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else // The bag is empty. { Console::WriteLine("\nThe tile bag is empty."); break; } } Console::WriteLine(); } } else { // A false return will indicate that the user has // changed his/her mind and may not want to pass. return false; } return true; } private: PlayType GetPlayType() { // Input the direction to play. Console::WriteLine( "Enter Direction to Play (A = across, D = down) or P to pass:"); String^ playTypeString = Console::ReadLine(); if (playTypeString == "P") { return PlayType::Pass; } if (playTypeString == "A") { return PlayType::Across; } else if (playTypeString == "D") { return PlayType::Down; } else { Console::WriteLine("Sorry, I didn't understand that input."); throw gcnew Exception(); } }

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With these three questions in mind, we can produce a very simple formula for the cost of a nested loop, which is as follows: Cost of acquiring data from first table + Cardinality of result from first table * Cost of single visit to second table For example, quoting a small section of a 10053 trace from 9i, we see the following on a typical nested loop join calculation: NL Join Outer table: cost: 14847 cdn: 4973 rcz: 35 Inner table: T2 Access path: tsc Resc: 51 Join: Resc: 268470 Resp: 268470 resp: 14847

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This translates as the serial cost (resc) of joining table t2 to the previous table (which may be an intermediate result set from a prior join) is 268,470. This comes from

// Get the position of the start of the play on the board. bool GetPlayStartPosition(int% row, int% col) { // Input the row and column of the first letter. Console::Write( "Enter Location to Play as [row][col]: 00 (top left) to EE (bottom right): "); String^ locString = Console::ReadLine(); // Parse as a hex number. int x = Int32::Parse(locString, System::Globalization::NumberStyles::HexNumber); row = x / 16; col = x % 16; if (row > 14 || col > 14 || row < 0 || col < 0) { Console::WriteLine("I did not understand that input."); Console::WriteLine("The first digit is the row (0 to E);" " the second is the column (0 to E)."); throw gcnew Exception(); } // Check to see that this is an unoccupied space. if (gameBoard[row, col] != nullptr) { Console::WriteLine("Sorry, that space is occupied by the tile: {0}", gameBoard[row, col]); return false; } return true; } // Return true if the play is successful. // Return false if the play is invalid and needs to be restarted. bool GetTilesForPlay(int row, int col, PlayType playType, List<Tile^>^ workingTiles, array<Tile^, 2>^ workingBoard ) { // Get the desired tiles to play from the user. Console::WriteLine( "Enter letters to play (_<letter> to play a blank as <letter>): "); int code;

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14,847 + 4973 * 51 -- cost of getting data from outer table -- number of rows in outer table -- cost of accessing inner table (t2) once

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You will notice the two resp figures these show the cost of using parallel execution, but since the trace file came from a query where the degree of parallelism was one, this matches the serial cost.

do { code = Console::Read(); wchar_t character = safe_cast<wchar_t>(code); Letter letter = Letter::_; if (character == safe_cast<wchar_t>(Characters::NEWLINE)) { Console::ReadLine(); break; } if (character == '_') { letter = Letter::_; // If a blank is entered, read the next character. code = Console::Read(); character = safe_cast<wchar_t>(code); } else if (Char::IsLetter(character)) { if (Char::IsUpper(character)) { letter = safe_cast<Letter>(character - 'A' + 1); } else // character is a lowercase letter. { letter = safe_cast<Letter>(character - 'a' + 1); } } // See if the letter is in the player's hand. Tile^ tile = gcnew Tile(letter); if (letter == Letter::_) { tile->BlankValue = character; } Tile^ tileToRemove = nullptr; bool tileFound = false; for each (Tile^ t in workingTiles) { if (t->LetterValue == tile->LetterValue) { tileToRemove = t; tileFound = true; } }

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