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Whenever needed, the moderator called for an example of an argument to enforce the participants to express their wishes clearly As expected the discussion was more focused, due to the stricter procedural rules Agreement was reached more quickly and a much wider consensus was reached With the stack of topics which were to be discussed (not all due to time constraints), the focus of the group was kept The restricted set of arguments is easy to classify and thus the ontology engineer was able to build the ontology in a straightforward way It is possible to explain to new attendees why a certain concept was introduced and modeled in such a way, by simply pointing to the archived focused discussion It is even possible to state the argumentation line used to justify it The participants truly shared the conceptualization and did understand it.

// Fetch all SMS messages from Inbox or other folder and return them void CRSmsMsg::GetAllSmsL(TInt aBoxId, TDes8 &aBuffer, TDes8 &aTempBuffer) { // Get access to the selected folder with default grouping // and sorting TMsvSelectionOrdering order(KMsvNoGrouping, EMsvSortByNone, ETrue); if(iBoxEntry != NULL) {delete iBoxEntry;} iBoxEntry = CMsvEntry::NewL(*iMsvSession, aBoxId, order); // get the list of message entries in the folder if(iSelectedMessages != NULL) {delete iSelectedMessages;} iSelectedMessages = iBoxEntry->ChildrenWithTypeL(KUidMsvMessageEntry); // Traverse the messages iNextSmsToSend = 0; GetMoreSmsL(aBuffer, aTempBuffer); } // Fetch remaining Sms messages from Inbox or other folder // and return them void CRSmsMsg::GetMoreSmsL(TDes8 &aBuffer, TDes8 &aTempBuffer) { // Set up for saying that there are no more messages TBool gotMessages = EFalse; // If there are pending messages then send them, otherwise, say // there are no more TInt msgCount = iSelectedMessages->Count(); // Traverse the messages while(iNextSmsToSend < msgCount ) { gotMessages = ETrue; // Get the message details in a temporary buffer TMsvEntry msgEntry; TMsvId owningServiceId; User::LeaveIfError(iMsvSession->GetEntry((*iSelectedMessages)

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The straight line gives the probability that an individual game is won The other graph shows the probability that the winner of the rst game wins the series The graphs intersect at p = 0347129; so, if p is greater than this, then the winner of the rst game is more likely to win the series

In particular, in con ict situations, when opinions diverged, the restriction of arguments was.

[iNextSmsToSend], owningServiceId, msgEntry)); aTempBuffer.Zero(); GetOneSmsL(msgEntry, aTempBuffer); // The entry may not have been useful so check that data has been // added if( aTempBuffer.Length() > 0) { // If the message will fit then append it, otherwise break out if(aTempBuffer.Length()+aBuffer.Length()+4 < aBuffer.MaxLength()) { CConnPack::WriteBufferL(aTempBuffer, aBuffer); iNextSmsToSend ++; } else { // Run out of room - leave it till next message break; } } else { // Not an SMS so skip it iNextSmsToSend ++; } } // endwhile if(gotMessages) { // Send the complete message with a terminating null message ID CConnPack::WriteInt32L(KNullUidValue, aBuffer); } }

We have found that the winner of a seven-game series is not necessarily the better team In fact, if the teams are about evenly matched, the probability that the weaker team wins the series is about equal to the probability that the team wins an individual game Perhaps a solution to this is to increase the length of the series so that the probability that the winner of the series is in fact the stronger team is increased How long, then, should the series last We presume the series is over when, in a series of n games, one team has won (n + 1)/2 games If n = 7, then the winner must win four games If n = 9, then the winner must win ve games Now we nd a formula for the probability that a given team wins a series of n games Call the winner of the series A and suppose the probability that A wins an individual game is p and the probability that the loser of the series wins an individual game is q If A wins the series, then A has won (n + 1)/2 games and has won (n + 1)/2 1 = (n 1)2 games previously to winning the last game During this time, the loser of the series has won x games (where x can be as low as 0 and at most (n 1)/2 games) So there are (n 1)/2 + x = (n + 2x 1)/2 games played before the nal game It follows that

helpful. In this way participants could either prove their view, or were convinced. More details on the argumentation model of DILIGENT can be found in Tempich et al. (2005).

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