悲催的科学匠人 - 冷水's blog

[zz]FORTRAN版非递归快速排序

在二叉树离散网格单元时,最初使用递归的,但是网格单元到百万量级时,排序程序就出错了。怀疑是递归耗尽系统堆栈的问题,搜到一个非递归版本:  http://www.fortran.com/fortran/quick_sort2.f   百万量级也不是问题了。

 

      PROGRAM test_nrqsort
      REAL,ALLOCATABLE :: array(:)
      INTEGER,ALLOCATABLE :: idx(:)
      INTEGER N,i
      INTEGER,PARAMETER :: seed = 86456
      
      !READ(*,*) N
      N = 5000000
      ALLOCATE(array(N),idx(N))
      CALL srand(seed)
      DO i=1,N; 
         array(i) = rand()
      ENDDO
      WRITE(*,*) 'sorting'
      CALL SORTRX(N,array,idx)
      !DO i=1,N
      !  WRITE(*,*) array(idx(i))
      !ENDDO
      DO i=2,N
      IF(array(idx(i)) .lt. array(idx(i-1)) ) THEN
        WRITE(*,*) 'error', array(idx(i)), array(idx(i-1))
      ENDIF
      ENDDO
      DEALLOCATE(array,idx)
      END

C From Leonard J. Moss of SLAC:

C Here's a hybrid QuickSort I wrote a number of years ago.  It's
C based on suggestions in Knuth, Volume 3, and performs much better
C than a pure QuickSort on short or partially ordered input arrays.  

      SUBROUTINE SORTRX(N,DATA,INDEX)
C===================================================================
C
C     SORTRX -- SORT, Real input, indeX output
C
C
C     Input:  N     INTEGER
C             DATA  REAL
C
C     Output: INDEX INTEGER (DIMENSION N)
C
C This routine performs an in-memory sort of the first N elements of
C array DATA, returning into array INDEX the indices of elements of
C DATA arranged in ascending order.  Thus,
C
C    DATA(INDEX(1)) will be the smallest number in array DATA;
C    DATA(INDEX(N)) will be the largest number in DATA.
C
C The original data is not physically rearranged.  The original order
C of equal input values is not necessarily preserved.
C
C===================================================================
C
C SORTRX uses a hybrid QuickSort algorithm, based on several
C suggestions in Knuth, Volume 3, Section 5.2.2.  In particular, the
C "pivot key" [my term] for dividing each subsequence is chosen to be
C the median of the first, last, and middle values of the subsequence;
C and the QuickSort is cut off when a subsequence has 9 or fewer
C elements, and a straight insertion sort of the entire array is done
C at the end.  The result is comparable to a pure insertion sort for
C very short arrays, and very fast for very large arrays (of order 12
C micro-sec/element on the 3081K for arrays of 10K elements).  It is
C also not subject to the poor performance of the pure QuickSort on
C partially ordered data.
C
C Created:  15 Jul 1986  Len Moss
C
C===================================================================
 
      INTEGER   N,INDEX(N)
      REAL      DATA(N)
 
      INTEGER   LSTK(31),RSTK(31),ISTK
      INTEGER   L,R,I,J,P,INDEXP,INDEXT
      REAL      DATAP
 
C     QuickSort Cutoff
C
C     Quit QuickSort-ing when a subsequence contains M or fewer
C     elements and finish off at end with straight insertion sort.
C     According to Knuth, V.3, the optimum value of M is around 9.
 
      INTEGER   M
      PARAMETER (M=9)
 
C===================================================================
C
C     Make initial guess for INDEX
 
      DO 50 I=1,N
         INDEX(I)=I
   50    CONTINUE
 
C     If array is short, skip QuickSort and go directly to
C     the straight insertion sort.
 
      IF (N.LE.M) GOTO 900
 
C===================================================================
C
C     QuickSort
C
C     The "Qn:"s correspond roughly to steps in Algorithm Q,
C     Knuth, V.3, PP.116-117, modified to select the median
C     of the first, last, and middle elements as the "pivot
C     key" (in Knuth's notation, "K").  Also modified to leave
C     data in place and produce an INDEX array.  To simplify
C     comments, let DATA[I]=DATA(INDEX(I)).
 
C Q1: Initialize
      ISTK=0
      L=1
      R=N
 
  200 CONTINUE
 
C Q2: Sort the subsequence DATA[L]..DATA[R].
C
C     At this point, DATA[l] <= DATA[m] <= DATA[r] for all l < L,
C     r > R, and L <= m <= R.  (First time through, there is no
C     DATA for l < L or r > R.)
 
      I=L
      J=R
 
C Q2.5: Select pivot key
C
C     Let the pivot, P, be the midpoint of this subsequence,
C     P=(L+R)/2; then rearrange INDEX(L), INDEX(P), and INDEX(R)
C     so the corresponding DATA values are in increasing order.
C     The pivot key, DATAP, is then DATA[P].
 
      P=(L+R)/2
      INDEXP=INDEX(P)
      DATAP=DATA(INDEXP)
 
      IF (DATA(INDEX(L)) .GT. DATAP) THEN
         INDEX(P)=INDEX(L)
         INDEX(L)=INDEXP
         INDEXP=INDEX(P)
         DATAP=DATA(INDEXP)
      ENDIF
 
      IF (DATAP .GT. DATA(INDEX(R))) THEN
         IF (DATA(INDEX(L)) .GT. DATA(INDEX(R))) THEN
            INDEX(P)=INDEX(L)
            INDEX(L)=INDEX(R)
         ELSE
            INDEX(P)=INDEX(R)
         ENDIF
         INDEX(R)=INDEXP
         INDEXP=INDEX(P)
         DATAP=DATA(INDEXP)
      ENDIF
 
C     Now we swap values between the right and left sides and/or
C     move DATAP until all smaller values are on the left and all
C     larger values are on the right.  Neither the left or right
C     side will be internally ordered yet; however, DATAP will be
C     in its final position.
 
  300 CONTINUE
 
C Q3: Search for datum on left >= DATAP
C
C     At this point, DATA[L] <= DATAP.  We can therefore start scanning
C     up from L, looking for a value >= DATAP (this scan is guaranteed
C     to terminate since we initially placed DATAP near the middle of
C     the subsequence).
 
         I=I+1
         IF (DATA(INDEX(I)).LT.DATAP) GOTO 300
 
  400 CONTINUE
 
C Q4: Search for datum on right <= DATAP
C
C     At this point, DATA[R] >= DATAP.  We can therefore start scanning
C     down from R, looking for a value <= DATAP (this scan is guaranteed
C     to terminate since we initially placed DATAP near the middle of
C     the subsequence).
 
         J=J-1
         IF (DATA(INDEX(J)).GT.DATAP) GOTO 400
 
C Q5: Have the two scans collided?
 
      IF (I.LT.J) THEN
 
C Q6: No, interchange DATA[I] <--> DATA[J] and continue
 
         INDEXT=INDEX(I)
         INDEX(I)=INDEX(J)
         INDEX(J)=INDEXT
         GOTO 300
      ELSE
 
C Q7: Yes, select next subsequence to sort
C
C     At this point, I >= J and DATA[l] <= DATA[I] == DATAP <= DATA[r],
C     for all L <= l < I and J < r <= R.  If both subsequences are
C     more than M elements long, push the longer one on the stack and
C     go back to QuickSort the shorter; if only one is more than M
C     elements long, go back and QuickSort it; otherwise, pop a
C     subsequence off the stack and QuickSort it.
 
         IF (R-J .GE. I-L .AND. I-L .GT. M) THEN
            ISTK=ISTK+1
            LSTK(ISTK)=J+1
            RSTK(ISTK)=R
            R=I-1
         ELSE IF (I-L .GT. R-J .AND. R-J .GT. M) THEN
            ISTK=ISTK+1
            LSTK(ISTK)=L
            RSTK(ISTK)=I-1
            L=J+1
         ELSE IF (R-J .GT. M) THEN
            L=J+1
         ELSE IF (I-L .GT. M) THEN
            R=I-1
         ELSE
C Q8: Pop the stack, or terminate QuickSort if empty
            IF (ISTK.LT.1) GOTO 900
            L=LSTK(ISTK)
            R=RSTK(ISTK)
            ISTK=ISTK-1
         ENDIF
         GOTO 200
      ENDIF
 
  900 CONTINUE
 
C===================================================================
C
C Q9: Straight Insertion sort
 
      DO 950 I=2,N
         IF (DATA(INDEX(I-1)) .GT. DATA(INDEX(I))) THEN
            INDEXP=INDEX(I)
            DATAP=DATA(INDEXP)
            P=I-1
  920       CONTINUE
               INDEX(P+1) = INDEX(P)
               P=P-1
               IF (P.GT.0) THEN
                  IF (DATA(INDEX(P)).GT.DATAP) GOTO 920
               ENDIF
            INDEX(P+1) = INDEXP
         ENDIF
  950    CONTINUE
 
C===================================================================
C
C     All done
 
      END

ExStream特性

 

大规模并行飞行器外流场分析工具 ExStream 特性简介

网格系统

  1. 支持1-1匹配式对接的多块结构化网格
  2. 支持重叠嵌套网格

数值方法

  1. MUSCL插值和多种梯度限制器
  2. JST格式与AUSM类格式
  3. 全梯度粘性通量与薄层假设粘性通量
  4. 几何多重网格与多级网格
  5. 显式Runge-Kutta时间推进与LUSGS隐式推进
  6. 双时间步长方法处理非稳态问题

流体属性

  1. 理想气体
  2. 基于Sutherland公式的粘性系数

 

湍流模型

  1. k-Omega SST RANS模型
  2. SA RANS模型
  3. SA DES 分离流算法
  4. SA DDES及其改进版本的分离流算法
  5. SA IDDES 分离流算法
  6. SA CLES 分离流算法

并行计算

  1. 基于MPI的对等式并行计算,主节点无显著额外负载及内存消耗
  2. 自动的网格切分和负载均衡
  3. 并行文件输入输出,支持2E63字节

输入

  1. 网格,mesh3d.dat,双精度三维多块PLOT3D格式,FORTRAN的unformat格式,只包含坐标数据,不包含IBLANK数据
  2. 边界信息,bc_in,边界类型与对接方式定义,可由ICEMCFD导出数据转换得到
  3. 基本求解控制参数,input,FORTRAN namelist格式
  4. 扩展模块的控制参数,usr.inp,FORTRAN namelist格式

输出

  1. 表面数据以文本tecpot格式输出,每个固体表面会标记其所属的部件名称,以便分类观察,p_surface.plt
  2. 体数据以PLOT3D格式给出,网格(mesh3dsp.dat)和数据(*.p3df)皆以单精度浮点数存储
  3. 各个部件的气动力以X-Y形式的tecplot格式提供,partforce.plt
  4. 残差收敛历史以X-Y形式的tecplot格式提供,res.dat
  5. 非稳态模拟的时间平均流场输出
  6. 非稳态下用户自定义监测点、监测块和面的输出

DPW-II表面压力分布

采用官方粗网格,单元数量约400万。

 

三角翼低速分离涡的DDES模拟

0.1马赫中等后掠角的三角翼,全模计算,网格单元约2000万

 

超声速底部流动的DDES模拟

来流马赫数为2.46,网格单元数约1000万。

天津超级计算中心天河-1A在linux下登录VPN的方法解释

原文见此链接

https://vpn.nscc-tj.cn/svpn/cn/support/OperationGuide%28Chinese%29.htm

 

在ubuntu下设置时,遇到几个问题

1 jre6的安装。ubuntu不再首选支持sun java,而推荐使用openjdk,因此软件库中没有sun java。为了保证兼容,还是选sun java吧。安装方法参考 https://help.ubuntu.com/community/Java 中关于Orancle(Sun) Java 6的部分

 

2 Firefox中设置JNLP文件打开方式。初始Firefox中没有JNLP文件选项,没法设置。方法是,自己建立一个空文件,后缀名为jnlp。把它拖到firefox中,firefox会提示打开方法,选择javaws。然后到首选项的application中就可以看到JNLP了,把它按照说明设置即可。

 

3 22端口问题。vpn客户端要占用22端口,只好把本机的ssh服务停了 

4 root身份的问题。vpn客户端必须用root身份,但是在root环境下工作有风险。解决方法是,登录一个root桌面,登录好vpn之后,再切换登录一个普通用户桌面,就可以ssh到th-1a的三个登录节点了。

 

 

 

一个有用的网页注释工具

http://webklipper.com

可以指定一个URL,它生成一个拷贝。然后可以在页面上做各种标记和注释。可以把注释版分享给别人浏览或者合作添加注释。这玩意还可以支持自行上载文件(pdf、图形和txt)进行注释。

缺点:

1 似乎标记的位置不是根据文本的内容来定位,而是根据网页区域的坐标。因此改变窗口大小后,标记位置就错了

2 pdf会被转成图片格式,不清楚。

 

不论如何,这玩意还是很好的。非常适合多人剖析源代码之类的活。




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