This is nothing to do with Python, or any other language. It's a simple matter of sorting the intial values into order and then searching for the two points closest to the one entered by the user. Could be speeded up by binary chop (Google for that).
I think Richard was suggesting that you google "Binary chop" not sorting.
Try a search of "python binary search closest value" - the first hit that came up for me (in Google) was an answer to the same homework question
Excuse me, first posted in the lounge, but Bill suggested I post here:
I have a range of values (voltage) over time (thousands of minutes, one value per minute). I am trying to chart these. Determining the length of my Y axis is quite a problem for me. If I take a minimum and maximum, and use that as the axis height, one or two zero values result in all the others being scrunched up at the top of the chart. If I remove zeroes, it looks much better, and for a chart, they aren't very important, I'll give all real values in a tabular report.
What I would like to do is determine the average height of the band of data points, sort of the space between the moving average of the low points and that of the heigh points. I figure to do that, I would need a median series, so I could determine a smoothed series of points above and below median, and make my Y axis 's' higher and 's' lower than those.
For my old computer I need an assembler which is able to take assembled code from a library and link it together in the smallest possible combination.
One 'speciality' of the old CDP1802 processor will force me to write the assembler and linker myself. There are two types of branching instructions: long branches and short branches. Long branches use full 16 bit addresses, but will cause timing issues with the graphics chip. This is an ancient hardware bug.
This is the reason why i must use short branches with short 8 bit addresses. The upper 8 bits are just assumed to remain the same as in the instruction's address. This way memory is segmented into 256 byte blocks. It's not a very strict segmentation as the code can run across the boundaries without any consequences, You just can't loop back with a short branch and long branches can't be used.
The linker will have to puzzle together snippets of code and data with this in mind. At the same time I must be sure that memory usage is as low as possible in the end. My old computer has only 4k RAM, and more than 16k is quite unusual.
The only thing I can think of is to make a memory map of each possible combination and take the one which needs the least amount of memory. There are easily hundreds of small code snippets to be linked and blindly testing every combination will be very slow and inefficient.
First thought: Build a tree with only valid options and then find the branch with the lowest byte count. This is alresy better than brute force, but I hope there is still a more elegant algorithm for this.
Assuming that you subdivide the code into N snippets, each terminated by an unconditional jump (e.g. procedures), you could test each possible sequence out of the N! possibilities.
Note that the maximum savings in bytes that you could achieve are the number of jumps that may be converted from 16-bit form to 8-bit form. If this number is smaller than the length of the smallest code snippet, you would not be able to use any sort of pruning of the search tree, but would be forced to evaluate all N! leaves of the tree, which might take a long time...
Borland's Turbo Assembler (for x86 processors) had an option whereby it attempted to optimize (conditional) jumps:
1. All jumps were written without qualifiers.
2. The assembler would make multiple passes through the code, applying the following algorithm:
a. If a jump target was within +127/-128 bytes, output a short (2-byte) jump.
b. If an unconditional jump target was outside that range, output a 3-byte jump.
b. If a conditional jump target was outside that range, output a 5-byte sequence - jump <inverted condition=""> over the following jump (2 bytes) / jump unconditional to the target (3 bytes).
This was applied in a loop until either no more jumps could be optimized or a predetermined number of loops was reached. Typically, only 2-3 loops were necessary.
In addition to the automatic method given above, I would try to write each procedure so that the jumps are all 8-bit forms. Optimizing a procedure by hand is likely to be much easier than attempting global optimization.
You have a misunderstanding about this -- the CLR doesn't understand any languages -- it's more like the languages understand the CLR, but even that is a misleading description.
Furthermore, most members here on CP (me anyway) are just ordinary developers who do not know (or care) anything about how the deep internals work. You would probbaly need to contact the developers at Microsoft to gain the level of detail you desire.
The things you are asking about are way beyond what is required for day-to-day development of commercial and enterprise applications. If you truly want to understand how it all works, you will likely need a doctorate degree.
I can't find anything wrong to understand things that have developed successfully already, i need your help if you already know it then just you can share it that will increase your's knowledge level also Mr PIEBALDconsult and i need to tell you one thing Sir Isaac Newton didn't have a doctorate degree when he found the gravitational force at all, it can mean you that doctorate degree is not necessary at all to become a big man in knowledge like Sir Isaac Newton.
Hi friends, need help!! i am absloute beginner and need advise. below algorithm is an extrat from a text book but when i try to apply and solve the problem on paper i see that this algorithm will fail. as i get a remainder 0 every digit i key in till 8 (i took number 8 as an examole and applied below)...please advise....also i found that applying this algorithm on number 2 would result in 0 as well and if it is 0 then not prime, then how is this algorithm correct!
2 read the number num
i <--2, flag <--1
4 repeatnsteps 4 through 6 unitl i
4th step - repeat steps 4 through 6 unitl i <num or flag =0 5) rem <--num mod i 6) if rem=0 then
flag<-- 0 else i<--i+1 7) if flag =0 then print number is not prime else prit number is prime 8) stop
in this step if i use number 2 as an example it would result in 0 which will result it number being non-prime.
-- modified 6-Sep-14 10:12am.
Last Visit: 31-Dec-99 18:00 Last Update: 3-Jul-22 6:03