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Binary Clustering with Missing Values (BCMV)

Binary Clustering with Missing Values (BCMV). 作者 : A. Figueroa, J. Borneman , T. Jiang 報告人 : 劉效飛. N. N. Problem Formulation. 1. 兩個 vector 怎 樣才叫相似 ? Def: 沒有任何一個位置是互斥的， 則稱為相似 i.e. 將 每一個 vector 看成指紋採樣， 則兩個指紋採樣可能是同一個犯人所留下 Ex: (1 0 N N 1) , (1 N 1 0 1) are similar

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Binary Clustering with Missing Values (BCMV)

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  1. Binary Clustering with Missing Values(BCMV) 作者: A. Figueroa, J. Borneman, T. Jiang 報告人: 劉效飛

  2. N N

  3. Problem Formulation 1. 兩個vector怎樣才叫相似? Def: 沒有任何一個位置是互斥的，則稱為相似 i.e. 將每一個 vector 看成指紋採樣， 則兩個指紋採樣可能是同一個犯人所留下 Ex: (1 0 N N 1) , (1 N 1 0 1) are similar ∵(1 0 1 0 1)可以同時留下這兩種指紋 (1 0 N N 1), (0 N 1 1 1) is not similar ∵沒人可以同時留下這兩種指紋

  4. 2. 一群兩兩相似的vectors，之後簡稱為群 3. BCMV Def: 將vectors分割成最少個群

  5. Topic 1: NP_complete? • Theorem 1: BCMV(1) is in P • Theorem 2: BCMV(k) is NPC for all k >= 3 but BCMV(2) is still unknown

  6. Proof of Theorem 2 Def 4: Minimum Clique Partition (MCP): 將一個graph 分割成最少個clique Lemma: MCP is NPC on graphs of bounded degree k，k >= 3 Claim: MCP(k) can be reduced to BCMV(k) k >= 3

  7. 2 3 1: 1 N 0 0 0 N 2: N 1 N 0 N 0 3: 0 N 1 0 0 N 4: 0 0 0 1 N 0 5: 0 N 0 N 1 N 6: N 0 N 0 N 1 Ex: 4 1 6 5 原因: Two vectors are similar iff the corresponding nodes are adjacent.

  8. Proof of Theorem 1 Lemma : Vertex Cover on bipartie graphs is in P Claim : BCMV(1) could be reduced to Vertex Cover on bipartie graphs

  9. BCMV的另一種解釋: 輸入: 每一個 0_N_1 vector 都是某個犯人的指紋採樣 目標: 這些犯人如何請最少的人頭，來幫所有人頂罪 EX: 警方資料 A. 0 0 1 1 N B. 1 N 1 N 1 • 0 0 1 N 1 • N 1 N 1 1 E. N N 1 1 1 最省錢的方式，只需請二個人頭 人頭甲 0 0 1 1 1 可以幫A, C, E 犯人頂罪 人頭乙 1 1 1 1 1 可以幫B, D, E 犯人頂罪

  10. 可以用k個人頭頂罪  可以將vectors 分割成k群 Pf: (=>) A. 0 0 1 1 N B. 1 N 1 N 1 C. 0 0 1 N 1 D. N 1 N 1 1 E. N N 1 1 1 人頭甲 0 0 1 1 1 可以幫{A, C, E} 頂罪 {A, C, E}, {B, D, E} 一定是群 (<=) ∵每個群至少存在一個人 頭可以幫 忙整個群頂罪

  11. Reduce 方法(犯人觀點): • 將所有可已用來頂罪的人頭列出， • 再用node代替人頭，含有奇數個1的人頭放一邊，偶數個1的人頭放另一邊 • 將所有犯人列出，再用edge代替犯人 • If 人頭甲可以替犯人A頂罪 then 將node 甲和edge A接起來

  12. Step 2 甲 丙 戊 (竒) 乙 丁 (偶) Ex: Step 1: A. 0 1 N 1 0 => 甲. 0 1 1 1 0 乙. 0 1 0 1 0 B. N 1 1 0 0 => 丙. 1 1 1 0 0 丁. 0 1 1 0 0 C. 0 1 1 N 0 => 甲. 丁. D. 0 1 N 0 0 => 丁. 戊. 0 1 0 0 0 Step 3 Step 4 甲 丙 戊 (竒) 乙 丁 (偶) A B C D A D B C

  13. 為何出來結果一定是bipartie graph? ∵同一邊的人頭一定差兩個bits以上 Ex: E 甲 丙 戊 (竒) 乙 丁 (偶) 人頭甲: 0 1 1 1 0 人頭丙: 1 1 1 0 0 犯人E: N …… N …矛盾 A D B C

  14. Topic 2: Approximation • Def : minimum set covering problem Input: Collection C of subsets of a finite set S. output: 從C中挑出最少的sets，cover S的每個element

  15. 1. 若每個element 最多出現k個 sets 中， 則 approximable within k Pf: 觀念: ∵ 每挑錯一個set，最後set cover的總數 最多加一個  只要確定每挑 k個sets，其中一定有個 是對的

  16. 觀察: 將每個人頭看成一個set， 則可以變成 minimum set covering的問題 犯人指紋 人頭 set A: 1 N 1 1 0 1 1 1 1 0: {A, B} 1 0 1 1 0: {A, C} B: N 1 1 1 N => 1 1 1 1 1: {B} 0 1 1 1 1: {B} 0 1 1 1 0: {B} C: 1 0 1 N 1 1 0 1 1 1: {C}  approximation ratio 一定可以到達2^p

  17. Algorithm 1. repeat if element f 只屬於 V這個set， 將 V選入set cover else go to 2 2. 選可以包含最多elements 的 set， go to 1 Notes: 1. Approximable within 1+ ln|S| 2. Not approximable within (1 - ) ln|S| for any , unless NP

  18. Conclusions • 1. Is BCMV(2) NPC? • 2. Better Approximation ratio?

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