) 代 表 處 理 器 (processor),邊(edge)代表通訊管道(communication channel)。我們對 k-valent 圖已完成下列的定義: 定義 1. k-valent 圖 Gk ,n 為一有 n(k 1) n 個節點之無向圖,且 n 和 k 是兩整數滿 足 n2;k3。 Gk ,n 的點(node) v = s0 s1...sm 1~sm sm 1...sn 1 為一長度為 n 的字 串, si {0,1,, k 2} ,並且字串裡恰有一個符號 ~sm 是被標記的,其 它的符號則是未標記。vm 表示一個被標記的符號是在位置 m 的節點 v。sm* 表示 si 或 ~ si。邊(edge)的形式為( v,δ ( v )),δ ∈{f,f-1,g1,g2,…gk-2}. 是一生成子(generator)定義如下: 1. f (u m ) v ( m 1) mod n ,其中 u m s0*s1*...sn* 1 , v ( m 1) mod n s1*s2*...sn* 1* and ( s0 1) mod(k 1) ;. 2. f 1 (u m ) v ( m 1) mod n ,其中 u m s0*s1*...sn* 1 , v ( m 1) mod n *s1*s2*...sn*2 and ( sn 1 1) mod(k 1) ;. 3. gi (u m ) v m ,其中 u m s0*s1*...sn* 1 , v m s0*s1*...sn* 2* and ( sn 1 i ) mod(k 1) 。. 方法二︰我們建立了 k-valent 圖和其它 Cayley 圖之間的關係,並藉此推導出. 3.

(6) k-valent 圖所具有的性質。我們發現 k-valent 圖和 trivalent Cayley 圖 TCn 有密 切的關係,TCn 的定義如下: 定義 2. 對 trivalent Cayley 圖 TCn﹙n≥2﹚而言,每一個節點對應到一個有 n 個符號並照字母順序的循環排列 t1,t2,…,tn,各符號可為本身或其 補集。TCn 的節點集合定義為 V(TCn)={ t *j t *j 1...tn*t1*t2*...t *j 1 | ti* ti or ti for . ( v )),δ ∈ all i,j {1,..., n} }。 j j 。每一個邊(edge)的形式是( v ,δ { fTC , fTC1 , gTC },定義如下: fTC ( t *j t *j 1...t n* t1*t 2* ...t *j 1 ) t *j 1...t n* t1*t 2* ...t *j 1 t *j fTC1 ( t *j t *j 1...t n* t1*t 2* ...t *j 1 ) t *j 1t *j ...t n* t1*t 2* ...t *j 2. gTC ( t *j t *j 1...t n*t1*t 2* ...t *j 1 ) t *j t *j 1...t n*t1*t 2* ...t *j 1. 方法三︰我們證明 k-valent 圖具節點對稱性(node-symmetric),所以,任兩點間 最短路徑長度等於某起始點(source node)v 到 identity node (n 個 symbols 皆為 0 , 被 標 記 的 symbol 是 第 一 個 0) 之 長 度 。 我 們 定 義 v 的 距 離 函 數 (distance-function): 定義 3. 在 Gk ,n 中,考慮一個任意的節點 v s 0 s1 ...s m1 s m~ s m1 ...s n 1。其中這個帶有記 號的符號 s m~ 將這個字串分成了兩個部分:左邊的部分 s0 s1 ...s m 1 和右邊 的部分 s m~ s m 1 ...s n 1 。我們也為這個節點 v 定義了以下的符號:  NSLv [x] | {si : s i x(mod k 1) for 0 i m 1} | 。  NSRv [x] | {si : s i x(mod k 1) for m i n 1} | 。  令 LZLv 為在左邊部分中,最長的「0」子字串的長度,亦即 LZLv max{l 0 : s j s j 1  s j l 1 0, 0 j m 1} 。而且令 j Lv 代. 4.

(7) 表在左邊部分中,最左邊那串最長「0」子字串的左端位置。如 果 LZLv 0 的話,我們定義 j L 0 。 v.  令 LZRv 為在右邊部分中,最長的「0」子字串的長度,亦即, LZRv max{l 0 : s j s j 1  s j l 1 0, m j n 1} 。而且令 j Rv 代. 表在右邊部分中,最左邊那串最長「0」子字串的左端位置。如 果 LZRv 0 的話,我們定義 j R m 。 v. 由上述定義我們可決定節點的距離函數如下: 定義 4. 在 Gk ,n 中,考慮一個任意的節點 v ,接著我們定義: 1. DIST1 (v)   2.. 2n m 2 LZL NSL[0] NSR[1] 2 if (m j L LZL) 0. otherwise.  2n m 2 LZL NSL[0] NSR[1]  3n m 2 LZR NSR[0] NSL[1] 2 if (n j R LZR) 0 DIST2 (v)  otherwise.  3n m 2 LZR NSR[0] NSR[1]. 3. DIST3 (v) 2n m NSL[2] NSR[1]. 4. DIST4 (v) 3n m NSR[2] NSL[1]. 節點 v 的距離函數 DIST (v) min 1i4  DISTi (v) 。 我們觀察 DIST1 - DIST4 代表四種從節點 v 到 identity node 之路徑長度。本計畫證 明了節點 v 到 identity node 之最短路徑即為上述四條路徑之一。下列四種演算 法可建構出由節點 v 到 identity node 長度為 DIST1 到 DIST4 之路徑:. Algorithm DIST_1(v,n,m,LZL,jL) 1: for i = 0 to (jL –1) do 2: v ← f(v) 3: end for 4: for i = 0 to (jL –1) + (n –m) do 5.

(8) 5: if v[n –1]≠1then 6:. v ← gk-v[n-1](v). 7: end if 8: v ← f-1(v) 9: end for 10: if (m –jL –LZL) > 0 then 11: for i = 0 to (m –jL –LZL –1) do 12:. v ← f-1(v). 13: end for 14: v ← gk-v[n-1]-1(v) 15: for i = 0 to (m –jL –LZL –1) do 16:. v ← f(v). 17:. if v[n –1]≠0 then. 18:. v ← gk-v[n-1]-1(v). 19:. end if. 20: end for 21: end if. 6.

(9) jL. m LZL. 1: 2: 3: 4: 5: :Bl o c kA;. :Bl oc kB;. :Bl o c kC;. : Block D.. Figure 1:An illustrate of Algorithm DIST_1. 3-valent Cayley graph G. 上圖說明 Algorithm DIST_1 是如何進行,圖中的任一字串表示將一個來源點 v 分割成 4 個區塊。在執行完演算法第 1 到第 3 行後,中間的節點就如「2:」所 示。在執行完第 4 到第 9 行後,中間的節點就會如「3:」所示,其中區塊 A 和 D 的所有符號都是 0。假如 m jL LZL 0 的話,第 10 到第 21 行將進一步地被 執行來產生我們的 identity node。Algorithm DIST_2~ DIST_4 的觀念與此相似。. Algorithm DIST_2(v,n,m,LZR,jR). 1: for i = 0 to (n –2 –jR –LZR) do 2: v ← f -1(v) 3: end for 4: if (n –jR –LZR) > 0 then 5: v ← gk-v[n-1]-1(v) 6: for i = 0 to (n –2 –jR –LZR) do. 7.

(10) 7:. v ← f(v). 8:. if v[n –1]≠0then. 9:. v ← gk-v[n-1]-1(v). 10:. end if. 11: end for 12: end if 13: for i = 0 to (m –1) do 14: v ← f(v) 15: if v[n –1]≠0then 16:. v ← gk-v[n-1]-1(v). 17: end if 18: end for 19: for i = 0 to jR –m –1 do 20: v ← f(v) 21: end for 22: for i = 0 to jR –m –1 do 23: if v[n –1]≠1then 24:. v ← gk-v[n-1](v). 25: end if 26: v ← f -1(v) 27: end for. 8.

(11) 1: 2: 3: 4: 5:. Figure 2:An illustrate of Algorithm DIST_2. 3-valent Cayley graph G. Algorithm DIST_3(v, n, m). 1: for i = 0 to (m –1) do 2: v ← f(v) 3: if v[n –1]≠k –2 then 4:. v ← gk–2–v[n–1](v). 5: end if 6: end for 7: for i = m to (n –1) do 8: v ← f(v) 9: if v[n –1]≠0then 10:. v ← gk–1–v[n–1](v). 11: end if 12: end for. 9.

(12) 13: for i = 0 to (m –1) do 14: v ← f(v) 15: end for. 1: 2: 3: 4:. Figure 3:An illustrate of Algorithm DIST_3. 3-valent Cayley graph G. Algorithm DIST_4(v, n, m). 1: for i = m to (n –1) do 2: if v[n –1]≠2 ( modk –1) then 3:. v ← gk–v[n–1]+1(v). 4: end if 5: v ← f –1(v) 6: end for 7: for i = 0 to (m –1) do 8: if v[n –1]≠1then 10.

(13) 9: v ← gk–v[n–1](v) 10: end if 11: v ← f –1(v) 12: end for 13: for i = m to (n –1) do 14: v ← f –1(v) 15: end for. 1: 2: 3: 4:. Figure 4:An illustrate of Algorithm DIST_4. 3-valent Cayley graph G. 方 法 四 ︰ 本 計 畫 證 明 k-valent 圖 可 以 嵌 入 節 點 互 斥 環 路 (node-disjoint cycles)。我們的方法是將 k-valent 圖的邊分成兩大類以簡化問題。環路(cycle) 是一個點和邊個數一樣多的子圖(subgraph)。如果兩個環路沒有共同的節點則 稱兩環路互斥。一個由生成子 f (generator f ) (或 f 1 )所定義的邊稱作 F-edge。 Gk ,n 中的任何環路只包含一個 F-cycle 中的 F-edge。例如:圖一中的. . . ~ ~~ ~ 環路 0 0,0 1 , 1 1,1 0 是一個 F-cycle。對於非負整數 i,我們遞迴地定義: 11.

(14) if i 0 v  f i (v)  i 1 f ( f (v)) if i 0. 符號 f i (v) 可以被類似的定義。. ~ 00. ~ 01. ~ 00. ~ 01. ~ 10. ~ 11. ~ 10. ~ 11 Figure 1:A 3-valent Cayley graph G3, 2. 對於兩個固定的整數 i  0,1,..., n 1跟 s  0,1,..., k 2 ,每一個 F-cycle 有一個 唯一的節點 v 使得 v[i] s 且 m 0 。因此,對於每個 F-cycle 我們可以定義唯 一的 F-leader 節點 v 使得 v[0] 0 且 m 0。每個 F-leader 唯一決定一個特殊的 F-cycle,用以簡化問題。對於任兩個 F-cycle C1 跟 C 2 ,如果存在一個節點 1,2,..., k 2, v1 g i (v2 ) ,則稱 v1 C1 跟另一個節點 v2 C 2 ,使得對於某個 i  C1 跟 C 2 相鄰。我們藉由上述定義來探討 k-valent 圖可以嵌入多少個節點互斥. F-cycles。. 方法五︰本計畫證明 k-valent 圖可以嵌入節點互斥 cliques (node-disjoint cliques)。我們的方法是將 k-valent 圖的邊分成兩大類以簡化問題。 Gk ,n 中一個由生成子 gi (generator gi )所定義的邊稱作 G-edge。一個圖中的. 12.

(15) clique 是一節點集合,集合中的任兩節點彼此相鄰(adjacent)。一個 clique 如果 不被任何其他的 clique 所包含,則稱該 clique 為 maximal。G-clique 是 Gk ,n 中.  . ~ ~ 讓每一對相鄰的節點是由一 G-edge 連接的 maximal clique。例如: 0 0, 0 1 是 G3, 2. 中的一個 G-clique。兩個 G-cliques 沒有共同節點則稱為互斥。我們藉由上述 定義來探討 k-valent 圖可以嵌入多少個節點互斥 G-cliques。. 方法六︰我們建構 k-valent 圖簡化圖來探討 k-valent 圖之連結度(connectivity) 及容錯性(fault-tolerance)。簡化圖的定義如下︰ 定義 5. 對於一個 k-valent 圖 Gk ,n,藉由下列兩個步驟可以建構出一個 Gk ,n 的簡 化圖 RGk , n : ~ 1. 對於每個 F-leader 0 s1 s 2  s n 2 的 F-cycle,把它縮成一個叫做超級. 節點的單一節點,並用 s1 s2  s n 2 標記。 2. 如果 Gk ,n 中的任意兩個 F-cycle 相鄰,則用一超級邊的無向邊連接 相關的兩個超級節點。. 3-2 進行步驟 本計畫依下列步驟進行: 步驟 1. 證明 Gk ,n ,n≥2 及 k≥3,是一個分支度為 k 的正則圖,且有. kn(n 1) n 個 2. 邊。 若有一個一對一且映成的函數 ,將 V (G1 ) 對應至 V (G2 ) ,使得 (u, v) E (G1 ) 若 且唯若 ((u ), (v)) E (G2 ) ,則我們稱圖 G1 和 G2 為同構 (isomorphism)。. 13.

(16) 步驟 2. 證明 G3, n 和 TCn 為同構。 步驟 3. 證明對於 Gk ,n 裡的任意一個節點 v ,我們有: DIST (v) DIST (v' ) 1 or 0,其中 v' (v) 而且 f , f 1 , g1 , g 2 , , g k 2 。. 步驟 4. 證明對於 G k ,n 中 identity node I,對於任意的. .  f ,. 1. f ,g ,g 1. 2. . I ) = 1。 ,, g k 2 ,DIST(I) = 0 而且 DIST( . 步驟 5. 在 Gk ,n 中給定一個任意節點 v ,設計最短路徑繞徑演算法 OPTIMAL_ROUTE,能夠正確建構從節點 v 到 identity node 之最短路 徑 DIST(v)。 步驟 6. 證明對於 G k ,n 中任何一個節點 v,DIST(v)等於 v 與 identity node 之間的距離。  5n  步驟 7. 證明對於 Gk ,n 中任何一個節點 v , DIST (v) max  2n, 2。此外當  2  5 n  k 6 且 n 4 時, diam   2 ;並且當 k 6 且 n=2,3 時, G k ,n  2  . diam G k ,n 2n 。. 步驟 8. 證明 Gk ,n 可以嵌入 (k 1) n 1 個長度為 n(k 1) 節點互斥的 F-cycles。 步驟 9. 證明當 n>2 時,每一個 Gk , n 中的 F-cycle 與 n(k 2) 個不同的 F-cycle 相 鄰。當 n=2 時,每一個 Gk ,n 中的 F-cycle 與 k 2 個不同的 F-cycle 相 鄰。 ~ 步驟 10. 證明一個 F-leader 為 v 0 x1 x2  xn2 的 F-cycle 是與 F-leader ~ 為 0 y1 y 2  y n 1 的 F-cycle 相鄰的條件是下列的某一性質成立:一、. 對於某整數 i, 1 i n 1,xi yi 而且對於所有的整數 j, 1 j n 1 14.

(17) 且 j i , x j y j 。二、 x1 y1 x2 y 2  xn 1 y n 1 (mod k-1)。. 對於一個 G-clique K 我們可以定義 G-leader 為 K 中 v[n 1] 0 的唯一節 點。為了方便,我們用這樣的一個節點來表示它所對應的 G-clique。如果兩個 G-clique 沒有共同的節點,則稱作不相交。 步驟 11. 證明 Gk ,n 可以嵌入 n(k 1) n 1 個長度為 n(k 1) 節點不相交大小為 k 1 的 G-clique。 兩個 G-clique K1 跟 K 2 相鄰若且為若存在一對點 u K1 跟 v K 2 使得如果 不是 v f (u ) 就是 v f 1 (u ) 。. 步驟 12. 證明當 n>2 時,每一個 Gk ,n 中的 G-clique 與 2(k 1) 個不同的 G-clique 相鄰。當 n=2 時,每一個 Gk ,n 中的 G-clique 與 k 1 個不同的 G-clique 相鄰。 步驟 13. 對於任意兩節點 u, v 而言,嵌入連接 u, v 兩節點的最長路徑(longest u,v-path)於 Gk ,n 。 步驟 14. 嵌入漢彌頓圓(Hamiltonian cycle)於 Gk ,n 。 步驟 15. 嵌入樹網路(tree networks)於 Gk ,n 。 步驟 16. 證明簡化圖的連結度至少為 k,亦即 ( RGk ,n ) k for n 3。 步驟 17. 利用在ξ-連通圖中給定兩個節點集合 V1 跟V2 ,其中 V1 V2 ,則 存在ξ條連接V1 跟 V2 的節點互斥路徑之特性證明 ( G k , n ) k 。 步驟 18. 證明 k-valent 圖的連結度為 k,亦即 ( G k , n ) k 。. 15.

(18) 3-3 執行進度 第一年:完成步驟 1~9。 第二年:完成步驟 10~18。. (四) 完成之工作項目及成果: 4.1. 完成之工作項目 第一年完成之工作項目: 1. 證明 Gk ,n ,n≥2 及 k≥3,是一個分支度為 k 的正則圖,且有. kn(n 1) n 個邊。 2. 2. 證明 G3, n 和 TCn 為同構。 3. 證明對於 Gk ,n 裡的任意一個節點 v ,我們有: DIST (v) DIST (v' ) 1 or 0, 其中 v' (v) 而且 f , f 1 , g1 , g 2 , , g k 2 。. . 4. 證明對於 G k ,n 中 identity node I,對於任意的  f ,. 1. f ,g ,g 1. 2. . ,, g k 2 ,. DIST(I) = 0 而且 DIST(  I ) = 1。 5. 在 Gk ,n 中給定一個任意節點 v ,設計最短路徑繞徑演算法 OPTIMAL_ROUTE,能夠正確建構從節點 v 到 identity node 之最短路徑. DIST(v)。 6. 證明對於 G k ,n 中任何一個節點 v,DIST(v)等於 v 與 identity node 之間的 距離。  5n  7. 證明對於 Gk ,n 中任何一個節點 v , DIST (v) max  2n, 2。此外當 k 6 且  2  5 n  n 4 時,diam   2 ;並且當 k 6 且 n=2,3 時,diam G k ,n 2n 。 G k ,n  2  . 8. 證明 Gk ,n 可以嵌入 (k 1) n 1 個長度為 n(k 1) 節點互斥的 F-cycles。 16.

(19) 9. 證明當 n>2 時,每一個 Gk , n 中的 F-cycle 與 n(k 2) 個不同的 F-cycle 相鄰; 當 n=2 時,每一個 Gk ,n 中的 F-cycle 與 k 2 個不同的 F-cycle 相鄰。. 第二年完成之工作項目: ~ 10. 證明一個 F-leader 為 v 0 x1 x2  xn2 的 F-cycle 是與 F-leader 為 ~ 0 y1 y 2  y n 1 的 F-cycle 相鄰的條件是下列的某一性質成立:一、對於某整. 數 i, 1 i n 1 , xi yi 而且對於所有的整數 j, 1 j n 1 且 j i , x j y j 。二、 x1 y1 x2 y 2  xn 1 y n 1 (mod k-1)。. 11. 證明 Gk ,n 可以嵌入 n(k 1) n 1 個長度為 n(k 1) 節點不相交大小為 k 1 的 G-clique。 12. 證明當 n>2 時,每一個 Gk ,n 中的 G-clique 與 2(k 1) 個不同的 G-clique 相鄰。 當 n=2 時,每一個 Gk ,n 中的 G-clique 與 k 1 個不同的 G-clique 相鄰。 13. 對於任意兩節點 u, v 而言,嵌入連接 u, v 兩節點的最長路徑(longest u,v-path) 於 Gk ,n 。 14. 嵌入漢彌頓圓(Hamiltonian cycle)於 Gk ,n 。 15. 嵌入樹網路(tree networks)於 Gk ,n 。 16. 證明簡化圖的連結度至少為 k,亦即 ( RGk ,n ) k for n 3。 17. 利用在ξ-連通圖中給定兩個節點集合 V1 跟 V2 ,其中 V1 V2 ,則存在 ξ條連接V1 跟V2 的節點互斥路徑之特性證明 ( G k , n ) k 。 18. 證明 k-valent 圖的連結度為 k,亦即 ( G k , n ) k 。. 17.

(20) 4.2. 對於學術研究方面之成果 本計劃對互連網路設計之相關研究提出一種嶄新的觀念:可調整分支度之 互連網路設計。先前針對“ 可調整分支度之互連網路”方面的研究相當缺乏。 這類的互連網路在科學計算上有其重要之應用:在某些只適用於有固定數目的 I/O 埠上之計算[7],我們可視計算量的多寡調整互連網路之分支度。 本計劃 提出一種新型可調整分支度之互連網路-k-valent 圖 Gk,n,並證明它有許多優異 之性質。除此之外,Gk,n 之最短路徑繞演算法也被實作出來。 我們也證明 Gk,n 可嵌進節點互斥環路(node-disjoint cycles)。眾所周知,環 路結構非常適合平行與分散式處理,例如,在區域網路中設計低溝通需求的平 行演算法[29]。另外,環路結構在任意的網路中也可作為流量/資料控制之用 [29]。我們的研究成果將可證明 Gk,n 圖可有效模擬(emulate)環路結構;所以, 先前許多已被發展出來的環路演算法[3]也可在 Gk,n 圖中執行。 這一系列研究成果可提供給日後從事“ 可調整分支度之互連網路”相關 研究者參考。. 參考資料 [1] S. B. Akers and Krishnamurthy, “ The star graph: an attractive alternative to n-cube,”in Proceedings of International Conference on Parallel Processing, St. Charles, IL, pp. 555-556, 1987. [2] S. B. Akers and Krishnamurthy, “ A group-theoretic model for symmetric interconnection networks,”IEEE Transactions on Computers, 38(4):555-556, 18.

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(25) [36] Junming Xu, “To p ol o gi c a ls t r uc t u r e an da n al y s i s of interconnection ne t wo r k s ”, Kluwer Academic, 2001. [37] H. G. Yeh and G. J. Chang, Weighted connected domination and Steiner trees in. distance-hereditary. graphs,. Discrete. Applied. Mathematics,. 87(1--3):245--253, 1998. [38] H. G. Yeh and G. J. Chang, Linear-time algorithms for bipartite distance-hereditary graphs, Manuscript. [39] H. G. Yeh and G. J. Chang, The path-partition problem in bipartite distance-hereditary graphs, Taiwaness Journal of Mathematics, 2(3):353--360, 1998.. (五) 發表之期刊論文: Sun-Yuan Hsieh and Tien-Te Hsiao, “ The k-degree Cayley graph and its topological properties, ”Networks, vol. 47, issue 1, pp. 26-36, 2006 (SCI; impact factor 0.742). A preliminary version of this paper appeared in Proceedings of the 2004 International Conference on Parallel Processing (ICPP), pp. 206-213, IEEE Computer Society Press (EI), under the title “ Topol o gi c a l properties, optimal Routing, and embedding on the k-valent gr a ph” .. 23.

(26) The k -Degree Cayley Graph and its Topological Properties. Sun-Yuan Hsieh, Tien-Te Hsiao Department of Computer Science and Information Engineering, National Cheng Kung University, No. 1, University Road, Tainan 701, Taiwan. This article introduces a new family of Cayley graphs, called k -degree Cayley graphs, for building interconnection networks. The k -degree Cayley graph possesses many valuable topological properties, such as regularity with degree k , logarithmic diameter, and maximal fault tolerance. We present an optimal shortest path routing algorithm for the k -degree Cayley graph. Cycleembedding and clique-embedding are also discussed. © 2005 Wiley Periodicals, Inc. NETWORKS, Vol. 47(1), 26–36 2006. Keywords: graph-theoretic interconnection networks; network topology; topological properties; embedding; shortest path routing algorithm. 1. INTRODUCTION The design of interconnection networks is an important issue in parallel processing or distributed systems, and many networks have been proposed in the literature [1, 2, 5, 7, 8, 10, 14–16, 18, 20, 22, 23, 25, 27]. However, the choice of network topology significantly affects the performance of such systems. In [29], several important measurements, such as connectivity, degree, diameter, symmetry, and fault tolerance for building interconnection networks are discussed. An interconnection network is often modeled as an undirected graph, in which the nodes correspond to the processors and the edges correspond to the communication channels. Thus, the terms graph and network, node and processor, and edge and channel are used interchangeably throughout this article. A class of Cayley graphs based on permutation groups has proven to be suitable for designing interconnection networks [2, 3]. Several well-known interconnection networks have been based on Cayley graphs, including hypercubes [5], Received January 2005; accepted October 2005 Correspondence to: S.-Y. Hsieh; e-mail: [email protected] Contract grant sponsor: National Science Council; Contract grant numbers: NSC93-2213-E-006-084 and NSC94-2213-E-006-025. DOI 10.1002/net.20096 Published online in Wiley InterScience (www.interscience.wiley. com). © 2005 Wiley Periodicals, Inc.. k-ary n-cubes [5], star graphs [1], and pancake graphs [2], all of which are regular and symmetric. However, the node degree (degree for short) increases with the number of nodes so that using these graphs is prohibitive in networks with a large number of nodes. There are applications where the computing nodes in an interconnection network only have a fixed number of I/O ports [7]. Moreover, fixed degree networks are important from the viewpoint of VLSI implementation [23]. Therefore, maintaining a fixed degree irrespective of the number of nodes is an important aspect of interconnection network design. Vadapalli and Srimani [25] proposed the trivalent Cayley graph with a fixed degree of 3. This class of graphs is known to have logarithmic diameter and maximal fault tolerance. Several topological properties and shortest path routing algorithms are discussed in [21, 24–26, 28]. Vadapalli and Srimani [27] also proposed a family of Cayley graphs of constant degree 4, which is isomorphic to the wrapped around butterfly graph [9]. However, both the degree and connectivity of the trivalent Cayley graph (respectively, the wrapped around butterfly graph) are restricted to 3 (respectively, 4), which may not be reliable when a large number of faulty nodes occur in a network. Also, low connectivity leads to high contention in communication. More recently, Fu and Chau [13] proposed cyclic-cubes (generalized butterfly graphs), which are Cayley graphs with fixed degrees being any even number greater than or equal to 4. These graphs have optimal fault tolerance and logarithmic diameters. Shortest path routing and embedding of a Hamiltonian cycle, meshes, and hypercubes are also discussed in [13]. In this article, we generalize the concept of constantdegree (even-degree) and propose a new family of Cayley graphs, called k-degree Cayley graphs, Gk,n . These graphs possess a useful property in that the degree of each node is fixed by a general positive integer k without regard to the number of nodes. We show that the trivalent graph defined in [25] is a subclass of k-degree Cayley graphs when k = 3. Furthermore, we show that Gk,n has logarithmic diameter when k ≥ 6, maximal fault tolerance, and can embed disjoint cycles or cliques. We also present a shortest path routing algorithm. Compared with the previously proposed Cayley. NETWORKS—2006—DOI 10.1002/net. 24.

(27) TABLE 1. Topological comparison of various Cayley graphs, where (a)–(g) mean the number of nodes, diameter, degree, connectivity, maximal fault tolerance, Hamiltonian, and disjoint cliques embedding, respectively. Topology. (a). (b). (c). (d). (e). (f). (g). n-Cycle Trivalent graph [25] Cube-connected-cycle [22] Wrapped around butterfly graph [17] n-Dimensional hypercube [5] k-Ary n-cube [5] n-Dimensional star graph [1]. n n2n n2n n2n 2n kn n!.  n2   5n −2 2 2n − 1 + max{1,  n−2 2 }  3n 2  n n 2k   3(n−1) 2 . 2 3 3 4 n 2n n−1. 2 3 3 4 n 2n n−1. yes yes yes yes yes yes yes. yes yes yes yes yes yes yes. no yes unknown unknown unknown unknown unknown. 2k k. 2k k. yes yes. yes unknown. unknown yes. Cyclic-cubes [13] k-Degree Cayley graph Gk,n. nk n n(k − 1)n.  3n 2  for k ≥ 6 and n ≥ 4; 2n for k ≥ 6 and n = 2, 3.  5n −2 2. graphs shown in Table 1, Gk,n possesses a much larger number of nodes for a given diameter. For example, with diameter 6 and fixed degree 6, the six-dimensional hypercube has 64 nodes, the 4-ary 3-cube has 64 nodes, the cyclic-cube has 324 nodes, but G6,3 possesses 375 nodes. We note here that Gk,n is generalized from the trivalent graph (which is isomorphic to G3,n ), not from the cube-connected-cycle. The trivalent graph and the cube-connected-cycle both have the same number of nodes (see Table 1) and are regular of constant degree 3. However, both graphs are not isomorphic. The remainder of this article is organized as follows. In the next section, we introduce the k-degree Cayley graph. In Section 3, a shortest path routing algorithm that gives the diameter is presented. Embedding and connectivity are discussed in Section 4 and Section 5, respectively. Finally, some concluding remarks are given in Section 6.. ∗ 3. gi (um ) = vm , where um = s0∗ s1∗ · · · sn−1 , vm = ∗ γ ∗ and γ = (sn−1 + i) mod (k − 1) for s0∗ s1∗ · · · sn−2 1 ≤ i ≤ k − 2.. ˜ 01, ˜ 10, ˜ 11, ˜ 00, ˜ 01, ˜ 10, ˜ 11} ˜ is the node set For example, {00, −1 of G3,2 . Because f (respectively, f ) is the inverse of f −1 (respectively f ), and the inverse of gi equals gk−i−1 for 1 ≤ i ≤ k − 2, the edges of Gk,n are bidirectional. Moreover, according to Definition 1, k-degree Cayley graphs form a class of Cayley graphs, a proof of which is given in Appendix A. Figure 1 illustrates a 3-degree Cayley graph G3,2 . Proposition 1. Gk,n , n ≥ 2 and k ≥ 3, is a regular graph n of degree k with kn(k−1) edges. 2 Proof.. Straightforward.. ■. 2. THE k -DEGREE CAYLEY GRAPH The following is a formal definition of the k-degree Cayley graph.. Vadapalli and Srimani [25] defined the trivalent Cayley graphs TCn as follows.. Definition 1. A k-degree Cayley graph Gk,n is an undirected graph with N = n(k − 1)n nodes for any integers n ≥ 2 and k ≥ 3. Each node v of Gk,n has the form s0 s1 · · · sm−1 s˜m sm+1 · · · sn−1 corresponding to a string of n symbols selected from {0, 1, . . . , k − 2} such that exactly one symbol s˜m is in marked form and the others are in unmarked form. We sometimes use vm to represent a node v with the marked symbol on position m; thus, the notations vm and v are used interchangeably throughout this article. Let si∗ = si or s˜i . Each edge is of type (v, δ(v)), where δ ∈ {f , f −1 , g1 , g2 , . . . , gk−2 } is a generator defined as follows:. Definition 2. Each node of a trivalent Cayley graph TCn , n ≥ 2, corresponds to a circular permutation in lexicographic order of n symbols, t1 , t2 , . . . , tn , complemented or uncomplemented. The node set of TCn is defined ∗ · · · t ∗ t ∗ t ∗ · · · t ∗ |t ∗ = t or ¯t for all as V (TCn ) = {tj∗ tj+1 i i n 1 2 j−1 i i, j ∈ {1, . . . , n}}. Let j = j. Each edge is of the type. ∗ 1. f (um ) = v(m−1) mod n , where um = s0∗ s1∗ · · · sn−1 , ∗ ∗ ∗ (m−1) mod n ∗ v = s1 s2 · · · sn−1 α and α = (s0 + 1) mod (k − 1); ∗ 2. f −1 (um ) = v(m+1) mod n , where um = s0∗ s1∗ · · · sn−1 , ∗ (m+1) mod n ∗ ∗ ∗ = β s0 s1 · · · sn−2 and β = (sn−1 − 1) mod v (k − 1);. FIG. 1.. A three-degree Cayley graph G3,2 .. NETWORKS—2006—DOI 10.1002/net. 25. 27.

(28) −1 (v, δ(v)), where δ ∈ {fTC , fTC , gTC }, which is defined in the following way: ∗ ∗ ∗ ∗ ¯∗ fTC (tj∗ tj+1 · · · tn∗ t1∗ t2∗ · · · tj−1 ) = tj+1 · · · tn∗ t1∗ t2∗ · · · tj−1 tj −1 ∗ ∗ ∗ fTC (tj tj+1 · · · tn∗ t1∗ t2∗ · · · tj−1 ). ∗ ∗ = ¯tj−1 tj∗ · · · tn∗ t1∗ t2∗ · · · tj−2. ∗ ∗ ∗ ∗ gTC (tj∗ tj+1 · · · tn∗ t1∗ t2∗ · · · tj−1 ) = tj∗ tj+1 · · · tn∗ t1∗ t2∗ · · · ¯tj−1. Two graphs, G1 and G2 , are isomorphic if there is a oneto-one function π from V (G1 ) onto V (G2 ) such that (u, v) ∈ E(G1 ) if and only if (φ(u), φ(v)) ∈ E(G2 ). We now show that k-degree Cayley graphs contain trivalent Cayley graphs as a subclass. Theorem 2.. G3,n and TCn are isomorphic.. Proof. Let t1 , t2 , . . . , tn be a lexicographic order of n symbols for TCn . For each node s0 s1 · · · sm−1 s˜m sm+1 · · · sn−1 of G3,n , where si ∈ {0, 1}, we define a function π mapping V (G3,n ) to V (TCn ) as follows: π(s0 s1 · · · sm−1 s˜m sm+1 · · · sn−1 )    = tn−m+1 tn−m+2 · · · tn t1 · · · tn−m ,. (1). where ti = ti if s(m+i−1) mod n = 0 and ti = t¯i if s(m+i−1) mod n = 1, for i ∈ {1, . . . , n}. The function π is obviously oneto-one and onto.  ∗ and vm are two adjacent Suppose that um = s0∗ s1∗ · · · sn−1    nodes in G3,n . Let π(um ) = tn−m+1 tn−m+2 · · · tn t1 · · · tn−m according to Equation (1). There are three cases according to the adjacency relation of G3,n . . . fore, because (um , vm ) ∈ E(G3,n ) if and only if (π(um ),  ■ π(vm )) ∈ E(TCn ), G3,n and TCn are isomorphic. 3. SHORTEST PATH ROUTING AND DIAMETER A graph is said to be simple if it contains neither loops nor multiple edges. A path is a simple graph whose nodes can be ordered so that two nodes are adjacent if and only if they are consecutive on the list of nodes. Define the distance between u and v as the length of a shortest path between u and v. Because Gk,n is a Cayley graph, it is node-symmetric [2]. Thus, the distance between two arbitrary nodes is equal to the distance between some node (source node) and the identity ˜ · · · 0. Our shortest path routing algorithm attempts node 00 to construct a shortest path from a given source node to the identity node. We begin with the following definitions. ∗ , let v[i] = s∗ , 0 ≤ i ≤ n − 1, and For v = s0∗ s1∗ · · · sn−1 i ∗ ∗ let v[i, j] = si si+1 · · · sj∗ , 0 ≤ i < j ≤ n − 1. If we view ∗ v = s0∗ s1∗ · · · sn−1 as a string, then v[i, j] is a substring. An 0-substring is a substring whose symbols are all zero. Definition 3. Consider an arbitrary node v = s0 s1 · · · sm−1 s˜m sm+1 · · · sn−1 in Gk,n . The marked symbol s˜m divides the string into two parts: the left part s0 s1 · · · sm−1 and the right part s˜m sm+1 · · · sn−1 . We also define the following notations for v: • Lv [x] = |{si : si = x (mod k − 1) for 0 ≤ i ≤ m − 1}|. (Throughout this article, the equality “=” is also used for the congruence modulo k − 1 such that a = b (mod k − 1) iff a − b is a multiple of k − 1.) That is, Lv [x] is used to denote the number of symbols in the left part of v such that the remainder of each symbol after division by k − 1 is the same as that of x after division by k − 1. • Rv [x] = |{si : si = x (mod k − 1) for m ≤ i ≤ n − 1}|. That is, Rv [x] is used to denote the number of symbols in the right part of v such that the remainder of each symbol after division by k − 1 is the same as that of x after division by k − 1. • Let LZLv be the length of a longest 0-substring in the left part of v. That is, LZLv = max{l ≥ 0 : sj = sj+1 = · · · = sj+l−1 = 0, 0 ≤ j ≤ m − 1}. Also, let jLv indicate the position of the left end of the left-most longest 0-substring in the left part. In particular, define jLv = 0 if LZLv = 0. • Let LZRv be the length of a longest 0-substring in the right part of v. That is, LZRv = max{l ≥ 0 : sj = sj+1 = · · · = sj+l−1 = 0, m ≤ j ≤ n − 1}. Also, let jRv indicate the position of the left end of the left-most longest 0-substring in the right part. In particular, define jRv = m if LZRv = 0.. . Case 1. f (um ) = vm . Then, vm = v(m−1) mod n = s1∗ s2∗ · · · ∗ α ∗ , where α = (s + 1) mod 2; and π(v(m−1) mod n ) = sn−1 0     tn−m+2 · · · tn t1 · · · tn−m tn−m+1 . Thus, π(um ) and π(vm ) are  adjacent in TCn , because fTC (π(um )) = π(vm ). . . Case 2. f −1 (um ) = vm . Then, vm = v(m+1) mod n = β ∗ s0∗ s1∗ · · · ∗ , where β = (s (m+1) mod n ) = sn−2 n−1 − 1) mod 2. So π(v     tn−m tn−m+1 · · · tn t1 · · · tn−m−1 . Thus, π(um ) and π(vm ) are  −1 (π(um )) = π(vm ). adjacent in TCn , because fTC . . Case 3. g1 (um ) = vm . Then, vm = vm = s0∗ s1∗ s2∗ · · · γ ∗ ,    tn−m+2 ··· where γ = (sn−1 + 1) mod 2. So π(vm ) = tn−m+1      m m tn t1 · · · tn−m−1 tn−m . Thus, π(u ) and π(v ) are adjacent in  TCn , because gTC (π(um )) = π(vm ). . On the other hand, if π(um ) and π(vm ) are adjacent in   −1 TCn , then fTC (π(um )) = π(vm ), or fTC (π(um )) = π(vm ), or  gTC (π(um )) = π(vm ). With proofs similar to those of Cases 1–3, we obtain the following results, which show that um   and vm are adjacent in G3,n : (1) If fTC (π(um )) = π(vm ), then    −1 f (um ) = vm ; (2) If fTC (π(um )) = π(vm ), then f −1 (um ) = vm ;   and (3) If gTC (π(um )) = π(vm ), then g1 (um ) = vm . There-. 28. Throughout this article, the subscript v can be omitted from the notations Lv [x], Rv [x], LZLv , LZRv , jLv , and jRv if no ambiguity arises. ˜ Example 1. Consider a node 1002100 of G4,7 . Then, L[0] = R[0] = 2; L[1] = R[1] = 1; L[2] = 1; LZL = LZR = 2; jL = 1 and jR = 5. Next, we define the distance function used in our routing algorithm.. NETWORKS—2006—DOI 10.1002/net. 26.

(29) Definition 4. Consider an arbitrary node v in Gk,n . Define:   2n + m − 2LZL    − L[0] − R[1] − 2 if (m − jL − LZL) > 0 1. D1 (v) = 2n + m − 2LZL    − L[0] − R[1] otherwise.   3n − m − 2LZR    − R[0] − L[−1] − 2 if (n − j − LZR) > 0 R 2. D2 (v) = 3n − m − 2LZR    − R[0] − L[−1] otherwise. 3. D3 (v) = 2n + m − L[−2] − R[−1]. 4. D4 (v) = 3n − m − R[2] − L[1].. The four terms Di (v), 1 ≤ i ≤ 4, defined in Definition 4 represent the lengths of four possible paths in Gk,n from v to the identity node. The paths are constructed using algorithms Dist_1–Dist_4, respectively. As will be shown later, the distance function D(v) computes the length of a shortest path from v to the identity node. For ease of understanding our shortest path routing algorithm, we first present Algorithm Dist_1, which we use to generate a path of length D1 (v) from v to the identity node.. Algorithm 1: 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: 21:. Dist_1 (v, n, m, LZL, jL ). for i = 0 to (jL − 1) do v ← f (v) end for for i = 0 to (jL − 1) + (n − m) do if v[n − 1] = 1 then v ← gk−v[n−1] (v) end if v ← f −1 (v) end for if (m − jL − LZL) > 0 then for i = 1 to (m − jL − LZL − 1) do v ← f −1 (v) end for v ← gk−v[n−1]−1 (v) for i = 1 to (m − jL − LZL − 1) do v ← f (v) if v[n − 1] = 0 then v ← gk−v[n−1]−1 (v) end if end for end if. Definition 5. The distance function of v is defined as D(v) = min1≤i≤4 {Di (v)}. Figure 2 illustrates how Algorithm Dist_1 works. The string representing the given source node v is divided into four blocks. After executing lines 1–3 of the algorithm, we. FIG. 2. An illustration of Algorithm Dist_1, which moves symbols from the given source node v to the identity node.. have the intermediate node shown in (2). After executing lines 4–9, we have the intermediate node shown in (3) in which the symbols of blocks A and D are all 0. If m − jL − LZL > 0, then lines 10–21 are further executed to form the resulting identity node. ˜ is the source node in Example 2. Suppose that v = 1002100 G4,7 . Then, by Example 1, LZL = 2 and jL = 1. The string v is divided into four blocks in which Blocks A, B, C, and D are 1, 00, 2, and 100, respectively. Algorithm Dist_1 works on v as follows: Because jL = 1, lines 1–3 are executed f ˜ −→ ˜ in one step such that v = 1002100 0021002. Moreover, because (jL − 1) + (n − m) = 3, the for-loop (lines 4–9) g2 ˜ is executed four times to generate a path 0021002 −→ −1 −1 f g4 f g4 ˜ ˜ ˜ ˜ −→ 0021004 −→ 0002100 −→ 0002101 −→ 0000210 f −1 f −1 ˜ ˜ −→ 0000021˜ −→ 0000002. As m − jL − LZL = 0000211 4 − 1 − 2 = 1 > 0, lines 11–13 need to be considered further. The two for-loops (one is described in lines 11–13 and the other in lines 15–20) need not be executed, because m − jL − LZL − 1 = 0 < 1. Only line 14 is executed to g1 ˜ ˜ arrive at the identity node by 0000002 −→ 0000000. Because the concepts of algorithms Dist_2–Dist_4 are similar to that of Algorithm Dist_1, we present them in Appendix B. Our shortest path routing algorithm uses the shortest path generated by Algorithms Dist_1, Dist_2, Dist_3, and Dist_4. The following lemma helps prove the correctness of our routing algorithm. Lemma 3. For an arbitrary node v in Gk,n , D(v) − D(v ) = ±1 or 0, where v = δ(v) and δ ∈ {f , f −1 , g1 , g2 , . . . , gk−2 }. Proof. Assume that v = s0 s1 · · · sm−1 s˜m sm+1 · · · sn−1 and δ(v) = v = t0 t1 · · · tm −1 tm˜  tm +1 · · · tn−1 . Let L , R , LZL  , and LZR be the corresponding parameters (defined in Definition 3) of v . By the definition of the distance function, we can utilize the fact that D(v) − Dp (v ) ≤ D(v) − D(v ) ≤ Dq (v) − D(v ), where p, q ∈ {1, 2, 3, 4}, to simplify the derivation.. NETWORKS—2006—DOI 10.1002/net. 27. 29.

(30) Case 1. δ ∈ {g1 , g2 , . . . , gk−2 }. In this case, we have m = m, L  [x] = L[x], and LZL  = LZL. Therefore, in accordance with Definition 4, we obtain the following results. (1) D1 (v) − D1 (v ) = −R[1] + R [1] = ±1 or 0. (2) Let q = −2LZR − R[0] + 2LZR + R [0]. • If v[n − 1] = 0 and v [n − 1] = 0, then q = 0 and D2 (v) − D2 (v ) = q = 0; • If v[n − 1] = 0, v [n − 1] = 0, and LZR = LZR , then q = 1 and D2 (v) − D2 (v ) = 1; • If v[n − 1] = 0, v [n − 1] = 0, and LZR = LZR , then q = 3 and D2 (v) − D2 (v ) = q − 2 = 1; • If v[n − 1] = 0, v [n − 1] = 0, and LZR = LZR , then q = −1 and D2 (v) − D2 (v ) = q = −1; • If v[n − 1] = 0, v [n − 1] = 0, and LZR = LZR , then q = −3 and D2 (v) − D2 (v ) = q + 2 = −1. (3) D3 (v) − D3 (v ) = −R[−1] + R [−1] = ±1 or 0. (4) D4 (v) − D4 (v ) = −R[2] + R [2] = ±1 or 0.. On the other hand, we have   3n − 1 − 2LZL     − L  [0] − R [1] − 2 D1 (v ) =  3n − 1 − 2LZL    − L  [0] − R [1]     2n + 1 − 2LZR   − R [0] − L [−1] − 2 D2 (v ) =  2n + 1 − 2LZR    − R [0] − L  [−1] (v ) = 3n. (1) Because D1 (v)−D4 (v ) = −1−R[1]+R [2]+L  [1], we can show that D1 (v) − D4 (v ) = −1 without considering whether v[0] = 1. (2) Because D2 (v) − D1 (v ) = 1 − 2LZR − R[0] + R [1] + 2LZL  + L  [0], we have  ±1 if v[0] = 0;  D2 (v) − D1 (v ) = 1 otherwise. (3) Clearly, D3 (v) − D2 (v ) =  −1 − R[−1] + 2LZR + R [0] + L  [−1] if v[0] = k − 2; 1 − R[−1] + 2LZR + R [0] + L  [−1] otherwise.. . (1) D1 (v) − D1 (v ) = 1 − 2LZL − L[0] − R[1] + 2LZL + L [0] + R [1]. Then,  1 − 2LZL  + 2LZL  = ±1 if v[0] = 0; D1 (v) − D1 (v ) =  1 otherwise.. Combining (1)–(4), we have D(v) − D(v ) = ±1 or 0. Case 2.2. m = 0. In this case, we have m = n − 1, L[x] = 0, and LZL = 0. Thus, D1 (v) = 2n − R[1].   3n − 2LZR − R[0] − 2 D2 (v) =  3n − 2LZR − R[0]. if (n − jR − LZR) > 0; otherwise.. D3 (v) = 2n − R[−1]. D4 (v) = 3n − R[2] ≥ 2n. [Note that we do not need to consider D4 (v) when computing D(v).]. 30. otherwise.. R [−1]. Then, the following results hold:. Case 2.1. m = 0. In this case, m = m − 1.. (3) D3 (v)−D3 (v ) = 1− L[−2]−R[−1]+L  [−2]+R [−1]. By discussing the situations where v[0] = k − 3 or v[0] = k − 3, we can conclude that D3 (v) − D3 (v ) = 1. (4) D4 (v) − D4 (v ) = −1 − R[2] − L[1] + R [2] + L  [1]. By discussing the situations where v[0] = 1 or v[0] = 1, we have D4 (v) − D4 (v ) = −1.. if (n − jRv − LZR )>0;. D4 (v ) = 2n + 1 − R [2] − L  [1].. Case 2. δ = f . By Definition 4, we have the following results.. (2) Let q = −1 − 2LZR − R[0] − L[−1] + 2LZR + R [0] + L [−1]. D2 (v) − D2 (v ) may equal q or q − 2 or q + 2 in this case. Then,   −1 − 2LZR + 2LZR = ±1 if v[0] = k − 2; D2 (v)−D2 (v ) =  −1 otherwise.. otherwise.. D3 −1− − ≥ 2n − 1 ≥ D2 (v ) [this term can be ignored when computing D(v )]; and. Combining (1)–(4), we get D(v) − D(v ) = ±1 or 0.. . L  [−2]. if (m − jLv − LZL  )>0;. It is not difficult to verify that D3 (v)−D2 (v ) = 1 without considering whether v[0] = k − 2.. From (1)–(3), D(v) − D(v ) = ±1 or 0. Case 3. δ = f −1 . Then, v = f −1 (v) ⇔ f (v ) = v. This is ■ symmetric to Case 2. Lemma 4. For the identity node I in Gk,n , D(I) = 0 and D(δ(I)) = 1 for any δ ∈ {f , f −1 , g1 , g2 , . . . , gk−2 }. Proof.. Straightforward.. ■. Theorem 5. Given an arbitrary node v in Gk,n , our shortest path routing algorithm correctly generates an optimal (shortest) path of length D(v) from v to the identity node. Proof. Let dist(v) be the distance (shortest path length) between v and the identity node I. Clearly, our algorithm constructs a path of length D(v) from v to the identity node, which implies that dist(v) ≤ D(v). On the other hand, assume that v, v1 , v2 , . . . , vq−1 , vq , I forms a shortest path from v to I. Because vq is adjacent to I, by Lemma 4, D(vq ) = 1. Moreover, because vi is adjacent to vi−1 , by Lemma 3, either D(vi−1 ) = D(vi ) or D(vi−1 ) = D(vi ) ± 1. Thus, we have D(vq−1 ) ≤ D(vq ) + 1 = 2, D(vq−2 ) ≤ D(vq−1 ) + 1 ≤ 3, . . . , and D(v) ≤ D(v1 ) + 1 = q + 1 = dist(v). Because dist(v) ≤ D(v) ≤ dist(v), we can conclude that D(v) = dist(v). ■ We can utilize our shortest path routing algorithm to compute the diameter diam(Gk,n ) for k ≥ 6.. NETWORKS—2006—DOI 10.1002/net. 28.

(31) Theorem 6. For an arbitrary node v in Gk,n , D(v) ≤ 5n max{2n, 5n 2 − 2}. Moreover, diam(Gk,n ) =  2  − 2 when k ≥ 6 and n ≥ 4; and diam(Gk,n ) = 2n when k ≥ 6 and n = 2, 3. Proof. By Definition 4, we have D(v) ≤ min{2n + m − 2, 3n − m − 2, 2n + m, 3n − m} = min{2n + m − 2, 3n − m−2} ≤ 5n 2 −2 for m > 0, and D(v) ≤ min{2n, 3n−2} ≤ 2n for m = 0. Hence, the upper bound on the diameter is 5n max{ 5n 2 − 2, 2n}. Note that it is 2 − 2 for n ≥ 4. We now show the lower bound. For k ≥ 6 and n ≥ 4, / {0, 1, k − 2, k − 3} consider the node v in Gk,n such that v[i] ∈ and m = n/2. Thus, L[0] = L[1] = L[−1] = L[−2] = R[0] = R[1] = R[−1] = R[−2] = 0 and LZL = LZR = 0. By Definition 4, D(v) = min{2n +  n2  − 2, 3n −  n2  − 2} = 5n 5n  5n 2  − 2. Therefore,  2  − 2 ≤ diam(Gk,n ) ≤ 2 − 2 and diam(Gk,n ) =  5n 2  − 2. The case where k ≥ 6 and n = 2, 3 ■ can be shown similarly. 4. EMBEDDING In this section, we show that the k-degree Cayley graph can embed two types of structure. A cycle is a sequence of distinct nodes v1 , v2 , . . . , vq such that for each i = 1, 2, . . . , q − 1, (vi , vi+1 ) and (vn , v1 ) are in E(G), and all the edges (v1 , v2 ), (v2 , v3 ), . . . , (vq , v1 ) are distinct. A clique in a graph is a set of pairwise adjacent nodes. Two cycles are node-disjoint if they have no common node. An edge defined by the generator f (or f −1 ) is an Fedge, and any cycle in Gk,n consisting of only F-edges is an ˜ 01, ˜ 11, ˜ 10) ˜ shown in Figure 1 is an F-cycle. The cycle (00, example of an F-cycle.For a nonnegative integer i, we recurv if i = 0, sively define f i (v) = The notation i−1 f ( f (v)) if i > 0. (f −1 )i (v) can be defined similarly. Theorem 7. Gk,n can embed (k − cycles of length n(k − 1).. 1)n−1. node-disjoint F-. Proof. Given an arbitrary node v in Gk,n , it is not difficult to verify that f n(k−1) (v) = v and f i (v) = f j (v), where 1 ≤ i, j ≤ n(k − 1) and i = j. Thus, from an arbitrary node v, a cycle of length n(k − 1) in Gk,n can be generated by a functional iteration f n(k−1) (v). Because Gk,n has n(k − 1)n nodes, totally (k − 1)n−1 F-cycles each of length n(k − 1) can be generated. These cycles are node-disjoint by the fact that f (u) = f (v) if and only if u = v. ■ Each F-cycle has a unique node v such that v[i] = s and m = 0 for two given integers 0 ≤ i ≤ n − 1 and 0 ≤ s ≤ k − 2. Therefore, for each F-cycle C, we can define the C-leader (leader for short if no ambiguity arises) to be the unique node v of C, thereby satisfying v[0] = 0 and m = 0. Note that each leader uniquely determines a specific F-cycle. Two F-cycles C1 and C2 are said to be adjacent if there exists a node v1 ∈ C1 and a node v2 ∈ C2 such that v1 = gi (v2 ) for some 1 ≤ i ≤ k − 2.. Theorem 8. Each F-cycle of Gk,n is adjacent to n(k − 2) different F-cycles for n > 2. In the case of n = 2, each F-cycle of Gk,2 is adjacent to k − 2 different F-cycles. Proof. We first consider n > 2. For an arbitrary F-cycle ˜ 1 s2 · · · sn−1 , C with the C-leader v = 0s ˜ 1 s2 · · · s · · · sn−1 , (f −1 )i ◦ (gj ◦ f i (v)) = 0s. (2). where ◦ means the function composition; s = (si−1 + j) mod (k − 1) for all i = 1; and 1 ≤ j ≤ k − 2. Thus, (n − 1)(k − 2) leaders of (n − 1)(k − 2) distinct F-cycles can be obtained from v. This implies that the given F-cycle is adjacent to (n − 1)(k − 2) distinct F-cycles. Moreover, the other (k − 2) leaders can be obtained from v by  ˜ 1 s2 · · · sn−1 , (f −1 )1+jn ◦ (gj ◦ f (v)) = 0s. (3). where si = (si − j) mod (k − 1). Therefore, each F-cycle is adjacent to n(k − 2) different F-cycles. The case of n = 2 can be proved using either (2) ■ or (3). The proof of the above theorem leads to the following result. ˜ 1 x2 · · · xn−2 Corollary 9. An F-cycle with leader v = 0x is adjacent to the following F-cycles with leaders ˜ 1 y2 · · · yn−1 such that one of the following conditions 0y holds. 1. xi = yi for some 1 ≤ i ≤ n − 1, and xj = yj for all 1 ≤ j ≤ n − 1 and j = i; 2. x1 − y1 = x2 − y2 = · · · = xn−1 − yn−1 (mod k − 1).. We next show that Gk,n can embed cliques. An edge of Gk,n , defined by the generator gi , is called a G-edge. A clique is said to be maximal if it is not contained in another clique. A G-clique is a maximal clique in Gk,n such that each pair of adjacent nodes is connected by a G-edge. For example, ˜ 01} ˜ is a G-clique in G3,2 . Note that a G-clique contains {00, k − 1 nodes. Given some fixed 0 ≤ s ≤ k − 2, each G-clique has a unique node v that satisfies v[n − 1] = s. Therefore, for a G-clique K, we can define the K-leader (leader for short if no ambiguity arises) to be the unique node v ∈ K that satisfies v[n − 1] = 0. For convenience, we use such a node to represent its corresponding G-clique. Two G-cliques are node-disjoint if they have no common node. Theorem 10. Gk,n can embed n(k − 1)n−1 node-disjoint G-cliques of size k − 1. Proof. Clearly, there are n(k − 1)n−1 leaders in Gk,n that correspond to n(k−1)n−1 G-cliques. It is not difficult to verify that these G-cliques are pairwise node-disjoint. Moreover, because each G-clique contains k − 1 nodes, the n(k − 1)n−1 G-cliques partition the nodes of Gk,n . ■. NETWORKS—2006—DOI 10.1002/net. 29. 31.

(32) Two G-cliques K1 and K2 are said to be adjacent if there exists a pair of nodes u ∈ K1 and v ∈ K2 such that v = f (u) or v = f −1 (u). Theorem 11. Each G-clique in Gk,n is adjacent to 2(k − 1) (respectively, k − 1) different G-cliques when n > 2 (respectively, n = 2). Proof. We first consider the case where n > 2. Given an ∗ 0∗ , we arbitrary G-clique whose leader is v = s0∗ s1∗ · · · sn−2 i can apply f ◦ gj on v for i = ±1 and 1 ≤ j ≤ k − 2, to reach 2(k − 2) distinct G-cliques, and apply f ±1 on v to reach another two distinct G-cliques. Therefore, each G-clique is adjacent to 2(k − 1) distinct G-cliques. When n = 2, we can apply f ◦gj and f (or f −1 ◦gj and f −1 ) on v to reach k − 1 distinct G-cliques. Note that the G-cliques reached by {f ◦ gj (v), f (v)} are the same as those reached by {f −1 ◦ gj (v), f −1 (v)}. ■ 5. MAXIMAL FAULT TOLERANCE The node fault tolerance of an undirected graph is measured by the connectivity of the graph [29]. The connectivity κ(G) of a graph G is the minimum size of a node set S such that G − S is disconnected or has only one node. A graph G is ξ -connected if its connectivity is at least ξ . Obviously, the connectivity of G cannot exceed the minimum degree of a node in G; thus κ(Gk,n ) ≤ k, because Gk,n is k-regular. A graph is said to have maximal fault tolerance if its connectivity equals the minimum degree of the given graph. A set of paths from u to v is pairwise internally disjoint if each pair of paths in the set have no internal nodes in common. In this section, we show that κ(Gk,n ) = k; hence, this class of graphs has maximal fault tolerance. Lemma 12.. κ(G3,n ) = 3 for n ≥ 2.. Proof. This was shown in [25]. Thus, in the following, we need only consider the case where k ≥ 4. ■ Definition 6. For a k-degree Cayley graph Gk,n , the reduced graph RGk,n of Gk,n can be constructed using the following two steps: ˜ 1 s2 · · · sn−1 , shrink it 1. For each F-cycle whose leader is 0s to a single node, namely, a super node, and label it with s1 s2 · · · sn−1 ; 2. If two F-cycles are adjacent in Gk,n , then connect the corresponding super nodes by an undirected edge, namely, a super edge.. Note that the nodes in RGk,n are precisely the super nodes as defined, and the edges in this graph are precisely the super edges as mentioned. For convenience, we use the notation P[x, y] to denote a path from x to y.. 32. Lemma 13.. κ(RGk,n ) ≥ k for k ≥ 4 and n ≥ 3.. Proof. Consider a subgraph Rk,n of RGk,n defined as follows. The node set of Rk,n is the same as RGk,n ; and two nodes, x1 x2 · · · xn−1 and y1 y2 · · · yn−1 of Rk,n , are adjacent iff there is exactly one i such that xi = yi . By induction on n, we prove the lemma by showing that there are k pairwise internally disjoint paths between two arbitrary distinct nodes in Rk,n . First, we prove the result for n = 3. Consider two arbitrary nodes, u and v, in Rk,3 . Case 1. u and v have exactly one different symbol. Without loss of generality, assume that u = 0x and v = 0y. According to Corollary 9 and the construction of Rk,n , k pairwise internally disjoint paths from u to v can be constructed as follows: (0x, 0y), (0x, 1x, 1y, 0y), . . . , (0x, (k − 2)x, (k − 2)y, 0y), and (0x, 0z, 0y) for z ∈ {0, 1, . . . , k − 2}\{x, y}. Case 2. u and v have two different symbols. Assume that u = u1 u2 and v = v1 v2 . Then, according to Corollary 9 and the construction of Rk,n , 2k − 4 (≥ k when k ≥ 4) pairwise internally disjoint paths from u to v can be constructed as follows: (u1 u2 , u1 v2 , v1 v2 ), (u1 u2 , v1 u2 , v1 v2 ), (u1 u2 , xu2 , xv2 , v1 v2 ), and (u1 u2 , u1 x, v1 x, v1 v2 ) for all x ∈ {0, 1, . . . , k−2}\{u1 , v1 }. It is not difficult to verify that Rk,n has the following recursive structure, which can be decomposed into k − 1 copies of Rk,n−1 . Each Rk,n−1 is a subgraph of Rk,n induced by nodes in x1 x2 · · · xn−2 y for some fixed symbol y, and xi ∈ {0, 1, . . . , k − 2}. Consider two arbitrary nodes, u and v. If u = u1 u2 · · · un−2 0 and v = v1 v2 · · · vn−2 0 are in the same Rk,n−1 , then, by the induction hypothesis, there are k pairwise internally disjoint paths from u to v; otherwise, u and v are in different Rk,n−1 s. Assume that u = u1 u2 · · · un−2 0 and v = v1 v2 · · · vn−2 1. By the induction hypothesis, there are k pairwise internally disjoint paths P1 , P2 , . . . , Pk from u to the node v = v1 v2 · · · vn−2 0, because u and v are in the same Rk,n−1 . Let n1 , n2 , . . . , nk be the neighbors of v that belong to paths P1 , P2 , . . . , Pk , respectively. According to Corollary 9, and because all ni s are in the same Rk,n−1 , i we have ni = t1i t2i · · · tn−2 0 in which tji = vj for some j, i tl = vl for all 1 ≤ l ≤ n − 2, and l = j. Let ni be the neighbor of ni which is located in the same Rk,n−1 as v; that i is, ni = t1i t2i · · · tn−2 1. Note that the ni , for all 1 ≤ i ≤ k, are adjacent to v. Also, let Pi [u, ni ] be the subpath of Pi from u to ni , 1 ≤ i ≤ k. Then, k pairwise internally from  disjoint paths )  , v) for u to v can be constructed as P [u, n ] (n , n (n i i i i i all i = 1, 2, . . . , k, where means the path-concatenation  operation. In the remainder of the article, the notation is used to mean path-concatenation. ■ We need two additional lemmas to prove our main result. The proof of the following lemma can be found in [11, 19]. Lemma 14. Given two sets of nodes V1 and V2 such that |V1 | = |V2 | = ξ in a ξ -connected graph, there are ξ nodedisjoint paths connecting the nodes from V1 to V2 .. NETWORKS—2006—DOI 10.1002/net. 30.

(33) Lemma 15.. κ(Gk,n ) ≥ k for k ≥ 4 and n ≥ 2.. Proof. We prove this lemma by showing that there are k pairwise internally disjoint paths between two arbitrary nodes, u and v, in Gk,n . There are two cases.. nodes are all different, except   u (respectively, v). Then,  ] (u , v ) Q[v1 , v], and P[u, u2 ] (u, v), P[u, u 1 1  1 (u2 , v2 ) Q[v2 , v] form the desired three paths.. Otherwise, if u and v are not adjacent, then the desired three paths can be constructed as follows:. Case 1. n = 2. Recall that in this case, RGk,2 (= Rk,2 ) is a complete graph with k − 1 nodes. Thus, there are totally k − 1 F-cycles in Gk,2 .. I. Find two nodes u1 and u2 (respectively, v1 and v2 ) different from u (respectively, v) in F1 (respectively, F2 ) such that gi (u1 ) = v, gj (u2 ) = v1 , and gl (u) = v2 , where 1 ≤ i, j, l ≤ k − 2 (see also Fig. 3b). II. Let P [u, u1 ] and P [u, u2 ] (respectively, Q [v1 , v] and Q [v2 , v]) be two paths in F1 (respectively, F2 ) whose nodes areall different, except  u (respectively, v). Then,  P [u, u1 ] (u1 , v), (u, v2 ) Q [v2 , v], and P [u, u2 ] (u2 , v1 ) Q [v1 , v] form the desired three paths.. Case 1.1. u and v are in the same F-cycle C. This cycle is a concatenation of two internally disjoint paths from u to v, denoted by P1 and P2 , which contribute two paths. Let C1 , C2 , . . . , Ck−2 be the other F-cycles of Gk,2 . Then, the other k − 2 paths can be constructed by the following three steps: 1. For each Ci , identify the entry node, gji (u) ∈ Ci , and the exit node, gli (v) ∈ Ci , where 1 ≤ ji , li ≤ k − 2. It is not difficult to show that there exist ji and li such that gji (u) and gli (v) are in the same F -cycle, Ci . 2. For each Ci , construct a path Qi from gji (u) to gli (v) using the F-edges of the corresponding F-cycle. 3. Construct the desired  k internally disjointpaths: P1 , P2 , (u, gj1 (u)) Q1 (gl1 (v), v), (u,  gj2 (u)) Q2 (gl2 (v), v), . . . , (u, gjk−2 (u)) Qk−2 (glk−2 (v), v).. Case 1.2. u and v are in different F-cycles. Let F1 and F2 be the F-cycles to which u and v belong, respectively. We first construct three pairwise internally disjoint paths from u to v using F1 and F2 . If u and v are adjacent, then the desired paths can be constructed as follows: A. Find two nodes u1 and u2 (respectively, v1 and v2 ) different from u (respectively, v) in F1 (respectively, F2 ) such that gi (u1 ) = v1 and gi (u2 ) = v2 for some 1 ≤ i ≤ k − 2 (see also Fig. 3a). B. Let P[u, u1 ] and P[u, u2 ] (respectively, Q[v1 , v] and Q[v2 , v]) be two paths in F1 (respectively, F2 ) whose. The other k − 3 internally disjoint paths from u to v can be constructed from the other k − 3 f -cycles using a method similar to that in Case 1.1. Case 2. n ≥ 3. If u and v are in the same F-cycle, there exist k internally disjoint paths from u to v. Two paths are then constructed in the same F-cycle, and the other k − 2 paths are constructed from adjacent k − 2 F-cycles based on a method similar to that in Case 1. If u and v are in different F-cycles, then the sets of nodes {g1 (u), g2 (u), . . . , gk−2 (u), g1 ( f (u)), g1 ( f −1 (u))} (respectively, {g1 (v), g2 (v), . . . , gk−2 (v), g1 ( f (v)), g1 ( f −1 (v))}), which are in k different F-cycles denoted by {C1 , C2 , . . . , Ck } (respectively, {C1 , C2 , . . . , Ck }), can be reached from u (respectively, v). By Lemma 13 and Lemma 14, there are k node-disjoint paths connecting {C1 , C2 , . . . , Ck } and {C1 , C2 , . . . , Ck } in RGk,n . This implies that there exist k pairwise internally disjoint paths from u to v in Gk,n . ■ We conclude this section with the following theorem. Theorem 16.. κ(Gk,n ) = k for k ≥ 3 and n ≥ 2.. Proof. By Lemma 12 and Lemma 15, we have κ(Gk,n ) ≥ k. Also, by the fact that Gk,n is a k-regular graph, we have κ(Gk,n ) ≤ k. Thus, the result holds. ■ 6. CONCLUDING REMARKS. FIG. 3.. An illustration of Case 1.2 in the proof of Lemma 15.. In this article, we generalize the concept of the trivalent graph and propose a new family Gk,n of k-degree Cayley graphs for building interconnection networks. The proposed graph is regular of degree k, where k can be any integer larger than 3. We also show that Gk,n for some fixed integer k ≥ 6 has diameter logarithmic in the number of nodes as well as maximal fault tolerance. A shortest path routing algorithm is also presented. Furthermore, we demonstrate that Gk,n can embed cycles. A cycle structure, which is a fundamental topology for parallel and distributed processing, is suitable for local area networks and for the development of simple parallel algorithms. NETWORKS—2006—DOI 10.1002/net. 31. 33.

數據

Figure 1:An illustrate of Algorithm DIST_1.
Figure 2:An illustrate of Algorithm DIST_2.
Figure 3:An illustrate of Algorithm DIST_3.
Figure 4:An illustrate of Algorithm DIST_4.
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