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March 09, 2012 matrix and fibonacci  

 
The following is a matrix implementation of the fibonacci algorithm: int fib(int n) { int M[2][2] = {{1,0},{0,1}} for (int i = 1; i < n; i++) M = M * {{1,1},{1,0}} return M[0][0]; } problem is I don't really understand how matrix multiplication works so i cannot translate it to an equivalent solution in D. Thank for you assistance in advance. newcomer[bob] 
March 09, 2012 Re: matrix and fibonacci  

 
Posted in reply to newcomer[bob]  On Friday, 9 March 2012 at 05:50:03 UTC, newcomer[bob] wrote:
> The following is a matrix implementation of the fibonacci
> algorithm:
>
> int fib(int n)
> {
> int M[2][2] = {{1,0},{0,1}}
> for (int i = 1; i < n; i++)
> M = M * {{1,1},{1,0}}
> return M[0][0];
> }
>
> problem is I don't really understand how matrix multiplication
> works so i cannot translate it to an equivalent solution in D.
> Thank for you assistance in advance.
>
> newcomer[bob]
I attempted the following but it does not work:
int fib(int n)
ulong[2][2] M = [[1,0], [0,1]];
ulong[2][2] C = [[1,1], [1,0]];
foreach (i; 0 .. n) {
M[0][0] = M[0][0] * C[0][0] + M[0][0] * C[0][1];
M[0][1] = M[0][1] * C[0][1] + M[0][1] * C[1][1];
M[1][0] = M[0][1] * C[0][0] + M[1][1] * C[0][1];
M[1][1] = M[1][1] * C[0][1] + M[1][1] * C[1][1];
}
return M[0][0];
}
any ideas what I'm doing wrong?
Thanks,
Bob

March 09, 2012 Re: matrix and fibonacci  

 
Posted in reply to newcomer[bob]  On Friday, 9 March 2012 at 09:22:47 UTC, newcomer[bob] wrote:
> On Friday, 9 March 2012 at 05:50:03 UTC, newcomer[bob] wrote:
>> The following is a matrix implementation of the fibonacci
>> algorithm:
>>
>> int fib(int n)
>> {
>> int M[2][2] = {{1,0},{0,1}}
>> for (int i = 1; i < n; i++)
>> M = M * {{1,1},{1,0}}
>> return M[0][0];
>> }
>>
>> problem is I don't really understand how matrix multiplication
>> works so i cannot translate it to an equivalent solution in D.
>> Thank for you assistance in advance.
>>
>> newcomer[bob]
>
> I attempted the following but it does not work:
>
> int fib(int n)
> long[2][2] M = [[1,0], [0,1]];
> ulong[2][2] C = [[1,1], [1,0]];
> foreach (i; 0 .. n) {
> M[0][0] = M[0][0] * C[0][0] + M[0][0] * C[0][1];
> M[0][1] = M[0][1] * C[0][1] + M[0][1] * C[1][1];
> M[1][0] = M[0][1] * C[0][0] + M[1][1] * C[0][1];
> M[1][1] = M[1][1] * C[0][1] + M[1][1] * C[1][1];
> }
> return M[0][0];
> }
>
> any ideas what I'm doing wrong?
>
> Thanks,
> Bob
Turns out that this this is an algorithm for calculating the
powers of two up to 2^63. I still cannot find how to modify it to
produce the fibonacci sequence though. Any advice would be
appreciated.
Thanks,
Bob

March 09, 2012 Re: matrix and fibonacci  

 
Posted in reply to newcomer[bob]  On 03/09/2012 10:51 AM, newcomer[bob] wrote: > On Friday, 9 March 2012 at 09:22:47 UTC, newcomer[bob] wrote: >> On Friday, 9 March 2012 at 05:50:03 UTC, newcomer[bob] wrote: >>> The following is a matrix implementation of the fibonacci >>> algorithm: >>> >>> int fib(int n) >>> { >>> int M[2][2] = {{1,0},{0,1}} >>> for (int i = 1; i < n; i++) >>> M = M * {{1,1},{1,0}} >>> return M[0][0]; >>> } >>> >>> problem is I don't really understand how matrix multiplication >>> works so i cannot translate it to an equivalent solution in D. >>> Thank for you assistance in advance. >>> >>> newcomer[bob] >> >> I attempted the following but it does not work: >> >> int fib(int n) >> long[2][2] M = [[1,0], [0,1]]; >> ulong[2][2] C = [[1,1], [1,0]]; >> foreach (i; 0 .. n) { >> M[0][0] = M[0][0] * C[0][0] + M[0][0] * C[0][1]; >> M[0][1] = M[0][1] * C[0][1] + M[0][1] * C[1][1]; >> M[1][0] = M[0][1] * C[0][0] + M[1][1] * C[0][1]; >> M[1][1] = M[1][1] * C[0][1] + M[1][1] * C[1][1]; >> } >> return M[0][0]; >> } >> >> any ideas what I'm doing wrong? >> >> Thanks, >> Bob > > Turns out that this this is an algorithm for calculating the > powers of two up to 2^63. I still cannot find how to modify it to > produce the fibonacci sequence though. Any advice would be > appreciated. > > Thanks, > Bob http://en.wikipedia.org/wiki/Matrix_multiplication#Nontechnical_example http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form I don't understand what you are trying to do above. 
March 09, 2012 Re: matrix and fibonacci  

 
Posted in reply to newcomer[bob]  On 03/09/2012 10:51 AM, newcomer[bob] wrote:
> On Friday, 9 March 2012 at 09:22:47 UTC, newcomer[bob] wrote:
>> On Friday, 9 March 2012 at 05:50:03 UTC, newcomer[bob] wrote:
>>> The following is a matrix implementation of the fibonacci algorithm:
>>>
>>> int fib(int n)
>>> {
>>> int M[2][2] = {{1,0},{0,1}}
>>> for (int i = 1; i < n; i++)
>>> M = M * {{1,1},{1,0}}
>>> return M[0][0];
>>> }
>>>
>>> problem is I don't really understand how matrix multiplication works so i cannot translate it to an equivalent solution in D. Thank for you assistance in advance.
>>>
>>> newcomer[bob]
>>
>> I attempted the following but it does not work:
>>
>> int fib(int n)
>> long[2][2] M = [[1,0], [0,1]];
>> ulong[2][2] C = [[1,1], [1,0]];
>> foreach (i; 0 .. n) {
>> M[0][0] = M[0][0] * C[0][0] + M[0][0] * C[0][1];
>> M[0][1] = M[0][1] * C[0][1] + M[0][1] * C[1][1];
>> M[1][0] = M[0][1] * C[0][0] + M[1][1] * C[0][1];
>> M[1][1] = M[1][1] * C[0][1] + M[1][1] * C[1][1];
>> }
>> return M[0][0];
>> }
>>
>> any ideas what I'm doing wrong?
>>
>> Thanks,
>> Bob
>
> Turns out that this this is an algorithm for calculating the powers of two up to 2^63. I still cannot find how to modify it to produce the fibonacci sequence though. Any advice would be appreciated.
>
> Thanks,
> Bob
>
The idea is not that bad but it contains a bug:
Once you modify M[0][1] in the second step of your loop, its changed content is plugged into the third step. You might simply save the contents of M in 4 additional variables and then use these to fill M again (with the result of M*C).
In fact it suffices to store three values as all matrices M (and C) are symmetric, that is, M[0][1] == M[1][0]. But I recommend to do this *after* you made the current version work.
Matthias

March 09, 2012 Re: matrix and fibonacci  

 
Posted in reply to newcomer[bob]  On 03/09/2012 06:50 AM, newcomer[bob] wrote: > The following is a matrix implementation of the fibonacci algorithm: > > int fib(int n) { int M[2][2] = {{1,0},{0,1}} for (int i = 1; i < n; > i++) M = M * {{1,1},{1,0}} return M[0][0]; } > > problem is I don't really understand how matrix multiplication works > so i cannot translate it to an equivalent solution in D. Thank for > you assistance in advance. > > newcomer[bob] // simple 2x2 matrix type alias int[2][2] Mat; // 2x2 matrix multiplication Mat matmul(Mat a, Mat b){ return [[a[0][0]*b[0][0]+a[0][1]*b[1][0], a[0][0]*b[0][1]+a[0][1]*b[1][1]], [a[1][0]*b[0][0]+a[1][1]*b[1][0], a[0][1]*b[0][1]+a[1][1]*b[1][1]]]; } // implementation of your algorithm int fib(int n)in{assert(n>=0 && n<46);}body{ int M[2][2] = [[1,0],[0,1]]; int F[2][2] = [[1,1],[1,0]]; foreach (i; 0..n) M = matmul(F,M); return M[0][0]; } // faster way of computing matrix power int fi(int n)in{assert(n>=0 && n<46);}body{ Mat M = [[1,0],[0,1]]; Mat F = [[1,1],[1,0]]; for(;n;n>>=1){ if(n&1) M = matmul(F,M); F = matmul(F,F); } return M[0][0]; } // closed form derived from basis transform to eigenbasis int f(int n)in{assert(n>=0 && n<46);}body{ enum sqrt5=sqrt(5.0); return cast(int)((((1+sqrt5)/2)^^(n+1)((1sqrt5)/2)^^(n+1))/sqrt5); } 
March 10, 2012 Re: matrix and fibonacci  

 
Posted in reply to Timon Gehr  On Friday, 9 March 2012 at 14:07:10 UTC, Timon Gehr wrote:
> On 03/09/2012 06:50 AM, newcomer[bob] wrote:
>> The following is a matrix implementation of the fibonacci algorithm:
>>
>> int fib(int n) { int M[2][2] = {{1,0},{0,1}} for (int i = 1; i < n;
>> i++) M = M * {{1,1},{1,0}} return M[0][0]; }
>>
>> problem is I don't really understand how matrix multiplication works
>> so i cannot translate it to an equivalent solution in D. Thank for
>> you assistance in advance.
>>
>> newcomer[bob]
>
> // simple 2x2 matrix type
> alias int[2][2] Mat;
> // 2x2 matrix multiplication
> Mat matmul(Mat a, Mat b){
> return [[a[0][0]*b[0][0]+a[0][1]*b[1][0], a[0][0]*b[0][1]+a[0][1]*b[1][1]],
> [a[1][0]*b[0][0]+a[1][1]*b[1][0], a[0][1]*b[0][1]+a[1][1]*b[1][1]]];
> }
> // implementation of your algorithm
> int fib(int n)in{assert(n>=0 && n<46);}body{
> int M[2][2] = [[1,0],[0,1]];
> int F[2][2] = [[1,1],[1,0]];
> foreach (i; 0..n) M = matmul(F,M);
> return M[0][0];
> }
> // faster way of computing matrix power
> int fi(int n)in{assert(n>=0 && n<46);}body{
> Mat M = [[1,0],[0,1]];
> Mat F = [[1,1],[1,0]];
> for(;n;n>>=1){
> if(n&1) M = matmul(F,M);
> F = matmul(F,F);
> }
> return M[0][0];
> }
> // closed form derived from basis transform to eigenbasis
> int f(int n)in{assert(n>=0 && n<46);}body{
> enum sqrt5=sqrt(5.0);
> return cast(int)((((1+sqrt5)/2)^^(n+1)((1sqrt5)/2)^^(n+1))/sqrt5);
> }
Thanks very much for the assist. All three of these methods
though, seem to have a bug. fib(), fi() and f() all produce the
same incorrect result which for lack of better word I'll call an
"off by one" bug. A call to any of these functions with any
integer value between 2 and 44 yields the return value of the
next higher integer, while 45 overflows. For example, 2 => 2, 10
=> 89, and 15 => 987 which are the actual values for 3, 11 and 16.
Thanks,
Bob

March 10, 2012 Re: matrix and fibonacci  

 
Posted in reply to newcomer[bob]  On 03/10/2012 11:00 PM, newcomer[bob] wrote: > On Friday, 9 March 2012 at 14:07:10 UTC, Timon Gehr wrote: >> On 03/09/2012 06:50 AM, newcomer[bob] wrote: >>> The following is a matrix implementation of the fibonacci algorithm: >>> >>> int fib(int n) { int M[2][2] = {{1,0},{0,1}} for (int i = 1; i < n; >>> i++) M = M * {{1,1},{1,0}} return M[0][0]; } >>> >>> problem is I don't really understand how matrix multiplication works >>> so i cannot translate it to an equivalent solution in D. Thank for >>> you assistance in advance. >>> >>> newcomer[bob] >> > > Thanks very much for the assist. All three of these methods > though, seem to have a bug. I just start the sequence from 1, because a quick glance at your implementation showed that it starts from 1. But it seems like actually it would produce the sequence 1,1,1,2,3,... > fib(), fi() and f() all produce the > same incorrect result which for lack of better word I'll call an > "off by one" bug. A call to any of these functions with any > integer value between 2 and 44 yields the return value of the > next higher integer, while 45 overflows. I don't observe any overflowing. But the assertions are a little too tight, 46 would be the first one that does not work. > For example, 2 => 2, 10 > => 89, and 15 => 987 which are the actual values for 3, 11 and 16. > > Thanks, > Bob > Probably this is what you want then: // simple 2x2 matrix type alias int[2][2] Mat; // 2x2 matrix multiplication Mat matmul(Mat a, Mat b){ return [[a[0][0]*b[0][0]+a[0][1]*b[1][0], a[0][0]*b[0][1]+a[0][1]*b[1][1]], [a[1][0]*b[0][0]+a[1][1]*b[1][0], a[0][1]*b[0][1]+a[1][1]*b[1][1]]]; } // implementation of your algorithm int fib(int n)in{assert(n>=0 && n<47);}body{ if(!n) return 0; int M[2][2] = [[1,0],[0,1]]; int F[2][2] = [[1,1],[1,0]]; foreach (i; 1..n) M = matmul(F,M); return M[0][0]; } // faster way of computing matrix power int fi(int n)in{assert(n>=0 && n<47);}body{ if(!n) return 0; Mat M = [[1,0],[0,1]]; Mat F = [[1,1],[1,0]]; for(n;n;n>>=1){ if(n&1) M = matmul(F,M); F = matmul(F,F); } return M[0][0]; } // closed form derived from basis transform to eigenbasis int f(int n)in{assert(n>=0 && n<47);}body{ enum sqrt5=sqrt(5.0); return cast(int)((((1+sqrt5)/2)^^n((1sqrt5)/2)^^n)/sqrt5); } 
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