How to compile VBscript to contain dll handlers?

I am using VBsedit to compile some VBs script to .exe The exe is created fine and shows MsgBox and so on. However, once it hits the part where liraries are used and tries to CreateObject it stops working. Obiously, I can run the script from VBsedit because I can add references to the libraries. But I don’t know how to compile the executable so it contais the libraries. Any help? Might gcc help?

Sub Main() Dim autECLConnMgr, autECLSession, sessionID, actCtx Dim myScreenStr, libPath

‘libPath = “C:\Program Files (x86)\IBM\Personal Communications\autcmgr.tlb” ‘Set autECLConnMgr = GetObject(libPath) ‘this does not work

Set autECLConnMgr = CreateObject(“PCOMM.autECLConnMgr”)

… ‘below here where some tests but it breaks on the lines above

End sub

The error message i get is ActiveX component can not create object.

Displays data arrays with appendrow where other columns contain data

I wrote a script to display a list of names and folder IDs.

The script that I wrote like this:

function listFolders() {  var sheet = SpreadsheetApp.getActiveSheet();   sheet.getRange('A2:B').clear();   //sheet.appendRow(["Name", "Folder-Id"]);  var folderID = DriveApp.getFolderById("1Hp_dM8WR0mGDT5q-a-VS-VVtl5dNCLfe"); var contents = folderID.getFolders();  var cnt = 0; var lFolder;  while (contents.hasNext()) {     var lFolder =;     cnt++;         data = [             lFolder.getName(),             lFolder.getId(),         ];           sheet.appendRow(data);     }; }; 

I want to add new data in column C. After I fill in a text in column C, then I run the script. But the second row is empty.

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CFG for all strings of a’s and b’s that contain a different number of a’s and b’s

I am trying to write CFG for all strings on {a,b} that contains different numbers of a’s and b’s? After two hours of brainstorming, I came up with this:


I tried running few strings on this, it worked, But then I run string “abbab” and it failed.
Kindly someone point out at my mistake or tell me the correct CFG for this as I don’t know how to correct this.

Consider a graph G = (V, E) that does not contain any cycle. Show that there is at least one node in G whose degree is at most one

Consider a graph G = (V, E) that does not contain any cycle. Show that there is at least one node in G whose degree is at most one.

I don’t really know how to formalise my logic with this proof. I understand it would be at the endpoint/standpoint of any path (as there are no cycles) but don’t know how to illustrate my logic in a proof like manner. Thank you.