# Dynamic programming: Fibonacci generator

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 ?

• Real programmers don't use loops – Dr. belisarius Aug 3 '15 at 19:03
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• 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 3 '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 3 '15 at 20:47
• @shi So you may post your formula ... – Dr. belisarius Aug 3 '15 at 20:50

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.

• Please, use the comment section just below the answer/question you are trying to respond to. This is strictly answer-only box. – Sektor Aug 3 '15 at 21:06
• sorry, first time – Francseco Insulla Aug 3 '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 3 '15 at 21:17
• @FrancsecoInsulla No worries :) – Sektor Aug 3 '15 at 21:19
• I fixed it, do you want me to put the code? I'll explain too – Francseco Insulla Aug 3 '15 at 21:23