## Todai Entrance Exam: Subject 2009 – Problem 1

(1) (2) (3) So: (4) Not sure if I’m doing this correctly…

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## Todai Entrance Exam: Subject 2009 – Problem 1

## Todai Entrance Exam: Subject 2010 – Problem 3

## Todai Entrance Exam: Subject 2010 – Problem 1

## Todai Entrance Exam: Subject 2011 – Problem 3

## Todai Entrance Exam: Subject 2011 – Problem 1

## Todai Entrance Exam: Subject 2013 – Problem 3

## Todai Entrance Exam: Subject 2012 – Problem 1

## Todai Entrance Exam: Subject 2013 – Problem 1

## Todai Entrance Exam: Subject 2014 – Problem 3

## Todai Entrance Exam: Subject 2014 – Problem 1

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39 Entries in: **Todai Entrance Exam**

(1) (2) (3) So: (4) Not sure if I’m doing this correctly…

(1) Skip (2) A worst-case example: (3) This way, the depth of the tree is decreased by half. Therefore, the worst-case time complexity is: (4) Reference: 1 & 2. We pick an edge and then we detect if nodes and are in the same set. If they are, then [...]

(1) (2) We see that: So: (3) (4) When the input signal is amplified into a wider range, its overall shape is less likely heavily distorted by the noise. It makes the input signal to be more tolerant against noise in communication. Reference: see slide 12-14.

(1) The variable is not resetted to when we start counting a new . Add right after the line to correct the code. (2) merge(L1, L2) { for (word1, freq1) in L1 { for (word2, freq2) in L2 { if (word1 is word2) { output_pair(word1, freq1 + freq2) break } } } } (3) We [...]

(1) There is a short-cut at the ammeter position (reference), so: (2) 2R (3) (reference). (4) (reference). (5) In Figure 1: In Figure 2: (6) An error on a resistor of the first circuit will affect the value of only one bit, the current value is guaranteed to be [...]

(1) Skip. (2) The adjacency matrix representation uses 36 memory units. The adjacency list representation uses 20 memory units. Therefore, the adjacency matrix representation uses more memory space than the list representation. (3) Let be the set of edges. The adjacency matrix representation uses memory units while the adjacency list representation uses memory units. So [...]

(1) I make a reference to the previous problem: I think using voltage divider would have been more straight-forward… (2) (3) I don’t understand this question. (4) From (2), together with (Laplace transform of unit step function): So:

(1) Reference: 1 & 2 & 3 There are 2 ways to attack this problem. For the first way, we attack the differential equation subject to . For the second way, we combine the series of capacitors and into one capacitor , and solve this circuit with the charge conservation equation (from the 3rd reference [...]

(1) func f(n): if n == 0: return 0 if n == 1: return 1 return f(n – 1) + f(n – 2) (2) func f(n): left = 1 right = 0 d = 1 if d > n: return right while d < n: temp = left left = left + right right = [...]

(1) (2) If the input impedance is not infinity then the input voltage will be decreased by a factor before getting amplified. If the input offset voltage is non-zero, then the input-output characteristics curve will also be offsetted. I don’t know about frequency response. (3) Check the last slide. (4) & (5) Skip [...]