SOS Dynamic Programming [Tutorial]

№	Пользователь	Рейтинг
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

№	Пользователь	Вклад
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

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.

$\text{[math]}$

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 $\text{[math]}$ . Therefore total iterations = $\text{[math]}$

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 $\text{[math]}$ . Now we will partition this set into non intersecting groups. $\text{[math]}$ , that is set of only those subsets of mask which differ from mask only in the first i bits (zero based).
For example $\text{[math]}$ . 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^th bit as it would mean that the numbers will have a 1 at i^th bit where 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. $\text{[math]}$ S(mask,i) = S(mask, i-1).
Consider that i^th bit of mask is 1. Now the numbers belonging to S(mask, i) can be divided into two non intersecting sets. One containing numbers with i^th bit as 1 and differing from mask in the next i-1 bits. Second containing numbers with i^th bit as 0 and differing from mask⊕2ⁱ in next i-1 bits. $\text{[math]}$ S(mask, i) = S(mask, i-1) ∪ S(mask⊕2ⁱ, i-1).

$\text{[math]}$

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.:

$\text{[math]}$

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.

//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]; }

F[0] = 0; for(int mask = 1; mask < (1 << n); mask++){ int badbit = 32 - 1 - __builtin_clz(mask); int nmask = mask - (1 << badbit); F[mask] = (1 << (__builtin_popcount(mask) - 1)) * a[badbit] + F[nmask] * 2; }

private int dp(int msk) { if (msk == (1 << n) - 1) return 0; if (memo[msk] != -1) return memo[msk]; int max = 0; int avalMsks = msk ^ ((1 << n) - 1); for (int i = avalMsks; i > 0; i = (i - 1) & (avalMsks)) { if (valid[i]) max = Math.max(max, dp(msk | i) + 1); } return memo[msk] = max; }

private int dp(int msk) { if (msk == (1 << n) - 1) return 1; if (memo[msk] != -1) return memo[msk]; int max = 0; for (int i = 0; i < n; i++) { if ((msk & (1 << i)) == 0) max = Math.max(max, dp(msk ^ (1 << i)) + (valid[msk] ? 1 : 0)); } return memo[msk] = max; }

Комментарии (139)

Показать архивные | Написать комментарий?

VastoLorde95

8 лет назад, # |

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

→ Ответить

usaxena95

8 лет назад, # ^ |

← Rev. 2 →

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

← Rev. 3 →

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 $\text{[math]}$

fchirica

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 :)

shubhamgoyal__

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

Thanks. Added :)

Prateek1996

Very well written blog.

P.S:The spelling of prerequisites is wrong

Thanks. Fixed.

byte_gambler

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

Thanks. Fixed now.

siyavash

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].

apoorv_kulsh

+13

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

asvikr

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

bluemmb

thanks great.

sorry, how we can prove that for(int i = mask; i > 0; i = (i-1) & mask) will pass over all subsets ?

← Rev. 6 →

+19

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: Operation M-1 switches OFF the first ON bit and switches ON the remaining prefix. Eg 10110000₂ - 1 = 10101111₂

Statement P(n) = Given an integer M, this algorithm will iterate over all subsets $\text{[math]}$ s.t. x differs from M only in the first n bits in strictly decreasing order. $\text{[math]}$ algorithm successfully iterates over all elements in S(M, n) in strictly decreasing order.

Base Case P(0):

Case 1: if M is even $\text{[math]}$ The first iteration i = M successfully visits S(M,0) = {M}
Case 2: if M is odd $\text{[math]}$ First iteration i = M, second iteration i = (i-1)&M switches off the 0^th bit thus visits $\text{[math]}$ . $\text{[math]}$ Algo visits $\text{[math]}$ in decreasing order

Hence P(0) is true.

Assumption: Assume P(n) to be true. Algo visits S(M, n) in descending order.

Inductive step: To prove P(n+1) is true. Since P(n) is true, algo visits all S(M, n) in descending order.
P(n+1) is trivially true if n + 1^th bit of M is OFF since S(M,n+1) = S(M,n).

Lets focus on case when n + 1^th bit of M is ON. Since the visits of S(M,n) as assumed by P(n) are in descending order, the last integer visited by this algo would be M with first n bits OFF. For example, if M = $\text{[math]}$ , n = 4 the last value of i would be $\text{[math]}$ .

After reaching this integer, we do i = (i-1)&M. The i-1 operation turns OFF the n + 1^th bit and turns ON first n bits. Taking bitwise AND with original M copies the first n bits of M into it.

Taking the example above, we have following transitions $\text{[math]}$ -> $\text{[math]}$ -> $\text{[math]}$ -> $\text{[math]}$ .
Thus what we final get is $\text{[math]}$ .

Since P(n) is true, we can now iterate over S( $\text{[math]}$ , n). But $\text{[math]}$ . Therefore we iterate over all elements of S(M,n+1).

Hence P(n+1) is true.

hpkt.pkt

7 лет назад, # ^ |

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 !!

DessertKid

I have the same doubt! Is it a typo or am I missing something?

UPD: I think it should be 2^(i+1)-1.

shivaram_22

4 года назад, # ^ |

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

Hepic_Antony_Skarlatos

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 !

← Rev. 5 →

Got the bug.

if(exist[0])
    --answer;

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 with A[i] -> it would just be a subset of one's complement of A[i]. So the answer is

Anyways 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_love_Captain_America

Given a fixed array A of 2^N integers

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

rumman_sust

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

anni11

great tutorial

In your memory optimized code shouldn't it be

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

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 is correct. Sorry for ignoring it.

Zuciso

Thanks for informing was stuck on that !

youthPaul

11 месяцев назад, # ^ |

Your comment inspired me!
if the $$$i^{th}$$$ bit of mask is $$$1$$$, then the $$$i^{th}$$$ bit of (mask $$$\bigoplus 2^i$$$) will be $$$0$$$, which will not change in this cycle!
In this case, the F[mask^(1<<i)] will remain unchanged.
So it doesn't matter which way the mask changes

anh1l1ator

-8

Hello ,

Can you explain the formula

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

Jima

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

Rezwan.Arefin01

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 $\text{[math]}$ . So we have base cases $\text{[math]}$ , for other masks initialize with 0.
Now, for each array elements you need to find out another number from the array, such that their AND is 0. Note that the number you need to find will always be a subset of $\text{[math]}$ . So you can just print the number stored at $\text{[math]}$ . If it is 0 then there is no solution. [ N = minimum number of bits needed to represent all the numbers of the array]

nmc

My idea was almost the same. But it is getting TLE on test 12. Here's the link to my submission http://mirror.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

dpsorari

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

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

-is-this-fft-

+11

Because this is pseudocode.

spellstaker

6 лет назад, # ^ |

-13

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

But, you know, unexpected behaviour.

yyyvvvyyy

7 лет назад, # |

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.

joker_in

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,

back_slash

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

vaibhav1997

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

dkp1024

← Rev. 4 →

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?

Accumulating F won't give you sum over all subsets of the array.
F[mask] is the sum of A[i] such that mask & i == i, that mean the on bits of i is subset of the on bits of mask.
Solve more problems, you'll find out yourself. Most of the times you'll just need to change the base case.

heello

Absolutely Perfect.

satyaki3794

Absolutely Pointless. :)

15_percent_vat

Absolutely perfectly pointless

abhikolar1512

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)];
}

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)];
}

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

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];

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.

2021

This reminds me of this: http://web.evanchen.cc/handouts/SOS_Dumbass/SOS_Dumbass.pdf

ashish1729

-6

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

generates all subsets of bitmask y.

How does this iteration works? Any intuitive explaination?

Reziba

x starts with the value y, and everytime you subtract 1 from x, 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 AND this with y (which was initially x), you get the common set bits between x and y ( definition of bitwise AND ).

Eg: x=10011 and y=10100

x&y = 10000

Everytime you AND x with y, it is making sure that x is always a subset of y.
Subtracting x by 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.

dush1729

TangHanh

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.

adarsh000321

5 лет назад, # ^ |

it should be 1011010 xor 1010000 = 0001010

skyhavoc

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?

Vedhachala

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.

forassk

Thanks

kunal_rai

Thanks really helped alot.

striver_79

Excellent editorial! Kudos...

abhayps

6 лет назад, # |

Approach for discussion problem?

stringray

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

xodiac

+12

recent one DANYANUM

fugazi

Thanks for such a nice blog

chinmay0906

One more problem from a recent contest: Or Plus Max

nsut_coder

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?

VLADO

I think it's because of compiler optimizations.

Pirate_joker

5 лет назад, # |

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 ??

kanishk779

How to solve the problem Pepsi Cola?

happypeople

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

KevinWan

+10

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.

ivan100sic

Yes, see this problem.

hielitos

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

XGptw

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

aneesh2312

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).

atom0410

The suboptimal solution is very clever

+32

There is another cool way 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:

for i in [1, N]: for j in [1, N]: a[i][j] += a[i][j-1]  
for i in [1, N]: for j in [1, N]: a[i][j] += a[i-1][j]

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!

saurabhs1206

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

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

blitzitout

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

dgupta0812

3 года назад, # ^ |

Link at the exact time if someone needs it.

rohan_ardulous

Thanks. Made my understanding of the situation clear.

2020

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.

itsmihir

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

DeadlyCritic

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)

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)];

}

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

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

zhang177048

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

Naithani

4 года назад, # |

Easy ques, just above code: link

soln : link SAME AS ABOVE

Dontgiveup20

thanks

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

issue solved

cote

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

Ahmed_Hosssam

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

clyring

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

costheta_z

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

dmkz

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: