# Inspirated

July 11, 2008

## Programming Challenges: Australian Voting

Filed under: Blog — krkhan @ 10:44 am

PC/UVa IDs: 110108/10142, Popularity: B, Success rate: low, Level: 1

I had 28 (that’s twenty-eight) “Time Limit Exceeded” tries on this problem. I was looking everywhere for a loop that would drag, for every possible input that would cause breakdown and I couldn’t find any such thing. So much so, that I considered labeling this post as Australian Nightmare. The issue, once I found it, had nothing to do with Australia and all to do with the `string` and `istringstream` objects being significantly heavy to be constructed/destructed on each iteration of the loop on lines #138-153.

The book recommends further reading on voting systems and has linked to some mathematical theorem that proves that no voting system can ever be perfect. Considering the democratic results around the world in recent elections, yeah, what an absolute shocker.

```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 #include <iostream> #include <iomanip> #include <string> #include <sstream> #include <vector> #include <iterator> #include <map>   using namespace std;   #define NBALLOTS 1000   typedef vector< vector< int > > twod_vector;   vector< int > calculate(vector< int > &votes, int ncand, bool atleast_50) { vector< int > winners;   int total = votes.size(); int max = 0; for(int i = 0; i < ncand; i++) { int vote_count = count(votes.begin(), votes.end(), i); bool cond = !atleast_50 || (vote_count > total / 2); if(vote_count > 0 && vote_count >= max && cond) { max = vote_count; winners.push_back(i); } }   return winners; }   bool compare(pair< int, int > a, pair< int, int > b) { return a.second < b.second; }   map< int, int > vote_map(vector< int > &votes) { map< int, int > table;   for(vector< int >::iterator p = votes.begin(); p < votes.end(); p++) { table[*p]++; }   return table; }   vector< int > least_votes(vector< int > &votes) { map< int, int > table = vote_map(votes); vector< int > min_keys; int min_value = min_element(table.begin(), table.end(), compare)->second;   map< int, int >::iterator p; for( p = table.begin(); p != table.end(); p++ ) { if(p->second == min_value) { min_keys.push_back(p->first); } }   return min_keys; }   void eliminate(twod_vector &ballots, int value) { for(int i = 0; i < (int) ballots.size(); i++) { for(int j = 0; j < (int) ballots.at(i).size(); j++) { if(ballots.at(i).at(j) == value) { for(int k = i; k < (int) ballots.size() - 1; k++) { ballots.at(k).at(j) = ballots.at(k + 1).at(j); } } } } }   void eliminate(twod_vector &ballots) { vector< int > min_values = least_votes(ballots.at(0));   for(vector< int >::iterator p = min_values.begin(); p != min_values.end(); p++) { eliminate(ballots, *p); } }   bool draw(vector< int > &votes) { if(!votes.size()) { return true; }   map< int, int > table = vote_map(votes); map< int, int >::iterator p = table.begin(); int n = p->second; p++; for(; p != table.end(); p++) { if( n != p->second ) { return false; } }   return true; }   int main(int argc, char *argv[]) { twod_vector ballots; int ncases; cin>>ncases; cin.ignore(); string dummy; getline(cin, dummy);   for(int i = 0; i < ncases; i++) { int ncand; cin>>ncand; cin.ignore();   vector< string > candidates(ncand); for(int j = 0; j < ncand; j++) { getline(cin, candidates.at(j)); }   twod_vector ballots(ncand); string str; istringstream iss; for(int j = 0; j < NBALLOTS; j++) { getline(cin, str);   if(str == "") { break; } iss.str(str);   for(int k = 0; k < ncand; k++) { int value = *istream_iterator< int >(iss); ballots.at(k).push_back( value - 1 ); } iss.clear(); }   vector< int > winners; int j = 0; while(winners.size() == 0) { if(draw(ballots.at(0))) { winners = calculate(ballots.at(0), ncand, false); break; }   winners = calculate(ballots.at(0), ncand, true); eliminate(ballots); j++; }   if(!ballots.at(0).size()) { for(int k = 0; k < ncand; k++) { winners.push_back(k); } }       sort(winners.begin(), winners.end()); if(i > 0) { cout<<endl; } vector< int >::const_iterator p; for(p = winners.begin(); p < winners.end(); p++) { cout<<candidates.at(*p)<<endl; } }   return 0; }```
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1. Very nice!!

Comment by Dymnallorymn — August 3, 2008 @ 4:22 pm

2. Hi krkhan,
I’m stuck with the Time Limit Exceeded, but I don’t understand if it’s a matter of algorithm efficiency, or my program falls inside an infinite loop.
Here are the steps of my solution:
1 count all the votes per candidate
2 find min/max number of votes per candidate
3a if min == max then all the remaining candidates have tied, and they are all winners. Then exit.
3b else if max/tot > 50 / 10 we have a single winner. Then exit.
3c find all the ballots starting with a losing candidate and shift the ballot until the head candidate is still in race for winning
4 come back to #1
This is the general idea (obviously we can avoid to count the votes at each iteration and also to shift the ballots).

Thank you so much.

Regards,
Matteo

Comment by Matteo Landi — November 15, 2010 @ 2:16 am

3. Hi again,
I forgot to post a link to the implementation of the algorithm, and one to both the input and the output I use to test things.

* code: http://pastebin.com/rhqYs1uZ
* input: http://pastebin.com/UU3Af4hh
* output: http://pastebin.com/q3hg5sD1

Regards,
Matteo

Comment by Matteo Landi — November 15, 2010 @ 2:18 am

4. I think I’ve found the bug.
I was initializing the `min’ variable to a wrong value(number of candidates minus one instead of number of ballots minus minus one): this makes the program enter in an infinite loop, and that’s why I got that LTE message.
As soon as Programmin-Challenges comes back on-line, I’ll try the fixed solution.

Regards,
Matteo

Comment by Matteo Landi — November 18, 2010 @ 1:37 am