## digitalmars.D - [OT] fastest fibbonacci

- Stefam Koch (17/17) Oct 23 2016 Hi Guys,
- Timon Gehr (14/31) Oct 23 2016 int computeFib(int n){
- Stefam Koch (3/43) Oct 23 2016 Wow, that looks intresting.
- Timon Gehr (5/48) Oct 23 2016 It uses a general technique to speed up computation of linear
- Andrei Alexandrescu (4/5) Oct 23 2016 Would be awesome to factor this out of the particular algorithm. I
- safety0ff (4/6) Oct 23 2016 Rosettacode is a good place to check for "floating around"
- Minas Mina (2/19) Oct 23 2016 You can even calculate Fibonacci in O(1).
- Stefam Koch (2/27) Oct 23 2016 An approximation of it.
- Matthias Bentrup (15/43) Oct 24 2016 The fibonacci sequence can be represented exactly as a linear
- Andrea Fontana (5/9) Oct 24 2016 You can simply write it as:
- Era Scarecrow (28/31) Oct 24 2016 Ran your example and it's perfect for 32bit code. But 64bit, not
- Era Scarecrow (9/15) Oct 25 2016 Hmm as an experiment I changed from doubles to reals, and got a
- Andrea Fontana (2/4) Oct 25 2016 Of course floating precision is not unlimited :)
- Timon Gehr (6/31) Oct 23 2016 The closed form does not give you an O(1) procedure that computes the
- Andrea Fontana (20/37) Oct 24 2016 import std.stdio;

Hi Guys, while brushing up on my C and algorithm skills, accidently created a version of fibbonaci which I deem to be faster then the other ones floating around. It's also more concise the code is : int computeFib(int n) { int t = 1; int result = 0; while(n--) { result = t - result; t = t + result; } return result; }

Oct 23 2016

On 23.10.2016 15:04, Stefam Koch wrote:Hi Guys, while brushing up on my C and algorithm skills, accidently created a version of fibbonaci which I deem to be faster then the other ones floating around. It's also more concise the code is : int computeFib(int n) { int t = 1; int result = 0; while(n--) { result = t - result; t = t + result; } return result; }int computeFib(int n){ int[2] a=[0,1],b=[1,2],c=[1,-1]; for(;n;n>>=1){ foreach(i;1-n&1..2){ auto d=a[i]*a[1]; a[i]=a[i]*b[1]+c[i]*a[1]; b[i]=b[i]*b[1]-d; c[i]=c[i]*c[1]-d; } } return a[0]; } (Also: you might want to use BigInt.)

Oct 23 2016

On Sunday, 23 October 2016 at 15:12:37 UTC, Timon Gehr wrote:On 23.10.2016 15:04, Stefam Koch wrote:Wow, that looks intresting. Can you explain how it computes fibbonacci ?Hi Guys, while brushing up on my C and algorithm skills, accidently created a version of fibbonaci which I deem to be faster then the other ones floating around. It's also more concise the code is : int computeFib(int n) { int t = 1; int result = 0; while(n--) { result = t - result; t = t + result; } return result; }int computeFib(int n){ int[2] a=[0,1],b=[1,2],c=[1,-1]; for(;n;n>>=1){ foreach(i;1-n&1..2){ auto d=a[i]*a[1]; a[i]=a[i]*b[1]+c[i]*a[1]; b[i]=b[i]*b[1]-d; c[i]=c[i]*c[1]-d; } } return a[0]; } (Also: you might want to use BigInt.)

Oct 23 2016

On 23.10.2016 17:42, Stefam Koch wrote:On Sunday, 23 October 2016 at 15:12:37 UTC, Timon Gehr wrote:It uses a general technique to speed up computation of linear recurrences, with a few additional optimizations. One iteration of your while loop multiplies the vector (result,t) by a matrix. I exponentiate this matrix using a logarithmic instead of a linear number of operations.On 23.10.2016 15:04, Stefam Koch wrote:Wow, that looks intresting. Can you explain how it computes fibbonacci ?

Oct 23 2016

On 10/23/16 12:32 PM, Timon Gehr wrote:It uses a general technique to speed up computation of linear recurrencesWould be awesome to factor this out of the particular algorithm. I recall SICP famously does that with a convergence accelerating technique for series. -- Andrei

Oct 23 2016

On Sunday, 23 October 2016 at 13:04:30 UTC, Stefam Koch wrote:created a version of fibbonaci which I deem to be faster then the other ones floating around.Rosettacode is a good place to check for "floating around" implementations of common practice exercises e.g.: http://rosettacode.org/wiki/Fibonacci_sequence#Matrix_Exponentiation_Version

Oct 23 2016

On Sunday, 23 October 2016 at 13:04:30 UTC, Stefam Koch wrote:

Oct 23 2016

On Sunday, 23 October 2016 at 19:59:16 UTC, Minas Mina wrote:On Sunday, 23 October 2016 at 13:04:30 UTC, Stefam Koch wrote:An approximation of it.

Oct 23 2016

On Sunday, 23 October 2016 at 23:17:28 UTC, Stefam Koch wrote:On Sunday, 23 October 2016 at 19:59:16 UTC, Minas Mina wrote:The fibonacci sequence can be represented exactly as a linear combination of two exponential functions, but the two bases of the exponentials and the linear multipliers of them are irrational numbers, which cannot be represented exactly on a computer. However the rounding error is so small, that rounding to int will give you always the correct answer as long as you stay within the precision limit of the floating point type you use, e.g. a real should give you 64-bit fibonacci in O(1), if the exponential function is O(1). PS: the exact formula is fib(n) = 1/sqrt(5) * (0.5 + 0.5sqrt(5))^n - 1/sqrt(5) * (0.5 - 0.5sqrt(5))^n. If you round to integer anyway, the second term can be ignored as it is always <= 0.5.On Sunday, 23 October 2016 at 13:04:30 UTC, Stefam Koch wrote:An approximation of it.

Oct 24 2016

On Monday, 24 October 2016 at 08:20:26 UTC, Matthias Bentrup wrote:PS: the exact formula is fib(n) = 1/sqrt(5) * (0.5 + 0.5sqrt(5))^n - 1/sqrt(5) * (0.5 - 0.5sqrt(5))^n. If you round to integer anyway, the second term can be ignored as it is always <= 0.5.You can simply write it as: round(phi^n/sqrt(5)); Check my example above :)

Oct 24 2016

On Monday, 24 October 2016 at 08:54:38 UTC, Andrea Fontana wrote:You can simply write it as: round(phi^n/sqrt(5)); Check my example above :)Ran your example and it's perfect for 32bit code. But 64bit, not so much. It's only good through 71 iterations (longs) then it starts having errors. Also for some odd reason the input is one off, so i had to add a -1 to the input for it to align. This makes it accurate to 41/64 bit results. for(int i = 1; i < 100; ++i) { auto cf = computeFib(i); auto cfm = computeFibMagic(i-1); //with magic numbers exampled writeln(i, ": ", cf, "\t", cfm, "\t", cf == cfm); } 64: 10610209857723 10610209857723 true 65: 17167680177565 17167680177565 true 66: 27777890035288 27777890035288 true 67: 44945570212853 44945570212853 true 68: 72723460248141 72723460248141 true 69: 117669030460994 117669030460994 true 70: 190392490709135 190392490709135 true 71: 308061521170129 308061521170129 true 72: 498454011879264 498454011879265 false 73: 806515533049393 806515533049395 false 74: 1304969544928657 1304969544928660 false 75: 2111485077978050 2111485077978055 false 76: 3416454622906707 3416454622906715 false 77: 5527939700884757 5527939700884771 false 78: 8944394323791464 8944394323791487 false 79: 14472334024676221 14472334024676258 false 80: 23416728348467685 23416728348467746 false

Oct 24 2016

On Monday, 24 October 2016 at 19:03:51 UTC, Era Scarecrow wrote:It's only good through 71 iterations (longs) then it starts having errors. Also for some odd reason the input is one off, so I had to add a -1 to the input for it to align. This makes it accurate to 41/64 bit results. 71: 308061521170129 308061521170129 true 72: 498454011879264 498454011879265 falseHmm as an experiment I changed from doubles to reals, and got a slightly higher result, up to 85 giving us a 58/64 83: 99194853094755497 99194853094755497 true 84: 160500643816367088 160500643816367088 true 85: 259695496911122585 259695496911122585 true 86: 420196140727489673 420196140727489674 false 87: 679891637638612258 679891637638612259 false 88: 1100087778366101931 1100087778366101933 false

Oct 25 2016

On Tuesday, 25 October 2016 at 07:05:20 UTC, Era Scarecrow wrote:Hmm as an experiment I changed from doubles to reals, and got a slightly higher result, up to 85 giving us a 58/64Of course floating precision is not unlimited :)

Oct 25 2016

On 23.10.2016 21:59, Minas Mina wrote:

Oct 23 2016

On Sunday, 23 October 2016 at 13:04:30 UTC, Stefam Koch wrote:

Oct 24 2016