Timeline for Maximum amount of the Letter-Number-game
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 26, 2020 at 10:17 | vote | accept | math scat | ||
| Jul 25, 2020 at 14:07 | history | edited | ManfP | CC BY-SA 4.0 |
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| Jul 25, 2020 at 14:02 | history | edited | ManfP | CC BY-SA 4.0 |
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| Jul 25, 2020 at 13:53 | history | edited | ManfP | CC BY-SA 4.0 |
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| Jul 25, 2020 at 12:27 | comment | added | Mark Murray | Let us continue this discussion in chat. | |
| Jul 25, 2020 at 12:23 | comment | added | ManfP | It does assume every number ends up at $4$. $\{4\}$ is finite. | |
| Jul 25, 2020 at 12:19 | comment | added | Mark Murray | That question does not imply there are only finitely many loops | |
| Jul 25, 2020 at 11:51 | comment | added | ManfP | I answered the question OP asked - "is there an upper limit to the number of times you have to apply the function to get 4"? (And $S_n$ definitely can't be empty, as 4 is a member of all $S_n$). It's also not that hard to prove OP's assumption - it boils down to showing $d(n)<n$ for all $n>4$, which is easy enough, just a bit tedious to do all the small cases. | |
| Jul 25, 2020 at 11:45 | comment | added | Mark Murray | This answer assumes that there are only finitely many loops and finitely many fixed points. Which is a lot to assume, it almost begs the question. If you do not assume this, your solution does not exclude the possibility that $S_n$ is empty. | |
| Jul 25, 2020 at 11:14 | comment | added | ManfP | (In any case, this proves that there is no number $x$ so that applying the function $x$ times yields $4$, which is what the question asked. In your example, it's still a valid proof that "there is no $x$ so that each number ends up at $2$ after $x$ steps" - which is still true, just not very interesting, as most numbers don't end up at $2$ at all) | |
| Jul 25, 2020 at 11:10 | comment | added | ManfP | @Mark Murray the assumption is that all numbers end up at 4, which is implicit in OP's question. For arbitrary languages, set $S_0$ to be the set of numbers that loop back to themselves after some finite amount of applications of $d$, to get a meaningful result. | |
| Jul 25, 2020 at 8:10 | comment | added | Mark Murray | There are almost certainly going to be infinitely many chains. For problems like this its helpful to try instead to construct a chain for an arbitrary length $l$. Rather than to try should that all chains apart from some exceptions will work. | |
| Jul 25, 2020 at 8:07 | comment | added | Mark Murray | I don't think this works. These sets are not defined properly. Even if they were, the claim that $S_n$ is finite for all $n$ does not give us the result. Consider the the following language, where $2$ is spelled "tu" and every other number is spelled like in English but with "a-boogoly" added on. Then $S_0=\{2\}$, but so is $S_n=\{2\}$ for all other $n$. This fits your conditions but does to show that we have arbitrarily large chains. | |
| Jul 24, 2020 at 19:32 | history | edited | ManfP | CC BY-SA 4.0 |
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| Jul 24, 2020 at 19:24 | history | edited | ManfP | CC BY-SA 4.0 |
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| Jul 24, 2020 at 19:19 | history | edited | ManfP | CC BY-SA 4.0 |
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| Jul 24, 2020 at 19:15 | review | First posts | |||
| Jul 24, 2020 at 19:19 | |||||
| Jul 24, 2020 at 19:13 | history | answered | ManfP | CC BY-SA 4.0 |