# Lexicographic Permutations: Euler Problem 24

June 14, 2017
By

(This article was first published on The Devil is in the Data, and kindly contributed to R-bloggers)

Euler Problem 24 asks to develop lexicographic permutations which are ordered arrangements of objects in lexicographic order. Tushar Roy of Coding Made Simple has shared a great introduction on how to generate lexicographic permutations.

## Euler Problem 24 Definition

A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012 021 102 120 201 210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

## Brute Force Solution

The digits 0 to 9 have $10! = 3628800$ permutations (including combinations that start with 0). Most of these permutations are, however, not in lexicographic order. A brute-force way to solve the problem is to determine the next lexicographic permutation of a number string and repeat this one million times.

nextPerm <- function(a) {
# Find longest non-increasing suffix
i <- length(a) while (i > 1 && a[i - 1] >= a[i])
i <- i - 1
# i is the head index of the suffix
# Are we at the last permutation?
if (i <= 1) return (NA)
# a[i - 1] is the pivot
# Find rightmost element that exceeds the pivot
j <- length(a)
while (a[j] <= a[i - 1])
j <- j - 1
# Swap pivot with j
temp <- a[i - 1]
a[i - 1] <- a[j]
a[j] <- temp
# Reverse the suffix
a[i:length(a)] <- rev(a[i:length(a)])
return(a)
}

numbers <- 0:9
for (i in 1:(1E6 - 1)) numbers <- nextPerm(numbers)


This code takes the following steps:

1. Find largest index $i$ such that $a_{i-1} < a_i$.
1. If no such index exists, then this is already the last permutation.
2. Find largest index $j$ such that $j \geq i$ and
3. Swap $a_j$ and $a_{i-1}$.
4. Reverse the suffix starting at $a_i$.

## Combinatorics

A more efficient solution is to use combinatorics, thanks to MathBlog. The last nine digits can be ordered in $9! = 362880$ ways. So the first $9!$ permutations start with a 0. By extending this thought, it follows that the millionth permutation must start with a 2.

$\lfloor (1000000 - 1) / 9! \rfloor = 2$

From this rule, it follows that the 725761st permutation is 2013456789. We now need 274239 more lexicographic permutations:

$(1000000 - 1) - (2 \times 9!) = 274239$

We can repeat this logic to find the next digit. The last 8 digits can be ordered in 40320 ways. The second digit is the 6th digit in the remaining numbers, which is 7 (2013456789).

$\lfloor 274239 / 8! \rfloor = 6$

$274239 - (6 \times 7!) = 32319$

This process is repeated until all digits have been used.

numbers <- 0:9
n <- length(numbers)
remain <- 1E6 - 1
for (i in 1:n) {
j <- floor(remain / factorial(n - i))
remain <- remain %% factorial(n - i)
numbers <- numbers[-(j + 1)]
}


R blogger Tony’s Bubble Universe created a generalised function to solve this problem a few years ago.

The post Lexicographic Permutations: Euler Problem 24 appeared first on The Devil is in the Data.

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