Chapter 3 Pentagon with Cycle Convergence
3.2 The Proof with Strong Induction
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Chapter 3 Pentagon with Cycle Convergence
3.1 Introduction
Let [a b c d e] be a cycle order pentagon, a, b, c, d, e are all nonnegative integers and not all equal, and let m = min{a, b, c, d, e}, then at least one element of [a－m, b－m, c－m, d－m, e－m] must be zero.
Since [a－m, b－m, c－m, d－m, e－m] and [a b c d e] have the same child, by Theorem 2.2, [a－m, b－m, c－m, d－m, e－m]
[a b c d e], and it is easy to see if one of them has a cycle convergence, the other must have the same cycle convergence, too. So we just need to prove [a－m, b－m, c－m, d－m, e－m] has a cycle convergence.3.2 The Proof with Strong Induction
We let [a b c d e] be a cycle order pentagon again, and assume at least one element of a, b, c, d, e be zero. By strong induction to prove cycle convergence, we check the following pentagons first:
1. [1 0 0 0 0] [1 0 0 0 1] = [0 0 0 1 1] (typeⅠ)
2. [1 1 0 0 0] = [0 0 0 1 1] (typeⅠ)
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3. [1 0 1 0 0] = [0 0 1 0 1] (typeⅡ)
4. [1 1 1 0 0] [0 0 1 0 1] (typeⅡ)
5. [1 1 0 1 0] [0 1 1 1 1] = [1 0 1 1 1] ( typeⅢ)
6. [1 1 1 1 0] = [1 0 1 1 1] ( typeⅢ)
And then, we assume for any pentagon must have cycle convergence if the value of max{a, b, c, d, e}
equal 1,2,3,〃〃〃,M－1.
Finally, we just need check if the value of max{a, b, c, d, e} equal M, then the pentagon still has cycle convergence, too.
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[M－c M－x |x－a| |a－b| |b－c|] must have cycle convergence.And then, [M 0 x y z] has cycle convergence, too.
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Therefore, [M－x M－x 0 |M－2x| |M－2x|] must have cycle convergence.
And then, [M 0 0 x y] has cycle convergence, too.
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So [M x |x－y| y M] = [M x a y M] [0 M－x |x－a| |y－a| M－y]
Since 0 ≦ a < M and 0 < x, y < M, by Theorem 2.4, 0 ≦ |x－a|, |y－a| < M
Since 0 < M－x, M－y < M, [0 M－x |x－a| |y－a| M－y] must have cycle convergence.
And then, [M 0 x y 0] has cycle convergence, too.
(4): [M x 0 0 y]
[M x 0 0 y] [M－x x 0 y M－y]
Since 0 < M－x, M－y, x, y < M, [M－x x 0 y M－y] must have cycle convergence.
And then, [M x 0 0 y] has cycle convergence, too.
■ Case 3.
Let n = 1, and t = 3, (1):[M 0 0 0 x]
[M 0 0 0 x] [M 0 0 x M－x]
Since 0 < M－x < M, by case 2 (1), [M 0 0 x M－x] has cycle convergence.
And then, [M 0 0 0 x] has cycle convergence, too.
(2): [M 0 0 x 0]
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(2):[M M x 0 y]
[M M x 0 y] [0 M－x x y M－y]
Since 0 < M－x, M－y, x, y < M, [0 M－x x y M－y] has cycle convergence.
And then, [M M 0 x y] has cycle convergence, too.
(3): [M 0 M x y]
[M 0 M x y] [M M M－x |x－y| M－y]
(a) If x = y : [M M M－x |x－y| M－y] = [M M M－x 0 M－x]
Since 0 < M－x < M, by case 5 (2), [M M M－x 0 M－x] has cycle convergence.
And then, [M 0 M x y] must have cycle convergence, too.
(b) If x≠y : Let k = min{ M－x, |x－y|, M－y }
Then [M M M－x |x－y| M－y]
[ M－k M－k M－x－k |x－y|－k M－y－k]Since at least one element of M－x－k, |x－y|－k, M－y－k is zero and all elements of M－k,
M－x－k, |x－y|－k, M－y－k are smaller then M.
It means [M－k M－k M－x－k |x－y|－k M－y－k] has cycle convergence.
And then, [M 0 M x y] has cycle convergence, too.
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Since at least one element of M－x－k, y－k, M－y－k is zero and all elements of M－x－k, M－k, y－k, M－y－k are smaller then M, it means [M－x－k M－x－k M－k y－k M－y－k]
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(3): [M 0 M 0 x]
[M 0 M 0 x] [M M M x M－x] [x 0 0 M－x |M－2x|]
Since 0 < x, M－x < M, by Theorem 2.4, 0 ≦ |M－2x| < M.
And since 0 ≦ x, M－x, |M－2x| < M, [x 0 0 M－x |M－2x|] has cycle convergence.
And then, [M 0 M 0 x] has cycle convergence, too.
(4): [M x M 0 0]
[M x M 0 0] [M－x M－x M 0 M]
Since 0 < M－x < M, by case 5 (3), [M－x M－x M 0 M] has cycle convergence.
And then, [M x M 0 0] has cycle convergence, too.
■ Case 7.
Let n = 2, and t = 3, (1):[M M 0 0 0]
[M M 0 0 0]
typeⅠSo [M M 0 0 0] has cycle convergence.
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(2):[M 0 M 0 0]
[M 0 M 0 0]
typeⅡSo [M 0 M 0 0] has cycle convergence.
■ Case 8.
Let n = 3, and t = 1, (1):[M M M 0 x]
[M M M 0 x] [0 0 M x M－x]
Since 0 < M－x < M, by case 2 (1), [0 0 M x M－x] has cycle convergence.
And then, [M M M 0 x] has cycle convergence, too.
(2):[M M 0 M x]
[M M 0 M x] [0 M M M－x M－x]
Since 0 < M－x < M, by case 5 (1), [0 M M M－x M－x] has cycle convergence.
And then, [M M 0 M x] has cycle convergence, too.
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■ Case 9.
Let n = 3, and t = 2, (1):[M M M 0 0]
[M M M 0 0] [0 0 M 0 M]
typeⅡ So [M M M 0 0] has cycle convergence.(2):[M M 0 M 0]
[M M 0 M 0]= [M 0 M M M]
typeⅢ So [M M 0 M 0] has cycle convergence.■ Case 10.
Let n = 4, and t = 1, (1): [M M M M 0]
[M M M M 0] = [M 0 M M M]
typeⅢ So [M M M M 0] has cycle convergence.‧
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Since all the pentagons have cycle convergences with the largest value of a, b, c, d, e is equal M, by strong induction, all the pentagons must have cycle convergences if all the elements of pentagons are nonnegative integers and not all equal.
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Chapter 4 Conclusion and Promotion
In our study, we find all the Diffy pentagons must have the following features if their elements are all nonnegative integers and not all equal:
1. All the Diffy pentagons must have cycle convergences.
2. All the cycle convergences must have the same type as the following figure for some integer n:
3. If all the values of vertices of pentagon are equal, this Diffy pentagon must converges to [0 0 0 0 0]
Next, if we take [a b c d e] to be a cycle order pentagon and assume at least one element of a, b, c, d, e is negative, we can subjoin one value which equal the absolute of the min{ a, b, c, d, e }. Then the new pentagon must be isomorphic to the first pentagon.
Also, if at least one element is rational, we can multiply each element by the least common multiple of denomicators such that all numbers of this pentagon become integers, then we can use the method above such that all numbers of this pentagon become nonnegative integers.
By these two ways, we can easily promote our conclusion to rational number.
During our study, we still find there are some rules of cycle convergences with hexagon, heptagon, and so on. I hope this article can help us to continue to study Diffy pentagon even further.
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The principle of mathematical induction asserts that P(k) being true implies P(k+1) is true, but sometimes is not enough. Strong induction is a variant on proof by induction. It comprises of the following steps:
Then we just need prove k+1 is the product of primes.
First, if k+1 is a prime:
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[1] A. Behn, C. Kribs-Zaleta, and V. Ponomarenko, The Convergence of Difference Boxes.
American Math. Monthly, volume 112, 426-438, (1995)
[2] F. Breuer, Ducci Sequences in Higher Dimensions. Electronic Journal of Combinatorial Number Theory 7, #A24, (2007)
[3] R. Brown and J. Merzel, The Length of Ducci’s Four-Number Game. Rocky Mountain Journal of Mathematics, volume37, 45-65, (2007)
[4] N. Calkin, J. Stevens, and D. Thomas, A Characterization for the Length of Cycles of the N-Number Ducci Game. Fibonacci Quarterly, volume 43, no1, 53-59, (2005).
[5] J. Creely, The Length of a Three-Number Game. Fibonacci Quarterly, volume 26, no2, 141-143, (1988).
[6] A. Ehrlich, Periods in Ducci’s n-Number Game of Differences. Fibonacci Quarterly, volume 28, no4, 302-305, (1990)
[7] A. Ludington-Young, Ducci-Processes of 5-Tuples. Fibonacci Quarterly, volume 36, no5, 419-434, (May 1998)
[8] A. Ludington-Young, Length of the 7-Number Game. Fibonacci Quarterly, volume 26, no3,195-204, (1998).
[9] A. Ludington-Young, Length of the n-Number Game. Fibonacci Quarterly, volume 28, no3, 259-265, (1990).
[10] W. Webb, The Length of the Four-Number Game. Fibonacci Quarterly, volume 20, no1, 33-35, (1982).
[11] F.B. Wong, Ducci Processes. Fibonacci Quarterly, volume 20, no2, 97-105, (1982).
[12] 蔡秀芬, Diffy Box (狄菲方塊). National Chengchi University, (2008)