留言板
- [1]林早
- The proof of Collatz problem: Construct the domain transformation: x = 3n + d, y = 3n-d, z = n / 2.
1. d = 0, n = 0, x = 3 × 0 + 0 = 0, y = 3 × 0-0 = 0, z = 0/2 = 0. 2. d = 1, n = 0, x = 3 = 0 + 1 = 1, y = 3 × 0-1 = -1, z = 0/2 = 0. 3. If d belongs to Z, n belongs to N+, x = 3 × 1 + d, y = 3 ×1-d, z=n/2; If d belongs to Z and n belongs to N-,x= 3× (-1) + d, y = 3 × (-1) -d, z = n / 2. (1) If d belongs to Z, n belongs to N+.【1】 (3 × 1 + d1-1) / 3 = (3 × 1-d1) / 2, d1 = 1; (3 × 1-d2-1) / 3 = (3 × 1 + d2) / 2 , d2 = -1. 【2】 (3 × 1 + d3) / 2 = 2 (3 × 1 + d3), d3 = 9/5; (3 × 1 + d4) / 2 = 2 (3 × 1-d4), d4 = 9/5. 【3】 3 (3 × 1 + d5) + 1 = 2 (3 × 1 + d5), d5 = 4/5; 3 (3 × 1 + d6) + 1 = 2 (3 × 1-d6) ,d6= -4 / 5. Take d = 1, then x = 4, y = 2, eavailable loop A = (4,2,1,4). According to the principle of transformation, when n = 1 satisfies 3 × 1 + 1 = 4,3 × 1-1 = 2, the circular circle F = (4,2,1,4) = A is obtained, so available loop circle(A , A). Therefore, the 3n + 1 transformation on the positive domain has only the loop A = (4,2,1,4). (2) If d belongs to Z, n belongs to N-. ＜1＞ Shows that n = -1 when the transformation is equivalent to (1), then d =1, x = -2, y = -4, the loop can be obtained B = (-1, -2, -1), because -4 does not belong to B, so n = -1 does not meet the transformation principle, so take n = -2. ＜2＞【1】[3×(2)-d7-1]/3= [3 × (-2) + d7] / 2, d7 = 4/5; [3 × (-2) + d8 - 1] / 3 = [3 × (-2) - d8] / 2, d8 = -4 / 5. 【2】[3 × (-2) -d9] / 2 = 2 [3 × (-2) + d9], d9 = 18/5; [3 × (-2) + d10] / 2 = 2 [3 × (-2) -d10], d10 = -18 / 5.【3】3 [3 × (-2) - d12] + 1 = 2 [3 × (-2) + d12], d12 = -1. Eavailable loop circle C= (-5, -14, -7, -20, -10, -5), according to the transformation principle, take d =1, x= -5 and y = -7. According to the transformation principle, take n=-14. ＜3＞【1】 [3 × (-14) - d13-1] / 3 = [3 × (-14) + d13] / 2, d13 = 8; [3 × (-14) + d14-1] / 3 = [3 × (-14) - d14] / 2, d14 = -8. 【2】[3 × (-14) + d16] / 2 = 2 [3 × (-14) + d15], d15 = 126/5; [3 × (-14) + d16] × (-14) -d16], d16 = -126 / 5. 【3】3 [3 × (-14) + d17] + 1 = 2[3 × (-14) -d17], d17 = 41/5; 3 [3 × (-14) -d18] +1=2[3× (-14) + d18], d18 = -41 / 5. Take d = 8, then x = -34, y = -50, eavailable loop circl D = (- 34, -17, -50, -25, -74, -37, -110, -55 ,-164, -82, -41, -122, -61, -182, -91, -272, -136, -68, -34), according to the transformation principle, take n = -17. ＜4＞【1】 [3×(-17)-d19-1]/3=[3 × (-17) + d19] / 2, d19 = 49/5; [3 × (-17) + d20- 1] / 3 = [3 × (-17) -d20] / 2, d20 = -49 / 5. 【2】 [3 × (-17) + d22] / 2 = 2 [3 × (-17) + d21], d21 = 153/5; [3 × (-17) + d22] × (-17) -d22], d22 = -153 / 5. 【3】3 [3 × (-17) + d2] + 1 = 2 [3 × (-17) -d23], d23 = 10; 3 [3 × (-17) -d24] + 1 = 2 [3 × (-17) +d24], d24 = -10. Vailable loop circle E= (-41, -122, -61, ..., -41) = D, so that the loop (D, D) can be obtained by taking d = 10, then x = -41, y = -61. So the transformation of each cycle on the negative domain ends in the loop D, and the 3n + 1 transformation on the negative domain has B, C, and D 3 cycles.
Conclusion: The 3n + 1 transformation on the whole domain has A, B, C, D 4 cycles.
