Deterministic Finite Automaton (DFA) can be used to check whether a number "num" is divisible by "k" or not. If the number is not divisible, remainder can also be obtained using DFA.
We consider the binary representation of 'num' and build a DFA with k states. The DFA has transition function for both 0 and 1. Once the DFA is built, we process ‘num’ over the DFA to get remainder.
Let us walk through an example. Suppose we want to check whether a given number 'num' is divisible by 3 or not. Any number can be written in the form: num = 3*a + b where 'a' is the quotient and 'b' is the remainder.
For 3, there can be 3 states in DFA, each corresponding to remainder 0, 1 and 2. And each state can have two transitions corresponding 0 and 1 (considering the binary representation of given 'num').
The transition function F(p, x) = q tells that on reading alphabet x, we move from state p to state q. Let us name the states as 0, 1 and 2. The initial state will always be 0. The final state indicates the remainder. If the final state is 0, the number is divisible.
In the above diagram, double circled state is final state.
1. When we are at state 0 and read 0, we remain at state 0.
2. When we are at state 0 and read 1, we move to state 1, why? The number so formed(1) in decimal gives remainder 1.
3. When we are at state 1 and read 0, we move to state 2, why? The number so formed(10) in decimal gives remainder 2.
4. When we are at state 1 and read 1, we move to state 0, why? The number so formed(11) in decimal gives remainder 0.
5. When we are at state 2 and read 0, we move to state 1, why? The number so formed(100) in decimal gives remainder 1.
6. When we are at state 2 and read 1, we remain at state 2, why? The number so formed(101) in decimal gives remainder 2.
The transition table looks like following:
state 0 1 _____________ 0 0 1 1 2 0 2 1 2
Let us check whether 6 is divisible by 3?
Binary representation of 6 is 110
state = 0
1. state=0, we read 1, new state=1
2. state=1, we read 1, new state=0
3. state=0, we read 0, new state=0
Since the final state is 0, the number is divisible by 3.
Let us take another example number as 4
state=0
1. state=0, we read 1, new state=1
2. state=1, we read 0, new state=2
3. state=2, we read 0, new state=1
Since, the final state is not 0, the number is not divisible by 3. The remainder is 1.
Note that the final state gives the remainder.
We can extend the above solution for any value of k. For a value k, the states would be 0, 1, .... , k-1. How to calculate the transition if the decimal equivalent of the binary bits seen so far, crosses the range k? If we are at state p, we have read p (in decimal). Now we read 0, new read number becomes 2*p. If we read 1, new read number becomes 2*p+1. The new state can be obtained by subtracting k from these values (2p or 2p+1) where 0 <= p < k.
Based on the above approach, following is the working code:
#include <bits/stdc++.h>
using namespace std;
// Function to build DFA for divisor k
void preprocess(int k, int Table[][2])
{
int trans0, trans1;
// The following loop calculates the
// two transitions for each state,
// starting from state 0
for (int state = 0; state < k; ++state)
{
// Calculate next state for bit 0
trans0 = state << 1;
Table[state][0] = (trans0 < k) ?
trans0 : trans0 - k;
// Calculate next state for bit 1
trans1 = (state << 1) + 1;
Table[state][1] = (trans1 < k) ?
trans1 : trans1 - k;
}
}
// A recursive utility function that
// takes a 'num' and DFA (transition
// table) as input and process 'num'
// bit by bit over DFA
void isDivisibleUtil(int num, int* state,
int Table[][2])
{
// process "num" bit by bit
// from MSB to LSB
if (num != 0)
{
isDivisibleUtil(num >> 1, state, Table);
*state = Table[*state][num & 1];
}
}
// The main function that divides 'num'
// by k and returns the remainder
int isDivisible (int num, int k)
{
// Allocate memory for transition table.
// The table will have k*2 entries
int (*Table)[2] = (int (*)[2])malloc(k*sizeof(*Table));
// Fill the transition table
preprocess(k, Table);
// Process ‘num’ over DFA and
// get the remainder
int state = 0;
isDivisibleUtil(num, &state, Table);
// Note that the final value
// of state is the remainder
return state;
}
// Driver Code
int main()
{
int num = 47; // Number to be divided
int k = 5; // Divisor
int remainder = isDivisible (num, k);
if (remainder == 0)
cout << "Divisible\n";
else
cout << "Not Divisible: Remainder is "
<< remainder;
return 0;
}
// This is code is contributed by rathbhupendra
#include <stdio.h>
#include <stdlib.h>
// Function to build DFA for divisor k
void preprocess(int k, int Table[][2])
{
int trans0, trans1;
// The following loop calculates the two transitions for each state,
// starting from state 0
for (int state=0; state<k; ++state)
{
// Calculate next state for bit 0
trans0 = state<<1;
Table[state][0] = (trans0 < k)? trans0: trans0-k;
// Calculate next state for bit 1
trans1 = (state<<1) + 1;
Table[state][1] = (trans1 < k)? trans1: trans1-k;
}
}
// A recursive utility function that takes a 'num' and DFA (transition
// table) as input and process 'num' bit by bit over DFA
void isDivisibleUtil(int num, int* state, int Table[][2])
{
// process "num" bit by bit from MSB to LSB
if (num != 0)
{
isDivisibleUtil(num>>1, state, Table);
*state = Table[*state][num&1];
}
}
// The main function that divides 'num' by k and returns the remainder
int isDivisible (int num, int k)
{
// Allocate memory for transition table. The table will have k*2 entries
int (*Table)[2] = (int (*)[2])malloc(k*sizeof(*Table));
// Fill the transition table
preprocess(k, Table);
// Process ‘num’ over DFA and get the remainder
int state = 0;
isDivisibleUtil(num, &state, Table);
// Note that the final value of state is the remainder
return state;
}
// Driver program to test above functions
int main()
{
int num = 47; // Number to be divided
int k = 5; // Divisor
int remainder = isDivisible (num, k);
if (remainder == 0)
printf("Divisible\n");
else
printf("Not Divisible: Remainder is %d\n", remainder);
return 0;
}
// Java program to implement the approach
import java.util.*;
class GFG {
// Function to build DFA for divisor k
static void preprocess(int k, int[][] Table)
{
int trans0, trans1;
// The following loop calculates the
// two transitions for each state,
// starting from state 0
for (int state = 0; state < k; ++state)
{
// Calculate next state for bit 0
trans0 = state << 1;
Table[state][0]
= (trans0 < k) ? trans0 : trans0 - k;
// Calculate next state for bit 1
trans1 = (state << 1) + 1;
Table[state][1]
= (trans1 < k) ? trans1 : trans1 - k;
}
}
// A recursive utility function that
// takes a 'num' and DFA (transition
// table) as input and process 'num'
// bit by bit over DFA
static int isDivisibleUtil(int num, int state,
int[][] Table)
{
// process "num" bit by bit
// from MSB to LSB
if (num != 0) {
state = isDivisibleUtil(num >> 1, state, Table);
state = Table[state][num & 1];
}
return state;
}
// The main function that divides 'num'
// by k and returns the remainder
static int isDivisible(int num, int k)
{
// Allocate memory for transition table.
// The table will have k*2 entries
int[][] Table = new int[k][2];
for (int i = 0; i < k; i++) {
Table[i][0] = 0;
Table[i][1] = 0;
}
// Fill the transition table
preprocess(k, Table);
// Process ‘num’ over DFA and
// get the remainder
int state = 0;
state = isDivisibleUtil(num, state, Table);
// Note that the final value
// of state is the remainder
return state;
}
// Driver Code
public static void main(String[] args)
{
int num = 47; // Number to be divided
int k = 5; // Divisor
int remainder = isDivisible(num, k);
if (remainder == 0)
System.out.println("Divisible");
else
System.out.println(
"Not Divisible: Remainder is " + remainder);
}
}
// This is code contributed by phasing17
# Python3 program to implement the approach
# Function to build DFA for divisor k
def preprocess(k, Table):
# The following loop calculates the
# two transitions for each state,
# starting from state 0
for state in range(k):
# Calculate next state for bit 0
trans0 = state << 1
if (trans0 < k):
Table[state][0] = trans0
else:
Table[state][0] = trans0 - k
# Calculate next state for bit 1
trans1 = (state << 1) + 1
if trans1 < k:
Table[state][1] = trans1
else:
Table[state][1] = trans1 - k
# A recursive utility function that
# takes a 'num' and DFA (transition
# table) as input and process 'num'
# bit by bit over DFA
def isDivisibleUtil(num, state, Table):
# process "num" bit by bit
# from MSB to LSB
if (num != 0):
state = isDivisibleUtil(num >> 1, state, Table)
state = Table[state][num & 1]
return state
# The main function that divides 'num'
# by k and returns the remainder
def isDivisible(num, k):
# Allocate memory for transition table.
# The table will have k*2 entries
Table = [None for i in range(k)]
for i in range(k):
Table[i] = [0, 0]
# Fill the transition table
preprocess(k, Table)
# Process ‘num’ over DFA and
# get the remainder
state = 0
state = isDivisibleUtil(num, state, Table)
# Note that the final value
# of state is the remainder
return state
# Driver Code
num = 47 # Number to be divided
k = 5 # Divisor
remainder = isDivisible(num, k)
if (remainder == 0):
print("Divisible")
else:
print("Not Divisible: Remainder is", remainder)
# This is code contributed by phasing17
// C# program to implement the approach
using System;
class GFG
{
// Function to build DFA for divisor k
static void preprocess(int k, int[, ] Table)
{
int trans0, trans1;
// The following loop calculates the
// two transitions for each state,
// starting from state 0
for (int state = 0; state < k; ++state) {
// Calculate next state for bit 0
trans0 = state << 1;
Table[state, 0]
= (trans0 < k) ? trans0 : trans0 - k;
// Calculate next state for bit 1
trans1 = (state << 1) + 1;
Table[state, 1]
= (trans1 < k) ? trans1 : trans1 - k;
}
}
// A recursive utility function that
// takes a 'num' and DFA (transition
// table) as input and process 'num'
// bit by bit over DFA
static int isDivisibleUtil(int num, int state,
int[, ] Table)
{
// process "num" bit by bit
// from MSB to LSB
if (num != 0) {
state = isDivisibleUtil(num >> 1, state, Table);
state = Table[state, num & 1];
}
return state;
}
// The main function that divides 'num'
// by k and returns the remainder
static int isDivisible(int num, int k)
{
// Allocate memory for transition table.
// The table will have k*2 entries
int[, ] Table = new int[k, 2];
for (int i = 0; i < k; i++) {
Table[i, 0] = 0;
Table[i, 1] = 0;
}
// Fill the transition table
preprocess(k, Table);
// Process ‘num’ over DFA and
// get the remainder
int state = 0;
state = isDivisibleUtil(num, state, Table);
// Note that the final value
// of state is the remainder
return state;
}
// Driver Code
public static void Main(string[] args)
{
int num = 47; // Number to be divided
int k = 5; // Divisor
int remainder = isDivisible(num, k);
if (remainder == 0)
Console.WriteLine("Divisible");
else
Console.WriteLine("Not Divisible: Remainder is "
+ remainder);
}
}
// This is code contributed by phasing17
// JavaScript program to implement the approach
// Function to build DFA for divisor k
function preprocess(k, Table)
{
let trans0, trans1;
// The following loop calculates the
// two transitions for each state,
// starting from state 0
for (let state = 0; state < k; ++state)
{
// Calculate next state for bit 0
trans0 = state << 1;
Table[state][0] = (trans0 < k) ?
trans0 : trans0 - k;
// Calculate next state for bit 1
trans1 = (state << 1) + 1;
Table[state][1] = (trans1 < k) ?
trans1 : trans1 - k;
}
}
// A recursive utility function that
// takes a 'num' and DFA (transition
// table) as input and process 'num'
// bit by bit over DFA
function isDivisibleUtil(num, state, Table)
{
// process "num" bit by bit
// from MSB to LSB
if (num != 0)
{
state = isDivisibleUtil(num >> 1, state, Table);
state = Table[state][num & 1];
}
return state;
}
// The main function that divides 'num'
// by k and returns the remainder
function isDivisible (num, k)
{
// Allocate memory for transition table.
// The table will have k*2 entries
Table = new Array(k);
for (let i = 0; i < k; i++)
Table[i] = [0, 0];
// Fill the transition table
preprocess(k, Table);
// Process ‘num’ over DFA and
// get the remainder
let state = 0;
state = isDivisibleUtil(num, state, Table);
// Note that the final value
// of state is the remainder
return state;
}
// Driver Code
let num = 47; // Number to be divided
let k = 5; // Divisor
let remainder = isDivisible (num, k);
if (remainder == 0)
console.log("Divisible");
else
console.log("Not Divisible: Remainder is " + remainder);
// This is code contributed by phasing17
Output:
Not Divisible: Remainder is 2
Time Complexity: O(k)
DFA based division can be useful if we have a binary stream as input and we want to check for divisibility of the decimal value of stream at any time.
Related Articles:
Check divisibility in a binary stream
Check if a stream is Multiple of 3
This article is compiled by Aashish Barnwal and reviewed by GeeksforGeeks team.