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The abstract and introduction don't explain this very well. My understanding is that the author wrote the digits 1 through 9 in ascending order, and then inser
by jlarocco 10y ago
The abstract and introduction don't explain this very well.
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.
And then he did the same thing using the digits 9 through 1 in descending order and did the same thing.
He was able to find a solution for every number except 10958 with 1-9 in ascending order.
EDIT:
The paper also doesn't explain much about how or why they did this. At first glance I don't think it would be too difficult to come up with a branch and bound algorithm to search through all the solutions.
- pavel_lishin 10y agoI wonder if any numbers have multiple solutions.
- mmanfrin 10y agoI am nearly certain many do.
- Beltiras 10y agoYour sibling comment has an example in 1+2+3=1* 2*3
- Cogito 10y agoAn easy way to try this is to consider any that start with 1+2+3. You can create a second version by replacing this with 1×2×3. For example, 10914 = 1 × 2 × 3 × (4^5 + 6 + 789) 10914 = (1 + 2 + 3) × (4^5 + 6 + 789)
- squidbot 10y agoI found this article about the paper and there are some comments from the author. Not much more enlightening that it was something of a "cocktail party game." https://nebusresearch.wordpress.com/2013/06/10/counting-from-52-to-11108/#more-1778 https://nebusresearch.wordpress.com/2013/06/10/counting-from...