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courses:cs211:winter2014:journals:shannon:chapter5 [2014/03/04 01:56] – nollets | courses:cs211:winter2014:journals:shannon:chapter5 [2014/03/12 01:47] (current) – [Chapter 5.5: Integer Multiplication] nollets | ||
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I would give this section a 7/10 as it was helpful but a little confusing for me. | I would give this section a 7/10 as it was helpful but a little confusing for me. | ||
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+ | ===== Chapter 5.2: Further Recurrence Relations===== | ||
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+ | This section generalizes recurrence relations and allows us to create a formula that we can use to determine the run time of later recurrence relations. The generalized formula (on page 215 of the text) must be broken down into 3 cases. The first when q=1, the second when q=2, and finally when q>2. (We went over the case where q>2 Wednesday in class). In order to determine the run time of the q>2 case, we unroll the recurrence relation //T(n) ≤ qT(n/2) + cn// and find that at any level j, we have q^j instances, each with size n/2^j, which will perform (q/2)^j*cn work. Then if we sum over these levels, we find that the run time of this recurrence relation is O(n^log2q). This means that as q increases so will the run time. This is the method we discussed in class. The book also walks through using partial substitution to determine the run time of the recurrence relation. Te section then looks at the case when q = 1. After unrolling the recurrence relation, we can see that the run time when q = 1 is O(n). This is the result of a geometric series with a decomposing exponent. | ||
+ | The section then looks at the recurrence relation //T(n) ≤ 2T(n/ | ||
+ | The motivation behind this section was to determine ways to figure out the run time of recurrence relations and to find a formula to calculate common recurrences. | ||
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+ | Practicing unrolling the recurrence relations in class was incredibly helpful as was having concrete examples of algorithms that have that particular recurrence relation. This section was a good follow up to our class discussion. I would give it a 8/10. | ||
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+ | ===== Chapter 5.3: Counting Inversions===== | ||
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+ | This section discusses the problem that occurs when trying to see how similar two people' | ||
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+ | We went over this algorithm pretty well in class and the section helped back it up. It was relatively helpful and clear. I would give it a 7/10. I felt that it was better explained and written out in the slide. | ||
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+ | ===== Chapter 5.4: Finding the Closest Pair of Points===== | ||
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+ | This algorithm BLEW MY MIND. | ||
+ | This section discussed the closet pair problem, where we want to find the closet pair of points in a plane. We first assume that no two points have the same x and y coordinate. The general idea of solving this problem is to divide the points into left and right sections, find the closest pair of points in each of these sections, and then determine if there are any pairs of points along the border of the two halves. In order to find the closest pairs of points in each half, we first sort the halves by their x and y coordinates (so that they are on a line) and look at the points closest to each other. Once we find the closest pair in each side, we can then create a distance delta. We then take that distance delta and look within the delta of the " | ||
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+ | This discussion in class was very helpful and the book was a good reinforcement of the discussion. | ||
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+ | I would give this section an 8/10 because this is a pretty interesting algorithm and a cool way to find the shortest path. | ||
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+ | ===== Chapter 5.5: Integer Multiplication===== | ||
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+ | In this section we look at a way to make the multiplication of two integers run in less than O(n^2) time. In order to do this, rather than using the traditional method, in which you multiply each digit in one integer by the other digits in the other integer and then adding the results together. | ||
+ | The change the algorithm, the section first looked at splitting the multiplication process into 4 parts of n/2 size. However, this runs in O(n^2) time, which is not an improvement. | ||
+ | If we instead divide the multiplication into 3 parts of n/2 size, we can have the algorithm run in (n^1.59) time. The algorithm (one page 233 in the text)finds a way to split the multiplication process into these three parts. We then write x1y1*2^n+(p-x1y1-x0y0)*2^n/ | ||
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+ | This section was short and sweet, which was nice. I did not know that was how you multiplied binary numbers (we learned a different way in 210). I would give it a 8/10. |