## Introduction

In this post, I am going to share my little knowledge on how to solve some problems involving calculation of **Sum over Subsets(SOS)** using dynamic programming. Thus the name **SOS DP**. I have chosen this topic because it appears frequently in contests as mediu2m-hard and above problems but has very few blogs/editorials explaining the interesting DP behind it. I also have a predilection for this since I came across it for the first time in ICPC Amritapuri Regionals 2014. Since then I have created many questions based on this concept on various platforms but the number of accepted solutions always seems to be disproportionate to the lucidity of the concept. Following is a small attempt to bridge this gap 😉

## Problem

I will be addressing the following problem: Given a fixed array **A** of 2^{N} integers, we need to calculate ∀ x function **F(x)** = Sum of all **A[i]** such that **x&i = i**, i.e., **i** is a subset of **x**.

## Prerequisite

- Basic Dynamic Programming
- Bitmasks

In no way this should be considered an introduction to the above topics.

## Solutions

#### Bruteforce

```
for(int mask = 0;mask < (1<<N); ++mask){
for(int i = 0;i < (1<<N); ++i){
if((mask&i) == i){
F[mask] += A[i];
}
}
}
```

This solution is quite straightforward and inefficient with time complexity of *O*(4^{N})

#### Suboptimal Solution

```
// iterate over all the masks
for (int mask = 0; mask < (1<<n); mask++){
F[mask] = A[0];
// iterate over all the subsets of the mask
for(int i = mask; i > 0; i = (i-1) & mask){
F[mask] += A[i];
}
}
```

Not as trivial, this solution is more efficient with time complexity of *O*(3^{N}). To calculate the time complexity of this algorithm, notice that for each mask we iterate only over its subsets. Therefore if a mask has *K* on bits, we do 2^{K} iterations. Also total number of masks with *K* *on* bits is . Therefore total iterations =

#### SoS Dynamic Programming solution

In this approach we will try to iterate over all subsets of mask in a smarter way. A noticeable flaw in our previous approach is that an index **A[x]** with **x** having **K** *off* bits is visited by 2^{K} **masks**. Thus there is repeated recalculation.

A reason for this overhead is that we are not establishing any relation between the **A[x]'s** that are being used by different **F[mask]'s**. We must somehow add another state to these masks and make semantic groups to avoid recalculation of the group.

Denote . Now we will partition this set into non intersecting groups. , that is set of only those subsets of **mask** which *differ* from **mask** only in the first **i** bits (zero based).

For example . Using this we can denote any set as a union of some non intersecting sets.

Lets try to relate these sets of numbers. S(mask, i) contains all subsets of *mask* which differ from it only in the first **i** bits.

Consider that **i ^{th}** bit of mask is

**0**. In this case no subset can differ from mask in the

**i**bit as it would mean that the numbers will have a

^{th}**1**at

**i**bit where

^{th}**mask**has a

**0**which would mean that it is not a subset of

**mask**. Thus the numbers in this set can now only differ in the first

**i-1**bits. S(mask,i) = S(mask, i-1).

Consider that

**i**bit of mask is

^{th}**1**. Now the numbers belonging to S(mask, i) can be divided into two non intersecting sets. One containing numbers with

**i**bit as

^{th}**1**and differing from

*mask*in the next

**i-1**bits. Second containing numbers with

**i**bit as

^{th}**0**and differing from

*mask⊕2*in next

^{i}**i-1**bits. S(mask, i) = S(mask, i-1) ∪ S(mask⊕2

^{i}, i-1).

The following diagram depicts how we can relate the **S(mask,i)** sets on each other. Elements of any set **S(mask,i)** are the **leaves** in its subtree. The red prefixes depicts that this part of mask will be common to all its members/children while the black part of mask is allowed to differ.

Kindly note that these relations form a directed acyclic graph and not necessarily a rooted tree (think about different values of **mask** and same value of **i**)

After realization of these relations we can easily come up with the corresponding dynamic programming.

```
//iterative version
for(int mask = 0; mask < (1<<N); ++mask){
dp[mask][-1] = A[mask]; //handle base case separately (leaf states)
for(int i = 0;i < N; ++i){
if(mask & (1<<i))
dp[mask][i] = dp[mask][i-1] + dp[mask^(1<<i)][i-1];
else
dp[mask][i] = dp[mask][i-1];
}
F[mask] = dp[mask][N-1];
}
```

```
//memory optimized, super easy to code.
for(int i = 0; i<(1<<N); ++i)
F[i] = A[i];
for(int i = 0;i < N; ++i) for(int mask = 0; mask < (1<<N); ++mask){
if(mask & (1<<i))
F[mask] += F[mask^(1<<i)];
}
```

The above algorithm runs in *O*(*N* 2^{N}) time.

## Discussion Problem

Now you know how to calculate Sum over Subsets for a **fixed** array **A**. What would happen if **A** and **F** are SOS functions of each other 😉 . Consider following modification to the problem. Assume H1, H2 to be 32 bit integer valued hash functions (just to avoid any combinatoric approach to circumvent this problem) and can be evaluated at any point in constant time.:

I enjoyed solving this with harshil. Lets discuss the approaches in comments :)

## Practice Problems

I hope you enjoyed it. Following are some problems built on SOS.

- Special Pairs
- Compatible Numbers
- Vowels
- Covering Sets
- COCI 2011/2012 Problem KOSARE
- Vim War
- Jzzhu and Numbers
- Subset
- Jersey Number
- Beautiful Sandwich
- Pepsi Cola(resembles above
*discussion problem*). Need to join this group. - Uchiha and Two Products(resembles above
*discussion problem*) - Strange Functions(Same as above
*discussion problem*) - Varying Kibibits

**EDIT**: Practice problems are now arranged in almost increasing order of difficulty.

Great tutorial! If only I knew about this before today's contest :P

Thankyou. Did a similar problem appear in yesterday's contest ?

363 Div 1 C. You can add it to the list.

Well I don't think this technique is required to solve this problem. The dp recurrence does not demand any summation over subsets.DP for solving this problem would be

where

Good job!

You can also add this problem to the list: http://hsin.hr/coci/archive/2011_2012/contest6_tasks.pdf (problem KOSARE).

Thankyou. Added :)

Some more problems that use a similar approach: http://codeforces.com/contest/165/problem/E https://www.hackerrank.com/contests/countercode/challenges/subset

Thanks. Added :)

Very well written blog.

P.S:The spelling of prerequisites is wrong

Thanks. Fixed.

In suboptimal solution: What is mask initialized to ? It is this line F[mask] = A[0]

Thanks. Fixed now.

in suboptimal solution -> first for : 'i' should be 'mask'! :)

and what's value of 'i' in this line ?! dp[mask][-1] = A[i]; //handle base case separately (leaf states)

Fixed. it should have been A[mask] not A[i].

Great tutorial. I find bitmask concepts hard to undestand. But got a clear understanding with this one. Kudos to the author. :)

I think value of 'i' in 2nd last row of diagram should be zero in all case.

thanks great.

sorry, how we can prove that

`for(int i = mask; i > 0; i = (i-1) & mask)`

will pass over all subsets ?I will give it a try. As of now I am not able to think about an easier proof. I will try to prove it by mathematical induction.

Note:OperationM-1switches OFF the first ON bit and switches ON the remaining prefix. Eg 10110000_{2}- 1 = 10101111_{2}Statement P(n)= Given an integerM, this algorithm will iterate over all subsets s.t.xdiffers fromMonly in the firstnbits in strictly decreasing order. algorithm successfully iterates over all elements inS(M,n) instrictly decreasing order.Base Case P(0):Case 1: if

MisevenThe first iteration i =Msuccessfully visits S(M,0) = {M}Case 2: if

MisoddFirst iteration i =M, second iterationi = (i-1)&Mswitches off the 0^{th}bit thus visits . Algo visits in decreasing orderHence

P(0)is true.Assumption: AssumeP(n)to be true. Algo visits S(M, n) in descending order.Inductive step:To proveP(n+1)is true. SinceP(n)is true, algo visits all S(M, n) in descending order.P(n+1)is trivially true ifn+ 1^{th}bit of M isOFFsince S(M,n+1) = S(M,n).Lets focus on case when

n+ 1^{th}bit of M isON. Since the visits of S(M,n) as assumed byP(n)are in descending order, the last integer visited by this algo would beMwith first n bits OFF. For example, if M = , n = 4 the last value of i would be .After reaching this integer, we do

i = (i-1)&M. Thei-1operation turnsOFFthen+ 1^{th}bit and turnsONfirstnbits. Taking bitwiseANDwith original M copies the firstnbits ofMinto it.Taking the example above, we have following transitions -> -> ->.

Thus what we final get is .

Since

P(n)is true, we can now iterate over S(, n). But . Therefore we iterate over all elements of S(M,n+1).Hence

P(n+1)is true.The example in 2nd last line of the 2nd paragraph under heading "SoS Dynamic Programming solution" doesn't makes sense to me.

This guy 101 0000 (last element in the set) when XORed with 101 1010 will produce 1010 which is no way <= (1<<3).

Also the statement that "set of only those subsets of mask which differ from mask only in the first i bits (zero based)" conflicts with x XOR mask being <= (1<<i). It should have been x XOR mask < (1<<i). I am assuming the numbering of positions from Right to Left.

UPD: I Figured the word it all starts from 0 out !!I have the same doubt! Is it a typo or am I missing something?

UPD: I think it should be2^(i+1)-1.Can we just say the following invariant:

let say j is the jth iteration then i is the jth largest value which is subset of mask

That is a great post! It really helped, thank you !!! I tryied the first problem (Special Pairs) source: http://pastebin.com/UXDiad27 But I get WA. My logic is the following: If we find 1 then because we need final result to be zero and we use AND bitwise, then we need to compare it with a number that has 0 at that position. So we go to dp[mask^(1<<i)][i-1] If we find 0 then we can have at that position 0 and 1 as well. So we are seeking in dp[mask][i-1] + dp[mask^(1<<i)][i-1]

Is the logic wrong, or just my source code ? Thanks in advance !

Got the bug.

Why man why yy yy ? And I was examining your recurrence all this time :P .It is nowhere written

i!=j.You can have a look at the diff. The correct answer was always just one less than the correct answer :P

Anyways have fun now :)

Yes! Removing that code gives AC! I thought i != j. I am sorry for your suffering checking my recurrence ! Thank you !

An easier way to solve this problem: for any

A[i], to find how many numbers are there which has bitwise AND zero withA[i]-> it would just be a subset ofone's complement of A[i]. So the answer isAnyways your modified recurrence shows your proper understanding of the concept :)

Nice way! Also even if I am not so strong in bitmask I managed to think a modification because your tutorial is so clear and good that made things easier ! Thanks again !

I'm unable to understand why 2^N integers must be there in A. Is this a typo?

Actually the technique was designed for 2

^{N}integers. But you can always make an arbitrary length of array to a 2^{N}sized array by adding extra elements with value "0".Shouldn't it be that the number of bitmasks=2^SIZE for any integer SIZE. If SIZE itself is 2^N, then overall complexity will be SIZE*2^SIZE = 2^N * 2^( 2^N ).

N = minimum number of bits required to represent any index of the array.

SIZE< 2^{N}Now it makes sense

great tutorial

In your memory optimized code shouldn't it be

mask is always greater than mask^(1<<i) if i'th bit is set. To calculate for i'th bit we need the value for (i-1)'th bit. In your code F[mask^(1<<i)] actually has the value for i'th bit because it was calculated for i'th bit before mask. If we start from (1<<N)-1 this won't be a problem because F[(mask^(1<<i))] will be calculated for i'th bit after the calculation of F[mask]. Please correct me if I'm wrong. Thanks :)

UPD:I just realized that for F[mask^(1<<i)] actually the calculated value for i'th bit and (i-1)'th bit is same because i'th bit is not set. So your code iscorrect. Sorry for ignoring it.Thanks for informing was stuck on that !

Hello ,

Can you explain the formula

used in Vim War.Seems like inclusion exclusion to me but I cannot understand it.

http://codeforces.com/contest/165/problem/E Can someone explain it ? please , can't understand dp-state.

First solve this problem — Special Pairs. Then this problem should be easy.

However, in my solution dp state was

dp[mask] = a number from the given array such that . So we have base cases , for other masks initialize with 0.Now, for each array elements you need to find out another number from the array, such that their

ANDis 0. Note that the number you need to find will always be a subset of . So you can just print the number stored at . If it is 0 then there is no solution. [N= minimum number of bits needed to represent all the numbers of the array]My idea was almost the same. But it is getting TLE on test 12. Here's the link to my submission http://codeforces.com/contest/165/submission/29478386

Can you please check and tell me what's wrong with it?

Change cin/cout to scanf/printf! The 12th Case has 1e6 numbers, cin/cout will definitely make it TLE.

Always use scanf/printf if you have some constrains > 5e5.

Thanks a lot. I never thought that it would make much of a mess..:) I got it accepted after using scanf and printf. Thanks a lot of pointing this out.

Added a nice problem to the set. Varying Kibibits

dp[mask][-1] = A[mask]; //handle base case separately (leaf states) .

why it doesn't give array index out of bound exception

Because this is pseudocode.

Well technically, c++ doesn't check for boundaries so it should work...

But, you know, unexpected behaviour.

It's a new problem in town guys. The problem is the editorial is not explained very well ,, i guess you guys should take a look and if anyone understands may be he can shed some light for us newbies.

https://www.hackerrank.com/contests/world-codesprint-11/challenges/best-mask/editorial

It's a recent codesprint problem from hackerrank.

hey can anyone explain how to approach question this. it is similar to discuss problem but i am unable to do it? thanks in advance,

In the question explained. What is the range of x ?? And the size of array is N or 2^N ???

I can't understand this completely, maybe this is too hard for me.

What does

`F`

will contain? Does accumulate of`F`

will give the sum over all subsets?I want to ask that... What is the meaning of

`F[mask]`

in the last implementation?If I need to find SoS then how should I proceed after calculation of

`F`

?`F`

won't give you sum over all subsetsof the array.F[mask] is the sum ofA[i] such that`mask & i == i`

, that mean the on bits of`i`

is subset of the on bits of`mask`

.Absolutely Perfect.

Absolutely Pointless. :)

Absolutely perfectly pointless

1)

2)

What is difference between above 2 codes? Do both codes give same result?

is equivalent to first version and not the second version. The ith iteration of 1) has F[mask] = dp[mask][i]. This is not true in 2).

So 2nd version will not count some dp states. Right?

It will visit every state but t will calculate it wrong. When you do

`F[mask] += F[mask^(1<<i)]`

. In the ith iteration for`F[mask]`

you actually add`dp[mask^(1<<i)][N-1]`

instead of`dp[mask^(1<<i)][i-1]`

. So all values are wrong.This reminds me of this: http://web.evanchen.cc/handouts/SOS_Dumbass/SOS_Dumbass.pdf

`for ( x = y; x > 0; x = ( y & (x-1) ) )`

generates all subsets of bitmask y.

How does this iteration works? Any intuitive explaination?

starts with the valuexy, and everytime you subtract 1 fromx, the lowest value bit that is set would become unset, and the bits ahead of the lowest set bit (now unset) would become set.`Eg: x = y = 10100`

`x-1 = 10011`

Now when you bitwise

ANDthis withy(which was initiallyx), you get the common set bits betweenxandy( definition of bitwiseAND).`Eg: x=10011 and y=10100`

`x&y = 10000`

Everytime you

ANDxwithy, it is making sure thatxis always a subset ofy.Subtracting

xby 1 after every iteration makes sure that you go through all combinations ( 2^{N}) of the mask y.Further proof for the same is given by usaxena95 above.

MAXOR

MONSTER

I see it is not clear with the example above. S(1011010, 3) contains 1010000.

Let take the XOR operator: 1011010 xor 101000 = 0001010. In decimal representation, this value should be equal to 2^3 + 2^1 > 2^3, contradicting to (1011010 xor 101000)_{10} <= 2^3.

I may misunderstand. Can someone help me explain this gap?

Thanks,

Hanh Tang

Thanks for pointing that out. I have fixed that. It is now < 2

^{i + 1}.Thank you, usaxena95.

it should be 1011010 xor 1010000 = 0001010

`In this approach we will try to iterate over all subsets of mask in a smarter way. A noticeable flaw in our previous approach is that an index A[x] with x having K off bits is visited by 2^K masks. Thus there is repeated recalculation`

Can someone explain me this line?

The mask: 4(100) has 2 off-bits, so 2^2=4 masks will visit the mask 4(100). How?

An index x with k off bits is a subset to 2^k masks (hence it is visited by 2^k masks).

In your example case 4(100) is visited by 100,101,110,111.

Thanks

Thanks really helped alot.

Excellent editorial! Kudos...

Approach for discussion problem?

this also uses SOS DP https://www.hackerearth.com/problem/algorithm/berland-programming-contests-9c8b5165/description/ .

recent one DANYANUM

Thanks for such a nice blog

One more problem from a recent contest: Or Plus Max

can someone explain what we can't use following for(int mask = 0; mask < (1<<N); ++mask) for(int i = 0;i < N; ++i) { if(mask & (1<<i)) F[mask] += F[mask^(1<<i)]; } how to identify which one to use?

I think it's because of compiler optimizations.

can we say that when ith bit is ON then simply S(mask,i) = 2*S(mask,i-1), Because all the subsets till (i-1)th bit now have 2 choices .Either they can pair up with ith bit as 0 or 1 ??

How to solve the problem Pepsi Cola?

I think the address of dp[0][-1] is undetectable, will it go wrong?

Great tutorial! Also, isn't the iterative solution akin to finding an N-dimensional prefix sum array with dimensions 2x2...x2? If this is the case, I think it could be possible to extend this idea to "bitmasks" with a different base.

Yes, see this problem.

When you do the partition, why it is a partition? Mask is in all those sets, right?

https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fd5e

Jersey Number is probably a better place to submit. There seems to be some problem with the testcases on ICPC Live Archive (AC on Codechef gets WA there; Noone has solved it).

The suboptimal solution is very clever

There is another

coolway of visualizing/memorizing SOS that I learnt from errichto's video:How do you transform a 1D array to its prefix sum? You do:

`a[i] += a[i - 1]`

.How do you transform a 2D array to its prefix sum? Normally it is done by

`p[i][j] = p[i-1][j] + p[i][j-1] - p[i-1][j-1] + a[i][j]`

. But notice that you can also apply the 1D prefix sum in rows and columns separately:Now, the sub over submasks problem can be imagined as doing a prefix sum on $$$2\times 2\times\ldots \times2$$$ hypercube! For example, lets say the mask has three bits, and you want sum over submasks for $$$101$$$. That is equivalent to taking the sum from cell $$$(0, 0, 0)$$$ to $$$(1, 0, 1)$$$ on a 3D cube.

So, you can just apply 1D prefix sum on each dimension separately. That is exactly what the final version of the code snippet is doing. It first iterates on the dimension, then does

`a[..][1][..] += a[..][0][..]`

for that dimension; in other words, takes prefix sum on that dimension. And after that, the initial array turns into the SOS!Very interesting visualization. Thanks!

Can you please post the link to that video of Errichto's?

Watch analysis of the first problem here: Innopolis Open 2018-19 analysis

Analysis of the 2nd problem (B: Cake Tasting) it is, actually.

Link at the exact time if someone needs it.

Thanks. Made my understanding of the situation clear.

Kindly note that these relations form a directed acyclic graph and not necessarily a rooted tree (think about different values of mask and same value of i)

can someone explain not a rooted tree more

Can anyone please help me out in the question — VIM WAR ? Couldn't understand the summation formula in the editorial. Thankyou.

Did you get it? I am also facing problem understanding it.

Can we use

`F[mask] = (1 << (__builtin_popcount(mask) - 1)) * a[badbit] + F[nmask] * 2;`

to update F[mask] in $$$O(1)$$$ where`badbit = 31 - __builtin_clz(mask)`

and`nmask = mask - (1 << badbit)`

?for(int i = 0; i<(1<<N); ++i)

for(int i = 0;i < N; ++i) for(int mask = 0; mask < (1<<N); ++mask){

}

let there be two numbers i and j such that i and j bit are set in mask then will f[mask^(1<<j)^(1<<i)] not added twice in f[mask] from above approach

wq

F[i] = A[i]????? why?, did u check your code with some examples?

its the memory optimized code in the blog i just copied it here.

f[i]=a[i] is initialization for all masks beacause every set is subset of itself ofc!

you are right, sorry for my wrong reply

Can I use FWT to solve this problem with the same complexity? F = FWT_AND(A,B),where B=[1,1,1,1...]

Easy ques, just above code: link

soln : link SAME AS ABOVE

thanks

can I get link of the same problem which discussed above Actually I want to check a little different approach

issue solved

This problem from csacademy uses SOS dp as a subroutine. Maybe checking it out.

If I need to iterate over the subsets of the mask's complement , How can I apply SoS DP approach? I'm able to apply the 3^N approach easily, However N is upto 20 which would lead to A TLE verdict.

usaxena95 any thoughts?

here is the problem if anyone is interested :101666G - Going Dutch

It seems isomorphic to this problem I ran into a few weeks ago. You'll have to change your DP from "How big is the largest valid partition of (the complement of) this mask?" to "How big is the largest valid partition of any subset of (the complement of) this mask?"

Edit : I get it now , here is the code in case someone needs it

Very well explained tutorial! it was very helpful. Thank you UwU

I have $$$Q \le 200000$$$ queries and set of $$$N \le 200000$$$ bitmasks. For each bitmask in query I have to calculate number of bitmasks in set that have a

`bitwise AND`

equal to $$$0$$$ with bitmask from query. How can I solve it, if bitmasks contains $$$50$$$ bits? I already can solve it for $$$20$$$ bits, but can't figure out sparse solution. I thought about something like:but it is directed acyclic graph...

Almost same article with lot more explanations — Link

The article linked has exactly same content. That one is blatantly copied, without even citing this blog as reference.

Cant get link for Special Pairs.

Quite an old problem I guess. They removed it.

Hey, I really liked your tutorial. Thanks for this blog. If you want, you can add this problem to your list ANDPREF

Here's problem from CSES Problemset using this technique: https://cses.fi/problemset/task/1654.

A small reminder, we can change

`if(mask & (1<<i))`

to`if (!(mask & (1<<i)))`

, then we can calculate the sum of super set (not subset).The correctness is simple, we can just change the dp formular above a bit.

And it can easily solve problem D in yesterday contest.

Thank you! I finally passed that problem!

or you can keep the if condition same and do F[mask&(1<<i)] += F[mask]; Just adding supersets to the set instead of adding subsets to the set.

How to prove complexity in the suboptimal solution? I haven't got how to come to 3^n through the sum.

https://cp-algorithms.com/algebra/all-submasks.html

omg, hadn't thought about binomial theorem here, thanks

Sorry for necroposting, but is there a way to fix the latex/markup sections? I keep on seeing

`Unable to parse markup [type=CF_TEX]`

. usaxena95View the cached one here : https://webcache.googleusercontent.com/search?q=cache:NZrcJXh-IVkJ:https://codeforces.com/blog/entry/45223+&cd=1&hl=en&ct=clnk&gl=in

Thanks. I'm investigating the reason. Will be fixed soon.

I was reading this and started to solve the problems and suddenly whole blog just started showing this. Now I can't see the blog.

Unable to parse markup [type=CF_MARKDOWN]

Thanks, fixed now.

I seem to have seen this trick a long time ago， But I forgot it in today's contest ；）