20161203, 01:56  #1 
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}·67 Posts 
Binary Multitasking
Hi,
Most of my life I have been struggling with coming up with a scheduling system to be a balanced approach to concentrating on high priority tasks as well as providing time allocation to lower priority tasks. One of the best methods I have come up with is binary multitasking. Essentially you go through a list of tasks as if they were binary digits, incrementing by 1. This way one would spend about 1/2 the total time on the 1st task, 1/4 on 2nd and so on. The problem with this method is that for lists longer than say 10 or so tasks virtually no time ends up being allocated to the tasks at the bottom. * Is there a way to increment a binary register with a constant value greater than one such that all binary digits are set at least once? If not, is there any formulaic simple parameter? Thanks in advance. ETA I assume any prime number greater than the number of the elements in the list would serve the purpose, but would such a solution end up being equivalent to incrementing 1 or would allocate more time to the items lower in the list? Last fiddled with by a1call on 20161203 at 02:28 
20161203, 03:25  #2 
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}×67 Posts 
I think it would be more accurate to phrase it am as:
ETA I assume any prime number equal to the number of the items in the list would serve the purpose, but would such a solution end up being equivalent to incrementing 1 or would allocate more time to the items lower in the list? 
20161203, 16:16  #3 
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}×67 Posts 
Here is a pseudocode for clarification:
My array tasks[26] forstep (j=0, 2197, step =13) . { . Print j in binary . Flag=0 . Exp=0 . While(flag=0) . { . Exp= exp +1 . If(j/(2^ exp)==j\(2^ exp)) . { . Tasks[Exp]=+ . Flag=1 . } . } . } For (k=1,26) . { . Print tasks[k] . } I would expect only tasks 1 to 13 to be nonzero and equal to the value if the step was equal to 1 Last fiddled with by a1call on 20161203 at 16:18 
20161203, 17:44  #4 
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}·67 Posts 
Too late to edit so posing with corrections:
Here is a pseudocode for clarification: My array tasks[26] forstep (j=0, 2197, step =13) . { . Print j in binary . Flag=0 . Exp=0 . While(flag=0) . { . Exp= exp +1 . If(j/(2^ exp)!=j\(2^ exp)) . { . Tasks[Exp]=+ . Flag=1 . } . } . } For (k=1,26) . { . Print tasks[k] . } I would expect only tasks 1 to 13 to be nonzero and equal in proportion to total, to the value if the step was equal to 1 
20161203, 18:30  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Code:
my(task=vector(26));forstep (j=0, 2197,13,print(binary(j));Flag=0;Exp=0;while(Flag==0,Exp+=1;if(j/(2^Exp)!=j\(2^Exp),task[Exp]+=1;Flag=1)));for(k=1,26,print(task[k]","k)) Last fiddled with by science_man_88 on 20161203 at 18:52 

20161203, 19:09  #6  
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}×67 Posts 
Quote:
The uppercase is inevitably substituted by my swipes keyboard. 

20161203, 19:30  #7 
Aug 2006
2^{2}×5×293 Posts 
More generally, choose any 0 < p < 1 and put 1  p, p(1  p), p^2(1  p), ... effort into each task. For example, with p = 2/3, you put a third of your effort into the first task, 22% into your second, 15% into your third, 9.9% into your fourth, 6.6% into your fifth, etc.
Last fiddled with by CRGreathouse on 20161203 at 19:34 
20161203, 19:34  #8  
Bamboozled!
May 2003
Down not across
5·1,999 Posts 
Quote:
If you find there is something you are not doing which ought to be done, find something else that needs to be done and don't do that instead. Works for me. Last fiddled with by xilman on 20161203 at 21:03 

20161203, 21:20  #9  
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}·67 Posts 
Quote:
I will need more time to wrap my head around that. 

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