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5 Pages

### Feb16

Course: FACULTY 475, Fall 2009
School: UMass Lowell
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Word Count: 758

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Tuesday: For read section 5.2 HOMEWORK Return assignments # through #3 Collect assignment #4 Arrange coordinates so that (0,0) is at upper left, with the first coordinate (n) indexing rows and the second coordinate (k) indexing columns: (0,0) (1,0) (1,1) (2,0) (2,1) (2,2) (3,0) (3,1) (3,2) (3,3) ... We have shown that the subsets of [n] of size k are in oneto-one correspondence with the paths from (0,0) to (n,k)...

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Tuesday: For read section 5.2 HOMEWORK Return assignments # through #3 Collect assignment #4 Arrange coordinates so that (0,0) is at upper left, with the first coordinate (n) indexing rows and the second coordinate (k) indexing columns: (0,0) (1,0) (1,1) (2,0) (2,1) (2,2) (3,0) (3,1) (3,2) (3,3) ... We have shown that the subsets of [n] of size k are in oneto-one correspondence with the paths from (0,0) to (n,k) in which each step is of type (i,j)(i+1,j) or (i,j)(i+1,j+1). That is, C(n,k) = p(n,k), where p(n,k) is defined to be the number of paths of this kind, starting at (0,0) and ending at (n,k). (Note: p(n,k) has nothing to do with P(n,k)! Brualdi uses this notation because `p' stands for `path', and I'm using this notation because Brualdi does.) Count paths: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 ... Each step is either downward or downward-and-rightward. Claim: p(n,k) = p(n-1,k) + p(n-1,k-1). Proof: Each path that ends at (n,k) either is a path from (0,0) to (n-1,k) followed by a downward edge or else is a path from (0,0) to (n-1,k-1) followed by a downward-andrightward edge. This gives Pascal's formula: C(n,k) = C(n-1,k) + C(n-1,k-1) Note that the case n=1 works fine too: C(1,k) = C(0,k) + C(0,k-1). We have C(0,k) = 1 if k=0, C(0,k) = 0 otherwise. We can use Pascal's formula to compute C(n,k) for specific values of n,k by reducing the problem to smaller values of n,k. E.g.: C(2,1) = C(1,1)+C(1,0) = (C(0,1)+C(0,0))+(C(0,0)+C(0,-1)) = (0+1)+(1+0) = 1+1 = 2. This method of computing is called ... recursion. Another name is ... dynamic programming. Another example of recursion is the Fibonacci recursion f(n) = f(n-1) + f(n-2) or the variant f(n) = f(n-1) + 2f(n-2). To start a recursion, we need initial conditions. E.g., if we use the variant of the Fibonacci recurrence given above and initial conditions f(1)=1 and f(2)=3, we get f(3) = 3+2 1 = 5 f(4) = 5+2 3 = 11 f(5) = 11+2 5 = 21 f(6) = 21+2 11 = 43 ... For the Pascal recursion, the initial conditions are C(0,k) = 1 if k=0, C(0,k) = 0 otherwise Pascal's triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 An idea related to recursion is proof by mathematical induction. Scientific induction: "The nth term of the sequence 1,3,5,11,21,43,... by given the initial conditions f(1)=1 and f(2)=3 and the recurrence relation f(n) = f(n-1)+2f(n-2) seems to be given by the formula f(n) = (2^(n+1)+(-1)^n)/3, Scientific induction is based on assumption that patterns persist. But not all patterns persist: e.g., 3, 5 and 7 are prime, but 9 isn't. Mathematical induction gives us a way to prove formulas that we guess by scientific induction. Theorem: If f(1) = 1, f(2) = 3, and f(n) = f(n-1)+2f(n-2) for all n 3, then f(n) = (2^(n+1)+(-1)^n)/3. Proof by induction: We can check that it's true for n=1 and n=2. Suppose we've proved the formula for n=1 through n=m-1. Then we can show that it's also true for n=m: f(m) = f(m-1) + 2f(m-2) = (2^((m-1)+1)+(-1)^(m-1))/3 + 2(2^((m-2)+1)+(-1)^(m-2))/3) = (2^m+(-1)^(m-1))/3 + 2(2^(m-1)+(-1)^(m-2))/3) = (2^m+(-1)^(m-1)+2^m+2(-1)^(m-2))/3 = (2^m+2^m-(-1)^(m)+2(-1)^(m))/3 = (2^(m+1)+(-1)^m)/3. Often we use a single symbol (n) instead of two (n and m). The supposition "we've proved the formula for smaller values of n" is called the induction hypothesis. Mathematicians refer to mathematical induction as simply "induction". Brualdi implicitly invokes proof-by-induction on page 127. Brualdi says: Let p(n,k) = the number of paths from (0,0) to (n,k). Since C(n,k) and p(n,k) satisfy the same recursion, and since they satisfy t...

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