2019-03-18
1. Two reducers are used in the algorithm, so it takes a total of 40 rounds are needed.
2. Tuples can be adjusted to hold the value of the accumulator which will be passed to the next processor.
3. LoadProcId would pass in the value k into the value parameter.
4. No, the processors perform their steps independently at time t
1. Two reducers are used in the algorithm, so it takes a total of 40 rounds are needed.
2. Tuples can be adjusted to hold the value of the accumulator which will be passed to the next processor.
3. LoadProcId would pass in the value k into the value parameter.
4. No, the processors perform their steps independently at time t
1. 20 steps PRAM algorithm means 40 map reduce steps are needed to finish the simulation.
2. add a tuple with necessary attributes including processor id, the number of the accumulator, list of locations of all accumulators.
3. Basically, LoadProcid k set the value of processor id into the tuple.
3. There is no definite relationship between PRAM processors, each of them can do mapper or reducer jobs.
1. 20 steps PRAM algorithm means 40 map reduce steps are needed to finish the simulation.
2. add a tuple with necessary attributes including processor id, the number of the accumulator, list of locations of all accumulators.
3. Basically, LoadProcid k set the value of processor id into the tuple.
3. There is no definite relationship between PRAM processors, each of them can do mapper or reducer jobs.
1. Each step uses 2 map reduce rounds, so 40 map reduce rounds are required to simulate a 20 step PRAM computation.
2. For each accumulator, pass (i,acc,n,v) where i is the processor, acc is tells if this is an accumulator tuple, n is the accumulator number, and v is the value.
3. LoadProcid k. Replace v with i in the accumulator tuple. (i,acc,k,v) -> (i,acc,k,i)
4. No, in a given timestep PRAM processors could all be doing different instructions.
1. Each step uses 2 map reduce rounds, so 40 map reduce rounds are required to simulate a 20 step PRAM computation.
2. For each accumulator, pass (i,acc,n,v) where i is the processor, acc is tells if this is an accumulator tuple, n is the accumulator number, and v is the value.
3. LoadProcid k. Replace v with i in the accumulator tuple. (i,acc,k,v) -> (i,acc,k,i)
4. No, in a given timestep PRAM processors could all be doing different instructions.
1. The number of rounds is 2 * 20 = 40. As there are 2 rounds of map-reduce simulate one step of PRAM computation.
2. we need to add tuple like is_acc, acc_num to global memory to present the accumulators.
3. Store the processor id value into the tuple.
4. No, as the PRAM processors might not to do the same instruction, can do the different instruction as they are independent.
1. The number of rounds is 2 * 20 = 40. As there are 2 rounds of map-reduce simulate one step of PRAM computation.
2. we need to add tuple like is_acc, acc_num to global memory to present the accumulators.
3. Store the processor id value into the tuple.
4. No, as the PRAM processors might not to do the same instruction, can do the different instruction as they are independent.
2019-05-15
1) since every PRAM includes two map/reduce rounds, hence total 40 rounds for 20 PRAM steps
2) for each accumulator, propose quadruple(i, 'acc' , K, v) and apply to each map/reduce step, where is processor number, K is accumulator number, v is value inside current accumulator
3) update uvalue v in the quadruple with i, from ( i, 'acc' , K, v) to (i, 'acc' , K,i)
4) no
(Edited: 2019-05-15) 2) for each accumulator, propose quadruple(i, 'acc' , K, v) and apply to each map/reduce step, where is processor number, K is accumulator number, v is value inside current accumulator
3) update uvalue v in the quadruple with i, from ( i, 'acc' , K, v) to (i, 'acc' , K,i)
4) no
1) since every PRAM includes two map/reduce rounds, hence total 40 rounds for 20 PRAM steps <br>
2) for each accumulator, propose quadruple(i, 'acc' , K, v) and apply to each map/reduce step, where is processor number, K
is accumulator number, v is value inside current accumulator<br>
3) update uvalue v in the quadruple with i, from ( i, 'acc' , K, v) to (i, 'acc' , K,i) <br>
4) no
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