seen from the outside of the outer object, the minimum slicing problem is to find the smallest numbers of ... Thin slices are made across the object, where every.
On The Minimum Slicing Problem Liawas Barukang1*, Rayner Alfred2 1 FRICS 2
Labs, Fakulti Kejuruteraan, Universiti Malaysia Sabah, Kota Kinabalu, Sabah MALAYSIA. Fakulti Komputeran dan Informatik, Universiti Malaysia Sabah, Kota Kinabalu, Sabah MALAYSIA.
Given an arbitrary small rigid object enclosed by another arbitrary rigid object and the inner object could not be seen from the outside of the outer object, the minimum slicing problem is to find the smallest numbers of slices that has to be made across the objects before the longest length between two points on the perimeter of the inner object is found. General, trivial solution to this problem is by brute-force method of infinite slice. Non-trivial solution is via Sylow’s Geometric Group Theorem. Keywords: LB Slicing Problem, Sylow’s Theorem, Group Theory.
1. INTRODUCTION Let an object A be a rigid 2-dimensional object, having an arbitrarily random shape. Inscribed within is another 2dimesional rigid object B, also having an arbitrarily random shape. The object B is not visible to observer from outside object A, and the shape of object B is not known to any observer. The objects A and B are depicted in Fig. 1.
Figure 2: Longest distance between two points.
Figure 1: Outer rigid shape and an inscribed inner rigid shape
2. BACKGROUND The Minimum Slicing Problem (MSP) is to determine the smallest numbers of cuts that has to be made across the outer object before the longest dimension of the inner object is found. The longest dimension of the inner object is the longest distance between two points on the perimeter of the inner object. The longest dimension of object B is illustrated in Fig. 2.
Let Cut 1 be a slice from a to a’, Cut 2 be the slice from b to b’ and Cut 3 be the slice from c to c’, and all (a, a’), (b, b’) and (c, c’) are points on the perimeter of Object B. The distance between the points a and a’, b and b’, and c and c’ is denoted as d(a, a’), d(b, b’) and d(c, c’). In Fig. 2, it can be observed that the distance d(a, a’) < d(c, c’) < d(b, b’). Therefore, the longest dimension of object B is the length bb’. The smallest numbers of cuts is referring to the numbers of cuts that has to be made on the outer object before the longest dimension of the inner object is found. The definition of the smallest numbers of cuts is illustrated in Fig. 3.
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RESEARCH ARTICLE
Adv. Sci. Lett. 4, 400–407, 2011
object. For the sake of arguments, lets assume that the maximum numbers of points parked on the perimeter on the outer object is 𝑘. Thin slices are made across the object, where every consecutive cuts are assumed parallel to each other. For every points, there 𝑛 ≈ +∞ numbers of slices to be made. ii. Solution based on geometric construction on a pair of integers (m,n)[1] Figure 3: Cutting through object A.
In the illustration in Fig. 3, the longest dimension of object A is found in the second cut (Cut 2). Therefore, within 3 cuts, the longest dimension is found. 2.1 General Minimum Slicing Problem (MSP) Definition Define the minimum numbers of slices or cuts that one needs to perform across an object before the longest dimension of the inner object is found.
General definition. Let 𝑛 ≥ 2 be an integer and 𝑛 defines the numbers of an equally spaced points distributed around a closed surface. For each integer relative to prime 𝑚, inscribe a figure in the point-marked closed surface by drawing a line from an initial point, P to another point Q, which is 𝑚 points away counter-clockwise from P; and continuously drawing lines from Q to another point 𝑚 points away counterclockwise from Q, until the line terminates at P. The line inscription is illustrated in Fig. 5.
2.1.1 Non-General Solutions There is no work found on this area up to this point of time. Initial perception by any applied mathematician to solve this problem is that the solution is finite, by mean of computable numbers. However, this perception is by far not supported and could be supported by any rigorous mathematical proof. It is just a perception. i.
Brute-force infinitesimally many cuts
The largest dimension can be found by marking the perimeter of object A with infinitely many points and slicing object A with an infinitesimally small slice each time, end to end, from every point marked on the perimeter. This solution is illustrated in Fig. 4.
Figure 5: MSP(3,8), n=8, m=3 Consider the following;-
Let 𝑥 be the length of PQ and 𝑦 be the length of P’Q’ where 𝑥 > 𝑦. Suppose that the points P’ and Q’ can be moved between 𝑦 = 0 and 𝑦 = 𝑥, illustrated in the following;
Where ∆𝑦1 ≠ ∆𝑦2 , Figure 4: Infinitely many cuts. This method of solving the MSP problem is exhaustive and extremely computational intensive. However, the solution will be precise and accurate. Lets assume that there are infinitely many points 𝑥𝑛 , 𝑛 ∈ ℤ+ , 𝑛 = {1, . . , 𝑖, . , 𝑗, . . 𝑘 ≈ +∞} marked as close as possible to each other on the perimeter of the outer
a. The greatest dimension that 𝑦 can have can be trivially computed as 𝑦 = max(𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 )
(1)
𝑖 = 1. . 𝑛 2
Where yi is the length y at 𝑖 𝑡ℎ slice and ∆𝑦1𝑖 and ∆𝑦2𝑖 is the change in length from 𝑦𝑖−1 to 𝑦𝑖 . b. Approximating the largest dimension
determine the largest dimension after n cuts. The dimension of the inner object can be determined using equation in (1) or in (2) with condition (3) in just n cuts.
Write 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 as 𝑑𝑦𝑖 and 𝑦 + ∆𝑦1 + ∆𝑦2 as 𝑑𝑦.
Example
Define 1
𝑑𝑦𝑎𝑣𝑔 = 𝑛 ∑𝑛1 𝑑𝑦𝑖
Assume that n=8 and m=3
(2)
𝑑𝑦𝑎𝑣𝑔 will provide the average length between all available 𝑦, and ∆𝑦1 and ∆𝑦2 . Using the knowledge of 𝑑𝑦𝑎𝑣𝑔 , the unwanted 𝑦 (and ∆𝑦1 and ∆𝑦2 ) can be filtered and select the most suitable 𝑦𝑖 that gives the greatest dimension using the following simple rule; 𝑑𝑦𝑖 ; 𝑑𝑦𝑖 > 𝑑𝑦𝑎𝑣𝑔 ,𝑖 = {1. . 𝑛} 𝑑𝑦𝑚𝑎𝑥 = { 0; 𝑑𝑦𝑖 = 0
(3)
Case 1: When no maximum value is found, the maximum dimension is zero; Case 2: If the 𝑖 𝑡ℎ length, or a set of length are greater than the average, then one of the dimension is the candidate. 2.1.2 Proposition It can be deduced, based on the general definition and for any given (m,n), that the maximum numbers of slices to search for the possible largest dimension is up to 𝑛 slices.
Figure 7: A (3,8) Cut
1. 𝑑𝑦𝑎𝑣𝑔 approximate the average length of all lines passing through B. 2. 𝑛 = 8, 𝑖 = 8 lines generated in the construction, thus 8
𝑑𝑦𝑎𝑣𝑔
1 = ∑ 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 8 1
Theorem
Note that at line i, ∆𝑦2 + ∆𝑦1 = (𝑦𝑖+1 − 𝑦𝑖 )
The maximum numbers of slices one can make to determine the largest dimension of the inner object in an MSP is up to 𝑛.
𝑦1 = length of 𝑷′𝑸𝟏 ′; 𝑑𝑦𝑎𝑣𝑔 = 1 ∑11 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 ;
Proof
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𝑦2 = 𝑦1 + (𝑦2 − 𝑦1 ), 1
Assume a geometric construction based on (3,8) construction (i.e. n=8, m=3), illustrated in Fig. 6.
length of 𝑸𝟏 ′𝑸𝟐 ′; 𝑑𝑦𝑎𝑣𝑔 = 2 ∑21 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 ; 𝑦3 = 𝑦2 + (𝑦3 − 𝑦2 ), 1
length of 𝑸𝟐 ′𝑸𝟑 ′; 𝑑𝑦𝑎𝑣𝑔 = 3 ∑31 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 ; 𝑦4 = 𝑦3 + (𝑦4 − 𝑦3 ), 1
length of 𝑸𝟑 ′𝑸𝟒 ′; 𝑑𝑦𝑎𝑣𝑔 = 4 ∑41 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 ; ⋮ 𝑦𝑛 = 𝑦𝑛−1 + (𝑦𝑛 − 𝑦𝑛−1 ), 1
Figure 6: (3,8) cuts on object A. Lets assume a (3,8) construction of a geometric shape. Every line will either cut through B or misses. For (m,n) construction, there will be n lines that will probably makes the cut. Since there are only n lines to search, then it is trivial to 3
length of 𝑸𝒊+𝟏 ′𝑸𝒊 ′; 𝑑𝑦𝑎𝑣𝑔 = 𝑛 ∑𝑛1 𝑦𝑖 + ∆𝑦1𝑖 + ∆𝑦2𝑖 ; In this example, 𝑦7 will be 𝑑𝑦𝑚𝑎𝑥 based on the rule in (3). The dimension of 𝑦7 is found after n slice and n comparisons.
RESEARCH ARTICLE
Adv. Sci. Lett. 4, 400–407, 2011 5. CONCLUSIONS
REFERENCES
This notes demonstrates the finiteness of the solution to the Minimum Slicing Problem. Even though the accuracy of the result is not of major concern here, the approximate solution can be found by adjusting the (m,n). Increasing n will increase the accuracy of the approximation, in expense of an increased numbers of slices that has to be made.
[1]
Sylow , M. L. (1872). Theorem on substitutions groups. Math. Ann. , Vol 5(1872), 584-594 .
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