Now we explore the tree structure of the problem and bring in the Branch and Bound idea. Notice that while 3 is invertible modulo 2g, 2 of course is not. In fact, what is happening as you go through the stages :1; > y- (2/3) ) y- (2/3)2 > ... > 231 is that each stage is a 2to1 map. For instance, in the rst step, the odd numbers y and (2'2'1 + y) mod 22 both map to 2y/ 3 mod 25, a number which is divisible by 2 but not by 4. In the very last stage, the two numbers 25*2 and 22*1 + 22'2 map to 23*1. Now you should \"turn this around\" and imagine a binary tree whose vertices are labeled with certain rational numbers. The root is labeled with 1/2. In general, if a vertex is labeled with 2, its children are labeled, one each, with the two values of 2' mod 1 satisfying the equation 22' = 32 mod 1. (\"Mod 1\" gives the fractional part of a number. Note there are indeed two distinct z' satisfying the equation 22' = 3z mod 1, offset by 1/2, since 2 - (1/2) = 0 mod 1.) For example, the children of the root are labeled with 1/4 and 3 / 4. The children of the vertex labeled 1/4 are labeled with 3/ 8 and 7/8; while the children of the vertex labeled 3/4 are labeled with 1/8 and 5/8. You will notice that at level I? of the tree (with the root being level 1), the labels have denominators = 245 and odd numerators. Let L; denote the set of vertices at level E of the tree; for vertex 1} let 14(1)) denote the ancestors of v (the vertices on the simple path from v to the root, inclusive); and let z(v) denote the rational number labeling vertex 1). You can now verify that if vertex v is such that 2(1)) = y/2E, 3; odd, then My) = is Z exp C by eN(y) = exp(27riy/N). For a. given If 2 1, let 21 mm = gate (2 ewe/w) m=0 Notice that the division by 3 is not in R; this is multiplication by 31 in Z/Z'E. Conjecture 1 There is a c