## Saturday, November 12, 2016

### Complete the sequence generated by a polynomial - GeeksforGeeks

Complete the sequence generated by a polynomial - GeeksforGeeks

Given a sequence with some of its term, we need to calculate next K term of this sequence. It is given that sequence is generated by some polynomial, however complex that polynomial can be. Notice polynomial is an expression of the following form:
P(x) = a0 + a1 x +a2 x^2 + a3 x^3 …….. + an x^n
The given sequence can always be described by a number of polynomials, among these polynomial we need to find polynomial with lowest degree and generate next terms using this polynomial only.

We can solve this problem using a technique called difference of differences method, which is derivable from lagrange polynomial.
The technique is simple, we take the difference between the consecutive terms, if difference are equal then we stop and build up next term of the sequence otherwise we again take the difference between these differences until they become constant.

In below code same technique is implemented, first we loop until we get a constant difference keeping first number of each difference sequence in a separate vector for rebuilding the sequence again. Then we add K instance of same constant difference to our array for generating new K term of sequence and we follow same procedure in reverse order to rebuild the sequence.

`void` `nextTermsInSequence(``int` `sequence[], ``int` `N, ``int` `terms)`
`{`
`    ``int` `diff[N + terms];`
`    ``//  first copy the sequence itself into diff array`
`    ``for` `(``int` `i = 0; i < N; i++)`
`        ``diff[i] = sequence[i];`
`    ``bool` `more = ``false``;`
`    ``vector<``int``> first;`
`    ``int` `len = N;`
`    ``// loop untill one difference remains or all`
`    ``// difference become constant`
`    ``while` `(len > 1)`
`    ``{`
`        ``// keeping the first term of sequence for`
`        ``// later rebuilding`
`        ``first.push_back(diff[0]);`
`        ``len--;`
`        ``// converting the difference to difference`
`        ``// of differences`
`        ``for` `(``int` `i = 0; i < len; i++)`
`            ``diff[i] = diff[i + 1] - diff[i];`
`        ``// checking if all difference values are`
`        ``// same or not`
`        ``int` `i;`
`        ``for` `(i = 1; i < len; i++)`
`            ``if` `(diff[i] != diff[i - 1])`
`                ``break``;`
`        ``// If some difference values were not same`
`        ``if` `(i != len)`
`           ``break``;`
`    ``}`
`    ``int` `iteration = N - len;`
`    ``//  padding terms instance of constant difference`
`    ``// at the end of array`
`    ``for` `(``int` `i = len; i < len + terms; i++)`
`        ``diff[i] = diff[i - 1];`
`    ``len += terms;`
`    ``//  iterating to get actual sequence back`
`    ``for` `(``int` `i = 0; i < iteration; i++)`
`    ``{`
`        ``len++;`
`        ``//  shifting all difference by one place`
`        ``for` `(``int` `j = len - 1; j > 0; j--)`
`            ``diff[j] = diff[j - 1];`
`        ``// copying actual first element`
`        ``diff[0] = first[first.size() - i - 1];`
`        ``// converting difference of differences to`
`        ``// difference array`
`        ``for` `(``int` `j = 1; j < len; j++)`
`            ``diff[j] = diff[j - 1] + diff[j];`
`    ``}`
`    ``//  printing the result`
`    ``for` `(``int` `i = 0; i < len; i++)`
`        ``cout << diff[i] << ``" "``;`
`    ``cout << endl;`
`}`