Re: Have you tried recursion with Ackerman?
- From: "Heinrich Wolf" <invalid@xxxxxxxxxxxxxxx>
- Date: Mon, 20 Nov 2006 19:57:18 +0100
"Maarten Wiltink" <maarten@xxxxxxxxxxxxxxxxxx> schrieb im Newsbeitrag
news:45617054$0$326$e4fe514c@xxxxxxxxxxxxxxxxx
"Heinrich Wolf" <invalid@xxxxxxxxxxxxxxx> wrote in message
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Ackerman(0, y) := y + 1;
Ackerman(x, 0) := Ackerman(x - 1, 1);
Ackerman(x, y) := Ackerman(x - 1, Ackerman(x, y - 1));
No. Why do you ask?
Groetjes,
Maarten Wiltink
In the context of recursion it just came to my mind.
Of course it is a based on a completely different problem.
On friday you wrote, the hailstone recursion can be substituted by a loop.
I'm afraid, the Ackermann recursion cannot be substituted totally by loops.
From playing and trying I found some partial solutions by loops.But I cannot proof those.
Viele Gruesse
Heiner
.
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