Find exponential value in micrologix

[BachPhi]

Given y = 2^x . Find x when y is known. IF x is not a whole number, reject it ( move -1 to result).

1. Given CPT, XPY, LOG available

2. Can you do it with less? like XPY or LOG is not avail. ML1000 do not have these instructions

I currently have a long routine to do it, but I would love to see other solutions. TIA.

 

[dweisser]

What range of values can Y be?

 

[BachPhi]

What range of values can Y be?

y range include 0 to infinite. y is a whole number

y = 2^x if y = 8 then x = 3
y = 12 then x = 3.585 => x = -1
y = 0 then x = -1

 

[dweisser]

Well, we know the formula is gonna be X = (log y/log 2). Log 2 we can do ourselves and change to 0.301029996. Now, the only times this is going to be a whole number, are multiplications of 8. 8, 16, 32, 64, 128, 256, etc. Try it. Map out the formula in Excel and look. 8, 16, 64, 128, 256, etc. It makes sense - your denominator is only going to evenly divide a numerator that is a multiple of 8 since 8 is the first time it divides evenly. Is a modulo function available? You could just divide your Y by 8 and see if the remainder is 0.

Edit: Just realized it goes by 8^1, 8^2, 8^3 etc and not multiples of 8. That changes things a bit. You could test if you continue to divide by 8 that it eventually reaches 1.

Edit 2: Jesus, why cant I math today? it's 8*2, 8*2*2, 8*2*2*2 etc so in reality, you'd have to divide by 2 until it reached 8, and make 8 itself an exception.

There's probably a trick with binary you could use that I haven't thought of, since that's how binary math works past the 3rd bit in the word. You could check above the 3rd LSB to see if only 1 bit is set as mentioned below me

 

[bernie_carlton]

I tried it by setting a single bit using indirect but the ML1000 doesn't support indirect. Too bad.

It works fine in a ML1100 for integers and long.

Edit - never mind - misread the OP question

 

[Peter Nachtwey]

Easy, count bits. If the log2 is not an integer there will be more than one bit set.
Bit hacks have been discussed before.

 

[dweisser]

Easy, count bits. If the log2 is not an integer there will be more than one bit set.
Bit hacks have been discussed before.

I believe his main problem is that you cannot use the log function with micrologix 1100/1000

 

[bernie_carlton]

ML1000 only supports integer - so y = 0 to 32767 (positive range)

If y is positive then ...

In 15 rungs examine the bits of y starting from the lowest. In each rung, if bit is set then increment a 'count' and set an 'index' value starting at 0 for the first rung, increasing by 1 in each rung.

After this ...

If the 'count' is greater than 1 then invalid (x is non integer) - set result to -1.

If 'count' is 1 then x = index.

If the 'count' is zero then Y was zero.

This is just an expansion of Peter's note.

You could also just use 15 rungs each comparing y to the result of 2^x where x is an integer (0 - 14)

 

[BachPhi]

X and Y can be [0,1,2,3,...]
x = log y / log 2 , but how do you handle the result without MOD function?

 

[Thomas_v2]

For a limited range of values, you could use the floating point representation bits of the exponent (8 bits) to get the log2 of your value.

load floating point value
shift left 1
shift right 24
subtract 127
get the log2 as integer value

 

[bernie_carlton]

The OP limited the discussion to the ML1000.
It doesn't support floats. It doesn't support Long (32 bit integer).

Therefore -32768 =< y =< 32767

 

[BachPhi]

The OP limited the discussion to the ML1000.
It doesn't support floats. It doesn't support Long (32 bit integer).

Therefore -32768 =< y =< 32767

I gave 2 options. in option 2 , some instructions may not be avail.

 

[bernie_carlton]

Sorry, did not recognize the separate options.

 

[proof]

Mov -1 x
equ y 1 mov 1 x
equ y 2 mov 2 x
equ y 4 mov 3 x
equ y 8 mov 4 x
equ y 16 mov 5 x
equ y 32 mov 6 x
equ y 64 mov 7 x
equ y 128 mov 8 x
equ y 256 mov 9 x
equ y 512 mov 10 x
equ y 1024 mov 11 x
equ y 2048 mov 12 x
equ y 4096 mov 13 x
equ y 8192 mov 14 x
equ y 16384 mov 15 x

 

[BachPhi]

@proof: I like your brute force approach, but what if y = 3, 12, 15, etc...? Actually x = 0 when y =1, x = 1 when y = 2, x =3 when y =8