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I have a user defined function h[i_,j_] and I would like to find its minimum value over all integers $1 \leq i \leq 100$ and $1 \leq j \leq 100$ and $j \leq i$. I would just like Mathematica to try all the possibilities rather than attempt any symbolic minimization. How can you do that?

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    $\begingroup$ Have a look at Table. $\endgroup$ – b.gates.you.know.what Apr 16 '13 at 18:20
  • $\begingroup$ Absolutely this way of minimization is not good. because if h take a lot of calculation you must evaluate all of them. i think its better to write your own numerical procedure minimization. $\endgroup$ – Rasoul-Ghadimi Mar 4 '17 at 2:27
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If you like loops, you could just go through all combinations and memorize the minimum value and position:

h[x_, y_] := ((x - 30)^2 + (y - 10)^2);

min = h[1, 1];
pos = {1, 1};
Do[If[h[i, j] < min, min = h[i, j]; pos = {i, j}], {i, 100}, {j, 100}];

And then min is 0 and pos is {30,10} as expected. With this approach you don't need to allocate the memory as in the Table example.

Not so fast for this specific example but definitely more beautiful is using Minimize although you don't wanted this

Minimize[{h[x, y], 1 <= x <= 100 && 1 <= y <= 100}, {x, y}, Integers]
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  • $\begingroup$ Thank you. The problem with Minimize is that it doesn't do anything in my case. $\endgroup$ – Anush Apr 16 '13 at 19:31
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As suggested in the comments, Table can be used to generate the matrix. To find the {i,j} pair corresponding to the minimum value, a combination of Min and Position can be used.

tbl = Table[h[i, j], {i, 1, 100}, {j, 1, i}];
Position[tbl, Min@tbl, {2}]
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If you don't need to find the parameters, but only the minimum value, you could save memory by finding the minimum for each row. Compare:

h[x_, y_] := ((x - 30)^2 + (y - 10)^2);
n = 1500;
Min @ Table[Min @ Array[h[i, #] &, i], {i, n}]
MaxMemoryUsed[]
0

15405976
Min @ Table[h[i, j], {i, n}, {j, i}];

MaxMemoryUsed[]
162985456
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