Ten children wait to get their gift. Each child gets its turn[i] by computing a number[i]....

 

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Ten children wait to get their gift. Each child gets its turn[i] by computing a number[i]. A clown has in his hat ten balls; each ball has a unique number, from one to ten. After all children have their turn set, the clown is ready to share the gifts. The clown picks a random ball, and gives a gift to the child(ren) whose turn value is the same as the number on the ball (after that, he throws away the selected ball). Once the clown has no balls left, he leaves. (Shared variables) child (int i) { number[i] =1+max(number[1], turn[i] = number[i]; while (!called[i]){}; get the gift; ball){ numBalls--; go home; } turn] = 0 number[] = 0; called]=0; numBalls = 10 number[2]...number[N]); // sleep //sleep i=1, 10 (N=10) clown() { while (numBalls > 0){ pick a ball; for(int j=1; j Ten children wait to get their gift. Each child gets its turn[i] by computing a number[i]. A clown has in his hat ten balls; each ball has a unique number, from one to ten. After all children have their turn set, the clown is ready to share the gifts. The clown picks a random ball, and gives a gift to the child(ren) whose turn value is the same as the number on the ball (after that, he throws away the selected ball). Once the clown has no balls left, he leaves. (Shared variables) child (int i) { number[i] =1+max(number[1], turn[i] = number[i]; while (!called[i]){}; get the gift; ball){ numBalls--; go home; } turn] = 0 number[] = 0; called]=0; numBalls = 10 number[2]...number[N]); // sleep //sleep i=1, 10 (N=10) clown() { while (numBalls > 0){ pick a ball; for(int j=1; j

Answer rating: 100% (QA)

A Yes it is possible for 2 children to have the same value for numberi This is pos... View the full answer