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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 276 0 "" }{TEXT 
277 163 "\nWorksheet for the cyclohexane computations in Section 24.4 \+
of \"Modern Computer Algebra\" by Joachim von zur Gathen and Juergen G
erhard, Cambridge University Press\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 25 "restart;\nprintlevel := 2:" }}}{EXCHG {PARA 256 "" 0 "" {TEXT 
256 39 "\nCreate the Gramian matrix G = (S[i,j])" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 431 "with(linalg):\nG :
= matrix(['['S[i,j]' $ 'j'=1..6]' $ 'i'=1..6]);\nfor i from 1 to 6 do
\n  S[i,i] := 1;\nod;\nfor i from 1 to 5 do\n  S[i,i+1] := 1/3;\nod;\n
S[1,6] := 1/3;\nfor i from 2 to 6 do\n  for j from 1 to i-1 do\n    S[
i,j] := S[j,i];\n  od;\nod;\neq := \{'S[i,1]+S[i,2]+S[i,3]+S[i,4]+S[i,
5]+S[i,6]=0' $ 'i'=1..6\};\nsol := solve(eq, \{S[1,4],S[2,4],S[2,5],S[
2,6],S[3,6],S[4,6]\});\nG := matrix(['['subs(sol,S[i,j])' $ 'j'=1..6]'
 $ 'i'=1..6]);" }}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 0 "" }}{PARA 257 
"" 0 "" {TEXT -1 69 "Rename variables and create the set F of all 4x4 \+
subdeterminants of G" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 206 "S[1,3] := x;\nS[3,5] := y;\nS[1,5] := z;\nA := \+
\{1,2,3,4,5,6\};\nF := \{\};\nfor i from 2 to 6 do\n  for j from 1 to \+
i-1 do\n    T := convert(A minus \{i,j\}, list);\n    F := F union \{d
et(submatrix(G, T, T))\};\n  od;\nod;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 260 0 "" }{TEXT 261 91 "\nTake a polynomial g[1] \+
in F which contains the variables x and y only, and solve it for y\n" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "g[1] := det(submatrix(G, [1,2,3,6
], [1,2,3,6]));\nsolve(g[1], y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }{TEXT 258 0 "" }{TEXT 259 64 "\nDetermine zeroes of the discrimi
nant of g[1] with respect to y\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 
"factor(discrim(g[1], y));\nsolve(%);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 262 0 "" }{TEXT 263 197 "\nFind a polynomial g[4]
 in the ideal I generated by F which contains z only linearly, and che
ck that all polynomials in F vanish modulo g[1] when we solve g[4] for
 z and substitute this everywhere\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
208 "g[2] := det(submatrix(G, [2,3,5,6], [2,3,5,6]));\ng[3] := det(sub
matrix(G, [1,3,4,6], [1,3,4,6]));\ng[4] := factor(g[3] - g[2]);\nw := \+
solve(g[4], z);\n'rem(numer(subs(z = w, F[i])), g[1], x)' $ 'i' = 1..n
ops(F);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 0 "" }
{TEXT 265 184 "\nSince g[4] contains the factor x-y, the above derivat
ion is only valid if x<>y, and we have to treat the latter case separa
tely.\nDetermine the intersecion of g[1]=0 with the line x=y\n" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "factor(subs(y = x, g[1]));\nsolve(%
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 266 0 "" }{TEXT 
267 58 "\nWhich points of g[1]=0 make the denominator of w vanish?\n" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "factor(subs(y = -x - 2/3, g[1]));
\nsolve(%);" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 0 "" }}{PARA 259 "" 
0 "" {TEXT -1 29 "Compute Groebner basis B of F" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "with(Groebner);\nB \+
:= gbasis(F, plex(z, y, x));" }}}{EXCHG {PARA 262 "" 0 "" {TEXT -1 0 "
" }}{PARA 261 "" 0 "" {TEXT -1 27 "Factor the polynomials in B" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "B :
= factor(B);" }}}{EXCHG {PARA 263 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 257 61 "Split into three new ideals and computer their Groe
bner bases" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 212 "F1 := \{B[1], B[3], B[4], op(1,B[2])\};\nF2 := \{B[1
], B[3], B[4], op(2,B[2])\};\nF3 := \{B[1], B[3], B[4], op(3,B[2])\};
\nB1 := gbasis(F1, plex(z, y, x));\nB2 := gbasis(F2, plex(z, y, x));\n
B3 := gbasis(F3, plex(z, y, x));" }}}{EXCHG {PARA 265 "" 0 "" {TEXT 
-1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 48 "Compute the solutions corresp
onding to B1 and B2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 45 "V1 := solve(\{op(B1)\});\nV2 := solve(\{op(B2)\});
" }}}{EXCHG {PARA 267 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {TEXT 
-1 43 "Check whether they are also solutions of B3" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "for v in V1 do\n  s
ubs(v, B3);\nod;\nfor v in V2 do\n  subs(v, B3);\nod;" }}}{EXCHG 
{PARA 269 "" 0 "" {TEXT -1 0 "" }}{PARA 268 "" 0 "" {TEXT -1 28 "Facto
r the polynomials in B3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "factor(B3);" }}}{EXCHG {PARA 271 "" 0 "" 
{TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT -1 104 "The first polynomial i
n B3, which is g[1] from above, does not contain the variable z, so le
t's plot it!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 0 "" }
{TEXT -1 0 "" }{TEXT 268 0 "" }{TEXT 269 42 "\nPlot preparation (pleas
e be patient ...)\n" }{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
190 "with(plots):\nplotsetup(inline):\n\nsmallegg := op(1, implicitplo
t(g[1], x = -5..5, y = -5..5, numpoints = 4000)):\nbigegg   := op(1, i
mplicitplot(g[1], x = -1..1, y = -1..1, numpoints = 4000)):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 270 0 "" }{TEXT 271 18 "\n
A plot of g[1]=0\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 1223 "stext := TE
XT([0.8,-1.2],`x+y+2/3=0`,ALIGNRIGHT,COLOUR(RGB, 0, 1, 0)),         TE
XT([-4,-2],`g[1]=0`,COLOUR(RGB, 1, 0, 0)),         TEXT([1.0,0.8],`x-y
=0`,ALIGNRIGHT,COLOUR(RGB,0,0,1)):\n\ntriangle := POLYGONS([[-1,-1],[-
1,1/3],[1/3,-1]], COLOUR(RGB, 1, 1, 0)):\n\nline := CURVES([[-5,5-2/3]
,[5-2/3,-5]], COLOUR(RGB, 0, 1, 0)), CURVES([[-5,-5],[5,5]],       COL
OUR(RGB, 0, 0, 1)):\n\npoints := POINTS([evalf(-1-sqrt(6)/3),0], [eval
f(-1+sqrt(6)/3),0], [-7/9,0], [1,0], COLOUR(RGB,1,0,1)),POINTS([evalf(
(-1-2*sqrt(6))/3),0], [evalf((-1+2*sqrt(6))/3),0], COLOUR(RGB,0,1,1)):
\n\nlabels := TEXT([0.4,5.0],`y`,FONT(TIMES,ITALIC,10)), TEXT([5.0,-0.
4],`x`,FONT(TIMES,ITALIC,10)), TEXT([-4,-0.3],`-4`), TEXT([-2,-0.3],`-
2`), TEXT([4,-0.3],`4`), TEXT([2,-0.3],`2`), TEXT([0.3,4],`4`), TEXT([
0.3,2],`2`), TEXT([0.3,-4],`-4`), TEXT([0.3,-2],`-2`), TEXT([0.25,-0.2
5],`0`):\n\naxes := CURVES([[-5,0],[5,0]],[[0,-5],[0,5]], '[[-0.04,i*0
.4],[0.04,i*0.4]]' $ 'i'=-12..12, '[[-0.1,i*2],[0.1,i*2]]' $ 'i'=-2..2
, '[[i*0.4,-0.04],[i*0.4,0.04]]' $ 'i'=-12..12, '[[i*2,-0.1],[i*2,0.1]
]' $ 'i'=-2..2):\n\nplot1 := PLOT(axes, smallegg, triangle, line, stex
t, labels, points, SYMBOL(CIRCLE), FONT(TIMES,ROMAN,10), AXESSTYLE(NON
E), SCALING(CONSTRAINED)):\n\nplot1;" }}}{EXCHG {PARA 273 "" 0 "" 
{TEXT -1 0 "" }}{PARA 272 "" 0 "" {TEXT -1 155 "The constraints that a
ll S[i,j] are cosines of angles and hence at most 1 in absolute value \+
imply that all admissable solutions are in the yellow triangle." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 543 "po
ints := POINTS([-1/3,-1/3], [-1/3,-7/9], [-7/9,-1/3]):\n\nlabels := TE
XT([-0.3,-0.04],`x`,FONT(TIMES,ITALIC,10)), TEXT([-0.04,-0.3],`y`,FONT
(TIMES,ITALIC,10)):\n\nbtext := TEXT([-0.7,-0.7],`g[1]=0`,COLOUR(RGB, \+
1, 0, 0)),           TEXT([-0.5,-0.5],`\"boat\"`,FONT(TIMES,ROMAN,10))
,         POLYGONS([[-0.75,-0.34],[-0.55,-0.48]], [[-0.35,-0.34],[-0.4
5,-0.48]],       [[-0.34,-0.75],[-0.45,-0.52]]), TEXT([-0.3,-0.3], `Q`
, FONT(TIMES,ROMAN,10)):\n\nplot2 := PLOT(bigegg, triangle, points, bt
ext, labels, SYMBOL(CIRCLE), FONT(TIMES,ROMAN,10)):\n\nplot2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "B3;" }}}{EXCHG {PARA 275 "" 
0 "" {TEXT -1 0 "" }}{PARA 274 "" 0 "" {TEXT -1 142 "How to obtain a t
hree-dimensional plot? The last two polynomials in B3 contain the vari
able z only linearly. Let's solve the second one for z:" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "ww := solve(B3
[2], z);" }}}{EXCHG {PARA 278 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 272 69 "Does this solution also satisfy the third equation in B3
 modulo g[1]?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "simplify(subs(z = ww, B3[3]));" }{TEXT 273 0 "" }
{TEXT -1 0 "" }}}{EXCHG {PARA 277 "" 0 "" {TEXT -1 0 "" }{TEXT 274 0 "
" }{TEXT 275 37 "\nYes, it is a multiple of g[1]=B3[1]." }}{PARA 276 "
" 0 "" {TEXT -1 51 "When does the denominator vanish on the curve B3=0
?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
46 "solve(\{g[1], denom(ww)\});\nconvert(%, radical);" }}}{EXCHG 
{PARA 279 "" 0 "" {TEXT -1 178 "\nNone of these points is inside the y
ellow triangle, so for each point of the curve insinde the triangle, t
here is precisely one solution for the z-coordinate. Here is the plot.
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1293 "lift := proc(p)\n  \+
if (type(p, list)) then\n    [[p[1][1], p[1][2], subs(x = p[1][1], y =
 p[1][2], ww)],\n     [p[2][1], p[2][2], subs(x = p[2][1], y = p[2][2]
, ww)]];\n  else\n    NULL;\n  fi;\nend:\n\nbigegg3D := map(lift, bige
gg):\n\naa := coeff(g[1], y, 2):\nbb := coeff(g[1], y, 1):\ncc := coef
f(g[1], y, 0):\nvv := simplify((-bb+sqrt(bb^2-4*aa*cc))/2/aa):\n\nu0 :
= -1/3:\nun := -7/9:\nanimpoints := []:\nnn := 7:\nfor i from 1 to nn \+
- 1 do\n  ui := evalf(u0 * (1 - sqrt(i / nn)) + un * sqrt(i / nn)):\n \+
 vi := evalf(subs(x = ui,vv)):\n  animpoints := [op(animpoints), [ui, \+
vi, evalf(subs(x = ui, y = vi, ww))]]:\nod:\nanimpoints := POINTS(op(a
nimpoints), SYMBOL(CIRCLE)):\n\ntext3D := TEXT([-0.5,-0.5,-0.5],`\"boa
t\"`,FONT(TIMES,ROMAN,10)), POLYGONS([[-0.35,-0.35,-0.75],[-0.48,-0.48
,-0.52]],[[-0.35,-0.75,-0.35],[-0.48,-0.52,-0.48]],[[-0.75,-0.35,-0.35
],[-0.52,-0.48,-0.48]]):\n\npoints3D := POINTS([u0 ,u0 ,un],[u0, un, u
0],[un, u0, u0], SYMBOL(CIRCLE)):\n\nlabels := TEXT([-0.77,-0.2,-0.65]
,`z`,FONT(TIMES,ITALIC,10)),\n\011TEXT([-0.78,-0.5,-0.76],`y`,FONT(TIM
ES,ITALIC,10)),\n\011TEXT([-0.5,-0.78,-0.76],`x`,FONT(TIMES,ITALIC,10)
):\n\nplot3 := PLOT3D(CURVES(op(bigegg3D), COLOUR(ZHUE), THICKNESS(2))
, points3D, animpoints, text3D, labels, COLOUR(RGB,0,0,0), AXESSTYLE(B
OX), ORIENTATION(-120,80), FONT(TIMES,ROMAN,10)):\n\nplot3;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 304 0 "" }{TEXT 305 102 "
\nCheck that the curve is extremely close to, but not identical, to a \+
circle in the plane x+y+z+13/9=0\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 302 "difference := proc(u)\n  local v,w;\n  v := max(solv
e(subs(x = u, g[1])));\n  w := solve(subs(x = u, y = v, B3[2]));\n  u \+
+ v + w + 13/9;\nend:\n\nL := []:\nDigits := 20:\nfor u from -20/100 b
y -5/100 to -75/100 do\n  L := [op(L), difference(u)]:\nod:\nevalf(L);
\n\nplot('difference(u)', 'u' = -7/9..-1 + sqrt(6)/3);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 1 0" 0 }{VIEWOPTS 1 
1 0 3 2 1804 }