Monday, October 22, 2007

A programmer's HP-35s review

Some thoughts on the HP-35s after been programming it for a while: Doing the HP-35s Euclid Pack implementation.

Two features I learned to love when programming the 35s: Vector type and equations.

Vector type

The vector type has been critizised for missing key features:
1. No cross product. Must make a program.
2. No native way to unpack or address vector components. Must use 'dot product mask vectors'.
3. No operation to tell a vector's dimension (dim or size).
Despite these drawbacks (of which the two first can be worked around), I find the vector type a very valuable feature because of it's data packing capabilities.

Some focus has been on using 3D vectors to augment the machine's 800 indirect registers. For the things I do and the fact that it is a closed system (no I/O), I can't imagine applications using a large fraction of the 800 registers even without packing 3 real numbers in one register!

But, I find the data packing offered by 3D vectors to be very valuable in the 'more valuable real estate territories' of near end user variable registers (A-Z) and working registers (stack and lastx).

For example, for the design of the euclid pack project I ended up with wanting to store definition of two lines (four 2D or 3D points), two planes (two 3D points and two normal vectors) and one triangle (three 3D points) in variables.

If all data entered is in 3D, that would require minimum use 33 registers! And this is data I would like to be close to the end user, so I would like to use variables: Unless had vector type would not have room. Since have the vector type minimum register use is 11.

And implementation actual ended up storing additional data in variables to make certain computations easy and results handy. And I still did not use up all variable registers...

Also now we can have 12 real numbers on the stack! That becomes usefull when using programs that take more than 4 numbers as input. For example to define a 3D triangle (the T program) end user must give in nine numbers (three 3D numbers), without the ability to just put the points on the stack and run the program, the program would had to prompt for the data.

Also the vector type, RPN paradigm and the use of equation is very powerfull when programming:

Equations in programs

I find programming complex calculations in need of much intermediate results to be easy on the 35s by combining the RPN paradigm, vector type and equations:
• The vector type allows 12 values to be stored on the stack.
• Equations in programs can use all these number while being evaluated, only when done is maximum three numbers lost on the stack: A 3D vector in the T stack register.
• The result of evaluating an equation may very well be a 3D vector: A single line in a program may execute three equation which result raises the stack only once.
I think second half of the last program (lines 28 to 35) I made in this series realy utilize this programming style, there is eight lines of equations in sequence!

A very nice feature of the HP 35s that helps programming this way is that when in programming mode and to insert a stack register as a variable in an equation using the RDN key one sees the content of the registers user side. If one has a worked example of the problem being programmed and is programming equation by equation and running the program before programming a new equation then when programming the equation it is easy to see where on the stack needed numbers are.

One drawback of this style of programming is that the evaluation of equation makes for much slower execution then a pure RPN version. For programs that intention is to be called a large (or small...) number of times as subroutines by other programs this style of programming may not be appropiate.

On an improved HP 35s (HP 35s+ :-) perhaps one could have some kind of equation compilation?

No need for me to get into any of the now well known short commings or bugs of the device documented by the community of the MoHPC here, except to say that the reported vector syntax bug is indeed real on some machines and makes this type of programming much harder than needed. I know, I got one... Still I have been able to do whatever I did without ever needing to reset the thing.

Sunday, October 21, 2007

HP-35s Euclid Implementation - XV : The inscribed and circumscribed circle of a triangle

The final program to fullfill the requirements! Problem 16 and 17: Finding the inscribed (incircle) and circumscribed (circumcircle) circles of a triangle.

Finding the inscribed circle is the same as solving the problem of finding the circle tangent to three lines (lines defined by the triangle edges) and finding the circumscribed circle is the same as solving the problem of finding the circle that passes through three points (triangle's vertexes).

The triangle in question has been entered using the T program.
`                  v3                 /  \                /    \               /      \                           v1------v2`
Refering to the above triangle I give the formulas used in the HP-35s program presented here:

Inscribed circle:

The barycentric coordinates for the incenter are:
`(l1/p, l2/p, l3/p)`
Where p is the perimeter of the triangle (sum of all edges lengths) and l1 is the length of the edge v3-v2, l2 is the length of the edge v1-v3 and l3 is the length of the edge v2-v1.

The radius of the inscribed circle is:
`(2A)/p`
Where A is the area of the triangle.

Circumscribed circle:

With the intermediate results
`d1=-e2*e3d2=-e3*e1d3=-e1*e2(e1=v3-v2, e2=v1-v3, e3=v2-v1, * is the dot product)c1=d2d3c2=d3d1c3=d1d2c=c1+c2+c3`
the barycentric coordinates for the circumcenter are:
`((c2+c3)/2c, (c3+c1)/2c, (c1+c2)/2c)`
The radius of the circumscribed circle is:
`sqrt((d1+d2)(d2+d3)(d3+d1)/c)/2`
These formulas are taken from the excelent book 3D Math Primer for Graphics and Game Development that also provide background for much of the geometric problems in the Euclid Pack specification.

Stack Input/Output:
`No input on stack-> XEQ O ->[[bc1, bc2, bc3], cr, [bi1, bi2, bi3], ir | L]`
Where (bc1, bc2, bc3) are the barycentric coordinates for the circumscribed circle's center (circumcenter) , cr is its radius, (bi1, bi2, bi3) are the barycentric coordinates for the inscribed circle's center (incenter) and ir is its radius.

Variables:

`K : First vertex (3D vector). Populated by the T program.L : Second vertex (3D vector). Populated by the T program.M : Third vertex (3D vector). Populated by the T program.`
Writes:
`Z : Scratch data.`
Program:
`O001 LBL OO002 5O003 STO IO004 STO(I)O005 4O006 STO IO007 LASTXO008 STO(I)O009 eq M-L0O10 eq M-KO011 eq [ABS(REGY),ABS(REGX),ABS(L-K)]O012 STO ZO013 eq REGX/([1,0,0]xREGX+[0,1,0]xREGX+[0,0,1]xREGX)O014 R^O015 R^O016 XEQ X001O017 R^O018 X<>YO019 ABSO020 eq REGX/([1,0,0]xZ+[0,1,0]xZ+[0,0,1]xZ)O021 X<>YO022 RDNO023 X<>YO024 RCL(I)O025 ABSO026 RDNO027 XEQ U001O028 eq M-LO029 eq K-MO030 eq L-KO031 eq [-REGYxREGX,-REGXxREGZ,-REGZxREGY]O032 eq [([0,1,0]xREGX)x([0,1,0]xREGX),         ([0,0,1]xREGX)x([1,0,0]xREGX),         ([1,0,0]xREGX)x([0,1,0]xREGX)]O033 eq [1,0,0]xREGX+[0,1,0]xREGX+[0,0,1]xREGXO034 eq 0.5xSQRT((([1,0,0]xREGZ+[0,1,0]xREGZ)x                  ([0,1,0]xREGZ+[0,0,1]xREGZ)x                  ([0,0,1]xREGZ+[1,0,0]xREGZ))/REGX)O035 eq [([0,1,0]xREGZ)+([0,0,1]xREGZ),         ([0,0,1]xREGZ)+([1,0,0]xREGZ),         ([1,0,0]xREGZ)+([0,1,0]xREGZ)]/(2xREGY)O036 XEQ U079O037 RTN`

Mnemonics: O since letter is a circle (or more or less depeding on font, rather circular on the 35s).

Reason for not converting to cartesian coordinates is that since the barycentric coordinates are relative to the vertices it is much easier to 'see' where the center points are relative to the triangle. If one where to be presented the cartesian coordinates one must remember the nine vertex coordinates to know where (relative to triangle) the center points are. Particular one see at once if the circumscribed circle's center is inside or outside the triangle.

Cartesian coordinates are easily obtained using the A program by the following key sequence:
`XEQ ARDNRDNXEQ ARDNRDN`
The last two RDN are only provided to get content of stack in same order as before except that barycentric positional vectors are now replaced with cartesian ones.

Thursday, October 18, 2007

HP-35s Euclid Implementation - XIV : Barycentric to Cartesian coordinates

Going from barycentric to cartesian coordinates (problem 15 in the requirements) is the easy part of Cartesian<->Barycentric 2D/3D conversion, this will be a much shorter blog entry than this one!

Stack Input/Output:
`3D case: [[b1, b2, b3], Y, Z, T | L]-> XEQ A ->[[x, y, z], Y, Z, T | [b1, b2, b3]]2D case: [[b1, b2, b3], Y, Z, T | L]-> XEQ A ->[[x, y], Y, Z, T | [b1, b2, b3]]`
Variables:

`K : First vertex (3D vector), in 3D case. Populated by the T program.L : Second vertex (3D vector), in 3D case. Populated by the T program.M : Third vertex (3D vector), in 3D case. Populated by the T program.U : First vertex of triangle projected to 2D (2D vector), in 2D case. Populated by the T program.V : Second vertex of triangle projected to 2D (2D vector), in 2D case. Populated by the T program.W : Third vertex of triangle projected to 2D (2D vector), in 2D case. Populated by the T program.`
Program:
`A001 LBL AA002 ABSA003 RDNA004 FS? 2A005 GTO A008A006 eq Kx([1,0,0]xLASTX)+Lx([0,1,0]xLASTX)+Mx([0,0,1]xLASTX)A007 RTNA008 eq Ux([1,0,0]xLASTX)+Vx([0,1,0]xLASTX)+Wx([0,0,1]xLASTX)A009 RTN`

Mnemonics: None realy, but letter is left to B, the program label for Barycentric to Cartesian conversion.

If we are working in 2D (flag 2 set) we are applying the weights (barycentric coordinates) to the 2D coordinate's of the projected triangle (line A008) else applying the weights to the vertexes 3D coordinates (line A006).

Wednesday, October 17, 2007

Fixing the HP-41CV - II : Sending it

Finaly got around to post the HP-41CV to Fix That Calc!

Monday, October 15, 2007

HP-35s Euclid Implementation - XIII : Cartesian to Barycentric coordinates

Arrived at problem 14 of the requirements: Computing barycentric coordinates given cartesian coordinates.

Barycentric coordinates [b1, b2, b3] of a point in a plane are coordinates defined by a triangle in the plane (defines the plane) where the triangle's first vertex's barycentric coordinates are [1,0,0], second vertex's are [0,1,0], third vertex's are [0,0,1] and one have that b1+b2+b3=1.

Coordinates of any point in the plane can be expressed as the weighted average of the triangle's vertices coordinates, these weights are the barycentric coordinates:
` Barycentric     Cartesian(b1, b2, b3)<=>b1v1+b2v2+b3v3`
Where vi is the positional vector for vertex i.

Barycentric coordinates has several uses and are of theoretical importance outside the scope of this blog entry. Here I give two practical uses:
1. Provides a robust test for if a point in triangle's plane with barycentric coordinates [b1, b2, b3] is inside the triangle, on one of the triangle's edge or outside the triangle: If all components are in the range[0,1] point is inside, if bi is 0 and the two others are in the [0,1] range then the point is on the edge oposite of triangle's vertice i, and finaly if at least one component is not in the [0,1] range then point is outside triangle.
2. If a function's values is known at the triangle's vertices one can perform linear interpolation inside the triangle at the point: v1*b1+v2*b2+v3*b3. Where vi is the value at vertex i and [b1, b2, b3] is the barycentric coordinates for point to evaluate at. Note that converting from barycentric coordinates to cartesian coordinates is a special case of this.
I choose to convert from cartesian coordinates to barycentric by programming the solution to the equations for the 2D cartesian to barycentric coordinates case:
`Q=(y1-y3)*(x2-x3)+(y2-y3)*(x2-x1)b1=((y-y3)*(x2-x3)+(y2-y3)*(x3-x))/Qb2=((y-y1)*(x3-x1)+(y3-y1)*(x1-x))/Qb3=((y-y2)*(x1-x2)+(y1-y2)*(x2-x))/Q`
Where (xi, yi) are cartesian coordinates for triangle's i'th vertex and (x, y) is the cartesian coordinates of the point in question. This is the solution to the equation set of three uknown (b1, b2, b3):
`x=b1*x1+b2*x2+b3*x3y=b1*y1+b2*y2+b3*y31=b1+b2+b3`
To solve 3D cartesian to barycentric I project the triangle and point in question to 2D and solve as a 2D problem.

Implementation is done in two steps:
1. Modify the T program so it: A) Also accept a 2D triangle (flag 2 set='2D mode'. B) Project the triangle to the cardinal plane that gives the largest area, this is the triangle used for computing barycentric coordinates. C) Computes the quotient Q in the above equations.
2. Implement a program labeled B to accepts a 2D or 3D cartesian tuple and use result from B and C above to compute barycentric coordinates.
Step 1 : The modified T program

Stack Input/Output:
`2D case: [[x2, y2], [x1, y1], [x0, y0] | L]->XEQ T->[[nx, ny, nz], [x2, y2], [x1, y1], [x0, y0] | L]3D case: [[x2, y2, z2], [x1, y1, z1], [x0, y0, z0] | L]->XEQ T->[[nx, ny, nz], [x2, y2, z2], [x1, y1, z1], [x0, y0, z0] | L]`
Where v0=(x0, y0, z0) is first, v1=(x1, y1, z1) is second, v2=(x2, y2, z2) is the third triangle vertex and [nx, ny, nz] is the cross product (v1-v0)x(v2-v0).

Variables:

Writes:
`K : First vertex (3D vector).L : Second vertex (3D vector).M : Third vertex (3D vector).Q : Quotient (y1-y3)*(x2-x3)+(y2-y3)*(x2-x1).U : First vertex of triangle projected to 2D (2D vector).V : Second vertex of triangle projected to 2D (2D vector).W : Third vertex of triangle projected to 2D (2D vector).X : Index of coordinate discarded in 3D->2D projection.`
Program:
`T001 LBL TT002 XEQ U001T003 FS? 2T004 XEQ X068T005 STO MT006 RDNT007 FS? 2T008 XEQ X068T009 STO LT010 RDNT011 FS? 2T012 XEQ X068T013 STO KT014 R^T015 X<>YT016 -T017 RDNT018 RDNT019 LASTXT020 -T021 X<>YT022 RDNT023 XEQ X001T024 ABST025 1E-6T026 X>Y?T027 GTO T065T028 LASTXT029 XEQ X025T030 STO XT031 RCL KT032 X<>YT033 XEQ X052T034 STO UT035 RDNT036 RCL LT037 X<>YT038 XEQ X052T039 STO VT040 RDNT041 RCL MT042 X<>YT043 XEQ X052T044 STO WT045 eq ([0,1]xU-[0,1]xW)x([1,0]xV-[1,0]xW)+        ([0,1]xV-[0,1]xW)x([1,0]xW-[1,0]xU)T046 STO QT047 LASTXT048 4T049 STO IT050 RDNT051 RCL(I)T052 ABSTO53 RDNT054 FS? 2T055 GTO T060T056 RCL KT057 RCL LT058 RCL MT059 GTO T063T060 RCL UT061 RCL VT062 RCL WT063 R^T064 RTNT065 SF 10T066 eq DEGENERATEDT067 PSET068 CF 10T069 XEQ U029T070 RTN`

Mnemonics: T for Triangle.

Uses the X068 subroutine to place 2D triangle in th XZ plane if flag 2 set and the X025 routine to find out which coordinate to discard for the 3D to 2D projection and finaly the X052 routine to actual do the 3D to 2D conversion. In the 2D case this is a rather roundabout way to get end user's input into the variables U, V and W! But I think this worked out quite well in order to have one program for both the 2D and 3D case:

One can compute barycentric coordinates in 3D using four cross product, four dot product and three scalar division operations avoiding projection from 3D to 2D, but involves more scalar math operations than the implemented method. On a vector architecture it may be a better choice, on the HP-35s architecture I think this is a good solution!

The index of the coordinate discarded to do the 3D to 2D projection is remembered in variabe X so the next program to actual compute barycentric coordinates know which coordinate to discard from input.

An unrelated change done to the T program is that it now check if the computed plane normal vector is considered to be of zero length: Then all points are the same or colinear, the message "DEGENERATED" is then shortly displayed and stack registers restored.

Note that one get a cross product calculated normal vector also in the 2D case (direction
[0, 1, 0]), length still twice triangle's area.

Step 2 : Program to convert from Cartesian to Barycentric Coordinates

Stack Input/Output:
`[[x, y, z], Y, Z, T | L]->XEQ B->[[b1, b2, b3], Y, Z, T | [x, y, z]]`
Where (x, y, z) are cartesian coordinates for point in question and (b1, b2, b3) are computed barycentric coordinates.

Variables:

`Q : Quotient (y1-y3)*(x2-x3)+(y2-y3)*(x2-x1). Populated by the P program.U : First vertex of triangle projected to 2D (2D vector). Populated by the P program.V : Second vertex of triangle projected to 2D (2D vector). Populated by the P program.W : Third vertex of triangle projected to 2D (2D vector). Populated by the P program.X : Index of coordinate discarded in 3D->2D projection. Populated by the P program. `
Program:
`B001 LBL BB002 XEQ U001B003 FS? 2B004 GTO B007B005 RCL XB006 XEQ X052B007 eq [([0,1]xREGX-[0,1]xW)x([1,0]xV-[1,0]xW)+         ([0,1]xV-[0,1]xW)x([1,0]xW-[1,0]xREGX),         ([0,1]xREGX-[0,1]xU)x([1,0]xW-[1,0]xU)+         ([0,1]xW-[0,1]xU)x([1,0]xU-[1,0]xREGX),         ([0,1]xREGX-[0,1]xV)x([1,0]xU-[1,0]xV)+         ([0,1]xU-[0,1]xV)x([1,0]xV-[1,0]xREGX)]/QB008 XEQ U050B009 RTN`

Mnemonics: B for Barycentric.

If the input is a 2D (flag 2 set) then goes straight to the equation that calculates baycentric coordinates using the data precomputed by the modified T program, else first project the 3D input to 2D using subroutine X052 and information found in the X variable.

HP-35s 2D to 3D vector conversion

Short program to convert a 2D vector to a 3D by placing it in the z=0 plane. Usefull when want to utilize 3D solutions to 2D data, for example taking cross product of two 2D vectors.

Stack Input/Output:
`[[u, v], Y, Z, T | L]-> XEQ 068 ->[[u, v, 0], Y, Z, T | [u, v]]`
Program:
`X068 ABSX069 RDNX070 eq [[1,0]xLASTX,[0,1]xLASTX,0]X071 RTN`

Placed in the X 'library' since a vector utlity, from before have projecting to a cardinal plane, finding the component of a 3D vector with max absolute value, vector unpack routines, cross product and normalization.

Sunday, October 14, 2007

HP-35s 3D to 2D projection utility subroutines - II

Subroutine to project a 3D vector to 2D by paralell projection to one of the cardinal planes; discards one coordinate.

Stack Input/Output
`[n, [u,v,w], Z, T | L]-> X047 ->[[x,y], n, [u,v,w], Z | L]`
Where n tells which coordinate to discard: [x=v, y=w] if n=1, [x=u, y=w] if n=2 and [x=u, y=v] if n=3.

Program:
`X052 1X053 X=Y?X054 GTO X062X055 RDNX056 2X057 X=Y?X058 GTO X065X059 RDNX060 eq [[1,0,0]xREGY,[0,1,0]xREGY]X061 RTNX062 RDNX063 eq [[0,1,0]xREGY,[0,0,1]xREGY]X064 RTNX065 RDNX066 eq [[1,0,0]xREGY,[0,0,0]xREGY]X067 RTN`

The parameter n is typical found using the X025 subroutine.

Placed in the X 'library' since a vector utlity, from before have finding the component of a 3D vector with max absolute value, vector unpack routines, cross product and normalization.

Change history:

20071014:1723UTC : Because of change in the X025 subroutine line numbers changed here. Entry point used to be X047 but is now X052.

Friday, October 12, 2007

HP-35s 3D to 2D projection utility subroutines - I

Utility 3D vector subroutine to tell which component that has the largest absolute value.

This is of use when projecting points in an arbitary plane to one of the three cardinal planes in order to reduce a 3D problem to a 2D problem (when possible). Use this subroutine on the arbitary plane's normal vector in order to decide which 3D coordinate to discard in order to avoid all points to be projected on a line if arbitary plane is perpendicular to the choosen cardinal plane or to avoid floating precision problems in case they are nearly perpendicular:

Examine the arbitary plane's normal vector (using this subroutine) and discard the coordinate that has the maximum absolute value.

Stack Input/Output:
`[[u, v, w], Y, Z, T | L]-> XEQ X025 ->[n, -, -, - | [u, v, w]]`
Where [u, v, w] is the vector in question and n is 1 if u has largest absolute value, 2 if v has and 3 if w has.

Program:
`X025 XEQ X008X026 LASTXX027 RDNX028 ABSX029 RDNX030 ABSX031 RDNX032 ABSX033 X<>YX034 ABSX035 X<>YX036 R^X037 X>=Y?X038 GTO X047X039 X<>YX040 R^X041 X<Y?X042 GTO X045X043 3X044 RTNX045 1X046 RTNX047 R^X048 X>=Y?X049 GTO X043X050 2X051 RTN`

Placed in the X 'library' since a vector utlity, from before have vector unpack routines, cross product and normalization.

Change history:

20071014:1723UTC : Problem that vector lost, changed so vector now can be found in LASTX register. This also effects the X047 subroutine now X052 subroutine and those using that: Entry point address changed to X052.

Thursday, October 11, 2007

HP-35s Euclid Implementation - XII : Plane Plane Intersection

The intersection between two non paralell planes is a line (problem 13 of the requirements).

The program here computes two points that defines the intersection line between the plane defined by the P program and a second plane given as input to this program.

Stack Input/Output:
`[[nx, ny, nz], [px, py, pz], Z, T | L]-> XEQ V ->[[x1, y1, z1], [x0, y0, z0], [nx, ny, nz], [px, py, pz] | L]`
The input of the second plane is similar to defining the first plane using the P program: [nx, ny, nz] in X stack register is a vector normal to the second plane and (px, py, pz) in Y stack register is a point in the second plane. The result in the case the planes are not paralell is two points defining the line: (x0, y0, z0) and (x1, y1, z1).

In the case the planes are paralell the message "PARALELL" is displayed shortly, stack and LASTX registers are then left unchanged.

Variables:

Writes:
`R : Normalized normal vector.S : Distance from origo (0,0,0) to plane.T : Defining point in the plane.Z : Scratch data.`
Program:
`V001 LBL VV002 XEQ U001V003 XEQ X004V004 STO RV005 X<>YV006 STO TV007 X<>YV008 xV009 +/-V010 STO SV011 RCL NV012 LASTXV013 XEQ X001V014 STO ZV015 ABSV016 1E-6V017 X>Y?V018 GTO M041V019 LASTXV020 XEQ X025V021 1V022 X=Y?V023 GTO V032V024 RDNV025 2V026 X=Y?V027 GTO V036V028 eq (Sx([0,1,0]xN)-Ox([0,1,0]xR))/([0,0,1]xZ)V029 eq (Ox([1,0,0]xR)-Sx([1,0,0]xN))/([0,0,1]xZ)V030 0V031 GTO VO39V032 0V033 eq (Sx([0,0,1]xN)-Ox([0,0,1]xR))/([1,0,0]xZ)V034 eq (Ox([0,1,0]xR)-Sx([0,1,0]xN))/([1,0,0]xZ)V035 GTO V039V036 eq (Ox([0,0,1]xR)-Sx([0,0,1]xN))/([0,1,0]xZ)V037 0V038 eq (Sx([1,0,0]xN)-Ox([1,0,0]xR))/([0,1,0]xZ)V039 eq [REGZ,REGY,REGX]V040 ENTERV041 ENTERV042 RCL+ ZV043 XEQ U079V044 RTN`

Mnemonic: None.

The program stores the normalized implicit equation for the second plane and it's defining point in the calculator the same way as the P program does for the first plane: The normalized normal vector is stored in variable R, distance to origo in variable S and defining point in variable T. Mnemonics for these are that the keys associated with these variables are below and to the left of the keys assigned to the variables N, O and P where the same values are stored for the plane defined using the program P. This makes the relation between the P program and V program similar to the relation between the L program and the M program.

This program needs to compute and store the cross product of the two plane's normal vector: If this vector is of zero length the planes are paralell else it is a directional vector for the intersection line. This vector is stored in the Z variable.

This shares the show "PARALELL" message and stack restore code with the M program in the no solution case.

Change history:

20071013:1415UTC : This program was implemented before the X025 routine that finds the component of a 3D vector with largest absolute value. This program did that inline, now gone back and modified this program to use X025.

Addition to the HP-35s stack save/restore program - III

Addition to the U program when in need to restore LASTX register and push two new values on the stack:
`[X, Y, Z, T | L]-> Program invoking U001 first and U070 last ->[X', Y', X, Y | L]`
Where X' is new content of X stack register present in the X stack register before U079 is executed and Y' is new content of Y stack register present in Y stack register before U079 is executed.

Of interest for programs that do not take stack register input but produces a result in the X and Y stack register.

Program:
`U079 4U080 STO IU081 RDNU082 RCL(I)U083 ABSU084 RDNUO85 XEQ U041U086 R^U087 R^U088 RTN`

So now we have the subroutines:
1. XEQ U001 to save the stack and LASTX.
2. XEQ U029 to restore the stack and LASTX given U001 has been used to save.
3. XEQ U050 to implement operation native unary operation stack and LASTX behaviour given U001 has been used to save.
4. XEQ U070 to push single value on the stack and preserve LASTX given U001 has been used to save.
5. XEQ U079 to push two values on the stack and preserve LASTX given U001 has been used to save.

Tuesday, October 9, 2007

Communicating RPN stack and LASTX register transitions, a notation

I will try using the following notation to communicate program's effect on the stack and LASTX register:
`[X, Y, Z, T | L]-> opr. desc. ->[X', Y', Z', T' | L']`
So for example to describe sqrt:
`[X, Y, Z, T | L]-> sqrt ->[sqrt(X), Y, Z, T | X]`
`[X, Y, Z, T | L]-> + ->[Y+X, Z, T, T | X]`

Sunday, October 7, 2007

HP-35s 3D and 2D vector unpack subroutines

Subroutines to unpack 3D and 2D vectors when need to work on individual components.

Stack Input/Output:
`3D case: [[u0, u1, u2], Y, Z, T | L]-> X008 ->[u2, u1, u0, u0 | [u0, u1, u2]]2D case: [[u0, u1], Y, Z, T | L]-> X018 ->[u1, u0, Y, Y | [u0, u1]]`
Program:
`XOO8 [1,0,0]X009 X<>YX010 xX011 [0,1,0]X012 LASTXX013 xX014 [0,0,1]X015 LASTXX016 xX017 RTNX018 [1,0]X019 X<>YX020 xX021 [0,1]X022 LASTXX023 xX024 RTN`

Uses the X label for vector utility programs/routines, from before have cross product and normalization.

Wednesday, October 3, 2007

HP-35s Euclid Implementation - XI : Line Plane Intersection

The point of intersection between a line and a plane (problem 12 in the requirements) is given by the parameter (to it's parametric equation) t=-(ax0+by0+cz0+d)/n*(p1-p0) (* is here the dot product) where p0=(x0, y0, z0) and p1 are the two points defining the line.

The following program finds the intersection between the 3D line defined using the L program (p0 stored in variable A and p1 stored in variable B) and the plane defined using the P program ([a, b, c] stored in variable N and d stored in variable O).

Stack Input/Output:
`No input on stack-> XEQ R ->[t, X, Y, Z | L]`
Where t is the parameter defining intersection point using line l's parametric equation.

If n*(p1-p0)=0 then the line and the plane are paralell and there is no intersection point. In that case the message "PARALELL" is displayed shortly, stack and LASTX registers are then left unchanged.

Variables:

`A : Line's first defining point. Populated by the L program.B : Line's second defining point. Populated by the L program.N : Plane's normalized normal vector. Populated by the P program.O : Plane's distance to origo (0, 0, 0). Populated by the P program.`
Program:
`R001 LBL RR002 XEQ U001R003 RCL NR004 RCL BR005 RCL AR006 -ROO7 xR008 ABSR009 1E-6R010 X>Y?R011 GTO M041R012 RDNR013 LASTXR014 eq -(AxN+0)/REGXR015 XEQ U070R016 RTN`

Mnemonic: None.

Again when the result is a point on a line we produce the parameter giving the point, not the coordinates. So for example one learns that the line segment A, B intersect the plane if result is in the range [0, 1]. To get the coordinates end user does XEQ E.

Uses the new subroutine U070 to raise the stack and preserve LASTX register content in the case there is a solution (line and plane not paralell).

This shares the show "PARALELL" message and stack restore code with the M program in the no solution case.

Change history:

20071004:1835UTC : Fixed typo in line R011, was GTO M014!

Addition to the HP-35s stack save/restore program - II

Extenstion to the U program for the case when a result has been produced and should be pushed on the stack:
`[X, Y, Z, T | L]->Program invoking U001 first and U070 last ->[X', X, Y, Z | L]`
Where X' is new content of X stack register present in the X stack register before U070 is executed.

Of interest for programs that do not take stack register input but produces a result in the X stack register.

Program:
`U070 4U071 STO IU072 RDNU073 RCL(I)U074 ABSU075 RDNU076 XEQ U037U077 R^U078 RTN`

So now we have the subroutines:
1. XEQ U001 to save the stack and LASTX.
2. XEQ U029 to restore the stack and LASTX given U001 has been used to save.
3. XEQ U050 to implement operation native unary operation stack and LASTX behaviour given U001 has been used to save.
4. XEQ U070 to push single value on stack and preserve LASTX given U001 has been used to save.

Tuesday, October 2, 2007

Disclaimer

Oh, since I'am posting code I guess I should have one of these....

You are welcome to use the programs you find on this site free of charge. While I hope these programs become rather bug free and robust I make no claim that they are. I am not liable for any damages, injuries, or losses that may result in using any of these programs. By using these programs or ones that you create based on them, you agree to these terms.

HP-35s Euclid Implementation - X : Closest point on a plane to a point

The closest point on a plane to a point P is P-(P*N+d)*N where N is the plane's normalized normal vector and d is the plane's distance to origo (* is here the dot product).

Since the program P stores N in variable N and d in variable O the following short program computes the closest point on the plane entered using program P if the point's positional vector is in the X stack register (problem 11 in the requirements).

Stack Input/Output:
`[[x, y, z], Y, Z, T | L]-> XEQ Q ->[[cx, cy, cz], [x, y, z], Y, Z | L]`
Where (x, y, z) is point p and (cx, cy, cz) is closest point in plane to p.

Variables:

`N : Plane's normalized normal vector. Populated by the P program.O : Plane's distance to origo (0, 0, 0). Populated by the P program.`
Program:
`Q001 LBL QQ002 eq REGX-(REGXxN+O)xNQ003 RTN`

Mnemonic: Sorry, run out :-(

Unless the input is a point in the plane (if this is the case the vectors in Y and X stack register are the same after execution) the line defined by the point in Y stack register and the point in X stack register defines the perpendicular line from point now in the Y stack register onto the plane entered using the program P. May be used as input to the L program or M program.

Monday, October 1, 2007

HP-35s Euclid Implementation - IX : Signed distance from a plane to a point

The plane is the hyperplane of 3D (like a line in 2D divides the space a plane divide 3D space) so since program P stores the plane's normalized implicit equation point p=(x, y, z)'s signed distance to the plane is ax+by+cz+d or (since [a, b, c] is stored in variable N and d is stored in variable O) N*p+O where * here is the dot product operator.

I choose to modify the S program so that it computes the signed distance in 2D (as before) if flag 2 is set and else computes signed distance in 3D (to the plane defined by the P program); problem 10 of the requirements.

Stack Input/Output:

2D case:
`[[x, y], Y, Z, T | L]-> XEQ S ->[d, Y, Z, T | [x, y]]`
3D case:
`[[x, y, z], Y, Z, T | L]-> XEQ S ->[d, Y, Z, T | [x, y, z]]`
Where (x, y) or (x, y, z) is point to find distance to and d is the signed distance.

Variables:

`C : [a, b, c], constants of line's normalized implicit equation. Populated by the L program. Reads in 2D case (flag 2 set).N : Plane's normalized normal vector. Populated by the P program. Reads in 3D case.O : Plane's distance to origo (0, 0, 0). Populated by the P program. Reads in 3D case. `
`S001 LBL SS002 ABSS003 RDNS004 FS? 2S005 GTO 008S006 eq LASTXxN+OS007 RTNS008 eq (([1,0,0]xC)x([1, 0]xLASTX))+(([0,1,0]xC)x([0,1]xLASTX))+[0,0,1]xCS009 RTN`