Dynamic programming: Fibonacci generator

Question

I have the following piece of code:

A[m_, k_] := For[i = 1, i <= m, i++,
  For[j = 1, j <= k, j++,
   A[i_, j_] = If[i <= 0, 0,
     If[[i == 1 || i == 2] && j == 2, 1,
      If[OddQ[i]; j == 1, 1,
       If[! OddQ[i]; j == 1, 0,
        If[j == 2, A[i - 2, 2] + A[i - 3, 2], 
         If[i <= j, Fibonacci[i], 
          False]
         ]
        ]
       ]
      ]
     ]
   ]
  ]

When executing A[2, 3] I should get an answer that is equal to Fibonacci[2], but it does not work that way. Any suggestions ?

Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the [faq]! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! — Dr. belisarius, Aug 03 '15 at 19:03
Not sure what you're trying to achieve here (and I'm sure there are better ways) but the problem may lie in your usage of i, j both as symbolic patterns in the second definition of A and as variables with values in the outer loop. It might be that removing the underscores in the second definition resolves the problem. — Sjoerd C. de Vries, Aug 03 '15 at 19:21
Actually, what I actually need is solving a recursive formula. Need to perform this without a recursion, for complexity reasons. And thus have this frustrating code. Is there a simple way to do this? (with two cases in the formula) — shi, Aug 03 '15 at 20:47

Francseco Insulla · Answer 1 · 2015-08-03T23:46:42.977

1

Ok, so this is the solution:

A[m_, k_] := For[i = 1, i <= m, i++,
 For[j = 1, j <= k, j++,
   A[i, j] =
    If[i <= 0, 0, 
     If[(i == 1 || i == 2) && j == 2, 1,
      If[OddQ[i]; j == 1, 1,
       If[! OddQ[i]; j == 1, 0,
        If[j == 2, A[i - 2, 2] + A[i - 3, 2], 
         If[i <= j, Fibonacci[i], 
          False]]]]]]]]

I found an error in one of the If statements:

[i == 1 || i == 2] && j == 2

should be

(i == 1 || i == 2) && j == 2

I subsequently changed

A[i_, j_] =

to

A[i, j] =

in this way, instead of defining the function again, the program is saving the value of whatever A[i,j] is each time.

To visualize this in a grid, the function Grid should do the job.

Grid[Table[Table[A[i,j],{i,1,10}],{j,1,10}]]

in which the tens can be changed to any number. And you can also add the optional

...Frame -> All]

to the Grid function to make it nicer.

edited Aug 03 '15 at 23:46

answered Aug 03 '15 at 21:03

Francseco Insulla

21
4

Please, use the comment section just below the answer/question you are trying to respond to. This is strictly answer-only box. – Sektor Aug 03 '15 at 21:06
sorry, first time – Francseco Insulla Aug 03 '15 at 21:17
This is the recursive formula I try to solve, not including all stopping conditions. I have plenty stopping condition, as you can see with all the if's above. A_{m,k} =\left{ \begin{array}{ll} F_m, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m\leq k,\cr %\cr \sum_{j=m-k}^{m-2} A_{j,k},\ \ \ \ \ \ \ \ m>k. \cr – shi Aug 03 '15 at 21:17
@FrancsecoInsulla No worries :) – Sektor Aug 03 '15 at 21:19
I fixed it, do you want me to put the code? I'll explain too – Francseco Insulla Aug 03 '15 at 21:23
sure, thank you! – shi Aug 03 '15 at 21:24
did it fix your problem? – Francseco Insulla Aug 03 '15 at 21:40
yes, thank you very much! but it's a little slow. Is this because of the implementation of the Fibonacci function or because of all those loops? – shi Aug 03 '15 at 21:55
the Fibonacci function is pretty fast, it's probably because of the loops and all the if statements – Francseco Insulla Aug 03 '15 at 22:02
Thanks. How can I enter this data calculated into a table, the moment it is calculated, which might have a better runtime? – shi Aug 03 '15 at 22:06
as soon as it is calculated, A of whatever i and j are is saved, because I put A[i,j]=... , this means that putting them into a table will not help. – Francseco Insulla Aug 03 '15 at 22:12
so I can print the whole A table? – shi Aug 03 '15 at 22:13
I'm not very good with optimization... – Francseco Insulla Aug 03 '15 at 22:14
ok thank you very much! – shi Aug 03 '15 at 22:15
Grid[Table[Table[A[j, i], {i, 1, 10}], {j, 1, 2}], Frame -> All] should work, changed the 10 and 2 – Francseco Insulla Aug 03 '15 at 22:18
@FrancsecoInsulla - your post has been flagged as "not an answer" by several users. But having reviewed it and the subsequent comments, I think you are on your way to having an answer. Please edit your post to incorporate some of the additional information you provided in comments, with a bit more explanation. – Verbeia Aug 03 '15 at 23:24
I tried to fix it, thanks – Francseco Insulla Aug 03 '15 at 23:48

Dynamic programming: Fibonacci generator

1 Answers1