Search results
Results From The WOW.Com Content Network
In the even-odd case, the ray is intersected by two lines, an even number; therefore P is concluded to be 'outside' the curve. By the non-zero winding rule, the ray is intersected in a clockwise direction twice, each contributing -1 to the winding score: because the total, -2, is not zero, P is concluded to be 'inside' the curve.
A curve (top) is filled according to two rules: the even–odd rule (left), and the non-zero winding rule (right). In each case an arrow shows a ray from a point P heading out of the curve. In the even–odd case, the ray is intersected by two lines, an even number; therefore P is concluded to be 'outside' the curve.
Assuming n even, therefore, the condition for π g to be an odd permutation, when g has order k, is that n/k should be odd, or that the subgroup <g> generated by g should have odd index. We will apply this to the group of nonzero numbers mod p , which is a cyclic group of order p − 1.
The odd–even sort algorithm correctly sorts this data in passes. (A pass here is defined to be a full sequence of odd–even, or even–odd comparisons. The passes occur in order pass 1: odd–even, pass 2: even–odd, etc.) Proof: This proof is based loosely on one by Thomas Worsch. [6]
Magic squares are generally classified according to their order n as: odd if n is odd, evenly even (also referred to as "doubly even") if n is a multiple of 4, oddly even (also known as "singly even") if n is any other even number. This classification is based on different techniques required to construct odd, evenly even, and oddly even squares.
Batcher's odd–even mergesort [1] is a generic construction devised by Ken Batcher for sorting networks of size O(n (log n) 2) and depth O((log n) 2), where n is the number of items to be sorted. Although it is not asymptotically optimal, Knuth concluded in 1998, with respect to the AKS network that "Batcher's method is much better, unless n ...
If k is odd, then put the number on the left end of the row k − 1 in the first position of the row k, and fill the row from the left to the right, with every entry being the sum of the number to the left and the number to the upper; At the end of the row duplicate the last number. If k is even, proceed similar in the other direction.
The number is taken to be 'odd' or 'even' according to whether its numerator is odd or even. Then the formula for the map is exactly the same as when the domain is the integers: an 'even' such rational is divided by 2; an 'odd' such rational is multiplied by 3 and then 1 is added.