Iterators CS

Chakrabarti What is an iterator?  Thus far, the only data structure over which we have iterated was the array for (int ix = 0; ix < arr.size(); ++ix) { … arr[ix] … // access as lhs or rhs }  In general there may not be such a simple notion of a single index; e.g., iterate over:  All dim  val map entries in sparse array  All permutations of n items  All non-attacking n-queen positions on an n by n chessboard

Chakrabarti Iterator as an abstraction  We initialize an iterator  Given any state of the iterator, we can ask if there is a next state or we have run out  If there is a next state, we can move the iterator from the current to the next state  We can fetch or modify data associated with the current state of the iterator

Chakrabarti Iterator on vector  Print all elements of a vector, using iterator vector vec; // suitably filled for (vector ::iterator vx = vec.begin(); vx != vec.end(); ++vx) { cout << (*vx) << endl; } Iterator type “++” overloaded to mean “advance the iterator” Access the current contents of the iterator

Chakrabarti Iterator lineage categorycharacteristicvalid expressions all categories Can be copied and copy-constructedX b(a); b = a; Can be incremented++a, a++, *a++ Random Access Bidirectional Forward Input Accepts equality/inequality comparisonsa == b, a != b Can be dereferenced as an rvalue*a, a->m Output Can be dereferenced to be the left side of an assignment operation *a = t, *a++ = t Can be default-constructedX a;, X() Can be decremented--a, a-- *a-- Supports arithmetic operators + and - a + n, n + a, a – n, a – b Supports inequality comparisons (, =) between iterators a b a = b Supports compound assignment operations += and -=a += n, a -= n Supports offset dereference operator ([])a[n] We will see other kinds of iterators after we study C++ collection types beyond vector

Chakrabarti Counting in any radix  Print all numbers in a given radix up to a given number of “digits”  E.g. nDigits = 2, radix = 3 gives 00, 01, 02, 10,11, 12, 20, 21, 22  (Note radix could be arbitrarily large) for dig n-1 = 0 to radix-1 for dig n-2 = 0 to radix-1 … for dig 0 = 0 to radix-1 print dig n-1, dig n-2, …, dig 1, dig 0 Don’t know how to do this for input variable nDigits

Chakrabarti Counting  Instead of writing nested loops…  Stick the loop indices into vector dig  Given any state of dig, to get the next state:  Locate least significant digit that can be incremented (i.e., is less than radix  1)  Increment this digit  Reset to zero all less significant digits

Chakrabarti Printing permutations without recursion  Given input n, print all n! permutations of n items (say characters a, b, …)  0 th item placed in 1 way, “way #0”  1 th item placed in 2 ways, “ways #0, #1” Left of first item or right of first item  Three gaps now, 2 th item in ways 0, 1, 2  … and so on to the (n-1) th item  Suppose we had integer variables way 0, way 1, …, way n-1

Chakrabarti Permutation example way 0 way 1 way 2 cba bcabaccabacbabc abba a

Chakrabarti Nested for loops for way 0 = 0 (up to 0) { for way 1 = 0 to 1 { for way 2 = 0 to 2 { … for way n-1 = 0 to n-1 { convert way 0, …, way n-1 into permutation } … } Unfortunately, we cannot read n as an input parameter and implement a variable number of nested for loops

Chakrabarti Variable number of nested loops?  Vector of wayi variables has n! possible tuple values  Is in 1-1 correspondence with the permutations of n items  Design two functions …  Given vector way that satisfies above properties, arrange way.size() items according to that correspondence  From one way find the next way (the iteration)

Chakrabarti bool nextPerm(vector & way) bool didChange = false; for (int ch=0; ch < way.size(); ++ch) { if (way[ch] < ch) { ++way[ch]; didChange = true; for (int zx = 0; zx < ch; ++zx) { way[zx] = 0; } break; } return didChange; nextPerm is an iterator --- it modifies way to the next permutation if any, returning true if it succeeded

Chakrabarti printPerm(vector way) vector arrange(way.size(), 0);//empty for (int px=way.size()-1; px >= 0; --px) { // skip way[px] empty slots in arrange int wx = 0; for (int sx = way[px]; ; --sx) { while (arrange[wx] != 0) { ++wx; } if (sx == 0) { break; } ++wx; } arrange[wx] = 'a' + px; } print(arrange);

Chakrabarti Semi-magic square  A semi-magic square contains non-negative integers  Each row and column adds up to the same positive integer c  A permutation matrix is a semi-magic square having c=1  Can always subtract a permutation matrix from a semi-magic square (but not greedily)  In time proportional to the number of nonzero elements in the square  Here we look at a brute force solution using iterators

Chakrabarti Iterator state  For each row, keep a (sparse) vector containing column indices with nonzero elements  vector  > nzCols;  For row rx, there are nzCols[rx].size() possible column options  Declare vector colsTaken, where colsTaken[rx] can range between 0 and nzCols[rx].size()-1

Chakrabarti Nested loop way of thinking for colsTaken[0] = 0 … nzCols[0].size()-1 { for colsTaken[1] = 0 … nzCols[1].size()-1 { … … … for colsTaken[n-1] = 0 … nzCols[n-1].size()-1 { check if colsTaken[0…n-1] define a permutation matrix if so { subtract from magic; update nzCols; return } } // end of colsTaken[n-1] loop … … … } // end of colsTaken[1] loop } // end of colsTaken[0] loop Can easily turn into iterator style

Chakrabarti Other applications Maze with walls  Starting point and destination  For every decision point, declare a “loop variable”, keep in vector  Earliest decision = lowest index of loop variable Game trees  Different possible moves = branch out  Represent with “loop variable”