I happened to write an answer on Quora yesterday for 'How to list all the subsets of a set where elements need not be necessarily unique'. http://qr.ae/fwOCc
Though several tutorials already exist, here is my 'simplified' version solving the above problem using bitmasking because personally, I was really confused when i learnt this first. Also, this is my first attempt at writing something. Please feel free to comment/criticize and please upvote if you liked the explanation :)
More than often, problems where you feel the answer can be found after brute forcing through all the subsets, have smarter and more efficient solutions using Dynamic Programming. Have a look at an Introduction to the Knapsack Problem and Dynamic Programming: http://www.cs.rit.edu/~zjb/courses/800/lec7.pdf
That aside, if n is reasonably small, you CAN use BruteForce and list down all the subsets in the process. As mentioned, we will use the technique of BitMasking.
Alright, so lets start by trying to solve the above problem and we'll learn the concept through that. The question essentially boils down to Finding the Power Set of a given Set. The set here may contain multiple elements. The term 'Multiset' is more appropriate...but well, lets just stick to set.
First of all, we Claim that :
"If we list down all the binary numbers from 0 to (2^n)-1 , we get ALL the possible combinations of selecting n items"
Lets verify for n=3
000: None of the values in the set chosen
001: 1st chosen, 2nd and 3rd items left out
010: 2nd chosen, 1st and 3rd items left out
011: 1st and 2rd item chosen, 3rd one left out.
111: All 3 items chosen
This way, we have listed the 2^n ways of obtaining all the subsets from a set of n numbers. Unique or not, does not matter because the index of every element we are dealing with is unique.
Now for the computation part, the core idea is to brute force through every bit of every number from 0-2^n-1 and check for the set bits of each number.
- Run a loop for 'i' for all numbers from 0 to 2^(n-1).
- When inside this loop, run a loop for 'j' from 0 to n-1 inclusive
- Inside this loop, check if the 'j'th bit is SET(equal to 1) for the number 'i'.
- If it is, then we include this element in our 'i'th subset
I was very confused when I learnt this the first time, so let me demonstrate with a small example: say for i=3, the binary representation is 011. When we run the inner loop from 1 to n for i which is currently 3, here's what we are actially doing.
- Is the right most (LSB) bit set? Yes. So Include it. [01x]
- Is the middle bit set? Yes. So Include it. [0x1]
- Is the last (MSB) bit set? No. So Leave it out. [x11]
What we are essentially doing is that For every iteration of 'j', we are 'masking' all the bits in 'i' except for the bit at the 'j'th position. Hence the name BitMasking.
We use the BitWise '<<' operator for shifting the bit to be checked each time in j. (We are taking the BitWise '&' operation of 1 and x. If it is 1, it is SET since in boolean algebra, only 1.1=1)
Here's a code snippet in Python.
n=eval(input("Enter n: ")) # keep sub 20-ish max for i in range(0,(2**n)):# loop from 0 to (2^n)-1 cursub="Current Subset Contains Elements: " for j in range(0,n): if((1<<j) & i >0): #Checking if jth bit in i is set cursub+=(str(j+1)+" ") print (cursub) print ("All ", (2**n)," Subsets printed")
Obviously, what the point of just listing down the subset? You would want to perform certain operations on say the selected bits/elements of that given set. For this, naturally, we maintain an arraay of n elements. So, for example, your task it to find the product of every element of each subset. In this case we maintain a number temp=1 before every ith iteration. Whenever you encounter a set bit in a number, multiply the element at that number, e.i array[i] with temp. At the end of every iteration, you get your desired result.
Code in action: http://ideone.com/8vfz4U
Problems you can try:
run your script from root and enter
Thank you very much,this is the best explain about Bit-masking I've read :)
"If we list down all the binary numbers from 0 to 2^(n-1) , we get ALL the possible combinations of selecting n items"
Do you mean (2^n) -1?
Is there any DP way of solving http://codeforces.com/problemset/problem/550/B ? I don't know of bitmasks but after reading this post I am feeling that both bitmasks and DP are similar to an extent.
In this problem you can just bruteforce all subsets.