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@ -0,0 +1,151 @@
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[Von Neumann's middle-square pseudo-random number generator, for Rosetta Code.
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EDSAC program, Initial Orders 2.]
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[Arrange the storage]
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T46K P56F [N parameter: library subroutine P7 to print integer]
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T47K P134F [M parameter: main routine]
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T51K P92F [G parameter: generator for pseudo-random numbers]
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[This version of von Neumann's PRNG uses values in the range 0..999999.
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Initialize: Call 0G with seed in 0D.
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Next term: Call 1G; term is returned in 0D.
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41 storage locations, load at even address. Workspace 4D.]
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E25K TG GK
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G10@ [jump to initialize the generator]
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G15@ [jump to return the next term in 0D]
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[Instructions to the loader - not executed at runtime]
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T2#Z PF [ensure sandwich bit between 2@ and 3@ is zero]
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T4#Z PF T6#Z PF [same for 4@ and 5@, 6@ and 7@]
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T2Z [resume normal loading at 2@]
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[Constants]
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[2] M1667D I1208F [2^29/10^9, near enough (see note at end)]
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[4] G1327D I393F [2^9/10^3, near enough (see note at end)]
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[10^9/2^34 can't be stored using pseudo-orders, so store its negative instead.]
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[6] D768F V140D [-10^9/2^34]
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[Variable]
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[8] PF PF [state of PRNG]
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[Initialize the PRNG. Caller passes seed in 0D.]
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[10] A3F T14@ [plant return link as usual]
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AD T8#@ [copy seed to state]
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[14] ZF [(planted) jump back to caller]
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[Return the next value in 0D.
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Outline: Let X = state, 0 <= X <= 999999, fits into 20 bits.
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Calculate Y = X^2 div 10^9.
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Deduce Z = X^2 mod 10^9 = X^2 - (10^9)*Y
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The next state is Z div 10^3.
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See note at end of program for details of integer division.]
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[15] A3F T40@ [plant return link as usual]
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A8#@ [acc := X/2^34]
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L32F L32F [shift 7 + 7 left, acc := X/2^20]
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T4D [store X/2^20 in 4D]
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H4D V4D [square, acc := (X^2)/2^40]
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[Here acc holds sign bit plus 40 binary places of (X^2)/2^40.]
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[On storing acc in a 35-bit location, the low 6 bits are lost.]
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[This doesn't matter, because 2^6 divides 10^9, so we can evaluate]
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[X^2 div 10^9 = (X^2 div 2^6) div (10^9/2^6).]
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TD [0D now represents X^2 div 2^6]
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H2#@ [mult reg := 2^29/10^9 nearly]
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VD
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[Commented out: code for 35-bit operations, following note at end.
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|R 1024 F| |R 512 F| shift 23 right, as in note
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|T D| 0D := Y/2^34 where Y = X^2 div 10^9
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|H neg_10_9 #@| |V D| times -10^9/2^34, acc := -10^9*Y/2^68
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We need to shift 34 left to restore the scaling after multiplication
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|L 1024 F| |L 1024 F| |L 4 F| first shift 28 left, acc := 10^9*Y/2^40]
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[More efficient code, possible because Y (= X^2 div 10^9) fits into a 17-bit location.
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Shifting is 18 less than the 35-bit version.]
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R8F [shift 5 right]
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TF [0F := Y/2^16]
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H6#@ VF [times -10^9/2^34, acc := -10^9*Y/2^50]
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[We need to shift 16 left to restore the scaling after multiplication]
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L256F [first shift 10 left, acc := 10^9*Y/2^40]
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H4D V4D [4D = X/2^20 from above, so acc := (X^2 - 10^9*Y)/2^40]
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L16F TD [shift 6 more left; 0D := Z/2^34 where Z = X^2 - 10^9*Y]
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H4#@ [mult reg := 2^9/10^3 nearly]
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VD
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R128F [shift 9 right]
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U8#@ [save next state u]
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TD [also return next state to caller in 0D]
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[40] ZF [(planted) jump back to caller]
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[------------------------------------------------------------------------]
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E25K TM GK [M parameter, main routine. Load at even address.]
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T#Z PF [clear 35-bit value at relative locations
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0 & 1, including the middle ("sandwich") bit]
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TZ [resume normal loading at relative location 0]
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[We could read the seed and number of terms from a separate tape,
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but that would require another subroutine.]
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[0] G296F V2046D [-675248 (negative of seed, cf -10^9 above)]
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[2] P211F [number of terms, in the address field]
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[3] PF [index of term]
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[4] #F [teleprinter, set figures mode]
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[5] !F [space]
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[6] &F [line feed]
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[7] @F [carriage return]
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[8] K4096F [null]
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[9] PF
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[Enter with acc = 0]
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[10] O4@ [set teleprinter to figures]
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S#@ TD [pass seed in 0D]
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[13] A13@ GG [call subroutine to initilaize PRNG]
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T3@ [index := 0]
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[Head of loop; here with acc = 0]
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[16] TD [clear parameter for print subroutine]
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A3@ S2@ [printed enough terms?]
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E34@ [if so, jump to exit]
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A2@ [restore acc after test]
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A2F U3@ [update index]
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RD TF [right-justify for printing; pass in 0D]
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[25] A25@ GN [call subroutine to print index]
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[27] A27@ G1G [call PRNG; returns next term in 0D]
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[29] A29@ GN [print term; clears acc.]
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O7@ O6@ [print CR, LF]
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E16@ [loop back for next term]
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[34] O8@ [print null to flush printer buffer]
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ZF [stop]
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[----------------------------------------------------------------------]
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E25K TN [N parameter, print subroutine]
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[Library subroutine P7, prints long strictly positive integer in 0D.]
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[10 characters, right justified, padded left with spaces.]
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[Even address; 35 storage locations; working position 4D.]
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GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSF
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L4FT4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@
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[----------------------------------------------------------------------]
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E25K TM GK [M parameter again]
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E10Z [start execution at relative address]
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PF
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[
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Note: Integer division by a constant on EDSAC.
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Some programs require the integer quotient N div D, where N is variable and
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D is constant throughout the program. Since EDSAC had hardware multiplication
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but not hardware division, it makes sense to store 1/D and multiply N by that.
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This note describes one way of doing this, and finds a sufficient condition on N
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for rounding errors not to affect the result. It's assumed that 35-bit integers are used,
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so that an integer N is represented by N*(2^-34); also that N and D are positive.
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Let k be the greatest integer such that 2^k < D. We store a multiplier 2^k/D,
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rounded up to an integer multiple of 2^-34. Let the stored multiplier be
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2^k/D + f*(2^-34), where 0 <= f < 1.
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Let N = q*D + r, where q is an integer and 0 <= r < D. To find the integer quotient q,
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multiply N*(2^-34) by the stored multiplier and shift right by k bits. This gives
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(2^-34)*(q + r/D + N*f*(2^(-k -34)))
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in the accumulator. If the top 35 bits of the accumulator are stored in memory,
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the value stored will be q*(2^-34), provided
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r/D + N*f*(2^(-k-34)) < 1.
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Since r <= D - 1, a sufficient condition for this is
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N*f*(2^(-k-34)) < 1/D
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which is equivalent to
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N < (2^k/D)*(2^34/f). (*)
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Examples from the middle-square program:
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(1) D = 10^9/2^6. Then k = 23 and 2^k/D = 0.536870912. Also
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2^(k+34)/D = 2^63/10^9 = 9223372036.854775808
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so that the stored multiplier is 9223372037*(2^-34) and f = 0.145224192.
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Since f < 2^k/D, the condition (*) is satisfied for all N < 2^34.
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(2) D = 10^3. Then k = 9 and 2^k/D = 0.512. Also
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2^(k+34)/D = 2^43/10^3 = 8796093022.208
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so that the stored multiplier is 8796093023*(2^-34) and f = 0.792.
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The condition (*) becomes N < 1.11*(10^10) (approx), which is satisfied
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in this program because N < 10^9. ]
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Ingy döt Net