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Matrix

Правка en83, от DanAlex, 2016-03-08 03:35:36

Yeah, yeah, I know you expect from me matrix jokes. What if I told you I have no jokes on that ? So, just take the blue pill and go into serious stuff like ...

Cutting to the chase

I personally find matrix multiplication as the guy who sells stolen phones at the corner of the street. I mean, you get stuff at lower price but it can break in two days and you can get busted by the cops. Or not. I really need to find better metaphors...

Matrix multiplication is a well known method. People wrote on it before quite good articles, but I think you might get stuff simpler just by looking over some problems. For the ones who are familiar with the topic, you can skip to the last two problems. ### Getting high fast

Now I need to find better subtitles... However, first of all to know logarithmic matrix multiplication, you have to know logarithmic multiplication.

Basically we have to compute xn considering the multiplication operation take O(1) time. Take it straight forward we get xn = x * x * .. * x , so O(N). Let's try to reduce it step by step. Let's take xn = x2 * x2 * .... And we multiply by x if n is odd. This should work fine and the constant is reduced at half. Right... Similarly we can go to xn = xsqrt(n) * xsqrt(n) * ... and this goes to O(sqrtN). This reasoning stops here.

To get it faster you have to simply observe that xn = xn / 2 * xn / 2 for n even and xn = xn / 2 * xn / 2 * x for n odd. The two terms are the same and the third is constant, so we really need to compute xn / 2 once. And xn / 4 once. And so on. Therefore the O(logN) complexity.

Now, notice that we did not specified that x is an integer or a number. The same rules hold for other mathematical associative structures such as matrices.

Don't get stuck with struct

If you sayin' Y U NO REMEMBER MATRIX, then let me refresh your maths knowledge. You don't really need to know much about matrices to use put recurrences in a matrix multiplication form. Multiplying squared matrices is straight forward. Given two matrices their product is

Each element in each row in M is multiplied by its correspondent in the columns of N. If you find it simpler to remember, just imagine horizontal rows splitting matrix M and vertical row splitting matrix N and then match each row with each column.

Let's make a structure in which we keep a matrix. If new to matrices you should get an idea how matrix multiplication works from the code below.

struct matrix {
  // N is the size of the matrix
  int m[N][N];
  matrix()
  {
     memset(m,0,sizeof(m));
  }
  matrix operator * (matrix b)
  {
     matrix c = matrix();
     for (int i = 0; i < N; ++i)
       for (int j = 0; j < N; ++j)
         for (int k = 0; k < N; ++k) 
           c.m[i][j] = (c.m[i][j] + 1LL * m[i][k] * b.m[k][j]) % M;
     return c;
  }
  ...
};

Notice that we define the multiplication operation. We specifically did this so we can use a matrix exactly as a number in the logarithmic multiplication algorithm. So the code will be the same for an int and a matrix. Pretty cool if I do say so. And I do.

matrix modPow(matrix m,int n)
{
  if ( n == 0 )
    return unit; // the unit matrix - that is 1 for principal diagonal , otherwise 0
  matrix half = modPow(m,n/2);
  matrix out = half * half;
  if ( n % 2 )
    out = out * m;
  return out; 
}

Note that we could have defined an operator for power multiplication or used a template that we could have applied for a general type, but I find the implementation above more clear due to the clarity of the recurrence.

N-th Fibonacci term

For starters, let's define:

Fn = Fn - 1 + Fn - 2 with F1 = 1, F2 = 1

We need to put this in the form of a matrix recurrence. Well, each term is dependent of other consecutive two. This is a good clue we need just a 2 row matrix. So, from Fn - 2 and Fn - 1 we need to compute Fn. To keep the recurrences squared we will compute from the pair (Fn - 2, Fn - 1) the pair (Fn - 1, Fn).

Fn = Fn - 1 * 1 + Fn - 2 * 1 Fn - 1 = Fn - 1 * 1 + Fn - 2 * 0

Or, as matrices:

Going one step backwards we got:

Finally:

Getting at power n takes logarithmic time , so that is just... fast.

Bits and pieces

As you can see, this technique can be used to calculate the n-th term of a linear recurrence. In the following example we need to find out how many arrays of length n with maximum k consecutive 0 bits are there. ( n ≤ 109, k ≤ 40 )

Let's suppose n is small enough so we can use dynamic programming to solve the problem. Denote Dn, k = number of arrays of length n which end in k number of 0s

As you can guess one can make two moves: add a 0 and add a 1. Therefore, from state (n, k) we can go to states (n + 1, k + 1) and (n + 1, 0). So and Dn, k = Dn - 1, k - 1. As we did before, let's write down all recurrences we are interested in.

Dn, 1 = Dn - 1, 0

Dn, 2 = Dn - 1, 1

...

Dn, k = Dn - 1, k - 1

The matrix recurrence will come straight away:

As you can see, now the complexity of the solution is reduces from O(N * K) to O(logN * K3), the K3 being the complexity of multiplying 2 K-size matrices.

How big can it get ?

Now seriously, I really need better subtitles. Problem Chimney from TopCoder can be solved similarly with the problems above.

Usually, when we got a big N, this is a hint in favor of logarithmic multiplication. So we have to find a recurrence. More specifically, we have to find a good way to represent a state.

You can also note that we are not really interested how many layers have been completed, but what are the last moves made. So we can have the next forms of completed blocks in the last layers:

+-----+--+  +-----+--+  +-----+--+  +-----+--+ 
|xxxxx|  |  |     |  |  |xxxxx|  |  |     |xx|  
+--+--|  |  +--+--|  |  +--+--|  |  +--+--|xx|    
|  |  |  |  |xx|  |  |  |  |  |  |  |xx|  |xx|    
|  +--+--+  |xx+--+--+  |  +--+--+  |xx+--+--+  
|  |     |  |xx|xxxxx|  |  |xxxxx|  |xx|xxxxx|  
+--+-----+  +--+-----+  +--+-----+  +--+-----+   
    1            2           3           4

+-----+--+  +-----+--+  +-----+--+  +-----+--+
|     |  |  |     |xx|  |     |xx|  |     |xx|
+--+--|  |  +--+--|xx|  +--+--|xx|  +--+--|oo| 
+--+--|  |  +--+--|xx|  +--+--|xx|  +--+--|oo|
|xx|  |  |  |xx|  |oo|  |xx|  |oo|  |xx|  |oo|
|xx+--+--+  |xx+--+oo+  |xx+--+oo+  |xx+--+oo+
|=====xxx|  |xx|xxxoo|  |===== oo|  |===@@@@@|  
+--+-----+  +--+-----+  +--+-----+  +--+-----+ 
    5            6           7           8
 

xx represent brick on the n-th layer. == and oo are bricks on the n+1-th layer. @@ are bricks on the n+2-th layer. The special thing is we do not care what is the orientation of the chimney as all its bricks are similar, so for one brick on n-th layer ( pic 1 ) we treat all 4 possible displays the same. Another important thing to notice is that after we place two bricks one next to the other we can complete another brick in the layer above. ( pics 5,6 and 7 ). Finally, we can just add another brick to the n+2-th layer. ( pic 8 ) For simplicity, we should also consider the free layer ( a layer that contains no bricks ).

Therefore, this problem can be solved by matrix multiplication, building a 9 per 9 matrix and logarithmic multiplying it. You can see my implementation here.

Summing up

What we looked at today can be a good tool, no doubt. But as a personal advice, use it only when is necessary outside contests. Some problems are intended to be solved in a different way and people can skip useful stuff by overusing the method described above.

I also recommend this blog post, which is really good.

PS

Thank you for reading and please state your opinion on my tutorial. ( or, more specifically, on my writing style and how useful you find the material presented ) Any suggestions for next tutorial are welcome.

You can find my previous article here.

Hope you enjoyed!

История

 
 
 
 
Правки
 
 
  Rev. Язык Кто Когда Δ Комментарий
en83 Английский DanAlex 2016-03-08 03:35:36 59 Tiny change: ' & ... & 0\\ 0 && 0 & ... ' -
en82 Английский DanAlex 2016-03-08 03:24:58 2 Tiny change: '= D_{n-1,k} $\n\nThe' -> '= D_{n-1,k-1} $\n\nThe'
en81 Английский DanAlex 2016-02-07 15:39:28 6 Tiny change: ' problems.\n\n### Ge' -> ' problems. [cut]\n\n### Ge'
en80 Английский DanAlex 2016-01-25 16:52:59 2 Tiny change: 't = out * x;\n retur' -> 't = out * m;\n retur'
en79 Английский DanAlex 2015-10-27 01:46:23 8
en78 Английский DanAlex 2015-10-27 01:45:38 4 Tiny change: '`==` and `&&` are bric' -> '`==` and `oo` are bric'
en77 Английский DanAlex 2015-10-27 01:45:03 559
en76 Английский DanAlex 2015-10-25 16:59:30 0 (published)
en75 Английский DanAlex 2015-10-25 16:54:33 748
en74 Английский DanAlex 2015-10-25 16:45:17 427
en73 Английский DanAlex 2015-10-25 16:41:18 31
en72 Английский DanAlex 2015-10-25 16:40:30 1120
en71 Английский DanAlex 2015-10-25 16:22:13 1 Tiny change: '-+ \n~~~~~' -> '-+ \n~~~~'
en70 Английский DanAlex 2015-10-25 16:21:53 797
en69 Английский DanAlex 2015-10-25 16:08:48 706
en68 Английский DanAlex 2015-10-25 15:57:10 66
en67 Английский DanAlex 2015-10-25 15:55:47 97
en66 Английский DanAlex 2015-10-25 15:54:38 8
en65 Английский DanAlex 2015-10-25 15:53:27 18
en64 Английский DanAlex 2015-10-25 15:52:45 7
en63 Английский DanAlex 2015-10-25 15:52:11 2 Tiny change: '. & 0 \\ ... \\ 0 & 0' -> '. & 0 \\ .&.&. \\ 0 & 0'
en62 Английский DanAlex 2015-10-25 15:51:21 376
en61 Английский DanAlex 2015-10-25 15:46:52 32
en60 Английский DanAlex 2015-10-25 15:45:55 20
en59 Английский DanAlex 2015-10-25 15:45:10 191
en58 Английский DanAlex 2015-10-25 15:43:23 12 Tiny change: 'here. ( $n<=10^9 , k <= 40$ )\n\' -> 'here. ( $n \le 10^9 , k \le 40$ )\n\'
en57 Английский DanAlex 2015-10-25 15:42:56 36
en56 Английский DanAlex 2015-10-25 15:42:01 76
en55 Английский DanAlex 2015-10-25 15:39:58 6 Tiny change: 'm. Denote `D_{(n,k)}' = number ' -> 'm. Denote $D_{n,k}$ = number '
en54 Английский DanAlex 2015-10-25 15:39:35 15
en53 Английский DanAlex 2015-10-25 15:38:28 143
en52 Английский DanAlex 2015-10-25 15:35:40 5
en51 Английский DanAlex 2015-10-25 15:35:21 452
en50 Английский DanAlex 2015-10-25 03:19:21 119
en49 Английский DanAlex 2015-10-25 03:18:05 4 Tiny change: 'matrix} ^ (n-2) = \begin{' -> 'matrix} ^ {n-2} = \begin{'
en48 Английский DanAlex 2015-10-25 03:17:48 182
en47 Английский DanAlex 2015-10-25 03:16:40 206
en46 Английский DanAlex 2015-10-25 03:14:12 7 Tiny change: 'matrix} 0 1 & 1 1 \end{bm' -> 'matrix} 0 & 1 \\ 1 & 1 \end{bm'
en45 Английский DanAlex 2015-10-25 03:13:45 258
en44 Английский DanAlex 2015-10-25 03:10:44 10
en43 Английский DanAlex 2015-10-25 03:10:10 347
en42 Английский DanAlex 2015-10-25 03:06:27 156
en41 Английский DanAlex 2015-10-25 03:01:27 37 Tiny change: 'more clear.' -> 'more clear due to the clarity of the recurrence.'
en40 Английский DanAlex 2015-10-25 03:00:37 186
en39 Английский DanAlex 2015-10-25 02:58:28 7 Tiny change: ' c;\n }\n};\n~~~~' -> ' c;\n }\n ...\n};\n~~~~'
en38 Английский DanAlex 2015-10-25 02:58:09 571
en37 Английский DanAlex 2015-10-25 02:53:18 584
en36 Английский DanAlex 2015-10-25 02:45:49 44
en35 Английский DanAlex 2015-10-25 02:44:10 23
en34 Английский DanAlex 2015-10-25 02:42:28 257
en33 Английский DanAlex 2015-10-25 02:36:49 8 Tiny change: 'ix} b_1 & b_2 \\ b_3 & b_4 \en' -> 'ix} b_1 & | & b_2 \\ b_3 & | & b_4 \en'
en32 Английский DanAlex 2015-10-25 02:36:17 5 Tiny change: '& a_2 \\ ---- \\ a_3 &' -> '& a_2 \\ - & - \\ a_3 &'
en31 Английский DanAlex 2015-10-25 02:35:58 3 Tiny change: '& a_2 \\ - - \\ a_3 &' -> '& a_2 \\ ---- \\ a_3 &'
en30 Английский DanAlex 2015-10-25 02:35:41 4 Tiny change: '& a_2 \\ ----- \\ a_3 &' -> '& a_2 \\ - - \\ a_3 &'
en29 Английский DanAlex 2015-10-25 02:35:24 9 Tiny change: ' & a_2 \\ a_3 & ' -> ' & a_2 \\ ----- \\ a_3 & '
en28 Английский DanAlex 2015-10-25 02:34:35 7 Tiny change: '{bmatrix} $ = \n$\begin{bma' -> '{bmatrix} = \begin{bma'
en27 Английский DanAlex 2015-10-25 02:34:14 4
en26 Английский DanAlex 2015-10-25 02:33:38 18
en25 Английский DanAlex 2015-10-25 02:32:46 5 Tiny change: '{bmatrix} = \n\begin{b' -> '{bmatrix} $ = $\n\begin{b'
en24 Английский DanAlex 2015-10-25 02:32:15 76
en23 Английский DanAlex 2015-10-25 02:31:14 68
en22 Английский DanAlex 2015-10-25 02:30:11 217
en21 Английский DanAlex 2015-10-25 02:27:25 43
en20 Английский DanAlex 2015-10-25 02:26:39 4
en19 Английский DanAlex 2015-10-25 02:25:45 11
en18 Английский DanAlex 2015-10-25 02:24:46 7
en17 Английский DanAlex 2015-10-25 02:24:31 3
en16 Английский DanAlex 2015-10-25 02:24:09 4
en15 Английский DanAlex 2015-10-25 02:23:44 20
en14 Английский DanAlex 2015-10-25 02:22:34 75
en13 Английский DanAlex 2015-10-25 02:20:26 7
en12 Английский DanAlex 2015-10-25 02:20:08 45
en11 Английский DanAlex 2015-10-25 02:19:17 54
en10 Английский DanAlex 2015-10-25 02:17:40 73
en9 Английский DanAlex 2015-10-25 02:15:46 1 Tiny change: ' \n\n$M = \[a & b & c' -> ' \n\n$M = [a & b & c'
en8 Английский DanAlex 2015-10-25 02:15:30 56
en7 Английский DanAlex 2015-10-25 02:14:55 95
en6 Английский DanAlex 2015-10-25 02:11:32 478
en5 Английский DanAlex 2015-10-25 02:02:29 333
en4 Английский DanAlex 2015-10-25 01:57:53 6 Tiny change: ' = x^{sqrt n} * x^{sqrt n} * ...$ a' -> ' = x^{sqrt(n)} * x^{sqrt(n)} * ...$ a'
en3 Английский DanAlex 2015-10-25 01:57:35 10
en2 Английский DanAlex 2015-10-25 01:56:21 268
en1 Английский DanAlex 2015-10-25 01:53:13 1004 Initial revision (saved to drafts)