321 CSE Worksheet 321 Informal Solutions
Thursday, November 30, 2003
1. procedure mult(n:positive integer,x:integer)
if n=1 then mult(n,x) := x else mult(n,x) := x + mult(n-1,x)
2. If n = 1 , then n = nx , and the algorithm correctly returns x . Assume that the algorithm
correctly computes kx . To compute (k + 1)x it recursively computes the product of k + 1 - 1 = k and x , and then adds x . By the inductive hypothesis, it computes that product correctly, so the answer returned is kx + x = (k + 1)x , which is correct.
3. If n = 0 , then the rst line of the program correctly returns the function value 1.
Assume that the program works correctly for n = k . If n = k + 1 , then the else clause is executed, and the value a times power(a,k) is returned. Since the latter is ak by inductive hypothesis, this equals a * ak = ak+1 , which is correct.
4. f (n) = n2 Let P(n) be f (n) = n2 .
Basis step: P(1) is true since f (1) = 1 = 12 , which follows form the denition of f. Inductive Step: Assume f (n) = n2 . Then f (n + 1) = f ((n + 1) 1) + 2(n + 1) 1 = f (n) + 2n + 1 = n2 + 2n + 1 = (n + 1)2 .
5. Let P(n) be a + (a + d) + . . . (a + nd) =
Basis Step: P(1) is true since a + (a + d) Inductive Step: Assume that P(k) is true. a+(a Then + d)+. . . (a + kd)+(a + (k + 1) d) = (k+1)(2a+kd) +a+(k + 1d) = 1 [2ak + 2a + k 2 d + 2a + 2kd + 2d] = 1 [2ak + 4a + k 2 d + 3kd + 2d] = 2 2 2 1 (k + 2) (2a + (k + 1)d) 2
6. Basis Step: When n = 1 there is one circle, and we can color the inside blue and the
(n+1)(2a+nd) . 2 = 2a + d = 2(2a+d) . 2
outside red to satisfy the conditions. Inductive Step: Assue the inductive hypothesis that if there are k circles, the the regions can be 2-colored such that no regions with a common boundary have the same color, and consider a situation with k + 1 circles. Remove one of the circles, producing a picture with k circles, and invoke the inductive hypothesis to color it in the prescribed manner. Then replace the removed circl...