1 Recursion, Recurrences and Induction Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong

2 e.g.1 (Page 7) Tower of Hanoi When n = 1 (i.e., no. of disks = 1) I want to move all n disks from i to j i j{i, j}

3 e.g.1 Tower of Hanoi When n = 1 (i.e., no. of disks = 1) I want to move all n disks from i to j i j{i, j} We move ONE disk from i to j. No. of disk moves = 1 Disk move:Moving ONE disk from i to j

4 e.g.1 Tower of Hanoi When n = 2 (i.e., no. of disks = 2) I want to move all n disks from i to j i j{i, j} Top disk Bottom disk

5 e.g.1 Tower of Hanoi When n = 2 (i.e., no. of disks = 2) I want to move all n disks from i to j i j{i, j}

6 e.g.1 Tower of Hanoi When n = 2 (i.e., no. of disks = 2) I want to move all n disks from i to j i j{i, j} We move ONE top disk from i to {i, j}. No. of disk moves = 1 Disk move: Moving ONE top disk from i to {i, j}

7 e.g.1 Tower of Hanoi When n = 2 (i.e., no. of disks = 2) I want to move all n disks from i to j i j{i, j} We move ONE bottom disk from i to j. No. of disk moves = 1 Disk move: Moving ONE top disk from i to {i, j} Moving ONE bottom disk from i to j 2

8 e.g.1 Tower of Hanoi When n = 2 (i.e., no. of disks = 2) I want to move all n disks from i to j i j{i, j} We move ONE top disk from {i, j} to j. No. of disk moves = 1 Disk move: Moving ONE top disk from i to {i, j} Moving ONE bottom disk from i to j 2 Moving ONE top disk from {i, j} to j 3

9 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} Top disk group containing n-1 disks Bottom disk Virtual top box

10 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j}

11 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} We move ONE top disk group from i to {i, j}. No. of disk moves = no. of disk moves for one top disk group Disk move: Moving ONE top disk group from i to {i, j}

12 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} We move ONE bottom disk from i to j. No. of disk moves = no. of disk moves for one top disk group Disk move: Moving ONE top disk group from i to {i, j} Moving ONE bottom disk from i to j + 1

13 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} Disk move: Moving ONE top disk group from i to {i, j} Moving ONE bottom disk from i to j We move ONE top disk group from {i, j} to j.Moving ONE top disk group from {i, j} to j No. of disk moves = no. of disk moves for one top disk group no. of disk moves for one top disk group

14 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} Disk move: Moving ONE top disk group from i to {i, j} Moving ONE bottom disk from i to j We move ONE top disk group from {i, j} to j. Moving ONE top disk group from {i, j} to j No. of disk moves = 2. no. of disk moves for one top disk group + 1 No. of disk moves for n disks No. of disk moves for n-1 disks

15 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j Disk move: Moving ONE top disk group from i to {i, j} Moving ONE bottom disk from i to j Moving ONE top disk group from {i, j} to j No. of disk moves = 2. no. of disk moves for one top disk group + 1 No. of disk moves for n disks No. of disk moves for n-1 disks The top disk group corresponds to top n-1 disks top n-1 disks The bottom disk corresponds to the largest disk the largest disk Let M(n) be the total number of disk moves needed for n disks. M(n)M(n-1) Note that M(1) = 1

16 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j Disk move: Moving ONE top disk group from i to {i, j} Moving ONE bottom disk from i to j Moving ONE top disk group from {i, j} to j No. of disk moves = 2. no. of disk moves for one top disk group + 1 No. of disk moves for n disks top n-1 disks the largest disk M(n)M(n-1) Why is it correct? This is because we are implicitly using induction. We assume that we can move top n-1 disks correctly. Let p(n) is a statement that the algorithm is correct for n. Step 1: Prove that p(1) (i.e., the base case) is true. Step 2: Prove that “ p(n-1)  p(n) ” is true for all n > 1. Step 2(a): Assume that p(n-1) is true for n > 1. Verify that p(1) is true Step 2(b): According to p(n-1), we deduce that p(n) is true. Inductive Hypothesis Inductive Step Note that M(1) = 1

17 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j Disk move: Moving ONE top disk group from i to {i, j} Moving ONE bottom disk from i to j Moving ONE top disk group from {i, j} to j No. of disk moves = 2. no. of disk moves for one top disk group + 1 No. of disk moves for n disks top n-1 disks the largest disk M(n)M(n-1) Why is it correct? This is because we are implicitly using induction. We assume that we can move top n-1 disks correctly. Let p(n) is a statement that the algorithm is correct for n. Step 1: Prove that p(1) (i.e., the base case) is true. Note that M(1) = 1

18 e.g.1 Tower of Hanoi When n = 1 (i.e., no. of disks = 1) I want to move all n disks from i to j i j{i, j} Step 1: Prove that p(1) (i.e., the base case) is true.

19 e.g.1 Tower of Hanoi When n = 1 (i.e., no. of disks = 1) I want to move all n disks from i to j i j{i, j} This case is obviously true. Step 1: Prove that p(1) (i.e., the base case) is true.

20 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} Step 2: Prove that “ p(n-1)  p(n) ” is true for all n > 1. Step 2(a): Assume that p(n-1) is true for n > 1. Step 2(b): According to p(n-1), we deduce that p(n) is true.

21 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} We move ONE top disk group from i to {i, j}. Step 2: Prove that “ p(n-1)  p(n) ” is true for all n > 1. Step 2(a): Assume that p(n-1) is true for n > 1. Step 2(b): According to p(n-1), we deduce that p(n) is true. We assume that moving n-1 disks is correct.

22 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} We move ONE bottom disk from i to j. Step 2: Prove that “ p(n-1)  p(n) ” is true for all n > 1. Step 2(a): Assume that p(n-1) is true for n > 1. Step 2(b): According to p(n-1), we deduce that p(n) is true.

23 e.g.1 Tower of Hanoi n can be any positive integer I want to move all n disks from i to j i j{i, j} We move ONE top disk group from {i, j} to j. Step 2: Prove that “ p(n-1)  p(n) ” is true for all n > 1. Step 2(a): Assume that p(n-1) is true for n > 1. Step 2(b): According to p(n-1), we deduce that p(n) is true. We assume that moving n-1 disks is correct.

24 e.g.2 (Page 11) Given M(1) = 1 and M(n) = 2M(n-1) + 1 for n > 1 What are the values of M(1), M(2), M(3), M(4) and M(5)? M(1) =1 M(2) =2M(1) + 1= = 3 M(3) =2M(2) + 1= = 7 M(4) =2M(3) + 1= = 15 M(5) =2M(4) + 1= = It “ seems ” that M(n) = 2 n -1

25 e.g.3 (Page 12) Illustration of “Proof by mathematical induction” (Weak Induction) We are going to prove the following claim C: statement P(n) is true for each positive integer n, namely 1, 2, … P(1)true P(2) P(3) P(4) P(5) … Step 1: Prove that P(1) (i.e., the base case) is true. Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 1. true Step 2(a): Assume that P(n-1) is true for n > 1. Verify that P(1) is true Step 2(b): According to P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step

26 e.g.4 (Page 17) Let S(n) be the number of subsets of a set X of size n. What are the values of S(0), S(1), S(2) and S(3)? When n = 0,X = {}There is a subset, namely an empty set (i.e., {}) When n = 1,X = {1}There are 2 subsets, namely {} and {1}. Thus, S(0) = 1 Thus, S(1) = 2 When n = 2,X = {1, 2}There are 4 subsets, namely {}, {1}, {2}, {1, 2}. Thus, S(2) = 4 When n = 3,X = {1, 2, 3}There are 8 subsets, namely {},{1},{2},{3},{1, 2},{1, 3},{2, 3},{1, 2, 3} Thus, S(3) = 8 It “ seems ” that S(n) = 2 n

27 e.g.4 Consider the 8 subsets of {1, 2, 3} {}{1}{2}{1, 2} {3}{3}{1, 3}{2, 3}{1, 2, 3} All 4 possible subsets of {1, 2} Each set is the same as the above set by adding a number 3 All possible subsets of a set of size 2. The total number of such subsets is equal to S(2). All possible subsets of a set of size 2 by adding a number 3 The total number of such subsets is equal to S(2). The total number of subsets of a set of size 3 (i.e., S(3) ) is equal to2. S(2) In general, we derive that S(n) = 2. S(n-1) We want to find a recursion formula for S(n).

28 e.g.5 (Page 21) time 0123 n-1n … Raymond borrows a loan with initial amount A (e.g., $100,000) from the bank. Raymond needs to return a monthly payment M (e.g., $1000) every month. Raymond returns a monthly payment M (e.g., $1000) to the bank. Raymond returns M Suppose T(n) is the amount of money in the bank at the n-th month. T(0)T(1)T(2)T(3)T(n-1)T(n)

29 e.g.5 time 0123 n-1n … Raymond borrows a loan with initial amount A (e.g., $100,000) from the bank. Raymond needs to return a monthly payment M (e.g., $1000) every month. Raymond returns a monthly payment M (e.g., $1000) to the bank. Raymond returns M Suppose T(n) is the amount of money in the bank at the n-th month. T(0)T(1)T(2)T(3)T(n-1)T(n) =A =T(0)*(1+0.01p/12) – M Suppose the interest rate is p% per year Suppose the interest rate is (p/12)% per month. Suppose the interest rate is (0.01p/12) per month. =T(1)*(1+0.01p/12) – M =T(n-1)*(1+0.01p/12) – M …

30 e.g.6 (Page 24) T(0) = b T(n) = rT(n-1) + a if n > 0 Derive a closed form for T(n) by a “ top-down ” approach (or called iterating recurrence) T(0) = b T(n) = rT(n-1) + a if n > 0

31 e.g.6 T(0) = b T(n) = rT(n-1) + a if n > 0 T(n) = rT(n-1) + a = r(rT(n-2) + a) + a = r 2 T(n-2) + ra + a = r 2 (rT(n-3) + a) + ra + a = r 3 T(n-3) + r 2 a + ra + a = r 3 (rT(n-4) + a) + r 2 a + ra + a = r 4 T(n-4) + r 3 a + r 2 a + ra + a = … = r n T(n-n) + r n-1 a + r n-2 a + … + ra + a = r n T(0) + a(r n-1 + r n-2 + … + r + 1) = r n T(0) + a(r n-1 + r n-2 + … + r + r 0 ) = r n b + a  r i Note that  r i = 1 – r n 1 – r if r  1 = r n b + a 1 – r n 1 – r

32 e.g.7 (Page 25) T(0) = b T(n) = rT(n-1) + a if n > 0 Derive a closed form for T(n) by a “ bottom-up ” approach T(0) = b T(n) = rT(n-1) + a if n > 0

33 e.g.6 T(0) = b T(n) = rT(n-1) + a if n > 0 T(0) = b T(1) = rT(0) + a= rb + a T(2) = rT(1) + a= r(rb+a) + a= r 2 b + ra + a T(3) = rT(2) + a= r(r 2 b + ra + a) + a= r 3 b + r 2 a+ ra + a … T(n) = r n b + r n-1 a + r n-2 a + … + ra + a = r n b + a(r n-1 + r n-2 + … + r + 1) = r n b + a(r n-1 + r n-2 + … + r + r 0 ) = r n b + a  r i Note that  r i = 1 – r n 1 – r if r  1 = r n b + a 1 – r n 1 – r

34 e.g.7 (Page 31) We want to prove the following corollary. Corollary 4.2: The formula for the sum of a geometric series with r  1 is  r i = 1 – r n 1 – r

35 e.g.7 Corollary 4.2: The formula for the sum of a geometric series with r  1 is  r i = 1 – r n 1 – r Theorem 4.1 is shown as follows. Theorem 4.1: If T(0) = b T(n) = rT(n-1) + a if n > 0 then = r n b + a 1 – r n 1 – r T(n) Let T(0) = 0 Let T(n) =  r i for n > 0 We want to show that T(n) = 1 – r n 1 – r by using Theorem 4.1 (which is related to recursion). We want to write T(n) in a recursion form. Consider T(n) = r n-1 + r n-2 + … + r + r 0 = (r n-1 + r n-2 + … + r) + r 0 = r(r n-2 + r n-3 + … + r 0 ) + 1 = r T(n-1) + 1 This is in a recursion form. We know that b = 0 We know that a = 1 According to Theorem 4.1, we have = r n b + a 1 – r n 1 – r T(n) = r n – r n 1 – r 1 – r n 1 – r = Done

36 e.g.8 (Page 32) We want to prove the following theorem. Lemma 4.3: Let r  1 be a positive value independent of n. Let t(n) be the largest term in the geometric series Then, the value of the geometric series is O(t(n)).  r i e.g.,If r = 0.5, then we have the value V of geometric series is equal to … n … n-1 The largest term t(n) in this series is equal to1 (= ) e.g.,If r = 2, then we have the value V of geometric series is equal to … + 2 n … + 2 n-1 The largest term t(n) in this series is equal to2 n-1 Lemma 4.3 states that V = O(t(n)) = O(1) Lemma 4.3 states that V = O(t(n)) = O(2 n-1 ) Why is it correct?

37 e.g.8 Lemma 4.3: Let r  1 be a positive value independent of n. Let t(n) be the largest term in the geometric series Then, the value of the geometric series is O(t(n)).  r i Consider two case. Case 1: r < 1 Case 2: r > 1 From our previous slide, we know that t(n) = r 0 = 1 From our previous slide, we know that t(n) = r n-1 Consider Case 1  r i We know that = 1 – r n 1 – r < 1 Note that r is a constant. 1 1 – r Thus, is a constant. = O(1) = O(t(n)) Note that we can show that =  (t(n))  r i

38 e.g.8 Lemma 4.3: Let r  1 be a positive value independent of n. Let t(n) be the largest term in the geometric series Then, the value of the geometric series is O(t(n)).  r i Consider two case. Case 1: r < 1 Case 2: r > 1 From our previous slide, we know that t(n) = r 0 = 1 From our previous slide, we know that t(n) = r n-1 Consider Case 2  r i We know that = 1 – r n 1 – r Note that r is a constant. r 1 – r Thus, is a constant. = O(r n-1 ) = O(t(n)) = r n – 1 r – 1 < r n r – 1 = r r n-1 Note that we can show that =  (t(n))  r i

39 e.g.9 (Page 35) First-order Linear Recurrence Dependent on one-step backward (i.e., T(n-1)) E.g., T(n) = f(n)T(n-1) + g(n) Second-order Linear Recurrence Dependent on two-steps backward (i.e., T(n-2)) E.g., T(n) = T(n-1) +2T(n-2) First-order Non-Linear Recurrence Linear means the power of T(n-1) is 1. Non-linear means the power of T(n-1) is not 1. E.g., T(n) = (T(n-1)) We are interested in this recurrence where f(n) is equal to a constant r.

40 e.g.10 (Page 37) Prove the following Theorem 4.5 by induction. Theorem 4.5: For any positive constants a and r, and any function g defined on the non-negative integers, the solution to the first-order linear recurrence is T(n) = rT(n-1) + g(n) if n > 0, a if n = 0,  r n-i T(n) = r n a + g(i) (*)

41 e.g.10 Theorem 4.5: For any positive constants a and r, and any function g defined on the non-negative integers, the solution to the first-order linear recurrence is T(n) = rT(n-1) + g(n) if n > 0, a if n = 0,  r n-i T(n) = r n a + g(i) (*) Step 1: Prove that P(0) (i.e., the base case) is true. Let P(n) be “ ”  r n-i T(n) = r n a + g(i) T(0) = a (by definition) = r 0 a We want to show that  r 0-i T(0) = r 0 a + g(i) That is, T(0) = r 0 a Thus, P(0) is true.

42 e.g.10 Theorem 4.5: For any positive constants a and r, and any function g defined on the non-negative integers, the solution to the first-order linear recurrence is T(n) = rT(n-1) + g(n) if n > 0, a if n = 0,  r n-i T(n) = r n a + g(i) (*) Let P(n) be “ ”  r n-i T(n) = r n a + g(i) Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 0. Step 2(a): Assume that P(n-1) is true for n > 0.  r (n-1)-i T(n-1) = r n-1 a + g(i) That is, Step 2(b): According to P(n-1), we deduce that P(n) is true. Consider T(n)= rT(n-1) + g(n) (by definition)  r (n-1)-i r n-1 a + g(i) = r( ) + g(n) We want to show that  r n-i T(n) = r n a + g(i)  r (n-1)-i r n a + r g(i) = + g(n)  r 1+(n-1)-i r n a + g(i) = + g(n)  r n-i r n a + g(i) = + r n-n g(n)  r n-i r n a + g(i) =

43 e.g.10 Theorem 4.5: For any positive constants a and r, and any function g defined on the non-negative integers, the solution to the first-order linear recurrence is T(n) = rT(n-1) + g(n) if n > 0, a if n = 0,  r n-i T(n) = r n a + g(i) (*) Let P(n) be “ ”  r n-i T(n) = r n a + g(i) Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 0. Step 2(a): Assume that P(n-1) is true for n > 0.  r (n-1)--i T(n-1) = r n-1 a + g(i) That is, Step 2(b): According to P(n-1), we deduce that P(n) is true. We want to show that  r n-i T(n) = r n a + g(i) Thus, P(n) is true. We prove that “ P(n-1)  P(n) ” is true for all n > 0 By Mathematical Induction,  r n-i T(n) = r n a + g(i)

44 e.g.11 (Page 39) We want to find the closed form of T(n) where T(0) = 6 T(n) = 4T(n-1) + 2 n if n > 0 T(0) = 6 T(n) = 4T(n-1) + 2 n if n > 0

45 e.g.11 T(0) = 6 T(n) = 4T(n-1) + 2 n if n > 0 Theorem 4.5: For any positive constants a and r, and any function g defined on the non-negative integers, the solution to the first-order linear recurrence is T(n) = rT(n-1) + g(n) if n > 0, a if n = 0,  r n-i T(n) = r n a + g(i) (*) We can write the closed form directly by using Theorem 4.5. We haver = 4g(n) = 2 n a= 6 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 4 n n-i. 2 i = 4 n n 4 -i. 2 i

46 e.g.11 T(0) = 6 T(n) = 4T(n-1) + 2 n if n > 0 We can write the closed form directly by using Theorem 4.5. We haver = 4g(n) = 2 n a= 6 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 4 n n-i. 2 i = 4 n n 4 -i. 2 i = 4 n n (2 2 ) -i. 2 i = 4 n n 2 -2i. 2 i = 4 n n 2 -i = 4 n n 1 2 ( ) i = 4 n n 1 2 ( ) i = 4 n n 1 2 ( ) i 1 2 = 4 n n – ( ½ ) n 1 – ½ = 4 n n – ( ½ ) n ½ = 4 n n. (1 – ( ½ ) n ) = 4 n n – 4 n ( ½ ) n = 4 n. 7 – 2 2n. 2 -n = 4 n. 7 – 2 n

47 e.g.12 (Page 40) We want to find the closed form of T(n) where T(0) = 10 T(n) = 3T(n-1) + n if n > 0 T(0) = 10 T(n) = 3T(n-1) + n if n > 0

48 e.g.12 T(0) = 10 T(n) = 3T(n-1) + n if n > 0 Theorem 4.5: For any positive constants a and r, and any function g defined on the non-negative integers, the solution to the first-order linear recurrence is T(n) = rT(n-1) + g(n) if n > 0, a if n = 0,  r n-i T(n) = r n a + g(i) (*) We can write the closed form directly by using Theorem 4.5. We haver = 3g(n) = na= 10 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 3 n n-i. i= 3 n n 3 -i. i= 3 n n i. 3 -i

49 e.g.12 T(0) = 10 T(n) = 3T(n-1) + n if n > 0 We can write the closed form directly by using Theorem 4.5. We haver = 3g(n) = na= 10 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 3 n n-i. i= 3 n n 3 -i. i= 3 n n i. 3 -i Theorem 4.6: For any real number x  1, i. x i = nx n+2 – (n+1)x n+1 + x (1-x) 2 Suppose that we know the following theorem. i. 3 -i =i. (1/3) i n(1/3) n+2 – (n+1)(1/3) n+1 + (1/3) (1-1/3) 2 = n. 3 -(n+2) – (n+1)3 -(n+1) (2/3) 2 =. (n. 3 -(n+2) – (n+1)3 -(n+1) ) = 32322

50 e.g.12 T(0) = 10 T(n) = 3T(n-1) + n if n > 0 We can write the closed form directly by using Theorem 4.5. We haver = 3g(n) = na= 10 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 3 n n-i. i= 3 n n 3 -i. i= 3 n n i. 3 -i Theorem 4.6: For any real number x  1, i. x i = nx n+2 – (n+1)x n+1 + x (1-x) 2 Suppose that we know the following theorem. i. 3 -i. (n. 3 -(n+2) – (n+1)3 -(n+1) ) = (n. 3 -(n+2)+2 – (n+1)3 -(n+1) ) = 12. (n. 3 -n – (n+1)3 -n ) = 1 4. (n. 3 -n – (n+1)3. 3 -n + 3) = 1 4

51 e.g.12 T(0) = 10 T(n) = 3T(n-1) + n if n > 0 We can write the closed form directly by using Theorem 4.5. We haver = 3g(n) = na= 10 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 3 n n-i. i= 3 n n 3 -i. i= 3 n n i. 3 -i Theorem 4.6: For any real number x  1, i. x i = nx n+2 – (n+1)x n+1 + x (1-x) 2 Suppose that we know the following theorem. i. 3 -i. (n. 3 -n – (n+1)3. 3 -n + 3) = 1 4. (n. 3 -n – (3n+3). 3 -n + 3) = 1 4. (n. 3 -n – 3n. 3 -n – n + 3) = 1 4. ( – 2n. 3 -n – n + 3) = 1 4

52 e.g.12 T(0) = 10 T(n) = 3T(n-1) + n if n > 0 We can write the closed form directly by using Theorem 4.5. We haver = 3g(n) = na= 10 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 3 n n-i. i= 3 n n 3 -i. i= 3 n n i. 3 -i Theorem 4.6: For any real number x  1, i. x i = nx n+2 – (n+1)x n+1 + x (1-x) 2 Suppose that we know the following theorem. i. 3 -i. ( – 2n. 3 -n – n + 3) = 1 4. [( – 2n – 3). 3 -n + 3] = 1 4 = – 2n – n i. 3 -i = – 2n – n + 3 4

53 e.g.12 T(0) = 10 T(n) = 3T(n-1) + n if n > 0 We can write the closed form directly by using Theorem 4.5. We haver = 3g(n) = na= 10 Thus, by using Theorem 4.5, we have  r n-i T(n) = r n a + g(i) = 3 n n-i. i= 3 n n 3 -i. i= 3 n n i. 3 -i Suppose that we know the following theorem. i. 3 -i = – 2n – n = 3 n n ( – 2n – n ) = 3 n – 2n – n. 3 4 = 3 n. (10 + ) – 2n = 3 n. – 2n