Programming Forums > Application Development > C++

veiga2 · Aug 13th, 2006, 6:27 AM

About the code in making the numbers from 1-42, in different combinations without repeatition, by 6 numbers, is there any function about that?, hmmmm... i cant find the answer on this... i know im going to use the for loop but i cant find the solution... pls help me hnx

example: collection of numbers from 1-7
answer:
123456
123457
123467
123567
124567
134567
234567

Thanks for helping a newbie prob

grumpy · Aug 13th, 2006, 9:35 AM

Use nested for loops, where the start or end conditions of inner loops are determined by the value controlling outer loops.

bl00dninja · Aug 15th, 2006, 7:31 PM

uh, what the hell are you trying to do? "making numbers"? w/o repitition there is the fundamental theorem of arithmetic, which might help you.

uman · Aug 16th, 2006, 2:41 AM

Next time, post in the proper forum: "Software design and algorithms".

(Toggle Plain Text)

std::vector<std::vector<value> >
get_combinations(std::vector<value> values, unsigned m)
{
	std::vector<std::vector<value> > ret;
	unsigned n = values.size();
	std::vector<value>::const_iterator* iters = 
		new std::vector<value>::const_iterator[m];
	for(unsigned i=0;i<m;i++)
		iters[i] = values.begin() + i;
	while(true)
	{
		unsigned i;
		std::vector<value> tmp;
		for(unsigned j=0;j<m;j++)
			tmp.push_back(*(iters[j]));
		ret.push_back(tmp);
		
		for(i=m;i>0;i--)
		{
			if(++(iters[i-1]) != values.end() - (m - i))
				break;
		}
		if(!i)
			break;
		for(unsigned j = i;j<m;j++)
		{
			iters[j] = iters[j-1]+1;
		}
	}
	return ret;
}

It was hard work forcing my brain to solve this in a non-recursive way.

uman · Aug 16th, 2006, 6:02 AM

By the way, that will use ten metric fucktons of memory if you run it with a lot of values. If you needed to find all the combinations of a large set of values, but you didn't need to hold them all in memory at once, it'd be far better to implement as find_first_combination() and find_next_combination() functions.

Aug 13th, 2006, 6:27 AM	#1
veiga2 Newbie Join Date: May 2006 Location: Philippines Posts: 21 Rep Power: 0	Newbie prob:how to make combinations without repeatition About the code in making the numbers from 1-42, in different combinations without repeatition, by 6 numbers, is there any function about that?, hmmmm... i cant find the answer on this... i know im going to use the for loop but i cant find the solution... pls help me hnx example: collection of numbers from 1-7 answer: 123456 123457 123467 123567 124567 134567 234567 Thanks for helping a newbie prob

Aug 15th, 2006, 7:31 PM	#3
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 6	uh, what the hell are you trying to do? "making numbers"? w/o repitition there is the fundamental theorem of arithmetic, which might help you. __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

Aug 16th, 2006, 2:41 AM	#4
uman Expert Programmer Join Date: Dec 2004 Posts: 794 Rep Power: 4	Next time, post in the proper forum: "Software design and algorithms". (Toggle Plain Text) std::vector<std::vector<value> > get_combinations(std::vector<value> values, unsigned m) { std::vector<std::vector<value> > ret; unsigned n = values.size(); std::vector<value>::const_iterator* iters = new std::vector<value>::const_iterator[m]; for(unsigned i=0;i<m;i++) iters[i] = values.begin() + i; while(true) { unsigned i; std::vector<value> tmp; for(unsigned j=0;j<m;j++) tmp.push_back((iters[j])); ret.push_back(tmp); for(i=m;i>0;i--) { if(++(iters[i-1]) != values.end() - (m - i)) break; } if(!i) break; for(unsigned j = i;j<m;j++) { iters[j] = iters[j-1]+1; } } return ret; } std::vector<std::vector<value> > get_combinations(std::vector<value> values, unsigned m) { std::vector<std::vector<value> > ret; unsigned n = values.size(); std::vector<value>::const_iterator iters = new std::vector<value>::const_iterator[m]; for(unsigned i=0;i<m;i++) iters[i] = values.begin() + i; while(true) { unsigned i; std::vector<value> tmp; for(unsigned j=0;j<m;j++) tmp.push_back(*(iters[j])); ret.push_back(tmp); for(i=m;i>0;i--) { if(++(iters[i-1]) != values.end() - (m - i)) break; } if(!i) break; for(unsigned j = i;j<m;j++) { iters[j] = iters[j-1]+1; } } return ret; } It was hard work forcing my brain to solve this in a non-recursive way. __________________ Few people deserve to be compared to (Rush) Limbaugh, most of them were convicted at the Nuremburg trials. --WilliamSChips on Slashdot

Aug 16th, 2006, 6:02 AM	#5
uman Expert Programmer Join Date: Dec 2004 Posts: 794 Rep Power: 4	By the way, that will use ten metric fucktons of memory if you run it with a lot of values. If you needed to find all the combinations of a large set of values, but you didn't need to hold them all in memory at once, it'd be far better to implement as find_first_combination() and find_next_combination() functions. __________________ Few people deserve to be compared to (Rush) Limbaugh, most of them were convicted at the Nuremburg trials. --WilliamSChips on Slashdot

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Aug 13th, 2006, 9:35 AM	#2
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,261 Rep Power: 5	Use nested for loops, where the start or end conditions of inner loops are determined by the value controlling outer loops.

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Programming Forums > Application Development > C++

Newbie prob:how to make combinations without repeatition