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Largest sum from absolute differences of N elements in an array

This isn't really a homework, but rather a practice and optimization, but this seemed like the best section for this type of questions.
It is a dynamical programming issue, and it's the following:
-Given an unsorted array of N elements, pick K number of elements from it, such that their absolute difference is the largest.
An absolute difference is calculated between adjacent elements here. So if we have an array of 5 elements: 1 5 3 2 1, and k = 3, the absolute differences would be:
1 5 3 = |5-1| + |3-5| = 6
1 5 2 = |5-1| + |2-5| = 7
1 5 1 = [5-1| + |1-5| = 8
etc
With 1 5 1 being the largest and needed one with 8
What i've tried so far is solving this by finding all possible combinations of K numbers with a recursion and then returning the biggest(brute force).
This showed as a terrible idea, because when tried with an array of N=50 and k=25 for example, there are 1.264106064E+14 combinations.
The recursion i used is a simple one used for printing all K-digit integers from an array, just instead of printing them, keeping them in an array:
static void solve(int[] numbers, int k, int startPosition, int[] result) {
if (k == 0) {
System.out.println(absoluteDifferenceSum(result));
return;
}
for (int i = startPosition; i <= numbers.length - k; i++) {
result[result.length - k] = numbers[i];
solve(numbers, k - 1, i + 1, result);
}
}
What I want to achieve is the optimal complexity (which i suppose can't be lower than O(n^2) here, but i'm out of ideas and don't know how to start. Any help is appreciated!

Generally, we can have a naive O(n^2 * k) formulation, where f(i, k) represents the best result for selecting k elements when the ith element is rightmost in the selection,
f(i, k) = max(
f(j, k - 1) + abs(Ai - Aj)
)
for all j < i
which we can expand to
max( f(j, k - 1) + Ai - Aj )
= Ai + max(f(j, k - 1) - Aj)
when Ai >= Aj
and
max( f(j, k - 1) + Aj - Ai )
= -Ai + max(f(j, k - 1) + Aj)
when Ai < Aj
Since the right summand is independent of Ai, we can build a tree with nodes that store both f(j, k - 1) - Aj, as well as f(j, k - 1) + Aj. Additionally, we'll store in each node both maxes for each subtree. We'll need O(k) trees. Let's skip to the examination of the tree for k = 2 when we reach the last element:
1
5 -> 4 -> (-1, 9)
3 -> 2 -> (-1, 5)
2 -> 3 -> (-1, 5)
3 -> (-3, 5)
tree for k = 2 so far:
3 (-1, 5)
/ \
2 (-1, 5) 5 (-1, 9)
1 is less than 3 so we first add -1
to the max for the right subtree
stored for f(j, k - 1) + Aj
-1 + 9 = 8
(note that this represents {1,5,1})
then we continue left in the tree
and compare 8 to a similar calculation
with node 2: -1 + 5 = 4
(note that this represents {5,2,1})
This way, we can reduce the time complexity to O(n log n * k) with space O(n * k).

Counting efficiently all the pairs with given sum

I came across the following question.
Given a list where each item represents the duration expressed in seconds of a song, return the total number of pairs of songs such their durations sum to minutes (e.g., 1m0s, 2m0s,..)
Example:
Input: [10,50,20,110,40]
Output: 3 (considering pairs at indexes (0,1),(0,3),(2,4))
I can only think to a brute force approach where I consider all pairs of songs. The time complexity of this approach is O(n^2).
Is there any better way of doing it?

The given problem can be reduced to the fact that we need to discover pairs (a,b) such that (a + b) mod 60 == 0 from the given list A.
Observation #1: For any integer x, (x mod 60) lies from o to 59.
Initialise an array of length 60 with default value set to 0 the index i of which will store the number of elements in the list A, such that x mod 60 = i for all x belonging to A
int freq[60] = {0};
for(int i = 0; i < A.size(); i++)
freq[(A[i] % 60)]++;
Now iterate over the array A again and for for every x, we need the count for the index 60 - (x mod 60) from our cumulative frequency map, which will corresponds to the number of elements it can form a pair with. The case where (x mod 60) == 30 would be a tricky one, which will require us to subtract 1 from the frequency count.
int ans = 0;
for(int i = 0; i < A.size(); i++) {
ans += freq[60 - (A[i] % 60)];
if(A[i] % 60 == 30) ans--;
}
The overall complexity of the solution is O(n).

Think on the lines of hashing, creating buckets and modulo division. All the possible minutes will go into one of 60 possible buckets. Then think how many possible combinations can there be when you choose two of second values from any two buckets. Then use nC2 to count. Here's my solution in Java.
public int numPairsDivisibleBy60(int[] time) {
int k = 60;
int[] mods = new int[k];
for (int i = 0; i < time.length; i++)
mods[time[i] % k]++;
// n(n-1)/2 pairs for multiples of k and numbers which leave remainder as half multiple of k
int count = ((mods[0] * (mods[0] - 1)) / 2) +
((mods[k / 2] * (mods[k / 2] - 1)) / 2);
for (int i = 1; i < k / 2; i++)
count += mods[i] * mods[k - i];
return count;
}

How to get the kth term in N Combination R [closed]

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How do I get the kth combination inNCR. without iterating through all possible outcomes. e.g. say I have 3C2 for 3 positions and 2identical-items. I am aware it's [011],[101] and [110]. how do I get e.g. the 2nd term(k=1) which is [101] using a method?
constraints(R < N k >= 0 and k < P where P = NCR).
NB:[101] is the 2nd term(in ascending/lexicographical order) because 011 = 3,101 = 5 ,110 = 6
in decimal. so basically the goal is to get what number k in NCR is,
because every kth output from NCR can be represented as a number.

Yes, you are correct when you say:
because every kth output from NCR can be represented as a number.
There is a bijection from the set of integers 1 to # of combs/perms to the entire set of combs/perms. Finding the specific index of a particular comb/perm is sometimes referred to as getting the rank. According to the example that you have in your question, these are ordinary permutations. Moreover when you mention ascending order, you are referring to the lexicographical order.
It is a straightforward exercise in counting to obtain the nth ordinary permutation of a given set. We first need to obtain the total number of permutations using the well established formula:
P(n, r) = n! / (n - r)!
This next part is the key observation that allows us to quickly obtain each element of our target permutation.
If we look at all permutations of our set of n choose r, there will be n groups that are only different by a permutation of the n elements.
For example, if we look at the first two group of the permutations of [0 1 2 3] choose 3, we have:
[,0] [,1] [,2]
[0,] 0 1 2
[1,] 0 1 3
[2,] 0 2 1
[3,] 0 2 3
[4,] 0 3 1
[5,] 0 3 2
[6,] 1 0 2
[7,] 1 0 3
[8,] 1 2 0
[9,] 1 2 3
[10,] 1 3 0
[11,] 1 3 2
Note that the last permutations are simply the first 6 permutations of the set [1 0 2 3].. that is, 0 is mapped to 1, 1 is mapped to 0, and the final 2 elements are mapped to themselves.
This pattern continues as we move to the right only instead of n identical groups, we will get n - 1 similar groups for the second column, n -2 for the third, and so on.
So to determine the first element of our permutation, we need to determine the 1st group. We do that by simply dividing the number of permutations by n. For our example above of permutations of 4 choose 3, if we were looking for the 15th permutation, we have the following for the first element:
Possible indices : [0 1 2 3]
P(4, 3) = 24
24 / 4 = 6 (elements per group)
15 / 6 = 2 (integer division) 2 means the 3rd element here (base zero)
Now that we have used the 3rd element, we need to remove it from our array of possible indices. How do we get the next element?
Easy, we get our next subindex by subtracting the product of the group we just found and the elements per group from our original index.
Possible indices : [0 1 3]
Next index is 15 - 6 * 2 = 3
Now, we just repeat this until we have filled all entries:
Possible indices : [0 1 3]
Second element
6 / 3 = 2 (elements per group)
3 / 2 = 1
Next index is 3 - 3 * 1 = 0
Possible indices : [0 3]
Third element
2 / 2 = 1
0 / 1 = 0
So our 15th element is : [2 1 0]
Here is a C++ implementation that should be pretty easy to translate to Java:
double NumPermsNoRep(int n, int k) {
double result = 1;
double i, m = n - k;
for (i = n; i > m; --i)
result *= i;
return result;
}
std::vector<int> nthPermutation(int n, int r, double myIndex) {
int j = 0, n1 = n;
double temp, index1 = myIndex;
std::vector<int> res(r);
temp = NumPermsNoRep(n, r);
std::vector<int> indexVec(n);
std::iota(indexVec.begin(), indexVec.end(), 0);
for (int k = 0; k < r; ++k, --n1) {
temp /= n1;
j = (int) std::trunc(index1 / temp);
res[k] = indexVec[j];
index1 -= (temp * (double) j);
indexVec.erase(indexVec.begin() + j);
}
}
These concepts extends to other types of combinatorial problems, such as finding the nth combination, or permutation with repetition, etc.

The time complexity is O(kn), space is O(n)
public static void main(String[] args) {
//n = 4, r = 2, k = 3
int[] ret1 = getKthPermutation(4, 2, 3);
//ret1 is [1,0,0,1]
//n = 3, r = 2, k = 1
int[] ret2 = getKthPermutation(3, 2, 1);
//ret2 is [1,0,1]
}
static int[] getKthPermutation(int n, int r, int k) {
int[] array = new int[n];
setLastN(array, r, 1);
int lastIndex = n - 1;
for(int count = 0; count < k; count++) {
int indexOfLastOne = findIndexOfLast(array, lastIndex, 1);
int indexOfLastZero = findIndexOfLast(array, indexOfLastOne, 0);
array[indexOfLastOne] = 0;
array[indexOfLastZero] = 1;
//shortcut: swap the part after indexOfLastZero to keep them sorted
int h = indexOfLastZero + 1;
int e = lastIndex;
while(h < e) {
int temp = array[h];
array[h] = array[e];
array[e] = temp;
h++;
e--;
}
}
return array;
}
//starting from `from`, and traveling the array forward, find the first `value` and return its index.
static int findIndexOfLast(int[] array, int from, int value) {
for(int i = from; i > -1; i--)
if(array[i] == value) return i;
return -1;
}
//set the last n elements of an array to `value`
static void setLastN(int[] array, int n, int value){
for(int i = 0, l = array.length - 1; i < n; i++)
array[l - i] = value;
}
This is an adaption of the very typical "find the kth permation" algorithm.
I will try to explain the general idea (yours is a special case as there are only two types of elements: 0 and 1).
Lets say I have [2,1,6,4,7,5]. What is the next smallest permutation that is bigger than the current one? Why do I concern the next smallest permutation bigger than current one? Because if you start with the smallest permutation [1,2,4,5,6,7] and you repeat the action (find the smallest bigger than current) k times, you will find k+1 th smallest permutation.
Now, since the one I am looking for needs to be bigger than current one, I need to increment the current one. To keep the incrementation as small as possible, I am going to try to modify 5 (last one). Now, I cannot just change 5 to a random value, I can only swap it with some digit before it.
If I swap 5 with a bigger number before it, say 7, then I will get [2,1,6,4,5,7], which is smaller than current one. Now obviously I need to swap 5 with some smaller digit before it, but which one? If I swap 5 with 2, I get [5,1,6,4,7,2], this increment is too big. I need to swap 5 with a "lower digit" to keep the increment as small as possible. Thats leads us to find the first(lowest) digit (from right to left) that is smaller than 5. In this case I would need to swap 5 with 4 and get [2,1,6,5,7,4]. This way, I can make the impact of "swap" small. Now the prefix is decided [2,1,6,5. There is no smaller prefix. We need to deal with suffix 7,4]. Clearly, if we sort the suffix and make it 4,7], then we are done.
In our case, there are two differences:
1. we need to swap the last 1, because you cannot make the permutation bigger by swapping the a zero with any digit before it.
2. we can always sort the suffix using a shortcut as shown in the code. I will leave it to you:)

public static String lexicographicPermutation(String str, long n) {
final long[] factorials = { 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600 };
n--;
char[] arr = str.toCharArray();
for (int i = 0; i < arr.length - 1; i++) {
long fact = factorials[arr.length - i - 2];
long p = i + n / fact;
n %= fact;
for (int j = i + 1; j <= p; j++)
swap(arr, i, j);
}
return new String(arr);
}
private static void swap(char[] arr, int i, int j) {
char tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
You can replace STR with required string. In the given example, 1st permutation is "abcdefghijklm" (this is a string with 13 chars), 13!st permutation is reverse string "mlkjihgfedcba" and 100st permutation is "abcfklgmeihjd".
To realise this soulution just google Factorial number system. This is a key to solve this problem. This is a Project Euler: Problem 24.
Demo:
for(int i = 1; i <= 6; i++)
System.out.println(lexicographicPermutation("110", i));
1 - 110
2 - 101
3 - 110
4 - 101
5 - 011
6 - 011
for(int i = 1; i <= 6; i++)
System.out.println(lexicographicPermutation("abc", i));
1 - abc
2 - acb
3 - bac
4 - bca
5 - cab
6 - cba

Finding a Prime Numbers

I have an array of length N=10^5 For each index 1<=i<=n I have to calculate the difference between A[j]-A[i] and (j-i) is prime and j>i
Here is my code:
for(int i=1;i<=n;i++){
for(int j=0;j<prime.size();j++){
int x = prime.get(j);
if(x+i>n) break;
ans+= A[x+i]-A[i];
}
}
How should i make this work even faster ? I think the time complexity is O(N*prime.size)

First, I will rephrase your question so that it states what I believe you want to achieve. You are probably looking for the sum of the differences of the form A[j]-A[i], where (j-i) is a "positive" prime and 1<=i<j<=N. With this statement in mind...
We count the number of times A[k] is added to the sum (denoted by p) and the number of times A[k] is subtracted from the sum (denoted by m). Well, m is equal to the number of primes in the interval [1,N-k], while p is equal to the number of primes in the interval [1,k-1]. If you don't believe me, simulate step-by-step what your code does. Then you can do:
S = 0
for k = 1 to N do
S = S + (p(k) - m(k)) * A[k]
endfor
Now, we need to find a way to determine p and m efficiently for each k in the interval [1,N]. I see you have already constructed what seems to be an ordered list of primes. So, to answer a query of the form 'how many primes in the interval [1,t]?' you could perform a binary search on that list for t. This would get the complexity down to O(N*log(prime.size)).
As an alternative, you can pre-compute the answers to the queries of the form 'how many primes in the interval [1,t]?'. You need an extra array nrPrimesLessThan of size N to keep the results, doing something like this to compute its values:
count = 0
for i = 1 to N do
if i < prime.get(count) then
nrPrimesLessThan[i] = count
else
count = count + 1
nrPrimesLessThan[i] = count
endif
endfor
The pre-computation part takes O(N) steps, but now one query takes O(1) steps, thus the calculating the sum takes O(N) steps. Overall, linear time in N.

Judging from your code example, you want to sum the differences of all value pairs in the array for which the difference of the indices is prime. You've got already an array of primes.
The diagram below shows how the elements get subtracted and added:
0 1 2 3 4 5 6 7 8 9
- + + + + 0
- + + + + 1
- + + + + 2
- + + + 3
- + + + 4
- + + 5
- + + 6
- + 7
- 8
- 9
A + means an element is added to the overall sum. A - means the element is subtracted from the sum. This is not a single subtraction; the subtraction happens for each addition to its left, so A[0] is subtracted 4 times. It is never added.
On the other hand, A[9] is never subtracted, but added four times. In general, each element is subtracted as many times as there are plusses in a row and it is added as many times as there are plusses in a columns. There is a symmetry here:
add[i] = sub[N - i - 1]
for zero-based indices. What is the value of add[i]? It is the number of primes that are smaller or equal to i.
Here's a code example where the add array is called m:
int psum2(const int A[], int n)
{
int sum = 0;
int m[n];
int j = 0;
int k = 0;
for (int i = 0; i < n; i++) {
if (i == prime[j]) {
k++;
j++;
}
m[i] = k;
}
for (int i = 0; i < n; i++) {
sum += (m[i] - m[n - i - 1]) * A[i];
}
return sum;
}
The array m is always the same and can be precalculated if you need to perform the sum more often. The algorithm is O(n).
The problem is also unrelated to primes at its core. The method above works for all conditional sums of differences where the difference of the indices must be conteined in a certain set of numbers.

In-place interleaving of the two halves of a string

Given a string of even size, say:
abcdef123456
How would I interleave the two halves, such that the same string would become this:
a1b2c3d4e5f6
I tried attempting to develop an algorithm, but couldn't. Would anybody give me some hints as to how to proceed? I need to do this without creating extra string variables or arrays. One or two variable is fine.
I just don't want a working code (or algorithm), I need to develop an algorithm and prove it correctness mathematically.

You may be able to do it in O(N*log(N)) time:
Want: abcdefgh12345678 -> a1b2c3d4e5f6g7h8
a b c d e f g h
1 2 3 4 5 6 7 8
4 1-sized swaps:
a 1 c 3 e 5 g 7
b 2 d 4 f 6 h 8
a1 c3 e5 g7
b2 d4 f6 h8
2 2-sized swaps:
a1 b2 e5 f6
c3 d4 g7 h8
a1b2 e5f6
c3d4 g7h8
1 4-sized swap:
a1b2 c3d4
e5f6 g7h8
a1b2c3d4
e5f6g7h8
Implementation in C:
#include <stdio.h>
#include <string.h>
void swap(void* pa, void* pb, size_t sz)
{
char *p1 = pa, *p2 = pb;
while (sz--)
{
char tmp = *p1;
*p1++ = *p2;
*p2++ = tmp;
}
}
void interleave(char* s, size_t len)
{
size_t start, step, i, j;
if (len <= 2)
return;
if (len & (len - 1))
return; // only power of 2 lengths are supported
for (start = 1, step = 2;
step < len;
start *= 2, step *= 2)
{
for (i = start, j = len / 2;
i < len / 2;
i += step, j += step)
{
swap(s + i,
s + j,
step / 2);
}
}
}
char testData[][64 + 1] =
{
{ "Aa" },
{ "ABab" },
{ "ABCDabcd" },
{ "ABCDEFGHabcdefgh" },
{ "ABCDEFGHIJKLMNOPabcdefghijklmnop" },
{ "ABCDEFGHIJKLMNOPQRSTUVWXYZ0<({[/abcdefghijklmnopqrstuvwxyz1>)}]\\" },
};
int main(void)
{
unsigned i;
for (i = 0; i < sizeof(testData) / sizeof(testData[0]); i++)
{
printf("%s -> ", testData[i]);
interleave(testData[i], strlen(testData[i]));
printf("%s\n", testData[i]);
}
return 0;
}
Output (ideone):
Aa -> Aa
ABab -> AaBb
ABCDabcd -> AaBbCcDd
ABCDEFGHabcdefgh -> AaBbCcDdEeFfGgHh
ABCDEFGHIJKLMNOPabcdefghijklmnop -> AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPp
ABCDEFGHIJKLMNOPQRSTUVWXYZ0<({[/abcdefghijklmnopqrstuvwxyz1>)}]\ -> AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz01<>(){}[]/\

Generically that problem is quite hard -- and it reduces to finding permutation cycles. The number and length of those varies quite a lot depending on the length.
The first and last cycles are always degenerate; the 10 entry array has 2 cycles of lengths 6 and 2 and the 12 entry array has a single cycle of length 10.
Withing a cycle one does:
for (i=j; next=get_next(i) != j; i=next) swap(i,next);
Even though the function next can be implemented as some relatively easy formula of N, the problem is postponed to do book accounting of what indices have been swapped. In the left case of 10 entries, one should [quickly] find the starting positions of the cycles (they are e.g. 1 and 3).

Ok lets start over. Here is what we are going to do:
def interleave(string):
i = (len(string)/2) - 1
j = i+1
while(i > 0):
k = i
while(k < j):
tmp = string[k]
string[k] = string[k+1]
string[k+1] = tmp
k+=2 #increment by 2 since were swapping every OTHER character
i-=1 #move lower bound by one
j+=1 #move upper bound by one
Here is an example of what the program is going to do. We are going to use variables i,j,k. i and j will be the lower and upper bounds respectively, where k is going to be the index at which we swap.
Example
`abcd1234`
i = 3 //got this from (length(string)/2) -1
j = 4 //this is really i+1 to begin with
k = 3 //k always starts off reset to whatever i is
swap d and 1
increment k by 2 (k = 3 + 2 = 5), since k > j we stop swapping
result `abc1d234` after the first swap
i = 3 - 1 //decrement i
j = 4 + 1 //increment j
k= 2 //reset k to i
swap c and 1, increment k (k = 2 + 2 = 4), we can swap again since k < j
swap d and 2, increment k (k = 4 + 2 = 6), k > j so we stop
//notice at EACH SWAP, the swap is occurring at index `k` and `k+1`
result `ab1c2d34`
i = 2 - 1
j = 5 + 1
k = 1
swap b and 1, increment k (k = 1 + 2 = 3), k < j so continue
swap c and 2, increment k (k = 3 + 2 = 5), k < j so continue
swap d and 3, increment k (k = 5 + 2 = 7), k > j so were done
result `a1b2c3d4`
As for proving program correctness, see this link. It explains how to prove this is correct by means of a loop invariant.
A rough proof would be the following:
Initialization: Prior to the first iteration of the loop we can see that i is set to
(length(string)/2) - 1. We can see that i <= length(string) before we enter the loop.
Maintenance. After each iteration, i is decremented (i = i-1, i=i-2,...) and there must be a point at which i<length(string).
Termination: Since i is a decreasing sequence of positive integers, the loop invariant i > 0 will eventually equate to false and the loop will exit.

The solution is here J. Ellis and M. Markov. In-situ, stable merging by way of perfect shuffle.
The Computer Journal. 43(1):40-53, (2000).
Also see the various discussions here:
https://cs.stackexchange.com/questions/332/in-place-algorithm-for-interleaving-an-array/400#400
https://cstheory.stackexchange.com/questions/13943/linear-time-in-place-riffle-shuffle-algorithm.

Alright, here's a rough draft. You say you don't just want an algorithm, but you are taking hints, so consider this algorithm a hint:
Length is N.
k = N/2 - 1.
1) Start in the middle, and shift (by successive swapping of neighboring pair elements) the element at position N/2 k places to the left (1st time: '1' goes to position 1).
2) --k. Is k==0? Quit.
3) Shift (by swapping) the element at N/2 (1st time:'f' goes to position N-1) k places to the right.
4) --k.
Edit: The above algorithm is correct, as the code below shows. Actually proving that it's correct is waaay beyond my capabilities, fun little question though.
#include <iostream>
#include <algorithm>
int main(void)
{
std::string s("abcdefghij1234567890");
int N = s.size();
int k = N/2 - 1;
while (true)
{
for (int j=0; j<k; ++j)
{
int i = N/2 - j;
std::swap(s[i], s[i-1]);
}
--k;
if (k==0) break;
for (int j=0; j<k; ++j)
{
int i = N/2 + j;
std::swap(s[i], s[i+1]);
}
--k;
}
std::cout << s << std::endl;
return 0;
}

Here's an algorithm and working code. It is in place, O(N), and conceptually simple.
Walk through the first half of the array, swapping items into place.
Items that started in the left half will be swapped to the right
before we need them, so we use a trick to determine where they
were swapped to.
When we get to the midpoint, unscramble the unplaced left items that were swapped to the right.
A variation of the same trick is used to find the correct order for unscrambling.
Repeat for the remaining half array.
This goes through the array making no more than N+N/2 swaps, and requires no temporary storage.
The trick is to find the index of the swapped items. Left items are swapped into a swap space vacated by the Right items as they are placed. The swap space grows by the following sequence:
Add an item to the end(into the space vacated by a Right Item)
Swap an item with the oldest existing (Left) item.
Adding items 1..N in order gives:
1 2 23 43 435 465 4657 ...
The index changed at each step is:
0 0 1 0 2 1 3 ...
This sequence is exactly OEIS A025480, and can be calculated in O(1) amortized time:
def next_index(n):
while n&1: n=n>>1
return n>>1
Once we get to the midpoint after swapping N items, we need to unscramble. The swap space will contain N/2 items where the actual index of the item that should be at offset i is given by next_index(N/2+i). We can advance through the swaps space, putting items back in place. The only complication is that as we advance, we may eventually find a source index that is left of the target index, and therefore has already been swapped somewhere else. But we can find out where it is by doing the previous index look up again.
def unscramble(start,i):
j = next_index(start+i)
while j<i: j = next_index(start+j)
return j
Note that this only an indexing calculation, not data movement. In practice, the total number of calls to next_index is < 3N for all N.
That's all we need for the complete implementation:
def interleave(a, idx=0):
if (len(a)<2): return
midpt = len(a)//2
# the following line makes this an out-shuffle.
# add a `not` to make an in-shuffle
base = 1 if idx&1==0 else 0
for i in range(base,midpt):
j=next_index(i-base)
swap(a,i,midpt+j)
for i in range(larger_half(midpt)-1):
j = unscramble( (midpt-base)//2, i);
if (i!=j):
swap(a, midpt+i, midpt+j)
interleave(a[midpt:], idx+midpt)
The tail-recursion at the end can easily be replaced by a loop. It's just less elegant with Python's array syntax. Also note that for this recursive version, the input must be a numpy array instead of a python list, because standard list slicing creates copies of the indexes that are not propagated back up.
Here's a quick test to verify correctness. (8 perfect shuffles of a 52 card deck restore it to the original order).
A = numpy.arange(52)
B = A.copy()
C =numpy.empty(52)
for _ in range(8):
#manual interleave
C[0::2]=numpy.array(A[:26])
C[1::2]=numpy.array(A[26:])
#our interleave
interleave(A)
print(A)
assert(numpy.array_equal(A,C))
assert(numpy.array_equal(A, B))