**Example 1:**

S = 4, 3, 3, 3, 1

Where n = 5 (no. of vertices)

**Step 1.** Degree of all vertices is less than or equal to n ( no.of vertices)

**Step 2.** Odd number vertices are four.

**Step 3. **There is no degree less than zero.

**Step 4. **Remove ‘4’ from the sequence and subtract 1 from the remaining new sequence and arrange again in non-increasing order to get

S = 2,2,2,0

**Step 5. **Again remove ‘2 ‘ from the sequence and subtracting 1 from the remaining new sequence and arrange in non-increasing order we get

S= 1,1,0

Repeating the above step

S= 0,0

**Step 6.** Since all the deg remaining in the sequence is zero, the given sequence is graphical.

**Example 2:**

Consider the degree sequence: S = 7, 5, 5, 4, 4, 4, 4, 3

Where n = 8 (no. of vertices)

**Step 1.** Degree of all vertices is less than or equal to n ( no.of vertices)

**Step 2.** Odd number vertices are four.

**Step 3. **There is no degree less than zero.

**Step 4. **Remove ‘7’ from the sequence and subtract 1 from the remaining new sequence and arrange again in non-increasing order to get

S = 4, 4, 3, 3, 3, 3, 2

**Step 5. **Now remove the first ‘4 ‘ from the sequence and subtract 1 from the remaining new sequence to get:

S = 3, 2, 2, 2, 3, 2

rearrange in non-increasing order to get:

S = 3, 3, 2, 2, 2, 2

Repeating the above step we get following degree sequences:

S = 2, 2, 2, 1, 1

S = 1, 1, 1, 1

S = 1, 1, 0

S = 0, 0

**Step 6.** Since all the deg remaining in the sequence is zero, the given sequence is graphical (or in other words, it is possible to construct a simple graph from the given degree sequence).

If during the recursive process I end up with a sequence of vertices that does not make a graph (checked by handshaking lemma) is the sequence non-graphical (i.e. fails the test)?