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Impara Egyptian Fraction Problem | Greedy Algorithms: Overview and Examples
Greedy Algorithms using Python
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Greedy Algorithms using Python

Greedy Algorithms using Python

1. Greedy Algorithms: Overview and Examples
2. Greedy on Arrays
3. Greedy on Graphs

book
Egyptian Fraction Problem

Ancient Egyptians represented each positive fraction as the sum of unique unit fractions. For example, 7/15 = 1/3 + 1/8 + 1/120, or 2/3 = 1/2 + 1/6, or 1/7 = 1/7.

So, your goal is to find such a representation for the number n/m, m, n>0.

That can be reached by using the Greedy Approach. Each time, try to “bite” the number as big as possible to reduce the current value. Let’s look at the 7/15:

  • N = 7/15 >= 1/3 – this is the maximum unit fraction we can reach, add it to the answer.

  • Now, update the number we’re solving problem for: N = 7/15 – 1/3 = 2/15.

  • N = 2/15 >= 1/8 – next maximum unit fraction, add to the answer.

  • Update N: N = 2/15 – 1/8 = 1/120.

  • N = 1/120 >= 1/120 - add to the answer.

  • Update N = 0 -> stop the algorithm.

So, to sum up:

  1. Check if the current N == 0. If it is, stop the algorithm.

  2. Find the biggest unit fraction less than N and add it to the ans

  3. Update value of N by reducing.

The answer is an array f of numbers f[0], f[1], ... , f[t], where f[i] is a divider for fraction 1/f[i]. For our example, answer is [3, 8, 120].

How to find the biggest possible unit fraction It can be easily done for N = n/m by calculating k = math.ceil(m/n). Greater values of k do not give the maximum unit fraction (since, for example, 1/k > 1/(k+1)).

Compito

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Add some code to the function and test it.

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Sezione 1. Capitolo 5
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book
Egyptian Fraction Problem

Ancient Egyptians represented each positive fraction as the sum of unique unit fractions. For example, 7/15 = 1/3 + 1/8 + 1/120, or 2/3 = 1/2 + 1/6, or 1/7 = 1/7.

So, your goal is to find such a representation for the number n/m, m, n>0.

That can be reached by using the Greedy Approach. Each time, try to “bite” the number as big as possible to reduce the current value. Let’s look at the 7/15:

  • N = 7/15 >= 1/3 – this is the maximum unit fraction we can reach, add it to the answer.

  • Now, update the number we’re solving problem for: N = 7/15 – 1/3 = 2/15.

  • N = 2/15 >= 1/8 – next maximum unit fraction, add to the answer.

  • Update N: N = 2/15 – 1/8 = 1/120.

  • N = 1/120 >= 1/120 - add to the answer.

  • Update N = 0 -> stop the algorithm.

So, to sum up:

  1. Check if the current N == 0. If it is, stop the algorithm.

  2. Find the biggest unit fraction less than N and add it to the ans

  3. Update value of N by reducing.

The answer is an array f of numbers f[0], f[1], ... , f[t], where f[i] is a divider for fraction 1/f[i]. For our example, answer is [3, 8, 120].

How to find the biggest possible unit fraction It can be easily done for N = n/m by calculating k = math.ceil(m/n). Greater values of k do not give the maximum unit fraction (since, for example, 1/k > 1/(k+1)).

Compito

Swipe to start coding

Add some code to the function and test it.

Switch to desktopCambia al desktop per esercitarti nel mondo realeContinua da dove ti trovi utilizzando una delle opzioni seguenti
Tutto è chiaro?

Come possiamo migliorarlo?

Grazie per i tuoi commenti!

Sezione 1. Capitolo 5
Switch to desktopCambia al desktop per esercitarti nel mondo realeContinua da dove ti trovi utilizzando una delle opzioni seguenti
Siamo spiacenti che qualcosa sia andato storto. Cosa è successo?
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