On Nov 22, 2010, at 4:17 PM, Ken Hegeland wrote: > The problem is 27.5.4 can be found at > http://htdp.org/2003-09-26/Book/curriculum-Z-H-34.html#node_sec_27.5 > > I am still on the beginning of this extended exercise and I am entirely lost. > I know how it works but can't get anything even remotely close. I was able to > create subtract from 27.5.2, but triangulate eludes me after a few days of > thinking. My first thought is to make something that takes either a > (listof(listof numbers)), or 3 (listof numbers), and have it output the first > part of triangulation. > > So for the lolon of '((2 2 3 10)(2 5 12 31)(4 1 -2 1)) I want to produce an > answer of > '((2 2 3 10)(3 9 21)(-3 -8 -19)) > > I feel like applying subtract to first and second in lolon produces (3 9 21), > but I am not sure how to make it continue on the rest of the list. I tried > (subtract2(subtract(first lolon)(second lolon))), which really doesn't work > and it seems obvious to me just from glancing at it. > > I am thinking back to how I did merge sort, create a function that merges 2 > lists, apply it to the first and second items in a lolon, then recursively > apply it to the rest of the list. > > In subtract2 my final goal is to apply subtract to the first and second item, > then recursively apply it to first and rest of the list, but I can't figure > out how to get this. > > I know the real question has to do with applying this over and over to any > lists of equal length, until you drop the first position to 0, and maybe I am > heading in the wrong direction, but it seems easier to break it into smaller > problems. I really want to complete this problem before moving on with the > book but I can't find anything that works, everything I think of causes > infinite loops and freezes my computer for 10 minutes. > > _________________________________________________ > For list-related administrative tasks: > http://lists.racket-lang.org/listinfo/users
Say you're starting from here: (a rectangle) > ((2 2 3 10) > (2 5 12 31) > (4 1 -2 1)) One step -- as described in the book gets you here: > ((2 2 3 10) > (0 3 9 21) > (0 -3 -8 -19)) If you cut the 0s out from row 2 through N, you get closer to the triangle: > ((2 2 3 10) > (3 9 21) > (-3 -8 -19)) This isn't quite a triangle yet. So how do you repeat this process? Eureka: find a sub-rectangle to which you could apply this trick again. Do so until you're out of naturally chosen sub-rectangles. Practically speaking, this means that you subtract the second row -1 times from the last: > > ((2 2 3 10) > (3 9 21) > (1 2)) Now you have got a triangle. Of course if the matrix has many more rows you need to repeat this step many more times. The good news: Typing in the ideas presented above -- assuming you have internalized Parts I - IV of HtDP -- will take you about as much time as typing the email. Well, perhaps I am slow typist -- Matthias _________________________________________________ For list-related administrative tasks: http://lists.racket-lang.org/listinfo/users
