With a given input of an expression, this code solves for the bracket-ing that results in the highest result.

max_noparens.rb

def getmax nums, ops, dbg=0, &f
  # f is action performed on numbers before comparing (like taking absolute value)
  maxn = nil
  pat = nil
  (0...ops.size).to_a.permutation do |p| # iterate over each possible pattern
    begin
      n = solve(nums, ops, p, dbg - 1) # attempt to solve (based on given pattern)
      if n.infinite?
        n = nil
        raise ZeroDivisionError, "divided by 0"
      end
    rescue StandardError => e # on error
      STDERR.puts "#{e.class}: #{e.message}" # log errors
    end
    if dbg > 0
      p([nums, ops, p, n, f.(n)]) # debug output
    end
    if maxn.nil? || (!n.nil? && f.(n) > f.(maxn)) # checks for new maximum
      maxn = n
      pat = p
    end
  end
  return [maxn, pat] # answer (number, brackets)
end

def solve nums, ops, pat, dbg=0
  if ops.size != pat.size || nums.size - 1 != ops.size # validate input
    raise
  end
  nums, ops, pat = nums.dup, ops.dup, pat.dup # copy input (ruby uses references)
  while i = pat.shift # iterates through the pattern
    nums[i, 2] = nums[i].send(ops.delete_at(i), nums[i + 1]) # flattening (perform operation)
    pat.map! { |x| x > i ? x - 1 : x } # fix pattern (so stuff doesn't break)
    if dbg > 0
      p([nums, ops, pat]) # debug output
    end
  end
  return nums[0] # final answer
end

# eg (15 ÷ 3 - 2 + 5 - 20 ÷ 2 × 7 - 19 + 9 + 6 ÷ 2 ^ 2):
#   getmax([15, 3, 2, 5, 20, 2, 7, 19, 9, 6, 2, 2], %i[/ - + - / * - + + / **]) { |num| num.abs }