Permutations of Binary Strings with Prime Length Subsequences

Given a binary string composed of 0s and 1s, the problem involves determining the number of permutations where every prime length contiguous subsequence contains a greater or equal number of 0s than 1s. This article will explore the underlying logic and provide a Java code solution that implements the required functionality.

Understanding the Problem

The challenge lies in ensuring that within the string, any prime length contiguous subsequence (for example, 111 and 101) has at least as many 0s as 1s. The key to solving this problem is realizing that the presence of 1s in the string must be separated by at least two 0s to meet the given condition.

Key Observation

The critical observation here is that the maximum number of 1s in the string can be determined by dividing the length of the string by 3. This is because, to ensure the condition is met, every 1 must be followed by at least two 0s. This division effectively maximizes the number of 1s while satisfying the given constraint.

Example and Pattern

A binary string can be structured in a pattern like 100100100100... to maximize the number of 1s. In a string of length 10, for instance, the string can be shaped as 1001001001, with a total of four 1s. This is determined by the formula:

Number of 1s floor(length of string / 3)

Recursive Function Implementation

The Java code provided below uses a recursive method to calculate the number of different combinations of such strings for a given length `n`:

public static void mainString args[]n{ nint n10sum0n//n is the length of the stringnforint l0lintMath.ceiln/3.0ln/l is the number of 1s that can be present in the string nSince there must be at least 2 0s between any two adjacent 1s the maximum number of ones will be when the string looks something like thisn1001001001001 etc..n/n sumfunnlsumn} //Computing the number of different combinations recursively public static int funint n int ln{ ifl0n return 1n else ifl1n return nn else ifl2n return n-2n-3/2n else n{n int sum0n/n-3l2 represents the number of positions that can be taken up by 1s. For e.g. - in a combination 10010000 the 2nd 1 can take any position occupied by the 0s at the end to give a new permutation which also satisfies the criterion/n forint i0in-3l2in sumfunn-3-il-1n return sumn}}

This recursive function `fun` calculates the number of valid permutations by dividing the possible positions into segments based on the number of 1s and then recursively finding the number of combinations for these segments.

Example Analysis

For a string of length 10:

Maximum number of 1s ceil(10 / 3) 4 The structure will be 1001001001, giving 4 ones.

The recursive function then calculates the valid permutations by considering all possible positions for the 1s within the 0s.

Conclusion

The problem of determining the number of permutations of binary strings with prime length subsequence constraints can be efficiently solved using a combination of logical observation and recursive computation. The provided Java code demonstrates an effective implementation of the solution, making full use of the critical property of separating 1s by at least two 0s.