30:T4a3,tdh_results = { const voters = ["T", "D", "H"]; const pro = "+"; const against = "-"; const combos = Math.pow(2, voters.length); const eval_fn = (r) => { if (r[pro].length >= (voters.length/2)) { return pro } else { return against } } let results = []; for (let i = 0; i < combos; i+= 1) { let result = {}; result[pro] = []; result[against] = []; for (let j = 0; j < voters.length; j += 1) { let voter = 1 << j; if (voter & i) { result[pro].push(voters[j]); } else { result[against].push(voters[j]); } } result['outcome'] = eval_fn(result); //result['v'] = i; results.push(result); } for (let i = 0; i < combos; i+= 1) { const tom_vote = 1 & i; // 000 --> 000 const flip_vote = (~tom_vote) & (combos - 1); // (111) & (111) --> 111 const tom_mask = (combos - 1) ^ 1; // 111 --> 110 set the Tom bit to 0; const flipped_mask = tom_mask | flip_vote; // 110 | 000 --> 110; const flipped = 1 ^ i; //~results[i]['flipped'] = flipped; results[i]['flippable'] = (results[i].outcome != results[flipped].outcome); } return results }31:T4b2,make_results = (voters) => { //const voters = ["T", "D", "H"]; const pro = "+"; const against = "-"; const combos = Math.pow(2, voters.length); const eval_fn = (r) => { if (r[pro].length >= (voters.length/2)) { return pro } else { return against } } let results = []; for (let i = 0; i < combos; i+= 1) { let result = {}; result[pro] = []; result[against] = []; for (let j = 0; j < voters.length; j += 1) { let voter = 1 << j; if (voter & i) { result[pro].push(voters[j]); } else { result[against].push(voters[j]); } } result['outcome'] = eval_fn(result); //result['v'] = i; results.push(result); } for (let i = 0; i < combos; i+= 1) { const tom_vote = 1 & i; // 000 --> 000 const flip_vote = (~tom_vote) & (combos - 1); // (111) & (111) --> 111 const tom_mask = (combos - 1) ^ 1; // 111 --> 110 set the Tom bit to 0; const flipped_mask = tom_mask | flip_vote; // 110 | 000 --> 110; const flipped = 1 ^ i; //~results[i]['flipped'] = flipped; results[i]['flippable'] = (results[i].outcome != results[flipped].outcome); } return results }32:T463,## Voting procedures When it comes to actual quadratic voting, the idea is that each voter gets a set number of tokens, and then each vote costs the square of the number. So if you want to cast 3 votes, it costs 9 tokens, 4 votes costs 16 tokens (or +7), 5 votes costs 25 tokens (or +11). To see why that's nice, consider 10 voters, 3 of which are die-hard for Candidate T, and 7 of which do not like Candidate T but are indifferent among 4 other candidates. With traditional voting, the results would look like: | Candidate | Votes | | --- | --- | | T | 3 | | A | 2 | | B | 2 | | C | 2 | | D | 1 | Here the voting doesn't allow the 7 voters to express that they'd prefer any other candidate aside from T. With quadratic voting though, they could. With 100 tokens and quadratic voting, a die-hard would cast 10 votes for T at a cost of 100 tokens, so T would have 30 votes. A non-die-hard would cast 5 tokens each for the other for candidates for ${tex`5^2 = 25`} tokens for each candidate (totaling to 100). That would mean all the other candidates would receive ${tex`7*5= 35`} votes, easily beating T. 33:[] 2e:{"title":"Quadratic Voting","cells":[{"id":4,"value":"# Quadratic Voting\n\nAfter reading an article in the Economist called [\"A mathematical method that could offer a fairer way to vote\"](https://www.economist.com/christmas-specials/2021/12/18/the-mathematical-method-that-could-offer-a-fairer-way-to-vote) about Quadratic Voting, I wanted to visualize what it meant.\n\nThe article talks about Lionel Penrose and his penchant for causing trouble in committees, starting off with an example of a 3 person committee of Tom, Dick and Harry. In the committee, a binary vote on an issue can go eight ways (${tex`2^3 = 8`}).\n\nThese 8 vote possibilities are listed below along with the resulting outcome of the vote, and whether if Tom had changed his vote, whether the outcome of the vote would have flipped. In this case, 4 out of 8 outcomes, Tom could have flipped. This puts his \"power\" index at 50%.\n","pinned":false,"hidden":false,"mode":"md"},{"id":12,"value":"$30","pinned":false,"hidden":false,"mode":"ojs"},{"id":21,"value":"viewof table = Inputs.table(tdh_results, {})","pinned":true,"hidden":false,"mode":"ojs"},{"id":66,"value":"results_9 = {\n return make_results(['T', 'D', 'H', 'a', 'b', 'c', 'd', 'e', 'f']);\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":77,"value":"## Committee of Size 9\n\nIn this example, with a committee that's 3 times bigger (9 members), there are 512 different outcomes(${tex `2^9 = 512`}). You might expect that Tom's \"power\" is cut by 3. However, he can still flip ${results_9.filter(x => x.flippable).length} of them, or ${(100* results_9.filter(x => x.flippable).length / 512).toLocaleString()}% which is still over half of his original power.","pinned":false,"hidden":false,"mode":"md"},{"id":74,"value":"viewof table2 = Inputs.table(results_9, {})","pinned":true,"hidden":false,"mode":"ojs"},{"id":80,"value":"results_9.filter(x => x.flippable).length / 512","pinned":true,"hidden":false,"mode":"ojs"},{"id":95,"value":"dataset = {\n let domain = Array.from(Array(20).keys()).filter(x => x %2 == 1).filter(x => x > 1);\n\n let committees = domain.map(c => Array.from(Array(c).keys()));\n \n \n return committees.map(make_results).map((rs, i) => {\n return {group_size: domain[i], voting_power:(rs.filter(r => r.flippable).length / rs.length), sqrt: 1/Math.sqrt(domain[i] + 1)}\n \n })\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":129,"value":"1/Math.sqrt(4)","pinned":true,"hidden":false,"mode":"ojs"},{"id":139,"value":"Below we can see how the voting power plays out for larger group sizes (in gold) compared with ${tex `\\frac{1}{\\sqrt{size}}`} (in blue). You can see that the voting power varies with the square root of the size of the group. This was the basis of one proposal for way to allocate voting tokens in the United Nations. It was not adopted.\n\nThe article doesn't directly draw the line from this relationship to that of Quadratic Voting, but it does introduce the idea of a square root bearing an important relationship for large groups.","pinned":false,"hidden":false,"mode":"md"},{"id":108,"value":"Plot.plot({\n y: {\n grid: true\n },\n marks: [\n Plot.line(dataset, {x: \"group_size\", y: \"voting_power\", stroke: \"goldenrod\"}),\n Plot.line(dataset, {x: \"group_size\", y: \"sqrt\", stroke: \"steelblue\"})\n\n ]\n})","pinned":false,"hidden":false,"mode":"ojs"},{"id":67,"value":"$31","pinned":false,"hidden":false,"mode":"ojs"},{"id":160,"value":"$32","pinned":true,"hidden":false,"mode":"md"}],"files":"$W33","stdlib":"1","titling":"auto"}