Related
I am trying to find the best algorithm for my particular application. I have searched around on SO, Google, read various articles about Levenshtein distances, etc. but honestly it's a bit out of my area of expertise. And most seem to find how similar two input strings are, like a Hamming distance between strings.
What I'm looking for is different, more of a fuzzy record search (and I'm sure there is a name for it, that I don't know to Google). I am sure someone has solved this problem before and I'm looking for a recommendation to point me in the right direction for my further research.
In my case I am needing a fuzzy search of a database of entries of music artists and their albums. As you can imagine, the database will have millions of entries so an algorithm that scales well is crucial. It's not important to my question that Artist and Album are in different columns, the database could just store all words in one column if that helped the search.
The database to search:
|-------------------|---------------------|
| Artist | Album |
|-------------------|---------------------|
| Alanis Morissette | Jagged Little Pill |
| Moby | Everything is Wrong |
| Air | Moon Safari |
| Pearl Jam | Ten |
| Nirvana | Nevermind |
| Radiohead | OK Computer |
| Beck | Odelay |
|-------------------|---------------------|
The query text will contain from just one word in the entire Artist_Album concatenation up to the entire thing. The query text is coming from OCR and is likely to have single character transpositions but the most likely thing is the words are not guaranteed to have the right order. Additionally, there could be extra words in the search that aren't a part of the album (like cover art text). For example, "OK Computer" might be at the top of the album and "Radiohead" below it, or some albums have text arranged in columns which intermixes the word orders.
Possible search strings:
C0mputer Rad1ohead
Pearl Ten Jan
Alanis Jagged Morisse11e Litt1e Pi11
Air Moon Virgin Records
Moby Everything
Note that with OCR, some letters will look like numbers, or the wrong letter completely (Jan instead of Jam). And in the case of Radiohead's OK Computer and Moby's Everything Is Wrong, the query text doesn't even have all of the words. In the case of Air's Moon Safari, the extra words Virgin Records are searched, but Safari is missing.
Is there a general algorithm that could return the single likeliest result from the database, and if none meet some "likeliness" score threshold, it returns nothing? I'm actually developing this in Python, but that's just a bonus, I'm looking more for where to get started researching.
Let's break the problem down in two parts.
First, you want to define some measure of likeness (this is called a metric). This metric should return a small number if the query text closely matches the album/artist cover, and return a larger number otherwise.
Second, you want a datastructure that speeds up this process. Obviously, you don't want to calculate this metric every single time a query is ran.
part 1: the metric
You already mentioned Levenshtein distance, which is a great place to start.
Think outside the box though.
LD makes certain assumptions (each character replacement is equally likely, deletion is equally likely as insertion, etc). You can obviously improve the performance of this metric by taking into account what faults OCR is likely to introduce.
E.g. turning a '1' into an 'i' should not be penalized as harshly as turning a '0' into an '_'.
I would implement the metric in two stages. For any given two strings:
split both strings in tokens (assume space as the separator)
look for the most similar words (using a modified version of LD)
assign a final score based on 'matching words', 'missing words' and 'added words' (preferably weighted)
This is an example implementation (fiddle around with the constants):
static double m(String a, String b){
String[] aParts = a.split(" ");
String[] bParts = b.split(" ");
boolean[] bUsed = new boolean[bParts.length];
int matchedTokens = 0;
int tokensInANotInB = 0;
int tokensInBNotInA = 0;
for(int i=0;i<aParts.length;i++){
String a0 = aParts[i];
boolean wasMatched = true;
for(int j=0;j<bParts.length;j++){
String b0 = bParts[j];
double d = levenshtein(a0, b0);
/* If we match the token a0 with a token from b0
* update the number of matchedTokens
* escape the loop
*/
if(d < 2){
bUsed[j]=true;
wasMatched = true;
matchedTokens++;
break;
}
}
if(!wasMatched){
tokensInANotInB++;
}
}
for(boolean partUsed : bUsed){
if(!partUsed){
tokensInBNotInA++;
}
}
return (matchedTokens
+ tokensInANotInB * -0.3 // the query is allowed to contain extra words at minimal cost
+ tokensInBNotInA * -0.5 // the album title should not contain too many extra words
) / java.lang.Math.max(aParts.length, bParts.length);
}
This function uses a modified levenshtein function:
static double levenshtein(String x, String y) {
double[][] dp = new double[x.length() + 1][y.length() + 1];
for (int i = 0; i <= x.length(); i++) {
for (int j = 0; j <= y.length(); j++) {
if (i == 0) {
dp[i][j] = j;
}
else if (j == 0) {
dp[i][j] = i;
}
else {
dp[i][j] = min(dp[i - 1][j - 1]
+ costOfSubstitution(x.charAt(i - 1), y.charAt(j - 1)),
dp[i - 1][j] + 1,
dp[i][j - 1] + 1);
}
}
}
return dp[x.length()][y.length()];
}
Which uses the function 'cost of substitution' (which works as explained)
static double costOfSubstitution(char a, char b){
if(a == b)
return 0.0;
else{
// 1 and i
if(a == '1' && b == 'i')
return 0.5;
if(a == 'i' && b == '1')
return 0.5;
// 0 and O
if(a == '0' && b == 'o')
return 0.5;
if(a == 'o' && b == '0')
return 0.5;
if(a == '0' && b == 'O')
return 0.5;
if(a == 'O' && b == '0')
return 0.5;
// default
return 1.0;
}
}
I only included a couple of examples (turning '1' into 'i' or '0' into 'o').
But I'm sure you get the idea.
part 2: the datastructure
Look into BK-trees. They are a specific datastructure to hold metric information. Your metric needs to be a genuine metric (in the mathematical sense of the word). But that's easily arranged.
Recently came across an interview question in glassdoor-like site and I can't find an optimized solution to solve this problem:
This is nothing like trapping water problem. Please read through the examples.
Given an input array whose each element represents the height of towers, the amount of water will be poured and the index number indicates the pouring water position.The width of every tower is 1. Print the graph after pouring water.
Notes:
Use * to indicate the tower, w to represent 1 amount water.
The pouring position will never at the peak position.No need to consider the divide water case.
(A Bonus point if you gave a solution for this case, you may assume that if Pouring N water at peak position, N/2 water goes to left, N/2 water goes to right.)
The definition for a peak: the height of peak position is greater than the both left and right index next to it.)
Assume there are 2 extreme high walls sits close to the histogram.
So if the water amount is over the capacity of the histogram,
you should indicate the capacity number and keep going. See Example 2.
Assume the water would go left first, see Example 1
Example 1:
int[] heights = {4,2,1,2,3,2,1,0,4,2,1}
It look like:
* *
* * **
** *** **
******* ***
+++++++++++ <- there'll always be a base layer
42123210431
Assume given this heights array, water amout 3, position 2:
Print:
* *
*ww * **
**w*** **
******* ***
+++++++++++
Example 2:
int[] heights = {4,2,1,2,3,2,1,0,4,2,1}, water amout 32, position 2
Print:
capacity:21
wwwwwwwwwww
*wwwwwww*ww
*www*www**w
**w***ww**w
*******w***
+++++++++++
At first I though it's like the trapping water problem but I was wrong. Does anyone have an algorithm to solve this problem?
An explanation or comments in the code would be welcomed.
Note:
The trapping water problem is asked for the capacity, but this question introduced two variables: water amount and the pouring index. Besides, the water has the flowing preference. So it not like trapping water problem.
I found a Python solution to this question. However, I'm not familiar with Python so I quote the code here. Hopefully, someone knows Python could help.
Code by #z026
def pour_water(terrains, location, water):
print 'location', location
print 'len terrains', len(terrains)
waters = [0] * len(terrains)
while water > 0:
left = location - 1
while left >= 0:
if terrains[left] + waters[left] > terrains[left + 1] + waters[left + 1]:
break
left -= 1
if terrains[left + 1] + waters[left + 1] < terrains[location] + waters[location]:
location_to_pour = left + 1
print 'set by left', location_to_pour
else:
right = location + 1
while right < len(terrains):
if terrains[right] + waters[right] > terrains[right - 1] + waters[right - 1]:
print 'break, right: {}, right - 1:{}'.format(right, right - 1)
break
right += 1
if terrains[right - 1] + waters[right - 1] < terrains[location] + waters[right - 1]:
location_to_pour = right - 1
print 'set by right', location_to_pour
else:
location_to_pour = location
print 'set to location', location_to_pour
waters[location_to_pour] += 1
print location_to_pour
water -= 1
max_height = max(terrains)
for height in xrange(max_height, -1, -1):
for i in xrange(len(terrains)):
if terrains + waters < height:
print ' ',
elif terrains < height <= terrains + waters:
print 'w',
else:
print '+',
print ''
Since you have to generate and print out the array anyway, I'd probably opt for a recursive approach keeping to the O(rows*columns) complexity. Note each cell can be "visited" at most twice.
On a high level: first recurse down, then left, then right, then fill the current cell.
However, this runs into a little problem: (assuming this is a problem)
*w * * *
**ww* * instead of **ww*w*
This can be fixed by updating the algorithm to go left and right first to fill cells below the current row, then to go both left and right again to fill the current row. Let's say state = v means we came from above, state = h1 means it's the first horizontal pass, state = h2 means it's the second horizontal pass.
You might be able to avoid this repeated visiting of cells by using a stack, but it's more complex.
Pseudo-code:
array[][] // populated with towers, as shown in the question
visited[][] // starts with all false
// call at the position you're inserting water (at the very top)
define fill(x, y, state):
if x or y out of bounds
or array[x][y] == '*'
or waterCount == 0
return
visited = true
// we came from above
if state == v
fill(x, y+1, v) // down
fill(x-1, y, h1) // left , 1st pass
fill(x+1, y, h1) // right, 1st pass
fill(x-1, y, h2) // left , 2nd pass
fill(x+1, y, h2) // right, 2nd pass
// this is a 1st horizontal pass
if state == h1
fill(x, y+1, v) // down
fill(x-1, y, h1) // left , 1st pass
fill(x+1, y, h1) // right, 1st pass
visited = false // need to revisit cell later
return // skip filling the current cell
// this is a 2nd horizontal pass
if state == h2
fill(x-1, y, h2) // left , 2nd pass
fill(x+1, y, h2) // right, 2nd pass
// fill current cell
if waterCount > 0
array[x][y] = 'w'
waterCount--
You have an array height with the height of the terrain in each column, so I would create a copy of this array (let's call it w for water) to indicate how high the water is in each column. Like this you also get rid of the problem not knowing how many rows to initialize when transforming into a grid and you can skip that step entirely.
The algorithm in Java code would look something like this:
public int[] getWaterHeight(int index, int drops, int[] heights) {
int[] w = Arrays.copyOf(heights);
for (; drops > 0; drops--) {
int idx = index;
// go left first
while (idx > 0 && w[idx - 1] <= w[idx])
idx--;
// go right
for (;;) {
int t = idx + 1;
while (t < w.length && w[t] == w[idx])
t++;
if (t >= w.length || w[t] >= w[idx]) {
w[idx]++;
break;
} else { // we can go down to the right side here
idx = t;
}
}
}
return w;
}
Even though there are many loops, the complexity is only O(drops * columns). If you expect huge amount of drops then it could be wise to count the number of empty spaces in regard to the highest terrain point O(columns), then if the number of drops exceeds the free spaces, the calculation of the column heights becomes trivial O(1), however setting them all still takes O(columns).
You can iterate over the 2D grid from bottom to top, create a node for every horizontal run of connected cells, and then string these nodes together into a linked list that represents the order in which the cells are filled.
After row one, you have one horizontal run, with a volume of 1:
1(1)
In row two, you find three runs, one of which is connected to node 1:
1(1)->2(1) 3(1) 4(1)
In row three, you find three runs, one of which connects runs 2 and 3; run 3 is closest to the column where the water is added, so it comes first:
3(1)->1(1)->2(1)->5(3) 6(1) 4(1)->7(1)
In row four you find two runs, one of which connects runs 6 and 7; run 6 is closest to the column where the water is added, so it comes first:
3(1)->1(1)->2(1)->5(3)->8(4) 6(1)->4(1)->7(1)->9(3)
In row five you find a run which connects runs 8 and 9; they are on opposite sides of the column where the water is added, so the run on the left goes first:
3(1)->1(1)->2(1)->5(3)->8(4)->6(1)->4(1)->7(1)->9(3)->A(8)
Run A combines all the columns, so it becomes the last node and is given infinite volume; any excess drops will simply be stacked up:
3(1)->1(1)->2(1)->5(3)->8(4)->6(1)->4(1)->7(1)->9(3)->A(infinite)
then we fill the runs in the order in which they are listed, until we run out of drops.
Thats my 20 minutes solution. Each drop is telling the client where it will stay, so the difficult task is done.(Copy-Paste in your IDE) Only the printing have to be done now, but the drops are taking their position. Take a look:
class Test2{
private static int[] heights = {3,4,4,4,3,2,1,0,4,2,1};
public static void main(String args[]){
int wAmount = 10;
int position = 2;
for(int i=0; i<wAmount; i++){
System.out.println(i+"#drop");
aDropLeft(position);
}
}
private static void aDropLeft(int position){
getHight(position);
int canFallTo = getFallPositionLeft(position);
if(canFallTo==-1){canFallTo = getFallPositionRight(position);}
if(canFallTo==-1){
stayThere(position);
return;
}
aDropLeft(canFallTo);
}
private static void stayThere(int position) {
System.out.print("Staying at: ");log(position);
heights[position]++;
}
//the position or -1 if it cant fall
private static int getFallPositionLeft(int position) {
int tempHeight = getHight(position);
int tempPosition = position;
//check left , if no, then check right
while(tempPosition>0){
if(tempHeight>getHight(tempPosition-1)){
return tempPosition-1;
}else tempPosition--;
}
return -1;
}
private static int getFallPositionRight(int position) {
int tempHeight = getHight(position);
int tempPosition = position;
while(tempPosition<heights.length-1){
if(tempHeight>getHight(tempPosition+1)){
return tempPosition+1;
}else if(tempHeight<getHight(tempPosition+1)){
return -1;
}else tempPosition++;
}
return -1;
}
private static int getHight(int position) {
return heights[position];
}
private static void log(int position) {
System.out.println("I am at position: " + position + " height: " + getHight(position));
}
}
Of course the code can be optimized, but thats my straightforward solution
l=[0,1,0,2,1,0,1,3,2,1,2,1]
def findwater(l):
w=0
for i in range(0,len(l)-1):
if i==0:
pass
else:
num = min(max(l[:i]),max(l[i:]))-l[i]
if num>0:
w+=num
return w
col_names=[1,2,3,4,5,6,7,8,9,10,11,12,13] #for visualization
bars=[4,0,2,0,1,0,4,0,5,0,3,0,1]
pd.DataFrame(dict(zip(col_names,bars)),index=range(1)).plot(kind='bar') # Plotting bars
def measure_water(l):
water=0
for i in range(len(l)-1): # iterate over bars (list)
if i==0: # case to avoid max(:i) situation in case no item on left
pass
else:
vol_at_curr_bar=min(max(l[:i]),max(l[i:]))-l[i] #select min of max heighted bar on both side and minus current height
if vol_at_curr_bar>0: # case to aviod any negative sum
water+=vol_at_curr_bar
return water
measure_water(bars)
I am suppose to code the snake game in java with processing for IT classes and since I had no idea how to do it I searched for a YouTube tutorial. Now I did find one but he used the keys 'w','s','d','a' to move the snake around - I on the other hand want to use the arrow keys. Could someone explain to me how I transform this code:
if (keyPressed == true) {
int newdir = key=='s' ? 0 : (key=='w' ? 1 : (key=='d' ? 2 : (key=='a' ? 3 : -1)));
}
if(newdir != -1 && (x.size() <= 1 || !(x.get(1) ==x.get(0) + dx[newdir] && y.get (1) == y.get(0) + dy[newdir]))) dir = newdir;
}
into something like this:
void keyPressed () {
if (key == CODED) {
if (keyCode == UP) {}
else if (keyCode == RIGHT) {}
else if (keyCode == DOWN) {}
else if (keyCode == LEFT) {}
}
This is my entire coding so far:
ArrayList<Integer> x = new ArrayList<Integer> (), y = new ArrayList<Integer> ();
int w = 900, h = 900, bs = 20, dir = 1; // w = width ; h = height ; bs = blocksize ; dir = 2 --> so that the snake goes up when it starts
int[] dx = {0,0,1,-1} , dy = {1,-1,0,0};// down, up, right, left
void setup () {
size (900,900); // the 'playing field' is going to be 900x900px big
// the snake starts off on x = 5 and y = 30
x.add(5);
y.add(30);
}
void draw() {
//white background
background (255);
//
// grid
// vertical lines ; the lines are only drawn if they are smaller than 'w'
// the operator ++ increases the value 'l = 0' by 1
//
for(int l = 0 ; l < w; l++) line (l*bs, 0, l*bs, height);
//
// horizontal lines ; the lines are only drawn if they are smaller than 'h'
// the operator ++ increases the value 'l = 0' by 1
//
for(int l = 0 ; l < h; l++) line (0, l*bs, width, l*bs);
//
// snake
for (int l = 0 ; l < x.size() ; l++) {
fill (0,255,0); // the snake is going to be green
rect (x.get(l)*bs, y.get(l)*bs, bs, bs);
}
if(frameCount%5==0) { // will check it every 1/12 of a second -- will check it every 5 frames at a frameRate = 60
// adding points
x.add (0,x.get(0) + dx[dir]); // will add a new point x in the chosen direction
y.add (0,y.get(0) + dy[dir]); // will add a new point y in the chosen direction
// removing points
x.remove(x.size()-1); // will remove the previous point x
y.remove(y.size()-1); // will remove the previous point y
}
}
It's hard to answer general "how do I do this" type questions. Stack Overflow is designed for more specific "I tried X, expected Y, but got Z instead" type questions. That being said, I'll try to answer in a general sense:
You're going to have a very difficult time trying to take random code you find on the internet and trying to make it work in your sketch. That's not a very good way to proceed.
Instead, you need to take a step back and really think about what you want to happen. Instead of taking on your entire end goal at one time, try breaking your problem down into smaller steps and taking on those steps one at a time.
Step 1: Can you store the state of your game in variables? You might store things like the direction the snake is traveling the location of the snake, etc.
Step 2: Can you write code that just prints something to the console when you press the arrow keys? You might do this in a separate example sketch instead of trying to add it directly to your full sketch.
Step 3: Can you combine those two steps and change the state of your sketch when an arrow key is pressed? Maybe you change the direction the snake is traveling.
The point is that you need to try something instead of trying to copy-paste random code without really understanding it. Break your problem down into small steps, and then post an MCVE of that specific step if you get stuck. Good luck.
You should take a look into Java API KeyEvent VK_LEFT.
And as pczeus already told you, you need to implement a capturing of the keystrokes! This can be checked here (Link from this SO answer).
Hi I am taking in data in real time where the value goes from 1009 , 1008 o 1007 to 0. I am trying to count the number of distinct times this occurs, for example the snippet below should count 2 distinct periods of change.
1008
1009
1008
0
0
0
1008
1007
1008
1008
1009
9
0
0
1009
1008
I have written a for loop as below but I can't figure out if the logic is correct as I get multiple increments instead of just the one
if(current != previous && current < 100)
x++;
else
x = x;
You tagged this with the LabVIEW tag. Is this actually supposed to be LabVIEW code?
Your logic has a bug related to the noise you say you have - if the value is less than 100 and it changes (for instance from 9 to 0), you log that as a change. You also have a line which doesn't do anything (x=x), although if this is supposed to be LV code, then this could make sense.
The code you posted here does not seem to make sense to me if I understand your goal. My understanding is that you want to identify this specific pattern:
1009
1008
1007
0
And that any deviation from this sequence of numbers would constitute data that should be ignored. To this end, you should be monitoring the history of the past 3 numbers. In C you might write this logic in the following way:
#include <stdio.h>
//Function to get the next value from our data stream.
int getNext(int *value) {
//Variable to hold our return code.
int code;
//Replace following line to get gext number from the stream. Possible read from a file?
*value = 0;
//Replace following logic to set 'code' appropriately.
if(*value == -1)
code = -1;
else
code = 0;
//Return 'code' to the caller.
return code;
}
//Example application for counting the occurrences of the sequence '1009','1008','1007','0'.
int main(int argc, char **argv) {
//Declare 4 items to store the past 4 items in the sequence (x0-x3)
//Declare a count and a flag to monitor the occurrence of our pattern
int x0 = 0, x1 = 0, x2 = 0, x3 = 0, count = 0, occurred = 0;
//Run forever (just as an example, you would provide your own looping structure or embed the algorithm in your app).
while(1) {
//Get the next element (implement getNext to provide numbers from your data source).
//If the function returns non-zero, exit the loop and print the count.
if( getNext(&x0) != 0 )
break;
//If the newest element is 0, we can trigger a check of the prior 3.
if(x0 == 0) {
//Set occurred to 0 if the prior elements don't match our pattern.
occurred = (x1 == 1007) && (x2 == 1008) && (x3 == 1009);
if(occurred) {
//Occurred was 1, meaning the pattern was matched. Increment our count.
count++;
//Reset occurred
occurred = 0;
}
//If the newest element is not 0, dont bother checking. Just shift the elements down our list.
} else {
x3 = x2; //Shift 3rd element to 4th position
x2 = x1; //Shift 2nd element to 3rd position
x1 = x0; //Shift 1st element to 2nd position
}
}
printf("The pattern count is %d\n", count);
//Exit application
return 0;
}
Note that the getNext function is just shown here as an example but obviously what I have implemented will not work. This function should be implemented based on how you are extracting data from the stream.
Writing the application in this way might not make sense within your larger application but the algorithm is what you should take away from this. Essentially you want to buffer 4 elements in a rolling window. You push the newest element into x0 and shift the others down. After this process you check the four elements to see if they match your desired pattern and increment the count accordingly.
If the requirement is to count falling edges and you don't care about the specific level, and want to reject noise band or ripple in the steady state then just make the conditional something like
if ((previous - current) > threshold)
No complex shifting, history, or filtering required. Depending on the application you can follow up with a debounce (persistency check) to ignore spurious samples (just keep track of falling/rising, or fell/rose as simple toggling state spanning a desired number of samples).
Code to the pattern, not the specific values; use constant or adjustable parameters to control the value sensitivity.
Out of curiosity, I was checking out the problem set to the 2009 ACM International Collegiate Programming Contest. The questions are pretty interesting. They're available at http://cm.baylor.edu/resources/pdf/2009Problems.pdf. I could not come up with an algorithm that solved problem 1, which I will reproduce here. It set off a lively discussion in the office, and we think we're pretty close to an answer, but we'd really appreciate it if somebody could find/work out a full solution (code not required).
I will reproduce problem here for your convenience:
Problem 1
Consider the task of scheduling the airplanes that are landing at an airport. Incoming airplanes report their positions, directions, and speeds, and then the controller has to devise a landing schedule that brings all airplanes safely to the ground. Generally, the more time there is between successive landings, the “safer” a landing schedule is. This extra time gives pilots the opportunity to react to changing weather and other surprises.
Luckily, part of this scheduling task can be automated – this is where you come in. You will be given scenarios of airplane landings. Each airplane has a time window during which it can safely land. You must compute an order for landing all airplanes that respects these time windows. Furthermore, the airplane landings should be stretched out as much as possible so that the minimum time gap between successive landings is as large as possible. For example, if three airplanes land at 10:00am, 10:05am, and 10:15am, then the smallest gap is five minutes, which occurs between the first two airplanes. Not all gaps have to be the same, but the smallest gap should be as large as possible.
Input
The input file contains several test cases consisting of descriptions of landing scenarios. Each test case starts with a line containing a single integer n (2 ≤ n ≤ 8), which is the number of airplanes in the scenario. This is followed by n lines, each containing two integers ai, bi, which give the beginning and end of the closed interval [ai, bi] during which the ith plane can land safely. The numbers ai and bi are specified in minutes and satisfy 0 ≤ ai ≤ bi ≤ 1440.
The input is terminated with a line containing the single integer zero.
Output
For each test case in the input, print its case number (starting with 1) followed by the minimum achievable time gap between successive landings. Print the time split into minutes and seconds, rounded to the closest second. Follow the format of the sample output.
Sample Input
3
0 10
5 15
10 15
2
0 10
10 20
0
Sample Output
Case 1: 7:30
Case 2: 20:00
I'll give a sketch of the algorithm.
First you binary search through the answer (minimal interval between flights). To do that, for each selected interval T you must be able to check whether it is possible to achieve it. If it is possible to achieve T, then you try making it smaller, if it is not - make it bigger.
To check whether you can achieve T, try all n! orders in which the planes may be landing (8! is small enough for this algo to work in time). For each permutation P1...Pn, you try assigning the times in a greedy manner:
int land = a[0];
for (int i = 1; i < n; i++) {
land = max(a[i], land + **T**);
if (land > b[i]) return "CAN NOT ACHIEVE INTERVAL T";
}
return "CAN ACHIEVE";
This optimization problem can be solved by linear programming http://en.wikipedia.org/wiki/Linear_programming
I would do something like this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef uint MASK;
#define INPUT_SCALE 60
#define MAX_TIME (1440 * 60)
void readPlaneData(int& endTime, MASK landingMask[MAX_TIME], int index)
{
char buf[128];
gets(buf);
int start, end;
sscanf(buf, "%d %d", &start, &end);
for(int i=start * INPUT_SCALE; i<=end * INPUT_SCALE; i++)
landingMask[i] |= 1 << index;
if(end * INPUT_SCALE > endTime)
endTime = end * INPUT_SCALE;
}
int findNextLandingForPlane(MASK landingMask[MAX_TIME], int start, int index)
{
while(start < MAX_TIME)
{
if(landingMask[start] & (1 << index))
return start;
start++;
}
return -1;
}
bool canLandPlanes(int minTime, MASK landingMask[MAX_TIME], int planeCount)
{
int next = 0;
for(int i=0; i<planeCount; i++)
{
int nextForPlane = findNextLandingForPlane(landingMask, next, i);
if(nextForPlane == -1)
return false;
next = nextForPlane + minTime;
}
return true;
}
int main(int argc, char* argv[])
{
while(true)
{
char buf[128];
gets(buf);
int count = atoi(buf);
if(count == 0)
break;
MASK landingMask[MAX_TIME];
memset(landingMask, 0, sizeof(landingMask));
int endTime = 0;
for(int i=0; i<count; i++)
readPlaneData(endTime, landingMask, i);
while((endTime > 0) && !canLandPlanes(endTime, landingMask, count))
endTime--;
printf("%d:%02d\n", endTime / 60, endTime % 60);
}
}
Here's some Ruby code that brute-forces the solution. Note that test_case_one actually fails because I have commented out the code that would make this work with seconds (instead of just whole minutes).
The brute-force strategy is to permute all the sequences in which the planes may land. For each landing sequence, create the product of all possible landing times. This is fine with whole minutes, brutal with seconds.
But of course premature optimization, evil, and all that, so this is a first step:
require 'test/unit'
class SampleTests < Test::Unit::TestCase
def test_case_one
problem = Problem.new
problem.add_plane(Plane.new(0, 10))
problem.add_plane(Plane.new(5, 15))
problem.add_plane(Plane.new(10, 15))
problem.solve()
minimum_gap = problem.minimum_gap()
assert_equal(7.5, minimum_gap)
end
def test_case_two
problem = Problem.new
problem.add_plane(Plane.new(0,10))
problem.add_plane(Plane.new(10, 20))
problem.solve()
minimum_gap = problem.minimum_gap()
assert_equal(20, minimum_gap)
end
def test_case_three
problem = Problem.new
problem.add_plane(Plane.new(0, 2))
problem.add_plane(Plane.new(7, 10))
problem.add_plane(Plane.new(4, 6))
minimum_gap = problem.minimum_gap()
assert_equal(5, minimum_gap)
end
def test_case_four
problem = Problem.new
problem.add_plane(Plane.new(1439, 1440))
problem.add_plane(Plane.new(1439, 1440))
problem.add_plane(Plane.new(1439, 1440))
assert_equal(0, problem.minimum_gap())
end
def test_case_five
problem = Problem.new
problem.add_plane(Plane.new(0, 10))
problem.add_plane(Plane.new(1, 2))
assert_equal(9, problem.minimum_gap())
end
def test_case_six
problem = Problem.new
problem.add_plane(Plane.new(8, 9))
problem.add_plane(Plane.new(0, 10))
assert_equal(9, problem.minimum_gap())
end
end
class Plane
def initialize(min, max)
#ts = Array.new
#This is a cheat to prevent combinatorial explosion. Just ignore 60 seconds in a minute!
#min = min * 60
#max = max * 60
min.upto(max) { | t | #ts << t}
end
#Array of times at which the plane might land.
def times
return #ts
end
end
#from 'permutation' gem
class Array
def permute(prefixed=[])
if (length < 2)
# there are no elements left to permute
yield(prefixed + self)
else
# recursively permute the remaining elements
each_with_index do |e, i|
(self[0,i]+self[(i+1)..-1]).permute(prefixed+[e]) { |a| yield a }
end
end
end
end
class Problem
def initialize
#solved = false
#maximum_gap = 0
#planes = Array.new
end
def add_plane(plane)
#planes << plane
end
#given a particular landing schedule, what's the minimum gap?
#A: Sort schedule and spin through it, looking for the min diff
#Note that this will return 0 for invalid schedules (planes landing simultaneously)
def gap_for(schedule)
schedule.sort!
min_gap = 1440
0.upto(schedule.length - 2) { | i |
gap = schedule[i + 1] - schedule[i]
if gap < min_gap
min_gap = gap
end
}
return min_gap
end
#Brute-force strategy
#Get every possible plane sequence (permute)
#Get every possible schedule for that sequence (brute_force_schedule)
#Check that schedule
def solve
#planes.permute { | sequence |
schedules = brute_force_schedule(sequence)
schedules.each { | schedule |
schedule.flatten!
gap = gap_for(schedule)
if gap > #maximum_gap
#puts "Found a new one: #{schedule.inspect}"
#maximum_gap = gap
end
}
}
end
#The list of all possible schedules associated with an array of planes
def brute_force_schedule(planes)
head = planes[0]
tail = planes[1..-1]
if tail.empty?
#Last element, return the times
return head.times.to_a
else
#Recurse and combine (product)
return head.times.to_a.product(brute_force_schedule(tail))
end
end
def minimum_gap
unless #solved
solve
end
return #maximum_gap
end
end