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(SLIGHTLY HARDER) CONTROL SIGNAL ATTENUATOR
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956.000 B
2 Dec, 2016 @ 12:11pm
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grasspound
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(SLIGHTLY HARDER) CONTROL SIGNAL ATTENUATOR

Description
Dividing a value by 2 has never been so difficult!
But seriously, this is just a harder version of CONTROL SIGNAL ATTENUATOR, in that it gives you all even numbers as inputs, not just multiples of 20.

My best is 5¥/660/11. What can you do?
10 Comments
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Verdammte Heinz 26 Jun, 2023 @ 1:30pm 
Improved costs to 7 / 780 / 14 but at the cost of 2 more lines of code
Verdammte Heinz 26 Jun, 2023 @ 1:11pm 
First attempt 9 / 660 / 12
slow down cadet 9 Oct, 2017 @ 2:15pm 
I got 5/660/11 (based on reusing the exact same solution that I made for the 'regular' Control Signal Attenuator.. which I didn't realise was 'easier' until I saw this 'Slightly Harder' version!).
Simon 29 Apr, 2017 @ 7:59am 
Hm, 8/721/12. I must be missing how to juggle the digits around in a single MC6000?
Razhan 25 Feb, 2017 @ 8:02pm 
3/357/7
All test inputs are multiples of 10. Treated 100 as a special case and for the rest only had to divide the central digit (dgt 1 plus mul 5 works)
ShadowCluster 22 Feb, 2017 @ 12:23pm 
A bit late, but whatever. 9/540/9. Used two ROMs to avoid using DST, but used DGT in this case.
konimaru 9 Dec, 2016 @ 1:32pm 
5/600/10 confirmed :)
Player1 3 Dec, 2016 @ 11:24am 
It seems I'm missing something. I got 5/8877/8 by counting up until I have half of the input signal and 9/720/13 when using one ROM per digit as lookup table . But I don't see any way to further reduce the power usage, at least not without signifcantly more code.
AVO 2 Dec, 2016 @ 6:25pm 
5/600/10. Use dst
The Slothful Pixie 2 Dec, 2016 @ 5:33pm 
5/660/11, exactly the same as what I had. Likely the same solution, it seemed the most obvious way to do it.