Julia @parallel for return语句
如何在函数中编写并行for循环，只要满足条件，就立即返回所有工作人员？
即这样的事情：
function test(n)
@sync @parallel for i in 1:1000
{... statement ...}
if {condition}
return test(n+1)
end
end
end
所有工作人员在for循环中停止工作，只有主进程返回？ （并且其他进程再次开始与下一个for循环一起工作？）
这个问题似乎是做“尴尬平行”搜索任务的基本模式。 @parallel for结构适用于分区工作，但没有for在单一流程中停止的break短路逻辑。
为了演示如何在茱莉亚做到这一点，考虑一个玩具问题，找出组合锁与几个轮子的组合。 可以使用某种方法检查车轮的每个设置是否正确（采用combodelay时间 - 请参阅下面的代码）。 在找到正确的车轮编号后，搜索下一个车轮。 高级别伪代码就像OP问题中给出的代码片段一样。
以下是运行代码（在0.5和0.6）来执行此操作。 一些评论解释了细节，并且代码在一个块中给出，以便于剪切和粘贴。
# combination lock problem parameters
const wheel_max = 1000 # size of wheel
@everywhere const magic_number = [55,10,993] # secret combination
const wheel_count = length(magic_number) # number of wheels
const combodelay = 0.01 # delay time to check single combination
# parallel short-circuit parameters
const check_to_work_ratio = 160 # ratio to limit short-circuit overhead
function find_combo(wheel,combo=Int[])
done = SharedArray{Int}(1) # shared variable to hold if and what combo
done[1] = 0 # succeded. 0 means not found yet
# setup counters to limit parallel overhead
@sync begin
@everywhere global localdone = false
@everywhere global checktime = 0.0
@everywhere global worktime = 0.0
end
# do the parallel work
@sync @parallel for i in 1:wheel_max
global localdone
global checktime
global worktime
# if not checking too much, look at shared variable
if !localdone && check_to_work_ratio*checktime < worktime
tic()
localdone = done[1]>0
checktime += toq()
end
# if no process found combo, check another combo
if !localdone
tic()
sleep(combodelay) # simulated work delay, {..statement..} from OP
if i==magic_number[wheel] # {condition} from OP
done[1] = i
localdone = true
end
worktime += toq()
else
break
end
end
if done[1]>0 # check if shared variable indicates combo for wheel found
push!(combo,done[1])
return wheel<wheel_count ? find_combo(wheel+1,combo) : (combo,true)
else
return (combo,false)
end
end
function find_combo_noparallel(wheel,combo=Int[])
found = false
i = 0
for i in 1:wheel_max
sleep(combodelay)
if i==magic_number[wheel]
found = true
break
end
end
if found
push!(combo,i)
return wheel<wheel_count ?
find_combo_noparallel(wheel+1,combo) : (combo,true)
else
return (combo,false)
end
end
function find_combo_nostop(wheel,combo=Int[])
done = SharedArray{Int}(1)
done[1] = 0
@sync @parallel for i in 1:wheel_max
sleep(combodelay)
if i==magic_number[wheel]
done[1] = i
end
end
if done[1]>0
push!(combo,done[1])
return wheel<wheel_count ?
find_combo_nostop(wheel+1,combo) : (combo,true)
else
return (combo,false)
end
end
result = find_combo(1)
println("parallel with short-circuit stopping: $result")
@assert result == (magic_number, true)
result = find_combo_noparallel(1)
println("single process with short-circuit stopping: $result")
@assert result == (magic_number, true)
result = find_combo_nostop(1)
println("parallel without short-circuit stopping: $result")
@assert result == (magic_number, true)
println("ntimings")
print("parallel with short-circuit stopping ")
@time find_combo(1);
print("single process with short-circuit stopping ")
@time find_combo_noparallel(1)
print("parallel without short-circuit stopping ")
@time find_combo_nostop(1)
nothing
可能会有更好的实现，而一些元编程可以隐藏一些短路机器。 但这应该是一个很好的开始。
结果应该如下所示：
parallel with short-circuit stopping: ([55,10,993],true)
single process with short-circuit stopping: ([55,10,993],true)
parallel without short-circuit stopping: ([55,10,993],true)
timings
parallel with short-circuit stopping 4.473687 seconds
single process with short-circuit stopping 11.963329 seconds
parallel without short-circuit stopping 11.316780 seconds
这是计算3个工作进程的演示。 真正的问题应该有更多的过程和每个过程的更多工作，然后短路的好处将是显而易见的。
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