WARNING: Numbers in this section are octal (and occasionally binary) unless followed by a decimal point. 105=69.. (And 105.=69 hexadecimal.) [PDP-10 Info]
0/ TLCA 1,1(1) 1/ see below 2/ ROT 1,9 3/ JRST 0This is a display hack (that is, it makes pretty patterns) with the low 9 bits = Y and the 9 next higher = X; also, it makes interesting, related noises with a stereo amplifier hooked to the X and Y signals. Recommended variations include:
CHANGE: GOOD INITIAL CONTENTS OF 1: none 377767,,377767; 757777,,757757; etc. TLC 1,2(1) 373777,,0; 300000,,0 TLC 1,3(1) -2,,-2; -5,,-1; -6,,-1 ROT 1,1 7,,7; A0000B,,A0000B ROTC 1,11 ;Can't use TLCA over data. AOJA 1,0
DATAI 2 ADDB 1,2 ROTC 2,-22 XOR 1,2 JRST .-42=X, 3=Y. Try things like 1001002 in data switches. This also does interesting things with operations other than XOR, and rotations other than -22. (Try IOR; AND; TSC; FADR; FDV(!); ROT -14, -9, -20, ...)
4000,,4 1000,,2002 2000,,4 0,,1002(Notation: <left half>,,<right half>)
Also try the FADR after the XOR, switches = 1001,,1.
NEW X = OLD X - epsilon * OLD Y
NEW Y = OLD Y + epsilon * NEW(!) X
This makes a very round ellipse centered at the origin with its size determined by the initial point. epsilon determines the angular velocity of the circulating point, and slightly affects the eccentricity. If epsilon is a power of 2, then we don't even need multiplication, let alone square roots, sines, and cosines! The "circle" will be perfectly stable because the points soon become periodic.
The circle algorithm was invented by mistake when I tried to save one register in a display hack! Ben Gurley had an amazing display hack using only about six or seven instructions, and it was a great wonder. But it was basically line-oriented. It occurred to me that it would be exciting to have curves, and I was trying to get a curve display hack with minimal instructions.
X(N+1) = (2 - epsilon^2) * X(N) - X(N-1)
Y(N+1) = (2 - epsilon) * Y(N) - Y(N-1).
These are just the Chebychev recurrence with cos theta (the angular increment) = 1-epsilon^2/2. Thus X(N) and Y(N) are expressible in the form R cos(N theta + phi). The phi's and R for X(N) and Y(N) can be found from N=0,1. The phi's will differ by less than pi/2 so that the curve is not really a circle. The algorithm is useful nevertheless, because it needs no sine or square root function, even to get started.
X(N) and Y(N) are also expressible in closed form in the algebra of ordered pairs described under linear recurrences, but they lack the remarkable numerical stability of the "simultaneous" form of the recurrence.
X^2 - epsilon X Y + Y^2 = constant,
where the constant is determined by the initial point. This ellipse has its major axis at 45 degrees (if epsilon > 0) or 135 degrees (if epsilon < 0) and has eccentricity
sqrt(epsilon/(1 + epsilon/2)).
let X = the sum of many powers of two = ...111111
now add X to itself; X + X = ...111110
thus, 2X = X - 1 so X = -1
therefore algebra is run on a machine (the universe) which is twos-complement.
HRLOI A,-1(A) EQVI A,ORIGIN
HRLOI A,-ORIGIN(A) EQVI A,ORIGINSlightly faster: change the HRLOIs to MOVSIs and the EQVI addresses to -ORIGIN-1. These two routines are clearly adjustable for BLKOs and other fenceposts.
COS: FADR A,[1.57079632679] ;pi/2 SIN: MOVM B,A ;ARGUMENT IN A CAMG B,[.00017] ; <= 3^(1/3) / 2^13 POPJ P, ;sin X = X, within 27. bits FDVRI A,(-3.0) PUSHJ P,SIN ;sin -X/3 FMPR B,B FSC B,2 FADRI B,(-3.0) FMPRB A,B ;sin X = 4(sin -X/3)^3 - 3(sin -X/3) POPJ P, ;sin in A, sin or |sin| in B ;|sin| in B occurs when angle is smaller than end testChanging both -3.0's to +3.0's gives sinh:
sinh X = 3 sinh X/3 + 4 (sinh X/3)^3.
Changing the first -3.0 to a +9.0, then inserting PUSHJ P,.+1 after PUSHJ P,SIN gains about 20% in speed and uses half the pushdown space (< 5 levels in the first 4 quadrants). PUSHJ P,.+1 is a nice way to have something happen twice. Other useful angle multiplying formulas are
tanh X = (2 tanh X/2) / (1 + (tanh X/2)^2)
tan X = (2 tan X/2) / (1 - (tan X/2)^2),
if infinity is handled correctly. For cos and cosh, one can use
cos X = 1 - 2 (sin X/2)^2, cosh X = 1 + 2 (sinh X/2)^2.
In general, to compute functions like e^X, cos X, elliptic functions, etc. by iterated application of double and triple argument formulas, it is necessary to subtract out the constant in the Taylor series and transform the range reduction formula accordingly. Thus:
F(X) = cos(X)-1
F(2 X) = 2 F*(F+2)
F(epsilon) = -epsilon^2/2
G(X) = e^X - 1
G(2 X) = G*(G+2)
G(epsilon) = epsilon
This is to prevent the destruction of the information in the range-reduced argument by the addition of a quantity near 1 upon the success of the epsilon test. The addition of such a quantity in the actual recurrences is OK since the information is restored by the multiply. In fact, a cheap and dirty test for F(epsilon) sufficiently small is to see if the addition step has no effect. People lucky enough to have a square root instruction can get natural log by iterating
X <- X/(sqrt(1+X) + 1) until 1+X = 1.
Then multiply by 2^(number of iterations). Here, a LSH or FSC would work.
The correct epsilon test in such functions as the foregoing SIN are generally the largest argument for which addition of the second term has no effect on the first. In SIN, the first term is x and the second is -x^3/6, so the answer is roughly the x which makes the ratio of those terms 1/2^27; so x = sqrt(3) / 2^13. But this is not exact, since the precise cutoff is where the neglected term is the power of 2 whose 1 bit coincides with the first neglected (28th) bit of the fraction. Thus, x^3/6 = 1/2^27 * 1/2^13, so x = 3^(1/3) / 2^13.
LOG2: LSHC A,-33 MOVSI C,-201(A) TLC C,211000 ;Speciner's bum MOVI A,200 ;exponent and sign sentinel LOGL: LSH B,-9 REPEAT 7, FMPR B,B ;moby flunderflo LSH B,2 LSHC A,7 SOJG A,LOGL ;fails on 4th try LSH A,-1 FADR A,C POPJ P, ;answer in ABasically you just square seven times and use the low seven bits of the exponent as the next seven bits of the log.
EXCH A,LOC1 EXCH A,LOC2 EXCH A,LOC1Note: LOC1 must not equal LOC2! If this can happen use MOVE-EXCH-MOVEM, clobbering A.
TRCE A,BITS TRCE A,BITS TRCE A,BITSNote (Nelson): last TRCE never skips, and used to be a TRC, but TRCE is less forgettable. Also, use TLCE or TDCE if the bits are not in the right half.
(SETQ X (PROG2 0 Y (SETQ Y X)))
ROTC A,6 CAMG A,B EXCH A,B ROTC A,-6
ROTC A,N ROT B,-N ROTC B,NThus M AC's can be ROTC'ed in 2M-3 instructions. (Stallman): For 73. > N > 35.:
ROTC A,N-36. EXCH A,C ROT B,36.-N ROTC A,N-72.
;B gets 7 bit character in A with even parity IMUL A, ;5 adjacent copies AND A, ;every 4th bit of left 4 copies + right copy IDIVI A,17<-7 ;casting out 15.'s in hexadecimal shifted 7 ;odd parity on 7 bits (Schroeppel) IMUL A, ;4 adjacent copies IOR A, ;leaves every 3rd bit+offset+right copy IDIVI A,9<-7 ;powers of 2^3 are +-1 mod 9 ;changing 7555555400 to 27555555400 gives even parity ;if A is a 9 bit quantity, B gets number of 1's (Schroeppel) IMUL A, ;4 copies AND A, ;every 4th bit IDIVI A,17 ;casting out 15.'s in hexadecimal ;if A is 6 bit quantity, B gets 6 bits reversed (Schroeppel) IMUL A, ;4 copies shifted AND A, ;where bits coincide with reverse repeated base 2^8 IDIVI A,377 ;casting out 2^8 - 1's ;reverse 7 bits (Schroeppel) IMUL A, ;4 copies sep by 000's base 2 (may set arith. o'flow) AND A, ;where bits coincide with reverse repeated base 2^8 IDIVI A,377 ;casting out 377's ;reverse 8 bits (Schroeppel) MUL A, ;5 copies in A and B AND B, ;where bits coincide with reverse repeated base 2^10 ANDI A,41 ;" DIVI A,1777 ;casting out 2^10 - 1's
foo, lat /DATAI switches adm a /ADDB and (707070 adm b iot 14 /output AC sign bit to a music flip-flop jmp fooMakes startling chords, arpeggios, and slides, with just the sign of the AC. This translates to the PDP-6 (roughly) as:
FOO: DATAI 2 ADDB 1,2 AND 2, ;or 171717171717, 363636363636, 454545454545, ... ADDB 2,3 LDB 0,[360600,,2] JRST FOOListen to the square waves from the low bits of 0.
LDB B,[014300,,A] ;or MOVE B,A then LSH B,-1 AND B,[333333,,333333] SUB A,B LSH B,-1 AND B,[333333,,333333] SUBB A,B ;each octal digit is replaced by number of 1's in it LSH B,-3 ADD A,B AND A,[070707,,070707] IDIVI A,77 ;casting out 63.'sThese ten instructions, with constants extended, would work on word lengths up to 62.; eleven suffice up to 254..
DPT: IDIVI F,12 HRLM G,(P) ;tuck remainder on pushdown list SKIPE F PUSHJ P,DPT LDB G,[220600,,(P)] ;retrieve low 6 bits of remainder TRCE G,"0 ;convert digit to character SETOM CCT ;that was no digit! TYO: .IOT TYOCHN,G ;or DATA0 or IDPB ... AOS G,CCT POPJ P,This is the standard recursive decimal print of the positive number in F, but with a LDB instead of a HLRZ. It falls into the typeout routine which returns in G the number of characters since the last carriage return. When called with a -36., DPT types carriage return, line feed, and resets CCT, the character position counter.
CONS: EXCH A,[EXCH A,[...[PUSH P,GC]...]] EXCH A,CONSOf course, the address-linked chain of EXCH's indicated by the nested brackets is concocted by the garbage collector. This method has the additional advantage of not constraining an accumulator for the free storage pointer.
UNCONS: HRLI A,(EXCH A,) EXCH A,CONS EXCH A,@CONSReturns cell addressed by A to free storage list; returns former cell contents in A.
MULI A,400 ;exponent to A, fraction to A+1 TSC A,A ;1's complement magnitude of excess 200 exponent ASH A+1,-200-27.-8(A) ;answer in A+1If number is known positive, you can omit the TSC.
On the PDP-10
UFA A,[+-233000,,] ;not in PDP-6 repertoire
TLC A+1,233000 ;if those bits really bother you
When you know the sign of A, and |A| < 2^26, you can
FAD A,[+-233400,,] ;or FADR for rounded fix!
TLC A,233400 ;if those bits are relevant
where the sign of the constant must match A's. This works on both
machines and doesn't involve A+1. On the 10, FADRI saves a cycle and
a constant, and rounds.
MOVE B,A MOVN C,B AND C,B ADD A,C MOVE D,A XOR D,B LSH D,-2 IDIVM D,C IOR A,C
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