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

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)).

Taak

Swipe to start coding

Add some code to the function and test it.

Switch to desktopSchakel over naar desktop voor praktijkervaringGa verder vanaf waar je bent met een van de onderstaande opties
Was alles duidelijk?

Hoe kunnen we het verbeteren?

Bedankt voor je feedback!

Sectie 1. Hoofdstuk 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)).

Taak

Swipe to start coding

Add some code to the function and test it.

Switch to desktopSchakel over naar desktop voor praktijkervaringGa verder vanaf waar je bent met een van de onderstaande opties
Was alles duidelijk?

Hoe kunnen we het verbeteren?

Bedankt voor je feedback!

Sectie 1. Hoofdstuk 5
Switch to desktopSchakel over naar desktop voor praktijkervaringGa verder vanaf waar je bent met een van de onderstaande opties
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