1. Binary Decomposition Approach: Considering R: Keys ={BEFG, CEFG, EFGH} F = {CD A, EC H, GHB AB, C D, EG A, H B, BE CD, EC B}

Transcription

1 Question 1: Consider R = {A, B, C, D, E, F, G, H} with set of FDs F = {CD A, EC H, GHB AB, C D, EG A, H B, BE CD, EC B} The cndidte keys re: {BEFG, CEFG, EFGH} Is R w.r.t. F in 3NF? If not decompose it into reltions in 3NF. No, R w.r.t. F is NOT in 3NF, becuse CD A violtes the 3NF requirements. i.e. CD A is not trivil FD CD is not superkey CD is not key, but A is not prt of ny key of R either 1. Binry Decomposition Approch: Considering R: Keys ={BEFG, CEFG, EFGH} F = {CD A, EC H, GHB AB, C D, EG A, H B, BE CD, EC B} Decomposition #1: CD A is violting FD R is decomposed into R 1 nd R 2 : R 1 (A,C,D): We need to project FDs F onto reltion R 1 : A+ = A C+ = CDA (C DA) D+ = D AC+ = ACD (AC D) AD+ = AD CD+ = CDA (CD A) So, F 1 = {C DA, AC D, CD A} R 2 ( B,C,D,E,F,G,H): In generl, we should project F onto R 2. However, if we look crefully, we cn esily see tht the only difference between R nd R 2 is ttribute A. Attribute A hs never ppered on LHS of ny FD. So, removing it won t mke ny chnge in F 2. So, F 2 ={EC H, GHB B, C D, H B, BE CD, EC B} FDs tht re lost in Decomposition #1 re: Lost = {GHB A, EG A} Do we need further decomposition? Consider R 1 (A,C,D): 1

2 F 1 = {C DA, AC D, CD A} Since C + =ACD, C is key. C (in C DA), CD (in CD A), nd AC (in AC D) re key/super keys. Therefore, we hve no violting FD. (So, we re done with this brnch.) Consider R 2 : F 2 ={EC H, C D, H B, BE CD, EC B} Keys of R 2 = Keys of R = {BEFG, CEFG, EFGH} EC H is not violting FD, since H is prt of key. C D is violting FD, since C is not super key nd D is not prt of ny key. So, further decomposition is needed. Decomposition #2: C D is violting FD R 2 is decomposed into R 21 nd R 22 : R 21 (C, D): We need to project F 2 onto reltion R 21 : C+=CD D+=D So, F 21 ={C D} R 22 (B,C,E,F,G,H): In generl, we should project F 2 onto R 22. However, if we look crefully, we cn esily see tht the only difference between R 2 nd R 22 is ttribute D. Attribute D hs never ppered on LHS of ny FD. So, removing it won t mke ny chnge in F 22. So, F 22 = {EC H, H B, BE C, EC B} FDs tht re lost in Decomposition #2 is: Lost = {BE D} So, overll, we ve lost the following FDs: Lost = {GHB A, EG A, BE D} Do we need further decomposition? Consider R 21 (C,D): F 21 ={C D} Since C + =CD, C is key. Therefore, we hve no violting FD. (So, we re done with this brnch.) Consider R 22 (B,C,E,F,G,H): Keys of R 22 = Keys of R 2 = {BEFG, CEFG, EFGH} 2

3 F 22 = {EC H, H B, BE C, EC B} EC H is not violting FD since H is prt of key. H B is not violting FD since B is prt of key. BE C is not violting FD since C is prt of key. EC B is not violting FD since B is prt of key. So, we re done with this brnch. Overll, we hve: R 1 (A, C, D) R 21 (C, D) R 22 (B, C, E, F, G, H) F 1 = {C DA, AC D, CD A} F 21 = {C D} F 22 = {EC H, H B, BE C, EC B} Since R 1 includes R 21 we might wnt to remove R 21. This is loss-less join decomposition, but it is not dependency preserving. To mke the decomposition dependency preserving, we need to dd the lost FDs s new reltions. The lost FDs re: Lost = {GHB AB, EG A, BE D} So, we dd three reltions: L 1 (A, B, G, H) F L1 = {GHB AB} L 2 (A, E, G) F L2 = {EG A} L 3 (B, D, E) F L3 = {BE D} 3

4 2. Synthesis Approch: R = {A, B, C, D, E, F, G, H} with set of FDs F = {CD A, EC H, GHB AB, C D, EG A, H B, BE CD, EC B} The cndidte keys re {BEFG, CEFG, EFGH} In the lst tutoril, we found tht the cnonicl cover for F is: F C = {C AD, EC H, GH A, EG A, H B, BE C} Now, we crete the reltions: R 1 = {A, C, D} F 1 = {C AD} R 2 = {E, C, H} F 2 = {EC H} R 3 = {A, G, H} F 3 = {GH A} R 4 = {A, E, G} F 4 = {EG A} R 5 = {B, H} F 5 = {H B} R 6 = {B, C, E} F 6 = {BE C} Now, we need to check if t lest one of the keys exists in the bove reltions. The cndidte keys re {BEFG, CEFG, EFGH} Since none of these keys is in the reltions, this decomposition is not lossless. So, we need to dd n extr reltion contining those ttributes tht form ny key of R: R 7 = {B, E, F, G} F 7 = {} 4

5 Question 2 Check if the decomposition of R(A, B, C, D, E, F, G) with the set of FDs F={C AD, E G, FG A, EF A, G B, BE C} into the following reltions is lossless join. R 1 = {A, C, D} R 2 = {E, C, G} R 3 = {A, F, G} R 4 = {A, E, F} R 5 = {B, G} R 6 = {B, C, E} Step 1- Tble initiliztion R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2a b2b b2d b2f R 3 ={A,F,G} b3b b3c b3d b3e R 4 ={A,E,F} b4b b4c b4d b4g R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6a b6d b6f b6g Round 1 Considering C AD R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} new b2b new b2f R 3 ={A,F,G} b3b b3c b3d b3e R 4 ={A,E,F} b4b b4c b4d b4g R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} new new b6f b6g Considering E G R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2b b2f R 3 ={A,F,G} b3b b3c b3d b3e R 4 ={A,E,F} b4b b4c b4d new R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f new 5

6 Considering FG A R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2b b2f R 3 ={A,F,G} b3b b3c b3d b3e R 4 ={A,E,F} b4b b4c b4d R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f Considering EF A R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2b b2f R 3 ={A,F,G} b3b b3c b3d b3e R 4 ={A,E,F} b4b b4c b4d R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f Considering G B R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} new b2f R 3 ={A,F,G} new b3c b3d b3e R 4 ={A,E,F} new b4c b4d R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f Considering BE C R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2f R 3 ={A,F,G} b3c b3d b3e R 4 ={A,E,F} new b4d R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f 6

7 Round 2 Considering C AD R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2f R 3 ={A,F,G} b3c b3d b3e R 4 ={A,E,F} new R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f We don t need to continue since we found one row in the tble with ll cells hving So, this is lossless join. R 1 ={A,C,D} b1b b1e b1f b1g R 2 ={E,C,G} b2f R 3 ={A,F,G} b3c b3d b3e R 4 ={A,E,F} R 5 ={B,G } b5a b5c b5d b5e b5f R 6 ={B,C,E} b6f 7