Question: For the following question, given a pseudorandom generator ( PRG ) G : { 0 , 1 } n { 0 , 1 } m

For the following question, given a pseudorandom generator(PRG)G:{0,1}n{0,1}m.
Define G1(s):=G(s)||G(s), i.e., the new function G1(*) runs G twice on the same seed and concatenates the outputs. Is G1(*) a PRG? Briefly explain why.
Edit Format TableFor the following question, given a pseudorandom generator(PRG)G:{0,1}n{0,1}m.
Define G1(s):=G(s)||G(s), i.e., the new function G1(*) runs G twice on the same seed and concatenates the outputs. Is G1(*) a PRG? Briefly explain why.
Edit Format Table
 For the following question, given a pseudorandom generator(PRG)G:{0,1}n{0,1}m. Define G1(s):=G(s)||G(s), i.e.,

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