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Jan
7
revised Why x = x doesn't cause an infinite loop, but f[x_] := f[x] does?
added 63 characters in body
Jan
7
accepted Forcing Mathematica to recognize a symbol without defining it
Jan
7
comment Forcing Mathematica to recognize a symbol without defining it
related: mathematica.stackexchange.com/q/39957/534
Jan
7
comment Why x = x doesn't cause an infinite loop, but f[x_] := f[x] does?
related to this answer: mathematica.stackexchange.com/a/39936/534
Jan
7
asked Why x = x doesn't cause an infinite loop, but f[x_] := f[x] does?
Jan
7
comment Forcing Mathematica to recognize a symbol without defining it
@Nasser Also, I'll accept your answer if you post it.
Jan
7
comment Forcing Mathematica to recognize a symbol without defining it
@Nasser After chrom = chrom has been evaluated, chrom is in the symbol table, pointing to itself. That's OK. After that I evaluate chrom. Since at this point chrom is pointing to itself, why it isn't replaced with chrom, which in turn is replaced with chrom, and so on?
Jan
7
comment Forcing Mathematica to recognize a symbol without defining it
@Nasser I don't understand why chrom = chrom doesn't produce an infinite loop when chrom is evaluated?
Jan
7
comment Forcing Mathematica to recognize a symbol without defining it
@heropup same as what you said... didn't see your comment
Jan
7
comment Forcing Mathematica to recognize a symbol without defining it
@Nasser ... seems to work, but I'm not sure I understand why. Isn't chrom = Unevaluated[chrom] the same as chrom = chrom?
Jan
6
asked Forcing Mathematica to recognize a symbol without defining it
Jan
6
comment Definitions inside ParallelTable?
@OleksandrR. I'm not very experienced. In fact, I did not know about SetShared*, nor had I done any MathLink-based method.
Jan
5
comment Definitions inside ParallelTable?
@OleksandrR. You're right. This doesn't answer the question as I phrased it. It does fix the problem I had, but that's irrelevant. I'm sure you understand I cannot post the detailed problem I am solving, since that would not help anyone. That's why I tried to post a simple toy case that represented my problem as closely as possible. I'll wait a couple of days in case someone else comes up with other ideas, and if there are no more answers, I'll Murta's accept answer since, as you say, it fits the question I asked better than my answer. I hope you don't mind if I don't delete my answer :)
Jan
5
answered Definitions inside ParallelTable?
Jan
4
comment Definitions inside ParallelTable?
@OleksandrR. I am making an effort to keep the question as simple as possible. I made a short edit, changing a = 4 to a[i] = i. I think that makes the issue clearer.
Jan
4
revised Definitions inside ParallelTable?
added 6 characters in body
Jan
4
comment Definitions inside ParallelTable?
@halirutan The example I used originally doesn't show why making the definition inside ParallelTable could be useful. I'll edit the question. Suppose that instead of a = 4 or a = i, I do a[i] = i. That's more similar to my actual problem. I do have several a's (one a[i] for each i), and their values are independent, so there should be no conflict between kernels.
Jan
3
comment Definitions inside ParallelTable?
@halirutan Because I compute the value to assign inside the ParallelTable.
Jan
3
asked Definitions inside ParallelTable?
Jan
3
answered Function that returns the binary string given its index in lexicographic ordering