多線程算法的工作速度要慢得多
[英]Multithreaded algorithms work much slower
問題描述
我曾嘗試使用 OpenMP 和 Cilk Plus。 結果是一樣的，多線程工作得更慢。 我不知道我做錯了什么。 我做了這個人在本教程中所做的
他的代碼並行運行效果更好，而我的情況是這樣的：
平行：斐波那契數 #42 是 267914296
使用 8 個工人在33.026 秒內計算
序列號：斐波那契數 #42 是 267914296
使用 8 個工人在2.110 秒內計算
我完全復制了教程的源代碼。
我也用 OpenMP 嘗試過，同樣的事情也在那里發生。 我在執行過程中檢查 CPU 內核的使用情況。 他們都工作，這很好。
我試圖用這個命令改變工人的數量：
export CILK_NWORKERS=4
看起來隨着worker數量的增加，算法運行得更慢。 但有時不會。 我在 C 和 C++ 上實現了 Cilk 代碼。 沒有不同。
這是順序斐波那契函數：
int fib_s(int n)
{
if (n < 2)
return n;
int x = fib_s(n-1);
int y = fib_s(n-2);
return x + y;
}
這是並行斐波那契函數：
int fib(int n)
{
if (n < 2)
return n;
int x = cilk_spawn fib(n-1);
int y = fib(n-2);
cilk_sync;
return x + y;
}
我在main()函數中像這樣計算運行時間：
clock_t start = clock();
int result = fib(n);
clock_t end = clock();
double duration = (double)(end - start) / CLOCKS_PER_SEC;
誰能幫我？
2 個解決方案
解決方案1
2 2020-01-10 10:58:12
解決方案2
2 2020-01-14 19:38:36
問：任何人都可以幫助我嗎？
是的。 你會看到， fib( 42 )在單線程解釋 (!) 代碼中可能需要不到25 [us]
鑒於上面的並行代碼已經報道花~33 [s]上處理，編譯代碼可以計算一個fib( ~ 1,700,000 )在同一~33 [s] ，如果右設計。
解決方案 ：
任何遞歸公式化的問題描述都是老數學家的罪過：
雖然在紙上看起來很酷，
它在堆棧上縮放丑陋，並為任何更深層次的遞歸阻塞了大量資源......
使所有“先前”級別的大部分時間都在等待，
直到return 2和return 1在它們的所有后代路徑中都發生了
並且遞歸公式化算法的累積階段開始增長，從深遞歸潛水的所有深度返回頂部。
這個依賴樹相當於一個純[SERIAL] （一個接一個）的計算進程，以及任何注入{ [CONCURENT] | [PARALLEL] }嘗試{ [CONCURENT] | [PARALLEL] } { [CONCURENT] | [PARALLEL] }處理編排只會增加處理成本（添加所有附加開銷），而不會對結果的依賴驅動累積的純[SERIAL]序列進行任何改進。
讓我們看看cilk_spawn fib( N )是多么糟糕：
f(42)
|
x=--> --> --> --> --> --> --> --> --> --> --> -- --> --> --> --> --> --> --> --> --> --> --> --> --> -->f(41)
| |
y=f(40) x=--> --> --> --> --> --> --> --> --> --> f(40)
~ | | |
~ x=--> --> --> --> --> --> --> --> --> f(39) y=f(39) x=--> --> --> --> --> --> --> --> --> f(39)
~ | | ~ | | |
~ y=f(38) x=--> --> --> --> --> --> f(38) ~ x=--> --> --> --> f(38) y=f(38) x=--> --> --> --> --> --> f(38)
~ ~ | | | ~ | | ~ | | |
~ ~ x=--> --> f(37) y=f(37) x=--> --> f(37) ~ y=f(37) x=--> --> --> f(37) ~ x=--> --> f(37) y=f(37) x=--> --> f(37)
~ ~ | | ~ | | | ~ ~ | | | ~ | | ~ | | |
~ ~ y=f(36) x=--> --> f(36) ~ x=--> --> f(36) y=f(36) x=-->f(36) ~ ~ x=--> --> f(36) y=f(36) x= ~ y=f(36) x=--> --> f(36) ~ x=--> --> f(36) y=f(36) x=--> --> f(36)
~ ~ ~ | | | ~ | | ~ | | | ~ ~ | | ~ | | ~ ~ | | | ~ | | ~ | | |
~ ~ ~ x=-->f y=f(35) x=-->f ~ y=f(35) x=-->f(35) ~ x=-->f y=f(35) x=-->f ~ ~ y=f(35) x= ~ x=-->f(35) y= ~ ~ x=-->f y=f(35) x=-->f(35) ~ y=f(35) x=-->f(35) ~ x=--> y=f(35) x=-->f(35)
~ ~ ~ | ~ | | ~ ~ | | | ~ | ~ | | ~ ~ ~ | | ~ | | ~ ~ ~ | ~ | | | ~ ~ | | | ~ | ~ | | |
~ ~ ~ y=f(34) ~ x=-->f y=f(34) ~ ~ x=-->f y=f(34) x= ~ y=f(34) ~ x= y=f(34) ~ ~ ~ x=-->f y= ~ y=f(34) x= ~ ~ ~ y=f(34) ~ x=-->f y=f(34) x= ~ ~ x=-->f y=f(34) x= ~ y=f(34) ~ x=-->f y=f(34) x=-->f
~ ~ ~ ~ | ~ | ~ | ~ ~ | ~ | | ~ ~ | ~ | ~ | ~ ~ ~ | ~ ~ ~ | | ~ ~ ~ ~ | ~ | ~ | | ~ ~ | ~ | | ~ ~ | ~ | ~ |
~ ~ ~ ~ x= ~ y=f(33) ~ x= ~ ~ y=f(33) ~ x= y= ~ ~ x= ~ y= ~ x= ~ ~ ~ y=f(33) ~ ~ ~ x= y= ~ ~ ~ ~ x= ~ y=f(33) ~ x= y= ~ ~ y=f(33) ~ x= y= ~ ~ x= ~ y=f(33) ~ y=f(33)
~ ~ ~ ~ | ~ ~ | ~ | ~ ~ ~ | ~ | ~ ~ ~ | ~ ~ ~ | ~ ~ ~ ~ | ~ ~ ~ | ~ ~ ~ ~ ~ | ~ ~ | ~ | ~ ~ ~ ~ | ~ | ~ ~ ~ | ~ ~ | ~ ~ |
~ ~ ~ ~ y= ~ ~ x= ~ y= ~ ~ ~ x= ~ y= ~ ~ ~ y= ~ ~ ~ y= ~ ~ ~ ~ x= ~ ~ ~ y= ~ ~ ~ ~ ~ y= ~ ~ x= ~ y= ~ ~ ~ ~ x= ~ y= ~ ~ ~ y= ~ ~ x= ~ ~ x=-->f
~ ~ ~ ~ ~ ~ ~ | ~ ~ ~ ~ ~ | ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ | ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ | ~ ~ ~ ~ ~ ~ | ~ ~ ~ ~ ~ ~ ~ ~ | ~ ~ |
: : : : :
: : : :
: : :
~ ~ --SYNC-----------f(36)+f(37)
~ ~ <--RET x+y // <-- f(38)
~ --SYNC----------------f(38)+f(39)
~ <--RET x+y // <-- f(40)
--SYNC---------------------f(40)+f(41)
<--RET x+y // <-- f(42)
只需計算一下， Fib( N )的自上而下運行的遞歸方法已經為N每個值重新計算了多少次 - 是的，您一次又一次地多次計算相同的事情，只是由於遞歸方法的“數學” -懶惰：
fib( N == 42 ) was during recursion calculated .........1x times...
fib( N == 41 ) was during recursion calculated .........1x times...
fib( N == 40 ) was during recursion calculated .........2x times...
fib( N == 39 ) was during recursion calculated .........3x times...
fib( N == 38 ) was during recursion calculated .........5x times...
fib( N == 37 ) was during recursion calculated .........8x times...
fib( N == 36 ) was during recursion calculated ........13x times...
fib( N == 35 ) was during recursion calculated ........21x times...
fib( N == 34 ) was during recursion calculated ........34x times...
fib( N == 33 ) was during recursion calculated ........55x times...
fib( N == 32 ) was during recursion calculated ........89x times...
fib( N == 31 ) was during recursion calculated .......144x times...
fib( N == 30 ) was during recursion calculated .......233x times...
fib( N == 29 ) was during recursion calculated .......377x times...
fib( N == 28 ) was during recursion calculated .......610x times...
fib( N == 27 ) was during recursion calculated .......987x times...
fib( N == 26 ) was during recursion calculated ......1597x times...
fib( N == 25 ) was during recursion calculated ......2584x times...
fib( N == 24 ) was during recursion calculated ......4181x times...
fib( N == 23 ) was during recursion calculated ......6765x times...
fib( N == 22 ) was during recursion calculated .....10946x times...
fib( N == 21 ) was during recursion calculated .....17711x times...
fib( N == 20 ) was during recursion calculated .....28657x times...
fib( N == 19 ) was during recursion calculated .....46368x times...
fib( N == 18 ) was during recursion calculated .....75025x times...
fib( N == 17 ) was during recursion calculated ....121393x times...
fib( N == 16 ) was during recursion calculated ....196418x times...
fib( N == 15 ) was during recursion calculated ....317811x times...
fib( N == 14 ) was during recursion calculated ....514229x times...
fib( N == 13 ) was during recursion calculated ....832040x times...
fib( N == 12 ) was during recursion calculated ...1346269x times...
fib( N == 11 ) was during recursion calculated ...2178309x times...
fib( N == 10 ) was during recursion calculated ...3524578x times...
fib( N == 9 ) was during recursion calculated ...5702887x times...
fib( N == 8 ) was during recursion calculated ...9227465x times...
fib( N == 7 ) was during recursion calculated ..14930352x times...
fib( N == 6 ) was during recursion calculated ..24157817x times...
fib( N == 5 ) was during recursion calculated ..39088169x times...
fib( N == 4 ) was during recursion calculated ..63245986x times...
fib( N == 3 ) was during recursion calculated .102334155x times...
fib( N == 2 ) was during recursion calculated .165580141x times...
fib( N == 1 ) was during recursion calculated .102334155x times...
快速和資源的高效處理 - 一個靈感：
雖然原始的遞歸計算調用了535,828,591次 (!!!) 相同的瑣碎fib() （通常是一個，已經在其他地方計算過）
----有的甚至數億多次已經〜 102,334,155x倍......作為fib( 3 )產卵多達267,914,295只是- [CONCURRENT]代碼執行塊，排隊等待，但8工人，所有的等待大多數情況下，要不是為了讓他們產生的孩子深入到return 1和return 2之后什么都不做，只是添加一對然后返回的數字並從昂貴的產生自己的過程中返回，一種“直接”方法處理是不可能的方式更聰明，方式更快：
int fib_direct( int n ) // PSEUDO-CODE
{ assert( n > 0 && "EXCEPTION: fib_direct() was called with a wrong parameter value" );
if ( n == 1
|| n == 2
) return n;
// ---------------------------- .ALLOC + .SET
int fib_[ max(4,n) ];
fib_[3] = 3;
fib_[4] = 5;
// ---------------------------- .LOOP LESS THAN N-TIMES
for( int i = 5; i <= n; i++ )
{ fib_[i] = fib_[i-2]
+ fib_[i-1];
}
// ---------------------------- .RET
return fib_[n];
}
更有效的實現（仍然只是一個線程並且仍然只是解釋）設法在不到2.1 [s]時間內輕松計算fib_direct( 230000 ) ，這是您編譯的代碼運行時僅fib( 42 ) 。
瀏覽器中的多線程 WebAssembly 比單線程慢，為什么？
[英]Multithreaded WebAssembly slower in browser than singlethreaded, why?
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