# Johnson-Trotter Algorithm

This is some of the first python code that I wrote. It's an implementation of the Johnson-Trotter permutations algorithm based on a Pascal program taken from the book Programming for Mathematicians by Raymond Seroul. I've made some improvements to my original code and adapted it to run on Python 2 and 3, but the basic approach is the same.

## Name rings a bell...

What wikipedia calls the Steinhaus-Johnson-Trotter Algorithm is a method of generating all permutations of a given sequence and was apparently known to 17th Century bell ringers! A feature of the algorithm is that each successive permutation differs from the previous permutation by a single transposition (one element is swapped with another), and these repeated transpositions follow a predictable pattern:

['A', 'B', 'C', 'D']     _ _ _ _
['A', 'B', 'D', 'C']     _ _ * *
['A', 'D', 'B', 'C']     _ * * _
['D', 'A', 'B', 'C']     * * _ _
['D', 'A', 'C', 'B']     _ _ * *
['A', 'D', 'C', 'B']     * * _ _
['A', 'C', 'D', 'B']     _ * * _
['A', 'C', 'B', 'D']     _ _ * *
['C', 'A', 'B', 'D']     * * _ _
['C', 'A', 'D', 'B']     _ _ * *
['C', 'D', 'A', 'B']     _ * * _
['D', 'C', 'A', 'B']     * * _ _
['D', 'C', 'B', 'A']     _ _ * *
['C', 'D', 'B', 'A']     * * _ _
['C', 'B', 'D', 'A']     _ * * _
['C', 'B', 'A', 'D']     _ _ * *
['B', 'C', 'A', 'D']     * * _ _
['B', 'C', 'D', 'A']     _ _ * *
['B', 'D', 'C', 'A']     _ * * _
['D', 'B', 'C', 'A']     * * _ _
['D', 'B', 'A', 'C']     _ _ * *
['B', 'D', 'A', 'C']     * * _ _
['B', 'A', 'D', 'C']     _ * * _
['B', 'A', 'C', 'D']     _ _ * *


So it's not hard to imagine how this pattern may have been discovered by Change Ringers ringing massive church bells in sequence:

The physical constraint of the mass of the bells means they can only be slightly delayed or advanced in the striking order, so they cannot be omitted from a sequence, and can only be made to change by one position in successive sequences.

—Wikipedia, Change Ringers

## The Algorithm

The algorithm itself is concerned with permutations of $$S(n)$$, the set of integers

\begin{equation*} 1, 2, 3, ..., n \end{equation*}

To fix ideas, let's say $$n = 4$$ and we want to generate all permutations of $${1, 2, 3, 4}$$.

Each element of the set is given a flag - 'left' or 'right' - which determines the direction that an element is "looking". You start off with the sequence:

['left',1], ['left',2], ['left',3], ['left',4]


So all elements are initially looking left. And if in the course of the algorithm you have the situation:

['left',3], ['right',2], ['right',1], ['left',4]


then you say:

• 1 sees 4
• 2 sees 1
• 3 sees nothing
• 4 sees 1

The direction flag is obviously a boolean quantity, but it simplifies calculations if it is represented by 1 or -1, rather than 0 or 1, because then you can determine which element another element "sees" by using element's index + element's flag.

An element is said to be mobile if it is looking at a smaller number. eg. in the sequence above, both 2 and 4 are mobile.

Then the algorithm is:

1. find the highest mobile; if none exists, stop
2. swap this mobile with the element it sees
3. reverse the direction flags of any element larger than the mobile
4. repeat

In coding the algorithm (following Seroul), sentinels with value $$n+1$$ are added at either end of the sequence, this means that any element which ends up at the beginning looking left, or at the end looking right, will always see a larger element and so will never be considered mobile. This removes the need to treat the left and rightmost elements as special cases in every loop.

## Python Implementation

A generator function.

def jpermute(iterable):
"""
Use the Johnson-Trotter algorithm to return all permutations of iterable.

The algorithm is applied to a 1-based set of integers representing the indices
of the given iterable, then a shallow copy of iterable is mutated and returned
for each successive permutation.
"""
# A shallow copy of 'iterable'. This is what is mutated and yielded for each perm.
sequence = list(iterable)
length = len(sequence)
indices = range(1, length+1)

# The list of directed integers: [-1, 1], [-1, 2], ...
state = [[-1, idx] for idx in indices]

# Add sentinels at the beginning and end
state = [[-1, length+1]] + state + [[-1, length+1]]

# The first permutation is the sequence itself
yield sequence

mobile_index = mobile_direction = direction = value = None
while True:
# 1. Find the highest mobile
mobile = -1
for idx in indices:
direction, value = state[idx]
if value > mobile and value > state[idx+direction][1]:
# value is mobile and greater than the previous mobile
mobile = value
mobile_index = idx
mobile_direction = direction
if mobile == length:
# no point in continuing as mobile is as large as it can be.
break
if mobile == -1:
break

# 2. Swap the mobile with the element it 'sees'
sees = mobile_index + mobile_direction
# ... first update the state
state[mobile_index], state[sees] = state[sees], state[mobile_index]
# ... then update the sequence
sequence[mobile_index-1], sequence[sees-1] = sequence[sees-1], sequence[mobile_index-1]

# 3. Switch the direction of elements greater than mobile
if mobile < length:
for idx in indices:
if state[idx][1] > mobile:
state[idx][0] = -state[idx][0]

yield sequence


### Notes

This is quicker than my original code but nowhere near competitive with the C code of the standard library's itertools.permutations.

$python2 -m timeit 'from jpermutation import jpermute;list(jpermute("ABC"))' 100000 loops, best of 3: 7.55 usec per loop$ python2 -m timeit 'from jpermutation import jpermute;list(jpermute("ABCD"))'
10000 loops, best of 3: 23.1 usec per loop
$python2 -m timeit 'from jpermutation import jpermute;list(jpermute("ABCDE"))' 10000 loops, best of 3: 108 usec per loop$ python2 -m timeit 'from jpermutation import jpermute;list(jpermute("ABCDEF"))'
1000 loops, best of 3: 658 usec per loop


Compare to:

$python2 -m timeit 'from itertools import permutations;list(permutations("ABC"))' 100000 loops, best of 3: 2.01 usec per loop$ python2 -m timeit 'from itertools import permutations;list(permutations("ABCD"))'
100000 loops, best of 3: 3.22 usec per loop
$python2 -m timeit 'from itertools import permutations;list(permutations("ABCDE"))' 100000 loops, best of 3: 8.88 usec per loop$ python2 -m timeit 'from itertools import permutations;list(permutations("ABCDEF"))'
10000 loops, best of 3: 44.9 usec per loop


The original code returned a new list for each permutation but there was a big speedup by returning the same mutated list each time.

There was also a minor speed improvement by writing:

direction, value = state[idx]


rather than the original:

direction = state[idx][0]
value = state[idx][1]