SASKiwi makes a good point and references another topic that I had actually posted: http://communities.sas.com/message/103760#103760 I have still not had an opportunity to look into using the write=immediate option for logparm. FriedEgg wrote: I am not currently able to test but I know in my current enviornment the following will not work: Curent Enviornment: LOGPARM=WRITE=BUFFERED ROLLOVER=NONE OPEN=REPLACE data _null_;   do i=1 to 10000000;    call ranuni(1234);    if mod(i,50000)=0 then putlog i=; end; run; If the write=immediate option doesn't resolve the issue my next thought is to call sysexec and cat a note onto the end of the logfile (if I even can incase there is a lock on the file held by sas while using write=buffered?). data _null_;   do i=1 to 10**6;    if mod(i,100000)=1 then call sysexec('x echo ' || put(i,comma.) || ' > /mylogs/tracker.dat'   end; run; ... View more

Sounds like you will need something that defines the heirarchy of conditions, here is an example. /* I have a set of diseases in a grouping with rank */ data diseases; infile cards dsd dlm='|'; input name $ group value rank; cards; cancer|1|16|1 metastic cancer|1|10|2 leukemia|1|20|2 ; run; /* I have people with a list of conditions */ data diseased; input person disease; cards; 1 10 1 16 2 20 2 16 3 20 4 10 1 2 2 70 5 1 ; run; /* I want a list of people with their most predominit disease by diagnosis group */ proc sql; create table want as select a.person, a.disease    from diseased a    left join diseases b on a.disease=b.value   group by a.person, b.group having b.rank=min(b.rank)   order by a.person, b.group, b.rank; quit; OUTPUT person    disease 1                    2 1                    16 2                    70 2                    16 3                    20 4                    10 5                    1 ... View more

Also, you can pass an altlog option to your sascmd statement in your signon statement. Such as: signon task1 sascmd="/usr/local/SAS/SASFoundation/9.2/sas -altlog /mylogs/mylog.log" wait=no; rdisplay is also useful as long as you are running you program interactivly. ... View more

I downloaded that dictionary from the internet, it claimed to be a distributable copy of the unix dictionary file but I guess it is not identical.  I am using dictionary file that is built into my OS and I guess it is slightly different.  I changed to use the same file and also fixed the issue with my program only finding solutions in odd number of steps in length. options fullstimer mlogic mprint; %let from=black; %let to=white; data _null_; if length("&from") ne length("&to") then   do;    put 'ERROR: FROM AND TO MUST BE SAME LENGTH';    abort return;   end; else if complev("&from","&to",'i')=1 then   do;    put 'ERROR: FROM AND TO ARE ALREADY 1 STEP APART';    abort return;   end; else   do;    call symput("len",length("&from"));    call symput("from",%upcase("&from"));    call symput("to",%upcase("&to"));   end; run; data dict; infile '/nas/sasbox/users/mkastin/unix-words'; input word :$upcase40.; if length(word)=&len and word=compress(word,,'ka') then output; run; proc sort data=dict nodupkey; by word; run; data step1; length start $40; set dict; where complev("&from",word,'i')=1; start="&from"; run; data _step1; length end $40; set dict; where complev("&to",word,'i')=1; end="&to"; run; /* loop */ %macro solve(in1,in2); data want;   merge &in1(in=b) &in2(in=a);   by word;   if a and b; run; proc sql noprint;   select nobs into :nobs     from sashelp.vtable    where libname='WORK' and memname='WANT'; quit; %if &nobs>0 %then %let done=Y; %mend; %macro ladder; %let done=N; %let i=1; %do %until (&done=Y);   %let n=%eval(&i+1);   data step&n;    set dict;     do i=1 to _nobs;      set step&i(rename=(word=step&i)) nobs=_nobs point=i;      if complev(step&i,word)=1 and word ne start %if &i>1 %then %do j=1 %to %eval(&i-1); and word ne step&j %end; then output;     end;   run;   %solve(_step&i,step&n);   %if &done=Y %then %goto exit;   data _step&n;    set dict;     do i=1 to _nobs;      set _step&i(rename=(word=_step&i)) nobs=_nobs point=i;      if complev(_step&i,word)=1 and word ne end %if &i>1 %then %do j=1 %to %eval(&i-1); and word ne _step&j %end; then output;     end;   run;   %solve(step&n,_step&n)   %let i=%eval(&i+1);   %exit: %end; %mend; %ladder BLACK 1. blank 2. blink or clank 3. clink 4. chink 5. chine 6. whine WHITE WHITE 1. whine 2. chine 3. chink 4. chick 5. check 6. cheek 7. creek 8. greek GREEN ... View more

data foo; input label $ start @@; retain fmtname 'myfmt'; cards; Yael 1 Hillary 2 Mechelle 3 Henry 4 Melinda 5 Josiah 6 Hyatt 7 Kalia 8 Phelan 9 Amal 10 Keith 11  Alea 12 Grady 13 Kaitlin 14 Damian 15 Berk 16 Timothy 17 Noelle 18 Carter 19 Libby 20 Kuame 21 Shafira 22 Philip 23 Uta 24 Hadassah 25 Cora 26 Holly 27 Kermit 28 Maia 29 Howard 30 Gail 31 Linus 32  Roth 33 Zephania 34 Abdul 35 Jescie 36  Jordan 37 Idola 38 Zelenia 39  Darrel 40 Jamalia 41 Kathleen 42  Dillon 43 Rudyard 44 Alice 45 Julie 46 Ulysses 47 Vivian 48 Colin 49 Baker 50 Kennan 51 Quynn 52 Velma 53 Ella 54 Nero 55 Erica 56 Karen 57 Blaze 58 Shaeleigh 59 Cadman 60 Katelyn 61 Chaney 62 Patricia 63 Gannon 64 Cedric 65 Paul 66 Ruby 67 Merrill 68 Carolyn 69 Roanna 70 Trevor 71 Madison 72  Fredericka 73 Chandler 74  Isadora 75 Graiden 76  Morgan 77 Grant 78 Erica 79 Tashya 80 Ivory 81 September 82 Micah 83 Harding 84 Brooke 85 Hadley 86 Mufutau 87  Martena 88 Jescie 89 Keelie 90 Raymond 91 Abbot 92 Noel 93 Kellie 94 Gareth 95 Tallulah 96 Dustin 97 Cara 98 Kirsten 99 Isadora 100 ; run; proc format library=work cntlin=foo; run; data bar; call streaminit( 12345 ); do i=1 to 100;   num=abs(mod(int(rand('NORMAL')*10**3),100))+1;   name=put(num,myfmt.);   output; end; drop i; run; ... View more

The update statement is also a good idea.  I often overlook the update and modify routines as there usefulness in my current tasks is limited. ... View more

Look into the coalescec function and doing a sql join instead of merge. http://support.sas.com/documentation/cdl/en/lrdict/64316/HTML/default/viewer.htm#a002518268.htm ... View more

This does not appear to be a contents of your actual data but rather a very small example probably taken from the data provided in this topic.  Can you please do contents of your real data or one of the sample sets you have been using for testing.  Also, it appears that your executing enviornment is solaris 64-bit, not windows.  You say the server has at least 200GB space, I assume you mean disk space, I am more concerned with memory space.  I will also ask again that you provide the output of proc options group=memory; run; proc options group=performance; run;  If you do not know about the memory space try running the following: filename ram pipe "/usr/sbin/prtconf | grep Memory | awk 'print $3, $4'"; data _null_;   infile ram;   input size unit $;   put 'Memory Size: ' size unit; run; ... View more

I did not thoroughly check my code but this is what I had come up with.  Very similar to KSharp's approach.  I am going to take some time later and convert this to using hash objects as that is the ultimate personal goal of the challenge for myself.  I am fairly certain that my example with only find odd number length ladders since I am approaching from both sides to save time with the size of the cartesian joins, which is obviously a problem to also fix. options fullstimer spool macrogen mlogic mprint; %let from=black; %let to=white; data _null_; /* rule 1: the from and to words must be the same length */ if length("&from") ne length("&to") then   do;    put 'ERROR: FROM AND TO MUST BE SAME LENGTH';    abort return;   end; else   do;    call symput("len",length("&from"));    call symput("from",%upcase("&from"));    call symput("to",%upcase("&to"));   end; run; data dict; infile '/usr/share/dict/words'; input word :$upcase40.; if length(word)=&len and word=compress(word,,'ka') then output; run; proc sort data=dict nodupkey; by word; run; /* streaminit */ data step1; length start $40; set dict; where complev("&from",word,'i')=1; start="&from"; run; data _step1; length end $40; set dict; where complev("&to",word,'i')=1; end="&to"; run; /* loop */ %macro ladder; %let done=N; %let i=1; %do %until (&done=Y);   %let n=%eval(&i+1);   data step&n;    set dict;     do i=1 to _nobs;      set step&i(rename=(word=step&i)) nobs=_nobs point=i;      if complev(step&i,word)=1 and word ne start %if &i>1 %then %do j=1 %to %eval(&i-1); and word ne step&j %end; then output;     end;   run;   data _step&n;    set dict;     do i=1 to _nobs;      set _step&i(rename=(word=_step&i)) nobs=_nobs point=i;      if complev(_step&i,word)=1 and word ne end %if &i>1 %then %do j=1 %to %eval(&i-1); and word ne _step&j %end; then output;     end;   run;   /* FIX: probably want to better name variables so order can be recognized */   /* start -> step1 -> stepn -> word -> _stepn -> _step1 -> end */   data want;    merge step&n(in=b) _step&n(in=a);    by word;    if a and b;   run;   proc sql noprint;    select nobs into :nobs      from sashelp.vtable     where libname='WORK' and memname='WANT';   quit;   %let i=%eval(&i+1);   %if &nobs>0 %then %let done=Y; %end; %mend; %ladder This code was able to find solution in 5 ladder steps.  The solution I had posted earlier as the example solution I had solved manually. BLACK 1. brack 2. brace 3. brice 4. brite 5. write WHITE ... View more