Discrete Math Solved Homework Solution Sample



Problem Set 1

4.4 Questions 6 and 12 only:

7.1 Questions 2, 6, and 20 only:


7.2 Questions 3 and 6 only:

7.3 Questions 3 and 43 Only:


Problem Set 2


8.1 Questions 3 and 10 only:

8.2 Questions 3, 5, 21, 29 Only:

8.3 Question 5 only:


8.4 Questions 2 and 7 only:

8.5 Question 3 only:

Section 8.4, page 410, exercise 3






Set 1

Ex 4.4-

6) for n=5,


  1. Start
  2. n!=1 and n!=2
  3. return
  4. walk(4) + walk (3)
  5. walk (4)
  6. n!=1 and n!=2
  7. return
  8. walk(3) + walk(2)
  9. walk(3)
  10. n!=1 and n!=2
  11. return
  12. walk(2) + walk (1)
  13. walk(2)
  14. n!=1 but n=2
  15. return 2 to line 12
  16. walk(3)=2+1=3 return 3 to line 8
  17. walk(4)=3+2=5 return 5 to line 4
  18. walk(5)=5+3=8
  19. return 8
  20. Stop


12) If there is single element, return it.

Else return minimum of following.

a) Last Element

b) Value returned by recursive call

for n-1 elements.

int findminimum(int S[], int n)

{ if (n == 1)

return S[0];

return min(S[n-1], findminimum(S, n-1));


Proof by mathematical induction:

Base case: for n=1, let Sn={2}

As n=1, S0=2 is returned, hence true for n=1.

Now, suppose, for n=k, let the minimum element be ‘m’, then for n=k+1, say if one more element ‘o’ is added, the last element will be ‘o’ and now min(o,m) will be returned, hence proved.


Ex 7.1

2) Given sequence: 3,6,9,15,24,39,……

The recurrence relation: an=an-1+an-2

Initial conditions: a0=3, a1=6.


6) P= $2000, r=0.14, An=P(1+r)n;





20) The sequence satisfies the recurrence relation Sn+1=Sn+Sn-1, with initial conditions, S1=2, S2=3.

To show that Sn=fn+1 , n=1,2,……, where f denotes the Fibonacci sequence, we will use induction.

The base case where n=1, S1=f1+1=f2=2 and n=2, S2=f2+1=f3=3 is true.

Now, suppose Sk=fk+1 for all k<n, where n>=3;

We see that for: Sk+1= Sk+Sk-1=fk+1+fk=fk+2.

Hence proved.


Ex 7.2-

3) No, an=2nan-1 is not a linear homogeneous recurrence relation with constant coefficients as the coefficients is 2n which is not constant.


6) Yes, an=7an-2-6n-3 is a linear linear homogeneous recurrence relation with constant coefficients. The order of the relation is 2.


Ex 7.3-

3) For sequence, sn={‘C’,’G’,’J’,’M’,’X’}

To find if ‘C’ is in the sequence or not.


i<j //i=1, j=5, key=’C’, n=5.

k=(i+j)/2 = 3


key!=sk or 3

key<sk or 3 //search left half


i<j //i=1,j=2,key=’C’,n=5.

k=(i+j)/2 = 1


key=sk or 1 //found.

Return k(1)




bn=the number of multiplications in line 5 required for compute an.

Using mathematical induction,

When, n=1, b1=0

When, n=k, we can see that the multiplication will be performed for every integers from k till 1, for 1 bk=0, so bk for k, will be k-1 (excluding the case of 1).

Similarly, for n=k+1, bk+1=bk+1(as only one interger increment is there so bn will only increase by 1)


Hence proved.


Set 2

Ex 8.1

Solution 3:


It is a kind of undirected (cyclic) graph because here we need to show no. of games played between the teams only.

Solution 10: Path from a to a that passes through each edge exactly one time is a→ b→ c→ e→ b→ d→ e→ f→ c→ a .


Ex 8.2

Solution 3:

  1. A simple path

Solution 5:

  1. A cycle

Solution 21:


Simple paths:

Solution 29 :

Above graph has an Euler cycle. Because it has all vertices with non-zero degree are connected and all vertices have even degree.

Euler cycle: a→b→f→g→j→h→i→e→j→f→e→d→b→c→d→i→c→h→a

Ex 8.3

Solution 5:

Lemma: if G is a graph having Hamiltonian cycle then for every non-empty subset S⊆ V(G) ,

C(G-S)<=|S|. where C(G-S) is the no. of connected components after removal of S subset and |S| is cardinality of set S.

Since it is a necessary condition for having Hamiltonian cycle.



Let S={g}

|S|=1 and C(G-S)=2

Since it doesn’t hold the necessary condition i.e. C(G-S)<=|S|.

Hence, given graph G doesn’t have Hamiltonian cycle.


Ex 8.4

Solution 2:

Shortest path from a to g : a→b→c→g with length=3+2+6=11

a→b→c→f→g with length =3+2+2+4=11

Solution 7:

Algorithm Steps:

1. Make an empty set of shortest path tree that will contain all shortest paths.
2.  Initially set all length values as ∞. Allocate length value equals to 0 for the source vertex.
3. While set  does not contain all the vertices repeat step4 and step5.
4. Choose a vertex u that does not belong to the set and has a minimum length value and include it in the set.
5. update length value of all the adjacent vertices of u by checking through all the adjacent vertices. For every adjacent vertex v, if total of length value from source and length of edge (u-v), is smaller than length value of v, then amend the length value of v.


Ex 8.5

Solution 3: adjacency matrix is given as follows:


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