5 ms·
It's a little unfortunate that they don't accept my somewhat elaborate but ultimately correct solutions, if I find a way to create a perfect rhombus within a re
by azeirah 10y ago
It's a little unfortunate that they don't accept my somewhat elaborate but ultimately correct solutions, if I find a way to create a perfect rhombus within a rectangle in Lwaytoomany Ealsowaytoomany, they should accept it at least, and give me 1 star for my effort!
Pretty fun so far though!
edit: So far, I've found two methods to find the center of a circle that they don't like. 5L E7 isn't that bad
>:(
- aisofteng 10y agoAs far as I can tell, solutions are recognized as long as they are valid, regardless of length. Can you post an example of a solution you feel is correct but that isn't being accepted?
- iamcreasy 10y agoHere[1] is one that's not being accepted. Here I have to draw an equilateral triangle given a straight line. I am drawing two circles, on the two terminal points, with radius as the length of the given line. The intersection of two circles is the third point of the triangle. [1] : http://i39.photobucket.com/albums/e179/iamcreasy/Screenshot_20161227-011845_zpsesvirsqw.png http://i39.photobucket.com/albums/e179/iamcreasy/Screenshot_...
- BlackFingolfin 10y agoIn principle, this is the right construction. However, you did not perform a proper construction, as can be seen by the two tiny red circles: Instead of constructing a circle passing through the end points of the given line, you "eyeballed" the circle diameter. Which may look right at this resolution, but if you were to zoom in sufficiently, you would inevitable end up with a circle that oh-so-slightly misses to go through one of the end points of the base line. In the smartphone app, to do this right you need to select the circle tool, start at one end point of the line, then drag your finger exactly to the other end point -- it'll "snap" to that, and create a circle through the second end point.
- iamcreasy 10y agoYes, I thought at too so I zoomed in to make sure that I am not making that mistake. I've found a solution though. The second point that indicates the radius of the circle also snaps to other points. I didn't know that. So now I can start dragging from one end point of the given line and end on the other point(not eye ball it) to make sure the circle is the exactly that radius.
- StavrosK 10y agoSolutions are accepted, no matter how many moves. It sounds to me you may be eyeing things. You need to start from a predefined point and end at another predefined point, you can't just end at random places, it won't give you the solution then.
- mcpherrinm 10y agoI've played this game and found solutions that I really believe are correct but aren't accepted, and need to redo the solution another way. The game is certainly not perfect. But I'd say overall it's a great game. It's been my commute puzzler for a while now. Trying to get optimal solutions can get really tricky.
- CarolineW 10y agoI've played this a lot and introduced others to it. When others make the claim you do here I have always found that their solutions look good and plausible, but are not exact. It may be that you are the exception, and that your constructions are correct, but as with compiler errors, the problem is statistically more likely to be in your construction. I'm sure the devs would love to see a construction that is proven to be correct, but is not recognised by the system. Having a proof is key.
- BigJono 10y agoKeeping in mind that I have absolutely no idea what I'm doing, was never taught any of this stuff at school, and am stone cold stuck on 1.5, why is this not a solution to the Rhombus in Rectangle problem? It seems to match perfectly to the diagram and doesn't have any red points. http://i.imgur.com/QDqptQ4.png http://i.imgur.com/QDqptQ4.png
- irishsultan 10y agoThat construction doesn't have 4 sides of equal length (although it's really close), which is required for the solution (you can see that by the 4 sides having a single mark through them, that indicates they have the same length)