Draw the diagram for an NFA for each of the following languages. Use as few states and as few transitions as possible. Don
’
t just give a DFA, unless you are convinced it is necessary. a
.

{
x
{
a

Question

Draw the diagram for an NFA for each of the following languages. Use as few states and as few transitions as possible. Don
’
t just give a DFA, unless you are convinced it is necessary. a
.
 
{
x 
{
a
,
 ﻿b
}
*
 
|
 ﻿x contains at least 
3
 ﻿as
}
 ﻿b
.
 
{
x 
{
a
,
 ﻿b
}
*
 
|
 ﻿x starts with at least 
3
 ﻿consecutive as
}
 ﻿c
.
 
{
x 
{
a
,
 ﻿b
}
*
 
|
 ﻿x ends with at least 
3
 ﻿consecutive as
}
 ﻿d
.
 
{
x 
{
a
,
 ﻿b
}
*
 
|
 ﻿x contains at least 
3
 ﻿consecutive as
}
 ﻿e
.
 
{
x 
{
a
,
 ﻿b
}
*
 
|
 ﻿x has no two consecutive as
}
 ﻿f
.
 
{
axb 
|
 ﻿x 
{
a
,
 ﻿b
}
*
}
 ﻿g
.
 
{
x 
{
0
,
 
1
}
*
 
|
 ﻿x ends in either 
0
0
0
1
 ﻿or 
1
0
0
0
}
 ﻿g
.
 
{
x 
{
0
,
 
1
}
*
 
|
 ﻿x ends in either 
0
0
0
1
 ﻿or 
1
0
0
0
}
 ﻿h
.
 
{
x 
{
0
,
 
1
}
*
 
|
 ﻿x either starts with 
0
0
0
 ﻿or ends with 
0
0
0
,
 ﻿or both
}

VIPSolutions · Accepted Answer

Here's a concise overview of the NFA diagrams for each language:

---

**a. Strings over {a, b} containing at least 3 'a's**

- States: `q0` (start, 0 'a's), `q1` (1 'a'), `q2` (2 'a's), `q3` (3+ 'a's, accepting)
- Transitions:
  - From `q0`: 'a' → `q1`, 'b' → `q0`
  - From `q1`: 'a' → `q2`, 'b' → `q1`
  - From `q2`: 'a' → `q3`, 'b' → `q2`
  - From `q3`: 'a'/'b' → `q3`

---

**b. Strings starting with at least 3 consecutive 'a's**

- States: `q0` (start), `q1` (1 'a'), `q2` (2 'a'), `q3` (3 'a's, accepting)
- Transitions:
  - `q0`: 'a' → `q1`, 'b' → `q0`
  - `q1`: 'a' → `q2`, 'b' → `q0`
  - `q2`: 'a' → `q3`, 'b' → `q0`
  - `q3`: 'a'/'b' → `q3`

---

**c. Strings ending with at least 3 consecutive 'a's**

- States: `q0` (start), `q1` (1 'a'), `q2` (2 'a'), `q3` (3 'a's, accepting)
- Transitions:
  - `q0`: 'a' → `q1`, 'b' → `q0`
  - `q1`: 'a' → `q2`, 'b' → `q0`
  - `q2`: 'a' → `q3`, 'b' → `q0`
  - `q3`: 'a'/'b' → `q3`

*(Note: For ending with 3 'a's, the automaton tracks the last characters; accept when in `q3` after processing the entire string.)*

---

**d. Strings containing at least 3 consecutive 'a's (anywhere)**

- Same as (a): states `q0` (start), `q1`, `q2`, `q3` (accept)
- Transitions same as (a)

---

**e. Strings with no two consecutive 'a's**

- States: `q0` (start, no 'a' just seen), `q1` (just saw an 'a', cannot see another 'a' now)
- Transitions:
  - `q0`: 'a' → `q1`, 'b' → `q0`
  - `q1`: 'a' → dead state (or no transition), 'b' → `q0`
- Accept states: `q0` only

---

**f. Strings of the form `a b x` where `x` ∈ {a, b}***

- Start with 'a', then 'b', then any combination
- States:
  - `q0` (start)
  - `q1` after 'a'
  - `q2` after 'a b'
  - `q3` accepting after 'a b' and any suffix
- Transitions:
  - `q0`: 'a' → `q1`
  - `q1`: 'b' → `q2`
  - `q2`: 'a'/'b' → `q3` (accept)
  - `q3`: 'a'/'b' → `q3`

---

**g. Strings over {0,1} ending with '0001' or '1000'**

- States track the last few bits:
  - For '0001':
    - `q0` (start)
    - `q1` (last '0')
    - `q2` (last '00')
    - `q3` (last '000')
    - `q_accept1` (if next bit is '1')
  - For '1000':
    - Similar states, starting with '1'

- Alternatively, combine into a minimal automaton with states representing recent suffixes.

---

**h. Strings over {0,1} ending with '0001' or '1000' (same as g)**

- Same as (g), minimal states for suffix detection.

---

**i. Strings over {0,1} either starting with '000'

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