First, practice the accumulation loop pattern, just as the sum of n numbers can be progressively computed as (· · · +x2) + x3) + x4) + · · · xn) using pairwise addition (to update the "sum so far" value), so can the maximum be computed as max(· · · max(max(max(x1, x2), x3), x4), · · · xn) using pairwise maximum (to update the "maximum so far" value). the minimum is also similar. second, practice with writing one loop that performs different accumulations over the same sequence of terms, "in lockstep"