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Numbers 0 to 11111 in terms of Increasing and Decreasing Orders of 1 to 9 (2014)

116 点作者 hardmath123超过 8 年前

11 条评论

jlarocco超过 8 年前
The abstract and introduction don&#x27;t explain this very well.<p>My understanding is that the author wrote the digits 1 through 9 in ascending order, and then inserted parenthesis, addition, subtraction, multiplication, division and exponentiation operators between them where appropriate to get every number from 0 to 11111.<p>And then he did the same thing using the digits 9 through 1 in descending order and did the same thing.<p>He was able to find a solution for every number except 10958 with 1-9 in ascending order.<p>EDIT:<p>The paper also doesn&#x27;t explain much about how or why they did this. At first glance I don&#x27;t think it would be too difficult to come up with a branch and bound algorithm to search through all the solutions.
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jwtadvice超过 8 年前
At my office we lost a lot of time because someone put up a challenge to represent all the numbers 0-99 with algebriac expressions containing exactly four digits &#x27;4&#x27;. The use of powers and radicals was allowed.<p>I myself solved several numbers before exhausting the tricks I could think of, and reverted to brute force&#x2F;search with Polish&#x2F;prefix notation, then realizing how massive the search space was for even just a few operators. Adding parentheses ordering further explodes the search space.<p>Finally, while this kind of concrete and explicit expression is &quot;toy math&quot;, it is related to Godel Sentences and lexigraphic ordering of proofs used through much of the establishment of logic, the foundations of proof systems, and computer science.
javajosh超过 8 年前
It wasn&#x27;t clear why this was interesting from the title. But here&#x27;s an example:<p><pre><code> 0 = 12 + 34 − 56 − 7 + 8 + 9. 1 = 1^23456789. 2 = 123+4−56−78+9. 3 = 123 − 45 − 6 − 78 + 9. </code></pre> And then in decreasing order<p><pre><code> 0 = 98−7−6−54−32+1. 1 = 98−76−54+32+1. 2 = 9+87−65+4−32−1. 3 = 98−76−5+4+3−21. </code></pre> That this is possible at all is counter-intuitive to me, and hence fascinating. Even more fascinating is that he does not have a solution for 10958 in the ascending case.
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adtac超过 8 年前
I just wrote a script to do this in, like, 5 minutes :) And the whole thing runs in 7 seconds on my machine.<p><a href="https:&#x2F;&#x2F;github.com&#x2F;adtac&#x2F;123456789&#x2F;blob&#x2F;master&#x2F;output" rel="nofollow">https:&#x2F;&#x2F;github.com&#x2F;adtac&#x2F;123456789&#x2F;blob&#x2F;master&#x2F;output</a><p>Admittedly, it doesn&#x27;t have every number. I suspect it&#x27;s because I haven&#x27;t included bracketed expressions.
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aaron695超过 8 年前
Would be more interested in the script than the list and how long it took to run.<p>And if this is the best so far? Seems surprising given it&#x27;s a popular kids game.<p>It&#x27;s not really maths IMO since I think using 98 is cheating for instance, it&#x27;s really (9 X 10 + 8), 10 not being allowed. So kids are not learning real maths, more arithmetic.
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iplaw超过 8 年前
So, who is going to find the increasing solution for 10958?
sp332超过 8 年前
Why did this paper need four revisions?<p>Edit: the revisions have different ranges of numbers, but they go up and down. Odd.
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ninju超过 8 年前
Maybe something neat to bring up as a &quot;drinking game&quot;...at my next Math Symposium<p>Pick at random 3 digit number and try to express using the digits 1 thru 9...loser has to drink
chriswarbo超过 8 年前
Seems related to this recent Numberphile video which does a similar thing for e and pi:<p><a href="https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=xgBGibfLD-U" rel="nofollow">https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=xgBGibfLD-U</a>
monochromatic超过 8 年前
This is way more numerology than math, and I don&#x27;t see any deep results coming out of this. Was this just a fun side project for the author?
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Grue3超过 8 年前
Interesting fact: every integer can be written with three 2&#x27;s.
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