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3 years ago

The complexity of bfs and dfs

Engineering

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

BFS uses Queue to find the shortest path. DFS uses Stack to find the shortest path. ... Time Complexity of BFS = O(V+E) where V is vertices and E is edges. Time Complexity of DFS is also O(V+E) where V is vertices and E is edges.

Explanation:

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worty [1.4K]

Uhm I’m not understanding the question

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3 years ago

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below

3 0

4 years ago

Tamara has received multiple bids from general contractors to install a new pool in her backyard. How should she go about select

vovikov84 [41]

Answer:

Option C

Explanation:

Tamara must choose the contractor who can provide best services among all and also has a fair pricing. Hence, option 3 is correct

If she chooses the contractor with low price, there are chances that the quality will not be good. Hence, option 2 is incorrect

Option 4 is incorrect because the contractor must take inputs of the contracting person or agency.

Option 1 is incorrect because selecting some one on the basis of reviews can be a biased selection of contractor.

7 0

3 years ago

nalin [4]

THE ANSWER IS B Beaycs i like BExplanation:

3 0

3 years ago

An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a pass

Finger [1]

Answer:

C_h = 0.166 nF

C_L = 0.153 nF

Explanation:

Given:

- Ideal frequency f_o = 500 KHz

- Bandwidth of frequency BW = 40 KHz

- The resistance identical to both low and high pass filter = 2 Kohms

Find:

Design a passive band-pass filter to do this by cascading a low and high pass filter.

Solution:

- First determine the cut-off frequencies f_c for each filter:

f_c,L for High pass filter:

f_c,L = f_o - BW/2 = 500 - 40/2

f_c,L = 480 KHz

f_c,h for Low pass filter:

f_c,h = f_o + BW/2 = 500 + 40/2

f_c,h = 520 KHz

- Now use the design formula for R-C circuit for each filter:

General design formula:

f_c = 1 /2*pi*R*C_i

C,h for High pass filter:

C_h = 1 /2*pi*R*f_c,L

C_h = 1 /2*pi*2000*480,000

C_h = 0.166 nF

C,L for Low pass filter:

C_L = 1 /2*pi*R*f_c,h

C_L = 1 /2*pi*2000*520,000

C_L = 0.153 nF

7 0

3 years ago