Algorithm to generate mountain ranges with upstrokes and down-strokes (java)
I tried to do the classical problem to implement an algorithm to print all valid combinations of n pairs of parentheses. And I found this program (which works perfectly) :
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Interesting task. The calculation can be done in parallel. I'll show the code within "not answer" tags, as it doesn't match the question's language criterion (made in the parallel array processing language Dyalog APL which in fact it does the job with one row of code). Please ignore that part as you wish. However, i'll show the data and what happens. It's rather intuitive.
< not answer >
fn←{(∧/(0≤+\a-~a),(⍵÷2)=+/a)⌿a←⍉(⍵⍴2)⊤⍳2*⍵} // Dynamic function, generates boolean matrix format←{⍉↑(-1+(0.5×⍴⍵)-+\⍵-0,¯1↓~⍵)↑¨'\/'[1+⍵]} // Dirty format function< /not answer >
Say the argument (the width of the mountains) is n=6.
Step 1. Generate all numbers between 0 and (2^6 - 1)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63Step 2: Grab the 2-base of each (they are vertically below. 0 is leftmost, then 1, etc):
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 11. In fact, one only needs to generate numbers from 32 to 63, since we only want 2-base's that begin with 1. See topmost row in data above. Columns (numbers) with zeroes at top shouldn't really even be generated.)
2. In fact one only needs to generate even numbers, since last bit must be 0. See downmost row in data above. Columns (numbers) with ones at bottom shouldn't really even be generated.)Step 3: Calculate the number of ones in each column:
0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 2 3 3 4 3 4 4 5 3 4 4 5 4 5 5 6and do a boolean marking = 1 where the sum is half of N, that is 3 (ie. in total we must have as many uphills as downhills). This is our 1st boolean result:
0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0Step 4: Ensure that we do not go "below horizon":
This means we must calculate a cumulative sum of each column, first:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 0 0 0 0 1 1 1 1 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 0 0 1 1 1 1 2 2 1 1 2 2 2 2 3 3 1 1 2 2 2 2 3 3 2 2 3 3 3 3 4 4 1 1 2 2 2 2 3 3 2 2 3 3 3 3 4 4 2 2 3 3 3 3 4 4 3 3 4 4 4 4 5 5 0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 2 3 3 4 3 4 4 5 3 4 4 5 4 5 5 6and then for the shifted bits (0's turned into 1's, and vice versa):
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 4 4 4 4 3 3 3 3 3 3 3 3 2 2 2 2 3 3 3 3 2 2 2 2 2 2 2 2 1 1 1 1 3 3 3 3 2 2 2 2 2 2 2 2 1 1 1 1 2 2 2 2 1 1 1 1 1 1 1 1 0 0 0 0 5 5 4 4 4 4 3 3 4 4 3 3 3 3 2 2 4 4 3 3 3 3 2 2 3 3 2 2 2 2 1 1 4 4 3 3 3 3 2 2 3 3 2 2 2 2 1 1 3 3 2 2 2 2 1 1 2 2 1 1 1 1 0 0 6 5 5 4 5 4 4 3 5 4 4 3 4 3 3 2 5 4 4 3 4 3 3 2 4 3 3 2 3 2 2 1 5 4 4 3 4 3 3 2 4 3 3 2 3 2 2 1 4 3 3 2 3 2 2 1 3 2 2 1 2 1 1 0then subtract the 2nd from the 1st, and get
¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ¯3 ¯3 ¯3 ¯3 ¯3 ¯3 ¯3 ¯3 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 1 1 1 1 1 1 1 1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 ¯1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 ¯4 ¯4 ¯4 ¯4 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 ¯2 0 0 0 0 ¯2 ¯2 ¯2 ¯2 0 0 0 0 0 0 0 0 2 2 2 2 ¯2 ¯2 ¯2 ¯2 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 4 4 4 4 ¯5 ¯5 ¯3 ¯3 ¯3 ¯3 ¯1 ¯1 ¯3 ¯3 ¯1 ¯1 ¯1 ¯1 1 1 ¯3 ¯3 ¯1 ¯1 ¯1 ¯1 1 1 ¯1 ¯1 1 1 1 1 3 3 ¯3 ¯3 ¯1 ¯1 ¯1 ¯1 1 1 ¯1 ¯1 1 1 1 1 3 3 ¯1 ¯1 1 1 1 1 3 3 1 1 3 3 3 3 5 5 ¯6 ¯4 ¯4 ¯2 ¯4 ¯2 ¯2 0 ¯4 ¯2 ¯2 0 ¯2 0 0 2 ¯4 ¯2 ¯2 0 ¯2 0 0 2 ¯2 0 0 2 0 2 2 4 ¯4 ¯2 ¯2 0 ¯2 0 0 2 ¯2 0 0 2 0 2 2 4 ¯2 0 0 2 0 2 2 4 0 2 2 4 2 4 4 6and see which columns have no negative values; this is the 2nd boolean result:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1Step 5: Grab an AND from the two boolean results above:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0These are the locations for columns of the binary data that create the good mountains. Below column-wise to left, then transposed (for readability) to right. 1 is uphills,
2edit:0 is downhills:1 1 1 1 1 1 0 1 0 1 0 // 1 0 1 0 1 0 means /\/\/\ 0 0 1 1 1 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 // means //\/\\ 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0That is the good answer. If needed, we can apply a formatting:
format [the boolean result] ┌──────┬──────┬──────┬──────┬──────┐ │ │ │ │ │ /\ │ │ │ /\ │ /\ │ /\/\ │ / \ │ │/\/\/\│/\/ \│/ \/\│/ \│/ \│ └──────┴──────┴──────┴──────┴──────┘A little bigger, n=10:
DISP format¨↓fn 10 ┌──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┬──────────┐ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ /\ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ /\ │ │ │ │ │ │ │ │ │ │ │ │ │ │ /\ │ │ │ │ │ │ │ │ │ /\ │ /\ │ /\ │ /\ │ /\/\ │ / \ │ │ │ │ │ │ /\ │ │ │ │ │ /\ │ /\ │ /\ │ /\/\ │ / \ │ │ │ │ │ /\ │ │ │ │ │ /\ │ /\ │ /\ │ /\/\ │ / \ │ /\ │ /\ │ /\ │ /\ │ /\ /\ │ /\/\ │ /\/\ │ /\/\/\ │ /\/ \ │ / \ │ / \ │ / \/\ │ / \ │ / \ │ │ │ /\ │ /\ │ /\/\ │ / \ │ /\ │ /\ /\ │ /\/\ │ /\/\/\ │ /\/ \ │ / \ │ / \/\ │ / \ │ / \ │ /\ │ /\ /\ │ /\ /\ │ /\ /\/\ │ /\ / \ │ /\/\ │ /\/\ /\ │ /\/\/\ │ /\/\/\/\ │ /\/\/ \ │ /\/ \ │ /\/ \/\ │ /\/ \ │ /\/ \ │ / \ │ / \ /\ │ / \/\ │ / \/\/\ │ / \/ \ │ / \ │ / \/\ │ / \ │ / \ │ / \ │ / \/\ │ / \ │ / \ │ / \ │ │/\/\/\/\/\│/\/\/\/ \│/\/\/ \/\│/\/\/ \│/\/\/ \│/\/ \/\/\│/\/ \/ \│/\/ \/\│/\/ \│/\/ \│/\/ \/\│/\/ \│/\/ \│/\/ \│/ \/\/\/\│/ \/\/ \│/ \/ \/\│/ \/ \│/ \/ \│/ \/\/\│/ \/ \│/ \/\│/ \│/ \│/ \/\│/ \│/ \│/ \│/ \/\/\│/ \/ \│/ \/\│/ \│/ \│/ \/\│/ \│/ \│/ \│/ \/\│/ \│/ \│/ \│/ \│ └──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┴──────────┘
Edit: Naturally one can do all this in a loop as well. Just take one number at time, and do the checks above (number of ones == half of n, does not go below horizon). Jump out if either check fails.
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