Given the list in A1:A6, where some entries consist of a single number and some of a grouping of numbers (where e.g. 13-16 represents 13, 14, 15 and 16), the challenge this week is to come up with a single formula in D1 which, when copied down an arbitrary number of rows, produces a list of all individual, ungrouped numbers from the list in A1:A6, as here:
The formula is to return a blank in rows beyond the expected number of returns. What’s more, this must be done via reference to a second formula, in C1, also to be derived and whose value is to equal the total expected number of non-blank entries to be returned in column D for any given dataset.
Last week I set readers the challenge which can be found here.
Three solutions were offered, two of which from the same person, and both of which were correct! So many congratulations to Bill on successfully solving what was quite a complex challenge!
Indeed, as Ben Schwartz pointed out, this challenge appears to have been set previously on the internet, and seems to have been only partially solved on those occasions. In any case, thanks also to Ben for his suggestion, which he confesses was cobbled together from those previous solutions he found, and which worked in all but a few exceptional cases.
For this, the 2nd in the series of Advanced Formula Challenges, readers are asked to come up with a solution to the following:
Given two lists of names in B1:B10 and E1:E10 (as above), a formula is to be entered into A1, such that, when copied down to A10, returns TRUE if, for the corresponding name in column B, there exists at least one name in the range E1:E10 which is an anagram of that name.
Last Sunday I set a challenge to readers to come up with a solution to the problem here.
Even though this site’s only been up for one week, I’m quite happy to have received the single solution that I did, even more so since that solution was a correct one, from John Jairo Vergara Domìnguez, whose offering you can see if you scroll down to the bottom of that link. Thanks again, John, and well done!
As excellent as John’s solution was, it would still require a little tweaking to work for other ranges (part of its construction is dependent on the array in question being in certain columns within the worksheet) and, in any case, I would now like to present the solution that I developed for this problem.
We saw in a previous post (here) an outline for a solution which, given a two-dimensional array, potentially containing some empty cells, generated a list of all non-blank entries from that array in a single column.
In that solution the returned entries were listed in an order which is consistent with the entries from an entire row from the original array being returned prior to moving onto those in the next row. The converse, in which entries are returned in a columns-first fashion, is the challenge I would like to set for any readers of this post willing to have a go.