3. We really care about the normal 3-5AT nad Subot Sum problems, not these add ones Make: the following changes to the SAT/Slut Sum reduction: Drop the wrique requirement on SAT, ie each clase can bare 1, 2, 3 of its clauses satisfied but not Drop the Left/Right requirement on Suhsat Sum, i.c. an instance Isg consists of a set of integers and a solution Sss chansen arbitrary mlect of them to sum in order to reach the target. . For the sache SAT instance C: - (au bord). C-vedore). Cs-1 or cure). C = leur dore) & 1 or eure) The Subext Sum instance becomes Pairs Tet Right Extra FI 1600 32000 01001 3000 10000 0000 100000000 10000 TODO O DIA 0000000000000 DKK10 0300 0101 04300 00100000000000000 11010 00030 00010 DOI30 0000 0000 0000 00000 0100100003 0011000003 0000000000000100000 . Before each integer in Island one digit indexed for each cluse C. Now it will also have one Ligit indexed for each variable 1. For both the Leftatui Right integer in this pair, this it digit will be 3 and the other y digits will be nema, . For each clause Cadd two new Integers to Sgs with all in the digit and a wo everywhere else. Change the Thrget to beling 3 in each of the cause digits and cach of the variable digits.in. Your Task: Suppose the crack gives you a satisfying solution Sgs to your instance Is Malisar) formed in this way, ie 5s is a subset of the integere in Iss that dels up exactly to T = 3333333333 Prove the following (a) There are still no carries in the sum (b) For each wrinhle row, the solution Sss still choces exactly one from buttesponding Left/Right peir of members in Iss. Hence, we can stil detine SAT = Map Sc) such that for mi variable, set it to be true if the Les Integer is chosen in the extresponding pair. (e) This solution Sgar sabies each cluse in Bar 3. We really care about the normal 3-SAT and Subset Sum problems, not these odd ones. Make the following changes to the SAT/Subset Sum reduction: a Drop the unique requirement on SAT, i.e. each clause can have 1, 2, or 3 of its clauses satisfied but not 0. . Drop the Left/Right requirement on Subset Sum, i.e. an instance Iss consists of a set of integers and a solution Sss chooses an arbitrary subset of them to sum in order to reach the target. . For the same SAT instance C1 = (a or bord), C2 = (a ord or e), Cg = (b or corne), C4 = (c or d or ne), & Cs= (na or core) The Subset Sum instance becomes Pairs Left Right Extra Extra 10000 30000 01001 30000 10000 00000 10000 00000 00100 03000 10000 03000 01000 00000 01000 00000 Ic 00010 00300 00101 00300 00100 00000 00100 00000 d 11010 00030 00010 00030 00010 00000 00010 00000 01001 00003 00110 00003 00001 00000 00001 00000 Before each integer in Iss had one digit indexed i for each clause C. Now it will also have one digit indexed for each variable xy. For both the Left and right integer in this jth pair, this jth digit will be 3 and the other jih digits will be zero. For each clause C, add two new integers to Sss with a l in the ith digit and a zero everywhere else. Change the Target to being 3 in each of the cause digits and each of the variable digits, i.e. T =3333333333 e Your Task: Suppose the oracle gives you a satisfying solution Sss to your instance Iss = MapIsar) formed in this way, Le. Sss is a subset of the integers in Iss that adds up exactly to T = 3333333333 Prove the following 3. We really care about the normal 3-5AT nad Subot Sum problems, not these add ones Make: the following changes to the SAT/Slut Sum reduction: Drop the wrique requirement on SAT, ie each clase can bare 1, 2, 3 of its clauses satisfied but not Drop the Left/Right requirement on Suhsat Sum, i.c. an instance Isg consists of a set of integers and a solution Sss chansen arbitrary mlect of them to sum in order to reach the target. . For the sache SAT instance C: - (au bord). C-vedore). Cs-1 or cure). C = leur dore) & 1 or eure) The Subext Sum instance becomes Pairs Tet Right Extra FI 1600 32000 01001 3000 10000 0000 100000000 10000 TODO O DIA 0000000000000 DKK10 0300 0101 04300 00100000000000000 11010 00030 00010 DOI30 0000 0000 0000 00000 0100100003 0011000003 0000000000000100000 . Before each integer in Island one digit indexed for each cluse C. Now it will also have one Ligit indexed for each variable 1. For both the Leftatui Right integer in this pair, this it digit will be 3 and the other y digits will be nema, . For each clause Cadd two new Integers to Sgs with all in the digit and a wo everywhere else. Change the Thrget to beling 3 in each of the cause digits and cach of the variable digits.in. Your Task: Suppose the crack gives you a satisfying solution Sgs to your instance Is Malisar) formed in this way, ie 5s is a subset of the integere in Iss that dels up exactly to T = 3333333333 Prove the following (a) There are still no carries in the sum (b) For each wrinhle row, the solution Sss still choces exactly one from buttesponding Left/Right peir of members in Iss. Hence, we can stil detine SAT = Map Sc) such that for mi variable, set it to be true if the Les Integer is chosen in the extresponding pair. (e) This solution Sgar sabies each cluse in Bar 3. We really care about the normal 3-SAT and Subset Sum problems, not these odd ones. Make the following changes to the SAT/Subset Sum reduction: a Drop the unique requirement on SAT, i.e. each clause can have 1, 2, or 3 of its clauses satisfied but not 0. . Drop the Left/Right requirement on Subset Sum, i.e. an instance Iss consists of a set of integers and a solution Sss chooses an arbitrary subset of them to sum in order to reach the target. . For the same SAT instance C1 = (a or bord), C2 = (a ord or e), Cg = (b or corne), C4 = (c or d or ne), & Cs= (na or core) The Subset Sum instance becomes Pairs Left Right Extra Extra 10000 30000 01001 30000 10000 00000 10000 00000 00100 03000 10000 03000 01000 00000 01000 00000 Ic 00010 00300 00101 00300 00100 00000 00100 00000 d 11010 00030 00010 00030 00010 00000 00010 00000 01001 00003 00110 00003 00001 00000 00001 00000 Before each integer in Iss had one digit indexed i for each clause C. Now it will also have one digit indexed for each variable xy. For both the Left and right integer in this jth pair, this jth digit will be 3 and the other jih digits will be zero. For each clause C, add two new integers to Sss with a l in the ith digit and a zero everywhere else. Change the Target to being 3 in each of the cause digits and each of the variable digits, i.e. T =3333333333 e Your Task: Suppose the oracle gives you a satisfying solution Sss to your instance Iss = MapIsar) formed in this way, Le. Sss is a subset of the integers in Iss that adds up exactly to T = 3333333333 Prove the following