I currently am working on a project for to find some new algebra structures. But I faced with a factorization problem which I cannot come out of it! I could find the lower dimensional of it. But I need a Mathematica codding algorithm for to find the higher dimensional cases. One of my easiest factorization is as what I explain bellow (Which I know its factorization, but I want to re-find it by using Mathematica!):
I need to factorize the following polynomial fraction by using Mathematica
F6[x1, y1, x2 , y2, x3 , y3, x4, y4, x5, y5, x6] = (2 x1 x2 x5 x6 y2 (x2 y1 + x3 (y1 + y2)) y3^2 y4 (x5 y5 + x4 (y4 + y5)))/((x2 y2 + x1 (y1 + y2)) (x3 y3+ x2 (y2 + y3))^2 (x4 y3 + x5 (y3 + y4))^2 (x5 y4 + x6 (y4 + y5)));
to the combination of the following polynomial fractions
K1[x1, y1, x2, y2, x3, y3] = ((x2 y2 + x1 (y1 + y2)) (x3 y3 + x2 (y2 + y3)))/(x2 y2 (x3 y3 + x2 (y2 + y3) + x1 (y1 + y2 + y3)));
K2[y1, x2, y2, x3, y3, x4] = ((x2 y1 + x3 (y1 + y2)) (x3 y2 + x4 (y2 + y3)))/(x3 y2 (x2 y1 + (x3 + x4) (y1 + y2) + x4 y3));
K3[x2, y2, x3, y3, x4, y4] = ((x3 y3 + x2 (y2 + y3)) (x4 y4 + x3 (y3 + y4)))/(x3 y3 (x4 y4 + x3 (y3 + y4) + x2 (y2 + y3 + y4)));
K4[ y2, x3, y3, x4, y4, x5] = ((x3 y2 + x4 (y2 + y3)) (x4 y3 + x5 (y3 + y4)))/(x4 y3 (x3 y2 + (x4 + x5) (y2 + y3) + x5 y4));
K5[ x3, y3, x4, y4, x5, y5] = ((x4 y4 + x3 (y3 + y4)) (x5 y5 + x4 (y4 + y5)))/(x4 y4 (x5 y5 + x4 (y4 + y5) + x3 (y3 + y4 + y5)));
K6[ y3, x4, y4, x5, y5, x6] = ((x4 y3 + x5 (y3 + y4)) (x5 y4 + x6 (y4 + y5)))/(x5 y4 (x4 y3 + (x5 + x6) (y3 + y4) + x6 y5));
But as I know the answer, so I will write it here and need a codding which can give me this factorization or any other factorization with the above mentioned factors:
The combination is as follows:
F6[x1, y1, x2 , y2, x3 , y3, x4, y4, x5, y5, x6] = ((K1[x1, y1, x2, y2, x3, y3] - 1) (K3[x2, y2, x3, y3, x4, y4] - 1) (K4[y2, x3, y3, x4, y4, x5] - 1)(K6[y3, x4, y4, x5, y5, x6] - 1))/(K1[x1, y1, x2, y2, x3, y3] K3[x2, y2, x3, y3, x4, y4] K4[y2, x3, y3, x4, y4, x5] K6[y3, x4, y4, x5, y5, x6]) = ((K1[x1, y1, x2, y2, x3, y3] - 1)/K1[x1, y1, x2, y2, x3, y3]) ((K3[x2, y2, x3, y3, x4, y4] - 1)/K3[x2, y2, x3, y3, x4, y4]) ((K4[y2, x3, y3, x4, y4, x5] - 1)/K4[y2, x3, y3, x4, y4, x5]) ((K6[y3, x4, y4, x5, y5, x6] - 1)/K6[y3, x4, y4, x5, y5, x6]);
I will appreciate a lot if someone helps me to write down a Mathematica codding which is able to give me such combinations of $K1, K2, K3, K4, K5 $ and $K6$.
Here there is another factorization problem in 9 variable (The easiest one!) which I still don't know how it should done!
The question is the same as previous one; We have the following polynomial fraction and wanna to decompose it
F9[x1, y1, z1, x2 , y2, z2, x3 , y3, z3, x4, y4, z4, x5, y5, z5, x6, y6, z6] = (2 x1 x2 x5 y2 y5 y6 z2 (x2 y1 y2 z1 + x2 y1 y3 z1 + x3 y1 y3 z1 + x2 y1 y3 z2 + x3 y1 y3 z2 + x3 y2 y3 z2) z3^2 z4 (x4 x5 y4 z4 + x4 x6 y4 z4 + x4 x6 y5 z4 + x4 x6 y4 z5 + x4 x6 y5 z5 + x5 x6 y5 z5))/((x1 y1 z1 + x1 y1 z2 + x1 y2 z2 + x2 y2 z2) (x2 y2 z2 + x2 y2 z3 + x2 y3 z3 + x3 y3 z3)^2 (x4 y4 z3 + x4 y5 z3 + x5 y5 z3 + x5 y5 z4)^2 (x5 y5 z4 + x5 y6 z4 + x6 y6 z4 + x6 y6 z5));
to the combination of the following polynomial fractions
K1[x1, y1, z1, x2, y2, z2, x3, y3, z3] = ((x1 y1 z1 + x2 y2 z2 + x1 (y1 + y2) z2) (x2 y2 z2 + x3 y3 z3 + x2 (y2 + y3) z3))/(x2 y2 z2 (x2 y2 z2 + x3 y3 z3 + x2 (y2 + y3) z3 + x1 (y2 z2 + (y2 + y3) z3 + y1 (z1 + z2 + z3))));
K2[y1, z1, x2, y2, z2, x3, y3, z3, x4] = (((x2 + x3) y1 z1 + x3 (y1 + y2) z2) ((x3 + x4) y2 z2 + x4 (y2 + y3) z3))/(x3 y2 z2 (x2 y1 z1 + (x3 + x4) (y2 z2 + y1 (z1 + z2)) + x4 (y1 + y2 + y3) z3));
K3[z1, x2, y2, z2, x3, y3, z3, x4, y4] = ((x2 (y2 + y3) z1 + x3 y3 (z1 + z2)) (x3 (y3 + y4) z2 + x4 y4 (z2 + z3)))/(x3 y3 z2 (x2 (y2 + y3 + y4) z1 + x3 (y3 + y4) (z1 + z2) + x4 y4 (z1 + z2 + z3)));
K4[x2, y2, z2, x3, y3, z3, x4, y4, z4] = ((x2 y2 z2 + x3 y3 z3 + x2 (y2 + y3) z3) (x3 y3 z3 + x4 y4 z4 + x3 (y3 + y4) z4))/(x3 y3 z3 (x3 y3 z3 + x4 y4 z4 + x3 (y3 + y4) z4 + x2 (y3 z3 + (y3 + y4) z4 + y2 (z2 + z3 + z4))));
K5[y2, z2, x3, y3, z3, x4, y4, z4, x5] = (((x3 + x4) y2 z2 + x4 (y2 + y3) z3) ((x4 + x5) y3 z3 + x5 (y3 + y4) z4))/(x4 y3 z3 (x3 y2 z2 + (x4 + x5) (y3 z3 + y2 (z2 + z3)) + x5 (y2 + y3 + y4) z4));
K6[z2, x3, y3, z3, x4, y4, z4, x5, y5] = ((x3 (y3 + y4) z2 + x4 y4 (z2 + z3)) (x4 (y4 + y5) z3 + x5 y5 (z3 + z4)))/(x4 y4 z3 (x3 (y3 + y4 + y5) z2 + x4 (y4 + y5) (z2 + z3) + x5 y5 (z2 + z3 + z4)));
K7[x3, y3, z3, x4, y4, z4, x5, y5, z5] = ((x3 y3 z3 + x4 y4 z4 + x3 (y3 + y4) z4) (x4 y4 z4 + x5 y5 z5 + x4 (y4 + y5) z5))/(x4 y4 z4 (x4 y4 z4 + x5 y5 z5 + x4 (y4 + y5) z5 + x3 (y4 z4 + (y4 + y5) z5 + y3 (z3 + z4 + z5))));
K8[y3, z3, x4, y4, z4, x5, y5, z5, x6] = (((x4 + x5) y3 z3 + x5 (y3 + y4) z4) ((x5 + x6) y4 z4 + x6 (y4 + y5) z5))/(x5 y4 z4 (x4 y3 z3 + (x5 + x6) (y4 z4 + y3 (z3 + z4)) + x6 (y3 + y4 + y5) z5));
K9[z3, x4, y4, z4, x5, y5, z5, x6, y6] = ((x4 (y4 + y5) z3 + x5 y5 (z3 + z4)) (x5 (y5 + y6) z4 + x6 y6 (z4 + z5)))/(x5 y5 z4 (x4 (y4 + y5 + y6) z3 + x5 (y5 + y6) (z3 + z4) + x6 y6 (z3 + z4 + z5)));
The question is the same as the previous one. We want to decompose $F9$ just in terms of $K1, K2, K3, K4, K5, K6, K7, K8, K9$, which will be our factors.
Please again let me know if the question is not clear.
Thank you very much for your help!
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then copy and paste. Indent four spaces for code block (or use{}
icon). See Markdown help $\endgroup$ – Bob Hanlon Aug 1 '16 at 22:26