Chapter 2 - Mathematical Review Functions Factorial - x!=1*2*3*…..*n Permutation - a rearrangement of a sequence Boolean variable - takes on 2 values TRUE, FALSE 0, 1 Floor - round down x Ceiling - round up x Modulus - mod - Remainder x mod y - remainder when divide x by y 10 mod 3 is 1

Logarithms logby=x  bx=y  blogby=y log mn = log m + log n XAXB=XA+B log m/n = log m - log n XA/XB=XA-B log nr = r log n (XA)B=XAB logan=logbn/logba 2n+2n=2*2n=2n+1

Recursion A recursive algorithm calls itself. Must have a base case to stop recursion Calls must pass a “smaller problem” to the function. This will cause the recursion to eventually stop. Factorial example: int fact(int n) { if (n<=1) return 1; else return n*fact(n-1) } Looks a bit like induction Better to use for’s and while’s when appropriate.

Summations Adding up a list of values f(1)+f(2)+f(3)+…+f(n) if f(i)=i then sum is n(n+1)/2 if f(i)=i2 then sum is n(n+1)(2n+1)/6 if f(i)=2i then sum is 2n+1–1 if f(i)=Ai then sum is (An+1 –1)/(A-1)

Recurrence Relations Many of our operations (functions) will be recursive. We will want to analyze the running time for a problem of size n. Recursion defines this in terms of some work plus the amount of time to solve a smaller problem (using the same algorithm). A mathematical representation of this time will look like: T(n) = T(n-1) + 1 Need the time for the base case T(1) = 0

Recurrence Relation Solving Many recurrence relations are solved using the brute force method. Expand it; see the pattern; solve it. For example: T(n) = T(n-1)+1 = T(n-1)+1+1 = T(1)+(n-1) = n-1 Another example: T(n) = T(n-1)+n; T(1)=1 T(n) = T(n-1)+n = T(n-2)+(n-1)+n = T(1)+2+…n = 1+2+3…+n = n(n+1)/2

Proof Techniques Proof by Contradiction Proof by induction Find a counterexample to prove theorem is false Assume the opposite of the conclusion is true. Use steps to prove a contradiction (for example 1=2), then the original theorem must be true. Proof by induction Show the theorem holds for a base case (n=0 or n=1) Assume the theorem holds for n=k-1 Prove the theorem is true for n=k Example: show 1+2+3+…+n=n(n+1)/2 for n=1 1=1(2)/2=1 assume 1+2+3+…+n-1=(n-1)(n)/2; show 1+2+…+n=n(n+1)/2 (n-1)(n)/2+n=[(n-1)(n)+2n]/2=[n(n-1+2)]/2=n(n+1)/2