Suppose the Euclidean algorithm is used to determine gcd2388

Suppose the Euclidean algorithm is used to determine gcd(238812,335946).

a) List the values of the variable num, using commas to separate values.

b) List the values of the variable denom, using commas to separate values.

c) gcd(238812,335946) =

Consider the graph given above. Use the FiberNet (Kruskal\'s) algorithm (p. 80, Maurer-Ralston) to find the minimum spanning tree.

a. What is the total weight of the spanning tree?

b. List the weights of the selected edges separated by commas in the order of selection.

Consider the cities {G, H, I, J, K, L}. The costs of the fiber optic link between cities are given below:

c(G,J)=10
c(G,K)=8
c(H,I)=15
c(H,K)=12
c(I,J)=13
c(I,L)=14
c(J,K)=9
c(J,L)=11

What is the minimum cost to build a fiber optic network that connects all the cities?

Question 1	Suppose the Euclidean algorithm is used to determine gcd(238812,335946). a) List the values of the variable num, using commas to separate values. b) List the values of the variable denom, using commas to separate values. c) gcd(238812,335946) =
Question 2	Consider the graph given above. Use the FiberNet (Kruskal\'s) algorithm (p. 80, Maurer-Ralston) to find the minimum spanning tree. a. What is the total weight of the spanning tree? b. List the weights of the selected edges separated by commas in the order of selection.
Question 3	Consider the cities {G, H, I, J, K, L}. The costs of the fiber optic link between cities are given below: c(G,J)=10 c(G,K)=8 c(H,I)=15 c(H,K)=12 c(I,J)=13 c(I,L)=14 c(J,K)=9 c(J,L)=11 What is the minimum cost to build a fiber optic network that connects all the cities?
Question 4

Solution

Post multiple question to get the remaining answer. Thanks

1) The GCD algorithm was used in the following manner to calculate the gcd(238812,335946)

335946 - (238812 x 1) = 97134
238812 - (97134 x 2) = 44544
97134 - (44544 x 2) = 8046
44544 - (8046 x 5) = 4314
8046 - (4314 x 1) = 3732
4314 - (`x 1) = 582
3732 - (582 x 6) = 240
582 - (240 x 2) = 102
240 - (102 x 2) = 36
102 - (36 x 2) = 30
36 - (30 x 1) = 6
30 - (6 x 5) = 0

a) Numerators are 335946 , 238812, 97134, 44544, 8046, 4314, 3732, 580, 240, 102, 36, 30, 6

b) Denominators are 238812, 97134, 44544, 8046, 4314, 3732, 580, 240, 102, 36, 30, 6

c) The final GCD of the numbers will be equal to 6