Timeline for The dessert problem (blind ballot remaining blind if non-unanimous)
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 31, 2022 at 23:37 | comment | added | 2012rcampion | @RalphJ You can avoid all that by doing the arithmetic modulo some large prime. | |
| Dec 31, 2022 at 23:06 | comment | added | Ralph J | ACTUALLY... if V3 hears products that are both "N" times larger than the number he heard & then passed on, the WILL know he's the sole "no" voter. Ah, heck! | |
| Dec 31, 2022 at 23:01 | comment | added | Ralph J | The reason for the multiplier is to obscure what the original "passed around" number was (since one might realize he was the sole "no" that way). If both the original value and the product have many factors, there's less risk that V3 learns he was the sole "no"... if V3 heard 51 & passed along as his "no" the value 59 and the products he hears are 102 and 118, he can infer that the multiplier was 2... nobody after him voted no. And knowing the multiplier, if the product of the original # is 2x the number he was first given, he knows nobody before him voted no either. | |
| Dec 31, 2022 at 21:59 | comment | added | 2012rcampion | @RalphJ I don't know what you mean about numbers with several factors or increments, isn't the best strategy just to choose a number uniformly at random from the available range? | |
| Dec 31, 2022 at 21:08 | comment | added | Ralph J | Agree - everyone using moderately large numbers each with several factors makes inferences about others' behavior less likely. Meaning that those who don't want their choice to be inferred, have motivation to use numbers /increments/multipliers like 120 - 240, instead of 3-5. | |
| Dec 31, 2022 at 16:57 | comment | added | 2012rcampion | @RalphJ Good suggestions! I think the second one works, but the first one leaks some (probabilistic) information about the number of 'no' voters. But if we choose large numbers then the probability of collision is negligible anyway. | |
| Dec 30, 2022 at 5:19 | comment | added | Ralph J | To remove the risk of the first-and-only No voter learning that he was it, have "voter 2" pick a multiplier at random, which he privately gives to First and Last. They each multiply their numbers by that multiplier, and "voter 3" hears, privately, the product from First and then from Last, and announces "equal" or "not equal". Not knowing the multiplier that V2 picked, he won't know if he's the sole No, and nobody else hears more than one number. | |
| Dec 30, 2022 at 5:12 | comment | added | Ralph J | To remove the risk of a second No voter randomly picking the initial number, you can instruct that any No voter must pick a higher number than the one they heard. | |
| Dec 27, 2022 at 16:14 | history | answered | 2012rcampion | CC BY-SA 4.0 |