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FIT2004 Algorithms and data structures - S1 2026

FIT2004 Algorithms and data structures - S1 2026

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Pierre's microwave has keys to input reheating time, with possible positive times described by the set K. For instance, we could have $K=\{1,10,60\}$ K=\{1,10,60\}.

Pierre is wondering what the minimum number of keys to press to input the time t. Each key can be pressed multiple times.

For example, the minimum number of keys to press for a time t=25 is 7: pressing 10 twice and pressing 1 five times.

In order to solve this problem, we decompose the problem into subproblems defined as

minPresses[t]= {the minimum number of key presses required to input a time t using keys from K}.

Give a recurrence relation for minPresses by giving all cases in the expression

$minPresses[t] = \begin{cases} \text{first case}\\ \dots\\ \text{last case}, \end{cases}$

minPresses[t] = \begin{cases} \text{first case}\\ \dots\\ \text{last case},

\end{cases}

(Recall that a recurrence relation needs at least one base case and one general case.)

$minPresses[t] = 1 \text{ if } t=30$ minPresses[t] = 1 \text{ if } t=30

$minPresses[t] = 0 \text{ if } t\leq0$ minPresses[t] = 0 \text{ if } t\leq0

$minPresses[t] = \min_{k \in K}(minPresses[t-k]) +1\text{ otherwise.}$ minPresses[t] = \min_{k \in K}(minPresses[t-k]) +1\text{ otherwise.}

$minPresses[t] = 30 \text{ if } t=30$ minPresses[t] = 30 \text{ if } t=30

$minPresses[t] = \max_{k \in K}(minPresses[t-k]) +1\text{ otherwise.}$ minPresses[t] = \max_{k \in K}(minPresses[t-k]) +1\text{ otherwise.}

$minPresses[t] = \max_{k \in K}(minPresses[t-k] + 1) \text{ otherwise.}$ minPresses[t] = \max_{k \in K}(minPresses[t-k] + 1) \text{ otherwise.}

$minPresses[t] = 1 \text{ if } t=0$ minPresses[t] = 1 \text{ if } t=0

$minPresses[t] = \sum_{k\in K}(minPresses[t-k] ) \text{ otherwise.}$ minPresses[t] = \sum_{k\in K}(minPresses[t-k] ) \text{ otherwise.}

$minPresses[t] = \infty \text{ if } t < 0$ minPresses[t] = \infty \text{ if } t < 0

$minPresses[t] = \infty \text{ if } t \leq 0$ minPresses[t] = \infty \text{ if } t \leq 0

$minPresses[t] = 0 \text{ if } t=0$ minPresses[t] = 0 \text{ if } t=0

$minPresses[t] = \sum_{k \in K}(minPresses[t-k] ) + 1 \text{ otherwise.}$ minPresses[t] = \sum_{k \in K}(minPresses[t-k] ) + 1 \text{ otherwise.}

$minPresses[t] = \min_{k \in K}(minPresses[t-k]) \text{ otherwise.}$ minPresses[t] = \min_{k \in K}(minPresses[t-k]) \text{ otherwise.}

$minPresses[t] = 1 \text{ if } t\leq0$ minPresses[t] = 1 \text{ if } t\leq0

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Solve, in big-θ, the following recurrence relation

T(n) = 4 * T(n/4), where n >= 4
T(n) = c, where n = 1
for a constant c.

θ(log n)

θ(1)

θ(n^3)

θ(n!)

θ(n^2)

θ(n)

θ(2^n)

θ(n log n)

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Given the following pseudocode, derive the recurrence relation that represents its time complexity.

def fibonacci(n):
    if (n==0 or n==1):
        return n 
    return fibonacci(n - 1) + fibonacci(n - 2)

Let b and c represent constant values. What is the base case and recurrence step?

T(n) = T(n-1) + T(n-2) + c
T(1) = T(0) = b

100%

T(n) = T(n-2) + c
T(1) = T(0) = b

T(n) = T(n-1) + c
T(1) = T(0) = b

T(n) = T(n-1) + T(n-2) + c
T(0) = T(1) = n

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