Please write 10 sentences for daily journal
Period: construction

)a grad student comes up with the following algorithm to sort an array a[1..n] that works by first sorting the first 2/3rds of the array, then sorting the last 2/3rds of the (resulting) array, and finally sorting the first 2/3rds of the new array. 1: function g-sort(a, n) . takes as input an array of n numbers, a[1..n] 2: g-sort-recurse(a, 1, n) 3: end function 4: function g-sort-recurse(a, `, u) 5: if u ⒠` ≤ 0 then 6: return . 1 or fewer elements already sorted 7: else if u ⒠` = 1 then . 2 elements 8: if a[u] < a[`] then . swap values 9: temp ↠a[u] 10: a[u] ↠a[`] 11: a[`] ↠temp 12: end if 13: else . 3 or more elements 14: size ↠u ⒠` + 1 15: twothirds ↠d(2 ◠size)/3e 16: g-sort-recurse(a, `, ` + twothirds ⒠1) 17: g-sort-recurse(a, u ⒠twothirds + 1, u) 18: g-sort-recurse(a, `, ` + twothirds ⒠1) 19: end if 20: end function first (5 pts), prove that the algorithm correctly sorts the numbers in the array (in increasing order). after showing that it correctly sorts 1 and 2 element intervals, you may make the (incorrect) assumption that the number of elements being passed to g-sort-recurse is always a multiple of 3 to simplify the notation (and drop the floors/ceilings).

When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2ă—, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5ă— slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?

The hybrid uses 144 to 158 volt batteries.