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101: after p1 completed, compute p2.
121: after p2 completed, compute p1.
100: after (* x x) is completed, compute p2 and set x to 11, then p1 set x to 100.
I think it is still possible to get 110 since that doesn't require splitting serialized p2 into multiple parts
I don't think 110 is a correct answer, there is s in the body of p1' (inner in p1, (s lambda() (* x x))), so it wouldn't be stop until (* x x) finish.