Here is the complete problem statement:
Given a rope of length n, you need to find the maximum number of pieces
you can make such that the length of every piece is in set {a, b, c} for
the given three values a, b, c
I know that the optimal solution can be achieved through Dynamic Programming, however, I have not learned that topic yet and I need to solve this problem recursively. With recursion, the main thing is to identify a subproblem and that's what I'm mainly having difficulty with doing. Can anyone give me an intuitive way to think of this problem? Sort of like a higher level description of the recursion if that makes sense. Is there an easier problem similar to this that I can try first that would help me solve this?
Thanks in advance.
It's already quite simple, with recursion we can just check all posibilities, in one step we can either cut away a piece of length a, b, or c so from problem of size n we get sup-problem of smaller size n-x
Of course we need a base case, so when n=0 we have succeeded so we can return 0, in case of n < 0 we have failed so we can return some negative infinity constant
Sample pseudo-code:
int solve(int n){
if(n < 0) return -123456789; //-Infinity
if(n == 0) return 0;
return 1 + max(solve(n-a), solve(n-b), solve(n-c));
}
going to dynamic programming is as simple as setting up memo lookup table
int solve(int n){
if(n < 0) return -123456789; //-Infinity
if(n == 0) return 0;
if(n in memo)return memo[n]
return memo[n] = 1 + max(solve(n-a), solve(n-b), solve(n-c));
}
int maxcut(int n, int a,int b,int c)
{
if(n==0) return 0;
if(n<0) return 1;
int result = max( maxcut(n-a,a,b,c), maxcut(n-b,a,b,c), maxcur(n-c,a,b,c));
if(res == -1) return -1;
return(result+1)
}
The way we should tackle the recursion problem is:
Finding the recursion case (Finding the subproblems)
Finding the base case (The last subproblem case we cannot break in subproblems)
Specific to this problem :
Recursion case: Cutting rope we all the possible values till we cannot break it further smaller subproblem.
Base case: a. It can be completely cut. (valid try)
b.It can't be completely cut. (invalid try)
int maxcut(int n, int a,int b,int c)
{
if(n==0) return 0; //base case a.
if(n<0) return -1; //base case b.
int result = max( maxcut(n-a,a,b,c), maxcut(n-b,a,b,c), maxcur(n-c,a,b,c)); //subproblems for all the cases
if(res == -1) return -1; // boundry coundtion
return(result+1); //to count the valid conditions and return to parent
}
here is the complete code for your problem
#include <iostream>
using namespace std;
int max(int a, int b, int c)
{
if (a > b)
{
if (a > c)
{
return a;
}
else
{
return c;
}
}
else
{
if (b > c)
{
return b;
}
else
{
return c;
}
}
}
int maxpiece(int l, int a, int b, int c)
{
int r;
if (l == 0)
{
return 0;
}
if (l<0)
{
return -1;
}
r = max(maxpiece(l-a, a, b, c), maxpiece(l-b, a, b, c), maxpiece(l-c, a, b, c));
if (r == -1)
return -1;
return r + 1;
}
int main()
{
int lenth;
cout << "enter rope lenth ";
cin >> lenth;
int p1, p2, p3;
cout << endl
<< "enter the only three parameters in which rope can be cut ";
cin >> p1 >> p2 >> p3;
cout << endl
<<"ans = "<< maxpiece(lenth, p1, p2, p3);
}
int cutRope(int n, int a, int b, int c){
// Base cases
if(n == 0) return 0;
if(n < 0) return -1;
int res = max(max(cutRope(n - a, a, b, c), cutRope(n - b, a, b, c)), cutRope(n - c, a, b, c));
if(res == -1) return -1;
return res + 1;
}
int main() {
cout << cutRope(23, 11, 9, 12) << endl;
return 0;
}
How do you calculate the least common multiple of multiple numbers?
So far I've only been able to calculate it between two numbers. But have no idea how to expand it to calculate 3 or more numbers.
So far this is how I did it
LCM = num1 * num2 / gcd ( num1 , num2 )
With gcd is the function to calculate the greatest common divisor for the numbers. Using euclidean algorithm
But I can't figure out how to calculate it for 3 or more numbers.
You can compute the LCM of more than two numbers by iteratively computing the LCM of two numbers, i.e.
lcm(a,b,c) = lcm(a,lcm(b,c))
In Python (modified primes.py):
def gcd(a, b):
"""Return greatest common divisor using Euclid's Algorithm."""
while b:
a, b = b, a % b
return a
def lcm(a, b):
"""Return lowest common multiple."""
return a * b // gcd(a, b)
def lcmm(*args):
"""Return lcm of args."""
return reduce(lcm, args)
Usage:
>>> lcmm(100, 23, 98)
112700
>>> lcmm(*range(1, 20))
232792560
reduce() works something like that:
>>> f = lambda a,b: "f(%s,%s)" % (a,b)
>>> print reduce(f, "abcd")
f(f(f(a,b),c),d)
Here's an ECMA-style implementation:
function gcd(a, b){
// Euclidean algorithm
while (b != 0){
var temp = b;
b = a % b;
a = temp;
}
return a;
}
function lcm(a, b){
return (a * b / gcd(a, b));
}
function lcmm(args){
// Recursively iterate through pairs of arguments
// i.e. lcm(args[0], lcm(args[1], lcm(args[2], args[3])))
if(args.length == 2){
return lcm(args[0], args[1]);
} else {
var arg0 = args[0];
args.shift();
return lcm(arg0, lcmm(args));
}
}
I would go with this one (C#):
static long LCM(long[] numbers)
{
return numbers.Aggregate(lcm);
}
static long lcm(long a, long b)
{
return Math.Abs(a * b) / GCD(a, b);
}
static long GCD(long a, long b)
{
return b == 0 ? a : GCD(b, a % b);
}
Just some clarifications, because at first glance it doesn't seams so clear what this code is doing:
Aggregate is a Linq Extension method, so you cant forget to add using System.Linq to your references.
Aggregate gets an accumulating function so we can make use of the property lcm(a,b,c) = lcm(a,lcm(b,c)) over an IEnumerable. More on Aggregate
GCD calculation makes use of the Euclidean algorithm.
lcm calculation uses Abs(a*b)/gcd(a,b) , refer to Reduction by the greatest common divisor.
Hope this helps,
I just figured this out in Haskell:
lcm' :: Integral a => a -> a -> a
lcm' a b = a`div`(gcd a b) * b
lcm :: Integral a => [a] -> a
lcm (n:ns) = foldr lcm' n ns
I even took the time to write my own gcd function, only to find it in Prelude! Lots of learning for me today :D
Some Python code that doesn't require a function for gcd:
from sys import argv
def lcm(x,y):
tmp=x
while (tmp%y)!=0:
tmp+=x
return tmp
def lcmm(*args):
return reduce(lcm,args)
args=map(int,argv[1:])
print lcmm(*args)
Here's what it looks like in the terminal:
$ python lcm.py 10 15 17
510
Here is a Python one-liner (not counting imports) to return the LCM of the integers from 1 to 20 inclusive:
Python 3.5+ imports:
from functools import reduce
from math import gcd
Python 2.7 imports:
from fractions import gcd
Common logic:
lcm = reduce(lambda x,y: x*y // gcd(x, y), range(1, 21))
Note that in both Python 2 and Python 3, operator precedence rules dictate that the * and // operators have the same precedence, and so they apply from left to right. As such, x*y // z means (x*y) // z and not x * (y//z). The two typically produce different results. This wouldn't have mattered as much for float division but it does for floor division.
Here it is in Swift.
// Euclid's algorithm for finding the greatest common divisor
func gcd(_ a: Int, _ b: Int) -> Int {
let r = a % b
if r != 0 {
return gcd(b, r)
} else {
return b
}
}
// Returns the least common multiple of two numbers.
func lcm(_ m: Int, _ n: Int) -> Int {
return m / gcd(m, n) * n
}
// Returns the least common multiple of multiple numbers.
func lcmm(_ numbers: [Int]) -> Int {
return numbers.reduce(1) { lcm($0, $1) }
}
Here is a C# port of Virgil Disgr4ce's implemenation:
public class MathUtils
{
/// <summary>
/// Calculates the least common multiple of 2+ numbers.
/// </summary>
/// <remarks>
/// Uses recursion based on lcm(a,b,c) = lcm(a,lcm(b,c)).
/// Ported from http://stackoverflow.com/a/2641293/420175.
/// </remarks>
public static Int64 LCM(IList<Int64> numbers)
{
if (numbers.Count < 2)
throw new ArgumentException("you must pass two or more numbers");
return LCM(numbers, 0);
}
public static Int64 LCM(params Int64[] numbers)
{
return LCM((IList<Int64>)numbers);
}
private static Int64 LCM(IList<Int64> numbers, int i)
{
// Recursively iterate through pairs of arguments
// i.e. lcm(args[0], lcm(args[1], lcm(args[2], args[3])))
if (i + 2 == numbers.Count)
{
return LCM(numbers[i], numbers[i+1]);
}
else
{
return LCM(numbers[i], LCM(numbers, i+1));
}
}
public static Int64 LCM(Int64 a, Int64 b)
{
return (a * b / GCD(a, b));
}
/// <summary>
/// Finds the greatest common denominator for 2 numbers.
/// </summary>
/// <remarks>
/// Also from http://stackoverflow.com/a/2641293/420175.
/// </remarks>
public static Int64 GCD(Int64 a, Int64 b)
{
// Euclidean algorithm
Int64 t;
while (b != 0)
{
t = b;
b = a % b;
a = t;
}
return a;
}
}'
And the Scala version:
def gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
def gcd(nums: Iterable[Int]): Int = nums.reduce(gcd)
def lcm(a: Int, b: Int): Int = if (a == 0 || b == 0) 0 else a * b / gcd(a, b)
def lcm(nums: Iterable[Int]): Int = nums.reduce(lcm)
Function to find lcm of any list of numbers:
def function(l):
s = 1
for i in l:
s = lcm(i, s)
return s
Using LINQ you could write:
static int LCM(int[] numbers)
{
return numbers.Aggregate(LCM);
}
static int LCM(int a, int b)
{
return a * b / GCD(a, b);
}
Should add using System.Linq; and don't forget to handle the exceptions ...
you can do it another way -
Let there be n numbers.Take a pair of consecutive numbers and save its lcm in another array. Doing this at first iteration program does n/2 iterations.Then next pick up pair starting from 0 like (0,1) , (2,3) and so on.Compute their LCM and store in another array. Do this until you are left with one array.
(it is not possible to find lcm if n is odd)
In R, we can use the functions mGCD(x) and mLCM(x) from the package numbers, to compute the greatest common divisor and least common multiple for all numbers in the integer vector x together:
library(numbers)
mGCD(c(4, 8, 12, 16, 20))
[1] 4
mLCM(c(8,9,21))
[1] 504
# Sequences
mLCM(1:20)
[1] 232792560
ES6 style
function gcd(...numbers) {
return numbers.reduce((a, b) => b === 0 ? a : gcd(b, a % b));
}
function lcm(...numbers) {
return numbers.reduce((a, b) => Math.abs(a * b) / gcd(a, b));
}
Just for fun, a shell (almost any shell) implementation:
#!/bin/sh
gcd() { # Calculate $1 % $2 until $2 becomes zero.
until [ "$2" -eq 0 ]; do set -- "$2" "$(($1%$2))"; done
echo "$1"
}
lcm() { echo "$(( $1 / $(gcd "$1" "$2") * $2 ))"; }
while [ $# -gt 1 ]; do
t="$(lcm "$1" "$2")"
shift 2
set -- "$t" "$#"
done
echo "$1"
try it with:
$ ./script 2 3 4 5 6
to get
60
The biggest input and result should be less than (2^63)-1 or the shell math will wrap.
i was looking for gcd and lcm of array elements and found a good solution in the following link.
https://www.hackerrank.com/challenges/between-two-sets/forum
which includes following code. The algorithm for gcd uses The Euclidean Algorithm explained well in the link below.
https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/the-euclidean-algorithm
private static int gcd(int a, int b) {
while (b > 0) {
int temp = b;
b = a % b; // % is remainder
a = temp;
}
return a;
}
private static int gcd(int[] input) {
int result = input[0];
for (int i = 1; i < input.length; i++) {
result = gcd(result, input[i]);
}
return result;
}
private static int lcm(int a, int b) {
return a * (b / gcd(a, b));
}
private static int lcm(int[] input) {
int result = input[0];
for (int i = 1; i < input.length; i++) {
result = lcm(result, input[i]);
}
return result;
}
Here is the PHP implementation:
// https://stackoverflow.com/q/12412782/1066234
function math_gcd($a,$b)
{
$a = abs($a);
$b = abs($b);
if($a < $b)
{
list($b,$a) = array($a,$b);
}
if($b == 0)
{
return $a;
}
$r = $a % $b;
while($r > 0)
{
$a = $b;
$b = $r;
$r = $a % $b;
}
return $b;
}
function math_lcm($a, $b)
{
return ($a * $b / math_gcd($a, $b));
}
// https://stackoverflow.com/a/2641293/1066234
function math_lcmm($args)
{
// Recursively iterate through pairs of arguments
// i.e. lcm(args[0], lcm(args[1], lcm(args[2], args[3])))
if(count($args) == 2)
{
return math_lcm($args[0], $args[1]);
}
else
{
$arg0 = $args[0];
array_shift($args);
return math_lcm($arg0, math_lcmm($args));
}
}
// fraction bonus
function math_fraction_simplify($num, $den)
{
$g = math_gcd($num, $den);
return array($num/$g, $den/$g);
}
var_dump( math_lcmm( array(4, 7) ) ); // 28
var_dump( math_lcmm( array(5, 25) ) ); // 25
var_dump( math_lcmm( array(3, 4, 12, 36) ) ); // 36
var_dump( math_lcmm( array(3, 4, 7, 12, 36) ) ); // 252
Credits go to #T3db0t with his answer above (ECMA-style code).
GCD needs a little correction for negative numbers:
def gcd(x,y):
while y:
if y<0:
x,y=-x,-y
x,y=y,x % y
return x
def gcdl(*list):
return reduce(gcd, *list)
def lcm(x,y):
return x*y / gcd(x,y)
def lcml(*list):
return reduce(lcm, *list)
How about this?
from operator import mul as MULTIPLY
def factors(n):
f = {} # a dict is necessary to create 'factor : exponent' pairs
divisor = 2
while n > 1:
while (divisor <= n):
if n % divisor == 0:
n /= divisor
f[divisor] = f.get(divisor, 0) + 1
else:
divisor += 1
return f
def mcm(numbers):
#numbers is a list of numbers so not restricted to two items
high_factors = {}
for n in numbers:
fn = factors(n)
for (key, value) in fn.iteritems():
if high_factors.get(key, 0) < value: # if fact not in dict or < val
high_factors[key] = value
return reduce (MULTIPLY, ((k ** v) for k, v in high_factors.items()))
We have working implementation of Least Common Multiple on Calculla which works for any number of inputs also displaying the steps.
What we do is:
0: Assume we got inputs[] array, filled with integers. So, for example:
inputsArray = [6, 15, 25, ...]
lcm = 1
1: Find minimal prime factor for each input.
Minimal means for 6 it's 2, for 25 it's 5, for 34 it's 17
minFactorsArray = []
2: Find lowest from minFactors:
minFactor = MIN(minFactorsArray)
3: lcm *= minFactor
4: Iterate minFactorsArray and if the factor for given input equals minFactor, then divide the input by it:
for (inIdx in minFactorsArray)
if minFactorsArray[inIdx] == minFactor
inputsArray[inIdx] \= minFactor
5: repeat steps 1-4 until there is nothing to factorize anymore.
So, until inputsArray contains only 1-s.
And that's it - you got your lcm.
LCM is both associative and commutative.
LCM(a,b,c)=LCM(LCM(a,b),c)=LCM(a,LCM(b,c))
here is sample code in C:
int main()
{
int a[20],i,n,result=1; // assumption: count can't exceed 20
printf("Enter number of numbers to calculate LCM(less than 20):");
scanf("%d",&n);
printf("Enter %d numbers to calculate their LCM :",n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
result=lcm(result,a[i]);
printf("LCM of given numbers = %d\n",result);
return 0;
}
int lcm(int a,int b)
{
int gcd=gcd_two_numbers(a,b);
return (a*b)/gcd;
}
int gcd_two_numbers(int a,int b)
{
int temp;
if(a>b)
{
temp=a;
a=b;
b=temp;
}
if(b%a==0)
return a;
else
return gcd_two_numbers(b%a,a);
}
Method compLCM takes a vector and returns LCM. All the numbers are within vector in_numbers.
int mathOps::compLCM(std::vector<int> &in_numbers)
{
int tmpNumbers = in_numbers.size();
int tmpMax = *max_element(in_numbers.begin(), in_numbers.end());
bool tmpNotDividable = false;
while (true)
{
for (int i = 0; i < tmpNumbers && tmpNotDividable == false; i++)
{
if (tmpMax % in_numbers[i] != 0 )
tmpNotDividable = true;
}
if (tmpNotDividable == false)
return tmpMax;
else
tmpMax++;
}
}
clc;
data = [1 2 3 4 5]
LCM=1;
for i=1:1:length(data)
LCM = lcm(LCM,data(i))
end
For anyone looking for quick working code, try this:
I wrote a function lcm_n(args, num) which computes and returns the lcm of all the numbers in the array args. The second parameternum is the count of numbers in the array.
Put all those numbers in an array args and then call the function like lcm_n(args,num);
This function returns the lcm of all those numbers.
Here is the implementation of the function lcm_n(args, num):
int lcm_n(int args[], int num) //lcm of more than 2 numbers
{
int i, temp[num-1];
if(num==2)
{
return lcm(args[0], args[1]);
}
else
{
for(i=0;i<num-1;i++)
{
temp[i] = args[i];
}
temp[num-2] = lcm(args[num-2], args[num-1]);
return lcm_n(temp,num-1);
}
}
This function needs below two functions to work. So, just add them along with it.
int lcm(int a, int b) //lcm of 2 numbers
{
return (a*b)/gcd(a,b);
}
int gcd(int a, int b) //gcd of 2 numbers
{
int numerator, denominator, remainder;
//Euclid's algorithm for computing GCD of two numbers
if(a > b)
{
numerator = a;
denominator = b;
}
else
{
numerator = b;
denominator = a;
}
remainder = numerator % denominator;
while(remainder != 0)
{
numerator = denominator;
denominator = remainder;
remainder = numerator % denominator;
}
return denominator;
}
int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a%b);
}
int lcm(int[] a, int n) {
int res = 1, i;
for (i = 0; i < n; i++) {
res = res*a[i]/gcd(res, a[i]);
}
return res;
}
In python:
def lcm(*args):
"""Calculates lcm of args"""
biggest = max(args) #find the largest of numbers
rest = [n for n in args if n != biggest] #the list of the numbers without the largest
factor = 1 #to multiply with the biggest as long as the result is not divisble by all of the numbers in the rest
while True:
#check if biggest is divisble by all in the rest:
ans = False in [(biggest * factor) % n == 0 for n in rest]
#if so the clm is found break the loop and return it, otherwise increment factor by 1 and try again
if not ans:
break
factor += 1
biggest *= factor
return "lcm of {0} is {1}".format(args, biggest)
>>> lcm(100,23,98)
'lcm of (100, 23, 98) is 112700'
>>> lcm(*range(1, 20))
'lcm of (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19) is 232792560'
This is what I used --
def greater(n):
a=num[0]
for i in range(0,len(n),1):
if(a<n[i]):
a=n[i]
return a
r=input('enter limit')
num=[]
for x in range (0,r,1):
a=input('enter number ')
num.append(a)
a= greater(num)
i=0
while True:
while (a%num[i]==0):
i=i+1
if(i==len(num)):
break
if i==len(num):
print 'L.C.M = ',a
break
else:
a=a+1
i=0
for python 3:
from functools import reduce
gcd = lambda a,b: a if b==0 else gcd(b, a%b)
def lcm(lst):
return reduce(lambda x,y: x*y//gcd(x, y), lst)
In Ruby, it's as simple as:
> [2, 3, 4, 6].reduce(:lcm)
=> 12
> [16, 32, 96].reduce(:gcd)
=> 16
(tested on Ruby 2.2.10 and 2.6.3.)