Timeline for Trapped in my Cellar
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| May 15, 2015 at 8:42 | comment | added | d'alar'cop | "other answer" = puzzling.stackexchange.com/a/14978/2484 | |
| May 15, 2015 at 3:21 | comment | added | d'alar'cop | if you mentiom that this is an open problem as in the other answer... and also do part 3. I'll give you the tick. | |
| May 14, 2015 at 20:16 | history | edited | f'' | CC BY-SA 3.0 |
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| May 14, 2015 at 1:46 | comment | added | d'alar'cop | @user12408 Maybe, but what I don't see is: if it does work, then why doesn't that same idea extend indefinitely? | |
| May 14, 2015 at 1:37 | comment | added | f'' | @d'alar'cop $1,2,4,8,...,2^{95},3*2^{96},5*2^{96},6*2^{96},7*2^{96}$ works if I didn't make another mistake. (So does the set I gave above, with $2^{96}$ removed from every element.) | |
| May 14, 2015 at 1:03 | comment | added | d'alar'cop | Anyway, I don't know - I'll need to see proofs maybe :p | |
| May 14, 2015 at 0:59 | comment | added | d'alar'cop | OK, yes, sorry. So, you are saying that the minimum can be higher in this way... so why doesn't that imply that you can slide the whole thing down? | |
| May 14, 2015 at 0:53 | comment | added | f'' | ant11's proof involves choosing the smallest possible weight every time, but this example shows that the same result can be achieved without selecting the smallest possible numbers. It doesn't use any weights less than $2^{98}$. | |
| May 14, 2015 at 0:49 | comment | added | d'alar'cop | how does that skip numbers? Isn't it the same thing? | |
| May 14, 2015 at 0:44 | comment | added | f'' | There are ways to 'skip' numbers while the highest weight remains the same, for example $2^{99}-2^{98},2^{99}-2^{97},2^{99}-2^{96},...,2^{99}-2^2,2^{99}-2^1,2^{99}-2^0,2^{99}$. | |
| May 14, 2015 at 0:42 | comment | added | ant11 | @user12408 I agree, I was not rigorous. Your proof of the minimum is very nice. | |
| May 14, 2015 at 0:36 | comment | added | f'' | However, the proof isn't rigorous. It doesn't show that there can't be some way to 'skip' a number that somehow makes the highest weight smaller. | |
| May 14, 2015 at 0:34 | history | edited | f'' | CC BY-SA 3.0 |
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| May 14, 2015 at 0:32 | comment | added | d'alar'cop | if it's right... then the lowest highest would be $2^{n-1}$. What do you reckon? | |
| May 14, 2015 at 0:30 | review | First posts | |||
| May 14, 2015 at 1:08 | |||||
| May 14, 2015 at 0:30 | comment | added | d'alar'cop | ant11 has an intuitive proof as to why it MUST be powers of 2. | |
| May 14, 2015 at 0:26 | history | answered | f'' | CC BY-SA 3.0 |