tag:blogger.com,1999:blog-2569892606315203117.post8219761360272401360..comments2021-09-22T07:32:53.639-04:00Comments on Essays & Endnotes: Observations from Jon Wild on the Partition Puzzlestephen soderberghttp://www.blogger.com/profile/17300056962479866094noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-2569892606315203117.post-88285062485490847472014-04-09T13:53:31.611-04:002014-04-09T13:53:31.611-04:00You wrote:
There is a quintuplet of Z-related hex...You wrote:<br /><br />There is a quintuplet of Z-related hexachords in 31-tone equal temperament whose interval vectors consist solely of 1s--they are all-interval hexachords, expressing 15 interval classes in just six pitches. This set of 5 hexachords can be packed into the 31-tone aggregate leaving only one pitch uncovered. The odds that this is by chance are astronomical--there is some other principle at work, related to the observations Stephen has started to make, but I don't know what that principle is. <br /><br />Here is a description of your example which may remove some of the mystery.<br />Modulo 31, the five sets can be written as <br />{3^0 , 3^10, 3^20, 3^3, 3^13, 3^23}<br />{3^2 , 3^12, 3^22, 3^5, 3^15, 3^25}<br />{3^4 , 3^14, 3^24, 3^7, 3^17, 3^27}<br />{3^6 , 3^16, 3^26, 3^9, 3^19, 3^29}<br />{3^8 , 3^18, 3^28, 3^11, 3^21, 3^31}<br /><br />In the language of abstract algebra, <br /> {3^0 , 3^10, 3^20} <br />constitutes a subgroup of the multiplicative group of the field with 31 elements and <br /> {3^3, 3^13, 3^23}<br />one of it's cosets.<br /><br />Moreover<br /> {3^0 , 3^6, 3^12, 3^18, 3^24} <br />constitutes a disjoint subgroup with 5 elements. The five sets come from the first by multiplication, as one see quickly if one keeps in mind the congruence of 3^30 with 1 modulo 31. Multiplication permutes intervals, but preserves all interval hexachords.<br /><br />One could still argue that {3^0 , 3^10, 3^20, 3^3, 3^13, 3^23} giving an all-interval hexachord seems miraculous,<br />but one surely gets all the intervals if no interval occurs twice. So write<br /> A = {3^0 , 3^10, 3^20} B = {3^3, 3^13, 3^23}<br />Obviously distinct pairs in A give distinct intervals (or the interval would have to be preserved by multiplication by 3^10); similarly distinct pairs in B; similarly a pair in A compared with a pair in B similarly two pairs each crossing A to B. <br /><br />Thus the only not obvious case would be a pair in A, say, and a pair crossing from A to B. One can also rule out some of these cases a priori, and reduce the total number by symmetry, but in any case one doesn't have so many case that the non-occurrence of an equality seems miraculous anymore. David Feldmanhttps://www.blogger.com/profile/08457408183558003996noreply@blogger.com