Notice: This page requires JavaScript to function properly.
Please enable JavaScript in your browser settings or update your browser.
Egyptian Fraction Problem | Greedy Algorithms: Overview and Examples
Greedy Algorithms using Python
course content

Conteúdo do Curso

Greedy Algorithms using Python

Greedy Algorithms using Python

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

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)).

Tarefa

Add some code to the function and test it.

Tarefa

Add some code to the function and test it.

Mude para o desktop para praticar no mundo realContinue de onde você está usando uma das opções abaixo

Tudo estava claro?

Seção 1. Capítulo 5
toggle bottom row

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)).

Tarefa

Add some code to the function and test it.

Tarefa

Add some code to the function and test it.

Mude para o desktop para praticar no mundo realContinue de onde você está usando uma das opções abaixo

Tudo estava claro?

Seção 1. Capítulo 5
toggle bottom row

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)).

Tarefa

Add some code to the function and test it.

Tarefa

Add some code to the function and test it.

Mude para o desktop para praticar no mundo realContinue de onde você está usando uma das opções abaixo

Tudo estava claro?

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)).

Tarefa

Add some code to the function and test it.

Mude para o desktop para praticar no mundo realContinue de onde você está usando uma das opções abaixo
Seção 1. Capítulo 5
Mude para o desktop para praticar no mundo realContinue de onde você está usando uma das opções abaixo
We're sorry to hear that something went wrong. What happened?
some-alt