By Sathish Govindarajan, Anil Maheshwari

ISBN-10: 331929220X

ISBN-13: 9783319292205

ISBN-10: 3319292218

ISBN-13: 9783319292212

This booklet collects the refereed court cases of the second one overseas convention on Algorithms and Discrete utilized arithmetic, CALDAM 2016, held in Thiruvananthapuram, India, in February 2016. the quantity comprises 30 complete revised papers from ninety submissions in addition to 1 invited speak provided on the convention. The convention makes a speciality of issues concerning effective algorithms and information constructions, their research (both theoretical and experimental) and the mathematical difficulties bobbing up thereof, and new purposes of discrete arithmetic, advances in current functions and improvement of recent instruments for discrete mathematics.

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**Additional info for Algorithms and Discrete Applied Mathematics: Second International Conference, CALDAM 2016, Thiruvananthapuram, India, February 18-20, 2016, Proceedings**

**Sample text**

By Remark 1, quotient matrix of the distance matrix is given by Q = k11 k12 k21 k22 and the largest eigenvalue of it is 21 (k11 + k22 ) + (k11 − k22 )2 + 4k12 k21 . Now by putting the values of kij , i, j = 1, 2 in the above expression and by applying Lemma 1. we get our result. Theorem 4. The distance spectral radius of the generalized Petersen graph P (n, 3), n ≥ 8, is given by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ λ1 (P (n, 3)) = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 (n2 12 + 28n − 60 + 1 (n2 12 + 28n − 89 + 1 (n2 12 + 28n − 96 + 1 (n2 12 + 28n − 69 + 1 (n2 12 + 28n − 80 + 1 (n2 12 + 28n − 93 + √ √ √ √ √ n4 + 40n3 + 416n2 − 4800n + 3600), n ≡ 0 (mod 6), n4 + 40n3 + 398n2 − 680n + 1681), n ≡ 1 (mod 6), n4 + 40n3 + 400n2 − 576n + 1088), n ≡ 2 (mod 6), n4 + 40n3 + 398n2 − 840n + 3681), n ≡ 3 (mod 6), n4 + 40n3 + 416n2 − 320n + 1600), n ≡ 4 (mod 6), √ n4 + 40n3 + 430n2 − 72n + 1985), n ≡ 5 (mod 6).

So, pj1 must lie either on H12 or on H22 . Let it lie on H22 . We want to ﬁnd the next maximum density segment with the same right end point pj . Let the left end point of this segment be pj2 . We need to ﬁnd it. Clearly, it lies either on the contour of or interior to H 1 . By construction of H, any pair of lch at the same level are mutually disjoint, Hji1 ∩ Hji2 = φ for all j1 and j2 with j1 = j2 . Since pj1 lies on the contour of H22 , pj2 can either be the point of contact of tangent from pj to H12 , or on the contour or interior of H22 .

N2 . The shortest path from u0 to u1 , u2 , u3 , or u4 is through the edges in cycle C. So one gets that d(u0 , u1 ), d(u0 , u2 ), d(u0 , u3 ), and d(u0 , u4 ) 4 are equal to 1, 2, 3, and 4 respectively. Let us take L = d(u0 , ui ) = i=1 n 10. , v2m , u2m ). So d(u0 , u2m ) = m + 2. , v2p+1 , u2p+1 ). So d(u0 , u2p+1 ) = p + 3. According to diﬀerent values of n, the range of 2m and 2p + 1 are given as below. For n ≡ 0, 1, 2, or 3 (mod 4) the range of 2m and 2p + 1 are respectively 4 < 2m ≤ n2 and 4 < 2p + 1 ≤ n2 − 1; 4 < 2m ≤ n2 and 4 < 2p + 1 < n2 ; 4 < 2m ≤ n2 − 1 and 4 < 2p + 1 ≤ n2 ; or 4 < 2m < n2 and 4 < 2p + 1 ≤ n2 .

### Algorithms and Discrete Applied Mathematics: Second International Conference, CALDAM 2016, Thiruvananthapuram, India, February 18-20, 2016, Proceedings by Sathish Govindarajan, Anil Maheshwari

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